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https://datascience.stackexchange.com/questions/57692/modeling-the-price-movement-what-analysis-should-be-used | [
"# Modeling the Price Movement- What analysis should be used\n\nI am trying to model the price of a hotel as the check-in date arrives. I have a data set which looks like-",
null,
"For e.g- if I am looking at the booking date of Dec 31st, I would want to analyze the movement of price for the hotels from 31st October onwards and model this movement with given variables-City, Star_rating, Accomodation_type, Chain_hotel, and Time before check-in.\n\nI started with checking which of the variable(Feature) was more important in deciding the price of the hotel on a given day and applied multiway ANOVA. However, I am still not confident on my analysis and I am really confused that what could be the best way to model such a problem.\n\nThe question is if you really want to treat this as a time series problem. I say this because there may be no obvious/persistent autocorrelation (meaning that $$y$$ is contingent on $$y_{t-1}$$).\n\nMy take (not knowing the data) would be that each hotel has an own unobserved \"identity\" (like location, reputation etc) apart from star rating. Bookings (and thus prices) may also be highly contingent on time. So you could give a fixed-effects model a try. The idea is that you model this in a way like:\n\n$$y=\\alpha + \\beta X+\\gamma Z + \\theta t + u.$$\n\nWhere $$X$$ are observed hotel characteristics, $$Z$$ are hotel fixed-effects (i.e. one indicator/dummy per hotel), $$t$$ is time (day of year or so), and $$u$$ is the error term.\n\nIn terms of model interpretation, $$X$$ are important (standard confounders), where $$Z$$ are not so relevant (and often omitted in FE regression) since $$Z$$ is only \"one intercept per hotel\". If you are interested in modeling the time aspect, clever encoding of $$t$$ is key. It is hard to tell how this could work out without knowing the data. There are two general options, a) dummy encoding (one dummy/indicator per time step) or continuous treatment (calender day?), maybe with a lagged component (price yesterday). In the latter case, you go in the direction of dynamic panel, which can be a little challanging.\n\nYou can try OLS first since it is very efficient in terms of computation. In order to get a better fit you may also try boosting, e.g. based on LightGBM. Here is a minimal example for boosting regression (but no fixed-effect model, just a normal one).\n\n• Hi Peter, this is really helpful, I am a little new to this domain and I have two concerns: 1. Since, all the features available to me are categorical(star rating, lead time, accommodation type, chain hotel) and my y variable is sales, a continuous variable, I am a little confused about can we use only these categorical variables to fit the continuous variable 2. At what level shall I be running the model, for e.g- at transaction level(each row is one data point) or shall I run at hotel level(I doubt this would be possible as each hotel has multiple bookings for multiple checkin dates) Thanks! – Shubham Malviya Aug 17 at 10:25\n• You can estimate $y$ (continuous) and $X$ and $t$ (all categorical), the question is how well it works (but with all methods, I guess). Run a regression on the transaction level and include one category (indicator/dummy) per hotel to model FEs (the $Z$) above. – Peter Aug 17 at 11:00\n• 1/2 - Thanks for the response. I tried running two models. Model-A= I ran the model at the transaction level and created dummy variables for each of 141 hotels(Now data has the same number of observation but 141 dummy cols for hotels+n dummy cols for star rating, accommodation type...) and I ran a simple regression. But now since I am running the model at trans level- 1. I think I am not able to capture the time component of the data. 2. Now since I have 141 dummy cols, does beta corresponding to each of this cols represent the effectiveness of individual heterogeneity? – Shubham Malviya Aug 17 at 22:59\n• 2/2 I also ran another Model B= I ran this model at (hotel, Lead time) and (check-in date), converting three-dimensional panel data into two by merging hotel_id and Lead time and ran Fixed Effect Estimator, however, this model does not fit at all. I am really confused here, is there anything I am missing? Thanks for your help and support. – Shubham Malviya Aug 17 at 23:16\n• I added some more details to my answer. However, without knowing the data I cannot do more in the moment. Regarding 2) if your model does not fit, you did something wrong, likely because of multicollinearity. – Peter Aug 18 at 11:56"
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https://help.tableau.com/current/pro/desktop/en-gb/calculations_calculatedfields_create.htm | [
"# Get Started with Calculations in Tableau\n\nThis article describes how to create and use calculated fields in Tableau using an example.\n\nYou'll learn Tableau calculation concepts, as well as how to create and edit a calculated field. You will also learn how to work with the calculation editor, and use a calculated field in the view.\n\nIf you're new to Tableau calculations or to creating calculated fields in Tableau, this is a good place to start.\n\n## Why Use Calculated Fields\n\nCalculated fields allow you to create new data from data that already exists in your data source. When you create a calculated field, you are essentially creating a new field (or column) in your data source, the values or members of which are determined by a calculation that you control. This new calculated field is saved to your data source in Tableau, and can be used to create more robust visualisations. But don't worry: your original data remains untouched.\n\nYou can use calculated fields for many, many reasons. Some examples might include:\n\n• To segment data\n• To convert the data type of a field, such as converting a string to a date.\n• To aggregate data\n• To filter results\n• To calculate ratios\n\n## Types of calculations\n\nYou create calculated fields using calculations. There are three main types of calculations you can use to create calculated fields in Tableau:\n\n• Basic calculations - Basic calculations allow you to transform values or members at the data source level of detail (a row-level calculation) or at the visualisation level of detail (an aggregate calculation).\n\n• Level of Detail (LOD) expressions - Just like basic calculations, LOD calculations allow you to compute values at the data source level and the visualisation level. However, LOD calculations give you even more control on the level of granularity you want to compute. They can be performed at a more granular level (INCLUDE), a less granular level (EXCLUDE), or an entirely independent level (FIXED) with respect to the granularity of the visualisation.\n\n• Table calculations - Table calculations allow you to transform values at the level of detail of the visualisation only. For more information, see Transform Values with Table Calculations(Link opens in a new window).\n\nThe type of calculation you choose depends on the needs of your analysis and the question you want to answer.\n\n## Create a calculated field\n\nOnce you have determined the type of calculation you want to use, it's time to create a calculated field. This example uses a basic calculation.\n\nNote: The example in this article uses the Sample-Superstore data source that comes with Tableau Desktop. To follow along with the steps in this article, connect to the Sample-Superstore saved data source and navigate to Sheet 1.\n\n1. In Tableau, select Analysis > Create Calculated Field.\n\n2. In the Calculation Editor that opens, do the following:\n\n• Enter a name for the calculated field. In this example, the field is called, Discount Ratio.\n\n• Enter a formula. This example uses the following formula:\n\n`IIF([Sales] !=0, [Discount]/[Sales],0)`\n\nThis formula checks if sales is not equal to zero. If true, it returns the discount ratio (Discount/Sales); if false, it returns zero.\n\n Tip: To see a list of available functions, click the triangle icon on the right-side of the Calculation Editor.",
null,
"Each function includes syntax, a description, and an example for your reference. Double-click a function in the list to add it to the formula. For more tips, see Tips for Working with Calculated Fields in Tableau.\n3. When finished, click OK.\n\nThe new calculated field is added to Measures in the Data pane because it returns a number. An equal sign (=) appears next to the data type icon. All calculated fields have equal signs (=) next to them in the Data pane.",
null,
"## Use a calculated field in the view\n\n### Step 1: Build the view\n\n1. From Dimensions, drag Region to the Columns shelf.\n\n2. From Dimensions, drag Category to the Rows shelf.\n\n3. On the Rows shelf, click the plus icon (+) on the Category field to drill-down to Subcategory.\n\nThe view updates to look like this:",
null,
"### Step 2: Add the calculated field to the view\n\n1. From Measures, drag Discount Ratio to Colour on the Marks card.\n\nThe view updates to highlight table.",
null,
"You can see that Binders are heavily discounted in the Central region. Notice that Discount Ratio is automatically aggregated as a sum.\n\n2. On the Rows shelf, right-click SUM(Discount Ratio) and select Measure (Sum) > Average.\n\nThe view updates with the average of discount ratio shown.",
null,
"## Edit a Calculated Field\n\nIf at any time you need to change a calculation, you can edit the calculated field and it will update across your entire workbook.\n\nTo edit a calculated field:\n\n1. In the Data pane, right-click the calculated field and select Edit.\n\n2. In the Calculation Editor that opens, you can do the following:\n\n• Edit the name of the calculated field.\n\n• Update the formula.\n\nFor this example, the formula is changed to return a discount ratio for orders over 2000 USD in sales:\n\n`IIF([Sales] > 2000, [Discount]/[Sales],0)`\n\n3. Click OK.\n\nThe view updates to reflect the changes automatically. You do not need to re-add the updated calculated field to the view.",
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"https://help.tableau.com/current/pro/desktop/en-gb/Img/calc_field3.png",
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"https://help.tableau.com/current/pro/desktop/en-gb/Img/calc_field4.png",
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"https://help.tableau.com/current/pro/desktop/en-gb/Img/calcs_create1.png",
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"https://help.tableau.com/current/pro/desktop/en-gb/Img/calcs_create2.png",
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"https://help.tableau.com/current/pro/desktop/en-gb/Img/calcs_create3.png",
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"https://help.tableau.com/current/pro/desktop/en-gb/Img/calcs_create4.png",
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https://kr.mathworks.com/matlabcentral/answers/580836-3d-temperature-distribution-plot | [
"# 3D Temperature distribution plot\n\n조회 수: 47(최근 30일)\nJoe Check 2020년 8월 17일\n편집: jonas 2020년 8월 18일\nHi there,\nI'm working on a tribology problem which looks into the temperature distribution in a thin oil film between a piston ring and a cylinder bore. I wrote a code which incorporates the finite difference method to give me the output of how the temperature varies throughout the film thickness with different crank angles (engine strokes).\nMy code works fine however I'm not quite sure how to plot my results correctly. I and my supervisor chose matlab to generate nice 3D plots of the temperature distribution however I don't have enough matlab experience to do it correctly and I don't understand much having read the long descriptions in the Matlab documents. (The original version of my code was written in Fortran).\nThe problem is that all my temperature values are stored in a 3D matrix of the size (721 x 21 x 101).\nWhere the number 721 is a number of degrees showing the engine crank angle going from 1 to 721. 21 is the number of grid points in the x direction and 101 is the number of grid points in the y direction.\nWhen I wrote a first version of my code which didn’t include the crank angle part I just ended up with a 2D temperature matrix t of the size of 21 x 101 then I was able to use the meshgrid and mesh commands to do the plots and my code looked like the following:\n%Plot results\nkmin = 1;\nkmax = 101;\nimin = 1;\nimax = 21;\n[kk,ii] = meshgrid(kmin:kmax,imin:imax);\nmesh(kk,ii,t);\nxlabel('Y-Coordinate (mm)')\nylabel('X-Coordinate (mm)')\nzlabel('Temperature (deg)')\nNow I don't know how to deal with this big 3D matrix t which stores 721 pages of seperate 2D temperature matrices.\nDo I need to use a for-loop to loop through the crank angles?\nAny help would be greatly appreciated.\n##### 댓글 수: 2표시숨기기 이전 댓글 수: 1\nJoe Check 2020년 8월 17일\nHi Jonas,\nFor the time being let's say that I want to see a single surface plot showing the temperature distribution at the crank angle of 175 degrees, how do I do that?\n\n댓글을 달려면 로그인하십시오.\n\n### 채택된 답변\n\njonas 2020년 8월 17일\n편집: jonas 2020년 8월 17일\nIt is convenient to have the variable you want to plot as layers in your 3d data. In other words you want a XYZ matrix but you have a ZXY. Use permute to rearrange your data, A\nA = permute(A,[2,3,1]);\nthen you can easily plot the i'th layer as\nsurf(A(:,:,i));\nIf you want to plot a single coordinate over crank angles, then you could use\nout = squeeze(A(x,y,:))\n##### 댓글 수: 2표시숨기기 이전 댓글 수: 1\njonas 2020년 8월 18일\nSure, you can have multiple surface objects in the same graph. I would use a for-loop to plot different layers. Remember to use \"hold on\" on the axes. There is a big risk the plot will be messy.\nfigure\nsurf(peaks); hold on\nsurf(peaks+10)\nsurf(peaks+20)\nAnother option is to use a flat surface plot, like pcolor() or contourf(), to avoid intersections between different surface objects. You can change the z-value to some arbitrary offset.\nfor i = 1:5\nh(i) = pcolor(peaks);hold on\nset(h(i),'zdata',repmat(i*5,size(peaks,2),size(peaks,1)),'facealpha',0.8) %first argument of repmat (i*5) is a variable offset\nend\nset(h,'facealpha',0.8) %if you want transparent surfaces\nview(3) %view in 3d\n\n댓글을 달려면 로그인하십시오.\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
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https://discourse.julialang.org/t/plotly-slider-how-to-change-data-values/72939 | [
"# Plotly Slider - how to change data values?\n\nHello Community,\n\nWith the support of @empet I was able to create nice choropleth plots to visualize runoffs in german communities during the flooding in last summer.\nMy next step would be to add a slider in order to scroll through the most interesting 48 hours to show the water runoff within the communities.\n\nI have found only the possibility to update values by creating 48 traces and to switch 47 to unvisible and the actual one to visible. But creating 48 traces with 2000 shapes, names and values would lead to unnecessary large html files whereas only the values z (for coloring and hover-text) would need different values.\n\nTherefore, I would prefer a possibility to update only the z values of the one choropleth trace.\n\nUntil now, I have not found any example in the web, where only e.g. y-values of a scatterplot are updated when changing the slider position (without making N-1 traces invisible and 1 visible)\n\nDoes one know, if and in the case of yes, how z values (of choropleth) or y values (of scatter traces) can be updated when updating the slider (without defining plenties of traces with redundant data for x coordinates for scatterplots or geojsons for choropleth)?\n\nThanks for any hints!\n\n``````using PlotlyJS\nusing DataFrames\n\ndf = DataFrame(iso=[\"8\",\"7\",\"9\",\"2\",\"1\",\"4\",\"3\",\"6\",\"5\"],\nnames=[\"Vorarlberg\", \"Tirol\", \"Wien\", \"Kärnten\",\"Burgenland\",\n\"Oberösterreich\",\"Niederösterreich\",\"Steiermark\",\"Salzburg\"])\nfor k in 1:48\ndf[!, \"Wert\\$k\"] = rand(9)\nend\n\nchoropl = choroplethmapbox(geojson = \"https://raw.githubusercontent.com/ginseng666/GeoJSON-TopoJSON-Austria/\"*\n\"master/2016/simplified-95/laender_95_geo.json\",\nfeatureidkey = \"properties.iso\",\nlocations = df.iso,\nz=df.Wert1,\nzmin=0, zmax=1,\ntext=df.names, colorscale=\"Viridis\",\nmarker=attr(opacity=0.85, line=attr(width=0.5,color=\"white\")))\n\nlayout = Layout(mapbox = attr(center=attr(lon =13.5,lat=47.76),\nzoom=5.67, style=\"open-street-map\"),\nsliders=[attr(active=0,\ncurrentvalue=attr(prefix= \"Nach \", suffix= \" Stunde\"),\nsteps=[attr(label=\"\\$k\",\nmethod = \"restyle\",\nargs = [attr(z=[df[!,\"Wert\\$k\"]])]) for k in 1:48])]\n)\npl= Plot(choropl, layout)\n``````\n\nIn each slider step are updated z-values in the choroplethmapbox, as `z= [df[!, \"Wert\\$k\"]]`, NOT `z=df[!, \"Wert\\$k\"]`!!!\nThat’s because we can pass to Plot a vector of traces, not just a single trace.\n`z= [df[!, \"Wert\\$k\"]]` means that here is updated z from the first trace in the vector of traces.\nSimilarly are updated x, y values in different trace types.\n\nBut if your plot contains two traces, for example, and in each step we want to update y-values in both traces, then the corresponding arg in the slider steps will be defined as:\n\n``````args = [attr(y=[[2,5,3,4], [7, 1, 2, 5]])]\n``````\n1 Like\n\nThank you @empet for this well working solution!"
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https://eevibes.com/mathematics/linear-algebra/how-to-find-the-basis-of-a-vector-space/ | [
"# How to find the basis of a vector space?\n\nHow to find the basis of a vector space V? In order to find the basis of a vector space , we need to check two properties:\n\n1. The vectors should be linearly independent.\n2. These vectors should span in that vector space.\n\nIf both of these properties hold, then it means the given set of vectors form the basis otherwise not.\n\n## What are the standard basis of R2?\n\nThe standard basis of R2 is composed of two vectors v1(1,0) and v2=(0,1). The reason is, these two vectors can be used for generating any vector having two components. These two vectors are actually the axes : x-axis and y-axis. The first component of v1 indicates the presence of x axis while y-axis is missing. Similarly in v2, the second component=1 indicating the presence of y axis and all x-component along it is equal to zero.\n\nLets say we want to see first property i.e., v1 and v2 are linearly independent or not. For this solve\n\nc1v1+c2v2=0\n\nc1(1,0)+c2(0,1)=(0,0)\n\nIt can be seen easily only c1=0 and c2=0. No other values of c1 and c2 satisfy the above equation. That is why v1 and v2 are linearly independent.\n\nNow, check either they span the vector space R2 or not?\n\nFor this we will try to generate any vector having components (a1,a2) using the linear combinations of these two vectors.\n\n(a1,a2)=c1v1+c2v2\n\nLets say\n\na1=5, a2=6 so we will have\n\n(5,6)=c1v1+c2v2\n\n(5,6)=c1(1,0)+c2(0,1)\n\n1.c1+0.c2=5\n\n0.c1+1.c2=6\n\nBy solving these equations, we will get c1=5 while c2=6\n\nAs\n\n(5,6)=5(1,0)+6(0.1)\n\nSo, it is clear that we can generate any vector in R2 using these vectors. So v1 and v2 span the vector space V.\n\nFrom the above explanation both properties of the basis are satisfied, that is why v1 and v2 form the basis in R2"
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https://www.maillot-bonsai.com/en/boutique/3_wind-bells-and-chimes/4650_japanese-cast-iron-grape-vine-wind-bell-g118 | [
"",
null,
"",
null,
"",
null,
"##### The Japanese Bonsai specialist\nDirect order Contact Help / Services Newsletter",
null,
"# Japanese cast iron grape vine wind bell g118\n\n› Wind bells and chimes",
null,
"",
null,
"ref. : 4650\n\n43,00\n\nAvailable quantity : 1",
null,
"Order\n\n###### Description\n\n12.5 cm. Green patina wind bell featuring a dragonfly and grapevine. A great gift for a wine lover. Well made and detailed, as is typical of Nambu cast iron products made in Morioka in Iwate Prefecture.\n\n#wind 5.4 #bell 3.4 #cast 3.4 #iron 3.4 #japanese 2.8 #chimes 2.6 #bells 2.5 #grape 2.5 #vine 2.4 #made 2.4\n\nFormule\n(( ROUND((CHAR_LENGTH(b.article_nom)-CHAR_LENGTH(REPLACE(b.article_nom, 'bell', '')))/LENGTH('bell')) + ROUND((CHAR_LENGTH(b.article_description)-CHAR_LENGTH(REPLACE(b.article_description, 'bell', '')))/LENGTH('bell')) ) * 3.4) + (( ROUND((CHAR_LENGTH(b.article_nom)-CHAR_LENGTH(REPLACE(b.article_nom, 'iron', '')))/LENGTH('iron')) + ROUND((CHAR_LENGTH(b.article_description)-CHAR_LENGTH(REPLACE(b.article_description, 'iron', '')))/LENGTH('iron')) ) * 3.4) + (( ROUND((CHAR_LENGTH(b.article_nom)-CHAR_LENGTH(REPLACE(b.article_nom, 'wind', '')))/LENGTH('wind')) + ROUND((CHAR_LENGTH(b.article_description)-CHAR_LENGTH(REPLACE(b.article_description, 'wind', '')))/LENGTH('wind')) ) * 3.4) + (( ROUND((CHAR_LENGTH(b.article_nom)-CHAR_LENGTH(REPLACE(b.article_nom, 'cast', '')))/LENGTH('cast')) + ROUND((CHAR_LENGTH(b.article_description)-CHAR_LENGTH(REPLACE(b.article_description, 'cast', '')))/LENGTH('cast')) ) * 3.4) + (( ROUND((CHAR_LENGTH(b.article_nom)-CHAR_LENGTH(REPLACE(b.article_nom, 'japanese', '')))/LENGTH('japanese')) + ROUND((CHAR_LENGTH(b.article_description)-CHAR_LENGTH(REPLACE(b.article_description, 'japanese', '')))/LENGTH('japanese')) ) * 2.8) + (( ROUND((CHAR_LENGTH(b.article_nom)-CHAR_LENGTH(REPLACE(b.article_nom, 'grape', '')))/LENGTH('grape')) + ROUND((CHAR_LENGTH(b.article_description)-CHAR_LENGTH(REPLACE(b.article_description, 'grape', '')))/LENGTH('grape')) ) * 2.5) + (( ROUND((CHAR_LENGTH(b.article_nom)-CHAR_LENGTH(REPLACE(b.article_nom, 'g118', '')))/LENGTH('g118')) + ROUND((CHAR_LENGTH(b.article_description)-CHAR_LENGTH(REPLACE(b.article_description, 'g118', '')))/LENGTH('g118')) ) * 2.4) + (( ROUND((CHAR_LENGTH(b.article_nom)-CHAR_LENGTH(REPLACE(b.article_nom, 'vine', '')))/LENGTH('vine')) + ROUND((CHAR_LENGTH(b.article_description)-CHAR_LENGTH(REPLACE(b.article_description, 'vine', '')))/LENGTH('vine')) ) * 2.4) + (( ROUND((CHAR_LENGTH(b.article_nom)-CHAR_LENGTH(REPLACE(b.article_nom, 'made', '')))/LENGTH('made')) + ROUND((CHAR_LENGTH(b.article_description)-CHAR_LENGTH(REPLACE(b.article_description, 'made', '')))/LENGTH('made')) ) * 2.4) + (( ROUND((CHAR_LENGTH(b.article_nom)-CHAR_LENGTH(REPLACE(b.article_nom, 'prefecture', '')))/LENGTH('prefecture')) + ROUND((CHAR_LENGTH(b.article_description)-CHAR_LENGTH(REPLACE(b.article_description, 'prefecture', '')))/LENGTH('prefecture')) ) * 2)\n\n## Secure payment",
null,
"## Delivery\n\nOur logistic partners :",
null,
"",
null,
"04 74 55 23 48\nPépinière MAILLOT-BONSAÏ\nLe Bois Frazy\n01990 RELEVANT - FRANCE\non appointment",
null,
""
]
| [
null,
"https://www.maillot-bonsai.com/images/front/info-close.png",
null,
"https://www.maillot-bonsai.com/images/logos/logo.png",
null,
"https://www.maillot-bonsai.com/images/logos/logo-bonsai.png",
null,
"https://www.maillot-bonsai.com/images/lang/en.png",
null,
"https://www.maillot-bonsai.com/images/articles/300/maillot-bonsai_1-4650.jpg",
null,
"https://www.maillot-bonsai.com/images/articles/60/maillot-bonsai_1-4650.jpg",
null,
"https://www.maillot-bonsai.com/images/front/add-panier.png",
null,
"https://www.maillot-bonsai.com/images/front/moyens-paiement-2021.png",
null,
"https://www.maillot-bonsai.com/images/front/logo-laposte-pied.png",
null,
"https://www.maillot-bonsai.com/images/front/logo-dpd-pied.png",
null,
"https://www.maillot-bonsai.com/images/logos/logo-bonsai.png",
null
]
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https://www.iue.tuwien.ac.at/phd/brech/ch_3_2.htm | [
"",
null,
"",
null,
"",
null,
"",
null,
"Next: 3.3 Determination of Capacitances by Quasi Static Approximation Up: 3 Measurement and Parameter Extraction Previous: 3.1 DC and RF Measurements\n\n3.2 Small Signal Equivalent Circuit Parameter Extraction\n\nThe HEMT can be described by an equivalent circuit model shown in Figure 3.1. To extract the model parameters from measured Sparameters a value for each parameter of the model is assumed and Sparameters are calculated based on the equivalent circuit. The chosen values are then optimized by minimizing the differences between the calculated and the measured Sparameters.",
null,
"Figure 3.1 Equivalent circuit model for parameter extraction. It includes the intrinsic device, the series resitances to all three terminals as well as the parasitic capacitances and inductances of the contacting network.\n\nThe transit frequencies fT and fmax can be traced back to equivalent circuit parameters. From the intrinsic device the well known approximation for the current gain cut-off frequency fT can be derived as",
null,
". (10)\n\nExtrinsic values of gm and CGS can be calculated by gm = gmi /(1+gmiRs) and CGS = CGSi (1+gmRs). In the active bias region of a HEMT (VDS > 0.5 V and VGS > VT) CGDi is only about 15 % of CGSi. Therefore fT can also be approximated using the extrinsic gm and the extrinsic total gate capacitance CG = CGS + CGDi in (10).\n\nAccording to [37, 38] fmax can also be approximated with equivalent circuit parameters by:",
null,
". (11)\n\nFor measuring purposes the devices are embedded in a network of coplanar lines and contacting pads. These will be not included in the simulation area as described in Chapter 5. The contacting network can be neglected in first order for DC measurements but must be taken into account when simulated and measured RF parameters such as fT and fmax are compared. Thus, the parasitic elements LG, LD, LS, CPG and CPD have to be included in the equivalent circuit model.\n\nBased on the equivalent circuit the parameters can be extracted from measured Sparameters. In the extraction procedure it is difficult to distinguish between the parasitic pad capacitances CPG and CPD and the gate source capacitance CGS because they are connected in parallel and therefore only their sum can be determined reliably. In first order CPG and CPD are the only capacitances which do not scale with the gate width Lw. This can be used to extract these values in a different way.\n\nThe measured value of fT includes all capacitances. The total capacitance can be determined rearranging (10) to",
null,
". (12)\n\nAs gm, CGS and CGD scale linearly with Lw and both CPG and CPD are constant for all the devices the sum CPG + CPD can be determined by extrapolating Ctot to Lw = 0. The calculated Ctot using measured gm and fT for different bias points is shown in Figure 3.2. The intercept for all linear fits is very close to about 25fF. This value is used for parameter extraction of all measured devices. More details of this deembedding procedure are described in .",
null,
"Figure 3.2 Total capacitance of HEMTs versus gate width at VDS=2.0V. The capacitances are calculated using (12) with measured gm and fT of devices with gate widths Lw= 80, 180, 360 µm. For all VGS a linear fit can be found. The intercept determines the constant part of Ctot, i. e. CPG + CPG.\n\nThe schematic of the extrinsic device used for simulation is shown in Figure 2.1 It corresponds to the intrinsic device indicated by the dashed box in Figure 3.1 and the parasitic resistances RS, RG, and RD. The deembedded device parameters will be the basis for comparison between measured and simulated data.\n\nHelmut Brech\n1998-03-11"
]
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null,
"https://www.iue.tuwien.ac.at/phd/brech/next_motif.gif",
null,
"https://www.iue.tuwien.ac.at/phd/brech/up_motif.gif",
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"https://www.iue.tuwien.ac.at/phd/brech/previous_motif.gif",
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"https://www.iue.tuwien.ac.at/phd/brech/contents_motif.gif",
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"https://www.iue.tuwien.ac.at/phd/brech/IMG00024.GIF",
null,
"https://www.iue.tuwien.ac.at/phd/brech/IMG00025.GIF",
null,
"https://www.iue.tuwien.ac.at/phd/brech/IMG00026.GIF",
null,
"https://www.iue.tuwien.ac.at/phd/brech/IMG00027.GIF",
null,
"https://www.iue.tuwien.ac.at/phd/brech/IMG00028.GIF",
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https://eel.is/c++draft/expr.static.cast | [
"# 7 Expressions [expr]\n\n## 7.6 Compound expressions [expr.compound]\n\n### 7.6.1 Postfix expressions [expr.post]\n\n#### 7.6.1.9 Static cast [expr.static.cast]\n\nThe result of the expression static_cast<T>(v) is the result of converting the expression v to type T.\nIf T is an lvalue reference type or an rvalue reference to function type, the result is an lvalue; if T is an rvalue reference to object type, the result is an xvalue; otherwise, the result is a prvalue.\nThe static_cast operator shall not cast away constness.\nAn lvalue of type “cv1 B”, where B is a class type, can be cast to type “reference to cv2 D”, where D is a class derived from B, if cv2 is the same cv-qualification as, or greater cv-qualification than, cv1.\nIf B is a virtual base class of D or a base class of a virtual base class of D, or if no valid standard conversion from “pointer to D” to “pointer to B” exists ([conv.ptr]), the program is ill-formed.\nAn xvalue of type “cv1 B” can be cast to type “rvalue reference to cv2 D” with the same constraints as for an lvalue of type “cv1 B.\nIf the object of type “cv1 B” is actually a base class subobject of an object of type D, the result refers to the enclosing object of type D.\nOtherwise, the behavior is undefined.\n[Example 1: struct B { }; struct D : public B { }; D d; B &br = d; static_cast<D&>(br); // produces lvalue denoting the original d object — end example]\nAn lvalue of type T1 can be cast to type “rvalue reference to T2” if T2 is reference-compatible with T1 ([dcl.init.ref]).\nIf the value is not a bit-field, the result refers to the object or the specified base class subobject thereof; otherwise, the lvalue-to-rvalue conversion is applied to the bit-field and the resulting prvalue is used as the expression of the static_cast for the remainder of this subclause.\nIf T2 is an inaccessible or ambiguous base class of T1, a program that necessitates such a cast is ill-formed.\nAn expression E can be explicitly converted to a type T if there is an implicit conversion sequence ([over.best.ics]) from E to T, if overload resolution for a direct-initialization ([dcl.init]) of an object or reference of type T from E would find at least one viable function ([over.match.viable]), or if T is an aggregate type ([dcl.init.aggr]) having a first element x and there is an implicit conversion sequence from E to the type of x.\nIf T is a reference type, the effect is the same as performing the declaration and initialization T t(E); for some invented temporary variable t ([dcl.init]) and then using the temporary variable as the result of the conversion.\nOtherwise, the result object is direct-initialized from E.\n[Note 1:\nThe conversion is ill-formed when attempting to convert an expression of class type to an inaccessible or ambiguous base class.\n— end note]\n[Note 2:\nIf T is “array of unknown bound of U”, this direct-initialization defines the type of the expression as U.\n— end note]\nOtherwise, the static_cast shall perform one of the conversions listed below.\nNo other conversion shall be performed explicitly using a static_cast.\nAny expression can be explicitly converted to type cv void, in which case it becomes a discarded-value expression.\n[Note 3:\nHowever, if the value is in a temporary object, the destructor for that object is not executed until the usual time, and the value of the object is preserved for the purpose of executing the destructor.\n— end note]\nThe inverse of any standard conversion sequence not containing an lvalue-to-rvalue, array-to-pointer, function-to-pointer, null pointer, null member pointer, boolean, or function pointer conversion, can be performed explicitly using static_cast.\nA program is ill-formed if it uses static_cast to perform the inverse of an ill-formed standard conversion sequence.\n[Example 2: struct B { }; struct D : private B { }; void f() { static_cast<D*>((B*)0); // error: B is a private base of D static_cast<int B::*>((int D::*)0); // error: B is a private base of D } — end example]\nThe lvalue-to-rvalue, array-to-pointer, and function-to-pointer conversions are applied to the operand.\nSuch a static_cast is subject to the restriction that the explicit conversion does not cast away constness, and the following additional rules for specific cases:\nA value of a scoped enumeration type ([dcl.enum]) can be explicitly converted to an integral type; the result is the same as that of converting to the enumeration's underlying type and then to the destination type.\nA value of a scoped enumeration type can also be explicitly converted to a floating-point type; the result is the same as that of converting from the original value to the floating-point type.\nA value of integral or enumeration type can be explicitly converted to a complete enumeration type.\nIf the enumeration type has a fixed underlying type, the value is first converted to that type by integral conversion, if necessary, and then to the enumeration type.\nIf the enumeration type does not have a fixed underlying type, the value is unchanged if the original value is within the range of the enumeration values ([dcl.enum]), and otherwise, the behavior is undefined.\nA value of floating-point type can also be explicitly converted to an enumeration type.\nThe resulting value is the same as converting the original value to the underlying type of the enumeration ([conv.fpint]), and subsequently to the enumeration type.\nA prvalue of type “pointer to cv1 B”, where B is a class type, can be converted to a prvalue of type “pointer to cv2 D”, where D is a complete class derived from B, if cv2 is the same cv-qualification as, or greater cv-qualification than, cv1.\nIf B is a virtual base class of D or a base class of a virtual base class of D, or if no valid standard conversion from “pointer to D” to “pointer to B” exists ([conv.ptr]), the program is ill-formed.\nThe null pointer value ([basic.compound]) is converted to the null pointer value of the destination type.\nIf the prvalue of type “pointer to cv1 B” points to a B that is actually a subobject of an object of type D, the resulting pointer points to the enclosing object of type D.\nOtherwise, the behavior is undefined.\nA prvalue of type “pointer to member of D of type cv1 T” can be converted to a prvalue of type “pointer to member of B of type cv2 T”, where D is a complete class type and B is a base class of D, if cv2 is the same cv-qualification as, or greater cv-qualification than, cv1.\n[Note 4:\nFunction types (including those used in pointer-to-member-function types) are never cv-qualified ([dcl.fct]).\n— end note]\nIf no valid standard conversion from “pointer to member of B of type T” to “pointer to member of D of type T” exists ([conv.mem]), the program is ill-formed.\nThe null member pointer value is converted to the null member pointer value of the destination type.\nIf class B contains the original member, or is a base or derived class of the class containing the original member, the resulting pointer to member points to the original member.\nOtherwise, the behavior is undefined.\n[Note 5:\nAlthough class B need not contain the original member, the dynamic type of the object with which indirection through the pointer to member is performed must contain the original member; see [expr.mptr.oper].\n— end note]\nA prvalue of type “pointer to cv1 void” can be converted to a prvalue of type “pointer to cv2 T”, where T is an object type and cv2 is the same cv-qualification as, or greater cv-qualification than, cv1.\nIf the original pointer value represents the address A of a byte in memory and A does not satisfy the alignment requirement of T, then the resulting pointer value is unspecified.\nOtherwise, if the original pointer value points to an object a, and there is an object b of type T (ignoring cv-qualification) that is pointer-interconvertible with a, the result is a pointer to b.\nOtherwise, the pointer value is unchanged by the conversion.\n[Example 3: T* p1 = new T; const T* p2 = static_cast<const T*>(static_cast<void*>(p1)); bool b = p1 == p2; // b will have the value true. — end example]"
]
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null
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https://www.jpost.com/defense/lindenstrauss-slams-police-defense-ministry-on-checkpoints | [
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.96811956,"math_prob":0.96919554,"size":2374,"snap":"2023-40-2023-50","text_gpt3_token_len":433,"char_repetition_ratio":0.118987344,"word_repetition_ratio":0.011527377,"special_character_ratio":0.17186184,"punctuation_ratio":0.08080808,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9789482,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-11T09:25:35Z\",\"WARC-Record-ID\":\"<urn:uuid:914cb0ab-002b-4684-8797-d2096944e590>\",\"Content-Length\":\"84804\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3ba16c81-a481-4f66-b084-12a639f02d14>\",\"WARC-Concurrent-To\":\"<urn:uuid:69be8fd6-b651-4ea7-be04-df9de23b8304>\",\"WARC-IP-Address\":\"34.149.213.158\",\"WARC-Target-URI\":\"https://www.jpost.com/defense/lindenstrauss-slams-police-defense-ministry-on-checkpoints\",\"WARC-Payload-Digest\":\"sha1:OP5RT2MZQIC3VZ6WWDKHT5K476THBOIY\",\"WARC-Block-Digest\":\"sha1:IJDGZPMYCTNBMZCOZHEKOS6XGQRFEA67\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679103810.88_warc_CC-MAIN-20231211080606-20231211110606-00790.warc.gz\"}"} |
https://en.unionpedia.org/1_%2B_2_%2B_3_%2B_4_%2B_⋯ | [
"Install",
null,
"Faster access than browser!\n\n# 1 + 2 + 3 + 4 + ⋯\n\nThe infinite series whose terms are the natural numbers is a divergent series. \n\n61 relations: A Disappearing Number, Additive identity, Alternating series, Analytic continuation, Asymptote, Bernoulli number, Bosonic string theory, Bounded function, Brady Haran, Cesàro summation, Complex analysis, David Leavitt, Dirichlet eta function, Dirichlet series, Divergent series, Edward Frenkel, Euler–Maclaurin formula, G. H. Hardy, Goddard–Thorn theorem, Grandi's series, Infinity, John C. Baez, John Edensor Littlewood, Limit of a sequence, Luboš Motl, Mollifier, Monstrous moonshine, Morris Kline, Natural number, One-sided limit, Power series, Pythagoreanism, Quantum field theory, Quantum harmonic oscillator, Ramanujan summation, Real analysis, Riemann zeta function, Scalar field, Scientific American, Sequence, Series (mathematics), Simon McBurney, Smithsonian (magazine), Smoothness, Srinivasa Ramanujan, String theory, Support (mathematics), Terence Tao, Term test, The Indian Clerk, ... Expand index (11 more) »\n\n## A Disappearing Number\n\nA Disappearing Number is a 2007 play co-written and devised by the Théâtre de Complicité company and directed and conceived by English playwright Simon McBurney.\n\nIn mathematics the additive identity of a set which is equipped with the operation of addition is an element which, when added to any element x in the set, yields x. One of the most familiar additive identities is the number 0 from elementary mathematics, but additive identities occur in other mathematical structures where addition is defined, such as in groups and rings.\n\n## Alternating series\n\nIn mathematics, an alternating series is an infinite series of the form with an > 0 for all n.\n\n## Analytic continuation\n\nIn complex analysis, a branch of mathematics, analytic continuation is a technique to extend the domain of a given analytic function.\n\n## Asymptote\n\nIn analytic geometry, an asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tends to infinity.\n\n## Bernoulli number\n\nIn mathematics, the Bernoulli numbers are a sequence of rational numbers which occur frequently in number theory.\n\n## Bosonic string theory\n\nBosonic string theory is the original version of string theory, developed in the late 1960s.\n\n## Bounded function\n\nIn mathematics, a function f defined on some set X with real or complex values is called bounded, if the set of its values is bounded.\n\nBrady John Haran (born 18 June 1976) is an Australian-born British independent filmmaker and video journalist who is known for his educational videos and documentary films produced for BBC News and his YouTube channels, the most notable being Periodic Videos and Numberphile.\n\n## Cesàro summation\n\nIn mathematical analysis, Cesàro summation (also known as the Cesàro mean) assigns values to some infinite sums that are not convergent in the usual sense.\n\n## Complex analysis\n\nComplex analysis, traditionally known as the theory of functions of a complex variable, is the branch of mathematical analysis that investigates functions of complex numbers.\n\n## David Leavitt\n\nDavid Leavitt (born June 23, 1961) is an American novelist, short story writer, and biographer.\n\n## Dirichlet eta function\n\nIn mathematics, in the area of analytic number theory, the Dirichlet eta function is defined by the following Dirichlet series, which converges for any complex number having real part > 0: This Dirichlet series is the alternating sum corresponding to the Dirichlet series expansion of the Riemann zeta function, ζ(s) — and for this reason the Dirichlet eta function is also known as the alternating zeta function, also denoted ζ*(s).\n\n## Dirichlet series\n\nIn mathematics, a Dirichlet series is any series of the form where s is complex, and a_n is a complex sequence.\n\n## Divergent series\n\nIn mathematics, a divergent series is an infinite series that is not convergent, meaning that the infinite sequence of the partial sums of the series does not have a finite limit.\n\n## Edward Frenkel\n\nEdward Vladimirovich Frenkel (sometimes spelled Э́двард Фре́нкель; born May 2, 1968) is a Russian-American mathematician working in representation theory, algebraic geometry, and mathematical physics.\n\n## Euler–Maclaurin formula\n\nIn mathematics, the Euler–Maclaurin formula provides a powerful connection between integrals (see calculus) and sums.\n\n## G. H. Hardy\n\nGodfrey Harold Hardy (7 February 1877 – 1 December 1947) was an English mathematician, known for his achievements in number theory and mathematical analysis.\n\n## Goddard–Thorn theorem\n\nIn mathematics, and in particular, in the mathematical background of string theory, the Goddard–Thorn theorem (also called the no-ghost theorem) is a theorem describing properties of a functor that quantizes bosonic strings.\n\n## Grandi's series\n\nIn mathematics, the infinite series 1 - 1 + 1 - 1 + \\dotsb, also written \\sum_^ (-1)^n is sometimes called Grandi's series, after Italian mathematician, philosopher, and priest Guido Grandi, who gave a memorable treatment of the series in 1703.\n\n## Infinity\n\nInfinity (symbol) is a concept describing something without any bound or larger than any natural number.\n\n## John C. Baez\n\nJohn Carlos Baez (born June 12, 1961) is an American mathematical physicist and a professor of mathematics at the University of California, Riverside (UCR) in Riverside, California.\n\n## John Edensor Littlewood\n\nJohn Edensor Littlewood FRS LLD (9 June 1885 – 6 September 1977) was an English mathematician.\n\n## Limit of a sequence\n\nAs the positive integer n becomes larger and larger, the value n\\cdot \\sin\\bigg(\\frac1\\bigg) becomes arbitrarily close to 1.\n\n## Luboš Motl\n\nLuboš Motl (born December 5, 1973) is a Czech theoretical physicist.\n\n## Mollifier\n\nIn mathematics, mollifiers (also known as approximations to the identity) are smooth functions with special properties, used for example in distribution theory to create sequences of smooth functions approximating nonsmooth (generalized) functions, via convolution.\n\n## Monstrous moonshine\n\nIn mathematics, monstrous moonshine, or moonshine theory, is the unexpected connection between the monster group M and modular functions, in particular, the ''j'' function.\n\n## Morris Kline\n\nMorris Kline (May 1, 1908 – June 10, 1992) was a Professor of Mathematics, a writer on the history, philosophy, and teaching of mathematics, and also a popularizer of mathematical subjects.\n\n## Natural number\n\nIn mathematics, the natural numbers are those used for counting (as in \"there are six coins on the table\") and ordering (as in \"this is the third largest city in the country\").\n\n## One-sided limit\n\nIn calculus, a one-sided limit is either of the two limits of a function f(x) of a real variable x as x approaches a specified point either from below or from above.\n\n## Power series\n\nIn mathematics, a power series (in one variable) is an infinite series of the form where an represents the coefficient of the nth term and c is a constant.\n\n## Pythagoreanism\n\nPythagoreanism originated in the 6th century BC, based on the teachings and beliefs held by Pythagoras and his followers, the Pythagoreans, who were considerably influenced by mathematics and mysticism.\n\n## Quantum field theory\n\nIn theoretical physics, quantum field theory (QFT) is the theoretical framework for constructing quantum mechanical models of subatomic particles in particle physics and quasiparticles in condensed matter physics.\n\n## Quantum harmonic oscillator\n\nThe quantum harmonic oscillator is the quantum-mechanical analog of the classical harmonic oscillator.\n\n## Ramanujan summation\n\nRamanujan summation is a technique invented by the mathematician Srinivasa Ramanujan for assigning a value to divergent infinite series.\n\n## Real analysis\n\nIn mathematics, real analysis is the branch of mathematical analysis that studies the behavior of real numbers, sequences and series of real numbers, and real-valued functions.\n\n## Riemann zeta function\n\nThe Riemann zeta function or Euler–Riemann zeta function,, is a function of a complex variable s that analytically continues the sum of the Dirichlet series which converges when the real part of is greater than 1.\n\n## Scalar field\n\nIn mathematics and physics, a scalar field associates a scalar value to every point in a space – possibly physical space.\n\n## Scientific American\n\nScientific American (informally abbreviated SciAm) is an American popular science magazine.\n\n## Sequence\n\nIn mathematics, a sequence is an enumerated collection of objects in which repetitions are allowed.\n\n## Series (mathematics)\n\nIn mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities, one after the other, to a given starting quantity.\n\n## Simon McBurney\n\nSimon Montagu McBurney, OBE (born 25 August 1957) is an English actor, writer and director.\n\n## Smithsonian (magazine)\n\nSmithsonian is the official journal published by the Smithsonian Institution in Washington, D.C. The first issue was published in 1970.\n\n## Smoothness\n\nIn mathematical analysis, the smoothness of a function is a property measured by the number of derivatives it has that are continuous.\n\n## Srinivasa Ramanujan\n\nSrinivasa Ramanujan (22 December 188726 April 1920) was an Indian mathematician who lived during the British Rule in India. Though he had almost no formal training in pure mathematics, he made substantial contributions to mathematical analysis, number theory, infinite series, and continued fractions, including solutions to mathematical problems considered to be unsolvable.\n\n## String theory\n\nIn physics, string theory is a theoretical framework in which the point-like particles of particle physics are replaced by one-dimensional objects called strings.\n\n## Support (mathematics)\n\nIn mathematics, the support of a real-valued function f is the subset of the domain containing those elements which are not mapped to zero.\n\n## Terence Tao\n\nTerence Chi-Shen Tao (born 17 July 1975) is an Australian-American mathematician who has worked in various areas of mathematics.\n\n## Term test\n\nIn mathematics, the nth-term test for divergenceKaczor p.336 is a simple test for the divergence of an infinite series.\n\n## The Indian Clerk\n\nThe Indian Clerk is a biographical novel by David Leavitt, published in 2007.\n\n## The New York Times\n\nThe New York Times (sometimes abbreviated as The NYT or The Times) is an American newspaper based in New York City with worldwide influence and readership.\n\n## Thomas John I'Anson Bromwich\n\nThomas John I'Anson Bromwich (1875–1929) was an English mathematician, and a Fellow of the Royal Society.\n\n## Transverse wave\n\nA transverse wave is a moving wave that consists of oscillations occurring perpendicular (right angled) to the direction of energy transfer (or the propagation of the wave).\n\n## Triangular number\n\nA triangular number or triangle number counts objects arranged in an equilateral triangle, as in the diagram on the right.\n\n## University of Nottingham\n\nThe University of Nottingham is a public research university in Nottingham, United Kingdom.\n\n## Zeta function regularization\n\nIn mathematics and theoretical physics, zeta function regularization is a type of regularization or summability method that assigns finite values to divergent sums or products, and in particular can be used to define determinants and traces of some self-adjoint operators.\n\n## 1 + 1 + 1 + 1 + ⋯\n\nIn mathematics,, also written \\sum_^ n^0, \\sum_^ 1^n, or simply \\sum_^ 1, is a divergent series, meaning that its sequence of partial sums does not converge to a limit in the real numbers.\n\n## 1 + 2 + 4 + 8 + ⋯\n\nIn mathematics, is the infinite series whose terms are the successive powers of two.\n\n## 1 − 2 + 3 − 4 + ⋯\n\nIn mathematics, 1 − 2 + 3 − 4 + ··· is the infinite series whose terms are the successive positive integers, given alternating signs.\n\n## 13 Reasons Why\n\n13 Reasons Why (stylized onscreen as TH1RTEEN R3ASONS WHY) is an American teen drama web television series developed for Netflix by Brian Yorkey, based on the 2007 novel Thirteen Reasons Why by Jay Asher.\n\n## References\n\nHey! We are on Facebook now! »"
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"https://en.unionpedia.org/static/images/tick.png",
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https://www.math-only-math.com/vertical-addition.html | [
"Now we will learn simple Vertical Addition of 1-digit number by arranging them one number under the other number.\n\nHow to add 1-digit number vertically?",
null,
"Here we added 2 plus 5? (Follow the steps along with the above picture)\n\n(i) Now we will write that as an addition problem arranging vertically.\n\n(ii) Beside the first number we will put two strokes since the first number is 2.\n\n(iii) Beside the second number we will put five strokes since the second number is 5.\n\n(iv) Count the strokes beside the two numbers together to get the answer.\n\n(v) The answer is 7.\n\nAddition Word Problems - 1-Digit Numbers\n\nSubtracting 1-Digit Number\n\nMissing Number in Addition\n\nMissing Number in Subtraction\n\nSubtraction Word Problems – 1-Digit Numbers\n\nMissing Addend Sums with 1-Digit Number\n\nAddition Fact Sums to 10\n\nAddition Fact Sums to 11\n\nAddition Fact Sums to 12\n\nAddition Fact Sums to 13\n\nAddition Facts of 14, 15, 16, 17 and 18\n\nSubtracting 1-Digit Number\n\nSubtracting 2-Digit Numbers"
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"https://www.math-only-math.com/images/xvertical-addition.jpg.pagespeed.ic.DKCXypIe4H.jpg",
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https://techoverflow.net/2019/07/29/volts-to-db%C2%B5v-online-calculator-ampamp-python-code/ | [
"# Volts to dBµV online calculator & Python code\n\nUse this online calculator to convert a voltage in Volts to a voltage in dBµV.\n\nTechOverflow calculators:\nYou can enter values with SI suffixes like 12.2m (equivalent to 0.012) or 14k (14000) or 32u (0.000032).\nThe results are calculated while you type and shown directly below the calculator, so there is no need to press return or click on a Calculate button. Just make sure that all inputs are green by entering valid values.\n\nV\n\n#### Formula:\n\n$$U_{\\text{dBµV}} = \\frac{20\\cdot\\log(1\\,000\\,000 \\cdot U_V)}{ \\log(2) + \\log(5)}$$\n\n#### Python code:\n\nimport math\ndef volts_to_dbuv(v):\n\"\"\"Convert a voltage in volts to a voltage in dBµV\"\"\"\nreturn (20*math.log(1e6 * v))/(math.log(2) + math.log(5))"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.672154,"math_prob":0.9982213,"size":1058,"snap":"2022-40-2023-06","text_gpt3_token_len":306,"char_repetition_ratio":0.116698295,"word_repetition_ratio":0.62352943,"special_character_ratio":0.30623817,"punctuation_ratio":0.110091746,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99859166,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-03T10:08:32Z\",\"WARC-Record-ID\":\"<urn:uuid:c7284c12-1905-42e5-a172-98cb8212c23e>\",\"Content-Length\":\"43527\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a4368baf-ccf5-4459-8fc3-e2e5ee5d7fa0>\",\"WARC-Concurrent-To\":\"<urn:uuid:2ae324d1-1fb3-4155-a1ec-8284623c03a2>\",\"WARC-IP-Address\":\"104.21.41.193\",\"WARC-Target-URI\":\"https://techoverflow.net/2019/07/29/volts-to-db%C2%B5v-online-calculator-ampamp-python-code/\",\"WARC-Payload-Digest\":\"sha1:5AVJJC5SVQWUGN5SVLHH5EQDO2OYR26L\",\"WARC-Block-Digest\":\"sha1:WWQZMKMK2CNYSSDATEREEBGX3CXSLUI3\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500044.66_warc_CC-MAIN-20230203091020-20230203121020-00446.warc.gz\"}"} |
http://www.927953.com/xt/zb/521076.html | [
"# A股市场：现阶段A股市场适合投资房产、股票还是留着现金？这才是真正的价值投资\n\n2020年04月02日 10:45:03 来源： 财富动力网综合",
null,
"如果没有交易纪律，技术等于零\n\n\"宁愿错过不要做错\"是一句著名的股市谚语。但能够真正做到这一点的人不多。究其原因，仍然是人性的弱点——贪婪所致。",
null,
"",
null,
"",
null,
"",
null,
"(一)、高位天量天价",
null,
"(二)、无量空跌",
null,
"(三)、高位量增价平",
null,
"(四)、高位量增价跌",
null,
"(1)高位的价跌量增，往往表示获利丰厚的长线筹码，开始不计成本地杀跌出局。虽然逢低入场的资金逐步增加，但是抛盘数量仍然超过了买盘数量。后市股价仍有较大的下跌空间。\n\n(2)价跌量增表明抛盘非常坚决，股价往往呈现连续下跌。因此投资者发现这种走势后，应该立即进行卖出操作，不可抱着“等反弹再卖”的想法。\n\n(3)在“价跌量增”时，投资者不可轻易地“逢低入场”，否则很容易成为机构杀跌出货的对象。\n\n1、低位放量买入定式\n\n(3)如发现巨量后成交量不能有效继续放大，应引起高度重视，最保险的做法是先清仓出局，观望后续走势。",
null,
"(1)突然放量时股价位置必须在相对低位，可从个股历史走势中确认，这样买入信号才可靠。\n\n(2)突然放量前的股价应在一个时间段内获得支撑，有跌不下去之感，股价呈平台整理形态，此区域成交量呈均匀缩量状态，突然放量才有效。\n\n(3)注意低位放量区域与前一轮行情高点的距离。与前期高点距离越远，空头力量消耗得越充分，多头力量确认的可靠性越强。\n\n2、量能突破买入定式",
null,
"(1)放出巨量时股价应处于相对低位，如果大盘已有较大涨幅，个股也有超过一倍升幅之后出现的放巨量现象，应引起高度重视，这有可能是主力在拉高出货。\n\n(2)股价在低位整理时间越长，出现巨量后股价上涨的概率越大、涨幅 越高。\n\n3、缩量横盘买入定式",
null,
"X_1:=IF(PERIOD=1,5,IF(PERIOD=2,15,IF(PERIOD=3,30,IF(PERIOD=4,60,IF(PERIOD=5,240,1)))));\n\nX_2:=MOD(FROMOPEN,X_1);\n\nX_3:=IF(X_2<0.5,X_1,X_2);\n\nX_4:=IF(CURRBARSCOUNT=1,VOL*X_1/X_3,DRAWNULL);\n\nSTICKLINE(CURRBARSCOUNT=1 AND (SETCODE=0 OR SETCODE=1),X_4,0,(-1),(-1)),COLOR00C0C0;\n\nVOLUME:VOL,VOLSTICK;\n\nX_5:=IF(CAPITAL=0,AMOUNT/100000000,VOL/CAPITAL*100);\n\nX_6:=EMA(X_5,5);\n\nX_7:=MA(X_5,13);\n\nX_8:=X_5\n\nX_9:=BARSLAST(X_8);\n\nSTICKLINE(五倍地量>0 AND CLOSE>OPEN,0,VOL,1,0),COLOR00CCFF;\n\nX_10:=VOL=HHV(VOL,30);\n\nX_11:=BARSLAST(FILTER(CROSS(0.9,X_10),2))+1;\n\nX_12:=REF(VOL,X_11);\n\nSTICKLINE(X_11<=30,X_12,X_12,1,0),COLOR00CCFF;\n\nX_13:=VOL=LLV(VOL,20);\n\nX_14:=BARSLAST(FILTER(CROSS(0.9,X_13),2))+1;\n\nX_15:=REF(VOL,X_14);\n\nSTICKLINE(X_14<=30,X_15,X_15,4,0),COLORFFAA00;",
null,
"(公式代码复制过程可能会有格式失误，如果导入不成功，可找我免费领取源代码，想了解更多目前A股阶段的操作技巧及完整公式代码，或有任何疑惑，可关注公众号越声投研(yslcwh)，第一时间获取最重要的投资情报和独创的股票技术分析方法，干货源源不断!)\n\n1. 为什么多数的散户亏钱或套牢?\n\n2. 股市上的赢利模式: 长线的价值投资和中短线的交易投机\n\n3. 交易技巧\n\n4. 股市哲理\n\n5. 止损和仓位配置\n\n6.选股\n\n(以上内容仅供参考，不构成操作建议。如自行操作，注意仓位控制和风险自负。)\n\n【免责声明】本文仅代表作者本人观点，与财富动力网无关。财富动力网站对文中陈述、观点判断保持中立，不对所包含内容的准确性、可靠性或完整性提供任何明示或暗示的保证。请读者仅作参考，并请自行承担全部责任。\n\n0条评论\n\n• 日排行\n• 周排行"
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https://stackoverflow.com/questions/16822016/write-multiple-variables-to-a-file/16822067 | [
"# Write multiple variables to a file\n\nI want to write two variable to a file using Python.\n\nBased on what is stated in this post I wrote:\n\n``````f.open('out','w')\nf.write(\"%s %s\\n\" %str(int(\"0xFF\",16)) %str(int(\"0xAA\",16))\n``````\n\nBut I get this error:\n\n``````Traceback (most recent call last):\nFile \"process-python\", line 8, in <module>\no.write(\"%s %s\\n\" %str(int(\"0xFF\", 16)) %str(int(\"0xAA\", 16)))\nTypeError: not enough arguments for format string\n``````\n• is this even syntactically correct? I would think it would have to be `\"%s %s\\n\" % (str(int(\"0xFF\", 16)), str(int(\"0xAA\", 16)))` – user35288 May 29 '13 at 19:15\n• Why are you using `str(int())`, anyway? `\"%i %i\\n\" % (int(\"0xFF\", 16), int(\"0xAA\",16))` would work just as well and, in my opinion, is a bit clearer. Also, if only hexadecimal strings are guaranteed to begin with `0x` then you can use `int(string, 0)`, as that will automatically convert properly-prefixed octal strings and handle decimal strings correctly as well. If all your strings are hex and might not be preceded by `0x` then using `int(string, 16)` is probably how you need to go, though. – JAB May 29 '13 at 20:26\n\nYou are not passing enough values to `%`, you have two specifiers in your format string so it expects a tuple of length 2. Try this:\n\n``````f.write(\"%s %s\\n\" % (int(\"0xFF\" ,16), int(\"0xAA\", 16)))\n``````\n• @mahmood yes I have, what problem are you having? – cmd May 29 '13 at 19:19\n• ok thanks. I had problems with opened and closed brackets. Now it is ok – mahmood May 29 '13 at 19:20\n\nThe % operator takes an object or tuple. So the correct way to write this is:\n\n``````f.write(\"%s %s\\n\" % (int(\"0xFF\", 16), int(\"0xAA\",16)))\n``````\n\nThere are also many other ways how to format a string, documentation is your friend http://docs.python.org/2/library/string.html\n\nBetter use `format` this way:\n\n``````'{0} {1}\\n'.format(int(\"0xFF\",16), int(\"0xAA\",16))\n``````\n\nAlso there is no need to wrap `int` with `str`.\n\n• yeah format is nice providing he is using python2.7 or greater – cmd May 29 '13 at 19:18\n• `format` is available in 2.6, although you must explicitly number the replacement fields (i.e., `'{0} {1}\\n'.format(...)`). – chepner May 29 '13 at 19:22\n\nFirstly, your opening the file is wrong `f.open('out', 'w')` should probably be:\n\n``````f = open('out', 'w')\n``````\n\nThen, for such simple formatting, you can use `print`, for Python 2.x, as:\n\n``````print >> f, int('0xff', 16), int('0xaa', 16)\n``````\n\nOr, for Python 3.x:\n\n``````print(int('0xff', 16), int('0xaa', 16), file=f)\n``````\n\nOtherwise, use `.format`:\n\n``````f.write('{} {}'.format(int('0xff', 16), int('0xaa', 16)))\n``````\n\nYou need to supply a tuple:\n\n``````f.open('out','w')\nf.write(\"%d %d\\n\" % (int(\"0xFF\",16), int(\"0xAA\",16)))\n``````\n\nThis should probably be written as:\n\n``````f.write(\"255 170\\n\")\n``````\n• Those are constant in the example. In reality they are variables. – mahmood May 29 '13 at 19:17"
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https://www.lrde.epita.fr/wiki/Publications/najman.13.ismm | [
"# Discrete set-valued continuity and interpolation\n\n## Abstract\n\nThe main question of this paper is to retrieve some continuity properties on (discrete) T0-Alexandroff spaces. One possible application, which will guide us, is the construction of the so-called \"tree of shapes\" (intuitively, the tree of level lines). This tree, which should allow to process maxima and minima in the same wayfaces quite a number of theoretical difficulties that we propose to solve using set-valued analysis in a purely discrete setting. We also propose a way to interpret any function defined on a grid as a \"continuous\" function thanks to an interpolation scheme. The continuity properties are essential to obtain a quasi-linear algorithm for computing the tree of shapes in any dimension, which is exposed in a companion paper.\n\n## Bibtex (lrde.bib)\n\n```@InProceedings{\t najman.13.ismm,\nauthor\t= {Laurent Najman and Thierry G\\'eraud},\ntitle\t\t= {Discrete set-valued continuity and interpolation},\nbooktitle\t= {Mathematical Morphology and Its Application to Signal and\nImage Processing -- Proceedings of the 11th International\nSymposium on Mathematical Morphology (ISMM)},\nyear\t\t= 2013,\neditor\t= {C.L. Luengo Hendriks and G. Borgefors and R. Strand},\nvolume\t= 7883,\nseries\t= {Lecture Notes in Computer Science Series},\npublisher\t= {Springer},\npages\t\t= {37--48},\nabstract\t= {The main question of this paper is to retrieve some\ncontinuity properties on (discrete) T0-Alexandroff spaces.\nOne possible application, which will guide us, is the\nconstruction of the so-called \"tree of shapes\"\n(intuitively, the tree of level lines). This tree, which\nshould allow to process maxima and minima in the same way,\nfaces quite a number of theoretical difficulties that we\npropose to solve using set-valued analysis in a purely\ndiscrete setting. We also propose a way to interpret any\nfunction defined on a grid as a \"continuous\" function\nthanks to an interpolation scheme. The continuity\nproperties are essential to obtain a quasi-linear algorithm\nfor computing the tree of shapes in any dimension, which is\nexposed in a companion paper.}\n}```"
]
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https://rd.springer.com/article/10.1007%2Fs11565-019-00318-1 | [
"Strassen’s $$2 \\times 2$$ matrix multiplication algorithm: a conceptual perspective\n\nOpen Access\nArticle\n\nAbstract\n\nThe main purpose of this paper is pedagogical. Despite its importance, all proofs of the correctness of Strassen’s famous 1969 algorithm to multiply two $$2 \\times 2$$ matrices with only seven multiplications involve some basis-dependent calculations such as explicitly multiplying specific $$2 \\times 2$$ matrices, expanding expressions to cancel terms with opposing signs, or expanding tensors over the standard basis, sometimes involving clever simplifications using the sparsity of tensor summands. This makes the proof nontrivial to memorize and many presentations of the proof avoid showing all the details and leave a significant amount of verifications to the reader. In this note we give a short, self-contained, basis-independent proof of the existence of Strassen’s algorithm that avoids these types of calculations. We achieve this by focusing on symmetries and algebraic properties. Our proof can be seen as a coordinate-free version of the construction of Clausen from 1988, combined with recent work on the geometry of Strassen’s algorithm by Chiantini, Ikenmeyer, Landsberg, and Ottaviani from 2016.\n\nKeywords\n\nMatrix multiplication Strassen’s algorithm Coordinate-free Elementary\n\nMathematics Subject Classification\n\n68W30 Symbolic computation and algebraic computation\n\n1 Introduction\n\nThe discovery of Strassen’s matrix multiplication algorithm was a breakthrough result in computational linear algebra. The study of fast (subcubic) matrix multiplication algorithms initiated by this discovery has become an important area of research (see for a survey and for the currently best upper bound on the complexity of matrix multiplication). Fast matrix multiplication has countless applications as a subroutine in algorithms for a wide variety of problems, see e.g. [7, §16] for numerous applications in computational linear algebra. In practice, algorithms more sophisticated than Strassen’s are rarely implemented, but Strassen’s algorithm is used for multiplication of large matrices (see [13, 19, 25] on practical fast matrix multiplication).\n\nThe core of Strassen’s result is an algorithm for multiplying $$2 \\times 2$$ matrices with only 7 multiplications instead of 8. It is a bilinear algorithm, which means that it arises from a decomposition of the form\n\nwhere $$u_k$$ and $$v_k$$ are cleverly chosen linear forms on the space of $$2 \\times 2$$ matrices and $$W_k$$ are seven explicit $$2 \\times 2$$ matrices. Because of this structure it can be applied to block matrices, and its recursive application results in an algorithm for the multiplication of two $$n \\times n$$ matrices using $$O(n^{\\log _2 7})$$ arithmetic operations (see [7, §15.2] or for details).\n\nBecause of the great importance of Strassen’s algorithm, our goal is to understand it on a deep level. In Strassen’s original paper, the linear forms $$u_k$$, $$v_k$$, and the matrices $$W_k$$ are given, but the verification of the correctness of the algorithm is left to the reader. Unfortunately, such a description does not yield many further immediate insights.\n\nShortly after Strassen’s paper, Gastinel published a proof of the existence of decomposition ($$\\star$$) using simple algebraic transformations that is much easier to follow and verify. Many other papers provide alternative descriptions of Strassen’s algorithm or proofs of its existence. Brent and Paterson present the algorithm in a graphical form using $$4 \\times 4$$ diagrams indicating which elements of the two matrices are used. A more formal version of these diagrams are matrices of linear forms, which are used, for example, by Fiduccia (the same proof appears in ), Brockett and Dobkin and Lafon . Makarov gives a proof that uses ideas of Karatsuba’s algorithm for the efficient multiplication of polynomials. Büchi and Clausen connect the existence of Strassen’s algorithm to the existence of special bases of the space of $$2 \\times 2$$ matrices in which the multiplication table has a specific structure (their results are more general and apply not only to matrix multiplication). Alexeyev describes several algorithms for matrix multiplication as embeddings of the matrix algebra into a 7-dimensional nonassociative algebra with a special properties.\n\nSometimes the clever use of sparsity makes a proof rather short (e.g. ), but usually the verification of these proofs requires simple but somewhat lengthy computations: expansion of explicit decompositions in some basis, multiplication of several matrices or following chains of algebraic transformations in which careful attention to details is required. To obtain a more conceptual proof of the existence of Strassen’s algorithm, we do not focus on the explicit algorithm, but on the algebraic properties of the $$2 \\times 2$$ matrices, their transformations and symmetries of Strassen’s algorithm. It is well-known that the decomposition ($$\\star$$) is not unique. Given one decomposition, we can obtain another one by applying the identity\n\\begin{aligned} XY = A^{-1} \\left[ (A X B^{-1}) (B Y C^{-1}) \\right] C \\end{aligned}\nand using the original decomposition for the product in the square brackets. Alternatively, we can talk about $$2 \\times 2$$ matrices as linear maps between 2-dimensional vector spaces. Any choice of bases in these vector spaces gives a new bilinear algorithm. De Groote proved that the algorithm with seven multiplications is unique up to these transformations (this result is also announced without a proof in , see also ). Thus, Strassen’s algorithm is unique in this sense and there should be a coordinate-free description of this algorithm which does not use explicit matrices. One such description is given in and the proof of its correctness uses the fact that matrix multiplication is the unique (up to scale) bilinear map invariant under the transformations described above. This is a nontrivial fact which requires representation theory to prove. Moreover, the verification of the correctness in is left to the reader.\nSymmetries of Strassen’s algorithm are also useful for its understanding. Clausen gives a description of Strassen’s algorithm in terms of special bases, as in , and notices that the elements of these bases form orbits under the action of the symmetric group $$S_3$$ on the space of $$2 \\times 2$$ matrices defined via conjugation with specific matrices, i. e., Strassen’s algorithm is invariant under this action. Clausen’s construction is also describled in [7, Ch.1]. Grochow and Moore [17, 18] generalize Clausen’s construction to $$n \\times n$$ matrices using other finite group orbits. Another symmetry is only apparent in the trilinear representation of the algorithm: the decompositions ($$\\star$$) are in one-to-one correspondence with decompositions of the trilinear form $$\\mathop {{\\text {tr}}}(XYZ)$$ of the form\n\\begin{aligned} \\mathop {{\\text {tr}}}(XYZ) = \\sum _{k = 1}^7 u_k(X) v_k(Y) w_k(Z) \\end{aligned}\nwhere $$u_k$$, $$v_k$$ and $$w_k$$ are linear forms. The decomposition corresponding to Strassen’s algorithm is then invariant under the cyclic permutation of matrices XYZ. This symmetry is exploited in the proof of Chatelin , which uses properties of polynomials invariant under this symmetry. He also notices the importance of a matrix which is related to the $$S_3$$ symmetry discussed above. The symmetries of Strassen’s algorithm are explored in detail in [8, 10]. Several earlier publications note their importance [16, 27]. The paper explores symmetries of algorithms for $$3 \\times 3$$ matrix multiplication.\nIn this paper we provide a proof of Strassen’s result which is\n• coordinate-free we do not use explicit matrices, which allows us to focus on the algebraic properties required to prove the correctness of the algorithm. We avoid all tedious explicit calculations, in particular any expansions of expressions and any verification of explicit sign cancellations. Our proof can be seen as a coordinate-free version of Clausen’s construction.\n\n• elementary our proof uses only simple facts from basic linear algebra and does not require knowledge of representation theory. This is also why we do not use tensor language. Proofs from and are based on more complicated mathematics and may offer other insights.\n\nFormally, the result that we prove is the following.\n\nTheorem 1\n\n(Strassen ) Fix any field $${\\mathbb {F}}$$. There exist fourteen linear forms $$u_1,\\ldots ,u_7, v_1,\\ldots ,v_7 :{\\mathbb {F}}^{2 \\times 2} \\rightarrow {\\mathbb {F}}$$ and seven matrices $$W_1,\\ldots , W_7 \\in {\\mathbb {F}}^{2 \\times 2}$$ such that for all pairs of $$2 \\times 2$$ matrices X and Y the product satisfies\n\n2 Preliminaries from linear algebra\n\nIf $$u_1, \\ldots , u_n$$ and $$v_1, \\ldots , v_m$$ form bases of the spaces of column vectors $$F^{n \\times 1}$$ and row vectors $$F^{1 \\times m}$$ respectively, then the nm products of the form $$u_i v_j$$ form a basis of the space of matrices $$F^{n \\times m}$$\n\nThe trace$$\\mathop {{\\text {tr}}}(A)$$ of a square matrix A is the sum of its diagonal entries. If $$\\mathop {{\\text {tr}}}(A)$$ is zero, then the matrix A is called traceless. Taking the trace of a product of (rectangular) matrices is invariant under cyclic shifts: $$\\mathop {{\\text {tr}}}(A_1 A_2 \\cdots A_n) = \\mathop {{\\text {tr}}}(A_2 \\cdots A_n A_1)$$. As a consequence, the trace of a matrix is invariant under conjugations: $$\\mathop {{\\text {tr}}}(B^{-1}AB) = \\mathop {{\\text {tr}}}(ABB^{-1}) = \\mathop {{\\text {tr}}}(A)$$. Another implication is that if u is a column vector and $$v^T$$ is a row vector, then $$v^T u = \\mathop {{\\text {tr}}}(v^T u) = \\mathop {{\\text {tr}}}(u v^T)$$.\n\nThe characteristic polynomial of a $$2 \\times 2$$ matrix A is $$\\lambda ^2 - \\mathop {{\\text {tr}}}(A)\\lambda + \\det (A)$$. The Cayley—Hamilton theorem says that substituting A for $$\\lambda$$ yields the zero matrix.\n\n3 Rotational symmetry\n\nIn this section we collect some standard facts about rotation matrices. We think of the $$2 \\times 2$$ matrix D as a rotation of the plane by $$120^\\circ$$, but to make our approach work over every field we use a more algebraic definition for D.\n\nLet D have determinant 1 and trace $$-1$$, that is, D has characteristic polynomial $$\\lambda ^2 + \\lambda + 1$$. We assume that D is not a multiple of the identity $$\\mathop {{\\text {id}}}$$ (this is implicitly satisfied if the characteristic is not 3). For example, we could choose $$D = \\begin{bmatrix} 0&-1 \\\\ 1&-1 \\end{bmatrix}$$, the matrix that cyclically permutes the three vectors $$\\begin{pmatrix}1\\\\ 0\\end{pmatrix}$$, $$\\begin{pmatrix}0\\\\ 1\\end{pmatrix}$$, $$\\begin{pmatrix}-1\\\\ -1\\end{pmatrix}$$.\n\nClaim 2\n\nFor the matrix D we have $$D^3 = \\mathop {{\\text {id}}}$$, $$D^{-1} = D^2$$, $$D^{-2} = D$$. Additionally, D has the following properties: $$\\mathop {{\\text {id}}}+ D + D^{-1} = 0$$ and $$\\mathop {{\\text {tr}}}(D^{-1})=-1$$.\n\nProof\n\nThe characteristic polynomial of D is $$\\lambda ^2 + \\lambda + 1$$. By the Cayley—Hamilton theorem $$D^2 + D + \\mathop {{\\text {id}}}= 0$$. Multiplying by D we obtain $$D+D^2+D^3=0=\\mathop {{\\text {id}}}+D+D^2$$ and hence $$D^3 = \\mathop {{\\text {id}}}$$. Consequently, $$D^{-1} = D^2$$ and $$D^{-2} = D$$. Using $$D^{-1}=D^2$$ we get $$\\mathop {{\\text {id}}}+ D + D^{-1} = 0$$. This implies $$\\mathop {{\\text {tr}}}(D^{-1}) = - \\mathop {{\\text {tr}}}(\\mathop {{\\text {id}}}) - \\mathop {{\\text {tr}}}(D) = -1$$. $$\\square$$\n\nFor every column vector u define $$u^{\\perp }$$ as the row vector satisfying conditions $$u^{\\perp } u = 0$$ and $$u^{\\perp } D u = 1$$. If u is not an eigenvector of D, then u and Du are linearly independent, so $$u^{\\perp }$$ is uniquely defined. If, on the other hand, u is an eigenvector of D, the two conditions are inconsistent and $$u^{\\perp }$$ does not exist.\n\nWe fix a vector u that is not an eigenvector of D and define $$u^{\\perp }$$ as above. In our example we could choose $$u=\\begin{pmatrix}1\\\\ 0\\end{pmatrix}$$, which is not an eigenvector of $$\\begin{bmatrix} 0&-1 \\\\ 1&-1 \\end{bmatrix}$$.\n\nA first simple observation relates $$u^\\perp$$ and $$(Du)^{\\perp }$$:\n\nClaim 3\n\n$$u^\\perp D^{-1} = (D u)^\\perp$$.\n\nProof\n\nWe need to verify the two defining properties for $$(D u)^\\perp$$. We have $$(u^\\perp D^{-1})(D u) = u^\\perp u = 0$$ and $$(u^\\perp D^{-1}) D (D u) = u^\\perp D u = 1$$ as required. $$\\square$$\n\nThe following observation complements the fact that $$u^\\perp D u =1$$.\n\nClaim 4\n\n$$u^\\perp D^{-1} u=-1$$.\n\nProof\n\nUsing Claim 2 we have $$\\mathop {{\\text {id}}}+D+D^{-1}=0$$ and thus\n\\begin{aligned} u^\\perp u+u^\\perp Du+u^\\perp D^{-1}u=0. \\end{aligned}\nSince $$u^\\perp u = 0$$ and $$u^\\perp D u = 1$$, the claim follows. $$\\square$$\n\n4 Seven multiplications suffice\n\nIn this section we apply structural properties from Sect. 3 to prove Theorem 1. We set $$M := u u^\\perp$$. Clearly $$\\mathop {{\\text {tr}}}(M) = u^\\perp u = 0$$ and we obtain the following identities that can be used to simplify products of M, D, and $$D^{-1}$$:\n\nClaim 5\n\n$$M^2 = 0$$ and $$MDM = M$$ and $$M D^{-1} M = -M$$.\n\nProof\n\n\\begin{aligned} M^2= & {} (u u^\\perp ) (u u^\\perp ) = u (u^\\perp u) u^\\perp = 0.\\\\ MDM= & {} (u u^\\perp ) D (u u^\\perp ) = u (u^\\perp D u) u^\\perp = u u^\\perp = M.\\\\ MD^{-1}M= & {} (u u^\\perp ) D^{-1} (u u^\\perp ) = u (u^\\perp D^{-1} u) u^\\perp = -u u^\\perp = -M, \\end{aligned}\nwhere in the last line we used Claim 4. $$\\square$$\n\nBy Claim 2, conjugation with D is a map of order 3 on the vector space of all $$2 \\times 2$$ matrices, i.e. for any matrix A there is a triple of conjugates $$A \\mapsto D^{-1}AD \\mapsto DAD^{-1} \\mapsto A$$. Moreover, if A is traceless, then so are its conjugates.\n\nClaim 6\n\nThe matrices M, $$D^{-1}MD$$, and $$DMD^{-1}$$ form a basis of the vector space of traceless matrices.\n\nProof\n\nSince M is traceless, its conjugates are also traceless. Hence it is enough to prove that M, $$D^{-1}MD$$ and $$DMD^{-1}$$ are linearly independent.\n\nSince u is not an eigenvector of D, the vectors u and Du are linearly independent and thus form a basis of the space of column vectors. The row vectors $$u^{\\perp }$$ and $$u^{\\perp } D^{-1} = (Du)^{\\perp }$$ (Claim 3) are orthogonal to u and Du, respectively. Therefore they form a basis of the space of row vectors. Thus, the four matrices\n\\begin{aligned} u \\cdot u^\\perp = M,\\quad u \\cdot u^\\perp D^{-1} = MD^{-1},\\quad D u \\cdot u^\\perp = DM, \\quad D u \\cdot u^\\perp D^{-1} = DMD^{-1} \\end{aligned}\nobtained as products of these basis vectors form a basis of the space of $$2 \\times 2$$ matrices. The matrices M and $$DMD^{-1}$$ are contained in this basis. Adding up all four matrices, we get $$(\\mathop {{\\text {id}}}+ D) M (\\mathop {{\\text {id}}}+ D^{-1})$$, which can be simplified to $$(-D^{-1}) M (-D) = D^{-1}MD$$ using Claim 2. Therefore the matrices M, $$DMD^{-1}$$, $$D^{-1}MD$$ are linearly independent. $$\\square$$\n\nSince D and $$D^{-1}$$ have trace $$-1 \\ne 0$$ (Claim 2), adding D or $$D^{-1}$$ to the basis in Claim 6 yields two bases for the full space of $$2 \\times 2$$ matrices: $$\\{ D, M, D^{-1}MD, DMD^{-1} \\}$$ and $$\\{ D^{-1}, M, D^{-1}MD, DMD^{-1} \\}$$.\n\nUsing the properties $$D^2 = D^{-1}$$, $$D^{-2} = D$$ and $$M^2 = 0$$ from Claim 2 and Claim 5, we can write down the multiplication table with respect to these two bases. We further simplify it using the identities $$MDM = M$$ and $$MD^{-1}M = -M$$ from Claim 5.\n\nProof of Theorem 1\n\nNotice that in the body of the table only (scalar multiples of) 7 matrices are used, and the entries are aligned in such a way that two occurrences of the same matrix are either in the same row or in the same column. At this point we are done proving Theorem 1, because the existence of such a pattern gives a simple way to construct a matrix multiplication algorithm as follows. To multiply matrices X and Y, represent them in the bases $$\\{ D, M, D^{-1}MD, DMD^{-1} \\}$$ and $$\\{ D^{-1}, M, D^{-1}MD, DMD^{-1} \\}$$, respectively:\n\\begin{aligned} X= & {} x_1 D + x_2 M + x_3 D^{-1}MD + x_4 DMD^{-1} \\nonumber \\\\ Y= & {} y_1 D^{-1} + y_2 M + y_3 D^{-1}MD + y_4 DMD^{-1} \\end{aligned}\n(4.1)\nNote that the $$x_i$$ are linear forms in the entries of X and the $$y_j$$ are linear forms in the entries of Y. We expand the product XY and group together summands according to the table:\nThis finishes the proof. $$\\square$$\n\nRemark\n\nTaking the trace in (4.1) and using the fact that M and its conjugates are traceless, we see that $$\\mathop {{\\text {tr}}}(X)=x_1 \\mathop {{\\text {tr}}}(D) = -x_1$$, and $$\\mathop {{\\text {tr}}}(Y)=-y_1$$. Thus the first of the 7 summands is $$\\mathop {{\\text {tr}}}(X)\\mathop {{\\text {tr}}}(Y)\\mathop {{\\text {id}}}$$.\n\nReferences\n\n1. 1.\nAlekseyev, V.B.: Maximal extensions with simple multiplication for the algebra of matrices of the second order. Discrete Math. Appl. 7(1), 89–102 (1996). Google Scholar\n2. 2.\nBallard, G., Ikenmeyer, C., Landsberg, J.M., Ryder, N.: The geometry of rank decompositions of matrix multiplication II: $$3 \\times 3$$ matrices. J. Pure Appl. Algebra 223, 3205–3224 (2019)\n3. 3.\nBläser, M.: Fast matrix multiplication. Theory Comput. Grad. Surv. 5, 1–60 (2013). Google Scholar\n4. 4.\nBrent, Richard P.: Algorithms for matrix multiplication. Technical Report STAN-CS-70-157, Stanford University, Department of Computer Science (1970).\n5. 5.\nBrockett, R.W., Dobkin, D.: On the optimal evaluation of a set of bilinear forms. In: Proceedings of the 5th ACM STOC, pp. 88–95 (1973).\n6. 6.\nBüchi, W., Clausen, M.: On a class of primary algebras of minimal rank. Linear Algebra Appl. 69, 249–268 (1985).\n7. 7.\nBürgisser, P., Clausen, M., Shokrollahi, M.A.: Algebraic Complexity Theory, volume 315 of Grundlehren der mathematischen Wissenschaften. Springer, Berlin (1997). Google Scholar\n8. 8.\nBurichenko, V.P.: On symmetries of the Strassen algorithm (2014). arXiv Preprint arXiv:1408.6273\n9. 9.\nChatelin, P.: On transformations of algorithms to multiply $$2 \\times 2$$ matrices. Inf. Process. Lett. 22(1), 1–5 (1986).\n10. 10.\nChiantini, L., Ikenmeyer, C., Landsberg, J.M., Ottaviani, G.: The geometry of rank decompositions of matrix multiplication I: $$2 \\times 2$$ matrices. Exp Math (2017). Google Scholar\n11. 11.\nClausen, M.: Beiträge zum Entwurf schneller Spektraltransformationen. Universität Karlsruhe, Habilitationsschrift (1988)Google Scholar\n12. 12.\nde Groote, H.F.: On varieties of optimal algorithms for the computation of bilinear mappings II: Optimal algorithms for $$2 \\times 2$$-matrix multiplication. Theor. Comput. Sci. 7(2), 127–184 (1978).\n13. 13.\nDumas, J.-G., Pan, V.Y.: Fast matrix multiplication and symbolic computation (2016). arXiv Preprint arXiv:1612.05766\n14. 14.\nFiduccia, CM.: On obtaining upper bounds on the complexity of matrix multiplication. In: Complexity of Computer Computations, pp. 31–40 (1972).\n15. 15.\nGastinel, N.: Sur le calcul des produits de matrices. Numer. Math. 17(3), 222–229 (1971).\n16. 16.\nGates, A.Q., Kreinovich, V.: Strassen’s algorithm made (somewhat) more natural: a pedagogical remark. Bull. EATCS 73, 142–145 (2001)\n17. 17.\nGrochow, J.A., Moore, C.: Matrix multiplication algorithms from group orbits (2016). arXiv Preprint arXiv:1612.01527\n18. 18.\nGrochow, J.A., Moore, C.: Designing Strassen’s algorithm (2017). arXiv Preprint arXiv:1708.09398\n19. 19.\nHuang, J., Rice, L., Matthews, D.A., van de Geijn, R.A.: Generating families of practical fast matrix multiplication algorithms. Proc. IPDPS 2017, 656–667 (2017). Google Scholar\n20. 20.\nLafon, J.-C.: Optimum computation of $$p$$ bilinear forms. Linear Algebra Appl. 10(3), 225–240 (1975).\n21. 21.\nLe Gall, F.: Powers of tensors and fast matrix multiplication. Proc. ISSAC 2014, 296–303 (2014).\n22. 22.\nMakarov, O.M.: The connection between two multiplication algorithms. USSR Comput. Math. Math. Phys. 15(1), 218–223 (1975).\n23. 23.\nPan, V.Y.: О схемах вычисления произведениЙ матриц и обратноЙ матрицы [On algorithms for matrix multiplication and inversion]. У с пм а тн а у к 27(5(167)), 249–250 (1972). Translation available in . http://mi.mathnet.ru/umn5125\n24. 24.\nPan, V.Y.: Better late than never: filling a void in the history of fast matrix multiplication and tensor decompositions (2014). arXiv Preprint arXiv:1411.1972\n25. 25.\nPan, V.Y.: Fast matrix multiplication and its algebraic neighbourhood. Sb. Math. 208(11), 1661–1704 (2017).\n26. 26.\nPaterson, M.: Complexity of product and closure algorithms for matrices. In: Proceedings of the ICM 1974, vol. 2, pp. 483–489 (1974). https://www.mathunion.org/fileadmin/ICM/Proceedings/ICM1974.2/ICM1974.2.ocr.pdf#page=491. cited 03 Feb 2018\n27. 27.\nPaterson, M.: Strassen symmetries. Presentation at Leslie Valiant’s 60th birthday celebration, 30.05.2009, Bethesda, Maryland, USA (2009). https://www.cis.upenn.edu/~mkearns/valiant/paterson.ppt. cited 03 Feb 2018\n28. 28.\nStrassen, V.: Gaussian elimination is not optimal. Numer. Math. 13(4), 354–356 (1969).\n29. 29.\nYuval, G.: A simple proof of Strassen’s result. Inf. Process. Lett. 7(6), 285–286 (1978)."
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https://www.ssccglapex.com/3-years-ago-the-average-of-a-family-of-5-members-was-17-years-a-baby-having-been-born-the-average-age-of-the-family-is-the-same-today-the-present-age-of-the-baby-is/ | [
"### 3 years ago the average of a family of 5 members was 17 years. A baby having been born, the average age of the family is the same today. The present age of the baby is:\n\nA. 1 years\n\nB.$\\Large\\frac{3}{2}$ years\n\nC. 2 years\n\nD. 3 years\n\nLet age of the baby is x. 3 years ago total age of the family = 5 × 17 = 85 years Total age of the 5 member at present time, = 85 + 3*5 = 100 years Total age of the family at present time including baby, = 100 + X The average of the family including baby at present time, = 17 years $\\begin{array}{l}\\left(\\Large\\frac{100+\\mathrm{x}}{6}\\right)=17\\\\ 100+\\mathrm{X}=102\\\\ \\mathrm{X}=102-100=2\\text{ years }\\end{array}$"
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https://stackoverflow.com/questions/3505701/grouping-functions-tapply-by-aggregate-and-the-apply-family/44593528 | [
"# Grouping functions (tapply, by, aggregate) and the *apply family\n\nWhenever I want to do something \"map\"py in R, I usually try to use a function in the `apply` family.\n\nHowever, I've never quite understood the differences between them -- how {`sapply`, `lapply`, etc.} apply the function to the input/grouped input, what the output will look like, or even what the input can be -- so I often just go through them all until I get what I want.\n\nCan someone explain how to use which one when?\n\nMy current (probably incorrect/incomplete) understanding is...\n\n1. `sapply(vec, f)`: input is a vector. output is a vector/matrix, where element `i` is `f(vec[i])`, giving you a matrix if `f` has a multi-element output\n\n2. `lapply(vec, f)`: same as `sapply`, but output is a list?\n\n3. `apply(matrix, 1/2, f)`: input is a matrix. output is a vector, where element `i` is f(row/col i of the matrix)\n4. `tapply(vector, grouping, f)`: output is a matrix/array, where an element in the matrix/array is the value of `f` at a grouping `g` of the vector, and `g` gets pushed to the row/col names\n5. `by(dataframe, grouping, f)`: let `g` be a grouping. apply `f` to each column of the group/dataframe. pretty print the grouping and the value of `f` at each column.\n6. `aggregate(matrix, grouping, f)`: similar to `by`, but instead of pretty printing the output, aggregate sticks everything into a dataframe.\n\nSide question: I still haven't learned plyr or reshape -- would `plyr` or `reshape` replace all of these entirely?\n\n• to your side question: for many things plyr is a direct replacement for `*apply()` and `by`. plyr (at least to me) seems much more consistent in that I always know exactly what data format it expects and exactly what it will spit out. That saves me a lot of hassle. – JD Long Aug 17 '10 at 18:40\n• Also, I'd recommend adding: `doBy` and the selection & apply capabilities of `data.table`. – Iterator Oct 10 '11 at 15:23\n• `sapply` is just `lapply` with the addition of `simplify2array` on the output. `apply` does coerce to atomic vector, but output can be vector or list. `by` splits dataframes into sub-dataframes, but it doesn't use `f` on columns separately. Only if there is a method for 'data.frame'-class might `f` get column-wise applied by `by`. `aggregate` is generic so different methods exist for different classes of the first argument. – IRTFM Jan 24 '13 at 21:18\n• Mnemonic: l is for 'list', s is for 'simplifying', t is for 'per type' (each level of the grouping is a type) – Lutz Prechelt Sep 16 '14 at 13:20\n• There also exist some functions in the package Rfast, like: eachcol.apply, apply.condition, and more, which are faster than R's equivalents – Stefanos Nov 17 '18 at 14:09\n\nR has many *apply functions which are ably described in the help files (e.g. `?apply`). There are enough of them, though, that beginning useRs may have difficulty deciding which one is appropriate for their situation or even remembering them all. They may have a general sense that \"I should be using an *apply function here\", but it can be tough to keep them all straight at first.\n\nDespite the fact (noted in other answers) that much of the functionality of the *apply family is covered by the extremely popular `plyr` package, the base functions remain useful and worth knowing.\n\nThis answer is intended to act as a sort of signpost for new useRs to help direct them to the correct *apply function for their particular problem. Note, this is not intended to simply regurgitate or replace the R documentation! The hope is that this answer helps you to decide which *apply function suits your situation and then it is up to you to research it further. With one exception, performance differences will not be addressed.\n\n• apply - When you want to apply a function to the rows or columns of a matrix (and higher-dimensional analogues); not generally advisable for data frames as it will coerce to a matrix first.\n\n``````# Two dimensional matrix\nM <- matrix(seq(1,16), 4, 4)\n\n# apply min to rows\napply(M, 1, min)\n 1 2 3 4\n\n# apply max to columns\napply(M, 2, max)\n 4 8 12 16\n\n# 3 dimensional array\nM <- array( seq(32), dim = c(4,4,2))\n\n# Apply sum across each M[*, , ] - i.e Sum across 2nd and 3rd dimension\napply(M, 1, sum)\n# Result is one-dimensional\n 120 128 136 144\n\n# Apply sum across each M[*, *, ] - i.e Sum across 3rd dimension\napply(M, c(1,2), sum)\n# Result is two-dimensional\n[,1] [,2] [,3] [,4]\n[1,] 18 26 34 42\n[2,] 20 28 36 44\n[3,] 22 30 38 46\n[4,] 24 32 40 48\n``````\n\nIf you want row/column means or sums for a 2D matrix, be sure to investigate the highly optimized, lightning-quick `colMeans`, `rowMeans`, `colSums`, `rowSums`.\n\n• lapply - When you want to apply a function to each element of a list in turn and get a list back.\n\nThis is the workhorse of many of the other *apply functions. Peel back their code and you will often find `lapply` underneath.\n\n``````x <- list(a = 1, b = 1:3, c = 10:100)\nlapply(x, FUN = length)\n\\$a\n 1\n\\$b\n 3\n\\$c\n 91\nlapply(x, FUN = sum)\n\\$a\n 1\n\\$b\n 6\n\\$c\n 5005\n``````\n• sapply - When you want to apply a function to each element of a list in turn, but you want a vector back, rather than a list.\n\nIf you find yourself typing `unlist(lapply(...))`, stop and consider `sapply`.\n\n``````x <- list(a = 1, b = 1:3, c = 10:100)\n# Compare with above; a named vector, not a list\nsapply(x, FUN = length)\na b c\n1 3 91\n\nsapply(x, FUN = sum)\na b c\n1 6 5005\n``````\n\nIn more advanced uses of `sapply` it will attempt to coerce the result to a multi-dimensional array, if appropriate. For example, if our function returns vectors of the same length, `sapply` will use them as columns of a matrix:\n\n``````sapply(1:5,function(x) rnorm(3,x))\n``````\n\nIf our function returns a 2 dimensional matrix, `sapply` will do essentially the same thing, treating each returned matrix as a single long vector:\n\n``````sapply(1:5,function(x) matrix(x,2,2))\n``````\n\nUnless we specify `simplify = \"array\"`, in which case it will use the individual matrices to build a multi-dimensional array:\n\n``````sapply(1:5,function(x) matrix(x,2,2), simplify = \"array\")\n``````\n\nEach of these behaviors is of course contingent on our function returning vectors or matrices of the same length or dimension.\n\n• vapply - When you want to use `sapply` but perhaps need to squeeze some more speed out of your code.\n\nFor `vapply`, you basically give R an example of what sort of thing your function will return, which can save some time coercing returned values to fit in a single atomic vector.\n\n``````x <- list(a = 1, b = 1:3, c = 10:100)\n#Note that since the advantage here is mainly speed, this\n# example is only for illustration. We're telling R that\n# everything returned by length() should be an integer of\n# length 1.\nvapply(x, FUN = length, FUN.VALUE = 0L)\na b c\n1 3 91\n``````\n• mapply - For when you have several data structures (e.g. vectors, lists) and you want to apply a function to the 1st elements of each, and then the 2nd elements of each, etc., coercing the result to a vector/array as in `sapply`.\n\nThis is multivariate in the sense that your function must accept multiple arguments.\n\n``````#Sums the 1st elements, the 2nd elements, etc.\nmapply(sum, 1:5, 1:5, 1:5)\n 3 6 9 12 15\n#To do rep(1,4), rep(2,3), etc.\nmapply(rep, 1:4, 4:1)\n[]\n 1 1 1 1\n\n[]\n 2 2 2\n\n[]\n 3 3\n\n[]\n 4\n``````\n• Map - A wrapper to `mapply` with `SIMPLIFY = FALSE`, so it is guaranteed to return a list.\n\n``````Map(sum, 1:5, 1:5, 1:5)\n[]\n 3\n\n[]\n 6\n\n[]\n 9\n\n[]\n 12\n\n[]\n 15\n``````\n• rapply - For when you want to apply a function to each element of a nested list structure, recursively.\n\nTo give you some idea of how uncommon `rapply` is, I forgot about it when first posting this answer! Obviously, I'm sure many people use it, but YMMV. `rapply` is best illustrated with a user-defined function to apply:\n\n``````# Append ! to string, otherwise increment\nmyFun <- function(x){\nif(is.character(x)){\nreturn(paste(x,\"!\",sep=\"\"))\n}\nelse{\nreturn(x + 1)\n}\n}\n\n#A nested list structure\nl <- list(a = list(a1 = \"Boo\", b1 = 2, c1 = \"Eeek\"),\nb = 3, c = \"Yikes\",\nd = list(a2 = 1, b2 = list(a3 = \"Hey\", b3 = 5)))\n\n# Result is named vector, coerced to character\nrapply(l, myFun)\n\n# Result is a nested list like l, with values altered\nrapply(l, myFun, how=\"replace\")\n``````\n• tapply - For when you want to apply a function to subsets of a vector and the subsets are defined by some other vector, usually a factor.\n\nThe black sheep of the *apply family, of sorts. The help file's use of the phrase \"ragged array\" can be a bit confusing, but it is actually quite simple.\n\nA vector:\n\n``````x <- 1:20\n``````\n\nA factor (of the same length!) defining groups:\n\n``````y <- factor(rep(letters[1:5], each = 4))\n``````\n\nAdd up the values in `x` within each subgroup defined by `y`:\n\n``````tapply(x, y, sum)\na b c d e\n10 26 42 58 74\n``````\n\nMore complex examples can be handled where the subgroups are defined by the unique combinations of a list of several factors. `tapply` is similar in spirit to the split-apply-combine functions that are common in R (`aggregate`, `by`, `ave`, `ddply`, etc.) Hence its black sheep status.\n\n• Believe you will find that `by` is pure split-lapply and `aggregate` is `tapply` at their cores. I think black sheep make excellent fabric. – IRTFM Sep 14 '11 at 3:42\n• Fantastic response! This should be part of the official R documentation :). One tiny suggestion: perhaps add some bullets on using `aggregate` and `by` as well? (I finally understand them after your description!, but they're pretty common, so it might be useful to separate out and have some specific examples for those two functions.) – grautur Sep 14 '11 at 18:54\n• @grautur I was actively pruning things from this answer to avoid it being (a) too long and (b) a re-write of the documentation. I decided that while `aggregate`, `by`, etc. are based on *apply functions, the way you approach using them is different enough from a users perspective that they ought to be summarized in a separate answer. I may attempt that if I have time, or maybe someone else will beat me to it and earn my upvote. – joran Sep 14 '11 at 23:03\n• also, `?Map` as a relative of `mapply` – baptiste Feb 16 '12 at 5:53\n• @jsanders - I wouldn't agree with that at all. `data.frame`s are an absolutely central part of R and as a `list` object are frequently manipulated using `lapply` particularly. They also act as containers for grouping vectors/factors of many types together in a traditional rectangular dataset. While `data.table` and `plyr` might add a certain type of syntax that some might find more comfortable, they are extending and acting on `data.frame`s respectively. – thelatemail Aug 20 '14 at 6:08\n\nOn the side note, here is how the various `plyr` functions correspond to the base `*apply` functions (from the intro to plyr document from the plyr webpage http://had.co.nz/plyr/)\n\n``````Base function Input Output plyr function\n---------------------------------------\naggregate d d ddply + colwise\napply a a/l aaply / alply\nby d l dlply\nlapply l l llply\nmapply a a/l maply / mlply\nreplicate r a/l raply / rlply\nsapply l a laply\n``````\n\nOne of the goals of `plyr` is to provide consistent naming conventions for each of the functions, encoding the input and output data types in the function name. It also provides consistency in output, in that output from `dlply()` is easily passable to `ldply()` to produce useful output, etc.\n\nConceptually, learning `plyr` is no more difficult than understanding the base `*apply` functions.\n\n`plyr` and `reshape` functions have replaced almost all of these functions in my every day use. But, also from the Intro to Plyr document:\n\nRelated functions `tapply` and `sweep` have no corresponding function in `plyr`, and remain useful. `merge` is useful for combining summaries with the original data.\n\n• When I started learning R from scratch I found plyr MUCH easier to learn than the `*apply()` family of functions. For me, `ddply()` was very intuitive as I was familiar with SQL aggregation functions. `ddply()` became my hammer for solving many problems, some of which could have been better solved with other commands. – JD Long Aug 17 '10 at 19:23\n• I guess I figured that the concept behind `plyr` functions is similar to `*apply` functions, so if you can do one, you can do the other, but `plyr` functions are easier to remember. But I totally agree on the `ddply()` hammer! – JoFrhwld Aug 17 '10 at 19:36\n• The plyr package has the `join()` function that performs tasks similar to merge. Perhaps it's more to the point to mention it in the context of plyr. – marbel Jan 2 '14 at 23:04\n• Let us not forget poor, forgotten `eapply` – JDL Mar 25 '19 at 13:48\n• Great answer in general, but I think it downplays the utility of `vapply` and the downsides of `sapply`. A major advantage of `vapply` is that it enforces output type and length, so you will end up with either the exact expected output or an informative error. On the other hand, `sapply` will try to simplify the output following rules that aren't always obvious, and fall back to a list otherwise. For instance, try to predict the type of output this will produce: `sapply(list(1:5, 6:10, matrix(1:4, 2)), function(x) head(x, 1))`. What about `sapply(list(matrix(1:4, 2), matrix(1:4, 2)), ...)`? – Alexey Shiklomanov May 6 '19 at 18:32",
null,
"(Hopefully it's clear that `apply` corresponds to @Hadley's `aaply` and `aggregate` corresponds to @Hadley's `ddply` etc. Slide 20 of the same slideshare will clarify if you don't get it from this image.)\n\n(on the left is input, on the top is output)\n\n• is there a typo in the slide? The top left cell should be aaply – JHowIX Sep 16 '16 at 18:16\n\nThen the following mnemonics may help to remember the distinctions between each. Whilst some are obvious, others may be less so --- for these you'll find justification in Joran's discussions.\n\nMnemonics\n\n• `lapply` is a list apply which acts on a list or vector and returns a list.\n• `sapply` is a simple `lapply` (function defaults to returning a vector or matrix when possible)\n• `vapply` is a verified apply (allows the return object type to be prespecified)\n• `rapply` is a recursive apply for nested lists, i.e. lists within lists\n• `tapply` is a tagged apply where the tags identify the subsets\n• `apply` is generic: applies a function to a matrix's rows or columns (or, more generally, to dimensions of an array)\n\nBuilding the Right Background\n\nIf using the `apply` family still feels a bit alien to you, then it might be that you're missing a key point of view.\n\nThese two articles can help. They provide the necessary background to motivate the functional programming techniques that are being provided by the `apply` family of functions.\n\nUsers of Lisp will recognise the paradigm immediately. If you're not familiar with Lisp, once you get your head around FP, you'll have gained a powerful point of view for use in R -- and `apply` will make a lot more sense.\n\nSince I realized that (the very excellent) answers of this post lack of `by` and `aggregate` explanations. Here is my contribution.\n\n### BY\n\nThe `by` function, as stated in the documentation can be though, as a \"wrapper\" for `tapply`. The power of `by` arises when we want to compute a task that `tapply` can't handle. One example is this code:\n\n``````ct <- tapply(iris\\$Sepal.Width , iris\\$Species , summary )\ncb <- by(iris\\$Sepal.Width , iris\\$Species , summary )\n\ncb\niris\\$Species: setosa\nMin. 1st Qu. Median Mean 3rd Qu. Max.\n2.300 3.200 3.400 3.428 3.675 4.400\n--------------------------------------------------------------\niris\\$Species: versicolor\nMin. 1st Qu. Median Mean 3rd Qu. Max.\n2.000 2.525 2.800 2.770 3.000 3.400\n--------------------------------------------------------------\niris\\$Species: virginica\nMin. 1st Qu. Median Mean 3rd Qu. Max.\n2.200 2.800 3.000 2.974 3.175 3.800\n\nct\n\\$setosa\nMin. 1st Qu. Median Mean 3rd Qu. Max.\n2.300 3.200 3.400 3.428 3.675 4.400\n\n\\$versicolor\nMin. 1st Qu. Median Mean 3rd Qu. Max.\n2.000 2.525 2.800 2.770 3.000 3.400\n\n\\$virginica\nMin. 1st Qu. Median Mean 3rd Qu. Max.\n2.200 2.800 3.000 2.974 3.175 3.800\n``````\n\nIf we print these two objects, `ct` and `cb`, we \"essentially\" have the same results and the only differences are in how they are shown and the different `class` attributes, respectively `by` for `cb` and `array` for `ct`.\n\nAs I've said, the power of `by` arises when we can't use `tapply`; the following code is one example:\n\n`````` tapply(iris, iris\\$Species, summary )\nError in tapply(iris, iris\\$Species, summary) :\narguments must have same length\n``````\n\nR says that arguments must have the same lengths, say \"we want to calculate the `summary` of all variable in `iris` along the factor `Species`\": but R just can't do that because it does not know how to handle.\n\nWith the `by` function R dispatch a specific method for `data frame` class and then let the `summary` function works even if the length of the first argument (and the type too) are different.\n\n``````bywork <- by(iris, iris\\$Species, summary )\n\nbywork\niris\\$Species: setosa\nSepal.Length Sepal.Width Petal.Length Petal.Width Species\nMin. :4.300 Min. :2.300 Min. :1.000 Min. :0.100 setosa :50\n1st Qu.:4.800 1st Qu.:3.200 1st Qu.:1.400 1st Qu.:0.200 versicolor: 0\nMedian :5.000 Median :3.400 Median :1.500 Median :0.200 virginica : 0\nMean :5.006 Mean :3.428 Mean :1.462 Mean :0.246\n3rd Qu.:5.200 3rd Qu.:3.675 3rd Qu.:1.575 3rd Qu.:0.300\nMax. :5.800 Max. :4.400 Max. :1.900 Max. :0.600\n--------------------------------------------------------------\niris\\$Species: versicolor\nSepal.Length Sepal.Width Petal.Length Petal.Width Species\nMin. :4.900 Min. :2.000 Min. :3.00 Min. :1.000 setosa : 0\n1st Qu.:5.600 1st Qu.:2.525 1st Qu.:4.00 1st Qu.:1.200 versicolor:50\nMedian :5.900 Median :2.800 Median :4.35 Median :1.300 virginica : 0\nMean :5.936 Mean :2.770 Mean :4.26 Mean :1.326\n3rd Qu.:6.300 3rd Qu.:3.000 3rd Qu.:4.60 3rd Qu.:1.500\nMax. :7.000 Max. :3.400 Max. :5.10 Max. :1.800\n--------------------------------------------------------------\niris\\$Species: virginica\nSepal.Length Sepal.Width Petal.Length Petal.Width Species\nMin. :4.900 Min. :2.200 Min. :4.500 Min. :1.400 setosa : 0\n1st Qu.:6.225 1st Qu.:2.800 1st Qu.:5.100 1st Qu.:1.800 versicolor: 0\nMedian :6.500 Median :3.000 Median :5.550 Median :2.000 virginica :50\nMean :6.588 Mean :2.974 Mean :5.552 Mean :2.026\n3rd Qu.:6.900 3rd Qu.:3.175 3rd Qu.:5.875 3rd Qu.:2.300\nMax. :7.900 Max. :3.800 Max. :6.900 Max. :2.500\n``````\n\nit works indeed and the result is very surprising. It is an object of class `by` that along `Species` (say, for each of them) computes the `summary` of each variable.\n\nNote that if the first argument is a `data frame`, the dispatched function must have a method for that class of objects. For example is we use this code with the `mean` function we will have this code that has no sense at all:\n\n`````` by(iris, iris\\$Species, mean)\niris\\$Species: setosa\n NA\n-------------------------------------------\niris\\$Species: versicolor\n NA\n-------------------------------------------\niris\\$Species: virginica\n NA\nWarning messages:\n1: In mean.default(data[x, , drop = FALSE], ...) :\nargument is not numeric or logical: returning NA\n2: In mean.default(data[x, , drop = FALSE], ...) :\nargument is not numeric or logical: returning NA\n3: In mean.default(data[x, , drop = FALSE], ...) :\nargument is not numeric or logical: returning NA\n``````\n\n### AGGREGATE\n\n`aggregate` can be seen as another a different way of use `tapply` if we use it in such a way.\n\n``````at <- tapply(iris\\$Sepal.Length , iris\\$Species , mean)\nag <- aggregate(iris\\$Sepal.Length , list(iris\\$Species), mean)\n\nat\nsetosa versicolor virginica\n5.006 5.936 6.588\nag\nGroup.1 x\n1 setosa 5.006\n2 versicolor 5.936\n3 virginica 6.588\n``````\n\nThe two immediate differences are that the second argument of `aggregate` must be a list while `tapply` can (not mandatory) be a list and that the output of `aggregate` is a data frame while the one of `tapply` is an `array`.\n\nThe power of `aggregate` is that it can handle easily subsets of the data with `subset` argument and that it has methods for `ts` objects and `formula` as well.\n\nThese elements make `aggregate` easier to work with that `tapply` in some situations. Here are some examples (available in documentation):\n\n``````ag <- aggregate(len ~ ., data = ToothGrowth, mean)\n\nag\nsupp dose len\n1 OJ 0.5 13.23\n2 VC 0.5 7.98\n3 OJ 1.0 22.70\n4 VC 1.0 16.77\n5 OJ 2.0 26.06\n6 VC 2.0 26.14\n``````\n\nWe can achieve the same with `tapply` but the syntax is slightly harder and the output (in some circumstances) less readable:\n\n``````att <- tapply(ToothGrowth\\$len, list(ToothGrowth\\$dose, ToothGrowth\\$supp), mean)\n\natt\nOJ VC\n0.5 13.23 7.98\n1 22.70 16.77\n2 26.06 26.14\n``````\n\nThere are other times when we can't use `by` or `tapply` and we have to use `aggregate`.\n\n`````` ag1 <- aggregate(cbind(Ozone, Temp) ~ Month, data = airquality, mean)\n\nag1\nMonth Ozone Temp\n1 5 23.61538 66.73077\n2 6 29.44444 78.22222\n3 7 59.11538 83.88462\n4 8 59.96154 83.96154\n5 9 31.44828 76.89655\n``````\n\nWe cannot obtain the previous result with `tapply` in one call but we have to calculate the mean along `Month` for each elements and then combine them (also note that we have to call the `na.rm = TRUE`, because the `formula` methods of the `aggregate` function has by default the `na.action = na.omit`):\n\n``````ta1 <- tapply(airquality\\$Ozone, airquality\\$Month, mean, na.rm = TRUE)\nta2 <- tapply(airquality\\$Temp, airquality\\$Month, mean, na.rm = TRUE)\n\ncbind(ta1, ta2)\nta1 ta2\n5 23.61538 65.54839\n6 29.44444 79.10000\n7 59.11538 83.90323\n8 59.96154 83.96774\n9 31.44828 76.90000\n``````\n\nwhile with `by` we just can't achieve that in fact the following function call returns an error (but most likely it is related to the supplied function, `mean`):\n\n``````by(airquality[c(\"Ozone\", \"Temp\")], airquality\\$Month, mean, na.rm = TRUE)\n``````\n\nOther times the results are the same and the differences are just in the class (and then how it is shown/printed and not only -- example, how to subset it) object:\n\n``````byagg <- by(airquality[c(\"Ozone\", \"Temp\")], airquality\\$Month, summary)\naggagg <- aggregate(cbind(Ozone, Temp) ~ Month, data = airquality, summary)\n``````\n\nThe previous code achieve the same goal and results, at some points what tool to use is just a matter of personal tastes and needs; the previous two objects have very different needs in terms of subsetting.\n\n• As I've said, the power of by arises when we can't use tapply; the following code is one example: THIS ARE THE WORDS YOU HAVE USED ABOVE. And you have given an example of computing the summary. Well lets say that the summary statistics can be computed only that it will need cleaning: eg `data.frame(tapply(unlist(iris[,-5]),list(rep(iris[,5],ncol(iris[-5])),col(iris[-5])),summary))` this is a use of tapply`. With the right splitting there is nothing you cant do with `tapply`. The only thing is it returns a matrix. Please be careful by saying we cant use `tapply` – Onyambu Dec 24 '17 at 20:23\n\nThere are lots of great answers which discuss differences in the use cases for each function. None of the answer discuss the differences in performance. That is reasonable cause various functions expects various input and produces various output, yet most of them have a general common objective to evaluate by series/groups. My answer is going to focus on performance. Due to above the input creation from the vectors is included in the timing, also the `apply` function is not measured.\n\nI have tested two different functions `sum` and `length` at once. Volume tested is 50M on input and 50K on output. I have also included two currently popular packages which were not widely used at the time when question was asked, `data.table` and `dplyr`. Both are definitely worth to look if you are aiming for good performance.\n\n``````library(dplyr)\nlibrary(data.table)\nset.seed(123)\nn = 5e7\nk = 5e5\nx = runif(n)\ngrp = sample(k, n, TRUE)\n\ntiming = list()\n\n# sapply\ntiming[[\"sapply\"]] = system.time({\nlt = split(x, grp)\nr.sapply = sapply(lt, function(x) list(sum(x), length(x)), simplify = FALSE)\n})\n\n# lapply\ntiming[[\"lapply\"]] = system.time({\nlt = split(x, grp)\nr.lapply = lapply(lt, function(x) list(sum(x), length(x)))\n})\n\n# tapply\ntiming[[\"tapply\"]] = system.time(\nr.tapply <- tapply(x, list(grp), function(x) list(sum(x), length(x)))\n)\n\n# by\ntiming[[\"by\"]] = system.time(\nr.by <- by(x, list(grp), function(x) list(sum(x), length(x)), simplify = FALSE)\n)\n\n# aggregate\ntiming[[\"aggregate\"]] = system.time(\nr.aggregate <- aggregate(x, list(grp), function(x) list(sum(x), length(x)), simplify = FALSE)\n)\n\n# dplyr\ntiming[[\"dplyr\"]] = system.time({\ndf = data_frame(x, grp)\nr.dplyr = summarise(group_by(df, grp), sum(x), n())\n})\n\n# data.table\ntiming[[\"data.table\"]] = system.time({\ndt = setnames(setDT(list(x, grp)), c(\"x\",\"grp\"))\nr.data.table = dt[, .(sum(x), .N), grp]\n})\n\n# all output size match to group count\nsapply(list(sapply=r.sapply, lapply=r.lapply, tapply=r.tapply, by=r.by, aggregate=r.aggregate, dplyr=r.dplyr, data.table=r.data.table),\nfunction(x) (if(is.data.frame(x)) nrow else length)(x)==k)\n# sapply lapply tapply by aggregate dplyr data.table\n# TRUE TRUE TRUE TRUE TRUE TRUE TRUE\n``````\n\n``````# print timings\nas.data.table(sapply(timing, `[[`, \"elapsed\"), keep.rownames = TRUE\n)[,.(fun = V1, elapsed = V2)\n][order(-elapsed)]\n# fun elapsed\n#1: aggregate 109.139\n#2: by 25.738\n#3: dplyr 18.978\n#4: tapply 17.006\n#5: lapply 11.524\n#6: sapply 11.326\n#7: data.table 2.686\n``````\n• Is it normal that dplyr is lower than the applt functions ? – Mostafa Jun 8 '16 at 9:35\n• @DimitriPetrenko I don't think so, not sure why it is here. It is best to test against your own data, as there are many factors that comes into play. – jangorecki Jun 8 '16 at 11:48\n\nDespite all the great answers here, there are 2 more base functions that deserve to be mentioned, the useful `outer` function and the obscure `eapply` function\n\nouter\n\n`outer` is a very useful function hidden as a more mundane one. If you read the help for `outer` its description says:\n\n``````The outer product of the arrays X and Y is the array A with dimension\nc(dim(X), dim(Y)) where element A[c(arrayindex.x, arrayindex.y)] =\nFUN(X[arrayindex.x], Y[arrayindex.y], ...).\n``````\n\nwhich makes it seem like this is only useful for linear algebra type things. However, it can be used much like `mapply` to apply a function to two vectors of inputs. The difference is that `mapply` will apply the function to the first two elements and then the second two etc, whereas `outer` will apply the function to every combination of one element from the first vector and one from the second. For example:\n\n`````` A<-c(1,3,5,7,9)\nB<-c(0,3,6,9,12)\n\nmapply(FUN=pmax, A, B)\n\n> mapply(FUN=pmax, A, B)\n 1 3 6 9 12\n\nouter(A,B, pmax)\n\n> outer(A,B, pmax)\n[,1] [,2] [,3] [,4] [,5]\n[1,] 1 3 6 9 12\n[2,] 3 3 6 9 12\n[3,] 5 5 6 9 12\n[4,] 7 7 7 9 12\n[5,] 9 9 9 9 12\n``````\n\nI have personally used this when I have a vector of values and a vector of conditions and wish to see which values meet which conditions.\n\neapply\n\n`eapply` is like `lapply` except that rather than applying a function to every element in a list, it applies a function to every element in an environment. For example if you want to find a list of user defined functions in the global environment:\n\n``````A<-c(1,3,5,7,9)\nB<-c(0,3,6,9,12)\nC<-list(x=1, y=2)\nD<-function(x){x+1}\n\n> eapply(.GlobalEnv, is.function)\n\\$A\n FALSE\n\n\\$B\n FALSE\n\n\\$C\n FALSE\n\n\\$D\n TRUE\n``````\n\nFrankly I don't use this very much but if you are building a lot of packages or create a lot of environments it may come in handy.\n\nIt is maybe worth mentioning `ave`. `ave` is `tapply`'s friendly cousin. It returns results in a form that you can plug straight back into your data frame.\n\n``````dfr <- data.frame(a=1:20, f=rep(LETTERS[1:5], each=4))\nmeans <- tapply(dfr\\$a, dfr\\$f, mean)\n## A B C D E\n## 2.5 6.5 10.5 14.5 18.5\n\n## great, but putting it back in the data frame is another line:\n\ndfr\\$m <- means[dfr\\$f]\n\ndfr\\$m2 <- ave(dfr\\$a, dfr\\$f, FUN=mean) # NB argument name FUN is needed!\ndfr\n## a f m m2\n## 1 A 2.5 2.5\n## 2 A 2.5 2.5\n## 3 A 2.5 2.5\n## 4 A 2.5 2.5\n## 5 B 6.5 6.5\n## 6 B 6.5 6.5\n## 7 B 6.5 6.5\n## ...\n``````\n\nThere is nothing in the base package that works like `ave` for whole data frames (as `by` is like `tapply` for data frames). But you can fudge it:\n\n``````dfr\\$foo <- ave(1:nrow(dfr), dfr\\$f, FUN=function(x) {\nx <- dfr[x,]\nsum(x\\$m*x\\$m2)\n})\ndfr\n## a f m m2 foo\n## 1 1 A 2.5 2.5 25\n## 2 2 A 2.5 2.5 25\n## 3 3 A 2.5 2.5 25\n## ...\n``````\n\nI recently discovered the rather useful `sweep` function and add it here for the sake of completeness:\n\nsweep\n\nThe basic idea is to sweep through an array row- or column-wise and return a modified array. An example will make this clear (source: datacamp):\n\nLet's say you have a matrix and want to standardize it column-wise:\n\n``````dataPoints <- matrix(4:15, nrow = 4)\n\n# Find means per column with `apply()`\ndataPoints_means <- apply(dataPoints, 2, mean)\n\n# Find standard deviation with `apply()`\ndataPoints_sdev <- apply(dataPoints, 2, sd)\n\n# Center the points\ndataPoints_Trans1 <- sweep(dataPoints, 2, dataPoints_means,\"-\")\n\n# Return the result\ndataPoints_Trans1\n## [,1] [,2] [,3]\n## [1,] -1.5 -1.5 -1.5\n## [2,] -0.5 -0.5 -0.5\n## [3,] 0.5 0.5 0.5\n## [4,] 1.5 1.5 1.5\n\n# Normalize\ndataPoints_Trans2 <- sweep(dataPoints_Trans1, 2, dataPoints_sdev, \"/\")\n\n# Return the result\ndataPoints_Trans2\n## [,1] [,2] [,3]\n## [1,] -1.1618950 -1.1618950 -1.1618950\n## [2,] -0.3872983 -0.3872983 -0.3872983\n## [3,] 0.3872983 0.3872983 0.3872983\n## [4,] 1.1618950 1.1618950 1.1618950\n``````\n\nNB: for this simple example the same result can of course be achieved more easily by\n`apply(dataPoints, 2, scale)`\n\n• Is this related to grouping? – Frank Jun 16 '17 at 16:55\n• @Frank: Well, to be honest with you the title of this post is rather misleading: when you read the question itself it is about \"the apply family\". `sweep` is a higher-order function like all the others mentioned here, e.g. `apply`, `sapply`, `lapply` So the same question could be asked about the accepted answer with over 1,000 upvotes and the examples given therein. Just have a look at the example given for `apply` there. – vonjd Jun 16 '17 at 17:03\n• sweep has a misleading name, misleading defaults, and misleading parameter name :). It's easier for me to understand it this way : 1) STATS is vector or single value that will be repeated to form a matrix of the same size as first input, 2) FUN will be applied on 1st input and this new matrix. Maybe better illustrated by : `sweep(matrix(1:6,nrow=2),2,7:9,list)` . It's usually more efficient than `apply`because where `apply` loops, `sweep` is able to use vectorised functions. – Moody_Mudskipper Mar 24 '18 at 15:04\n\nIn the collapse package recently released on CRAN, I have attempted to compress most of the common apply functionality into just 2 functions:\n\n1. `dapply` (Data-Apply) applies functions to rows or (default) columns of matrices and data.frames and (default) returns an object of the same type and with the same attributes (unless the result of each computation is atomic and `drop = TRUE`). The performance is comparable to `lapply` for data.frame columns, and about 2x faster than `apply` for matrix rows or columns. Parallelism is available via `mclapply` (only for MAC).\n\nSyntax:\n\n``````dapply(X, FUN, ..., MARGIN = 2, parallel = FALSE, mc.cores = 1L,\nreturn = c(\"same\", \"matrix\", \"data.frame\"), drop = TRUE)\n``````\n\nExamples:\n\n``````# Apply to columns:\ndapply(mtcars, log)\ndapply(mtcars, sum)\ndapply(mtcars, quantile)\n# Apply to rows:\ndapply(mtcars, sum, MARGIN = 1)\ndapply(mtcars, quantile, MARGIN = 1)\n# Return as matrix:\ndapply(mtcars, quantile, return = \"matrix\")\ndapply(mtcars, quantile, MARGIN = 1, return = \"matrix\")\n# Same for matrices ...\n``````\n1. `BY` is a S3 generic for split-apply-combine computing with vector, matrix and data.frame method. It is significantly faster than `tapply`, `by` and `aggregate` (an also faster than `plyr`, on large data `dplyr` is faster though).\n\nSyntax:\n\n``````BY(X, g, FUN, ..., use.g.names = TRUE, sort = TRUE,\nexpand.wide = FALSE, parallel = FALSE, mc.cores = 1L,\nreturn = c(\"same\", \"matrix\", \"data.frame\", \"list\"))\n``````\n\nExamples:\n\n``````# Vectors:\nBY(iris\\$Sepal.Length, iris\\$Species, sum)\nBY(iris\\$Sepal.Length, iris\\$Species, quantile)\nBY(iris\\$Sepal.Length, iris\\$Species, quantile, expand.wide = TRUE) # This returns a matrix\n# Data.frames\nBY(iris[-5], iris\\$Species, sum)\nBY(iris[-5], iris\\$Species, quantile)\nBY(iris[-5], iris\\$Species, quantile, expand.wide = TRUE) # This returns a wider data.frame\nBY(iris[-5], iris\\$Species, quantile, return = \"matrix\") # This returns a matrix\n# Same for matrices ...\n``````\n\nLists of grouping variables can also be supplied to `g`.\n\nTalking about performance: A main goal of collapse is to foster high-performance programming in R and to move beyond split-apply-combine alltogether. For this purpose the package has a full set of C++ based fast generic functions: `fmean`, `fmedian`, `fmode`, `fsum`, `fprod`, `fsd`, `fvar`, `fmin`, `fmax`, `ffirst`, `flast`, `fNobs`, `fNdistinct`, `fscale`, `fbetween`, `fwithin`, `fHDbetween`, `fHDwithin`, `flag`, `fdiff` and `fgrowth`. They perform grouped computations in a single pass through the data (i.e. no splitting and recombining).\n\nSyntax:\n\n``````fFUN(x, g = NULL, [w = NULL,] TRA = NULL, [na.rm = TRUE,] use.g.names = TRUE, drop = TRUE)\n``````\n\nExamples:\n\n``````v <- iris\\$Sepal.Length\nf <- iris\\$Species\n\n# Vectors\nfmean(v) # mean\nfmean(v, f) # grouped mean\nfsd(v, f) # grouped standard deviation\nfsd(v, f, TRA = \"/\") # grouped scaling\nfscale(v, f) # grouped standardizing (scaling and centering)\nfwithin(v, f) # grouped demeaning\n\nw <- abs(rnorm(nrow(iris)))\nfmean(v, w = w) # Weighted mean\nfmean(v, f, w) # Weighted grouped mean\nfsd(v, f, w) # Weighted grouped standard-deviation\nfsd(v, f, w, \"/\") # Weighted grouped scaling\nfscale(v, f, w) # Weighted grouped standardizing\nfwithin(v, f, w) # Weighted grouped demeaning\n\n# Same using data.frames...\nfmean(iris[-5], f) # grouped mean\nfscale(iris[-5], f) # grouped standardizing\nfwithin(iris[-5], f) # grouped demeaning\n\n# Same with matrices ...\n``````\n\nIn the package vignettes I provide benchmarks. Programming with the fast functions is significantly faster than programming with dplyr or data.table, especially on smaller data, but also on large data."
]
| [
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"https://i.stack.imgur.com/UMzZ4.png",
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http://www.geekinterview.com/question_details/28878 | [
"# 10 digit number puzzle\n\nA 10 digit number has it first digit equals to the numbers of 1's, second digit equals to the numbers of 2's, 3rd digit equals to the numbers of 3's .4th equals number of 4’s…till 9th digit equals to the numbers of 9's and 10th digit equals to the number of 0's. what is the number?\nThis question is related to Infosys Interview\n\n#### vidya sagar\n\n• May 23rd, 2006\n\nthe no is 6000100012\n\n#### aruna_sakhile",
null,
"Profile",
null,
"Answers by aruna_sakhile\n\n• May 24th, 2006\n\nAns:\n\n1 2 3 4 5 6 7 8 9 0\n\n1 0 0 0 0 0 1 0 0 7 (The no of Zeros in 10th digit place)\n\n#### k srujana",
null,
"Profile",
null,
"Answers by k srujana",
null,
"Questions by k srujana\n\n• May 24th, 2006\n\ni think the answer is 1000000008\n\n#### dangerous",
null,
"Profile",
null,
"Answers by dangerous\n\n• May 25th, 2006\n\nthe no is 6000100012.let me clear ur doubt\n\ntotal no.of 1's in above no=2(in 1st digit );no.of 2's=1(in 2nd digit);no.of 3's,4's,5's=0;no.of 6's=1(6th digit);no.of 7's,8's nd 9's=0;in 10th digit no.of 0's=6;\n\nvidya sagar.\n\n#### CA Alpesh Tated\n\n• Apr 19th, 2013\n\n2100010006\n\n#### Gowda Shivamurthy\n\n• Oct 25th, 2014\n\n6210001000\n\n#### pranjay sagar\n\n• Dec 24th, 2014\n\n9000000000\n\n#### soubhik\n\n• Dec 29th, 2014\n\n9000000000...only 10th digit contains no.of total zeros dats 9\n\n#### soubhik\n\n• Dec 29th, 2014\n\nSorry the no is 1000000008...coz 10th digit resembles unit place...\n\n#### Sumit\n\n• Dec 30th, 2014\n\n2100010006\n\n#### sumit\n\n• Dec 30th, 2014\n\n1st digit is last most digit\n\n#### Dinesh\n\n• Jan 8th, 2015\n\nYou can understand it as\n1 2 3 4 5 6 7 8 9 10\n2 1 0 0 0 1 0 0 0 6\nHere total no. of 1s=2, total no. of 2s=1, total no. of 6s=1 & total no. of 3s,4s 5s,7s,8s,9s=0\n\n#### Mano\n\n• Aug 22nd, 2018\n\nThis is when first digit is number of zeros. The question for your answer is\nFind a 10-digit number where the first digit is how many zeros in the number, the second digit is how many 1s in the number etc. until the tenth digit which is how many 9s in the number.",
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"",
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http://www.godreams.cn/?post=108 | [
" C++ primer plus Day03\n\n#",
null,
"会飞的鱼\n\n2020\nGodam",
null,
"2020-6-11\n\n## C++ primer plus Day03",
null,
"手机扫码查看\n\n# 第5章 循环和关系表达式\n\n## P152 循环和文本输入\n\n### 使用原始的cin进行输入\n\n``````#include<iostream>\nusing namespace std;\n\nint main() {\nchar ch;\nint count = 0;\ncout << \"请输入字符以#结束:\" << endl;\ncin >> ch;\nwhile (ch != '#') {\ncout << ch;\n++count;\ncin >> ch;\n}\ncout << endl << count << \"个字母读入\\n\";\nreturn 0;\n}\n``````\n\n### 使用cin.get(char)进行补救\n\n``````#include<iostream>\nusing namespace std;\n\nint main() {\nchar ch;\nint count = 0;\ncout << \"请输入字符以#结束:\" << endl;\ncin.get(ch);\nwhile (ch != '#') {\ncout << ch;\n++count;\ncin.get(ch);\n}\ncout << endl << count << \"个字母读入\\n\";\nreturn 0;\n}\n``````\n\n### 文件尾条件\n\n``````#include<iostream>\nusing namespace std;\n\nint main() {\nchar ch;\nint count = 0;\ncout << \"请输入字符以#结束:\" << endl;\ncin.get(ch);\nwhile (cin.fail()==false) {\ncout << ch;\n++count;\ncin.get(ch);\n}\ncout << endl << count << \"个字母读入\\n\";\nreturn 0;\n}\n``````\n• 注意在使用EOF标记后，cin将不再读取输入，这对于文件来说是很合理的因为一个文件的尾之后一定不会再有内容了，但在输入的过程中可能会还有，这时可以在之后使用cin.clear()清除EOF标记，使输入继续进行。\n• cin.get(char)的返回值是一个cin对象，但是在需要其发生转换的地方时会自动转换为bool类型（如在while循环中）：`while(cin)`\n\n``````#include<iostream>\nusing namespace std;\n\nint main() {\nint ch;\nint count = 0;\ncout << \"请输入字符以:\" << endl;\nwhile ((ch = cin.get())!=EOF) {\ncout.put(char(ch));\n++count;\n}\ncout << endl << count << \"个字母读入\\n\";\nreturn 0;\n}\n``````\n\n# 第6章 分支语句和逻辑运算符\n\n## P195 读取文本文件\n\n• is_open()函数可以判断文件是否被打开，成功打开将返回true。\n\n### 文件尾的判断\n\neof()只能判断是否达到EOF，而fail()可以用于检测EOF和类型不匹配，请看下面代码段：\n``````if (inFile.eof())\ncout << \"文件已经结束\";\nelse if (inFile.fail())\ncout << \"读取数据不正确\";\nelse\ncout << \"未知错误\";\n``````\n这样就可以对不同的错误进行处理。P197页有关于good()的描述。\n\n# 第7章 函数 C++的编程模块\n\n使用交替乘除可防止中间操作数超过最大浮点数。\n\n请看如下代码：\n```\n\n# includeusing namespace std;\n\nint test(int a[]) {\nint c = sizeof(a);\nreturn c;\n\nint main() {\nint a3 = { 1,2,3 };\nint b = test(a);\ncout << sizeof(a)<<endl;\ncout << b;\n}\n```",
null,
"## const的说明\n\n`const int * a = &b`指向的内容不可变，但指针指向可以变化\n`int *const a = &b`b内容可以边，但指针指向b这一内容不可变化\n\n### 评论",
null,
"游客\n\n### 注册",
null,
"sitemap"
]
| [
null,
"http://www.godreams.cn/content/templates/FYS/img/logo.png",
null,
"http://www.godreams.cn/content/templates/FYS/images/random/6.jpg",
null,
"http://qr.liantu.com/api.php",
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"http://img.godreams.cn/images/2020/06/11/OBmS.png",
null,
"http://www.godreams.cn/content/templates/FYS/img/avatar.png",
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"http://www.godreams.cn/include/lib/checkcode.php",
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https://sco.m.wikipedia.org/wiki/Octahedron | [
"# Octahedron\n\nRegular octahedron",
null,
"Type Platonic solid\nElements F = 8, E = 12\nV = 6 (χ = 2)\nFaces by sides 8{3}\nConway notation {{{O-conway}}}\nSchläfli seembols {3,4}\nr{3,3}\nFace confeeguration {{{O-ffig}}}\nWythoff seembol 4 | 2 3\nCoxeter diagram",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"Symmetry Oh, BC3, [4,3], (*432)\nRotation group O, [4,3]+, (432)\nReferences U05, C17, W2\nProperties regular, convexdeltahedron\nDihedral angle 109.47122° = arccos(-1/3)",
null,
"3.3.3.3\n(Vertex feegur)",
null,
"Cube\n(dual polyhedron)",
null,
"Net\n\nIn geometry, an octahedron (plural: octahedra) is a polyhedron wi aicht faces. A regular octahedron is a Platonic solid componed o aicht equilateral triangles, fower o which meet at each vertex.\n\nA regular octahedron is the dual polyhedron o a cube. It is a rectified tetrahedron. It is a square bipyramid in ony o three orthogonal orientations. It is an aa a triangular antiprism in ony o fower orientations.\n\nAn octahedron is the three-dimensional case o the mair general concept o a cross polytope."
]
| [
null,
"https://upload.wikimedia.org/wikipedia/commons/thumb/0/07/Octahedron.svg/280px-Octahedron.svg.png",
null,
"https://upload.wikimedia.org/wikipedia/commons/5/5e/CDel_node.png",
null,
"https://upload.wikimedia.org/wikipedia/commons/8/8c/CDel_4.png",
null,
"https://upload.wikimedia.org/wikipedia/commons/5/5e/CDel_node.png",
null,
"https://upload.wikimedia.org/wikipedia/commons/c/c3/CDel_3.png",
null,
"https://upload.wikimedia.org/wikipedia/commons/b/bd/CDel_node_1.png",
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"https://upload.wikimedia.org/wikipedia/commons/thumb/a/ab/Octahedron_vertfig.png/100px-Octahedron_vertfig.png",
null,
"https://upload.wikimedia.org/wikipedia/commons/thumb/3/33/Hexahedron.png/120px-Hexahedron.png",
null,
"https://upload.wikimedia.org/wikipedia/commons/thumb/b/b3/Octahedron_flat.svg/240px-Octahedron_flat.svg.png",
null
]
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http://www-personal.umich.edu/~weath/lda/topic51.html | [
"## 2.51 Mathematics\n\nCategory: Logic and Mathematics\n\nKeywords: mathematics, arithmetic, numbers, mathematical, proofs, infinity, axioms, counting, infinitely, theorems, proved, prime, proof, hilbert, numerical\n\nNumber of Articles: 420\nPercentage of Total: 1.3%\nRank: 23rd\n\nWeighted Number of Articles: 333.3\nPercentage of Total: 1%\nRank: 40th\n\nMean Publication Year: 1978.7\nWeighted Mean Publication Year: 1974.9\nMedian Publication Year: 1982\nModal Publication Year: 1993\n\nTopic with Most Overlap: Sets and Grue (0.0422)\nTopic this Overlaps Most With: Sets and Grue (0.0335)\nTopic with Least Overlap: Emotions (0.00019)\nTopic this Overlaps Least With: Duties (0.00025)",
null,
"Figure 2.119: Mathematics",
null,
"Figure 2.120: Mathematics Articles in Each Journal\n\nWhen I did a new model run, one of the first things I checked was whether the model had found the philosophy of mathematics topic. And the quickest way to check for that was whether it had clearly put the two big Benacerraf papers in the same topic. Most models passed this test, but some of them did so less clearly than others. This model run just passes the test. Here are the probability distributions over the 90 topics for the two articles.12\n\nTable 2.22: Paul Benacerraf (1965) “What Numbers Could Not Be” Philosophical Review 74:47-73.\nSubject Probability\nMathematics 0.3073\nOrdinary Language 0.2166\nSets and Grue 0.1680\nMeaning and Use 0.0633\nDenoting 0.0370\nPersonal Identity 0.0226\nUniversals and Particulars 0.0220\nVerification 0.0208\nNorms 0.0202\n\nThat makes sense - it is a Philosophy of Mathematics article, but it is largely about sets, and it can’t escape the fact that it’s written during the era of Ordinary Language Philosophy. But the second table is a closer run thing.\n\nTable 2.23: Paul Benacerraf (1973) “Mathematical Truth” Journal of Philosophy 70:661-679.\nSubject Probability\nMathematics 0.2001\nTruth 0.1713\nOrdinary Language 0.1198\nNorms 0.0744\nKnowledge 0.0637\nConcepts 0.0539\nDefinitions 0.0465\nTheories and Realism 0.0347\nAnalytic/Synthetic 0.0246\nVerification 0.0245\nJustification 0.0216\nPropositions and Implications 0.0207\n\nAgain, this sort of makes sense - the article is about truth. But the model is much less sure how to classify this article, as evidenced by the number of topics that get a probability between 4.6% and 7.4%.\n\nWhat the model is working with is a conception of philosophy of mathematics that is centered around two debates. One is the nature of infinity, the other is the nature of proof. That’s not a terrible take on twentieth century philosophy of mathematics, but it does mean that papers like Mathematical Truth get treated as less than fully paradigmatic.\n\nI mostly look at how big a topic is by eyeballing the 12 journal graph. But in this case that would be highly misleading. Because this topic is spread across many journals, and many years, those twelve lines look like they barely have a pulse. But the tables show that, depending on how you measure size, this is the 23rd or 40th biggest topic. So many other topics are concentrated in a small number of journals or times, and they make a bigger visual impact than topics like this that keep chugging along.\n\n1. Remember that these tables are cropped at 2%; the model assigns probabilities to all 90 topics but I’m not showing them all here.↩︎"
]
| [
null,
"http://www-personal.umich.edu/~weath/lda/lda-bookdown_files/figure-html/t51b-1.png",
null,
"http://www-personal.umich.edu/~weath/lda/lda-bookdown_files/figure-html/t51c-1.png",
null
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https://zh.wikibooks.org/wiki/%E5%9F%BA%E7%A1%80%E6%95%B0%E5%AD%A6 | [
"# 基础数学\n\n## 分数\n\n### 分数的加减法\n\n${\\frac {a}{b}}+{\\frac {c}{d}}={\\frac {ad+bc}{bd}}$",
null,
"${\\frac {3}{2}}-{\\frac {3}{4}}={\\frac {6-3}{4}}={\\frac {3}{4}}$",
null,
"### 分数的乘除法\n\n${\\frac {4}{3}}\\times {\\frac {27}{11}}={\\frac {4}{1}}\\times {\\frac {9}{11}}={\\frac {36}{11}}$",
null,
"### 分数的四則運算\n\n${\\frac {4}{3}}\\times {\\frac {27}{11}}-1={\\frac {4}{1}}\\times {\\frac {9}{11}}-1={\\frac {36}{11}}-{\\frac {11}{11}}={\\frac {36-11}{11}}={\\frac {25}{11}}$",
null,
"## 几何\n\n### 平面图形\n\n#### 导言\n\n 平面图形：所表示的各个部分都在同一平面内的图形被称为平面图形。\n\n#### 图形的分类\n\n 圆：由曲线构成的、看起来非常匀称完美的平面图形.\n 多边形：由几条直边首尾相连构成的平面图形.\n\n### 线\n\n#### 导言\n\n 线构成多边形，有直和曲的两种。\n\n## 代數\n\n### 用字母表示数\n\n……\n\n……\n\n……\n\n#### 正文\n\n 例题 题目: 用一句话来表示刚才的这首儿歌 解答: $?$",
null,
"隻青蛙$?$",
null,
"张嘴，2×?只眼睛4×?条腿，扑通扑通跳下水用代數表示：$n$",
null,
"只青蛙$n$",
null,
"张嘴，$2n$",
null,
"只眼睛$4n$",
null,
"条腿，扑通扑通跳下水 O\n\n### 方程\n\n#### 正文",
null,
"现在，你已经懂得怎样用字母表示数了。那么我们来看一道小题，检验你的学习成果。\n\n 例题 题目: 小明有6个苹果，吃掉三个，问：小明还有几个苹果？ 解答: …… O\n\n 例题 题目: 小明有6个苹果，吃掉三个，问：小明还有几个苹果？ 解答: $6-3=3$",
null,
"(个) O\n\n 例题 题目: 小明有6个苹果，吃掉三个，问：小明还有几个苹果？ 解答: 用英文字母表示小明现在的苹果数量 假设它有$?$",
null,
"苹果（用字母表示成$x$",
null,
"），由题意得$?=6-3$",
null,
"（字母表示$x=6-3$",
null,
"），$6-3=3$",
null,
"，解得$?=3$",
null,
"（$x=3$",
null,
"） O\n\n 例题 题目: 葡萄和苹果共有36个，苹果有11个，问：葡萄有几个？ 解答: …… O\n\n 例题 题目: 葡萄和苹果共有36个，苹果有11个，问：葡萄有几个？ 解答: $36-11=25$",
null,
"(个) O\n\n 例题 题目: 葡萄和苹果共有36个，苹果有11个，问：葡萄有几个？ 解答: 用英文字母表示小明现在的苹果数量 假设它有$?$",
null,
"葡萄（用字母表示成$x$",
null,
"），由题意得$?=36-11$",
null,
"（字母表示$x=36-11$",
null,
"），$36-11=25$",
null,
"，解得$?=25$",
null,
"（$x=25$",
null,
"） O\n\n 例题 题目: 小明在菜市场买了三个西瓜，两串香蕉，西瓜8元，总共34元，问香蕉多少块钱？ 解答: 假设香蕉单价$x$",
null,
"元，由题意得$3*8+2*x=34$",
null,
"。 O"
]
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null,
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null,
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null,
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null,
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null
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https://arvimal.blog/2017/06/27/recursion-algorithm-study/ | [
"# Recursion – Algorithm Study\n\nRecursion is a technique by which a function calls itself until a condition is met.\n\n## Introduction\n\nLoops or repetitive execution based on certain conditions are inevitable in programs. Usual loops include `if`, `while` and `for` loops. `Recursion` is an entirely different way to deal with such situations, and in many cases, easier.\n\n`Recursion` is a when a function calls itself in each iteration till a condition is met. Ideally, the data set in each iteration gets smaller until it reach the required condition, after which the recursive function exists.\n\nA typical example of recursion is a `factorial` function.\n\n## How does Recursion work?\n\nA `recursive` function ideally contains a `Base` case and a `Recursive` case.\n\n`Recursive` case is when the function calls itself, until the `Base` case is met. Each level of iteration in the `Recursive` case moves the control to the next level.\n\nOnce a specific level finishes execution, the control is passed back to the previous level of execution. A `Recursive` function can go several layers deep until the `Base` condition is met. In short, a `Recursive` case is a loop in which the function calls itself.\n\nThe `Base` case is required so that the function doesn’t continue running in the `Recursive` loop forever. Once the `Base` case is met, the control moves out of the `Recursive` case, executes the conditions in the `Base` case (if any), and exits.\n\nAs mentioned in the `Introduction`, a factorial function can be seen as an example of recursion.\n\n### NOTE:\n\nThe `Base` case for a factorial function is when `n == 1`\n\nConsider `n!`:\n\n`n!` can be written as:\n\nn x (n – 1) x (n – 2) x (n – 3) x …. x 1\n\n`n!` can also be represented as:\n\n``` n! = n * (n - 1)! ---> [Step 1]\n(n - 1)! = (n - 1) * (n - 2)! ---> [Step 2]\n(n - 2)! = (n - 2) * (n - 3)! ---> [Step 3]\n.\n..\n...\n(n - (n - 1)) = 1 ---> [Base case]\n```\n\nEach level/step is a product of a value and all the levels below it. Hence, `Step 1` will end up moving to `Step 2` to get the factorial of elements below it, then to `Step 3` and so on.\n\nie.. the control of execution move as:\n\n[Step 1] -> [Step 2] -> [Step 3] -> ….. [Step n]\n\nIn a much easier-to-grasp example, a `5!` would be:\n\n```5! = 5 * 4! ---> [Step 1]\n4! = 4 * 3! ---> [Step 2]\n3! = 3 * 2! ---> [Step 3]\n2! = 2 * 1! ---> [Step 4]\n1! = 1 ---> [Step 5] / [Base case]\n```\n\nThe order of execution will be :\n\n[Step 1] -> [Step 2] -> [Step 3] -> [Step 4] -> [Step 5]\n\nAs we know, in `Recursion`, each layer pause itself and pass the control to the next level. Once it reach the end or the `Base` case, it returns the result back to the previous level one by one until it reaches where it started off.\n\nIn this example, once the control of execution reaches `Step 5 / Base case` , the control is returned back to its previous level `Step 4` . This level returns the result back to `Step 3` which completes its execution and returns to `Step 2` , so on and so forth until it reach `Step 1` .\n\nThe return control flow would be as:\n\n[Base case / Step 5] -> [Step 4] -> [Step 3] -> [Step 2] -> [Step 1] -> Result.\n\nThis can be summed up using an awesome pictorial representation, from the book `Grokking Algorithms` by Adit. Please check out the `References` section for the link for more information about this awesome book.",
null,
"Figure 1: Recursion, Recursive case and Base case (Copyright Manning Publications, drawn by adit.io)\n\n## Code\n\n### Example 1:\n\n• A `factorial` function in a `while` loop\n```def fact(n):\nfactorial = 1\nwhile n > 1:\nfactorial = factorial * n\nn = n - 1\nreturn factorial\n\nprint(\"Factorial of {0} is {1}\".format(10, fact(10)))\nprint(\"Factorial of {0} is {1}\".format(20, fact(20)))\n```\n• The same function above, in a `recursive` loop\n```def factorial(n):\nif n == 0:\nreturn 1\nelse:\nreturn n * factorial(n - 1)\n\nprint(\"Factorial of {0} is {1}\".format(10, factorial(10)))\nprint(\"Factorial of {0} is {1}\".format(20, factorial(20)))\n```\n\n### Example 2:\n\n• A function to sum numbers in a normal `for` loop.\n```def my_sum(my_list):\nnum = 0\nfor i in my_list:\nnum += i\nreturn num\n\nprint(my_sum([10, 23, 14, 12, 11, 94, 20]))\n```\n• The same function to add numbers, in a `recursive` loop\n```def my_sum(my_list):\nif my_list == []:\nreturn 0\nelse:\nreturn my_list + my_sum(my_list[1:])\n\nprint(my_sum([10, 23, 14, 12, 11, 94, 20]))\n```\n\n## Code explanation\n\nBoth `Example 1` and `Example 2` are represented as an iterative function as well as a recursive function.\n\nThe iterative function calls the `next()` function on the iterator `sum.__iter__()` magic method iterate over the entire data set. The recursive function calls itself to reach a base case and return the result.\n\n## Observations:\n\nWhile a recursive function does not necessarily give you an edge on performance, it is much easier to understand and the code is cleaner.\n\nRecursion has a disadvantage though, for large data sets. Each loop is put on a call stack until it reaches a `Base` case. Once the `Base` case is met, the call stack is rewound back to reach where it started, executing each of the previous levels on the way. The examples above showed a `sum` function and a `factorial` function. In large data sets, this can lead to a large call stack which in turns take a lot of memory.\n\n## References:\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
]
| [
null,
"https://arvimal.files.wordpress.com/2017/02/factorial_recursion.png",
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8251584,"math_prob":0.962634,"size":5029,"snap":"2021-04-2021-17","text_gpt3_token_len":1382,"char_repetition_ratio":0.160199,"word_repetition_ratio":0.04761905,"special_character_ratio":0.2998608,"punctuation_ratio":0.13697319,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99480444,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-17T18:05:04Z\",\"WARC-Record-ID\":\"<urn:uuid:83a182dc-52f2-4608-bda4-8b5b44aca8b5>\",\"Content-Length\":\"154065\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ed75f324-e730-4642-9876-a7bfd231836d>\",\"WARC-Concurrent-To\":\"<urn:uuid:7030414e-0ef8-47fc-a129-df8d4fa1b1af>\",\"WARC-IP-Address\":\"192.0.78.25\",\"WARC-Target-URI\":\"https://arvimal.blog/2017/06/27/recursion-algorithm-study/\",\"WARC-Payload-Digest\":\"sha1:WWEHRXSQ66C6JTD2OCQ3X4UHLX5HSA4A\",\"WARC-Block-Digest\":\"sha1:JWJASLEZOUAUWZH5V2W5G6XPEEP2HFZ4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038461619.53_warc_CC-MAIN-20210417162353-20210417192353-00344.warc.gz\"}"} |
https://sikademy.com/answer/computer-science/discrete-mathematics/solution-ihzy/ | [
"is below this banner.\n\nCan't find a solution anywhere?\n\nNEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?\n\nYou will get a detailed answer to your question or assignment in the shortest time possible.\n\n## Here's the Solution to this Question\n\nLet $N = 250$ be the total members of the society voted for $X, Y, Z$. Let $n(X), n(Y), n(Z)$ denote the number of candidates voted for $X, Y, Z$ respectively. Then,\n\n\\begin{aligned} N &= n(X \\cup Y \\cup Z) + n(X \\cup Y \\cup Z)'\\qquad (\\text{Voted + Not voted})\\\\ &= n(X) + n(Y) + n(Z) - n(X \\cap Y)-n(X \\cap Z)- n(Y \\cap Z) \\\\ &\\qquad\\qquad\\qquad\\qquad\\qquad+ n(X \\cap Y \\cap Z)+n(X \\cup Y \\cup Z)'\\\\ 250 &= n(X) + n(Y) + n(Z) - n(X \\cap Y)-n(X \\cap Z) - n(Y \\cap Z)\\\\ &\\qquad\\qquad\\qquad\\qquad\\qquad + n(X \\cup Y \\cup Z)' \\qquad\\qquad\\qquad\\qquad\\qquad(1)\\\\ &(\\text{Since each member may vote for either one or two candidates}\\\\ &~~~~~~~~~~~~n(X \\cap Y \\cap Z)=0)\\\\ \\end{aligned}\n\n12~ \\text{voted for X and Y, ~} 14~ \\text{voted for X and Z, i.e.,}\\\\ \\begin{aligned} n(X \\cap Y) &=12 \\qquad\\qquad\\quad (2)\\\\ n(X \\cap Z) &=14\\qquad\\qquad\\quad (3)\\\\\\\\ \\end{aligned}\\\\ 59~ \\text{voted for Y only and~ } 37~ \\text{voted for Z only, i.e.,}\\\\ \\begin{aligned} n(Y) - n(X \\cap Y) - n(Y \\cap Z) &=59 \\qquad\\qquad\\quad (4)\\\\ n(Z) - n(X \\cap Z) - n(Y \\cap Z) &=37\\qquad\\qquad\\quad (5)\\\\\\\\ \\end{aligned}\\\\ 147~ \\text{voted for X or Y or both but not for Z }\\\\ 102~ \\text{voted for Y or Z or both but not for X, i.e.,}\\\\ \\begin{aligned} n(X) + n(Y) - n(X \\cap Y) - n(X \\cap Z) - n(Y \\cap Z) &=147 \\qquad\\qquad (6)\\\\ n(Y) + n(Z) - n(Y \\cap Z) - n(X \\cap Z) - n(X \\cap Y) &=102\\qquad\\qquad(7)\\\\\\\\ \\end{aligned}\\\\\n\nAdding (4) and (5), we get\n\n$n(Y) + n(Z) - n(X \\cap Y) - n(X \\cap Z) - 2\\cdot n(Y \\cap Z) = 96 \\qquad\\quad (8)$\n\nSubtracting (8) from (7), we get $n(Y \\cap Z) = 6$.\n\nFrom (4), (5), (6)\n\n\\begin{aligned} n(Y) &= 59 + n(X\\cap Y) + n(Y \\cap Z) = 59+12+6=77\\\\ n(Z) &= 37 + n(X\\cap Z) + n(Y \\cap Z) = 37+14+6=57\\\\ n(X) & = 147- n(Y) + n(X \\cap Y)+ n(X \\cap Z) + n(Y \\cap Z)\\\\ &=147 -77+12+14+6 = 102 \\end{aligned}\n\ni)",
null,
"ii) Number of candidates who did not vote = $n(X \\cup Y \\cup Z)'$\n\nFrom (1),\n\n\\begin{aligned} n(X \\cup Y \\cup Z)' &= 250 - (n(X) + n(Y) + n(Z) - n(X \\cap Y)-n(X \\cap Z) - n(Y \\cap Z))\\\\ &= 250-(102+77+57-12-14-6)\\\\ & = 46 \\end{aligned}\n\niii) Number of members voted for X only = $n(X) - n(X \\cap Y) - n(X \\cap Z) = 102-12-14 = 76$\n\niv) Number of votes for X only = 76,\n\nNumber of votes for Y only = 59,\n\nNumber of votes for Z only = 37.\n\nTherefore, X won the election."
]
| [
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"https://www.assignmentexpert.com/image",
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https://palaeo-electronica.org/2003_2/egypt/methods.htm | [
"### METHODS\n\nFour common morphometric methodologies were used: 1) canonical variate analysis; 2) eigenshape analysis; 3) relative warps analysis; and 4) the thin plate spline method. In the following section, a brief description of these techniques is introduced.\n\n#### Canonical Variate Analysis\n\nCanonical variate analysis is one of the more interesting morphometric applications of multivariate statistics. The technique is used to examine the interrelationships between a number of populations simultaneously with a goal of objectively representing the interrelationships graphically in few dimensions (ideally only two or three dimensions). The axes of variation are chosen to maximize the separation between the populations relative to the variation within each of the populations.\n\nIn algebraic terms, the first canonical variate is the linear combination of variables that maximizes the ratio of between-groups sums of squares to the within-groups sums of squares for a one-way multivariate analysis of variance of the canonical variate scores.\n\nFor k groups (characters) and p variables (populations), the canonical variate scores are obtained from the following equation:\n\nYij = cTxij ,\n\nwhere cT is the transpose of the canonical vectors and xij denotes the i-th of N observations for the j-th group. The first canonical vector c is derived so as to maximize the ratio:\n\nF = cTBc/ cTWc\n\nwhere B is the between-groups matrix of sums of squares and cross products and W is the within-groups matrix of sums of squares and cross products.\n\nThe canonical vectors c and the canonical roots f satisfy the following equations:\n\n(B - fW)c = 0\n\nand\n\n|B - fW| = 0\n\nThe canonical vectors are usually scaled so that\n\ncTWc = Nw\n\nwhere Nw is the within-groups degrees of freedom.\n\nFor more information see Blackith and Reyment (1971), Reyment et al. (1984), and Reyment and Savazzi (1999).\n\n#### Eigenshape Analysis\n\nIn 1983, Lohmann developed a Q-mode principal component technique for analyzing changes in the shape of an organism that he called eigenshape analysis. The observations consist of coordinate pairs determined at definite points around the circumference of the shell. In the present study, 11 landmarks were selected around the outline of each studied ostracod specimen to express these points.\n\nThe technique is a simple ordinating procedure that has as its starting point the x, y coordinates of a set of p points along outlines of N objects of interest (here outline of the ostracod carapace). A transformation procedure is followed in the same manner as principal component analysis.\n\n#### Thin Plate Spline and Relative Warps Analyses\n\nRohlf (1996) stated that there are two methods of comparison of shapes in geometric morphometrics. One of them is based on the least square method and is most efficient if overall similarity depends largely on few landmarks. The other method is thin plate spline analysis, which works best if the similarity depends on many landmarks.\n\nThe thin plate spline method is based on analogy of a two-dimensional morphological object to a thin homogeneous deformable metallic plate (Bookstein 1989, 1991); thus one specimen is fitted to another by stretching, and the numerical estimate of degree of such a smooth deformation is the bending energy coefficient.\n\nThe shape variation encompasses two components, an affine (uniform) part and non-affine (non-uniform) part (Bookstein 1991). In the affine change, the orthogonality of principal axes is preserved, and parallel lines remain parallel, like the deformation of a square into a parallelogram or a circle into an ellipse. The non-affine change is represented by the residual of size-free change that remains after the difference due to any affine change has been subtracted from the total change in shape. An example is when an initially flat object is twisted or warped. For some examples of the method, see Reyment (1993, 1995a, 1995b, and 1997), Reyment and Bookstein (1993), and Reyment and Elewa (2002).",
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""
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"https://palaeo-electronica.org/2003_2/egypt/images/next.jpg",
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https://xianblog.wordpress.com/tag/rare-events/ | [
"## rare events for ABC\n\nPosted in Books, Mountains, pictures, Statistics, Travel, University life with tags , , , , , , , on November 24, 2016 by xi'an\n\nDennis Prangle, Richard G. Everitt and Theodore Kypraios just arXived a new paper on ABC, aiming at handling high dimensional data with latent variables, thanks to a cascading (or nested) approximation of the probability of a near coincidence between the observed data and the ABC simulated data. The approach amalgamates a rare event simulation method based on SMC, pseudo-marginal Metropolis-Hastings and of course ABC. The rare event is the near coincidence of the observed summary and of a simulated summary. This is so rare that regular ABC is forced to accept not so near coincidences. Especially as the dimension increases. I mentioned nested above purposedly because I find that the rare event simulation method of Cérou et al. (2012) has a nested sampling flavour, in that each move of the particle system (in the sample space) is done according to a constrained MCMC move. Constraint derived from the distance between observed and simulated samples. Finding an efficient move of that kind may prove difficult or impossible. The authors opt for a slice sampler, proposed by Murray and Graham (2016), however they assume that the distribution of the latent variables is uniform over a unit hypercube, an assumption I do not fully understand. For the pseudo-marginal aspect, note that while the approach produces a better and faster evaluation of the likelihood, it remains an ABC likelihood and not the original likelihood. Because the estimate of the ABC likelihood is monotonic in the number of terms, a proposal can be terminated earlier without inducing a bias in the method.",
null,
"This is certainly an innovative approach of clear interest and I hope we will discuss it at length at our BIRS ABC 15w5025 workshop next February. At this stage of light reading, I am slightly overwhelmed by the combination of so many computational techniques altogether towards a single algorithm. The authors argue there is very little calibration involved, but so many steps have to depend on as many configuration choices.\n\n## an extension of nested sampling\n\nPosted in Books, Statistics, University life with tags , , , , , , , on December 16, 2014 by xi'an\n\nI was reading [in the Paris métro] Hastings-Metropolis algorithm on Markov chains for small-probability estimation, arXived a few weeks ago by François Bachoc, Lionel Lenôtre, and Achref Bachouch, when I came upon their first algorithm that reminded me much of nested sampling: the following was proposed by Guyader et al. in 2011,\n\nTo approximate a tail probability P(H(X)>h),\n\n• start from an iid sample of size N from the reference distribution;\n• at each iteration m, select the point x with the smallest H(x)=ξ and replace it with a new point y simulated under the constraint H(y)≥ξ;\n• stop when all points in the sample are such that H(X)>h;\n• take",
null,
"$\\left(1-\\dfrac{1}{N}\\right)^{m-1}$\n\nas the unbiased estimator of P(H(X)>h).\n\nHence, except for the stopping rule, this is the same implementation as nested sampling. Furthermore, Guyader et al. (2011) also take advantage of the bested sampling fact that, if direct simulation under the constraint H(y)≥ξ is infeasible, simulating via one single step of a Metropolis-Hastings algorithm is as valid as direct simulation. (I could not access the paper, but the reference list of Guyader et al. (2011) includes both original papers by John Skilling, so the connection must be made in the paper.) What I find most interesting in this algorithm is that it even achieves unbiasedness (even in the MCMC case!).\n\n## computational methods for statistical mechanics [day #3]\n\nPosted in Mountains, pictures, Running, Statistics, Travel, University life with tags , , , , , , , , , , , , , , on June 6, 2014 by xi'an",
null,
"The third day [morn] at our ICMS workshop was dedicated to path sampling. And rare events. Much more into [my taste] Monte Carlo territory. The first talk by Rosalind Allen looked at reweighting trajectories that are not in an equilibrium or are missing the Boltzmann [normalizing] constant. Although the derivation against a calibration parameter looked like the primary goal rather than the tool for constant estimation. Again papers in J. Chem. Phys.! And a potential link with ABC raised by Antonietta Mira… Then Jonathan Weare discussed stratification. With a nice trick of expressing the normalising constants of the different terms in the partition as solution(s) of a Markov system",
null,
"$v\\mathbf{M}=v$\n\nBecause the stochastic matrix M is easier (?) to approximate. Valleau’s and Torrie’s umbrella sampling was a constant reference in this morning of talks. Arnaud Guyader’s talk was in the continuation of Toni Lelièvre’s introduction, which helped a lot in my better understanding of the concepts. Rephrasing things in more statistical terms. Like the distinction between equilibrium and paths. Or bias being importance sampling. Frédéric Cérou actually gave a sort of second part to Arnaud’s talk, using importance splitting algorithms. Presenting an algorithm for simulating rare events that sounded like an opposite nested sampling, where the goal is to get down the target, rather than up. Pushing particles away from a current level of the target function with probability ½. Michela Ottobre completed the series with an entry into diffusion limits in the Roberts-Gelman-Gilks spirit when the Markov chain is not yet stationary. In the transient phase thus.\n\n## Split Sampling: expectations, normalisation and rare events\n\nPosted in Books, Statistics, University life with tags , , , , , , on January 27, 2014 by xi'an\n\nJust before Christmas (a year ago), John Birge, Changgee Chang, and Nick Polson arXived a paper with the above title. Split sampling is presented a a tool conceived to handle rare event probabilities, written in this paper as",
null,
"$Z(m)=\\mathbb{E}_\\pi[\\mathbb{I}\\{L(X)>m\\}]$\n\nwhere π is the prior and L the likelihood, m being a large enough bound to make the probability small. However, given John Skilling’s representation of the marginal likelihood as the integral of the Z(m)’s, this simulation technique also applies to the approximation of the evidence. The paper refers from the start to nested sampling as a motivation for this method, presumably not as a way to run nested sampling, which was created as a tool for evidence evaluation, but as a competitor. Nested sampling may indeed face difficulties in handling the coverage of the higher likelihood regions under the prior and it is an approximative method, as we detailed in our earlier paper with Nicolas Chopin. The difference between nested and split sampling is that split sampling adds a distribution ω(m) on the likelihood levels m. If pairs (x,m) can be efficiently generated by MCMC for the target",
null,
"$\\pi(x)\\omega(m)\\mathbb{I}\\{L(X)>m\\},$\n\nthe marginal density of m can then be approximated by Rao-Blackwellisation. From which the authors derive an estimate of Z(m), since the marginal is actually proportional to ω(m)Z(m). (Because of the Rao-Blackwell argument, I wonder how much this differs from Chib’s 1995 method, i.e. if the split sampling estimator could be expressed as a special case of Chib’s estimator.) The resulting estimator of the marginal also requires a choice of ω(m) such that the associated cdf can be computed analytically. More generally, the choice of ω(m) impacts the quality of the approximation since it determines how often and easily high likelihood regions will be hit. Note also that the conditional π(x|m) is the same as in nested sampling, hence may run into difficulties for complex likelihoods or large datasets.\n\nWhen reading the beginning of the paper, the remark that “the chain will visit each level roughly uniformly” (p.13) made me wonder at a possible correspondence with the Wang-Landau estimator. Until I read the reference to Jacob and Ryder (2012) on page 16. Once again, I wonder at a stronger link between both papers since the Wang-Landau approach aims at optimising the exploration of the simulation space towards a flat histogram. See for instance Figure 2.\n\nThe following part of the paper draws a comparison with both nested sampling and the product estimator of Fishman (1994). I do not fully understand the consequences of the equivalence between those estimators and the split sampling estimator for specific choices of the weight function ω(m). Indeed, it seemed to me that the main point was to draw from a joint density on (x,m) to avoid the difficulties of exploring separately each level set. And also avoiding the approximation issues of nested sampling. As a side remark, the fact that the harmonic mean estimator occurs at several points of the paper makes me worried. The qualification of “poor Monte Carlo error variances properties” is an understatement for the harmonic mean estimator, as it generally has infinite variance and it hence should not be used at all, even as a starting point. The paper does not elaborate much about the cross-entropy method, despite using an example from Rubinstein and Kroese (2004).\n\nIn conclusion, an interesting paper that made me think anew about the nested sampling approach, which keeps its fascination over the years! I will most likely use it to build an MSc thesis project this summer in Warwick.\n\n## Special Issue of ACM TOMACS on Monte Carlo Methods in Statistics\n\nPosted in Books, R, Statistics, University life with tags , , , , , , , , , , , , on December 10, 2012 by xi'an\n\nAs posted here a long, long while ago, following a suggestion from the editor (and North America Cycling Champion!) Pierre Lécuyer (Université de Montréal), Arnaud Doucet (University of Oxford) and myself acted as guest editors for a special issue of ACM TOMACS on Monte Carlo Methods in Statistics. (Coincidentally, I am attending a board meeting for TOMACS tonight in Berlin!) The issue is now ready for publication (next February unless I am confused!) and made of the following papers:\n\n * Massive parallelization of serial inference algorithms for a complex generalized linear model MARC A. SUCHARD, IVAN ZORYCH, PATRICK RYAN, DAVID MADIGAN Abstract *Convergence of a Particle-based Approximation of the Block Online Expectation Maximization Algorithm SYLVAIN LE CORFF and GERSENDE FORT Abstract * Efficient MCMC for Binomial Logit Models AGNES FUSSL, SYLVIA FRÜHWIRTH-SCHNATTER, RUDOLF FRÜHWIRTH Abstract * Adaptive Equi-Energy Sampler: Convergence and Illustration AMANDINE SCHRECK and GERSENDE FORT and ERIC MOULINES Abstract * Particle algorithms for optimization on binary spaces CHRISTIAN SCHÄFER Abstract * Posterior expectation of regularly paved random histograms RAAZESH SAINUDIIN, GLORIA TENG, JENNIFER HARLOW, and DOMINIC LEE Abstract * Small variance estimators for rare event probabilities MICHEL BRONIATOWSKI and VIRGILE CARON Abstract * Self-Avoiding Random Dynamics on Integer Complex Systems FIRAS HAMZE, ZIYU WANG, and NANDO DE FREITAS Abstract * Bayesian learning of noisy Markov decision processes SUMEETPAL S. SINGH, NICOLAS CHOPIN, and NICK WHITELEY Abstract\n\nHere is the draft of the editorial that will appear at the beginning of this special issue. (All faults are mine, of course!) Continue reading"
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https://hackernoon.com/how-to-solve-unique-path-problem-zj4qt30z3 | [
"# How to solve Unique path problem",
null,
"Dynamic programming approach.\nBefore getting started, let’s discuss about the problem.\nA robot is located at the top-left corner of a grid (m*n) (marked ‘Start’ in the diagram below).\nThe robot can only move either down or right each time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).",
null,
"How many possible unique paths are there?\nUnderstand the problem:\nThe problem asks for how many unique paths start from top-left of the grid to the bottom-right.\nNow let’s take a scenario. Suppose a 3*2 grid is given. From the top-left corner, our target is to reach the bottom-right corner.\nSo based on the condition lets see what we can generate.\n``````Path-1: Right → Right → Down\nPath-2: Right → Down → Right\nPath-3: Down → Right → Right``````",
null,
"Hold it right there. That is the solution to this scenario. There are three unique paths. The approach we tried is not Dynamic Programming. Now a fancy word has arrived. what is it? Let’s see…\nDynamic Programming:\n(DP) is a technique in computer programming that helps to efficiently solve a class of problems that have overlapping sub-problems and optimal substructure property.\nSuch problems involve repeatedly calculating the value of the same sub-problems to find the optimum solution.\nLet’s take the factorial problem as an example. In mathematics, the factorial of a positive integer n, denoted by n!. Which is the multiplication of all positive integers less than or equal to n.\nFor example,\n``5! = 5*4*3*2*1 = 120``\nNow the parameter of our function to find the factorial is 6. So our factorial function will execute again as,\n``6! = 6*5*4*3*2*1 = 720``\nwe can divide this factorial of 6 in some sub-problem.\n``6! = 6* 5!``\nSo if we save the 5! in a position then the calculation of 6! will be easy for us. All we need to multiply the number with its previous calculated factorial. As a result, we are reducing the number of operations. This is the base of dynamic programming.\nNow let’s head to our problem. If we try to visualize all the ways that will be tough. So our first task is to divide the problem into sub-problems. Instead of the goal, assume our finish point is the right adjacent square. There is only one way to go there, just have to move right, Same way for the cell below the start, Have to move down. What about diagonal square? well, we have two ways.\n``````1. From start to right, then down\n2. From start to down, then right``````\nHow can we calculate it in numbers? Aa..ha. That relies on those two steps. If we store the number of ways to move adjacent right and down from the start point, all we need to do is add those two ways for the diagonal one.\nLet’s expand our view now if we have a grid of size m*n. what is the number of ways to reach the right squares from the start. It should be 1 because we are not changing our direction, same for the below squares of the start point.\nThe first row will be filled with 1 because there is only one way to reach every square. same for the first column. Now we have to fill the rest of the squares. We need two loops to iterate through each square. which will fill them with the sum of its top square value and left square value. The final result will be stored in the bottom-right corner.\nNow finally we got four steps to complete the problem.\n``````1. Fill the first row with 1\n2. Fill the first column with 1\n3. Fill the rest of the squares with the sum of its top square value and left square value.\n4. Return the bottom- right corner square value.``````\nAssume we have a grid 4*3.\nFirst things first, let’s make a matrix dp of 4*3 size. and fill the first row and column with 1.",
null,
"The second columns first empty place will be filled with its top and left square’s sum. If the place is dp[i][j], then it will be filled with dp[i-1][j]+dp[i][j-1]",
null,
"we will fill the rest of the empty places of the second row by following the same process.",
null,
"same goes for the last row",
null,
"Our final result will be stored at the bottom right corner square, which is\n``````dp[i][j] = dp[i-1][j] + dp[i]dp[j-1]\ndp = dp+dp\ndp = 4 + 6 = 10``````",
null,
"So, the ruby approach to solve this problem should be like this\n``````def unique_paths(m, n)\n0 if m == 0 || n == 0\ndp = Array.new(n) { Array.new(m,0) }\nfor i in 0...n\ndp[i] = 1\nend\nfor j in 0...m\ndp[j] = 1\nend\nfor i in 1...n\nfor j in 1...m\ndp[i][j] = dp[i-1][j] + dp[i][j-1]\nend\nprint dp\nend\ndp[n-1][m-1]\nend``````\nTime Complexity\nThe time complexity of this algorithm is O(m*n). As the iteration goes for the full size of the grid.\n\n## Space Complexity\n\nThe main factor of this algorithm is the 2D Array. And its size is dependent on its row and column. So the space complexity for this algorithm is O(m*n) as well.\nFor further information:\n\n## Tags",
null,
""
]
| [
null,
"https://hackernoon.com/drafts/g1143xdi.png",
null,
"https://hackernoon.com/photos/tRi82kbKnOcpcbOfCiFn8dN9Dju1-ga2fa309k",
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"https://hackernoon.com/photos/tRi82kbKnOcpcbOfCiFn8dN9Dju1-kt3bv30ov",
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"https://hackernoon.com/photos/tRi82kbKnOcpcbOfCiFn8dN9Dju1-es3is302x",
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"https://hackernoon.com/photos/tRi82kbKnOcpcbOfCiFn8dN9Dju1-hu3nk30xi",
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"https://hackernoon.com/photos/tRi82kbKnOcpcbOfCiFn8dN9Dju1-tu3p030qj",
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"https://hackernoon.com/photos/tRi82kbKnOcpcbOfCiFn8dN9Dju1-ry4m030jl",
null,
"https://hackernoon.com/photos/tRi82kbKnOcpcbOfCiFn8dN9Dju1-924ng30xh",
null,
"https://ucarecdn.com/57662ddc-da7c-407b-afcb-cb45184b2705/",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8830898,"math_prob":0.99238133,"size":4691,"snap":"2019-51-2020-05","text_gpt3_token_len":1218,"char_repetition_ratio":0.114358865,"word_repetition_ratio":0.027617952,"special_character_ratio":0.26369643,"punctuation_ratio":0.11153119,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9985222,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-20T17:12:26Z\",\"WARC-Record-ID\":\"<urn:uuid:e92fbca2-7797-48ce-a92e-bfb1c29b0f65>\",\"Content-Length\":\"27028\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d33530be-6aeb-45d6-a39b-e1aa725e0a77>\",\"WARC-Concurrent-To\":\"<urn:uuid:d8e6046f-9798-44fa-aa5c-d0f83502c0d1>\",\"WARC-IP-Address\":\"104.197.241.218\",\"WARC-Target-URI\":\"https://hackernoon.com/how-to-solve-unique-path-problem-zj4qt30z3\",\"WARC-Payload-Digest\":\"sha1:YVMM5CMRBTJKFZRTGDVUZMQP55Z4SAKI\",\"WARC-Block-Digest\":\"sha1:X5LRD32OQHNG5JR3YIMRPDGU4TZBTOE5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250599718.13_warc_CC-MAIN-20200120165335-20200120194335-00030.warc.gz\"}"} |
https://www.semesprit.com/70083/5-4-slope-as-a-rate-of-change-worksheet/ | [
"# 5.4 Slope as A Rate Of Change Worksheet\n\nThe slope of a hill is an important factor in determining the overall slope of the slope. Thus, this is the 5.4 Slope As A Rate Of Change Worksheet. If you are analyzing the slope of a slope and you would like to apply the term that was used earlier. Then the slope rate of change must be calculated. This slope rate of change is called the slope factor.\n\nThe slope factor is the rate of change in the slope, the angle of slope. It is not a simple linear relation between the slope and the rate of change of slope. Therefore, it is called slope rate of change.\n\nSlopes are usually defined as numbers that are shown in a graph. These numbers indicate the slopes of the slopes. If the slopes are plotted on a graph, then the slope factor is the number that is applied to the graph. Thus, if you want to calculate the slope factor for a slope graph, then you have to convert the number of degrees that are on the graph to radians and use this to make the calculation. You must use the slope factor when calculating slope factors.\n\nYou can also compare the slopes of two slopes to find out the slope factor. Therefore, it is quite important for you to find out the slope factor. In this article, we will discuss the importance of finding the slope factor.\n\nThere are many ways in which you can use the slope factor. These include the concept of the slope ratio, the first order approximation of the slope, and the weighted-slope approximation. There are many other methods that are used to compute slope factors. For example, there is the “effective slope”, or the sum of the slopes. Thus, you must know the factors that are used to compute these.\n\nYou must consider the fact that the slope is the slope itself. If the slope is constant, then the slope factor is also constant. You may even want to check the fact that the slope factor may increase with an increase in slope."
]
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https://www.shaalaa.com/question-bank-solutions/another-condition-quadrilateral-be-parallelogram-in-parallelogram-abcd-two-points-p-q-are-taken-diagonal-bd-such-that-dp-bq-see-given-figure-show-that_6724 | [
"Share\n\n# In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see the given figure). Show that: - CBSE Class 9 - Mathematics\n\nConceptAnother Condition for a Quadrilateral to Be a Parallelogram\n\n#### Question\n\nIn parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see the given figure). Show that:\n\n(i) ΔAPD ≅ ΔCQB\n\n(ii) AP = CQ\n\n(iii) ΔAQB ≅ ΔCPD\n\n(iv) AQ = CP\n\n(v) APCQ is a parallelogram",
null,
"#### Solution\n\n(i) In ΔAPD and ΔCQB,\n\nAD = CB (Opposite sides of parallelogram ABCD)\n\nDP = BQ (Given)\n\n∴ ΔAPD ≅ ΔCQB (Using SAS congruence rule)\n\n(ii) As we had observed that ΔAPD ≅ ΔCQB,\n\n∴ AP = CQ (CPCT)\n\n(iii) In ΔAQB and ΔCPD,\n\n∠ABQ = ∠CDP (Alternate interior angles for AB || CD)\n\nAB = CD (Opposite sides of parallelogram ABCD)\n\nBQ = DP (Given)\n\n∴ ΔAQB ≅ ΔCPD (Using SAS congruence rule)\n\n(iv) As we had observed that ΔAQB ≅ ΔCPD,\n\n∴ AQ = CP (CPCT)\n\n(v) From the result obtained in (ii) and (iv),\n\nAQ = CP and\n\nAP = CQ\n\nSince opposite sides in quadrilateral APCQ are equal to each other, APCQ is a parallelogram.\n\nIs there an error in this question or solution?\n\n#### APPEARS IN\n\nNCERT Solution for Mathematics Class 9 (2018 to Current)"
]
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null,
"https://www.shaalaa.com/images/_4:ae388aaace844c09b4d58256fac7e82c.png",
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https://majanto.com/tag/statement/ | [
"# Tag Archives: statement\n\n## Which two are true about a SQL statement using SET operators such as UNION? (Choose two.)\n\nWhich two are true about a SQL statement using SET operators such as UNION? (Choose two.) A. The number, but not names, of columns must be identical for all SELECT statements in the query. B. The data type of each column returned by the second query must exactly match the data type of the corresponding column returned by… Read More »\n\n## Given: (Exhibit) Which statement is true?\n\nGiven:Which statement is true? A. Only s is accessible by obj. B. p, r, and s are accessible by obj. C. Both p and s are accessible by obj. D. Both r and s are accessible by obj. Correct Answer: D\n\n## Which two statements are true about the ORDER BY clause when used with a SQL statement containing a SET\n\nWhich two statements are true about the ORDER BY clause when used with a SQL statement containing a SET operator such as UNION? (Choose two.) A. The first column in the first SELECT of the compound query with the UNION operator is used by default to sort output in the absence of an ORDER BY clause B. Only… Read More »\n\n## Given: (Exhibit) Which statement is equivalent to line 1?\n\nGiven:Which statement is equivalent to line 1? A. double totalSalary = list.stream().map(e > e.getSalary() * ratio).reduce (bo).ifPresent (p > p.doubleValue()); B. double totalSalary = list.stream().mapToDouble(e > e.getSalary() * ratio).sum; C. double totalSalary = list.stream().map(Employee::getSalary * ratio).reduce (bo).orElse(0.0); D. double totalSalary = list.stream().mapToDouble(e > e.getSalary() * ratio).reduce(starts, bo); Correct Answer: C\n\n## Given: (Exhibit) Which statement is equivalent to line 1?\n\nGiven:Which statement is equivalent to line 1? A. double totalSalary = list.stream().map(e -> e.getSalary() * ratio).reduce (bo).ifPresent (p -> p.doubleValue()); B. double totalSalary = list.stream().mapToDouble(e -> e.getSalary() * ratio).sum; C. double totalSalary = list.stream().map(Employee::getSalary * ratio).reduce (bo).orElse(0.0); D. double totalSalary = list.stream().mapToDouble(e -> e.getSalary() * ratio).reduce(starts, bo); Correct Answer: C\n\n## Given: (Exhibit) Which statement on line 1 enables this code fragment to compile?\n\nGiven:Which statement on line 1 enables this code fragment to compile? A. Function function = String::toUpperCase; B. UnaryOperator function = s > s.toUpperCase(); C. UnaryOperator<String> function = String::toUpperCase; D. Function<String> function = m > m.toUpperCase(); Correct Answer: C\n\n## Given: (Exhibit) Which statement on line 1 enables this code to compile?\n\nGiven:Which statement on line 1 enables this code to compile? A. Function<Integer, Integer> f = n > n * 2; B. Function<Integer> f = n > n * 2; C. Function<int> f = n > n * 2; D. Function<int, int> f = n > n * 2; E. Function f = n > n * 2; Correct… Read More »\n\n## Given: Acc.java: (Exhibit) Which statement is true?\n\nGiven:Acc.java:Which statement is true? A. Both p and s are accessible by obj. B. Both r and s are accessible by obj. C. p, r, and s are accessible by obj. D. Only s is accessible by obj. Correct Answer: D\n\n## Given the content of three files: (Exhibit) Which statement is true?\n\nGiven the content of three files:Which statement is true? A. The A.Java and B.java files compile successfully. B. Only the A.Java file compiles successfully. C. Only the C.java file compiles successfully. D. The A.Java and C.java files compile successfully. E. Only the B.java file compiles successfully. F. The B.java and C.java files compile successfully. Correct Answer: B\n\n## Given: (Exhibit) You want to use the myResource class in a try-with-resources statement. Which change\n\nGiven:You want to use the myResource class in a try-with-resources statement. Which change will accomplish this? A. Implement AutoCloseable and override the close method. B. Extend AutoCloseable and override the autoClose method. C. Extend AutoCloseable and override the close method. D. Implement AutoCloseable and override the autoClose method. Correct Answer: A"
]
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https://www.biologyonline.com/dictionary/dimensions | [
"Dictionary > Dimensions\n\n# Dimensions\n\nDimension\n1. Measure in a single line, as length, breadth, height, thickness, or circumference; extension; measurement; usually, in the plural, measure in length and breadth, or in length, breadth, and thickness; extent; size; as, the dimensions of a room, or of a ship; the dimensions of a farm, of a kingdom. Gentlemen of more than ordinary dimensions. (W. Irving) space of dimension, extension that has length but no breadth or thickness; a straight or curved line. Space of two dimensions, extension which has length and breadth, but no thickness; a plane or curved surface. Space of three dimensions, extension which has length, breadth, and thickness; a solid. Space of four dimensions, as imaginary kind of extension, which is assumed to have length, breadth, thickness, and also a fourth imaginary dimension. Space of five or six, or more dimensions is also sometimes assumed in mathematics.\n2. Extent; reach; scope; importance; as, a project of large dimensions.\n3. (Science: mathematics) The degree of manifoldness of a quantity; as, time is quantity having one dimension; volume has three dimensions, relative to extension.\n4. (Science: mathematics) a literal factor, as numbered in characterising a term. The term dimensions forms with the cardinal numbers a phrase equivalent to degree with the ordinal; thus, a^2b^2c is a term of five dimensions, or of the fifth degree.\n5. (Science: physics) The manifoldness with which the fundamental units of time, length, and mass are involved in determining the units of other physical quantities. Thus, since the unit of velocity varies directly as the unit of length and inversely as the unit of time, the dimensions of velocity are said to be length <divby/ time; the dimensions of work are mass <times/ (length)^2 <divby/ (time)^2; the dimensions of density are mass <divby/ (length)^3. Dimension lumber, dimension scantling, or dimension stock, lumber for building, etc, cut to the sizes usually in demand, or to special sizes as ordered. Dimension stone, stone delivered from the quarry rough, but brought to such sizes as are requisite for cutting to dimensions given.\nOrigin: L. Dimensio, fr. Dimensus, p. P. Of dimetiri to measure out; di- = dis- – metiri to measure: cf. F. Dimension. See measure.\n\n## Related Articles...\n\nNo related articles found"
]
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https://physics.com.hk/2018/08/09/ | [
"# The Jacobian of the inverse of a transformation\n\nThe Jacobian of the inverse of a transformation is the inverse of the Jacobian of that transformation\n\n.\n\nIn this post, we would like to illustrate the meaning of\n\nthe Jacobian of the inverse of a transformation = the inverse of the Jacobian of that transformation\n\nby proving a special case.\n\n.\n\nConsider a transformation",
null,
"$\\mathscr{T}: \\bar{x}^i=\\bar{x}^i (x^1,x^2)$, which is an one-to-one mapping from unbarred",
null,
"$x^i$‘s to barred",
null,
"$\\bar{x}^i$ coordinates, where",
null,
"$i=1, 2$.\n\nBy definition, the Jacobian matrix J of",
null,
"$\\mathscr{T}$ is",
null,
"$J= \\begin{pmatrix} \\displaystyle{\\frac{\\partial \\bar{x}^1}{\\partial x^1}} & \\displaystyle{\\frac{\\partial \\bar{x}^1}{\\partial x^2}} \\\\ \\displaystyle{\\frac{\\partial \\bar{x}^2}{\\partial x^1}} & \\displaystyle{\\frac{\\partial \\bar{x}^2}{\\partial x^2}} \\end{pmatrix}$\n\n.\n\nNow we consider the the inverse of the transformation",
null,
"$\\mathscr{T}$:",
null,
"$\\mathscr{T}^{-1}: x^i=x^i(\\bar{x}^1,\\bar{x}^2)$\n\nBy definition, the Jacobian matrix",
null,
"$\\bar{J}$ of this inverse transformation,",
null,
"$\\mathscr{T}^{-1}$, is",
null,
"$\\bar{J}= \\begin{pmatrix} \\displaystyle{\\frac{\\partial x^1}{\\partial \\bar{x}^1}} & \\displaystyle{\\frac{\\partial x^1}{\\partial \\bar{x}^2}} \\\\ \\displaystyle{\\frac{\\partial x^2}{\\partial \\bar{x}^1}} & \\displaystyle{\\frac{\\partial x^2}{\\partial \\bar{x}^2}} \\end{pmatrix}$\n\n.\n\nOn the other hand, the inverse of Jacobian",
null,
"$J$ of the original transformation",
null,
"$\\mathscr{T}$ is",
null,
"$J^{-1}=\\displaystyle{\\frac{1}{ \\begin{vmatrix} \\displaystyle{\\frac{\\partial \\bar{x}^1}{\\partial x^1}} & \\displaystyle{\\frac{\\partial \\bar{x}^1}{\\partial x^2}} \\\\ \\displaystyle{\\frac{\\partial \\bar{x}^2}{\\partial x^1}} & \\displaystyle{\\frac{\\partial \\bar{x}^2}{\\partial x^2}} \\end{vmatrix} }} \\begin{pmatrix} \\displaystyle{\\frac{\\partial \\bar{x}^2}{\\partial x^2}} & \\displaystyle{-\\frac{\\partial \\bar{x}^1}{\\partial x^2}} \\\\ \\displaystyle{-\\frac{\\partial \\bar{x}^2}{\\partial x^1}} & \\displaystyle{\\frac{\\partial \\bar{x}^1}{\\partial x^1}} \\end{pmatrix}$\n\n.\n\nIf",
null,
"$\\bar{J} = J^{-1}$, their (1, 1)-elementd should be equation:",
null,
"$\\displaystyle{\\frac{\\partial x^1}{\\partial \\bar{x}^1}}\\stackrel{?}{=}\\displaystyle{\\frac{1}{\\displaystyle{\\frac{\\partial \\bar{x}^1}{\\partial x^1}}\\displaystyle{\\frac{\\partial \\bar{x}^2}{\\partial x^2}}-\\displaystyle{\\frac{\\partial \\bar{x}^1}{\\partial x^2}}\\displaystyle{\\frac{\\partial \\bar{x}^2}{\\partial x^1}} }} \\bigg( \\displaystyle{\\frac{\\partial \\bar{x}^2}{\\partial x^2}} \\bigg)$\n\nLet’s try to prove that.\n\n.\n\nConsider equations",
null,
"$\\bar{x}^1 = \\bar{x}^1(x^1,x^2)$",
null,
"$\\bar{x}^2 = \\bar{x}^2(x^1,x^2)$\n\nDifferentiate both sides of each equation with respect to",
null,
"$\\bar{x}^1$, we have:",
null,
"$A := 1=\\displaystyle{\\frac{\\partial \\bar{x}^1}{\\partial \\bar{x}^1}=\\frac{\\partial \\bar{x}^1}{\\partial x^1}\\frac{\\partial x^1}{\\partial \\bar{x}^1}+\\frac{\\partial \\bar{x}^1}{\\partial x^2}\\frac{\\partial x^2}{\\partial \\bar{x}^1}}$",
null,
"$B := 0 = \\displaystyle{\\frac{\\partial \\bar{x}^2}{\\partial \\bar{x}^1}=\\frac{\\partial \\bar{x}^2}{\\partial x^1}\\frac{\\partial x^1}{\\partial \\bar{x}^1}+\\frac{\\partial \\bar{x}^2}{\\partial x^2}\\frac{\\partial x^2}{\\partial \\bar{x}^1}}$\n\n.",
null,
"$A \\times \\displaystyle{\\frac{\\partial \\bar{x}^2}{\\partial x^2}}:~~~~~C := \\displaystyle{\\frac{\\partial \\bar{x}^2}{\\partial x^2}=\\frac{\\partial \\bar{x}^1}{\\partial x^1}\\frac{\\partial x^1}{\\partial \\bar{x}^1}\\frac{\\partial \\bar{x}^2}{\\partial x^2}+\\frac{\\partial \\bar{x}^1}{\\partial x^2}\\frac{\\partial x^2}{\\partial \\bar{x}^1}\\frac{\\partial \\bar{x}^2}{\\partial x^2}}$",
null,
"$B \\times \\displaystyle{\\frac{\\partial \\bar{x}^1}{\\partial x^2}}:~~~~~D := \\displaystyle{0=\\frac{\\partial \\bar{x}^2}{\\partial x^1}\\frac{\\partial x^1}{\\partial \\bar{x}^1}\\frac{\\partial \\bar{x}^1}{\\partial x^2}+\\frac{\\partial \\bar{x}^2}{\\partial x^2}\\frac{\\partial x^2}{\\partial \\bar{x}^1}\\frac{\\partial \\bar{x}^1}{\\partial x^2}}$\n\n.",
null,
"$D-C:$",
null,
"$\\displaystyle{ \\frac{\\partial \\bar{x}^2}{\\partial x^2}= \\bigg( \\frac{\\partial \\bar{x}^1}{\\partial x^1}\\frac{\\partial \\bar{x}^2}{\\partial x^2} - \\frac{\\partial \\bar{x}^2}{\\partial x^1}\\frac{\\partial \\bar{x}^1}{\\partial x^2}\\bigg) \\frac{\\partial x^1}{\\partial \\bar{x}^1}}$,\n\nresults",
null,
"$\\displaystyle{ \\frac{\\partial x^1}{\\partial \\bar{x}^1}}=\\frac{\\displaystyle{\\frac{\\partial \\bar{x}^2}{\\partial x^2}}}{\\displaystyle{\\frac{\\partial \\bar{x}^1}{\\partial x^1}\\frac{\\partial \\bar{x}^2}{\\partial x^2} - \\frac{\\partial \\bar{x}^1}{\\partial x^2}\\frac{\\partial \\bar{x}^2}{\\partial x^1}}}$\n\n— Me@2018-08-09 09:49:51 PM\n\n.\n\n.\n\n2018.08.09 Thursday (c) All rights reserved by ACHK"
]
| [
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.86796916,"math_prob":1.0000062,"size":913,"snap":"2019-13-2019-22","text_gpt3_token_len":220,"char_repetition_ratio":0.21342134,"word_repetition_ratio":0.10191083,"special_character_ratio":0.25301206,"punctuation_ratio":0.17391305,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000097,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-24T23:00:57Z\",\"WARC-Record-ID\":\"<urn:uuid:42fdeb5a-2e9f-4d45-a7e3-7eb507eae703>\",\"Content-Length\":\"84127\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8d42b382-b809-4e75-bfc2-a84543ec9a4f>\",\"WARC-Concurrent-To\":\"<urn:uuid:f7c43c17-c450-4e42-90a7-3e0f98556875>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://physics.com.hk/2018/08/09/\",\"WARC-Payload-Digest\":\"sha1:5I6C43BKNEBPLND6JHOEBDO24JXJGSDY\",\"WARC-Block-Digest\":\"sha1:FFCCQWTQ53NTZDWMDLQCC2GGTQBYCPPE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232257781.73_warc_CC-MAIN-20190524224619-20190525010619-00513.warc.gz\"}"} |
http://www.actforlibraries.org/origins-and-use-of-the-kelvin-temperature-scale/ | [
"# Origins and use of the Kelvin Temperature Scale\n\nOrigins and Use of the Kelvin Temperature Scale\n\nHumankind is constantly seeking to improve its understanding of the fundamental properties of nature. As a result of this curiosity, mankind has devised numerous methods of taking quantitative measurements. As the years passed, we abandoned many of these systems, and only the most fundamental and true remain. Of all of the ways we can measure temperature, the Kelvin scale is the most natural. To understand why, we must look at some history.\n\nBefore the Kelvin scale, scientific measurements were taken using the Celsius scale, which is centered around the boiling-point of water. The Celsius scale worked well, and is still used in many scientific reports. However, the boiling point of water is arbitrary, and William Thompson desired a scale that was based upon something much more definite. As temperature is the measure of thermal energy, zero should delineate a thermal-energy free system. This led William Thompson to devise his scale.\n\nBy far, the most difficult step in designing the Kelvin scale was to find the temperature of something with no energy. The solution turned out to be gasses. As gasses decrease in temperature, they decrease in volume by a certain amount. An extrapolation of the graphs of differing gasses shows that the linear functions of temperature vs. volume converge at a point. This point, at -273.16C is absolute zero, or 0K.\n\nBefore the invention of the Kelvin temperature scale, the following question would be difficult if not impossible to answer:\n\nHow many times more thermal energy does mercury have at 70C than at -10C?\n\nAs the Celsius scale does not begin at absolute zero, it is difficult to use to answer this question. However, if the temperatures are converted into Kelvin, the question is much simpler:\n\nHow many times more thermal energy does mercury have at 343.16K than at 263.16K?\n\nAs zero on the Kelvin scale is an absence of thermal energy, the two temperatures can simply be divided to yield the result:\n\n343.16K / 263.16K = 1.3040\n\nThe mercury at 343.16K has 1.3040 times more thermal energy than the mercury at 263.16K.\n\nThis is the reason that the Kelvin temperature scale is useful to scientists. Because of the properties of the Kelvin scale, it is the only scale that can be used with the ideal gas law. Many scientific equations require temperatures to be input in Kelvin form. This natural scale of measurement is a brilliant invention, and has led the way to the development of modern science."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9393302,"math_prob":0.9567253,"size":2497,"snap":"2022-27-2022-33","text_gpt3_token_len":518,"char_repetition_ratio":0.1456077,"word_repetition_ratio":0.02905569,"special_character_ratio":0.2122547,"punctuation_ratio":0.11983471,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97439027,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-14T01:27:43Z\",\"WARC-Record-ID\":\"<urn:uuid:72e3cd07-43f7-42cc-a50d-18817a13ff19>\",\"Content-Length\":\"20948\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6d10d10b-f0f0-4113-9ee4-605eb3a589ff>\",\"WARC-Concurrent-To\":\"<urn:uuid:2f3edb02-70d0-4f2f-9371-ce238d92ee76>\",\"WARC-IP-Address\":\"192.138.210.104\",\"WARC-Target-URI\":\"http://www.actforlibraries.org/origins-and-use-of-the-kelvin-temperature-scale/\",\"WARC-Payload-Digest\":\"sha1:HX2F2UNWOKFSQFENW34RDNJMSFINGWWT\",\"WARC-Block-Digest\":\"sha1:PV6GHKL7CY2ORN6ZQ6CIL35V62U6Y5LR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571989.67_warc_CC-MAIN-20220813232744-20220814022744-00203.warc.gz\"}"} |
https://www.lehman.edu/mathematics/learning-goals-and-objective.php | [
"Give\n\n# Lehman College",
null,
"",
null,
"## Learning Goals and Objective\n\n### B.A. in Mathematics\n\nUpon completion of a B.A. in Mathematics, a graduate will be able to do the following:\n\n• Perform numeric and symbolic computations\n• Construct and apply symbolic and graphical representations of functions\n• Model real-life problems mathematically\n• Use technology appropriately to analyze mathematical problems\n• State and apply mathematical definitions and theorems\n• Prove fundamental theorems\n• Construct and present a rigorous mathematical argument\n\nTo achieve these goals, courses required for the major are targeted for specific areas as outlined in the Lehman Math Majors Goals Site. Final exams in relevant courses will include problems directly testing this knowledge."
]
| [
null,
"https://www.lehman.edu/2014-redesign-images/template-assets/lehman-logo.png",
null,
"https://www.lehman.edu/mathematics/images/l2-header.jpg",
null
]
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https://www.mathexpression.com/coordinates.html | [
"### Graphs: Coordinates of a Point",
null,
"#### Lesson Objective\n\nLearn how to read & write the coordinates of a point.",
null,
"#### Why Learn This?\n\nTo use the coordinate plane, the first thing you must learn is to read & write the position (coordinates) of a point on the plane.\n\nThe picture on the right helps to illustrate this further.\n\nNow, let's proceed to the study tips below.",
null,
"### Study Tips\n\nIf you are not sure what are 'x-coordinate' or 'y-coordinate', just keep the study tips in mind. Things will be much clearer after you watch the video.",
null,
"#### Tip #1\n\nRead the x-coordinate first, then the y-coordinate. This minimizes mistake when you write down the coordinates.",
null,
"#### Tip #2\n\nDouble check the coordinates that you have written down. Make sure the x-coordinate and y-coordinate are in the correct order and not swapped.\n\nNow, watch the following math video to understand more.\n\n### Math Video",
null,
"#### Click Play to Watch",
null,
"#### Math Video Transcript\n\n00:00:02.110\nNow that you are already familiar with the coordinate plane, let's move on to learn how to read the co-ordinates of a point.\n\n00:00:12.010\nConsider this point right over here. When it is placed on the plane, it will have it's own co-ordinates that shows it's position.\n\n00:00:22.000\nNow, how to read the co-ordinates of this point? First read the x-coordinate of this point from the x-axis. You see that you'll get 2.0.\n\n00:00:35.190\nNow read the y-coordinate of this point from the y-axis. You'll get 3.0. By combining x and y-coordinate together, you'll get the co-ordinates of the point as (2.0,3.0)\n\n00:00:54.000\nYou can write the co-ordinates beside this point like this. Bear in mind, the 2.0 here corresponds to the x-coordinate and the 3.0 corresponds to the y-coordinate.\n\n00:01:09.040\nAlso, you must write the co-ordinates of the point in this order... 2.0,... 3.0. These numbers must be written in the this order; and this order also known as 'Ordered Pair'.\n\n00:01:22.110\nLets experiment with what we have learned. I'm going to move this point around. Observe how the co-ordinates changes accordingly. (Moves from (2.0,3.0) - ( 2.0,2.0) - (2.0,-3.0) - (-2.0,-3.0) - (-2.0,3.0) stops for about 3s for each co-ordinates).\n\n00:01:31.140\nx-coordinate: 2.0 , y-coordinate: 2.0.\n\n00:01:50.020\nx-coordinate: 2.0 , y-coordinate: negative 3.0.\n\n00:02:07.190\nx-coordinate: 2.0 negative 2.0 , y-coordinate: negative 3.0.\n\n00:02:26.110\nx-coordinate: negative 2.0 , y-coordinate: 3.0.\n\n00:02:35.070\nAlright, that is all for this lesson. You can move on to the practice questions to test your understanding.\n\n### Practice Questions & More",
null,
"#### Multiple Choice Questions (MCQ)\n\nNow, let's try some MCQ questions to understand this lesson better.\n\nYou can start by going through the series of the Coordinates of a point questions here or pick your question below.",
null,
"#### Site-Search and Q&A Library\n\nPlease feel free to visit the Q&A Library. You can read the Q&As listed in any of the available categories such as Algebra, Graphs, Exponents and more. Also, you can submit math question, share or give comments there.",
null,
""
]
| [
null,
"https://www.mathexpression.com/image-files/xlessoniconsmall.png.pagespeed.ic.rhUQ31kjea.jpg",
null,
"https://www.mathexpression.com/image-files/whyiconsmall.png.pagespeed.ce.XS7atGd14e.png",
null,
"https://www.mathexpression.com/image-files/coordinateofapoint.jpg",
null,
"https://www.mathexpression.com/image-files/xstudytipsiconsmall.png.pagespeed.ic.KB5qvIRoz9.jpg",
null,
"https://www.mathexpression.com/image-files/xstudytipsiconsmall.png.pagespeed.ic.KB5qvIRoz9.jpg",
null,
"https://www.mathexpression.com/image-files/xlessoniconsmall.png.pagespeed.ic.rhUQ31kjea.jpg",
null,
"https://www.mathexpression.com/image-files/xlessoniconsmall.png.pagespeed.ic.rhUQ31kjea.jpg",
null,
"https://www.mathexpression.com/image-files/xpracticeiconsmall.png.pagespeed.ic.vqB1-4smwb.jpg",
null,
"https://www.mathexpression.com/image-files/xsearch-icon-small.png.pagespeed.ic.GPjdZlfEzo.png",
null,
"https://www.mathexpression.com/image-files/xpracticeiconsmall.png.pagespeed.ic.vqB1-4smwb.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8861047,"math_prob":0.7931325,"size":3140,"snap":"2019-51-2020-05","text_gpt3_token_len":848,"char_repetition_ratio":0.18431123,"word_repetition_ratio":0.027559055,"special_character_ratio":0.2888535,"punctuation_ratio":0.19759679,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95883363,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,null,null,null,null,4,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-23T08:10:14Z\",\"WARC-Record-ID\":\"<urn:uuid:d7edc1bb-d681-4d56-999e-ff2a457e821a>\",\"Content-Length\":\"44043\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8f985a8b-d42b-4761-b247-25a5ef1e80e4>\",\"WARC-Concurrent-To\":\"<urn:uuid:5a821900-e68c-4484-af82-cb5188d7b9e5>\",\"WARC-IP-Address\":\"173.247.219.12\",\"WARC-Target-URI\":\"https://www.mathexpression.com/coordinates.html\",\"WARC-Payload-Digest\":\"sha1:63C3Y57DWNZ4LVUGK3E3GNXE7IQRDS54\",\"WARC-Block-Digest\":\"sha1:42ZNQB3WL7GO4UTPDANL65MEZEAKYQKG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250609478.50_warc_CC-MAIN-20200123071220-20200123100220-00131.warc.gz\"}"} |
http://gainmode.org/can-we-now-predict-when-a-neutron-star-will-give-birth-to-a-black-hole/2/ | [
"# Can We Now Predict When A Neutron Star Will Give Birth To A Black Hole?\n\nThe reason for this is because such a maximum value is dependent on the equation of state of the matter composing the star. This thermodynamic equation describes the state of matter under a given set of physical conditions – i.e. temperature, pressure, volume, or internal energy. And while astronomers have been able to ascertain within a degree of certainty what the maximum mass of a nonrotating neutron stars would be, they have been less successful in calculating what the maximum mass is for those that are rotating.",
null,
"Cross-section of a neutron star. Credit: Wikipedia Commons/Robert Schulze\n\nIn short, they have been unable to determine how much mass is needed before a rotating neutron star will surpass its maximum speed of rotation and finally form a new black hole. As Rezzolla explained:\n\n“What made it difficult in the past to calculate M_max is its value will differ from what composes the neutron star (i.e. its “equation of state”) and this is something we don’t really know. Neutron-star matter is so different from the one we know that we can only make educated guesses; and unfortunately, there are many guesses because there are several different ways to compute the properties of the equation of state. So one ended up up with a situation in which not only the maximum mass was different for different equations of state, but even the maximum rotation speed was different for different equations of state.”\n\nHowever, in their study, titled “Maximum mass, moment of inertia and compactness of relativistic stars” – which appeared recently in the Monthly Notices of the Royal Astronomical Society – Rezzolla and Cosima Breu (a Masters student in theoretical physics at Goethe University and co-author of the study) argue that it may now be possible to infer what the maximum mass of a rapidly rotating star would be."
]
| [
null,
"https://www.universetoday.com/wp-content/uploads/2016/04/800px-Neutron_star_cross_section.svg_-580x334.png",
null
]
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https://semanti.ca/blog/?modern-ai-for-executives | [
"# Modern AI for Executives\n\n## A concise bullshit-free guide to machine learning and data science for top managers (the only guide top managers really need)\n\nPosted by Josh on 11-08-2018\n\nWe, at semanti.ca, are constantly monitoring online publications on Artificial Intelligence and Machine Learning. What we see is that there are three major groups of online publications:\n\n• Too low-level ones, oriented on initiated readers;\n• Too high-level ones, which are kind of trying to explain AI to a non-technical reader, but rather confuse the reader even more;\n• Marketing bullshit.\n\nNone of the three groups of publications really helps executives to gain confidence in this complex topic. So, below is our guide to Machine Learning for top managers. We believe that this is the guide every top manager desperately looks for, and you are the lucky one to find it. Congratulations!\n\nFirst of all, let's clarify the terminology: AI, Machine Intelligence, Machine Learning, Deep Learning, Data Science, Data Mining, Neural Networks, Predictive Analytics, Cognitive Computing... We understand your pain when you read white papers full of buzzwords and try to get an idea on what the hell is the difference between them? Did you just read the same text yesterday or it's something new? Is there a difference? If not, why people use so many different terms to say the same?\n\n## Terminology\n\nArtificial Intelligence, AI, or sometimes also Machine Intelligence, is the science of making machines that act similarly to living species. It's very vague as a definition, so even a calculator can be called an AI because it mimics what only humans could do just two centuries ago.\n\nMachine Learning (ML) is one of the ways of building an AI. Today, it's the most popular and effective way of making a machine that acts similarly to an animal brain on some very specific problems. When we say that it acts similarly to we mean how it looks like to an outside observer and NOT how it actually functions.\n\nDeep Learning (DL) is a family of loosely related Machine Learning algorithms that employ the concept of an artificial neural network. This has nothing in common with animal brain neurons, just like Artificial Intelligence has nothing in common with the animal intelligence.\n\nSo, if you read an article on Neural Networks/Deep Learning and it has an illustration as below, know that the author is a layman. Do yourself a favor: close the browser tab. You're welcome.",
null,
"Animal Neuron. Source: Wikipedia.\n\nDon't worry, we will look at artificial neural networks more closely soon. Let's finish with the definitions.\n\nData Science is not a science. It's a set of methods and skills necessary to analyze data in the era of Big Data. Those methods and skills usually include a significant body of Machine Learning algorithms, hence the confusion between the two.\n\nBig Data is the data that is hard to work with on a conventional computer. The notion of a conventional computer is quite vague, but we can approximate it with the best gamer's PC available on the market. It can be hard to work with the data because of one of the following reasons:\n\n• the data is too big to fit in memory or to be stored on a hard drive;\n• the data comes too fast to be processed by a CPU;\n• the data comes from too many heterogeneous sources to handle it in one user-friendly desktop application.\n\nThose three reasons are often cited as the three \"V\": volume, velocity, and variety. If you read an article on Big Data or Data Science and it mentions four \"V\" (for \"veracity\") or five \"V\" (for \"value\"), know that you are reading marketing crap. Do yourself a favor: close the browser tab. You're welcome.\n\nPredictive Analytics is a strict subset of Machine Learning and statistical techniques focusing on predicting anything based on the data from the past. From statistics, it takes regression analysis; from Machine Learning it takes classification.\n\nData Mining is mostly a retired buzzword. One of the best books on Machine Learning called Data Mining: Practical Machine Learning Tools and Techniques was originally to be named just \"Practical Machine Learning\", and the term \"data mining\" was only added for marketing reasons. As of 2018, the term \"data mining\" as a buzzword lost the lead to \"data science\".\n\nCognitive Computing is a buzzword promoted by IBM to stand out from the competition. It roughly means using Machine Learning to build AI systems.\n\n## Modern AI\n\nThe research on artificial intelligence has been carried out since the 1960s. Artificial neural networks in a form similar to the modern one were proposed and trained in the mid-1980s. So, what's so different now? The main difference is that during the first two decades of the 2000s, several important breakthroughs have been made that made it possible to train very big neural networks (with billions of parameters). The most important such breakthroughs were:\n\n• Rectified Linear Unit (ReLU);\n• Long short-term memory (LSTM);\n• Dropout.\n\nReLU and LSTM are types of neural network units that made it possible to train very deep neural networks. Dropout made it possible to train neural networks that generalize well. Generalization is the quality of a machine learning system that shows how well the system performs on previously unseen examples.\n\nOther important factors that differ the modern AI from the AI of the XX century are:\n\n• the availability of very big training datasets (thanks to the Internet);\n• accessibility of very high computing power (thanks to GPUs and the cloud computing), and\n• the availability of high-quality high-level, accessible machine learning software (thanks to the Internet and open source movement).\n\n## What's So Special About Neural Networks?\n\nNeural Networks are the only known machine learning algorithms that can significantly benefit from bigger datasets. Almost any machine learning algorithm becomes better with more data. See for example the article \"The Unreasonable Effectiveness of Data\" written by the lead research scientists from Google. However, neural networks, thanks to their multi-layer architecture can reach almost perfect performance on virtually any machine learning task, given enough data.\n\n## Machine Learning Tasks\n\nMachine learning doesn't try to (and most likely could not) solve any problem the humanity and the business face. There're certain tasks well suited for solving using machine learning:\n\n• Classification,\n• Regression,\n• Clustering.\n\nMany real-world engineering problems can be reduced to those three tasks.\n\n### Classification\n\nClassification aims to find a mathematical function $$f$$ such that given an example $$x$$ as input it produces an output $$y$$. The output $$y$$ belongs to a finite set of labels. The example $$x$$ usually is a vector of features; the quantity of possible values of $$x$$ is usually infinite. A classical example of a classification task is spam detection. $$y$$, in this problem, can take only two values (depending on the machine learning algorithm the values are zero and one or minus one and plus one).\n\nThe definition of $$x$$ depends on the machine learning engineer. Because $$x$$ has to represent an email message somehow, and it has to be a vector of features, one can decide that the vector will have $$10,000$$ dimensions; every dimension will correspond to the presence of absence, in the email message, of some specific English word. If a word is present, then the value of the feature would be $$1$$, otherwise, it would be $$0$$. How to convert an email message into a vector totally depends on the machine learning engineer. The process is called feature engineering. There's no one correct way to do feature engineering, and it's rather an art than a science.\n\nAn engineer has to try different ways to extract features from email messages and see which one gives the best result on the problem of spam detection. Other types of features, not based on words, are also possible and can be combined with word-based features:\n\n• Whether the sender is in the contact book of the recipient;\n• Whether the email contains an attachment;\n• Whether the email or subject contain exclamation marks;\n• Whether the email contains a link;\n• Etc. The creativity of the engineer is very important here.\n\nExamples of business problems that can be solved as a classification problem:\n\n• Churn Modeling: $$x$$ represents what is known about the customer and $$y$$ will be the prediction of whether the customer will stop engaging with your business in, let's say, the next 6 months.\n\n• Customer Segmentation: $$x$$ represents what is known about the customer and $$y$$ will be the segment of customers; The definition of segments, what they are, how many of them should be, etc, is the preliminary task.\n\n• Sentiment Analysis: $$x$$ represents a customer's email or a tweet about a product and $$y$$ represents whether the customer is satisfied or dissatisfied with the product.\n\n### Regression\n\nRegression deals with the situations when we want to predict a certain quantity about an example. Here we aim to find a mathematical function $$f$$ such that given an example $$x$$ as input it produces an output $$y$$, which is a real number. As in classification, $$x$$ usually is a vector of features and the number of possible values of $$x$$ is usually also infinite.\n\nThe feature engineering task in regression is also similar to that of classification.\n\nBusiness problems that can be solved using regression:\n\n• Customer lifetime value modeling, that is predicting the future revenue that an individual customer will bring to your business in a given period. Here, $$x$$ represents everything we know about the customer and $$y$$ is a positive real number.\n\n• Demand analysis, that is predicting how many units of a product consumers will purchase. Here, $$x$$ could represent the state of the market and $$y$$ is a positive real number. For a company that produces soft drinks, $$x$$ could contain such features as the average temperature for the past week and past months, the number of bottles sold during the past week and past months, the number of bottles sold during the same period last year, etc.\n\n• Goods price prediction: for example, given everything we know about the house (the number of rooms, the area, the year of construction, zip code, and so on) as $$x$$, to predict what would be the price $$y$$ of this house.\n\n• Stock price prediction: given everything we know about the company (the number of employees, the volume investments in R&D, the last-year revenue, the total number of clients, the percentage of clients that renew subscription, what is the state of the stock market compared to last month (up or down), the current price of the stock, and so on) as $$x$$, to predict what would be the price $$y$$ of the stock of this company in one month.\n\n### Clustering\n\nThe two previous tasks, classification and regression, assume that we can make our machine learning system predict either the correct label (spam/not spam) or the correct real number (the house price). To make this happen, we have to train our system by giving examples of what is a correct prediction is. For example, if we want to train our system to predict spam, we have to give examples of what is spam and what's not spam. So, to enable the learning, we have to provide our machine learning algorithms with a sufficiently large collection of correct pairs $$x, y$$. Such kind of machine learning tasks are called supervised learning tasks.\n\nClustering, on the other hand, is an example of an unsupervised machine learning task. In the unsupervised settings, we still have a sufficiently large collection of example, however, every example is only $$x$$; we don't have $$y$$s.\n\nClustering can be helpful, for example, in the Customer Segmentation. Remember that we said above that the definition of segments, what they are, how many of them should be, etc, is the preliminary task. This can be done using clustering.\n\nThe task of clustering is defined like this: given a collection of examples $$X$$ and the number of clusters n, find a mathematical function $$f$$ that assigns to every examples $$x$$ from collection $$X$$ a cluster id $$i \\in 1..n$$.\n\nIdeally, we would like that $$f$$ puts similar examples to the same cluster (or, assigns the same $$i)$$ to similar values of $$x$$. Therefore, the job of the machine learning engineer here is not just feature engineering but also metric engineering. Metric engineering is the task of crafting a mathematical function $$m$$ which takes two vectors $$x$$ and $$x'$$ as input and returns a small value if $$x$$ is similar to $$x'$$ or a high value if $$x$$ and $$x'$$ are dissimilar.\n\nTwo popular choices of metric functions are:\n\n• Euclidian distance;\n• Invert cosine similarity.\n\nEuclidian distance computes the straight line distance between two points in space ($$x$$ and $$x'$$); the invert cosine similarity computes one minus the cosine of the angle between two vectors ($$x$$ and $$x'$$). The idea is that the bigger the angle between the vectors, the more dissimilar are two examples. As with feature engineering, metric engineering is an art and each machine learning engineer could develop their own metrics for every practical business problem.\n\nOne of the primary use of clustering in business is segmentation: customer, product or store. Similar products can be clustered together into groups based on their attributes like size, brand, use, flavor, etc; similar stores can be grouped together based on characteristics as sales, size of the store and size of customer base, availability of in-store services, etc.\n\n## How Machine Learning Algorithms Learn\n\nThis is the most important question: how does the machine learn? How, given a collection of pairs $$x, y$$ the machine figures out what the function $$y = f(x)$$ should look like and how this function then can be applied to new values of $$x$$, previously unseen by the machine such the predicted value of $$y$$ is most of the time correct?\n\nWe will show you only three pictures you need to have in mind to understand how most machine learning algorithms learn.\n\n### Linear Regression\n\nLinear regression is an algorithm that solves the regression problem by making one assumption about the function $$f$$. It assumes that $$f$$ has the following form: $$y = f(x) = ax + b$$\n\nSo the problem of the machine learning algorithm is to find two values: $$a$$ and $$b$$, where $$a$$ is a vector and $$b$$ is scalar.\n\nThe machine uses the collection of examples $$x, y$$ and finds such values for $$a$$ and $$b$$ that the following loss function is minimized: $$loss(y, f(x)) = \\sum_{(x, y)} (y - f(x))^2,$$ where $$\\Sigma$$ means the sum over all training examples. Basically, the loss function defines how good our prediction $$f(x)$$ is compared to the correct value $$y$$ that we know.\n\nThe choice of the loss function is one of the choices the machine learning engineer makes, along with the choice of features (in feature engineering) and the choice of similarity metric (in clustering). If we assume that $$x$$ is one-dimensional, then the function $$f$$ would look like the blue line on the below plot:",
null,
"Linear Regression. Source: Towards Data Science.\n\nOn the above plot, grey lines indicate the loss (before the square is applied).\n\nNothing fundamentally changes if $$x$$ has many dimensions. The only difference is that instead of the straight line, $$f$$ will have the form of a plane (if $$x$$ is two dimensional) or a hyperplane (if $$x$$ has more than three dimensions).\n\nNow you know how the simples form of regression works. Different algorithms can make different assumptions on the form of $$f$$ (different from $$ax + b$$), but the principle remains the same:\n\n1. First, make an assumption on the form of a function $$f$$, then\n2. Define the loss function appropriate for your business problem, finally\n3. Make the machine find the parameters that minimize the loss on the training data.\n\nThe machine will find the parameters by using some learning algorithm. Many of them exist.\n\n### Support Vector Machine\n\nSupport vector machine is a classification machine learning algorithm. It illustrates the best how classification algorithms work. Let's assume that our x is two dimensional ($$x = (x_1, x_2)$$) and our $$y$$ can take values either $$-1$$ or $$+1$$. Then the Support Vector Machine algorithm will find a straight line that separates the best the examples that have label $$-1$$ from the examples that have label $$+1$$:",
null,
"Support Vector Machine. Source: OpenCV Documentation.\n\nOn the above plot, the examples with label $$-1$$ are represented by the squares and those with label $$+1$$ are represented by circles. The green line separates the two kinds of examples the best, where best means that the distance from the closest member of each group to the green line is maximized.\n\nAgain, as with the regression task, nothing fundamentally changes if $$x$$ has three or more dimensions. Instead of the line, the machine will find a plane or a hyperplane that separates the best the examples belonging to different groups.\n\nDifferent classification algorithms differ from one another on how this line (the one that separates groups of examples with different labels from one another) is drawn. It's not always possible to draw a straight line to separate examples, so more complex algorithms find more complex mathematical functions to draw this line.\n\n### K-means Clustering\n\nTo illustrate clustering, we will take the example of an algorithm called K-means. In this algorithm, the machine learning engineer has to make the following three choices:\n\n1. How to convert examples into vectors (feature engineering),\n2. Define the similarity metric (metric engineering), and\n3. Guess how many clusters exist in the data.\n\nThe machine learning will do the rest.\n\nK-means clustering algorithm works as follows:\n\n1. Randomly (or according to a heuristics) put cluster centers in the proximity of the examples;\n2. Assign each example to the most similar cluster center (according to the similarity metrics);\n3. Compute the new position of each cluster center as the average of all examples assigned to this cluster center.\n4. Iteratively repeat steps 2 and 3 until the cluster center positions don't change significantly enough.\n\nTo illustrate the algorithm, let's assume that our $$x$$ is two-dimensional and there are three clusters. The following animation represents iterations of the clustering algorithm:",
null,
"K-means Clustering. Source: Source: Towards Data Science.\n\n## Generalization\n\nWhen working on a supervised learning problem, one important concept to understand is that of generalization. We say that a machine learning algorithm generalizes well to previously unseen examples (the examples we didn't use to train the algorithm) if it doesn't make much more prediction errors on the unseen examples compared to the number of errors the algorithm makes on the data seen during training.\n\nThe learning algorithm can predict perfectly all the $$y$$s from the data used for training but predict very poorly the $$y$$s from new data. This problem is called overfitting or the problem of high bias.\n\nUsually, every machine learning comes with several hyperparameters that the machine learning can tweak to reduce overfitting. It's generally impossible to completely remove overfitting, so the learning algorithm almost always performs better on the data seen during the training. However, such techniques as regularization that add a penalty to the loss function for overly complex shapes of the function $$f$$ allows reducing overfitting.\n\n## Special Cases\n\nSupervised learning has several important special taks:\n\n• In multi-label classification, $$y$$ is also a vector. So multiple labels could be predicted at once for an $$x$$. For example, in image segmentation, the goal is to predict the boundaries of an object on an image, so $$y$$ is four-dimensional;\n• In sequence labeling, $$x$$ is a sequence of vectors and $$y$$ is a sequence of labels of the same length. Sequence labeling algorithms are frequently used in Natural Language Processing to assigns different labels to different words in a sentence;\n• In Sequence-to-Sequence learning, $$y$$ is a sequence that can have different length than $$x$$. This is a scenario in machine translation or spelling correction.\n\n## Artificial Neural Networks\n\nBelow, we quote a short part of our blog post on neural networks called How Neural Networks Work. The post is very accessible and high-level but gives just enough details so that the interested reader could get a clear idea about neural networks:\n\nArtificial neural networks make another assumption about the function $$f$$. They assume that $$f$$ is a nested function. You have probably heard of neural network layers. So, for a 4-layer neural network, $$f$$ will look like this:\n\n$$y = f(x) = f_4(f_3(f_2(f_1(x)))),$$\n\nwhere $$f_1$$, $$f_2$$, $$f_3$$, and $$f_4$$ are simple functions like this:\n\n$$f_i = f_i(z) = nonlinear_i(a_i*z + b_i),$$\n\nwhere $$i$$ is called the layer index and can span from 1 to any number of layers. The function $$nonlinear_i$$ is a fixed mathematical function chosen by the neural network designer (a human); it doesn't change once chosen. The coefficients $$a_i$$ and $$b_i$$, to the contrary, for every $$i$$, are learned using an optimization algorithm called gradient descent.\n\nThe gradient descent algorithm finds the best values for all $$a_i$$ and $$b_i$$ for all layers $$i$$ at once. What is considered \"best\" is also defined by the neural network designer by choosing the loss function.\n\nOther important details to know by a senior executive about neural networks are:\n\n• They require less feature engineering and can be successfully applied to raw data. Text can be \"fed\" to the neural network as a sequence of words; an image could be \"seen\" by the neural network as a matrix of pixels;\n• Neural networks learn (during training) and compute (on prediction time) multiple representations of the input before producing an output; each layer learns different levels of representation: from low-level (straight lines and curves) to high-level (ears, eyes, and noses).\n• Since they require less feature engineering, neural networks require significantly more data to train those different levels of input representation;\n• Since they require significantly more data to train, GPUs (Graphical Processing Units, graphical cards) are used to accelerate training. GPUs are used for training neural networks because their specialization is fast operations on matrices (need in 3D-graphics). Coincidentally, neural network training also requires making a lot of mathematical operations on matrices.\n• Having GPUs is not critical at the classification time; usual CPUs can be used;\n• Neural networks have different architectures depending on the tasks:\n• Convolutional Neural Networks are usually used to process images;\n• Recurrent Neural Networks are applied to sequences (of words, sound frequencies or stock market prices);\n• Feed Forward Neural Networks are used for classical classification and regression tasks.\n• Dropout is used as a regularization technique that prevents overfitting.\n\nThat's it. Now you know everything an executive should know about machine learning and the modern AI.\n\nRead our previous post \"An Introduction to Approximate String Matching\" or subscribe to our RSS feed.\n\nFound a mistyping or an inconsistency in the text? Let us know and we will improve it.\n\nLike it? Share it!"
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https://oipapio.com/keywords-sequences-and-series-1 | [
" oipapio\n\n### proof writing - If $x_k ≥ 0\\;\\forall \\in \\mathbb N$, and $y_k$ a bounded sequence, then the series $\\sum_{k=1}^\\infty x_ky_k$ converges\n\nHi I'm really struggling with this proof. For a start I'm struggling to believe it's true:For example, if we take $x_k = \\dfrac{1}{k^2}$ and $y_k = -k^3$ (which is bounded above by any positive number), then the series $x_ky_k$ does not converge? What am I doing wrong? I feel like I'm being insanely stupid.Thank you!...Read more\n\n### optimization - The longest sequence of numbers with a certain divisibility property\n\nEDIT- result. westzynthius(1931) showed that we can create a $p_x$ denizen longer than $p_x \\times log(log(log(p_x)))$... Meaning for large enough prime numbers, the maximum denizen is much larger than $2p_x, 3p_x$ etc.Definition - Denizen A sequence $\\lbrace a_k \\rbrace$ is a denizen if all of it's members are prime numbers, i.e $a_0, a_1, ... a_n \\in \\mathbb{P}$; and it satisfies the following condition; if \"$a_{x_1} =y_1$\", \"$a_{x_2} =y_2$\", \"$x_1 \\pm m_1y_1 \\neq x_2 \\pm m_2 y_2$ when $y_2<y_1$\" and \"$m_3$ isn't divisible by $y_1$...Read more\n\n### summation - Sum of a series close to geometric\n\nI'm stuck in finding the sum:\\begin{align}S = \\sum_{k=2}^n \\frac{1}{2^k-1}\\end{align}It seems to be close to geometric without being geometric. I wanted to try to transform it to a geometric series but I don't seem to come up with a valid transformation to do it. Anyone has an idea?...Read more\n\n### self learning - necessary and sufficient condition for convergence of power series\n\nShow that $\\sum_{n=1}^{\\infty} a^{\\ln~n}$ is convergent if and only if $0<a<\\frac{1}{e}$.Proof:$\\ln~n<n$ for all $n \\geq 1$.Hence the necessary condition for the series to be convergent will be $\\lim_{n \\rightarrow \\infty} a^{\\ln~n}=0$.And this will happen if $0<a<1$.Please give me some hint about how to proceed further....Read more\n\n### convergence - (Dis-)proving the series $\\sum\\limits_n\\left( 1+ \\frac{1}{n} \\right)^n$ converges\n\nI am trying to prove that the series:$$\\sum^\\infty_{n=1}\\left( 1+ \\frac{1}{n} \\right)^n$$converges.Now I know that$$\\lim_{n\\rightarrow\\infty} \\left( 1+ \\frac{1}{n} \\right)^n=e$$But how can I use that knowledge to prove the convergance ?Intuitively I would say that the series diverges since it doesn't approach zero but how can I formally prove this?...Read more\n\n### Limit of series with exponent\n\nI want to calculate the limit of the following series:$$\\sum^{\\infty}_{k=0} \\frac{(-3)^k +5}{4^k}$$My first step would be to split the term into these parts:$\\sum^{\\infty}_{k=0} \\frac{(-3)^k}{4^k}$ $\\sum^{\\infty}_{k=0} \\frac{5}{4^k}$If both of them have a limit I can just add them together, right ?I have looked through my notes on limits and convergence but I dont know how to get rid of the exponent so I can determine the limit.I have used various online calculators but I could not understand their result....Read more\n\n### summation - How to explain the formula for the sum of a geometric series without calculus?\n\nHow to explain to a middle-school student the notion of a geometric series without any calculus (i.e. limits)? For example I want to convince my student that$$1 + \\frac{1}{4} + \\frac{1}{4^2} + \\ldots + \\frac{1}{4^n} = \\frac{1 - (\\frac{1}{4})^{n+1} }{ 1 - \\frac{1}{4}}$$at $n \\to \\infty$ gives 4/3?...Read more\n\n### soft question - Interesting explicit convergent subsequence for not converging bounded sequence\n\nTo illustrate the (power of) Bolzano-Weierstrass theorem I am searching for an example of a bounded but not convergent sequence and an explicit convergent subsequence. I would like it to be non trivial in the sense that (cyclic) sequences like $(-1)^n$ or $1,2,3, 1,2,3, 1,2,3$ or $1,2,3,2,1,2,3,2,1$, where one can easily \"see\" the (constant) convergent subsequence don't count.I wanted to use $\\sin(n)$, but the construction of a convergent subsequence isn't very explicit....Read more\n\n### sequences and series - Which one is the correct notation?\n\nI know this notation is correct:$a_1,a_2, a_3,\\cdots,a_n=\\left\\{a_k\\right\\}_{k=1}^n$Now, we have a function $f(n)$.I want to write this sequence in correct notation:$\\left\\{ f(1),f(2),f(3),\\cdots ,f(n)\\right\\}$ I have two notations. Which one is correct? $\\bigcup_{k=1}^nf(k)=\\left\\{ f(1),f(2),f(3),\\cdots , f(n)\\right\\}$ $\\left\\{f(k)\\right\\}_{k=1}^n=\\left\\{ f(1),f(2),f(3),\\cdots, f(n)\\right\\}$Thank you....Read more\n\n### sequences and series - Are these statements always true?\n\nI haven't found an answer in my books. Although the question seems very simple, I want to ask.Are these statements always true? a) For any infinity non-negative integer sequence, if there is an exist $n-$th term closed form expression formula, for this sequence, we have always a recurrence formula. b) For any infinity non-negative integer sequence, if there is an exist recurrence formula,for this sequence, we have always $n-$th term closed form expression formula. c) For any infinity non-negative integer sequence, if there is not an exi...Read more\n\n### sequences and series - Can the entropy of a random variable with countably many outcomes be infinite?\n\nConsider a random variable $X$ taking values over $\\mathbb{N}$. Let $\\mathbb{P}(X = i) = p_i$ for $i \\in \\mathbb{N}$. The entropy of $X$ is defined by$$H(X) = \\sum_i -p_i \\log p_i.$$Is it possible for $H(X)$ to be infinite?...Read more\n\n### sequences and series - Extension of the Jacobi triple product identity\n\nThe Jacobi triple product and the mathematical identity of it is:$$\\prod\\limits_{n=1}^{ \\infty }(1-q^{2n})(1+zq^{2n-1})(1+z^{-1}q^{2n-1})=\\sum\\limits_{n = - \\infty }^ \\infty z^n q^{n^2}$$I would like to extend the idea for $\\sum\\limits_{n = - \\infty }^ \\infty z^n q^{n^2} h^{n^3}$My idea is below for extension:Let's assume we define $G(z,q,h)$ as$$G(z,q,h)\\prod\\limits_{n=1}^{ \\infty }(1+zq^{2n-1}h^{3n^2-3n+1})(1+z^{-1}q^{2n-1}h^{-3n^2+3n-1})=\\sum\\limits_{n = - \\infty }^ \\infty z^n q^{n^2} h^{n^3}$$$z=ZQ^{2}h^{3}$$q=Qh^{3}$$$G(ZQ^{2}h^{3},Qh...Read more\n\n### sequences and series - Angle between two given vector is small. Can we permute coordinates of them such that new vectors be orthogonal?\n\nLet $x\\in\\mathbb{R}^n$ and $y\\in\\mathbb{R}^n$ be two unit vectors such that $\\sum_{i}{x_i}=\\sum_{i}{y_i}=0$ $$x_{1}y_{1}+x_{2}y_{2}+\\cdots +x_{n}y_{n} \\gt 1-\\frac{1}{n} .$$ Can we prove that$$x_{1}y_{\\sigma(1)}+x_{2}y_{\\sigma(2)}+\\cdots +x_{n}y_{\\sigma(n)} \\neq 0$$ for all permutations $\\sigma$?...Read more\n\n### Adding the harmonic sequence and a permutation of it\n\nLet $\\pi:\\mathbb{N}\\to\\mathbb{N}$ be a bijection. Then does there exist another bijection $\\nu:\\mathbb{N}\\to\\mathbb{N}$ and a constant $C$ such that $$\\frac{1}{n} + \\frac{1}{\\pi(n)} \\leq \\frac{C}{\\nu(n)}$$ for all $n$? If so, can the constant be chosen independent of $\\pi$?While the harmonic sequence $(\\frac{1}{n})_{n\\in\\mathbb{N}}$ is what comes up in my application, I imagine that a good answer will be able to make a much more general statement about a suitable class of sequences. But I'd be perfectly happy with an answer to the question ab...Read more\n\n### sequences and series - recursive to standard convertion\n\nI have been trying to find an equation for a sequence, and got interested on how to convert any recursive sequence ex: $F_n=F_{n-1}+F_{n-2},\\space F_0=1,\\space F_1=1$ into a standard equationex: $F_n=\\frac 1{\\sqrt 5}(\\frac {1+\\sqrt 5}2)^{n+1}-\\frac 1{\\sqrt 5}(\\frac {1-\\sqrt 5}2)^{n+1}$ I decided to search around but only got beginners algebra stuff, in fact the only helpful thing I found was a video on how to do it with the Fibonacci sequence which doesn't help me with the equation I have. if anyone could give me a link to something helpful, th...Read more"
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https://numbermatics.com/n/605006742/ | [
"605006742\n\n605,006,742 is an even composite number composed of five prime numbers multiplied together.\n\nWhat does the number 605006742 look like?\n\nThis visualization shows the relationship between its 5 prime factors (large circles) and 32 divisors.\n\n605006742 is an even composite number. It is composed of five distinct prime numbers multiplied together. It has a total of thirty-two divisors.\n\nPrime factorization of 605006742:\n\n2 × 3 × 41 × 307 × 8011\n\nSee below for interesting mathematical facts about the number 605006742 from the Numbermatics database.\n\nNames of 605006742\n\n• Cardinal: 605006742 can be written as Six hundred five million, six thousand, seven hundred forty-two.\n\nScientific notation\n\n• Scientific notation: 6.05006742 × 108\n\nFactors of 605006742\n\n• Number of distinct prime factors ω(n): 5\n• Total number of prime factors Ω(n): 5\n• Sum of prime factors: 8364\n\nDivisors of 605006742\n\n• Number of divisors d(n): 32\n• Complete list of divisors:\n• Sum of all divisors σ(n): 1243718784\n• Sum of proper divisors (its aliquot sum) s(n): 638712042\n• 605006742 is an abundant number, because the sum of its proper divisors (638712042) is greater than itself. Its abundance is 33705300\n\nBases of 605006742\n\n• Binary: 1001000000111110101011100101102\n• Base-36: A07E6U\n\nSquares and roots of 605006742\n\n• 605006742 squared (6050067422) is 366033157865454564\n• 605006742 cubed (6050067423) is 221452528304150340114670488\n• The square root of 605006742 is 24596.8848027549\n• The cube root of 605006742 is 845.7721974939\n\nScales and comparisons\n\nHow big is 605006742?\n• 605,006,742 seconds is equal to 19 years, 12 weeks, 2 days, 9 hours, 25 minutes, 42 seconds.\n• To count from 1 to 605,006,742 would take you about twenty-eight years!\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 605006742 cubic inches would be around 70.5 feet tall.\n\nRecreational maths with 605006742\n\n• 605006742 backwards is 247600506\n• The number of decimal digits it has is: 9\n• The sum of 605006742's digits is 30\n• More coming soon!\n\nMLA style:\n\"Number 605006742 - Facts about the integer\". Numbermatics.com. 2022. Web. 22 January 2022.\n\nAPA style:\nNumbermatics. (2022). Number 605006742 - Facts about the integer. Retrieved 22 January 2022, from https://numbermatics.com/n/605006742/\n\nChicago style:\nNumbermatics. 2022. \"Number 605006742 - Facts about the integer\". https://numbermatics.com/n/605006742/\n\nThe information we have on file for 605006742 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!\n\nKeywords: Divisors of 605006742, math, Factors of 605006742, curriculum, school, college, exams, university, Prime factorization of 605006742, STEM, science, technology, engineering, physics, economics, calculator, six hundred five million, six thousand, seven hundred forty-two.\n\nOh no. Javascript is switched off in your browser.\nSome bits of this website may not work unless you switch it on."
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http://orientdb.com/docs/3.0.x/sql/Sequences-and-auto-increment.html | [
"# Sequences and auto increment\n\nStarting from v2.2, OrientDB supports sequences like most of RDBMS. What's a sequence? It's a structure that manage counters. Sequences are mostly used when you need a number that always increments. Sequence types can be:\n\n• ORDERED: each call to `.next()` will result in a new value.\n• CACHED: the sequence will cache N items on each node, thus improving the performance if many `.next()` calls are required. However, this may create holes.\n\nTo manipulate sequences you can use the Java API or SQL commands.\n\n## Create a sequence\n\n``````CREATE SEQUENCE idseq\nINSERT INTO account SET id = sequence('idseq').next()\n``````\n\n### Using a sequence from SQL\n\nYou can use a sequence from SQL with the following syntax:\n\n``````sequence('<sequence>').<method>\n``````\n\nWhere:\n\n• `method` can be:\n• `next()` retrieves the next value\n• `current()` gets the current value\n• `reset()` resets the sequence value to it's initial value\n\nExample\n\n``````INSERT INTO Account SET id = sequence('mysequence').next()\n``````\n\n## Alter a sequence\n\n``````ALTER SEQUENCE idseq START 1000\n``````\n\n## Drop a sequence\n\n``````DROP SEQUENCE idseq\n``````\n\n# OrientDB before v2.2\n\nOrientDB before v2.2 doesn't support sequences (autoincrement), so you can manage your own counter in this way (example using SQL):\n\n``````CREATE CLASS counter\nINSERT INTO counter SET name='mycounter', value=0\n``````\n\nAnd then every time you need a new number you can do:\n\n``````UPDATE counter INCREMENT value = 1 WHERE name = 'mycounter'\n``````\n\nThis works in a SQL batch in this way:\n\n``````BEGIN\nlet \\$counter = UPDATE counter INCREMENT value = 1 return after \\$current WHERE name = 'mycounter'\nINSERT INTO items SET id = \\$counter.value, qty = 10, price = 1000\nCOMMIT\n``````"
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https://journals.ametsoc.org/view/journals/phoc/32/3/1520-0485_2002_032_0870_ct_2.0.co_2.xml | [
"• Andersen, O. B., , P. L. Woodworth, , and R. A. Flather, 1995: Intercomparison of recent ocean tide models. J. Geophys. Res., 100 , 2526125282.\n\n• Export Citation\n• Aref, H., 1984: Stirring by chaotic advection. J. Fluid Mech., 143 , 121.\n\n• Beigie, D., , A. Leonard, , and S. Wiggins, 1991: Chaotic transport in the homoclinic and heteroclinic tangle regions of quasiperiodically forced two-dimensional systems. Nonlinearity, 4 , 775819.\n\n• Export Citation\n• Cartwright, D. E., 1977: Oceanic tides. Rep. Prog. Phys., 40 , 665.\n\n• Davies, A. M., 1986: A three-dimensional model of the northwest European continental shelf, with application to the M4 tide. J. Phys. Oceanogr., 16 , 797813.\n\n• Export Citation\n• Defant, A., 1961: Physical Oceanography. Vol. II,. Pergamon, 598 pp.\n\n• Frison, T. W., , H. D. I. Abarbanel, , M. D. Earle, , J. R. Schultz, , and W. Scherer, 1999: Chaos and predictability in ocean water levels. J. Geophys. Res., 104 , 79357951.\n\n• Export Citation\n• Garrett, C., 1975: Tides in gulfs. Deep-Sea Res., 22 , 2335.\n\n• Godin, G., 1991: The analysis of tides and currents. Tidal Hydrodynamics, B. B. Parker, Ed., Wiley and Sons, 675–709.\n\n• Golmen, L. G., , J. Molvaer, , and J. Magnusson, 1994: Sea level oscillations with super-tidal frequencies in a coastal embayment of western Norway. Cont. Shelf Res., 14 , 14391454.\n\n• Export Citation\n• Green, T., 1992: Liquid oscillations in a basin with varying surface area. Phys. Fluids A, 4 , 630632.\n\n• Guckenheimer, J., , and P. Holmes, 1983: Nonlinear Oscillations, Dynamical Systems, and Bifurcations of Vector Fields. Springer-Verlag, 459 pp.\n\n• Export Citation\n• Gutiérrez, A., , F. Mosetti, , and N. Purga, 1981: On the indetermination of the tidal harmonic constants. Nuovo Cimento, 4 , 563575.\n\n• Honda, K., , T. Terada, , Y. Yoshida, , and D. Isitani, 1908: Secondary undulations of oceanic tides. J. Coll. Sc. Imp. Univ. Tokyo, 24 , 1113. and 95 plates.\n\n• Export Citation\n• Kjerfve, B., , and B. A. Knoppers, 1991: Tidal choking in a coastal lagoon. Tidal Hydrodynamics, B. B. Parker, Ed., Wiley and Sons, 169–182.\n\n• Export Citation\n• Krümmel, O., 1911: Handbuch der Ozeanographie. Verlag von J. Engelhorns, 766 pp.\n\n• LeBlond, P. H., 1991: Tides and their interactions with other oceanographic phenomena in shallow water. Tidal Hydrodynamics, B. B. Parker, Ed., Wiley and Sons, 675–709.\n\n• Export Citation\n• Lynch, D. R., , and F. E. Werner, 1991: Three-dimensional velocities from a finite-element model of English Channel/Southern Bight tides. Tidal Hydrodynamics, B. B. Parker, Ed., Wiley and Sons, 183–200.\n\n• Export Citation\n• Maas, L. R. M., 1997: On the nonlinear Helmholtz response of almost-enclosed tidal basins with a sloping bottom. J. Fluid Mech., 349 , 361380.\n\n• Export Citation\n• Maas, L. R. M., . 1998: On an oscillator equation for tides in almost enclosed basins of non-uniform depth. Physics of Estuaries and Coastal Seas, J. Dronkers, and M. Scheffers, Eds., A. A. Balkema, 127–132.\n\n• Export Citation\n• Munk, W., , and C. Wunsch, 1998: Abyssal recipes II: Energetics of tidal and wind mixing. Deep-Sea Res. I, 45 , 19772010.\n\n• Nakano, M., 1932: Preliminary note on the accumulation and dissipation of energy of the secondary oscillations in a bay. Proc. Phys.-Math. Soc. Japan, Ser. 3,, 14 , 4456.\n\n• Export Citation\n• Nayfeh, A. H., , and D. T. Mook, 1979: Nonlinear Oscillations. Wiley-Interscience, 704 pp.\n\n• Platzman, G., 1971: Ocean tides and related waves. Mathematical Problems in the Geophysical Sciences. Vol. 1, Geophysical Fluid Dynamics, W. H. Reid, Ed., American Mathematical Society, 239–291.\n\n• Export Citation\n• Prudnikov, A. P., , Yu A. Brychkov, , and O. I. Marichev, 1986: Integrals and Series. Vol. I, Gordon and Breach, 798 pp.\n\n• Pugh, D. T., 1987: Tides, Surges and Mean Sea-Level. John Wiley and Sons, 472 pp.\n\n• Ruelle, D., 1980: Strange attractors. Math. Intelligencer, 2 , 126137. [Reprinted in Cvitanović, P., 1984: Universality in Chaos, Adam-Hilger, 37–48.].\n\n• Export Citation\n• Schwiderski, E. W., 1980: On charting global ocean tides. Rev. Geophys. Space Phys., 18 , 243268.\n\n• Seim, H. E., , and J. E. Sneed, 1988: Enhancement of semidiurnal tidal currents in the tidal inlets to Mississippi Sound. Hydrodynamics and Sediment Dynamics of Tidal Inlets, D. G. Aubrey and L. Weishar, Eds., Lecture Notes on Coastal and Estuarine Studies, Vol. 29, Springer-Verlag, 157–168.\n\n• Export Citation\n• Smith, N. P., 1994: Water, salt and heat balance of coastal lagoons. Coastal Lagoon Processes, B. Kjerfve, Ed., Elsevier Oceanographic Series, Vol. 60, 69–101.\n\n• Export Citation\n• van Ette, A. C. M., , and H. J. Schoemaker, 1966: Harmonic analysis of tides. Essential features and disturbing influences. Hydrographic Letter, Vol. 1, Netherlands' Hydrographer Special Publication 2, 1–35.\n\n• Export Citation\n• Vittori, G., 1992: On the chaotic structure of tide elevation in the Lagoon of Venice. Proc. 23d Int. Conf. on Coastal Engineering, Venice, Italy, American Society of Civil Engineers, 1826–1839.\n\n• Export Citation\n• Wiggins, S., 1990: Introduction to Applied Nonlinear Dynamical Systems and Chaos. Springer-Verlag, 672 pp.\n\n• Yagasaki, K., 1990: Second-order averaging and chaos in quasiperiodically forced weakly nonlinear oscillators. Physica D, 44 , 445458.\n\n• Export Citation\n• Yagasaki, K., . 1993: Chaotic motions near homoclinic manifolds and resonant tori in quasiperiodic perturbations of planar Hamiltonian systems. Physica D, 69 , 232269.\n\n• Export Citation\n• Zimmerman, J. T. F., 1992: On the Lorentz linearization of a nonlinearly damped tidal Helmholtz oscillator. Proc. K. Ned. Akad. Wet., 95 , 127145.\n\n• Export Citation\n•",
null,
"Observed sea level in Moldefjord and currents through connecting strait; from Golmen et al. (1994)\n•",
null,
"Sketch showing a side view of two basins that have a channel, connecting it to the sea, at their right-hand sides (Maas 1998). It demonstrates that the elevation change brought about by two equal, consecutively entering packages of water (see arrows) is the same when the basin has a cross-sectional area that is independent of height above the bottom (left), while it is different when this cross-sectional area increases with height (right). In the latter case, the response depends on the preexisting water level and is thus nonlinearly related to the incoming flux of water\n•",
null,
"(a) Example of time (t) evolution of volume (x) and current (y) over two tidal periods T, together with three projections, of the unforced (ζe = 0), damped system (5). (b) Projection of helix in (a) on the x–y plane (solid curve), and stroboscopic (tidal) sampling thereof (dots), representing intersections in (a) with horizontal planes at t/T = 0, 1, 2, · · · , 6\n•",
null,
"Stroboscopic plot at t/T = 0, 1, 2, · · · , 0, where T = 2π/ω, and forcing ζe = −f cos(ωt), ω = 1.01, and forcing amplitude f and damping parameter C given by (a) f = 0, C = 0.01, (b) f = 0.001, C = 0, (c) f = 0.001, C = 0.01, and (d) f = 0.001, C = 0.001\n•",
null,
"Amplitude response R against detuned frequency σ. Near resonancy, 1 < σ < 5, it has a bent “resonance horn,” indicating multiple equilibria and an increased response. Here forcing F = 3 and frictional parameter C = 0.4. Near σ ≈ 1.5 (vertical line) three equilibria, A, B, and C, are obtained. Two of these are stable and correspond to a choked (C) and amplified (A) regime\n•",
null,
"(a) Poincaré plot of numerical integration of the inviscid (c = 0) evolution equation (5) for forcing frequency ω = 1.01, close to resonance. Forcing f = −10−3, θ = 0, and initial conditions are y = 0, x ∈ {−0.3, −0.25, −0.2, 0.05, 0.2, 0.3, 0.35, 0.4}, as well as y = 0.01, x = 0.3185, which is close to the saddle point. (b) Trajectories of modulation equations (11) and (12) for the corresponding parameters and initial conditions, with ε = (ω − 1)1/2 = 0.1 (and, hence, σ = 1), after a rescaling x = εX, and employing a rescaled forcing F = −1, which is ε3 times the original f in (a). The two figures are comparable, taking into account the scale factor ε = 0.1\n•",
null,
"(a) The quartic P4(S), Eq. (21), as a function of S = [(X2 + Y2)/12 − 1], for F = −1/2 and three “energy” levels K = 0.2, 0.2832, · · · , 0.4. The central K value corresponds to the presence of a double zero at S1, which, in the X–Y phase-plane is related to the homoclinic orbits (separatrices) γin,out emanating from the saddle at (X1, 0). (b) X–Y phase plane of modulation equations (11)–(12), for F = −1/2 and energy levels as in (a). The three zeros in (a), Sk for k ∈ {1, 2, 3}, in this plane correspond to circles that intersect the Y = 0 axis at Xk, and are related by Sk = X2k/12 − 1. These circles correspond to the maximum and minimum distances that points on the two homoclinic orbits, γin,out, have with respect to the “special” radius Rs = 23\n•",
null,
"Surfaces of critical damping as a function of frequency difference ΔΩ and forcing amplitude F ∈ (−8/3, 0) along (a) the outer and (b) the inner homoclinic orbit. For damping values below these surfaces, chaotic states exist in the modulation equations\n•",
null,
"Numerical integration of modulation equations (11)–(12) when δC = 0.1 gives an indication of chaos as observed here in current velocity Y vs (a) slow time T and (b) vs excess volume X. The integration is started from the origin (X(0) = Y(0) = 0) for forcing F = 2 cosΩ′t, and frequency Ω′ = ΔΩ/2 = 0.5\n•",
null,
"Stroboscopic (Poincaré) plot of numerically integrated orbit of modulation equations, calculated as in Fig. 9, here extended to an integration over 3000 dimensionless units in time. Sampling is at the long period rate 2π/Ω′, where Ω′ = ΔΩ/2\n•",
null,
"Poincaré plot of numerically integrated exact equations (5) in case c = 0.001 gives an indication of chaos as observed here in current velocity y vs time t (a) and versus excess volume x (b). Integration is started from the origin (x(0) = y(0) = 0) for forcing values f1 = f2 = 0.001, and frequencies ω1 = 1 and ω2 = 1.01. Sampling is at the average forcing period 2π/ω\n•",
null,
"Stroboscopic (Poincaré) plot of numerically integrated exact orbit calculated as in Fig. 11, here extended to an integration over 3000 dimensionless units. Sampling is at the (long) modulation period 2π/Δω\n•",
null,
"Numerical integration of (2) for a forcing (35) that consists of a low-frequency tide and a resonant perturbation. (a) A trajectory in x–y phase space and (b) the corresponding Poincaré plot, subsampled on the tidal period. Parameters read ε = 0.1, a = 3ε2, b = 3ε3, c = 0.4ε2, and ω = 1 + 1.5ε2 (and θ = 0)\nAll Time Past Year Past 30 Days\nAbstract Views 0 0 0\nFull Text Views 47 47 22\n\n# Chaotic Tides\n\nView More View Less\n• 1 Netherlands Institute for Sea Research, Texel, Netherlands\n• 2 Korteweg-de Vries Instituut, Universiteit van Amsterdam, Amsterdam, Netherlands\nFull access\n\n## Abstract\n\nPersistent reports exist that the tides in coastal basins are often accompanied by regular or irregular oscillations of periods ranging from minutes up to several hours. A conceptual model relating the two is proposed here. It employs an almost-enclosed basin, connected to an open (tidal) sea by a narrow strait. Such a basin is a Helmholtz resonator, which is dominated by the “pumping” mode. Its response is governed by an ordinary differential equation that is forced by the tide, damped by friction and wave radiation, and whose restoring term is nonlinear due to the sloping bottom. When forced resonantly by a single frequency tide, due to this nonlinearity, the basin may exhibit multiple equilibria. Its response can either be amplified or choked, depending on the precise initial conditions. The presence of a second forcing term may, on a slow timescale, kick the system irregularly from the amplified into the choked regime, yielding a chaotic response. This may happen when either two nearby frequencies, for example, a combination of semidiurnal lunar and solar tides, are near resonance (and the frequency difference provides the beat), or when a small-amplitude, resonant perturbation is modulated by a large-amplitude, low-frequency tide. The aforementioned observations of irregular tides are discussed in the light of analytical and numerical results obtained with this model for these two regimes.\n\nCorresponding author address: Dr. Leo Maas, Netherlands Institute for Sea Research, P.O. Box 59, 1790 AB Texel, Netherlands. Email: [email protected]\n\n## Abstract\n\nPersistent reports exist that the tides in coastal basins are often accompanied by regular or irregular oscillations of periods ranging from minutes up to several hours. A conceptual model relating the two is proposed here. It employs an almost-enclosed basin, connected to an open (tidal) sea by a narrow strait. Such a basin is a Helmholtz resonator, which is dominated by the “pumping” mode. Its response is governed by an ordinary differential equation that is forced by the tide, damped by friction and wave radiation, and whose restoring term is nonlinear due to the sloping bottom. When forced resonantly by a single frequency tide, due to this nonlinearity, the basin may exhibit multiple equilibria. Its response can either be amplified or choked, depending on the precise initial conditions. The presence of a second forcing term may, on a slow timescale, kick the system irregularly from the amplified into the choked regime, yielding a chaotic response. This may happen when either two nearby frequencies, for example, a combination of semidiurnal lunar and solar tides, are near resonance (and the frequency difference provides the beat), or when a small-amplitude, resonant perturbation is modulated by a large-amplitude, low-frequency tide. The aforementioned observations of irregular tides are discussed in the light of analytical and numerical results obtained with this model for these two regimes.\n\nCorresponding author address: Dr. Leo Maas, Netherlands Institute for Sea Research, P.O. Box 59, 1790 AB Texel, Netherlands. Email: [email protected]\n\n## 1. Introduction\n\n“Chaotic tides” sounds like an oxymoron. As the etymology of the word “tide,” being time, season, or period according to Webster's (1913) Revised Unabridged Dictionary implies, it has, from ancient times on (Pugh 1987), often been held to be synonymous with periodicity. Indeed, this regularity made tidal heights one of the earliest known and best predictable phenomena in physical oceanography, with relative accuracies of prediction often exceeding 90%. The predictive capacities of the corresponding tidal currents is, admittedly, less impressive, but is still often in excess of 50%. Less known, however, is the fact that spurious, but persistent, reports on the irregularity of the tides exist for over a century (Honda et al. 1908; Krümmel 1911). It is this irregularity in the tide that we here aim to identify as being possibly due to the chaotic response of certain coastal basins to the tide at the open sea.\n\nTidal predictions have traditionally been made for individual ports on the basis of previously observed tidal elevation records at those locations. Amplitudes and phases of a number of precisely known “tidal frequencies,” stemming from the gravitational potential determined by celestial mechanics, are estimated, which can then be used to sum the corresponding harmonic series and thus to predict the future occurrence of tides at that location; see, for example, the review by Godin (1991).\n\nAlthough the gravitational (or tidal) potential carries only about 5 relevant, independent frequency components, which, through an expansion of the tidal potential in harmonic terms leads to perhaps 11 principal tidal components (Platzman 1971), in the shallow, coastal areas, local nonlinear effects lead to leakage of energy to higher harmonics and sum and difference (combination) frequencies. In this way tidal energy fills in entire spectral bands surrounding the principal components (Pugh 1987, p. 188). As there are, in principle, no difficulties in also resolving the amplitudes and phases of these combination frequencies, running harmonic analysis with 150 frequency components or more has become routine procedure. This empirical local analysis, however, does not automatically guarantee that the amplitudes and phases of these independently determined frequencies also show a spatially coherent and stationary picture. Quite the contrary.\n\nA spatially coherent description of the tides can be obtained by solving the governing dynamical equations, the Laplace tidal equations (see, e.g., Cartwright 1977). In a realistic ocean domain this is achieved by numerical methods (Schwiderski 1980). Outside continental margins, tidal predictions for these principal components, particularly when constrained by observations from deep-sea tidal stations, are in fair agreement with observations obtained at different seaports and deep sea tidal stations (global root-mean-square difference between modeled and observed elevations of about 10 cm). With the advent of satellite altimetry providing tidal observations, this agreement has in recent years actually been dramatically improved (global root-mean-square difference of about 3 cm); see Andersen et al. (1995).\n\nExcept for these directly driven, principal components, together perhaps with their first few harmonics (Davies 1986; Lynch and Werner 1991), spatial coherency is cumbersome and agreement with locally observed tidal components poor, particularly in the shallow, coastal areas. Waves at combination frequencies are generated locally, get amplified, or damped, to a degree depending on the particular resonance properties of that locality, and do not persist as free waves so that they decorrelate quickly. Treating the nonlinearly generated compound waves locally as linearly independent (as harmonic, or Fourier analysis wants it) is therefore somewhat contorted but still defendable when the local nonlinear transfer of energy is frozen so that harmonics appear as the result of a stationary process. However, when analyzing different, independent datasets from the same location, variations in tidal amplitudes and phases often still occur, to the extent that some of the minor components appear, in fact, unresolvable (Gutiérrez et al. 1981; Godin 1991). Although such variations in locally estimated tidal amplitudes and phases are usually attributed to nonstationary effects due to wind (van Ette and Schoemaker 1966; Gutiérrez et al. 1981), this need not necessarily be its only source. Nonstationarity may also be due to nonlinearity of the hydrodynamic system itself (Pugh 1987), which may not only change the tidal elevation profile, by giving rise to superharmonics (overtides, which stay fixed in time), but may also modulate its amplitude, by giving rise to subharmonics. Finally, strong nonlinearity may provoke a cascade of such subharmonic bifurcations, giving rise to chaotic behavior.\n\nIs there any observational basis for chaotic behavior of the tides, other than the aforementioned unresolvability, or noisiness, of tidal “constants”? In an analysis of the tides in Venice Lagoon, at the head of the Adriatic Sea, where the tides seem to pick up because of near-resonancy of the basin, Vittori (1992) observed that consecutive tidal maxima are highly irregular. She argued this to be indicative of low-dimensional chaos. Whether the low-order dynamics to which this is due is either inherited from the dynamics of the local wind fields or of a genuinely oceanographic nature is not clear. Similar changes in consecutive maxima also appear in long-term, tidal elevation records in the Wadden Sea basins (J. T. F. Zimmerman 2000, personal communication), which are again close to resonance (see Maas 1997, hereafter referred to as M). Frison et al. (1999) find evidence of nonlinear tides in U.S. coastal estuaries from the nontrivial shape of the attractor obtained in a diagram where observed elevation at some moment is plotted against that observed at some previous instant.\n\nSurprisingly, direct observations of irregular oscillations, accompanying the tides, have been reported for over a century (see Krümmel 1911, 157–185; Defant 1961, p. 187). The suggestion that these “secondary oscillations” were actually related to the “primary,” tidal oscillation was firmly put forward in an impressive study by Honda et al. (1908), who had intensively studied the seas around Japan. This study eventually led to a complete classification of bays in terms of “periodic,” “quasiperiodic,” etc. (Honda et al. 1908; Nakano, 1932). In the absence of a dynamical framework, such as recently offered by the field of nonlinear dynamics, this classification was, however, not further interpreted, and these observations seem, if not completely forgotten, neglected in contemporary tidal literature. One reason for this might be that these secondary oscillations generally seem inconspicuous, as they consist of high-frequency, low-amplitude waves superimposed on a low-frequency, high-amplitude (primary) tide. Typical periods range from some minutes to several hours, and amplitudes do not exceed a few percent of the tidal range. However, an interesting twist is given to this interpretation by recent observations of such irregular tidal elevations in a Norwegian fjord (Golmen et al. 1994). By making simultaneous observations of the currents in the strait connecting the fjord to the sea they found that these irregular small-amplitude elevations were accompanied by irregular O(1) variations of the tidal current (see Fig. 1). As velocities are determined by elevation differences over the strait, by inference the velocity field can be amplified when, at any moment, the difference in tidal elevation also amounts to just a few percent of the elevation itself, thereby becoming of comparable magnitude to the difference between the (small amplitude) high-frequency elevation that is resonantly excited within the basin but is practically absent at sea. Such O(1) velocity variations are not only important for nautical reasons, but are clearly equally relevant for the flushing of the fjord: the exchange of water, sediments, and dissolved gases or nutrients with the connected sea. It is believed (LeBlond 1991) that this property will transcend the global relevance that both dissipation as well as resonance of tides in the coastal environment may have in setting the boundary conditions for the global tide (Garrett 1975), although Munk and Wunsch (1998) recently made the interesting suggestion that tides might be playing a key role in ocean circulation.\n\nIn order to illustrate these issues we will consider the resonant response of a short, deep basin having a sloping bottom, to the tide in the adjacent open sea to which it is connected by a narrow strait—a nonlinear Helmholtz resonator (see M). This choice of geometry differs from the mostly wide bays with their quarter-wave resonancies, encountered by Honda et al. (1908), but it ensures that the tide within the basin takes the simplest form possible, as it is governed by the pumping, or Helmholtz mode. This mode is the lowest and generally also the most important frequency in the basin's spectrum, and is characterized by a spatially uniform response. It can thus be represented by a single state variable (the excess volume of water, related to the free surface elevation), whose evolution is therefore described simply by an ordinary differential equation. Although nonlinearity can also be present in the frictional damping term (Zimmerman 1992), it is the nonlinearity in the restoring term that gives rise to multiple equilibria and chaos here. The latter nonlinearity is due to the sloping bottom (Green 1992) and can be understood as follows. The current through the strait is driven simply by the elevation difference over the strait. However, the elevation change within the basin, affected by the transport through the strait, clearly depends on the surface area that the incoming water has to cover, which is nonuniform with depth when the walls of the basin are not vertical. The time needed to produce a particular elevation change therefore depends on the preexisting sea level; see Fig. 2. In M its free, forced, and forced-and-damped response was discussed in the case where the tidal forcing is “pure,” that is, contains one frequency component only. In inviscid circumstances, a single frequency forcing can provoke a chaotic response in a small parameter range. However, when damping is added, the chaos fails to persist. Since the inclusion of damping terms is mandatory in any realistic physical context, this “Hamiltonian chaos” appears to have no physical relevance. It is one of the goals of the present paper to determine under what conditions tides, within basins of the kind considered here, can become chaotic, even in the presence of damping. Multiple frequency forcing seems to be a prerequisite for this. Therefore, in light of the observations discussed above, we will consider forcing with either two nearby frequencies, as is the case for a combination of lunar and solar tides, a “mixed” type of tide (see Defant 1961), or with a single low-frequency tide, accompanied by a small-amplitude, high-frequency, resonant perturbation, as is the case in a fjord.\n\nIn section 2 the nonlinear tidal Helmholtz resonator will be introduced. It represents the response of a bay to tidal variations at the open sea, to which it is connected by a narrow channel. Its governing ordinary differential equation will be derived. This section also introduces the Poincaré phase plane, which gives a more comprehensive way to follow its evolution than that obtained by direct numerical integration. The tidal response of the bay typically shows a gradual modulation of amplitude and phase, which can be captured by averaging the original governing equation (section 3). In section 4 this modulation equation is applied to the case that forcing is at two nearby frequencies. When these are close to resonance, the response of the basin may, under certain conditions, be chaotic. The case that a single, low-frequency tide is accompanied by a small-amplitude, high-frequency resonant perturbation is very similar to this, and is discussed in section 5. In section 6 the results will be summarized.\n\n## 2. Nonlinear Helmholtz resonator\n\nAn almost-enclosed, short basin, connected to a tidal sea by a narrow strait (modeled here as a pipe), is generally governed by the pumping (Helmholtz) mode, in which the water level within the basin rises and sinks in unison. Elevation is scaled with (maximum) depth H, volume with A0H (where A0 is the surface area of the basin at rest), while time t is scaled with the Helmholtz frequency of the basin σH,\nσHgHBA0L1/2\nwith B and L denoting the strait's width and length, respectively, and g the acceleration of gravity. In M it is shown that the evolution of the (nondimensional) amount of water in excess of that present in the basin in the absence of tides—the excess volume, υ—is (neglecting quadratic damping terms) governed by",
null,
"Here ζ and ζe denote the surface elevation within the basin and in the exterior sea, respectively. The latter is a prescribed function of time. At any time their difference provides the pressure difference that drives the flow u along the connecting strait: du/dt = ζeζ. This flow, multiplied by the (unit) cross-sectional area of the strait, equals the rate at which the volume within the basin changes, u = /dt. From these two equations the inviscid form (c = 0) of (2) follows. The basin hypsometry (horizontal area A as a function of vertical coordinate z) determines the excess volume within the basin. With respect to mean sea level z = 0, this is defined as",
null,
"The inverse of this relation, needed in (2), yields the nonlinear restoring term ζ(υ), which can be made explicit for some simple basin shapes. For a basin with vertical sidewalls, A = 1, for instance, it follows from (3) that the restoring term is linear, ζ = υ. However, when the basin area increases linearly with depth, A = 1 + z (the deepest point of the basin being at z = −1), this restoring term is given by\nζυυ1/2\nThis shape of the basin is the only one addressed here and its hypsometric behavior is thought to be characteristic of a typical basin with shoaling sides (see M). Other basin shapes and the effect of this shape on the dynamics of the basin are addressed elsewhere (Doelman et al. 2001, submitted to Phys. D, hereafter DKM). A nonempty basin (and analyticity of the solution) requires υ > −1/2. In the right hand of (2) we find forcing by the prescribed external tide, ζe(t), and damping, respectively. Although this linear damping formally stands for radiative damping only, we may also consider it to incorporate a linearized version of bottom friction and form drag (see M). For further analysis we treat (2) as a dynamical system by writing for the basin's excess volume x = υ, and for the strait's flow rate y = /dt, and obtain, with (4),",
null,
"As usual, a dot indicates a time derivative. Although the emphasis will here be on the response to quasiperiodic (two frequency) forcing, some results obtained without forcing and with “single frequency” forcing, pertinent to the discussion to follow, will be recapitulated first.\n\n### a. Poincaré plane, free response, and periodic forcing\n\nA numerical example of an integration of (5) is given in Fig. 3a for the case without forcing. It presents the evolution over two tidal periods of volume and current, (x(t), y(t)), as a function of time, increasing upward. Projections of this helical curve at the back (xt) and side (yt) planes give the time evolution of volume and current proper, such as they are commonly presented (except that time now points upward). The third projection, on the bottom xy plane, suppresses time, which now acts as a parameter along such orbits, and is called the phase plane. In this strongly viscous example time can still be discerned in the gradual decay of the oscillations. However, when damping is also absent, solution curves will “circle” around a common center point, which is the only fixed point present in this case. Each solution curve is egg shaped and is described by elliptic functions in terms of which the inviscid, unforced version of (5) can be solved (M). Different curves represent different initial conditions (volume and current). The phase plane is limited by an outer boundary (the limiting orbit) because the argument of the square root in the restoring term of (5) needs to be nonnegative. It can be found in parametric form as\nxtt2t2yttt2\nfor t ∈ (−3, 3), which is periodic with period 6. Owing to the nonlinearity of the oscillator, the period of such a free, finite-amplitude oscillation is a function of its amplitude. For the present geometry this period spans just a narrow range, decreasing from 2π, at the center, down to 6 at the outer boundary. The frequency range of other geometries is considered in DKM.\nA comprehensive way of monitoring gradual changes of the tidal response due to forcing, damping, or nonlinearity is obtained by determining subsequent intersections of the helical curve with a sequence of periodically placed, horizontal planes at t/T = 0, 1, 2, … , with T indicating the forcing period (so-called Poincaré planes of section). These subsequent intersections define a map, called the Poincaré map, P",
null,
". A strictly periodic signal (of period T) thus yields the same intersection. Compiling each of these subsequent intersections on one and the same horizontal (phase) plane, a stroboscopic impression of the slow evolution of the nearly periodic behavior is obtained. An example of such a Poincaré plane is presented by the dots in Fig. 3b. Here, the spiral is the projection of the continuous helical curve of Fig. 3a on the bottom plane. The phase at which the forcing is sampled can still be varied but is in this study taken to correspond to the moment that the external tide is at low water. From here on we will mostly use this phase-plane representation of the evolution, accepting the suppression of time information. This is illustrated in Fig. 4 with some stroboscopically sampled solutions of (5). When forcing is absent, damping brings any initial motion to rest, in this Poincaré plane represented by a sequence of points approaching the origin (Fig. 4a). When damping is absent and a single forcing component,\nζefωtθ\nis near resonance (ω ≈ 1), the amplitude of the basin tide will grow because of the nonlinearity of the restoring term. However, doing so, it changes its period so that the resonator is detuning itself, and the tide at sea will actually damp the basin response. Once it is small again, this process may start afresh, and the whole cycle takes a time span, which one calls the modulation period, that is large compared to the tidal period. This slow growth and decay thus reveals itself in the Poincaré plane as a single closed loop that is traversed in one modulation period. Other confocal loops represent other amplitudes of oscillation. However, when the forcing period is close to resonance, it is clear from the three curves of Fig. 4b that there may be different types of responses, which will be discussed further in the next section. With forcing and relatively strong damping, a regular oscillation of definite amplitude and phase eventually results, so stroboscopic intersections will approach one single spot, away from the orgin (Fig. 4c). When damping is weak and forcing is near resonance, depending on the precise initial state, the system may lock into different states (Fig. 4d).\n\nTo summarize, in this Poincaré plane, a strictly periodic oscillation will be represented by a single (fixed) point, a quasiperiodic oscillation by multiple points lying on a closed curve (limit cycle), and an aperiodic state (encountered in later sections) by an irregularly located set of points (strange attractor).\n\n### b. Qualitative effect of quasiperiodic forcing\n\nThe multiple equilibria near the Helmholtz resonance correspond to an amplified and a choked tidal response (see the two centers in Fig. 4d, at large and small volumes |x|, respectively). These states are stable and attracting when damping is present (Fig. 4d). Without damping, however, these states would retain their original distances to the fixed points (as the curves in Fig. 4b), and each state can be characterized by its energy level and period (with which it is traversed). The qualitatively different regimes, recognizable in Fig. 4b, are separated from each other by particular special orbits, the separatrices. One outer separatrix separates the outermost orbit from the other two, and one inner separatrix separates the banana-shaped orbit and small circular orbit. The period needed to traverse these separatrices approaches infinity, as the intersection of the two separatrices acts like a saddle. A state can creep up to this relative maximum in potential energy by exchanging some of its “kinetic” energy, allowing it to move forward along its orbit, in favor of “potential” energy.\n\nWhen friction is introduced, it pulls the state down into one of the equilibria, except for the odd point sitting right at the saddle and the two orbits that limit exactly on the saddle point in forward time. Note that these orbits merge with the aforementioned separatrices as the friction decreases to 0. Apart from these exceptions, all initial states thus belong to one of two domains of attraction (amplification or choking). When one starts to weakly and slowly “shake” the system (i.e., the saddle and the separatrices) by introducing a large-period perturbation, a new &ldquo=uilibrium” may arise. This happens because it is clearly quite a sensitive matter near the separatrices to which equilibrium their state will recede. Therefore, when, due to shaking, the separatrices move a little, while the state is slowly receding to one equilibrium, it may find itself a little later in the attraction domain of the other. Making again its way to the other equilibrium, something similar may happen, and, in effect, the neighborhoods of the separatrices may trap the state.\n\nThis new “steady” state might turn out, however, to be chaotic. This is because, close to the separatrices, the period of the inviscid orbits (the modulation period of the Helmholtz oscillator) varies greatly, and thus, when the state of the system varies due to the weak forcing, neighboring states fastly diverge from each other.\n\nThe way to verify that such a new dynamical equilibrium state turns into existence under the addition of perturbative damping and extra forcing is by performing an energy budget study. For this, one picks one of the original inviscid orbits, characterized by a particular period, and one then evaluates, along that orbit, the net increase or decrease of energy over one period due to both the work done by the perturbative forcing as well as the energy dissipation by the perturbative damping term. Only when these two are in balance (and there is thus no net energy increase or decrease) will the point on average stay near the original, unperturbed orbit. The single orbits considered in the present study are the previously introduced separatrices (homoclinic orbits), along which the period turns to infinity. The function determining the net energy change on the homoclinic orbits is called the Melnikov function. This function is usually given the alternative interpretation of measuring the distance of the perturbed orbits leaving and approaching the saddle. A (traditional) search for its zeros—implying intersections of the two perturbed orbits so that points both belong to the sets of points approaching and leaving the (perturbed) saddle, and are therefore “stuck”—is thus equivalent to the requirement that there is no net change in energy content along its trajectory. Mathematically, all we can show for the moment is that certain points (that belong to this invariant set) get trapped near the unperturbed separatrices, and that they traverse that region in an unpredictable, chaotic way (section 4a). However, we lack the ability to rigorously show that this set is actually attracting (Wiggins 1990): Suggestions that it is follow from numerical analysis (section 4b).\n\nSimilar such phenomena appear at other resonances, like near ω ≈ 1/2 + ε, and in particular near ω ≈ 2 ± ε (see M).\n\n## 3. Modulation equations\n\nMuch of the behavior obtained by numerically integrating the exact equation can, to a remarkable degree, also be found analytically in the solutions of the modulation equations that will be considered now. The advantage of using modulation equations is that fixed points, homoclinic orbits, Melnikov functions, etc. can be obtained explicitly, which is very helpful in addressing the main issue, namely, whether (and under what conditions) chaos can occur in the nonlinear Helmholtz resonator in the presence of damping. The stroboscopic intersections in Fig. 4 trace out continuous curves, whose locations can be obtained approximately when averaging the governing equations over the forcing period. Doing so, one obtains modulation equations that govern the amplitude dynamics on a slow timescale, which reveal where and under what conditions the dynamical response can become “complex.” When the excess-volume fluctuation υ(t) is of small amplitude (and when also the forcing and frictional parameters are small), (2) can be approximated by rescaling υ = εV, c = ε2C, f = ε3F (with ε ≪ 1). With sinusoidal forcing (6), this leads to",
null,
"This equation is solved by a multiple-scale perturbation expansion and has an amplified response at rational values of ω (Nayfeh and Mook 1979, p. 196). By requiring the absence of secular terms in the equations governing the subsequent orders of the perturbation expansion near the primary resonance (which, for consistency, requires ω = 1 + ε2σ) the solution is, up to second order, given by",
null,
"The response is thus frequency locked to the forcing and exhibits a “drift” term (the constant term), providing a net displacement of the mean state. Amplitude R and phase Φ vary on a slow timescale and are governed by",
null,
"where a dot indicates differentiating with respect to the slow time variable T = ε2t. Equating the right-hand sides to zero determines the steady states and allows one to obtain the frequency response curve for these equations σ(R; F, C); see Fig. 5. However, by rescaling tt/σ, Rσ1/2R, CσC, and Fσ3/2F, the detuning frequency σ can be scaled out and can, therefore, without loss of generality, be set equal to one. We may use this property to argue that the actual detuning defines the small parameter ε ≡ (ω − 1)1/2, which would otherwise be left arbitrary. This is consistent with the two rescalings, employed previously. In Cartesian coordinates, X = R cosΦ and Y = R sinΦ, these modulation equations read",
null,
"where R2 = X2 + Y2.\n\nFor the single-frequency forcing one can assume, without loss of generality, θ = 0 (which amounts to a rotation of our coordinate system). Its explicit presence is retained here however for later use, when, in the double-frequency forcing case, both F and θ will be varying on the slow timescale. Note that these modulation equations are the same as those found for the Duffing equation [in polar coordinates by the method of multiple scales, see Nayfeh and Mook (1979), and in both polar and Cartesian coordinates by the averaging method, see Guckenheimer and Holmes (1983)]. The extension to two forcing frequencies, discussed later, should thus have relevance to the quasiperiodically forced Duffing equation (Wiggins 1990; Yagasaki 1990, 1993), and extends it to the “double resonant” case of two nearby frequencies, both approximately equal to the natural frequency.\n\n### a. Comparison of exact and modulated systems\n\nIt is useful to understand the system of modulation equations because this local approximation has a strong correspondence with the Poincaré plot of the exact equation. To appreciate this, compare some of its numerically obtained solution curves with Poincaré plots of the exact equation (Fig. 6). Not only is the topology of fixed points and separatrices preserved but, indeed, so is much of their location (when giving proper notice to the rescaling factor ε, which is present in between these two systems). The modulation equations, however, have no restrictions on the region attainable in XY phase space, while the exact equations do. It is probably this restricted access of the original phase plane that makes a (formally) local solution perform so well in approximating its “global” dynamics. We will find that all of the behavior found near resonance in the original phase space is mimicked in the phase plane of the modulation equations (including, as we will see, the occurrence of chaos), which therefore forms a useful substitute.\n\n### b. Hamiltonian description\n\nIn the next section, particular attention will be given to the inviscid (C = 0) limit of (11)–(12) since these equations then turn into Hamiltonian form:",
null,
"with (for θ = 0) Hamiltonian",
null,
"Here a constant has arbitrarily been included to simplify the expression. Multiple equilibria, in this case, exist for |F| < 8/3. Because of the symmetry of the modulation equations (11)–(12), X → −X, when F → −F, it is sufficient to consider F to be single signed only. From here on we take this to be the interval: −8/3 < F < 0. Equilibria are, of course, more readily identified from (11)–(12). With θ ≡ 0, they have Y = 0, while X(X2/12 − 1) = F/2. With auxiliary variable α ∈ (−π/2, π/2), defined through F = −8 sin(α)/3, these steady states are in this case given by",
null,
"and k ∈ {−1, 0, 1}, which implies X−1 < X0 < X1.\n\n### c. Homoclinic orbits\n\nThe motivation for studying the inviscid case (C = 0) is related to the fact that the modulation equations can then be solved explicitly. This offers the means to calculate Melnikov functions explicitly, once, at a later stage, weak (slow timescale) forcing and damping are added to the modulation equations. Zeros of the Melnikov functions determine the presence of invariant chaotic sets and suggest the presence of chaotic solutions. The conditions under which zeros of these functions appear will map out regions in parameter space where one may find “chaotic tides.”\n\nIn the inviscid case (C = 0) explicit solutions of (11)–(12) are obtained by introducing\nSR2\nwhich measures the radial “distance” to the “special” radius Rs = 23. Note that it obeys the equation",
null,
"where we have used θ ≡ 0. Hence, by multiplying (11) by F/6 one obtains, upon integration,",
null,
"where K is an integration constant related to the energy level. By using Y = ±(R2X2)1/2 in (19) and then eliminating X from (20), and R2 from (18), one obtains",
null,
"where explicit expressions for S1,2,3, in terms of F and K, are found by identification of both expressions (see appendix A). Plots of the graphic P4(S) appearing in (21) are given in Fig. 7a. These correspond, for any given energy level, to the orbits in the XY phase plane, given in Fig. 7b. Of particular interest is the case that the central zeros of the quartic coalesce, and the corresponding orbits represent the homoclinic orbits γin,out. These homoclinic orbits correspond to the separatrices described in section 2b. Equation (21) can be integrated and its solutions can be expressed in terms of elliptic functions. However, we will have no need here for the general expressions and will only describe the homoclinic orbits coming from the saddle point (X1, 0), which will be used in section 5a to explicitly calculate Melnikov functions.\nThe solution of (21) for the outer homoclinic orbit, γout, for which S2 < S < S1 and K = −4S1 − 3S21, is obtained in appendix A as",
null,
"where S2,3 = S2,3S1. This shows that SS1 as T → ±∞ and S = S2 for T = 0. Similarly an expression for the inner separatrix can be obtained by simply interchanging S2 and S3, which has S = S3 for T = 0 and which also approaches the saddle point SS1 as T → ±∞. With these expressions for S(T), the trajectories γ(T) = (X(T), Y(T)) of the inner and outer homoclinic orbits follow from (20) and Y(T) = ±(R2X2)1/2; see Fig. 7b.\n\n## 4. Forcing at two nearly resonant frequencies\n\nThe tidal forcing term ζe(t) is of course not a perfect cosine. Here we consider a somewhat more realistic double-frequency forcing:\nζetf1ω1tθ1f2ω2tθ2\nThe first component of this external tidal forcing might represent the semidiurnal lunar M2 tide (dimensionally, ω1 = 1.4056 × 10−4 s−1); the second component the semidiurnal solar S2 tide (ω2 = 1.4544 × 10−4 s−1). Of course this means that |ω1ω2|/ω1 = 3.47 × 10−2 ≪ 1. As a consequence we cannot apply the ideas for the analysis of weakly, quasiperiodically forced oscillators, developed in, for instance, Beigie et al. (1991). However, we will see that the closeness of the two frequencies ω1 and ω2 can provide an additional periodic forcing term in the modulation equations, on the slow timescale. This periodic forcing term will break up the homoclinic orbits of the modulation equations (in the integrable limit) and thus might lead to (transversely) intersecting stable and unstable manifolds for the Poincaré map associated to this new forcing term.\n\nThus, taking into account an additional solar, and therefore almost resonant, tidal forcing term gives a mechanism for the construction of chaotic solutions (see also Yagasaki 1990, 1993).\n\nIn section 3 we defined the small parameter ε as a detuning. Analogously, we set here (now again employing nondimensional frequencies),\n2ω1\nThe above-described mechanism can be constructed if we assume that ω1ω2 = O2); that is, if\nωω2ω12\nand ΔΩ = O(1). The forcing amplitude and the phase now vary on the slow timescale T = ε2t:\nζefTω1tθT\nwhere the combined amplitude and phase are defined by",
null,
"The derivation of the modulation equations is now identical to the derivation given for the single-frequency case in section 3 (now one averages over 2π/ω1 instead of 2π/ω) and the modulation equations are again of the form (9)–(10), or, in Cartesian coordinates, (11)–(12), but now with slowly modulating forcing F(T) = f(T)/ε3 and phase θ(T), Eqs. (27) and (28) replacing their constant counterparts (see Nayfeh and Mook 1979, section 4.4). The right-hand sides of the modulation equations (11)–(12) now depend on the slow time explicitly, and, in principle, this offers the opportunity for the occurrence of aperiodic solutions.\n\nWhen the two components are of equal magnitude, f2 = f1, the phase is simply θ = (ΔΩ T + θ1 + θ2)/2, which redefines the effective frequency to be equal to the average frequency. The amplitude gradually changes sign over a long timescale f = 2f1 cos[(ΔΩ T + Δθ)/2]. The block version of this, f = 4f1(1/2 − Θ{cos[(ΔΩ T + Δθ)/2]}, is piecewise integrable. Here Θ(x) = 0 (1) for x < 0 (>0) denotes the Heaviside function. It has a saddle point on the X axis, whose position alternates with respect to the center of the XY phase plane with subsequent phases of the forcing. It is thus reminiscent of Aref's (1984) blinking vortex with its ensuing chaos. This suggests similar results for a sinusoidal forcing, which is confirmed by numerical analysis (see section 4b), although this cannot be substantiated analytically.\n\nWhen the second component is but a small perturbation with respect to the first one f2/f1δ ≪ 1, then ff1[1 + δ cos(ΔΩ T + Δθ)] and θθ1 + δ sin(ΔΩ T + Δθ). Note that this occurs quite naturally in the context of M2 and S2 tides. In contrast to the previous case, this limit is more amendable to further analysis and will be addressed now. Without loss of generality we set θ1 = 0, fixing the origin of the “fast” time t, and Δθ = −π/2, fixing the origin of the “slow” time T, whence the amplitude f = f1[1 + δ sin(ΔΩ T)] and phase θ = −δ cos(ΔΩ T), so that, with f = ε3F, for small values of δ, apart from a factor 2, the forcing terms in (11) and (12) are approximated as F sinθ ≈ −δF cos(ΔΩ T) and F cosθF[1 + δ sin(ΔΩ T)], respectively.\n\n### a. Perturbative second forcing component and Melnikov analysis\n\nWith a weak perturbative second forcing component (δ ≪ 1) and weak damping (CδC) the averaged Eqs. (11) and (12) become, after an appropriate shift in the origin of the slow and fast time,",
null,
"with H as in (15). These equations can be written symbolically as\ngXδhXT\nHere a dot denotes differentiation with respect to the slow time T. Note that it is assumed implicitly that 0 < ε ≪ δ. One has to include higher-order terms in ε and perform a much more detailed perturbation analysis when δ = O(ε) [see Guckenheimer and Holmes (1983); one has to be careful with applying Melnikovs method in the averaged system in that case; see also DKM for a detailed discussion]. System (31) has a hyperbolic periodic orbit, with period 2π/ΔΩ, that is O(δ) close to the saddle point P0 = (X1, 0) of the unperturbed (δ = 0) problem (see Fig. 7b), or, equivalently, the stroboscopic or Poincaré map PT0 associated with (31) has a fixed point PT0δ of saddle type O(δ) close to P0 (see also Guckenheimer and Holmes 1983, chapter 4, for more details). Time T0 defines which section is taken to construct this map. The stable and unstable manifolds of PT0δ are also O(δ) close to those of P0 in the unperturbed system. Thus, the perturbation will in general break open both homoclinic loops of the unperturbed system.\n\nNeglecting for the moment the additional forcing term [last term in Eq. (30)], the effect of the damping is to draw trajectories inward across separatrices. Multiplying (30) with ∇H, the rate of change of the Hamiltonian is then obtained, on the separatrix (where K = −3S21 − 4S1), as dH/dt = −3C(SS1)(3(S + S1) + 4)/2. Since on γout, SS1, it has dH/dt ≤ 0, while on γin, the inner separatrix, dH/dt ≥ 0, because S3SS1 and 3(S3 + S1) + 4 = 4 − 3[−8S1(1 + S1)]1/2 ≥ 0, for −2/3 ≤ S1 ≤ 0. Since outside γout the Hamiltonian has relatively large (K) values (see Fig. 7b) and since its variation is strictly negative (dH/dt < 0), this implies that all states outside it are drawn inward. Since the Hamiltonian is again relatively large inside γin, any trajectories that reach up to that separatrix are further drawn inward, to the choked state. All other states end up in the banana-shaped area in between γin and γout, approaching the amplified state (with relatively low values of the Hamiltonian K). With the additional forcing term, multiplication of (30) with ∇H gives the general expression for the rate of change of the energy (i.e., that of the perturbed Hamiltonian, dH/dt). If we evaluate that at the separatrices, we specify X(T) and Y(T) to satisfy the parametric relation corresponding to the particular separatrix (section 3c). The net change of the Hamiltonian, following this trajectory over one period (which is infinite on the separatrices), then amounts to the integral given below. Its vanishing would imply that the net change in energy is zero, and hence forcing and damping are in equilibrium and the state does not drift, representing a new, dynamic equilibrium.\n\nA geometric interpretation of the Melnikov method is that it measures the splitting distance between the stable and unstable manifolds of PT0δ that used to be part of a homoclinic loop to P0 in the unperturbed case. In this case there are two Melnikov functions: one that measures the splitting distance in the perturbed, outer loop, Mout(T0), and one for the inner loop, Min(T0). In both situations this distance is for instance measured at the intersections of the stable/unstable manifolds with the Y = 0 axis; Fig. 7b. Following, once again Guckenheimer and Holmes (1983, chapter 4) we see that Mout,in(T0) are, at leading order, given by",
null,
"where γ0(T) = (X0(T), Y0(T)) is either the outer or the inner homoclinic orbit of the unperturbed system (see section 3c); g = (g1, g2) and h = (h1, h2) are defined by (30) and (31).\n\nA (nondegenerate) zero of, for example, Mout(T0) corresponds to a transversal intersection of the “outer” stable and unstable manifolds of PT0δ for the map PT0. It is a standard procedure to associate a Smale horseshoe map to this situation (Guckenheimer and Holmes 1983). Thus, it follows that system (31) has chaotic solutions when either Mout(T0) or Min(T0) has zeros. Note, however, that the existence of chaotic solutions does not automatically imply the existence of a chaotic attractor; see section 4b.\n\nIn this paper we only consider the “classical way” to construct chaotic solutions, namely due to intersections of either the inner stable and unstable homoclinic orbits, or the outer ones. Other possibilities, like the intersection of the outer unstable manifold of PT0δ with its inner stable manifold, are discussed elsewhere (DKM).\n\nFor the outer homoclinic orbit, Eqs. (30), (31), and the definition of the unperturbed homoclinic orbit γ0(T) = (X0(T), Y0(T)), the Melnikov function (32) is calculated in appendix B. It is obtained as",
null,
"where auxiliary variables ψ and k are related to forcing F and frequency difference ΔΩ by F = −83 cos2ψ/(1 + 2 cos2ψ)3/2 and k = −ΔΩ(3 + tan2ψ)/2 tan2ψ, where F ∈ (−8/3, 0) for ψ ∈ (π/2, π). The derivation of the equivalent of Mout(T0) for the inner homoclinic orbit, Min(T0), is completely similar (see appendix B). There is a critical value of the damping C = Cout = Cout(ΔΩ, ψ) = Cout(ΔΩ, F) for which the Melnikov function (33) vanishes, thus implying that there will be transverse homoclinic intersections for the outer orbit and chaotic solutions, when",
null,
"depicted in parameter space in Fig. 8a. A similar critical surface Cin is obtained in appendix B for the inner orbit; see Fig. 8b.\n\nNote that both Cout and Cin vanish when the two basic frequencies are equal (i.e., ΔΩ = 0). In this case both Mout(T0) and Min(T0) are equal to M1 (for definition of M1,2,3 see appendix B), which does not depend on T0, and thus, in this case there can be no (transversal) intersections of the stable and unstable manifolds for the Poincaré map PT0. As a consequence there can also be no chaotic solutions. This has been observed earlier in the monofrequency case to which it reduces. The same is true for tidal frequencies that are not both close to the same resonance or, in other words, when |ΔΩ| is too large. This follows from (34) and its inner equivalent since the denominator sinhπk = sinh[πΔΩ/(2ν)] dominates the numerator when |ΔΩ| ≫ 1 (the decay rate decreases to zero when ψπ (F → −8/3 and ΔΩ > 0; see Fig. 8a and below). Here an upward (downward) directed arrow implies the limit is approached from below (above). For ΔΩ < 0 the critical damping curve of the outer orbit is very small, O(0.01), and hence not visibly different from zero on the left-hand side of Fig. 8a.\n\nThere are two other physically relevant limits: F ↑ 0 (i.e., ψπ/2 and ϕπ/2) and F ↓ −8/3 (ψπ and ϕ ↓ 0). In the first case (”no forcing”) one expects once again the disappearance of (possibly) chaotic solutions. Indeed we find that\nMoutT0MinT0Cπ,\nwhen F ↑ 0 [by (B9) and (B11): M2 and M3 vanish exponentially fast]. In a sense Mout(T0) and Min(T0) measure the same splitting distance in this case since the outer and inner homoclinic orbits merge in this limit. The difference in sign is explained by the fact that the stable manifold of the inner orbit merges with the unstable manifold of the outer orbit (and vice versa). A similar geometrical point of view gives insight in the limit F ↓ −8/3. In that case the inner homoclinic orbit shrinks to a point, while the outer orbit reaches a well-defined limit. Thus, nothing “singular” will happen to the outer orbit in this limit (see Fig. 8a). However, for the inner orbit, all components M1,2,3 will approach 0 as ϕ ↓ 0. The behavior of M1 is quite degenerate: M1 = O(ϕ5) for small ϕ, but both M2 and M3 will approach 0 with an exponential decay rate, for a fixed value of ΔΩ. Thus, we conclude that Cin ↓ 0 for F ↓ −8/3. However, the situation changes if we take the limit ϕ ↓ 0 differently, namely along any straight oblique line through the “origin” (ΔΩ, ϕ) = (0, 0): ΔΩ = qϕ, (q ≠ 0, ∞), as the exponentially fast decay is then absent and the behavior of Cin is dominated by the singularity in it due to the ϕ-dependent factor in M1. This implies that the critical damping magnitude may increase indefinitely in this limit. This singularity also dominates the magnitude of Cin farther away from the origin, in the interior of the ΔΩ–ϕ plane, and explains the two lobes visible in Fig. 8b on either side of the line ΔΩ = 0. It may be the reason why chaos can become important, even in quite strongly damped, realistic circumstances.\n\n### b. Numerical solutions of modulation and exact equations\n\nProving that the modulation equations possess a chaotic invariant set in the case when the Melnikov function vanishes is as far as we go analytically. As for the (closely related) Duffing equation, more work is needed to show that this set becomes attractive in some parameter range (Wiggins 1990, 612–613). In the following we will only give some numerical support that the (periodically forced) modulation equations possess such a chaotic (strange) attractor. This, in turn, suggests that the original, quasiperiodically forced Helmholtz oscillator should likewise have a strange attractor in the corresponding parameter regime; a result that we confirm by numerical integration below.\n\nNumerical integration of the modulation equations, (11)–(12), with a fourth-order Runge–Kutta scheme, employing a double-frequency forcing (23), shows that for two nearby frequencies within the frequency range over which multiple equilibria exist, the solution curves appear to be chaotic (Fig. 9). This is suggested both in the (slow) time domain T, by the strait's current speed Y (Fig. 9a), as well as in the phase space of current velocity Y versus excess volume X (Fig. 9b). (Recall that the original x and averaged quantities X are related by x ∝ εX, while the fast t and slow T timescales are related by t = ε−2T.) Here we take forcing amplitudes of the two external tidal components to be equal. One of these components is supposed to be at the (linear) resonance frequency, 1 (the Helmholtz frequency), while the second is at a slightly higher frequency (1.01). The latter frequency is setting the small parameter ε = 0.1. Within the resonance band the theory requires the forcing amplitudes to be small, O3), so that we take them f1 = f2 = 10−3. This leads, from (27), to F(t) = f3 = 2 cos[(ΔΩ T + Δθ)/2] in the modulation equations (12). In striking contrast to the monofrequency case, where the Hamiltonian chaos disappears with the introduction of even the smallest amount of damping (see M), the present calculations show that chaos survives the addition of damping. Commensurate with the theoretical requirements we need to take cO2) and, in fact, take c = ε3 here. This amounts to δC = 0.1. The erratic solution curves suggest that the modulation equations may possess a strange attractor, which is indeed revealed by making a Poincaré plot, by sampling at the modulation period (Fig. 10). Its shape is very similar to the “Japanese attractor” (see Ruelle 1980), which has been encountered for the forced and damped Duffing equation. Notice, in particular, the familiar Cantor-like division of the attractor's branches.\n\nChaos in the modulation equations suggests the occurrence of chaos in the original equations (5) under double-frequency forcing (23). In the following numerical experiments, the same values for forcing amplitudes and frequencies, as well as damping, were taken as were used in the previous numerical integration of the modulation equations. In order to compare the slow modulation of the numerically computed orbits directly to those obtained from the modulation equations, we sample the former at the tidal period. So, in Fig. 11 Poincaré plots of damped (c = 0.001) flow rate y versus time t (panel a) and versus excess volume x (panel b) are shown. Sampling is done once every average period 2π/ω, where ω = (ω1 + ω2)/2. Both figures suggest that chaos is also present in the exact equations. Indeed, this is again made manifest by making a Poincaré plot, but now by sampling at the long-period timescale, 2πω. Figure 12 shows that it reveals the presence of a strange attractor, not unlike that encountered for the modulation equations; Fig. 10.\n\n## 5. Forcing at two widely separate frequencies\n\nEstimating the natural frequency of a number of basins occurring in nature leads to a wide range of periods varying from a few minutes to several hours, up to periods of the tides, with a predominancy in the range of several tens of minutes (Honda et al. 1908). For this reason it is useful to consider here the response of the nonlinear Helmholtz resonator to a forcing of the form",
null,
"with 0 < ba: the first term represents the dominant, low-frequency tidal component and the second term a relatively weak perturbation that is almost in resonance with the Helmholtz frequency of the basin (ω ≈ 1). Thus, in this setting, ε2 ≪ 1 measures the ratio of tidal to Helmholtz frequency. Note that we have for simplicity neglected all “intermediate” Fourier components of the external tide: the main purpose of this section is to show that chaos can occur in basins with a high Helmholtz frequency. The mechanism is quite similar to the one described and studied in the previous sections. We will first discuss the effect of such a forcing in qualitative terms and continue this with a more detailed treatment.\n\nConsider first the case in which the perturbation is absent altogether, b = 0. Then, when damping is relatively weak because the frequency of the tide ε2 ≪ 1, the elevation within the basin will be able to follow the external tide: ζZ. Therefore, in the presence of a high-frequency perturbation (0 ≠ ba), on the fast timescale the dominant component of the tide, Z, acts as a slow, quasi-adiabatic change of the mean depth of the basin. However, the basin depth (among other parameters) determines the Helmholtz frequency; see (1). Therefore, the frequency of the perturbation, assumed to be fixed in absolute measure, is slowly varying when scaled with this Helmholtz frequency. This will provoke the quasi-stationary response to drift along the response curve (see Fig. 5). When this frequency drift, induced by the apparent modulation of water depth, is large enough—covering the frequency range over which multiple equilibria exist, this will lead to a sequence of “catastrophes”: rapid changes in the range of the (near) resonant, high-frequency response, which is subject to periodical collapse and expansion. There will thus be a hysteretic change in the amplitude of the basin elevation. This change may, however, be chaotic because the moments in time at which the amplitude jumps can become randomized. This is perhaps so because the high-frequency response, if not rapidly constrained to a particular phase by strong damping, will still be in a transient adapting phase, recovering from the previous catastrophe, when it undergoes the next one.\n\nIt is ironic that this explanation, based on the possibility of undergoing a hysteretic curve requires fairly large, but not too large, damping magnitudes: if damping gets too small, the frequency range over which multiple equilibria exist becomes broader than the range of the modulating, detuned frequency, and the response is “stuck” to a particular branch; if, on the other hand, damping is too strong, the response curve has no multiple equilibria at all, again lacking the ability to shift branches.\n\nIn practice, the conditions under which chaos appears may become weaker because not only will equilibria shift location under slow variation of the mean depth, but so, and probably more importantly, will the corresponding domains of attraction. This is perhaps the reason for the apparently chaotic behavior found in Fig. 13 under a mild modulation of the forcing frequency. This figure shows the result of a numerical integration of (2) in phase space with a forcing of type (35). Both results of direct integration (Fig. 13a), as well as a sampling of these on the long, tidal timescale (Fig. 13b) are shown. The direct integration shows that the dominant resonant response on the short Helmholtz timescale vaccilates on the long timescale, between a small- and large-scale response. These small- and large-scale responses correspond to the amplified and choked regimes in the amplitude response diagram of Fig. 5 to which these solutions would tend in the absence of the tide (a = 0). Under modulation with a tide of amplitude a = 3ε2 this provokes a modulation in apparent frequency σ, of approximately 1.5 dimensionless units, which is not big enough to span the entire frequency range over which multiple equilibria exist. Both this figure, as well as the tidally subsampled version in Fig. 13b (with period 2π2), however, testify about the irregular nature of the response, the latter figure in particular bearing some resemblance to the earlier Poincaré plots. In order to verify these qualitative ideas one may proceed with a more quantitative analysis. For the sake of brievity, this will however just be sketched here.\n\nAgain consider a basin in which the area increases linearly with the depth so that, with (4) and (35), the evolution of the excess volume υ is described by",
null,
"[see (2)]. The homogeneous problem (b = c = 0) has a “quasi-stationary” equilibium at υ2t) = Z2t) + ½Z2t)2, so it is natural to make the following shift in υ:",
null,
"which yields",
null,
"where dZ/dt = ε2dZ/dT = ε2Z′. As in section 3 we rescale υ̃ = εV, c = ε2C, and b = ε3B so that we can start the derivation of a modulation equation for V:",
null,
"[see (7)]. Thus, as predicted by the above qualitative arguments, the leading-order Helmholtz frequency 1/1 + Z2t) varies slowly and periodically in time ε2t = T. Since we intend to proceed along the lines of the preceeding sections, we need to assume that the perturbation b cos(ωt + θ) is almost in resonance with the Helmholtz frequency of the basin. Hence we write (again) ω = 1 + ε2σ and assume that Z = a cosε2t = ε2A cosε2t, so that 1/1 + Z = 1 − ½ε2A cosε2t + O4). This last assumption is especially motivated by the mathematical approach, but may be quite realistic in the case of the tides in fjords too. Note that the scalings of a = ε2A and b = ε3B are consistent with the assumption ba.\nAs motivated by the above arguments we now introduce a new timescale = t/1 + Z = t(1 − ½ε2A cosε2t + O4)), so that (39) indeed reduces to (7) with t replaced by and",
null,
"at leading order on a timescale of O(1/ε2). The derivation of the modulation equations is now completely similar to that of (9)–(10) in section 3 (as in section 4) and results in (9)–(10) with σ replaced by the periodically varying function σα(T). As in section 3 we set θ = 0 and scale σ to 1 so that the modulation equations are now given by",
null,
"The presence of the periodically oscillating term can indeed be interpreted as a drift along the response curve of Fig. 5. This system can again be analyzed by the Melnikov approach of section 4a if we assume that C, α = O(δ) with 0 < ε ≪ δ ≪ 1. Note that in this case the drift along the response curve is also only of order δ. The earlier described emergence of chaos, qualitatively attributed to hysteretic changes, which can only exist for δ = O(1), is apparently overrestrictive, as a simple “jitter” of the saddle point seems to be sufficient for the emergence of crossing of stable and unstable homoclinic orbits, and thus for the presence of a chaotic invariant set. For small δ the Melnikov analysis proceeds along the lines of section 4a. The only difference is that the perturbation term h(X, T) in (31) has changed. It is a straightforward procedure to write down the equivalents of (B1) and redo the calculations of section 4b for the “new” inner and outer Melnikov functions, by which equivalents of (33) can be obtained. Note that (by construction) the structures of Mout,in(T0) will be the same as in section 4a: C × (a constant term M1) + (a T0 periodic term); M1 will even be exactly as in (33) and its inner equivalent. We do not present the details of the computations here. Our main observation is that one can thus again show the existence of a Smale horseshoe: it is the same mechanism as that of the quasiperiodically forced case of sections 3 and 4, which creates chaotic solutions to systems with two widely separated frequencies.\n\nSeveral aspects of the numerical integration in Fig. 13 are still unrealistic. First, the response is dominated by the local resonance and thus the elevation shows a nearly periodic change on the fast timescale, instead of on the (long) tidal timescale. Second, the response is much too big, covering (almost) the entire basin depth. More work is therefore needed to bring the response closer to the ordering observed in nature, as for example, in Moldefjord (see Fig. 1), where Helmholtz response, O(0.1 m) ≪ tidal range, O(1 m) ≪ total depth, O(80 m), while currents on the fast Helmholtz timescale are comparable to tidal currents. It should be noticed that the correct ordering can partly be obtained quite simply by choosing model parameters such that all fixed points of the modulation equations are close to the origin. With this choice the homoclinic orbits, and therefore the actual state of the oscillator, are close to the origin of the phase space. Hence, corresponding elevations will be small compared to water depth. Notwithstanding the remaining issues that need clarification, it is believed that the mechanism described in this section may act as a “building block” for the explanation of the appearance of chaotic tides in basins with high Helmholtz frequency, particularly when the tide is extended with additional tidal components and harmonics.\n\nFrom a practical point of view, any observed persistence of observed secondary oscillations might point at a non-meteorological origin. When their (average) period compares to estimates of the Helmholtz frequency, the model advanced here may be appropriate. Additional support may be obtained by plotting observed elevation (volume) against a strait current. Particularly, a stroboscopic sampling thereof (at the Helmholtz and tidal periods) may elucidate the structure of the underlying attractor. By comparison to predictions from the present model, an estimate of the most elusive factor, the damping, may perhaps be inferred.\n\n## 6. Summary\n\nBays and estuaries may co-oscillate with the tides in the adjacent seas (Defant 1961). The extent to which they do generally depends on the geometry of the basin. Two competing mechanisms exist that mainly determine the final state of the coastal tides: the proximity of one of the more prolific tidal frequencies to basin resonances and the amount of friction. Depending on the balance between forcing and damping the coastal tide may be amplified or choked. This twofold nature of the response may actually occur for one and the same set of geometric and frictional parameters, as was shown in the case of the Helmholtz resonator for an almost-enclosed, relatively deep tidal basin with sloping bottom (see M). The slope in the bottom provokes a nonlinear restoring term (Green 1992) that, in turn, is responsible for the occurrence of multiple (stable) equilibria near the (linear) resonance frequency. When forced at the entrance by a singlefrequency tide (located within the resonance band), depending on the initial state, the tide within the basin may either have a small or large range that, after the decay of transients, is stationary.\n\nHere we have shown that the presence of a second tidal frequency component, also within (or near to) the resonance band, may, in contrast, lead to the occurrence of a strange (chaotic) attractor, even with relatively strong damping. The basin tide may thus exhibit a perpetual change, manifested by unstable estimates of the tidal “constants” (amplitude and phase), when based on the analysis of a finite length time series, which may be of relevance to the observation by Gutiérrez et al. (1981) that some tidal components appear “unresolvable.” It may also be relevant to chaos in consecutive tidal maxima, observed in Venice Lagoon (Vittori 1992) or in coastal tides (Frison et al. 1999).\n\nSurprisingly, there have been many observations of “irregular” secondary oscillations, accompanying the tide, which date back long before the notion of chaos appeared in the literature. Already in 1908, Honda et al. devoted an extensive study to ascertain the presence of both regular and irregular secondary oscillations in vertical elevation records in some 50, mostly Japanese, basins, usually based on observations covering just a few days. The nature of the basin response is typically, however, a superposition of the tide and a high-frequency oscillation, rendering the previous model, which predicts secondary oscillations on the tidal timescale, less applicable. Periods of these oscillations are observed to be in the range from several minutes up to several hours, depending on the size of the basin. Depending on the basin shape, this high-frequency oscillation can either be identified as the Helmholtz or quarter-wave resonance. The amplitude of the secondary oscillations in vertical elevation is, as the name suggests, observed to be an order of magnitude less than that of the (primary) tide. This may probably explain why, until fairly recently, the interest in secondary oscillations seems to have waned, and why the topic is held as a curiosity. A recent observation by Golmen et al. (1994), in a Norwegian fjord connected to the sea by a narrow strait, has, however, put the possible relevance of secondary oscillations into a new perspective by noticing that the associated secondary, irregular currents through the connecting strait reach a magnitude comparable to that of the tidal current. The reason for this is that, whereas the tide is present both outside and (with some delay) inside of the strait, the secondary elevation is present basically only within the fjord. The elevation differences, responsible for the associated currents, may therefore, at any time be of similar magnitude when comparing tidal and secondary oscillations. Hence, also the corresponding currents may be expected to be equally important. The same process may perhaps explain similar highly irregular current observations through lagoon entrances by Kjerfve and Knoppers (1991) and by Smith (1994), and the observed preferential enhancement of current harmonics (relative to that in the elevation field) by Seim and Sneed (1988). The relevance of this observation is that secondary oscillations may greatly alter the corresponding currents and therefore seriously modify the transfer of matter and dissolved substances (the “flushing” of the bay or fjord).\n\nAs remarked, these observations require a different kind of model in which tidal and resonance frequency—in our present model the Helmholtz frequency—are widely disparate. By adding a perturbative, but resonant, second forcing term to the primary tidal forcing it was shown that modulation equations are obtained that, to some extent, mimic those derived for the previous case with two tidal frequencies. In both cases, the modulation equations are driven at the long timescale. In the case of two nearby and resonant tidal frequencies this forcing stems from the beat (difference) frequency. In the case when the tidal frequency is much less than that of the resonant perturbative forcing term the tide itself is directly providing a modulation on the long timescale. The presence of a chaotic invariant set that could be ascertained analytically in these case, unfortunately does not guarantee that this set is also attractive. Therefore further support for the presence of a strange attractor was offered by (stroboscopic) Poincaré plots stemming from the numerical integration of the nonlinear Helmholtz resonator, when appropriately forced. The presence of a chaotic response may show up in a numerical simulation (in a basin with sloping shoreline) in much the same way as it reveals itself in nature: by irregularity in the elevation and, particularly, current fields (either within one tidal period, or in between subsequent tidal periods), by broadening of spectral peaks (with poorly resolved phases), and by showing a sensitive dependence on the initial tidal phase (neighboring orbits evolving to different states). The implication is that irregularity in the current and elevation fields is predictable as a phenomenon, but not its exact development. It requires further effort to quantify the resulting enhanced exchange between sea and basin.\n\nThus, despite the fact that, with the introduction of satellite-altimetry-derived tidal observations, tidal predictions have reached new unprecedented accuracies, it now seems that, at the same time, the tides may to some extent and in certain locations be intrinsically unpredictable, an unpredictability that reflects the nonlinear nature of the response of such locations.\n\nAcknowledgments\n\nDiscussions with Marko Wilpshaar, and comments and suggestions by Jef Zimmerman, Huib de Swart, and Ferdinand Verhulst are greatly appreciated.\n\n## REFERENCES\n\n• Andersen, O. B., , P. L. Woodworth, , and R. A. Flather, 1995: Intercomparison of recent ocean tide models. J. Geophys. Res., 100 , 2526125282.\n\n• Export Citation\n• Aref, H., 1984: Stirring by chaotic advection. J. Fluid Mech., 143 , 121.\n\n• Beigie, D., , A. Leonard, , and S. Wiggins, 1991: Chaotic transport in the homoclinic and heteroclinic tangle regions of quasiperiodically forced two-dimensional systems. Nonlinearity, 4 , 775819.\n\n• Export Citation\n• Cartwright, D. E., 1977: Oceanic tides. Rep. Prog. Phys., 40 , 665.\n\n• Davies, A. M., 1986: A three-dimensional model of the northwest European continental shelf, with application to the M4 tide. J. Phys. Oceanogr., 16 , 797813.\n\n• Export Citation\n• Defant, A., 1961: Physical Oceanography. Vol. II,. Pergamon, 598 pp.\n\n• Frison, T. W., , H. D. I. Abarbanel, , M. D. Earle, , J. R. Schultz, , and W. Scherer, 1999: Chaos and predictability in ocean water levels. J. Geophys. Res., 104 , 79357951.\n\n• Export Citation\n• Garrett, C., 1975: Tides in gulfs. Deep-Sea Res., 22 , 2335.\n\n• Godin, G., 1991: The analysis of tides and currents. Tidal Hydrodynamics, B. B. Parker, Ed., Wiley and Sons, 675–709.\n\n• Golmen, L. G., , J. Molvaer, , and J. Magnusson, 1994: Sea level oscillations with super-tidal frequencies in a coastal embayment of western Norway. Cont. Shelf Res., 14 , 14391454.\n\n• Export Citation\n• Green, T., 1992: Liquid oscillations in a basin with varying surface area. Phys. Fluids A, 4 , 630632.\n\n• Guckenheimer, J., , and P. Holmes, 1983: Nonlinear Oscillations, Dynamical Systems, and Bifurcations of Vector Fields. Springer-Verlag, 459 pp.\n\n• Export Citation\n• Gutiérrez, A., , F. Mosetti, , and N. Purga, 1981: On the indetermination of the tidal harmonic constants. Nuovo Cimento, 4 , 563575.\n\n• Honda, K., , T. Terada, , Y. Yoshida, , and D. Isitani, 1908: Secondary undulations of oceanic tides. J. Coll. Sc. Imp. Univ. Tokyo, 24 , 1113. and 95 plates.\n\n• Export Citation\n• Kjerfve, B., , and B. A. Knoppers, 1991: Tidal choking in a coastal lagoon. Tidal Hydrodynamics, B. B. Parker, Ed., Wiley and Sons, 169–182.\n\n• Export Citation\n• Krümmel, O., 1911: Handbuch der Ozeanographie. Verlag von J. Engelhorns, 766 pp.\n\n• LeBlond, P. H., 1991: Tides and their interactions with other oceanographic phenomena in shallow water. Tidal Hydrodynamics, B. B. Parker, Ed., Wiley and Sons, 675–709.\n\n• Export Citation\n• Lynch, D. R., , and F. E. Werner, 1991: Three-dimensional velocities from a finite-element model of English Channel/Southern Bight tides. Tidal Hydrodynamics, B. B. Parker, Ed., Wiley and Sons, 183–200.\n\n• Export Citation\n• Maas, L. R. M., 1997: On the nonlinear Helmholtz response of almost-enclosed tidal basins with a sloping bottom. J. Fluid Mech., 349 , 361380.\n\n• Export Citation\n• Maas, L. R. M., . 1998: On an oscillator equation for tides in almost enclosed basins of non-uniform depth. Physics of Estuaries and Coastal Seas, J. Dronkers, and M. Scheffers, Eds., A. A. Balkema, 127–132.\n\n• Export Citation\n• Munk, W., , and C. Wunsch, 1998: Abyssal recipes II: Energetics of tidal and wind mixing. Deep-Sea Res. I, 45 , 19772010.\n\n• Nakano, M., 1932: Preliminary note on the accumulation and dissipation of energy of the secondary oscillations in a bay. Proc. Phys.-Math. Soc. Japan, Ser. 3,, 14 , 4456.\n\n• Export Citation\n• Nayfeh, A. H., , and D. T. Mook, 1979: Nonlinear Oscillations. Wiley-Interscience, 704 pp.\n\n• Platzman, G., 1971: Ocean tides and related waves. Mathematical Problems in the Geophysical Sciences. Vol. 1, Geophysical Fluid Dynamics, W. H. Reid, Ed., American Mathematical Society, 239–291.\n\n• Export Citation\n• Prudnikov, A. P., , Yu A. Brychkov, , and O. I. Marichev, 1986: Integrals and Series. Vol. I, Gordon and Breach, 798 pp.\n\n• Pugh, D. T., 1987: Tides, Surges and Mean Sea-Level. John Wiley and Sons, 472 pp.\n\n• Ruelle, D., 1980: Strange attractors. Math. Intelligencer, 2 , 126137. [Reprinted in Cvitanović, P., 1984: Universality in Chaos, Adam-Hilger, 37–48.].\n\n• Export Citation\n• Schwiderski, E. W., 1980: On charting global ocean tides. Rev. Geophys. Space Phys., 18 , 243268.\n\n• Seim, H. E., , and J. E. Sneed, 1988: Enhancement of semidiurnal tidal currents in the tidal inlets to Mississippi Sound. Hydrodynamics and Sediment Dynamics of Tidal Inlets, D. G. Aubrey and L. Weishar, Eds., Lecture Notes on Coastal and Estuarine Studies, Vol. 29, Springer-Verlag, 157–168.\n\n• Export Citation\n• Smith, N. P., 1994: Water, salt and heat balance of coastal lagoons. Coastal Lagoon Processes, B. Kjerfve, Ed., Elsevier Oceanographic Series, Vol. 60, 69–101.\n\n• Export Citation\n• van Ette, A. C. M., , and H. J. Schoemaker, 1966: Harmonic analysis of tides. Essential features and disturbing influences. Hydrographic Letter, Vol. 1, Netherlands' Hydrographer Special Publication 2, 1–35.\n\n• Export Citation\n• Vittori, G., 1992: On the chaotic structure of tide elevation in the Lagoon of Venice. Proc. 23d Int. Conf. on Coastal Engineering, Venice, Italy, American Society of Civil Engineers, 1826–1839.\n\n• Export Citation\n• Wiggins, S., 1990: Introduction to Applied Nonlinear Dynamical Systems and Chaos. Springer-Verlag, 672 pp.\n\n• Yagasaki, K., 1990: Second-order averaging and chaos in quasiperiodically forced weakly nonlinear oscillators. Physica D, 44 , 445458.\n\n• Export Citation\n• Yagasaki, K., . 1993: Chaotic motions near homoclinic manifolds and resonant tori in quasiperiodic perturbations of planar Hamiltonian systems. Physica D, 69 , 232269.\n\n• Export Citation\n• Zimmerman, J. T. F., 1992: On the Lorentz linearization of a nonlinearly damped tidal Helmholtz oscillator. Proc. K. Ned. Akad. Wet., 95 , 127145.\n\n• Export Citation\n\n# APPENDIX A\n\n## Integrating Eq. (21)\n\nIn order for the right-hand of (21) to be real, the argument of the square root, the quartic",
null,
"needs to be positive somewhere, which sets a bound on K, whose particular value is of no relevance here. Then, there can either be one or two intervals over which the quartic is positive, corresponding to the cases that the quartic has two or four real zeros, respectively; see the upper and lower curves in Fig. 7a. For fixed forcing F, these curves have differing values of K. In the former case this value is relatively low; hence the corresponding region in the XY parameter plane, in between γin and γout, is labeled with a minus sign; see Fig. 7b. In the latter case, it has a relatively high value of K, and there are two corresponding curves in the XY plane, one inside γin and one outside γout, which regions are therefore labeled with a plus sign. These regions are separated from each other by separatrices γin,out that emanate from the third fixed point: the saddle. These separatrices, or homoclinic orbits, are characterized by the fact that the two central zeros of the quartic coalesce, in which case (21) can be solved algebraically. Equating (A1) to the quartic −(SS1)2(SS2)(SS3)/4, and requiring equivalence of each of the coefficients of the fourth-degree polynomial, leads to four equations from which S1,2,3 and K can be determined. Because in the XY plane the central, double zero of the quartic corresponds with the saddle point (X1; 0), whose position is already determined [see (16)–(17)], we find that one of these equations is redundant. The remaining equations yield the two other zeros,",
null,
"and the energy level of the separatrix",
null,
"where\nS1X21\n(Y1 = 0) and X1 = 4 sinν1, where ν1 = (α + 2π)/3 and α = −sin−1(3F/8). Over the range of F values considered, −8/3 < F < 0, one finds −2/3 < S1 < 0.\nReplacing the quartic in (21) with its expression in terms of zeros requires the solution of",
null,
"where we have assumed S3 < S < S2. Now S can still be on either side of S1, which itself lies in between the lower and upper limits S3 and S2, respectively. By taking S either smaller or larger than S1 we obtain the inner or outer separatrix, respectively.\nFor definiteness assume S1SS2 so that we are constructing the outer separatrix. Introducing\nSSS1\nand similarly S2 = S2S1 > 0 and S3 = S3S1 < 0, then",
null,
"Introducing σ = 1/S′ > 0 yields",
null,
"Multiplying by S2 and defining σ′ = S2σ > 1 and η = −S2/S3 > 0 gives",
null,
"Let σ′ = 1 + P2 and η = Q2 − 1 (with Q2 > 1), then",
null,
"with ν ≡ (−S2S3)1/2/4. Hence, P = ±Q sinhνT, or, following back the substitutions, we find that (21) is solved by (22).\n\n# APPENDIX B\n\n## Computing the Melnikov Function\n\nWith g, h from (30) and (31), the Melnikov function (32) is, for the outer homoclinic orbit, defined as",
null,
"with a suffix 0 referring to spatial coordinates along an unperturbed homoclinic orbit. Since (18)–(20) apply in general, S0 = R20/12 − 1, FX0/6 = S20K, and FY0/12 = dS0/dT. Thus, along these orbits, the Melnikov function simplifies to",
null,
"We will denote the three integrals in (B2) as M1(T0), M2(T0), and M3(T0), respectively. Note that S0 approaches the saddle S1 for T → ±∞ along the separatrices, so that the integrands vanish in these limits (recall that K = −3S21 − 4S1). We now define",
null,
"so that we can rewrite the Mi(T0):",
null,
"The expressions for M2,3(T0) were obtained by expanding cosΔΩ(T + T0) and sinΔΩ(T + T0) and by partial integration; all integrands involving sin(ΔΩT0) “average out” due to the odd/even symmetries of S0(T) and its derivative. The Jn integrals can be computed explicitly by expressing them in an integral I(a, k) that can be evaluated by a contour integral in the complex plane (see appendix C):",
null,
"where a = cosϕ. For J1(ΔΩ) this is a rather straightforward procedure, since, by section 3c",
null,
"where A, a, and ν can be expressed, in terms of S2 and S3, as",
null,
"It follows that −1 < a < 0, and therefore we introduce ψ by",
null,
"and hence, from (43),",
null,
"so that",
null,
"where",
null,
"Note that the limit ΔΩ → 0, or k → 0, is well defined. Integral J2(ΔΩ) can be obtained by taking the a derivative of I(a, k):",
null,
"Once again, the limit k → 0 is well defined. Compiling expressions, the Melnikov function for the outer homoclinic orbit, Mout(T0) = M1 + M2 + M3, is given by (33). The derivation of the equivalent of Mout(T0) for the inner homoclinic orbit, Min(T0), is similar to the above analysis. The main difference is that the roles of indices 2 and 3 in the description of the homoclinic orbit [(B6) and section 3c] are exchanged and thus that S0 < S1. This amounts to a replacement in all formulas of ψ → −ϕ, where ϕ ∈ (0, π/2), which is significant, because it now opens the possibility of a singularity in the surface of critical damping; Fig. 8b.\n\n# APPENDIX C\n\n## Evaluation of an Integral\n\nIn order to evaluate Melnikov's distance function, integral I(a, k), Eq. (B5), needs to be obtained for 0 < |a| < 1. Although this integral can be found in the literature (Prudnikov et al. 1986, p. 470) we here give a short sketch of its derivation by writing it as",
null,
"for p = exp(x). Hence, by separating the denominator,",
null,
"where, with a ≡ cosϕ, p± = exp(±). Recasting the integration in terms of x, the integral reads",
null,
"For definiteness assume k > 0. Then, by extending the integral over the real axis to a complex integration in the upper half-plane over a region enclosed also by a semicircle of infinite radius, this integral can be evaluated by means of Cauchy's theorem. It determines it as 2πi times the sum of the residues at the poles x± = ± + (2n + 1)πi, for n = 0, 1, 2, · · · . Here, minus and plus signs refer to the poles appearing in the first and second integral, respectively. This yields",
null,
"where SN = ΣNn=0 q2n+1 with q ≡ exp(−πk), whence S = q/(1 − q2) = 1/(2 sinhπk), as 0 < q < 1. Therefore the expression in (B5) follows. Note that this expression holds also when k = 0."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.91301876,"math_prob":0.9705008,"size":44415,"snap":"2021-04-2021-17","text_gpt3_token_len":10219,"char_repetition_ratio":0.14667538,"word_repetition_ratio":0.00924083,"special_character_ratio":0.22033097,"punctuation_ratio":0.15064563,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98408353,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-22T00:27:31Z\",\"WARC-Record-ID\":\"<urn:uuid:ec569ca9-cd6b-45ea-b016-abe982189d4e>\",\"Content-Length\":\"606800\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8290285b-730a-4e22-92d3-f1f90512e59d>\",\"WARC-Concurrent-To\":\"<urn:uuid:a24726f9-4edc-4459-9890-4fc2d2ee30c8>\",\"WARC-IP-Address\":\"34.243.57.88\",\"WARC-Target-URI\":\"https://journals.ametsoc.org/view/journals/phoc/32/3/1520-0485_2002_032_0870_ct_2.0.co_2.xml\",\"WARC-Payload-Digest\":\"sha1:EDSYOJYNV3GIS26RQ35YWYKNUMLJMBD5\",\"WARC-Block-Digest\":\"sha1:OJDPGQOPYD6TMAUH77GV3VOAZ7ZVVSMC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703528672.38_warc_CC-MAIN-20210121225305-20210122015305-00141.warc.gz\"}"} |
https://socratic.org/questions/how-do-you-solve-6m-n-21-and-m-8n-28-using-substitution#579209 | [
"# How do you solve 6m + n = 21 and m - 8n = 28 using substitution?\n\nMar 22, 2018\n\n$n = - 3 , \\text{ } m = 4$\n\n#### Explanation:\n\nTake the second equation:\n\n$m - 8 n = 28$\n\nIf we add $8 n$ to both sides, we get:\n\n$m = 8 n + 28$\n\nNow we have two equivalent expressions: $\\textcolor{red}{m}$ and $\\textcolor{b l u e}{8 n + 28}$.\n\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n\nNow, look back at the first equation:\n\n$6 m + n = 21$\n\nTo solve by substitution, we need to replace (or substitute) one equivalent expression for the other so that we can solve this equation. Let's use the expressions from earlier:\n\n$6 \\textcolor{red}{m} + n = 21$\n\nRemember that $\\textcolor{red}{m} = \\textcolor{b l u e}{8 n + 28}$, so we can do this:\n\n$6 \\left(\\textcolor{b l u e}{8 n + 28}\\right) + n = 21$\n\nNow, we can use simple algebra to solve for $n$.\n\n$6 \\left(8 n + 28\\right) + n = 21$\n\n$6 \\cdot 8 n + 6 \\cdot 28 + n = 21$\n\n$48 n + 168 + n = 21$\n\n$49 n + 168 = 21$\n\n$49 n = - 147$\n\n$n = - 3$\n\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n\nFinally, let's plug $n$ back into one of the equations and solve for $m$ to finish the problem. I'm using equation 1 to solve for $m$, but either one will work just fine.\n\n$6 m + n = 21$\n\nRemember from above that $\\textcolor{\\mathmr{and} a n \\ge}{n} = \\textcolor{\\lim e g r e e n}{- 3}$\n\n$6 m + \\textcolor{\\mathmr{and} a n \\ge}{n} = 21$\n\n$6 m + \\textcolor{\\lim e g r e e n}{\\left(- 3\\right)} = 21$\n\n$6 m - 3 = 21$\n\n$6 m = 24$\n\n$m = 4$\n\nSo now we have both parts of our solution! Here's the final solution:\n\n$n = - 3 , \\text{ } m = 4$"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7283344,"math_prob":1.0000064,"size":1008,"snap":"2023-40-2023-50","text_gpt3_token_len":349,"char_repetition_ratio":0.1992032,"word_repetition_ratio":0.035532996,"special_character_ratio":0.45833334,"punctuation_ratio":0.07526882,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000043,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T08:40:33Z\",\"WARC-Record-ID\":\"<urn:uuid:a151208b-73b3-45c6-bceb-5b9c8a504c2f>\",\"Content-Length\":\"36910\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9384b6c0-f4ee-4d1c-b09d-9855754779e8>\",\"WARC-Concurrent-To\":\"<urn:uuid:2dc5496d-1670-4cb5-98e6-9bd207923018>\",\"WARC-IP-Address\":\"216.239.38.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-solve-6m-n-21-and-m-8n-28-using-substitution#579209\",\"WARC-Payload-Digest\":\"sha1:VK56Q5XRQXBI6MRVO5FDKPEL6O3GZSAY\",\"WARC-Block-Digest\":\"sha1:NUBBHEBHAW3IQQ2ATESQBMVA73PCFXKC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510983.45_warc_CC-MAIN-20231002064957-20231002094957-00764.warc.gz\"}"} |
https://sonichours.com/40-ounces-is-equal-to-how-many-pounds/ | [
"# 40 Ounces Is Equal To How Many Pounds\n\nHow many pounds are 40 ounces? To find out, multiply 40 ounces by 0.4. Although it is not difficult to do, these examples will help you better understand the process.\n\nOne pound equals 16 ounces, and eight ounces are half a pound. Then, 40 ounces are equal to 2.5 pounds. However, there’s an important distinction between dry and liquid ounces. A 16 oz bottle will weigh less than an 8-ounce. When figuring out how many pounds 40 ounces equals, divide it by 16 to get the answer.\n\nThe avoirdupois weight is 0.45359237 kgs. The troy ounce is 31.1 grams, and one sixteenth of an avoirdupois pound weighs 28.3 grams. The ounce is the weight unit used in most British derived customary systems. An ounce weighs 0.06225 pounds, so 40 ounces equals about 4.6 pounds.\n\nThe imperial system uses the pound as a unit for mass. It is the most widely used unit of mass. It is often abbreviated as #, lbm and lbm. Depending on your metric system, a pound may be represented as 48 ounces. The same goes for grams. You will have approximately the same amount of pounds if you have 40 ounces in each hand.\n\nAn ounces-to-pound calculator is required to convert ounces into pounds. It will tell you how many pounds a given mass is in ounces. You can also use a table to find out how many pounds a given mass weighs. This table includes examples and formulas. You’ll be amazed at the results if you remember that the pounds are equal. You’ll be glad you found the right information."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.93167835,"math_prob":0.9913877,"size":1497,"snap":"2022-27-2022-33","text_gpt3_token_len":366,"char_repetition_ratio":0.15003349,"word_repetition_ratio":0.014869888,"special_character_ratio":0.24716099,"punctuation_ratio":0.12576687,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.993981,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-01T08:28:25Z\",\"WARC-Record-ID\":\"<urn:uuid:5dcea9d4-0b1b-4d43-aa74-4aa65e26613c>\",\"Content-Length\":\"87407\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:18e14c24-99e3-4a76-812c-d4d058737051>\",\"WARC-Concurrent-To\":\"<urn:uuid:2f789fd5-e5cb-4e47-bafd-6e561d7e1d50>\",\"WARC-IP-Address\":\"162.0.217.81\",\"WARC-Target-URI\":\"https://sonichours.com/40-ounces-is-equal-to-how-many-pounds/\",\"WARC-Payload-Digest\":\"sha1:MDH27R7CW772KOWHAZ3SL6KJ4DSTTTRU\",\"WARC-Block-Digest\":\"sha1:T4DPZ6KERDZFBCMMTCQSSANISW4H4XHR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103922377.50_warc_CC-MAIN-20220701064920-20220701094920-00233.warc.gz\"}"} |
http://www.freepatentsonline.com/4559713.html | [
"Title:\nAzimuth determination for vector sensor tools\nUnited States Patent 4559713\n\nAbstract:\nThis invention relates to mapping or survey apparatus and methods, and more particularly concerns derivation of the azimuth output indications for such apparatus in a borehole from the outputs or output indications of either an inertial angular rate vector sensor (or sensors) and an acceleration vector sensor (or sensors), or a magnetic field vector sensor (or sensors), and from the outputs of an acceleration vector sensor (or sensors). At least one of such sensors in any instrument may be canted relative to the borehole axis. Borehole tilt is also derived.\n\nInventors:\nEngebretson, Harold J. (Yorba Linda, CA)\nLahue, Philip M. (West Lake Village, CA)\nVan Steenwyk, Brett H. (San Marino, CA)\nApplication Number:\n06/558598\nPublication Date:\n12/24/1985\nFiling Date:\n12/06/1983\nExport Citation:\nAssignee:\nApplied Technologies Associates (San Marino, CA)\nPrimary Class:\nOther Classes:\n33/304\nInternational Classes:\nE21B47/022; (IPC1-7): G01C19/38\nField of Search:\n33/302, 33/304, 33/312, 33/313, 33/324\nView Patent Images:\nUS Patent References:\n 4293046 Survey apparatus, method employing angular accelerometer 1981-10-06 Van Steenwyk 33/304 4244116 Devices for measuring the azimuth and the slope of a drilling line 1981-01-13 Barriac 33/304 4238889 Devices for the azimuth and slope scanning of a drilling line 1980-12-16 Barriac 33/304 4199869 Mapping apparatus employing two input axis gyroscopic means 1980-04-29 Van Steenwyk 33/302 4021774 Borehole sensor 1977-05-03 Asmundsson et al. 33/313 3896412 Method and apparatus for logging the course of a borehole 1975-07-22 Rohr 33/304 3894341 Rapid resetting gyroscope 1975-07-15 Kapeller 33/324 3753296 WELL MAPPING APPARATUS AND METHOD 1973-08-21 Van Steenwyk 33/304 3561129 N/A 1971-02-09 Johnston 33/302 3308670 Gyro platform arrangement 1967-03-14 Granqvist 74/53.4 3241363 Navigation instruments 1966-03-22 Alderson et al. 73/178 3137077 Drill-hole direction indicator 1964-06-16 Rosenthal 33/302 3052029 Automatic teeth separators 1962-09-04 Wallshein 33/302 3037295 Process and means for determining hole direction in drilling 1962-06-05 Roberson 33/302 2806295 Electrical borehole surveying device 1957-09-17 Ball 33/302 2681657 Apparatus for steam cleaning and liquid cleaning internal-combustion engine cooling systems 1954-06-22 Widess 33/302 2674049 Apparatus for subsurface exploration 1954-04-06 James, Jr. 33/302 2635349 Well-surveying inclinometer 1953-04-21 Green 33/302 2309905 Device for surveying well bores 1943-02-02 Irwin et al. 33/302\n\nPrimary Examiner:\nLittle, Willis\nAttorney, Agent or Firm:\nHaefliger, William W.\nParent Case Data:\nBACKGROUND OF THE INVENTION\n\nThis application is a continuation-in-part of our prior application Ser. No. 351,744, filed Feb. 24, 1982 now U.S. Pat. No. 4,433,491.\n\nClaims:\nWe claim:\n\n1. In mapping or survey apparatus using a single inertial angular rate sensor and a single acceleration sensor, one or both with input axes of sensitivity nominally canted relative to a well or borehole axis, rotated about the borehole axis and both sensors having outputs, the combination comprising\n\n(a) means responsive to the accleration sensor output together with the inertial angular rate sensor output to compute from the rate sensor output the components thereof in the horizontal plane and the vertical plane, and\n\n(b) means to derive azimuth as the Arcsin of the component of sensed inertial angular rate in the horizontal plane normal to the plane containing the borehole axis and the gravity vector divided by the horizontal plane component of the earth's angular velocity vector.\n\n2. The apparatus of claim 1 including\n\n(c) means responsive to the acceleration sensor output to compute the component of the earth's rotation rate in the horizontal plane at its intersection with the vertical plane containing the gravity vector and the borehole axis from the sensed angular rate, and\n\n(d) means to derive azimuth as the Arcos of the component computed in (c) divided by the horizontal plane component of the earth's angular velocity vector.\n\n3. The apparatus of claim 2 including\n\n(e) means to compute azimuth as the Arctan of the argument computed in (b) divided by the argument computed in (c).\n\n4. In mapping or survey apparatus using two axis inertial angular rate sensor means and two-axis acceleration sensor means both with their two input axes of sensitivity nominally normal to a well or borehole axis, both sensors having outputs,\n\n(a) means responsive to acceleration sensor outputs together with the inertial angular rate sensor outputs to compute from the rate sensor outputs the components thereof in the horizontal and vertical plane, and\n\n(b) means to derive azimuth as the Arcsin of the component of sensed inertial angular rate in the horizontal plane normal to the plane containing the borehole axis and the gravity vector divided by the horizontal plane component of the earth's angular velocity vector.\n\n5. The apparatus of claim 4 including\n\n(c) means responsive to the acceleration sensor output to compute the component of the earth's rotation rate in the horizontal plane at its intersection with the vertical plane containing the gravity vector and the borehole axis from the sensed angular rate, and\n\n(d) means to compute azimuth as the Arccos of the component computed in (c) divided by the horizontal plane component at the earth's angular velocity vector.\n\n6. The apparatus of claim 5 including\n\n(e) means to compute azimuth as the Arctan of the argument computed in (b) divided by the argument computed in (c).\n\n7. The combination of claim 1 where the inertial angular rate sensor is replaced by a magnetic field vector sensor, the earth's angular velocity vector is replaced by the earth's magnetic field vector, and magnetic azimuth is computed.\n\n8. The combination of claim 1 wherein the inertial angular rate sensors are replaced by magnetic field vector sensors, the earth's angular velocity vector is replaced by the earth's magnetic field vector, and magnetic azimuth is computed.\n\n9. The apparatus of claim 2 in which one of the sensors has its input axis of sensitivity canted so as to provide sensing of a component along the borehole axis, and employing a canted accelerometer output component along the borehole axis to increase the accuracy of tilt or inclination angle for near horizontal boreholes.\n\n10. The apparatus of claim 9 includes means to compute tilt or inclination as the Arccos of the gravity component along the borehole axis, divided by the known magnitude of the earth's gravity vector.\n\n11. The apparatus of claim 9 including means to compute tilt or azimuth as the Arctan of the gravity component normal to the borehole axis in the vertical plane divided by the component along the borehole axis.\n\n12. The apparatus of claim 9 including means responsive to the acceleration sensor output together with the canted inertial angular rate sensor output component along the borehole axis to compute a second estimate of the component of the earth's rotation rate in the horizontal plane at its intersection with the vertical plane containing the gravity vector and the borehole axis.\n\n13. The apparatus of claim 9 including means to compute azimuth as the Arccos of the component computed in (d) divided by the horizontal plane component of the earth's angular velocity vector.\n\n14. The apparatus of claim 9 including means to compute azimuth as the Arctan of the argument computed in (b) of claim 1 above divided by the argument computed in (d) above.\n\n15. In borehole survey apparatus wherein angular rate sensor means and acceleration sensor means are suspended and effectively rotated in a borehole, at least one of said sensors canted at an angle γ relative to the borehole axis, the angular rate sensor means having amplitude output GA, and rotation related phase output GP, and the acceleration sensor means having amplitude output AA and rotation related phase output AP, there also being means supplying a signal value Ωv proportional to earth's angular rate of rotation, and means supplying a value derived from γ, the improvement which comprises\n\n(a) first means for combining AA, AP, GA, GP, said value derived from γ, and Ωv, to derive a value ψ for borehole azimuth at the level of said sensor means in the borehole.\n\n16. The apparatus of claim 15 including\n\n(b) second means operatively connected with said first means for employing AA modified by said value derived from γ to derive a value φ for borehole tilt from vertical at the level of said sensor means in the borehole.\n\n17. The apparatus of claim 15 wherein said first means includes (c) means responsive to GA modified by said value derived from γ, GP and AP to derive\n\n(i) a first component Ωx of the angular rate sensor output, and\n\n(ii) a second component Ωy of the angular rate sensor output.\n\n18. The apparatus of claim 17 wherein said (c) means to derive Ωx and Ωy includes (d) means responsive to GP and AP to produce a phase angle value α representative of the difference in phase of said GP and AP outputs, (e) means responsive to α to produce sin α and cos α values, (f) means to multiply modified GA and said sin α value to produce Ωy, and (g) means to multiply modified GA and said cos α value to produce Ωx.\n\n19. The apparatus of claim 17 wherein said first means includes (h) means responsive to Ωx, AA modified by said value derived from γ and Ωv to derive a value ΩB ', and (j) means responsive to Ωy and Ω'B to derive said value ψ for borehole azimuth.\n\n20. The apparatus of claim 19 wherein said (h) means includes:\n\n(h1) an arc sin generator responsive to modified AA to generate an output,\n\n(h2) sin and cos generator means responsive to said output of the arc sin generator to generate an output sin φ and an output cos φ,\n\n(h3) multiplier means responsive to sin φ and Ωv to produce a product thereof,\n\n(h4) substractor means responsive to said product and Ωx to obtain a difference value,\n\n(h5) divider means to divide said difference value by said output cos φ to obtain said value Ω'B.\n\n21. The apparatus of either one of claims 19 and 20 wherein said (i) means includes an arc tangent generator responsive to Ωy and ΩB to produce an output proportional to arc tan -Ωy /Ω'B which is representative of azimuth ψ.\n\n22. The apparatus of claim 20 wherein said elements (h1)-(h5) are operatively interconnected.\n\n23. In well bore survey apparatus wherein angular rate sensor means and accelerometer means are located in a borehole and at a cant angle γ relative to the borehole axis, the angular rate sensor means having amplitude output GA and phase output GP, and the accelerometer means having amplitude output AA and phase output AP, there being means providing a value Ωv proportional to earth's angular velocity vector, and means supplying the combination comprising values derived from γ,\n\n(a) means operatively connection to said sensors to be responsive to GA, GP and AP and a value derived from γ to derive a first component Ωx of the angular rate sensor output,\n\n(b) means operatively connected to said sensors to be responsive to GA, GP and AP and a value derived from γ to derive a second component Ωy of the angular rate sensor output,\n\n(c) means operatively connected to said (a) means to be responsive to Ωx, AA and Ωv to derive a value Ω'B, and\n\n(d) means operatively connected to said (b) and (c) means to derive ψ from Ωy and Ω'B\n\nwherein ψ is an azimuth value indicative of the azimuth angle of the borehole relative to the true North at the location of said sensor means.\n\n24. The combination of claim 23 including\n\n(a) means responsive to AA to derive a value φ for borehole tilt at the location of said sensor means in the borehole.\n\n25. The apparatus of either one of claims 15 and 23 including means suspending said rate sensor means and accelerometer sensor means in the borehole at an elevation at which said derivation of ψ is carried out.\n\n26. In borehole survey apparatus wherein magnetic sensor means and acceleration sensor means are suspended and effectively rotated in a borehole and at a cant angle γ relative to the borehole axis, the magnetic sensor means having amplitude output GA and rotation related phase output GP, and the acceleration sensor means having amplitude output AA and rotation related phase output AP, there also being means supplying a signal value Ωv proportional to earth's angular rate of rotation, and means supplying a value derived from γ the improvement which comprises\n\n(a) first means for combining AA, AP, GA, GP said value derived from γ and Ωv to derive a value ψ for borehole azimuth at the level of said sensor means in the borehole.\n\n27. The apparatus of claim 26 including\n\n(b) second means operatively connected with said first means for employing AA modified by said value derived from γ to derive a value φ for borehole tilt from vertical at the level of said sensor means in the borehole.\n\n28. The apparatus of claim 26 wherein said first means includes (c) means responsive to GA modified by said value derived from γ, GP and AP to derive\n\n(i) a first component Ωx of the magnetic sensor output, and\n\n(ii) a second component Ωy of the magnetic sensor output.\n\n29. The apparatus of claim 28 wherein said (c) means to derive Ωx and Ωy includes (d) means responsive to GP and AP to produce a phase angle value α representative of the difference in phase of said modified GP and AP outputs, (e) means responsive α to to produce sin α and cos α values, (f) means to multiply modified GA and said sin α value to produce Ωy, and (g) means to multiply modified GA and said cos α value to produce Ωx.\n\n30. The apparatus of claim 28 wherein said first means includes (h) means responsive to Ωx, modified AA and Ωx to derive a value Ω'B and (j) means responsive to Ωy and Ω'B to derive said value ψ for borehole azimuth.\n\n31. The apparatus of claim 30 wherein said (h) means includes:\n\n(h1) an arc sin generator responsive to modified AA to generate an output,\n\n(h2) sin and cos generator means responsive to said output of the arc sin generator to generate an output sin φ and an output cos φ,\n\n(h3) multiplier means responsive to sin φ and Ωv to produce a product thereof,\n\n(h4) substractor means responsive to said product and Ωx to obtain a difference value,\n\n(h5) divider means to divide said difference value by said output cos φ to obtain said value Ω'B.\n\n32. The apparatus of one of claims 15, 23 and 26 wherein said value derived from γ is a value for cos γ and said first means (a) includes means for dividing GA by said cos γ value to derive a modified value of GA, and for dividing AA by said cos γ value to derive a modified vlaue of AA.\n\n33. The apparatus of claim 16 wherein said angular rate sensor has a steady output component GAV and said acceleration sensor means has a steady output component AAV, and said first means receives said steady outputs for combination with AA, AP, GA, GP, values derived from γ, and Ωv to derive said value ψ.\n\n34. The apparatus of claim 33 wherein said values derived from γ correspond to cos γ and sin γ, and said first means includes (c) means responsive to GA divided by cos γ to produce a value GA', GP and AP to derive\n\n(i) a primary component Ωy of the angular rate sensor output.\n\n35. The apparatus of claim 34 wherein said (c) means to derive Ωy includes (d) means responsive to GP and AP to produce a phase angle value α representative of the difference in phase of said GP and AP outputs, (e) means responsive to α to produce sin α and cos α values, and (f) means to multiply GA' and said sin α value to produce Ωy.\n\n36. The apparatus of claim 35 wherein said first means includes (g) means responsive to GAV divided by sin γ to produce a value Ωz, and including (h) means responsive to said values AA and AAV, and said cos α and sin α values and v to produce a value Ωy cos φ for addition to Ωz and subsequent division by sin φ to produce a value Ω'B.\n\n37. The apparatus of claim 36 including (i) means responsive to Ωy and Ω'B to produce said value ψ.\n\n38. The apparatus of claim 36 wherein φ is the borehole tilt angle, and said (h) means includes (j) means responsive to AA, AAV, cos α and sin α to generate φ.\n\n39. The apparatus of claim 15 wherein the angular rate sensor is canted at said angle γ.\n\n40. The apparatus of claim 15 wherein the acceleration sensor means is canted at said angle γ.\n\n41. Apparatus as defined in claim 9 wherein the inertial angular rate sensor is replaced by a magnetic field sensor, the earth's angular velocity vector is replaced by the earth's magnetic field vector and magnetic azimuth is computed.\n\nDescription:\n\nThis invention relates generally to mapping or survey apparatus and methods, and more particularly concerns derivation of the azimuth output indications for such apparatus from the outputs or output indications of either an inertial angular rate vector sensor (or sensors) and an acceleration vector sensor (or sensors), or a magnetic field vector sensor (or sensors), and from the outputs of an acceleration vector sensor (or sensors).\n\nThis invention also relates to methods used to compute tilt and azimuth from outputs of canted sensors which have a component or components of their input axis or axes of sensitivity along the axis of rotation, for survey apparatus having an axis or axes of rotation for one or more of the sensors.\n\nU.S. Pat. No. 3,753,296 describes the use of a single inertial angular rate sensor, or \"rate-of-turn gyroscope\", and a single acceleration sensor, both having their input axes of sensitivity nominally normal to the direction of travel in a borehole and parallel to each other for survey in a well or borehole. In this case, both sensors are rotated about an axis parallel to the borehole by either the carrying structure and container or by a rotatable frame internal to the survey tool. U.S. Pat. No. 4,199,869 describes the use of one or two dual axis inertial angular rate sensors in combination with a dual axis acceleration sensor for survey in a well or borehole. Again in this case, the sensors are rotated about an axis parallel to the borehole by either the carrying structure or by a rotatable frame internal to the survey tool. U.S. Pat. No. 4,244,116 describes the of a one dual axis inertial angular rate sensor having its spin axis parallel to the borehole axis and one dual axis accelerometer for survey in a well of borehole. In this case no provision is made for rotation of the sensors about the borehole axis. U.S. patent application Ser. No. 338,261, filed Jan. 11, 1982 describes the use of one or more magnetic field vector sensors in combination with one or more acceleration sensors for survey with respect to the earth's magnetic field vector in a way related to the sensors of U.S. Pat. Nos. 3,753,296 and 4,199,869 which survey with respect to the earth's inertial angular rate vector.\n\nThe referenced patents and application describe the sensing equipments and show provisions to compute the output desired azimuth indication, but none of them show or teach the method and means described herein for obtaining the desired output, nor do they show the essential use of the output of the acceleration sensor (or sensors) to resolve the output (or outputs) of either the inertial angular rate sensor (or sensors) or the magnetic field sensor (or sensors) into a known coordinate system. For example, U.S. Pat. No. 4,244,116 shows a computation of: ##EQU1## Where E=Vector of the earth rotation\n\nΩz =Component of E along the borehole\n\nΩx, Ωy =Gyro outputs normal to the borehole,\n\nand states \"the measurement Ωx and Ωy and the calculation of Ωz give then the azimuth of the drilling line\". This is in general not true since Ωx and Ωy are known only to be perpendicular to the drilling line, but are not known in a known earth fixed coordinate set.\n\nIt may be shown that the apparatus of the above cited patents and applications, and with respect to described methods of computation, have associated geometric regions in which poor accuracy can result. Thus, for the determination of tilt or inclination, poor accuracy results when the plane containing the accelerometer input axis approaches the gravity vector. There is then only a small angle between the plane containing the accelerometer input axis (or axes) and the vector to be sensed, this condition being reached whenever the borehole axis approaches horizontal. For the determination of azimuth, the region of poor accuracy is that in which the plane containing the input sensitive axis (or axes) of the acceleration vector sensor and the direction reference vector (either earth rotation or earth magnetic field) approaches parallelism to the plane containing both the earth's gravity vector and the direction reference vector. When using an angular rate sensor and true azimuth is to be computed, this region of poor accuracy exists for a borehole axis that is near true East-true West and near horizontal. When using a magnetic field sensor and magnetic azimuth is to be computed, the region of poor accuracy exists for a borehole axis near magnetic East-magnetic West and near horizontal. In these regions of poor accuracy, small sensor errors will lead to large errors in the desired inclination and/or azimuth.\n\nTo overcome these regions of poor accuracy and avoid large errors in such cases, U.S. Pat. Nos. 4,265,028 and 4,197,654 and U.S. patent application Ser. No. 338,261 show that a single vector sensor device can be used to obtain knowledge of three orthogonal components of a reference vector quantity. The method shown in these cited patents and application is that of rotating a sensor about an axis of rotation that has the sensor input sensitive axis canted or inclined relative to a normal to the rotation axis by some angle γ. U.S. Pat. No. 4,265,028 describes the use of a canted accelerometer to measure three orthogonal components of the earth's gravity vector at a fixed (but moveable) location in a well or borehole. U.S. Pat. No. 4,197,654 describes the use of a canted gyroscope (or angular rate sensing device) to measure three orthogonal components of the earth's angular velocity vector. The referenced patent application describes the use of a canted magnetic field sensing device to measure three orthogonal components of the earth's magnetic field vector. Such provision of a cant to the sensor input axis of sensitivity provides a component of the sensed vector along the rotation or borehole axis and this is sufficient to eliminate the geometric regions of poor accuracy which all apparatus having sensing axes only normal to the borehole will have. The present invention discloses apparatus and methods to use this third component of sensed data to be combined with the computation previously described in parent application Ser. No. 351,744 for deriving inclination and azimuth.\n\nSUMMARY OF THE INVENTION\n\nIt is a major purpose of this invention to provide method and means to use data from angular rate and acceleration sensors in a mapping or survey tool, one or more of the sensors being canted, to determine the orientation of the inertial angular rate or magnetic field vector sensor (or sensors) with respect to a known earth fixed coodinate set so that correct azimuth determination can be made. It is a second purpose of this invention to provide method and means for azimuth determination in a completely explicit non-ambiguous manner once the sensor data has been resolved to a known earth fixed coordinate set.\n\nThe determination of azimuth with respect of either the earth's inertial angular velocity vector (so called true azimuth) or earth's magnetic field vector (so called magnetic azimuth) requires that one first determine at least one (but for complete all azimuths two orthogonal) component of the desired reference vector (angular velocity or magnetic) in a plane parallel to the earth's surface and in a known orientation to the desired unknown azimuth direction. In mapping or survey apparatus of the types cited as previously used in wells or boreholes, the reference direction vector sensors, either inertial angular rate or magnetic, provide outputs proportional to the vector dot product of vectors along their input sensitive axes and the reference vectors. Such outputs of themselves provide no means to know the components of the reference direction vector in a horizontal plane. However, an acceleration sensor (sensors) at a fixed location in the well or borehole provides direct knowledge of the relation of its input axis of sensitivity with respect to the local gravity vector which by definition is normal to the horizontal plane. Since the orientation of the input axis of sensitivity of the reference direction vector sensor, either angular rate or magnetic, is known with respect to the input axis of sensitivity of the acceleration sensor, the output of the acceleration sensor (or sensors) thus may be used to process the output (or outputs) of the reference direction vector sensor (or sensors) to determine one or more components in the horizontal plane.\n\nWhen the reference direction vector sensor is canted to sense a component along the borehole axis, the third component can be used in computation of azimuth along with the previously stated components resolvable into the horizontal plane. For example, when the acceleration sensor is canted, the component of gravity along the borehole axis may be used in computation with the gravity component in the vertical plane to compute improved accuracy values for the tilt or inclination angle.\n\nAccordingly, it is a major object of the invention to provide borehole survey apparatus wherein angular rate sensor means and acceleration sensor means are suspended and effectively rotated in a borehole, at least one of the sensors may be canted at one angle γ relative to the borehole axis, the angular rate sensor means having amplitude output GA and rotation related phase output GP, and the acceleration sensor means having amplitude output AA and rotation related phase output AP, there also being means supplying a signal vaue Ωv proportional to the local vertical component of the earth's angular rate of rotation, and there being means supplying a value derived from γ, the improvement which comprises\n\n(a) first means for combining AA, AP, GA, GP, said value derived from γ, and Ωv to derive a value ψ for borehole azimuth at the level of the sensor means in the borehole.\n\nIn addition, the invention provides means operatively connected with said first means for employing AA modified by the value derived from γ to derive a value θ for borehole tilt from vertical at the level of said sensor means in the borehole.\n\nThe basis method of the invention involves the method of borehole mapping or surveying typically using a single angular rate sensor and a single acceleration sensor, both with input axis of sensitivity, one or both sensors being typically canted, the sensors being effectively rotated about the borehole axis, the sensors having outputs.\n\nThe outputs of the angular rate sensor and the acceleration sensor are typically employed to derive, from the rate sensor, two or three components respectively in a horizontal plane, one normal to the plane containing the borehole axis and the gravity vector, and the other two or three in that plane, borehole azimuth being derived form the components in a horizontal plane. Three components are formed when the sensors are canted.\n\nThese and other objects and advantages of the invention, as well as the details of an illustrative embodiment, will be more fully understood from the following specification and drawings, in which:\n\nDRAWING DESCRIPTION\n\nFIG. 1 is a geometrical depiction of a reference coordinate system established at the start of borehole drilling;\n\nFIG. 1a relates the FIG. 1 co-ordinate system to an instrument level co-ordinate system in a borehole;\n\nFIGS. 2 and 2a show plots of single axis accelerometer and gyroscope outputs vs instrument rotation angle;\n\nFIG. 3 is a geometrical showing of vector relationships in a borehole;\n\nFIG. 4 is a circuit block diagram;\n\nFIG. 5 is a coordinate system diagram;\n\nFIG. 6 is a circuit block diagram;\n\nFIG. 7 shows instrumentation in a borehole (single axis angular rate sensor, and single axis accelerometer);\n\nFIG. 8 shows instrumentation in a borehole (dual axis angular rate sensor, and dual axis accelerometer);\n\nFIG. 9 is an elevation taken in section to show one form of instrumentation employing the invention;\n\nFIG. 10 is an elevation showing use of the FIG. 9 instrumentation in multiple modes, in a borehole;\n\nFIG. 11 is a vertical section showing further details of the FIG. 9 apparatus as used in a borehole; and\n\nFIG. 12 is a circuit block diagram.\n\nDETAILED DESCRIPTION\n\nReferring first to FIG. 9, a carrier such as elongated housing 10 is movable in a borehole indicated at 11, the hole being cased at 11a. Means such as a cable to travel the carrier lengthwise in the hole is indicated at 12. A motor or other manipulatory drive means 13 is carried by and within the carrier, and its rotary output shaft 14 is shown as connected at 15 to an angular rate sensor means 16. The shaft may be extended at 14a, 14b and 14c for connection to first acceleration sensor means 17, second acceleration sensor means 18, and a resolver 19. The accelerometers 17 and 18 can together be considered as means for sensing tilt. These devices have terminals 16a - - - 19a connected via suitable slip rings with circuitry indicated at 29 carried within the carrier (or at the well surface, if desired).\n\nThe apparatus operates for example as described in U.S. Pat. No. 3,753,296 and as described above to determine the azimuthal direction of tilt of the borehole at a first location in the borehole. See for example first location indicated at 27 in FIG. 2. Other U.S. patents describing such operation are U.S. Pat. Nos. 4,199,869, 4,192,077 and 4,197,654. During such operation, the motor 13 rotates the sensor 16 and the accelerometers either continuously, or incrementally.\n\nThe angular rate sensor 16 may for example take the form of one or more of the following known devices, but is not limited to them:\n\n1. Single degree of freedom rate gyroscope\n\n2. Tuned rotor rate gyroscope\n\n3. Two axis rate gyroscope\n\n4. Nuclear spin rate gyroscope\n\n5. Sonic rate gyroscope\n\n6. Vibrating rate gyroscope\n\n7. Jet stream rate gyroscope\n\n8. Rotating angular accelerometer\n\n9. Integrating angular accelerometer\n\n10. Differential position gyroscopes and platforms\n\n11. Laser gyroscope\n\n12. Combination rate gyroscope and linear accelerometer\n\nEach such device may be characterized as having a \"sensitive\" axis, (or axes) which is the axis about which rotation occurs to produce an output which is a measure of rate-of-turn, or angular rate ω. That value may have components ω1, ω2 and ω3 in a three axis co-ordinate system. The sensitive axis may be generally normal to the axis 20 of instrument travel in the borehole, or canted at an angle γ.\n\nThe acceleration sensor means 17 may for example take the form of one or more of the following known devices; however, the term \"acceleration sensor means\" is not limited to such devices:\n\n1. one or more single axis accelerometers\n\n2. one or more dual axis accelerometers\n\n3. one or more triple axis accelerometers\n\nExamples of acceleration sensors include the accelerometers disclosed in U.S. Pat. Nos. 3,753,296 and 4,199,869, having the functions disclosed therein. Such sensors may be supported to be orthogonal to the carrier axis. They may be stationary or carouseled, or may be otherwise manipulated, to enhance accuracy and/or gain an added axis or axes of sensitivity. The axis of sensitivity is the axis along which acceleration measurement occurs.\n\nFIG. 11 shows in detail dual input axis rate sensor means and dual output axis accelerometer means, and associated surface apparatus. In FIG. 11, well tubing 110 extends downwardly in a well 111, which may or may not be cased. Extending within the tubing is a well mapping instrument or apparatus 112 for determining the direction of tilt, from vertical, of the well or borehole. Such apparatus may readily be traveled up and down in the well, as by lifting and lowering of a cable 113 attached to the top 114 of the instrument. The upper end of the cable is turned at 115 and spooled at 116, where a suitable metter 117 may record the length of cable extending downwardly in the well, for logging purposes.\n\nThe apparatus 112 is shown to include a generally vertically elongated tubular housing or carried 118 of diameter less than that of the tubing bore, so that well fluid in the tubing may readily pass, relatively, the instrument as it is lowered in the tubing. Also, the lower terminal of the housing may be tapered at 119, for assisting downward travel or penetration of the instrument through well liquid in the tubing. The carrier 118 supports first and second angular sensors such as a rate gyroscopes G1 and G2, and accelerometers 120 and 121, and drive means 122 to rotate the latter, for travel lengthwise in the well. Bowed springs 170 on the carrier center it in the tubing 110.\n\nThe drive means 122 may include an electric motor and speed reducer functioning to rotate a shaft 123 relatively slowly about a common axis 124 which is generally parallel to the length axis of the tubular carrier, i.e. axis 124 is vertical when the instrument is vertical, and axis 124 is tilted at the same angle form vertical as is the instrument when the latter bears sidewardly against the bore of the tubing 110 when such tubing assumes the same tilt angle due to borehole tilt from vertical. Merely as illustrative, for the continuous rotation case, the rate of rotation of shaft 124 may be within the range 0.5 RPM to 5 RPM. The motor and housing may be considered as within the scope of means to support and rotate the gyroscope and accelerometers.\n\nDue to rotation of the shaft 123, and lower extensions 123a, 123b and 123c thereof, the frames 125 and 225 of the gyroscopes and the frames 126 and 226 of the accelerometers are typically all rotated simultaneously about axis 124, within and relative to the sealed housing 118. The signal outputs of the gyroscopes and accelerometers are transmitted via terminals at suitable slip ring structures 125a, 225a, 126a and 226a, and via cables 127, 127a, 128 and 128a, to the processing circuitry at 129 within the instrument, such circuitry for example including that to be described, and multiplexing means if desired. The multiplexed or nonmultiplexed output from such circuitry is transmitted via a lead in cable 113 to a surface recorder, as for example include pens 131-134 of a strip chart recorder 135, whose advancement may be synchronized with the lowering of the instrument in the well. The drivers 131a - - - 134a for recorder pens 131-134 are calibrated to indicate borehole azimuth, degree of tilt and depth, respectively, and another strip chart indicating borehole depth along its length may be employed, if desired. The recorder can be located at the instrument for subsequent retrieval and read-out after the instrument is pulled from the hole.\n\nThe angular rate sensor 16 may take the form of gyroscope G1 or G2, or their combination, as described in U.S. Pat. No. 4,199,869. Accelerometers 126 and 226 correspond to 17 and 18 in FIG. 9.\n\nConsider now a reference coordinate system is established at the start of the borehole such that X is parallel to the earth surface and North, Y is parallel to the earth surface and East, and Z is perpendicular to the earth surface and Down. The starting point is at latitude λ and FIG. 1 shows the basic geometry.\n\nFrom this starting reference, the bore axis is defined as rotated through an azimuth angle ψ clockwise about Z, followed by a rotation φ about the new Y axis to obtain a bore axis reference coordinate set in the bore such that z is downward along the bore axis, y is parallel to the earth surface, and x lies perpendicular to y and z. Also, as will be seen, x is in the vertical plane containing the gravity vector and the borehole axis z. See also FIG. 1a.\n\nIt may be shown that the direction cosine matrix relating this bore axis reference set to the starting reference set is as shown below:\n\n ______________________________________ co-ord set at surface -- --X -- --Y Z ______________________________________ co-ord. set x CψCφSψCφ-Sφin Bore y -Sψ Cψ 0Hole z CψXφSψSφCφ ______________________________________\n\nIn the above C represents Cosine, S represents Sine.\n\nThere is no direct way to measure all three direction cosines relating z to the fixed (starting) reference set. However, gyroscopic and acceleration sensing devices can be used to sense quantities from which the required coefficients can be calculated.\n\nThe earth rate rotation vector, Ω, in reference coordinates is ##EQU2## where ΩH =ΩCλ, the horizontal compoment,\n\nΩv =ΩSλ, the vertical component,\n\nand the earth's gravity vector, g, in reference coordinates is ##EQU3##\n\nIn the above expression the symbol 1n is a unit vector in the N direction.\n\nThe components of Ω and gin the bore axis reference set can be found by forming the dot products 1x Ω, 1y Ω,1z Ω and 1x g, 1y g, and 1z g The results of these operations are: ##EQU4##\n\nIdeally, three accelerometers and three gyro sensing axes could determine all of the required information with no ambiguities or unusual sensitivities other than the classical and well known increased sensitivity to gyro error as latitude increases toward the polar axis.\n\nAs shown by the earlier cited patents, to reduce the size and system complexity, sufficient information may be obtained by either a single axis gyro and single axis accelerometer rotated such that their input axes are swept about the x, y plane (normal to the bore axis) or by a two axis gyro and two axis accelerometer having their input axes in the x, y plane. FIG. 7 shows such a single axis gyro G and single axis accelerometer A (see also 16 and 17 in FIG. 9); and FIG. 8 shows such a two axis gyro G1 and G2 and two axis accelerometer A1 and A2 (see also FIG. 11).\n\nIf single axis instruments are used, the plot of their outputs vs rotation angle will appear as (in the absence of sensor errors) shown in FIG. 2.\n\nIn this figure, it is evident that the accelerometer output is a sinusoid having its peak output at the point where the input sensitive axis is parallel to the x axis, where x was as previously defined to be in the vertical plane containing the gravity vector and the borehole axis. If the phase angle α, between the accelerometer peak output and the gyro peak output is measured by suitable signal processing, it is then possible to compute Ωx, the component of the earth rotation vector in the vertical plane containing the borehole axis and the earth's gravity vector, and Ωy, the component of the earth rotation vector in the horizontal plane (normal to the gravity vector). Such components are: ##EQU5## From the previously shown mechanization that Ωy was equivalent to: Ωy =-ΩH sin ψ (11)\n\nit would be possible to compute azimuth as: ##EQU6## using the value of Ωy computed from the gyro output and the phase angle between the gyroscope output and the accelerometer output. This displays the essential usage of the accelerometer output to determine a component of the earth's inertial angular rate vector in a horizontal plane.\n\nThe method shown above is suitable except that for azimuths near 90° (East) or 270° (West) the arcsin function provides very poor sensitivity since the rate of change of sin ψ with ψ is very low in these regions. This would lead to large errors in azimuth from small sensor errors. It is, therefore, desirable to find another component of the earth's inertial angular rotation vector in the horizontal plane. FIG. 3 shows a side view of the borehole along the previously defined y axis.\n\nThe value of the measured component Ωx is by inspection: Ωxv sin φ+ΩB cos φ(13)\n\nSince Ωx has been determined from the gyro output and the accelerometer to gyro phase angle, and since φ can be determined from the amplitude of the accelerometer signal as: ##EQU7## Then ##EQU8## But as previously defined; ##EQU9## so that ##EQU10## From this it is possible to compute: ##EQU11##\n\nThis mechanization also provides a value of ψ and since it is based on a arccos vs the previously cited arcsin function, the region of poor sensitivity is near azimuth of 0° (North) or 180° (South). This again shows the essential use of the accelerometer output to properly resolve the gyro output into the horizontal plane. If one desires, these two functions can be combined into one which has no regions of poor sensitivity. Such a form is: ##EQU12##\n\nFIG. 4 shows a complete block diagram of the described mechanization. As the combination of sensing devices is rotated about its rotation axis in a borehole, both the inertial angular rate sensing and acceleration sensing devices will produce variable output indications proportional to the vector dot product of a unit vector along the respective input axis and the local earth rotation vector and gravity vector respectively. For continuous rotation operation at a fixed location in the borehole these signals will be sinusoidal in nature. For discrete step rotation, the sensor outputs will be just the equivalent of sampling points on the above mentioned sinusoidal signals. Thus, from a knowledge of sample point amplitudes and position along the sinusoid, the character of an equivalent sinusoid in amplitude and phase may be determined. For either continuous rotation or discrete positioning, the quantities that must be determined are the gyro signal amplitude GA (GAMPLITUDE), the accelerometer signal amplitude AA (AAMPLITUDE), and the phase angle between the peak values of these two signals, α. FIG. 4 shows the two sensor signals, after required scaling, and a reference time or angle signal as inputs to the two blocks labeled \"Sinusoid Amplitude and Phase.\" Each of these blocks finds the amplitude of the input sinusoid and the phase angle between the input signal and the reference derived from the rotation drive function. The outputs of the upper block are gyro amplitude and phase, labelled GA (GAMPLITUDE) and GP (GPHASE). The outputs of the lower block are accelerometer amplitude and phase, labelled AA (AMPLITUDE) and AP (APHASE). These amplitude functions are then directly input to subsequent elements and the required phase difference α, is shown, GPHASE minus APHASE.\n\nIf a two axis gyro and two axis accelerometer are used, allowance must be made for unknown rotation β about the bore axis. FIG. 5 shows a view looking at the x, y plane from the positive z side.\n\nThe sensed accelerometer outputs in terms of the gravity components gx =g sin φ and gy =0 are: g1 =gx cos β (21) g2 =-gx sin β (22)\n\nThe sensed angular rates for the two gyro outputs are: Ω1x cos β+Ωy sin β(23) Ω2 =-Ωx sin β+Ωy cos β(24)\n\nFrom the accelerometer data ##EQU13##\n\nUsing the value of β determined above from the accelerometer data and the sensed outputs of the two gyros, two components of the gyro output Ωx and Ωy in a known coordinate set may be computed as: Ωx1 cos β-Ω2 sin β(27) Ωy1 sin β+Ω2 cos β(28)\n\nThese values are then identical to the AAMPLITUDE, Ωx, and Ωy previously described for the rotated single axis case and the value of azimuth may be determined in the same way. FIG. 6 shows a complete block diagram of circuitry to perform this determination. In FIG. 6, the differences in signal processing for the two angular rate inputs and the two acceleration inputs compared to one rotated sensor of each kind are as shown in FIG. 4. The portion to the left of the dotted line in FIG. 6 would be substituted for the corresponding portion of FIG. 4. Note that since there are two nominally orthogonal signals of each kind, no reference time or angle input is required. Again, the essential use of acceleration sensor outputs to resolve the angular rate sensor data to a known coordinate system is shown.\n\nAlthough the previous description has used the earth's inertial angular rate vector as the reference direction vector, the earth's magnetic field vector can be used as a reference if magnetic vector sensors replace angular velocity sensors, as in the drawings. All that is necessary is to substitute MH and MV for ΩH and Ωv, Mx and My for Ωx and Ωy, and MB for ΩB. In these formulations the various components of the earth's magnetic field vector are used and the resulting azimuth is the magnetic azimuth measured with respect to the horizontal component of the earth's magnetic field. The same essential dependence on the acceleration sensors, for the resolution of the magnetic sensor outputs into a horizontal plane, is evident in this usage.\n\nReferring now in detail to FIG. 4, the angular rate sensor (gyroscope) amplitude and phase outputs are indicated at GA and GP. These are typically in voltage signal form. Similarly, the accelerometer amplitude and phase outputs are indicated at AA and AP. A synchronizing reference time or angle signal is supplied at 150 to the amplitude and phase detectors 148 and 149 which respond to the gyroscope and accelerometer outputs to produce GA, GP, AP and AA. Means is also provided to supply at 151 a signal corresponding to earth's rotation rate Ω, and to supply at 152 a signal corresponding to the borehole latitude λ. A sin/cos generator 153 operates on signal 152 to produce the output sin λ at 154. The latter and signal 151 are supplied to multiplier 155 whose output Ω sin λ=Ωv appears on lead 156.\n\nIn accordance with the invention, (a) first means is provided for combining (or operating upon) AA, AP, GA, GP and Ωv to derive a value ψ for borehole azimuth at the level of the sensors suspended in the borehole. The azimuth signal ψ appears on lead 157 at the right of the circuitry shown. In addition, (b) second means is operatively connected with the referenced first means for employing AA to derive a signal value φ representative of borehole tilt from vertical, at the level of the sensor means in the borehole.\n\nMore specifically, such (a) first means include (c) means responsive to GA, GP and AP to derive:\n\n(i) a first component Ωx of the angular rate sensor output, and\n\n(ii) a second component Ωy of the angular rate sensor output. See Ωv on lead 158, and Ωy on lead 159. Such (c) means may typically include:\n\n(d) means responsive to GP and AP to produce a phase angle value or signal α representative of the difference in phase of the GP and AP signals (see for example the subtractor 159 connected with the output sides of 148 and 149, the subtractor output α appearing on lead 160),\n\n(e) means responsive to α to produce signal values sin α and cos α (see for example the sin/cos generator 161 whose input side is connected with lead 160, and whose outputs sin α and cos α appear on leads 162 ad 163),\n\n(f) means responsive to GA and cos α to multiply same and produce the signal value Ωx (see for example the multiplier 164 whose inputs are connected with GA lead 165 and cos α lead 163),\n\n(g) means responsive to GA and sin α to multiply same and produce the signal value Ωy (see for example multiplier 166 whose inputs are connected with the GA input and with sin α lead 112).\n\nThe (a) first means also includes (h) means responsive to Ωx, AA and Ωv to derive a value ΩB and (j) means responsive to Ωy and ΩB to derive the said value ψ for borehole azimuth. For example, the (h) means may include:\n\n(h1) an arcsin generator 170 responsive to AA (supplied on lead 171 from detector 149) to generate output at 172,\n\n(h2) sin/cos generator 173 connected with output 172 to produce output sin φ, on lead 174 and output cos φ on lead 175,\n\n(h3) multiplier 176 responsive to sin φ on lead 174 and Ωv on lead 156 to produce their product on output lead 177,\n\n(h4) subtractor 178 connected with leads 177 and 158 to produce the value (Ωxv sin φ) on lead 179\n\n(h5) a divider 180 to divide the values on leads 179 and 175 and produce the desired values ΩB on lead 181.\n\nThe (i) means referred to above is shown in FIG. 4 to include an arc tangent generator 182 connected with leads 159 and 181 to be responsive to Ωy and ΩB to produce the ψ output proportional to arctan ##EQU14## The tilt output signal φ is produced on lead 184 connected with the output of arcsin generator 170.\n\nFIG. 6 shows similar connections and circuit elements responsive to inputs Ω1 and Ω2 from two gyroscopes (or dual axis gyroscope) and inputs 1 and 2 from two accelerometers (or from a dual axis accelerometer), to produce the values Ωy, Ωx and AA, which are then processed as in FIG. 4. See also FIG. 8.\n\nThe above operational devices as at 148, 149, 159, 178, 155, 164, 166, 180, 153, 161, 173, 170 and 182 may be analog or digital devices, or combinations thereof.\n\nReferring to FIG. 10, the determinations of azimuth ψ and tilt φ are carried out at multiple locations in a borehole, as at 27, 27' and 27\"; and they may be carried out at each such location during cessation of elevation or lowering by operation of cable 12, or during such elevation or lowering.\n\nWhen canted sensors are used, the computations are modified to use the components of the reference direction and gravity vectors along the borehole axis. For the case of single axis sensors, FIG. 2 would be modified to incorporate steady outputs for both the gyro and accelerometer. These appear at 210 and 211 in FIG. 2a. Each steady component, in the absence of sensor errors, is proportional to the component along the borehole rotation axis. Also, the amplitudes of the sinusoids would be reduced. Specifically, if the canted angle is designated as gamma, γ, then as shown in FIG. 2a, ##EQU15##\n\nIf the values of AAMPLITUDE and GAMPLITUDE obtained from the canted sensors are modified by a value derived from γ, as for example divided by cos γ, then the previously discussed values gx and √Ωx2y2 are obtained and computations can proceed as previously described to compute the resolved gyro components Ωx and Ωz, the tilt or inclination angle φ, and the azimuth ψ by any of the three indicated methods. See for example in FIG. 4 the optional provision of the cant angle γ signal source 190, the cos γ generator 191, divider 192 to divide GA by cos γ, and the quotient GA' (which is the modified GA) on lead 193. Associated switches are shown. Similarly, in FIG. 4, AA is divided at 194 by cos γ to produce modified AA' on lead 195 to produce AA' (the modified AA).\n\nWhen the values of AAVERAGE and GAVERAGE are divided by sin γ, then gz and Ωz are obtained. Since as previously shown: gx =g sin φ (33) gz =-g cos φ (34)\n\na value of φ may be computed either as: ##EQU16##\n\nEither of these methods or values is free of the reduced sensitivity for near horizontal boreholes of the previously (FIG. 4) shown: ##EQU17##\n\nOf the two new forms shown, the Arctan form of Equation (36) has the additional benefit that it is not in error due to either a scale factor error, or an error in the knowledge of gravity. Also, as previously shown: ΩzH cos ψ sin φ-Ωv cos φ(8)\n\nReturning to FIG. 3, it is again possible to compute the shown vector ΩB from Ωz rather than from Ωx. By inspection Ωz =Ω'B sin φ-Ωv cos φ (38)\n\nSolving for Ω'B : ##EQU18## But as previously defined, ΩzH cos ψ sin φ-Ωv cos φ(40)\n\nSo that: Ω'BH cos ψ (41)\n\nFrom this it would be possible to compute azimuth as: ##EQU19## or as previously shown, using Ωy and Ω'B, ##EQU20## FIG. 12 shows a block diagram of electrical apparatus for performing these computations.\n\nReferring to detail to FIG. 12, elements thereof also found in FIG. 4 carry the same identifying numerals. Also shown is means supplying a value derived from γ, the cant angle of the gyroscope and accelerometer relative to the borehole axis. Such means is shown for example to include a voltage at 210 corresponding to cant angle γ, and a sine/cosine generator 211 to produce sine γ and cosine γ outputs at 212 and 213. These are, of course, trigonometric values drived from γ.\n\nIn accordance with the invention, (a) first means is provided for combining AA, AP, GA, GP, a value or values derived from γ, (i.e. sin γ and cos γ, for example) and Ωv to derive a value ψ for borehole azimuth at the level of the sensor means in the borehole. In addition, (b) second means is operatively connected with the first means for employing AA modified by a value derived from γ to derive a value φ for borehole tilt from vertical at the sensor level in the hole.\n\nMore specifically, the first means include (c) means such as divider 215 responsive to GA and cos γ to divide GA by cos γ and produce a value GA', indicated at lead 216. Also, the first means (c) typically is responsive to GP and AP to derive a primary component Ωy of the angular rate sensor output. See for example sin α generator 161 and multiplier 217 for multiplying sin α and GA' to produce Ωy, to be used in the derivation of ψ.\n\nThe first means is also shown as including means responsive to GAV output 217 from device 148, divided by sin γ at 218 to produce a value Ωz on lead 217. Multiplier 220 then multiplies Ωz by the value Ωv cos φ to produce an output on lead 221, the latter then being divided by sin φ at 222 to produce Ω'B. Generator 223 then responds to Ω'B and Ωy to produce ψ by computing arctan Ωy /Ω'B, as shown in FIG. 12.\n\nIn the above, φ is the borehole tilt angle, and is computed by the devices shown. Thus, output AA at 171 is divided at 230 by cos α to produce gx ; AAV output from device 149 is divided by sin γ at 231 to produce output gz, and arctangent generator 224 receives gx and gy to compute arctan -gx /gz, which produces φ. Sin/cosine generator 232 receives φ to produce sin φ at 233, and cos φ at 234, for use as described above. Multiplier 235 multiplies cos φ and Ωv to produce Ωv cos φ on line 237.\n\nIt is of course possible to have more than one axis of sensing of each kind, canted. If, in the previously described approach using a two axis gyro and a two axis accelerometer, both sensor axes of each kind are canted, then two independent estimates of the component of the sensed reference vector along the borehole axis are obtained.\n\nAlthough the previously described computations used the example of a canted accelerometer and a canted angular rate sensor, it is clear that either sensor could be canted alone or that a magnetic field sensor could be substituted for the angular rate sensor if magnetic azimuth were desired."
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https://www.datasciencemadesimple.com/multinomial-function-in-excel/ | [
"# Multinomial Function in Excel\n\nMultinomial function in Excel calculates the ratio of the factorial of a sum of supplied values to the product of factorials of those values.\n\ni.e. Multinomial (x1,x2,x3….xn) = (x1+x2+x3….+xn) / x1!x2!x3!….xn!\n\n#### Syntax of Multinomial Function in Excel\n\nMultinomial(number1, [number2], …)\n\nWhere number is the Positive number that you want to calculate the Multinomial Function of.\n\n#### Example of Multinomial Function in Excel\n\n##### Formula",
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"In Row number 1 values 3,4and 5 are passed as input to multinomial functions so it calculates as follows (3+4+5)! / (3!*4!*5!) so the Resultant value will be 27720.\n\nSimilarly multinomial Function calculates for other values too\n\nSo the result will be\n\n##### Result",
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"#NUM! Error occurs if the supplied number argument is negative.",
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http://math.eretrandre.org/tetrationforum/showthread.php?tid=1043&pid=8176 | [
"• 0 Vote(s) - 0 Average\n• 1\n• 2\n• 3\n• 4\n• 5\n Derivative of exp^[1/2] at the fixed point?",
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"sheldonison",
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"Long Time Fellow",
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"Posts: 679 Threads: 24 Joined: Oct 2008 12/23/2015, 04:39 PM (This post was last modified: 12/23/2015, 04:43 PM by sheldonison.) Let exp^[1/2](x) be the half iterate of exp(x), generated by Kneser's Riemann mapping. Is the derivative defined of exp^[1/2](L) defined at the fixed point of L~=0.318132 + 1.33724i, where exp(L)=L? From an old 2010 post#3..post#8 http://math.eretrandre.org/tetrationforu...4&pid=5400, we see that L is the closest singularity to the real axis for the half iterate, but it is a mild singularity, and that exp^[1/2](L)=L, so that exp^[1/2](L) is continuous at L. Is the derivative is also continuous at this singularity, and if so, then what is its value? If the derivative is continuous at the singularity, how many of the higher derivatives are also continuous? - Sheldon",
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"sheldonison",
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"Long Time Fellow",
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"Posts: 679 Threads: 24 Joined: Oct 2008 12/24/2015, 03:25 AM (This post was last modified: 12/29/2015, 11:02 PM by sheldonison.) (12/23/2015, 04:39 PM)sheldonison Wrote: Is the derivative is also continuous at this singularity, and if so, then what is its value? If the derivative is continuous at the singularity, how many of the higher derivatives are also continuous? In particular, there is a formal half iterate that is not real valued that can be developed at the fixed point L. And the formal half iterate begins with Numerical experiments suggest that the derivative of Knesser's real valued half iterate is continuous at L, and the first derivative at the singularity at L is , and that the 2nd derivative may also match the formal half iterate. - Sheldon",
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"sheldonison",
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"Long Time Fellow",
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"Posts: 679 Threads: 24 Joined: Oct 2008 12/25/2015, 04:05 PM (This post was last modified: 12/25/2015, 04:24 PM by sheldonison.) (12/24/2015, 03:25 AM)sheldonison Wrote: Numerical experiments suggest that the derivative of Knesser's real valued half iterate is continuous at L, and the first derivative at the singularity at L is , and that the 2nd derivative may also match the formal half iterate. I think the key equations in understanding this behavior are the slog, the sexp, the Schroeder equation at L, and the Abel equation generated from the Schroeder equation, and the theta mapping from the Abel equation to the slog. Kneser's exp^{0.5}: Here is the formal exp^{0.5} at L, generated from the formal Schroeder equation developed at the fixed point L where: Then the formal exp^{0.5} at L is exactly the same as: The next step is to show the Abel equation, developed from the Schroeder equation; where And Kneser's slog(z) developed from the Abel equation is: where is a 1-cyclic function decaying to zero at - Sheldon",
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"andydude",
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"Long Time Fellow",
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"Posts: 509 Threads: 44 Joined: Aug 2007 12/27/2015, 11:15 AM (12/24/2015, 03:25 AM)sheldonison Wrote: And the formal half iterate begins with Ok, so I replaced y with 1/2 and log(L) with L in the regular iteration power series to get this: as expected it's the same power series. I wanted to highlight one of my findings in this paper (page 12) that is related but separate from this, which is a power series for for any analytic function with a parabolic fixed point at 0. Substituting in we get which I realize is a different base, but still interesting. Using a similar technique, we might be able to find a comparable power series for , at which point would at the very least intersect with this function.",
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"sheldonison",
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"Long Time Fellow",
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"Posts: 679 Threads: 24 Joined: Oct 2008 12/27/2015, 11:40 PM (This post was last modified: 12/27/2015, 11:41 PM by sheldonison.) (12/27/2015, 11:15 AM)andydude Wrote: Ok, so I replaced y with 1/2 and log(L) with L in the regular iteration power series to get this: as expected it's the same power series.Hey Andy, Thanks for your reply. Oops; I had a typo in my 2nd derivative which I fixed. I have a pari-gp program, that calculate the coefficients iteratively. Quote:I wanted to highlight one of my findings in this paper (page 12) that is related but separate from this, which is a power series for for any analytic function with a parabolic fixed point at 0. Substituting in we get which I realize is a different base, but still interesting. The parabolic case is hugely interesting. I usually work with iterating which is equivalent to iterating base . Anyway, the cool thing about the parabolic case is that the fixed point of zero for the fractional iterate is a singularity, and the formal power series is divergent at zero. References on mathoverflow: http://mathoverflow.net/questions/4347/f...ar-and-exp For the case at hand, , my new conjecture is that the first four derivatives are continuous, but the fifth derivative at the fixed point has a singularity. And the first four derivatives would match the first four derivatives of the formal half iterate at the fixed point. I'm still not totally comfortable it yet, so I haven't posted the justification. - Sheldon",
null,
"sheldonison",
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"Long Time Fellow",
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"",
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"Posts: 679 Threads: 24 Joined: Oct 2008 12/29/2015, 10:25 PM (This post was last modified: 01/01/2016, 04:57 PM by sheldonison.) I wanted to post what I've found out so far; which is not complete, and explain why only four derivatives are defined for Kneser's half iterate at the fixed point of L. First, let S(z) be the Schroeder function, is the Abel function, and be the formal half iterate which is not real valued at the real axis. Then let be the real valued Kneser half iterate. Here's the closest I've gotten, where is a new 1-cyclic function whose coefficients can be derived from the 1-cyclic mapping used for the . The constant term for , so that Since , then the , a taylor series in (z-L). So what we have for the individual theta(z) terms is p is the pseudo period of sexp multiplier for S(z); also a function of (z-L) update I think this is a complete form for the Kneser half iterate. I left out many details including all of details about how to derive from . I haven't yet used these equation to calculate values for Kneser's half iterate in terms of the formal half iterate, and verify they match the expected values generated through other means. But the first 4 derivatives are zero at L, and the 5th derivative of has a singularity at L. That's all for now; next I would like to calculate some of the terms to test this equation out. - Sheldon",
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"andydude",
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"Long Time Fellow",
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"Posts: 509 Threads: 44 Joined: Aug 2007 12/29/2015, 10:51 PM (12/27/2015, 11:40 PM)sheldonison Wrote: Thanks for your reply. Oops; I had a typo in my 2nd derivative which I fixed. I have a pari-gp program, that calculate the coefficients iteratively. I don't think there was a typo, if you multiply the numerator and denominator by and distribute, I think it's the same.",
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"tommy1729",
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"Ultimate Fellow",
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"Posts: 1,480 Threads: 354 Joined: Feb 2009 12/30/2015, 01:27 PM Putting the issue in limit form (exp^[1/2](L+h i) - exp^[1/2](L-hi)) / h^n Where h is infinitesimal. Regards Tommy1729",
null,
"sheldonison",
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"Long Time Fellow",
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"Posts: 679 Threads: 24 Joined: Oct 2008 12/31/2015, 11:02 AM (This post was last modified: 01/01/2016, 04:34 PM by sheldonison.) I updated some of the equations in post#6 So then we have a conjectured equation for the real valued Kneser half iterate in terms of the formal half iterate which is as follows edit: fixed typos p is the pseudo period of sexp I think this is a complete form for the Kneser half iterate. Next I would like to calculate some of the terms to test this equation out, as well as the term. The value and the first four derivatives of this equation are zero; the adder delta equation for the Kneser half iterate in terms of the formal half iterate. - Sheldon",
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"tommy1729",
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"Ultimate Fellow",
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"Posts: 1,480 Threads: 354 Joined: Feb 2009 12/31/2015, 01:25 PM (12/29/2015, 10:25 PM)sheldonison Wrote: I wanted to post what I've found out so far; which is not complete, and explain why only four derivatives are defined for Kneser's half iterate at the fixed point of L. First, let S(z) be the Schroeder function, is the Abel function, and be the formal half iterate which is not real valued at the real axis. Then let be the real valued Kneser half iterate. Here's the closest I've gotten, where is a new 1-cyclic function whose coefficients can be derived from the 1-cyclic mapping used for the . The constant term for , so that Since , then the , a taylor series in (z-L). So what we have for the individual theta(z) terms is p is the pseudo period of sexp multiplier for S(z); also a function of (z-L) update I think this is a complete form for the Kneser half iterate. I left out many details including all of details about how to derive from . I haven't yet used these equation to calculate values for Kneser's half iterate in terms of the formal half iterate, and verify they match the expected values generated through other means. But the first 4 derivatives are zero at L, and the 5th derivative of has a singularity at L. That's all for now; next I would like to calculate some of the terms to test this equation out. The 5 th derivative of is equal to of ?? No singularity ? Im sure you make sense , but it is not clear what you are doing to me. Regards Tommy1729 « Next Oldest | Next Newest »\n\n Possibly Related Threads... Thread Author Replies Views Last Post On the first derivative of the n-th tetration of f(x) Luknik 0 68 10/16/2021, 08:30 PM Last Post: Luknik tetration from alternative fixed point sheldonison 22 50,973 12/24/2019, 06:26 AM Last Post: Daniel Semi-exp and the geometric derivative. A criterion. tommy1729 0 2,772 09/19/2017, 09:45 PM Last Post: tommy1729 How to find the first negative derivative ? tommy1729 0 2,874 02/13/2017, 01:30 PM Last Post: tommy1729 Are tetrations fixed points analytic? JmsNxn 2 6,058 12/14/2016, 08:50 PM Last Post: JmsNxn A calculus proposition about sum and derivative tommy1729 1 3,825 08/19/2016, 12:24 PM Last Post: tommy1729 Derivative of E tetra x Forehead 7 14,697 12/25/2015, 03:59 AM Last Post: andydude [MSE] Fixed point and fractional iteration of a map MphLee 0 3,643 01/08/2015, 03:02 PM Last Post: MphLee A derivative conjecture matrix 3 7,579 10/25/2013, 11:33 PM Last Post: tommy1729 attracting fixed point lemma sheldonison 4 14,757 06/03/2011, 05:22 PM Last Post: bo198214\n\nUsers browsing this thread: 1 Guest(s)"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.928303,"math_prob":0.8211434,"size":8782,"snap":"2021-43-2021-49","text_gpt3_token_len":2360,"char_repetition_ratio":0.14467987,"word_repetition_ratio":0.57189757,"special_character_ratio":0.27009794,"punctuation_ratio":0.1306426,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9828022,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-24T02:44:48Z\",\"WARC-Record-ID\":\"<urn:uuid:ffc6bfcb-872d-4862-9dc4-75c1a7c6d8dc>\",\"Content-Length\":\"68287\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:229e9f9d-f55a-4e35-97d1-aba772cbb9e2>\",\"WARC-Concurrent-To\":\"<urn:uuid:6c13f281-d13f-4bda-881d-c9d5f5562e66>\",\"WARC-IP-Address\":\"109.237.132.18\",\"WARC-Target-URI\":\"http://math.eretrandre.org/tetrationforum/showthread.php?tid=1043&pid=8176\",\"WARC-Payload-Digest\":\"sha1:CFDUQKJP7LNGL7R4F4LMTQVZYACB2YV2\",\"WARC-Block-Digest\":\"sha1:FNDZAR5IM6VTXCOIS5KNSNRKLB6XL4JT\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585837.82_warc_CC-MAIN-20211024015104-20211024045104-00378.warc.gz\"}"} |
https://researchoutput.csu.edu.au/en/publications/lax-representation-for-a-triplet-of-scalar-fields | [
"# LAX REPRESENTATION FOR A TRIPLET OF SCALAR FIELDS\n\nDemskoi D.K., A.G. Meshkov\n\nResearch output: Contribution to journalArticlepeer-review\n\n8 Citations (Scopus)\n\n## Abstract\n\nWe construct a 3×3 matrix zero-curvature representation for the system of three two-dimensional relativistically invariant scalar fields. This system belongs to the class described by the Lagrangian L =[g_{ij}(u)u i_x u j_t]/2 + f(u), where g_{ij} is the metric tensor of a three-dimensional reducible Riemannian space. We previously found all systems of this class that have higher polynomial symmetries of the orders 2, 3, 4, or 5. In this paper, we find a zero-curvature representation for one of these systems. The calculation is based on the analysis of an evolutionary system u_t = S(u), where S is one of the higher symmetries. This approach can also be applied to other hyperbolic systems. We also find recursion relations for a sequence of conserved currents of the triplet of scalar fields under consideration.\nOriginal language English 351-364 14 Theoretical and Mathematical Physics 134 3 https://doi.org/10.1023/A:1022649405488 Published - 2003\n\n## Fingerprint\n\nDive into the research topics of 'LAX REPRESENTATION FOR A TRIPLET OF SCALAR FIELDS'. Together they form a unique fingerprint."
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https://answers.everydaycalculation.com/subtract-fractions/1-2-minus-1-35 | [
"Solutions by everydaycalculation.com\n\n## Subtract 1/35 from 1/2\n\n1/2 - 1/35 is 33/70.\n\n#### Steps for subtracting fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 2 and 35 is 70\n\nNext, find the equivalent fraction of both fractional numbers with denominator 70\n2. For the 1st fraction, since 2 × 35 = 70,\n1/2 = 1 × 35/2 × 35 = 35/70\n3. Likewise, for the 2nd fraction, since 35 × 2 = 70,\n1/35 = 1 × 2/35 × 2 = 2/70\n4. Subtract the two like fractions:\n35/70 - 2/70 = 35 - 2/70 = 33/70\n\nMathStep (Works offline)",
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"Download our mobile app and learn to work with fractions in your own time:"
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https://itprospt.com/num/3131797/q-amctal-strip-that-is-2-00-wide-and-0-05-20t-cmnt-thick | [
"5\n\n# Q) Amctal strip that is 2.00 _ wide and 0.05 . 20T cmnt thick caries cumcnnedi 30A in a Is shown in Figure 26-41. The Hall region with uniforn magnetic field of vel...\n\n## Question\n\n###### Q) Amctal strip that is 2.00 _ wide and 0.05 . 20T cmnt thick caries cumcnnedi 30A in a Is shown in Figure 26-41. The Hall region with uniforn magnetic field of velocity _ voltage elettor- cnsund Curr Zurt be 85 #V: Fxplain how arises. You Cun fows from kefi think of itas # right.hich dinections do the clectrons 6) Find the number_ flow? Calculate the drift spced of the frec electrons density 0f the frce clectrons strip. the strip_ In which dineclion (ic: nietl dils pushed? noimt tnc trort and .\n\nQ) Amctal strip that is 2.00 _ wide and 0.05 . 20T cmnt thick caries cumcnnedi 30A in a Is shown in Figure 26-41. The Hall region with uniforn magnetic field of velocity _ voltage elettor- cnsund Curr Zurt be 85 #V: Fxplain how arises. You Cun fows from kefi think of itas # right. hich dinections do the clectrons 6) Find the number_ flow? Calculate the drift spced of the frec electrons density 0f the frce clectrons strip. the strip_ In which dineclion (ic: nietl dils pushed? noimt tnc trort and . point \"hthc is at the bxatck: higher potential? Explain yOur answer Lnu carefully. Which nowadays thit mcgitive electrons the positive edge? Jntigm CM the cunent_ chalrge differemt? Again . explain }Our Curricre Wen nosilive how inshet carcfully. would B-field 2.OOcm",
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"#### Similar Solved Questions\n\n##### G.C.-3 Precipitation Reactions Solution9. CuSo KIO;- Balanced Eq_ CLENLECoCl NaOH + Balanced Eq:CLE NLECoCl; NaCro,> Balanced Eq: CLENLE12 CoCl KIO ,+ Balanced Eq; CLENLENaOH NaCro + Balanced Eq; CLE NLENaOH Kio; Balanced Eq: CLESNLE NuCro KI0,+ Bulanced Eq: CLE: N.LECettun Kreol Ra No+ €Clhz Asci +CWo,\nG.C.-3 Precipitation Reactions Solution 9. CuSo KIO;- Balanced Eq_ CLE NLE CoCl NaOH + Balanced Eq: CLE NLE CoCl; NaCro,> Balanced Eq: CLE NLE 12 CoCl KIO ,+ Balanced Eq; CLE NLE NaOH NaCro + Balanced Eq; CLE NLE NaOH Kio; Balanced Eq: CLES NLE NuCro KI0,+ Bulanced Eq: CLE: N.LE Cettun Kreol Ra...\n##### Question15pusFor the following multistep synthesis of the target struclure on right frorn the starting the left choose the correct order of structure on reagents: Ignore stereochemiatry: To preview the image clickhere &Step Step 2 Slep 3Step . StepStep Reagent(s) Select ]Step 2 Reagent(s) Select ]Step 3 Reagent(s) Select ]Step 4 Reagent(s) Select ]Step 5 Reagent(s [ Select ] CH3OH NaCN HOCHZCHZOHIHCI (cat ) H30+ CH3CHZOHITsOH (cat ) SoCi2 Give curved arri CHBCHZOH Zn(Hg), HCIH2O Skow WHERE\nQuestion 15pus For the following multistep synthesis of the target struclure on right frorn the starting the left choose the correct order of structure on reagents: Ignore stereochemiatry: To preview the image clickhere & Step Step 2 Slep 3 Step . Step Step Reagent(s) Select ] Step 2 Reagent(s...\n##### (Ipt) Recall that c is isometrically isomorphic to (1, call this isom- etry 1. Show that F 01 7R given by F(T1,T2, Cii Ti is norm continuous but F 0 is not wcak-* continuous_\n(Ipt) Recall that c is isometrically isomorphic to (1, call this isom- etry 1. Show that F 01 7R given by F(T1,T2, Cii Ti is norm continuous but F 0 is not wcak-* continuous_...\n##### Question number 10Suppose that the probability that a new medication will cause bad side effect is 0.03. If this medication is given to 150 people, what is the probability that exactly three of them will experience bad side effect? Place your answer; rounded to decimal places_ the blank:\nQuestion number 10 Suppose that the probability that a new medication will cause bad side effect is 0.03. If this medication is given to 150 people, what is the probability that exactly three of them will experience bad side effect? Place your answer; rounded to decimal places_ the blank:...\n##### Use the Extreme Value Theorem to find the absolute extrema of the function on the closed interval. State the conditions that have been met in order t0 use that theorem: (8 points}f(r) =x' 3r' on [1,3]\nUse the Extreme Value Theorem to find the absolute extrema of the function on the closed interval. State the conditions that have been met in order t0 use that theorem: (8 points} f(r) =x' 3r' on [1,3]...\n##### 1 Your 1 1 havl uinino] ploak 0l distidulion 1 1 Usino Youi L Preview My Answers L 1 1 1 1 3 Submit Answers pinous UMC cumdanci\n1 Your 1 1 havl uinino] ploak 0l distidulion 1 1 Usino Youi L Preview My Answers L 1 1 1 1 3 Submit Answers pinous UMC cumdanci...\n##### Point) In a study of redlgreen color blindness, 850 men and 3000 women are randomly selected and tested. Among the men, 79 have redlgreen color blindness. Among the women, 9 have redlgreen color blindness Test the claim that men have higher rate of redlgreen color blindness_ (Note: Type 'p_m\" for the symbol Pm for example p_mn not = P_w for the proportions are not equal, p_m > p_W for the proportion of men with color blindness is larger; p_m <p_ for the proportion of men is small\npoint) In a study of redlgreen color blindness, 850 men and 3000 women are randomly selected and tested. Among the men, 79 have redlgreen color blindness. Among the women, 9 have redlgreen color blindness Test the claim that men have higher rate of redlgreen color blindness_ (Note: Type 'p_m&qu...\n##### Solve the recurrence relation an 5an-1 6an-2, n > 2, with the initial [email protected] = 2 and 01 = 5.\nSolve the recurrence relation an 5an-1 6an-2, n > 2, with the initial conditions @0 = 2 and 01 = 5....\n##### Evaluatc thc doublc intcgralLlfh- ~Inlx2+y2+1)dxdy by convcrting to doublc intcgral in polar coordinotcs; 1 -y\nEvaluatc thc doublc intcgral Llfh- ~Inlx2+y2+1)dxdy by convcrting to doublc intcgral in polar coordinotcs; 1 -y...\n##### Determine the horizontal and vertical components of the pin reaction at B on member ABC of the structure shown: Draw the shear force and bending moment diagrams for member BDF. Neglect the weights of all members_Coble300 Lb/ft6p0 Lb\nDetermine the horizontal and vertical components of the pin reaction at B on member ABC of the structure shown: Draw the shear force and bending moment diagrams for member BDF. Neglect the weights of all members_ Coble 300 Lb/ft 6p0 Lb...\n##### Write a structural formula for the principal organic product formed by treating each compound in Problem 17.37 with $\\mathrm{NaBH}_{4}$ followed by $\\mathrm{H}_{2} \\mathrm{O}$.\nWrite a structural formula for the principal organic product formed by treating each compound in Problem 17.37 with $\\mathrm{NaBH}_{4}$ followed by $\\mathrm{H}_{2} \\mathrm{O}$....\n##### 225 ftbuilding that is 225 feet tall casts a shadow of various lengths as the day goes by: An angle of elevation 0 is formed by lines from the top and bottom of the building to the tip of the shadow, as seen in the figure above: de Find the rate of change of the angle of elevation when € = 156. dx0' (156)radians per foot. Round to five decimal places\n225 ft building that is 225 feet tall casts a shadow of various lengths as the day goes by: An angle of elevation 0 is formed by lines from the top and bottom of the building to the tip of the shadow, as seen in the figure above: de Find the rate of change of the angle of elevation when € = 15...\n##### 12. [~I4 Points]DETAILSSCALCET8 8.1.019. Find the exact length of the curve. Y = In( 1 ~ x 0 sxs 4Need Help?RandILWk\n12. [~I4 Points] DETAILS SCALCET8 8.1.019. Find the exact length of the curve. Y = In( 1 ~ x 0 sxs 4 Need Help? RandIL Wk...\n##### Determine whether the given ordered pair is a solution of the equation.Is $(-9,-9)$ a solution of $-3 x+4 y=-11 ?$\nDetermine whether the given ordered pair is a solution of the equation. Is $(-9,-9)$ a solution of $-3 x+4 y=-11 ?$...\n##### Will $(-a)^{n}$ ever equal $a^{n-2}$ If so, when?\nWill $(-a)^{n}$ ever equal $a^{n-2}$ If so, when?..."
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https://fr.slideserve.com/Patman/electron-transfer | [
"",
null,
"Download",
null,
"Download Presentation",
null,
"ELECTRON TRANSFER\n\n# ELECTRON TRANSFER\n\nTélécharger la présentation",
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"## ELECTRON TRANSFER\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. ELECTRON TRANSFER Reduction-Oxidation RX (redox) A reaction in which electrons are transferred from one species to another. Combustion reactions are redox reactions - oxidation means the loss of electrons - reduction means the gain of electrons - electrolyte is a substance dissolved in water which produces an electrically conducting solution - nonelectrolyte is a substance dissolved in water which does not conduct electricity. Rusting is a redox reaction: 4Fe(s) + 302(g) 2Fe2O3(s) Electrochemistry involves redox reactions: Cu(s) + 2AgNO3(aq) 2Ag(s) + Cu(NO3)2(aq)\n\n2. IDENTIFING REDOX RX Element + compound New element + New compound A + BC B + AC Element + Element Compound A + B AB Check oxidation state (charges) of species A change in oxidation # means redox reaction Identify the Redox Rx: Cu + AgNO3 Cu(NO3)2 + Ag NO + O2 NO2 K2SO4 + CaCl2 KCl + CaSO4 C2H4O2 + O2 CO2 + H2O\n\n3. LABELING COMPONENTS OF REDOX REACTIONS The REDUCING AGENT is the species which undergoes OXIDATION. The OXIDIZING AGENT is the species which undergoes REDUCTION. CuO + H2 Cu + H2O\n\n4. A summary of redox terminology. OXIDATION One reactant loses electrons. Zn loses electrons. Reducing agent is oxidized. Zn is the reducing agent and becomes oxidized. Oxidation number increases. The oxidation number of Zn increases from x to +2. REDUCTION Other reactant gains electrons. Hydrogen ion gains electrons. Oxidizing agent is reduced. Hydrogen ion is the oxidizing agent and becomes reduced. Oxidation number decreases. The oxidation number of H decreases from +1 to 0.\n\n5. Key Points About Redox Reactions • Oxidation (electron loss) always accompanies reduction (electron gain). • The oxidizing agent is reduced, and the reducing agent is oxidized. • The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent.\n\n6. ACTIVITY SERIES OF SOME SELECTED METALS A brief activity series of selected metals, hydrogen and halogens are shown below. The series are listed in descending order of chemical reactivity, with the most active metals and halogens at the top (the elements most likely to undergo oxidation). Any metal on the list will replace the ions of those metals (to undergo reduction) that appear anywhere underneath it on the list. METALSHALOGENS K (most oxidized F2 (relatively stronger oxidizing agent) Ca Cl2 Na Br2 Mg l2 (relatively stronger reducing agent) Al Zn Fe Ni Sn Pb H Cu Ag Hg Au(least oxidized) Oxidation refers to the loss of electrons and reduction refers to the gain of electrons\n\n7. Oxidizing/Reducing Agents Strongest oxidizing agent Most positive values of E° red Increasing strength of reducing agent F2(g) + 2e- 2F-(aq) • • 2H+(aq) + 2e- H2(g) • • Li+(aq) + e- Li(s) Increasing strength of oxidizing agent Strongest reducing agent Most negative values of E° red\n\n8. REDOX REACTIONS For the following reactions, identify the oxidizing and reducing agents. MnO4- + C2O42- MnO2 + CO2 acid: Cr2O72- + Fe2+ Cr3+ + Fe3+ base: CO2+ + H2O2 CO(OH)3 + H2O As + ClO3- H3AsO3 + HClO Which of the following species is the strongest oxidizing agent: NO3-(aq), Ag+(aq), or Cr2O72-(aq)?\n\n9. Standard Reduction Potentials in Water at 25°C Standard Potential (V)Reduction Half Reaction 2.87 F2(g) + 2e- 2F-(aq) 1.51 MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l) 1.36 Cl2(g) + 2e- 2Cl-(aq) 1.33 Cr2O72-(aq) + 14H+(aq) + 6e- 2Cr3+(aq) + H2O(l) 1.23 O2(g) + 4H+(aq) + 4e- 2H2O(l) 1.06 Br2(l) + 2e- 2Br-(aq) 0.96 NO3-(aq) + 4H+(aq) + 3e- NO(g) + H2O(l) 0.80 Ag+(aq) + e- Ag(s) 0.77 Fe3+(aq) + e- Fe2+(aq) 0.68 O2(g) + 2H+(aq) + 2e- H2O2(aq) 0.59 MnO4-(aq) + 2H2O(l) + 3e- MnO2(s) + 4OH-(aq) 0.54 I2(s) + 2e- 2I-(aq) 0.40 O2(g) + 2H2O(l) + 4e- 4OH-(aq) 0.34 Cu2+(aq) + 2e- Cu(s) 0 2H+(aq) + 2e- H2(g) -0.28 Ni2+(aq) + 2e- Ni(s) -0.44 Fe2+(aq) + 2e- Fe(s) -0.76 Zn2+(aq) + 2e- Zn(s) -0.83 2H2O(l) + 2e- H2(g) + 2OH-(aq) -1.66 Al3+(aq) + 3e- Al(s) -2.71 Na+(aq) + e- Na(s) -3.05 Li+(aq) + e- Li(s)\n\n10. Half-Reaction Method for Balancing Redox Reactions • Summary: This method divides the overall redox reaction into oxidation and reduction half-reactions. • Each reaction is balanced for mass (atoms) and charge. • One or both are multiplied by some integer to make the number of electrons gained and lost equal. • The half-reactions are then recombined to give the balanced redox equation. • Advantages: • The separation of half-reactions reflects actual physical separations in electrochemical cells. • The half-reactions are easier to balance especially if they involve acid or base. • It is usually not necessary to assign oxidation numbers to those species not undergoing change.\n\n11. The guidelines for balancing via the half-reaction method are found below: • 1. Write the corresponding half reactions. • 2. Balance all atoms except O and H. • 3. Balance O; add H2O as needed. • 4. Balance H as acidic (H+). • 5. Add electrons to both half reactions and balance. • 6. Add the half reactions; cross out “like” terms. • 7. If basic or alkaline, add the equivalent number of hydroxides (OH-) to counterbalance the H+ (remember to add to both sides of the equation). Recall that • H+ + OH- H2O.\n\n12. Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq) Cr2O72- Cr3+ I-I2 Cr2O72- Cr3+ 6e- + 14H+(aq) + Cr2O72- Cr3+ 2 + 7H2O(l) Balancing Redox Reactions in Acidic Solution Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq) 1. Divide the reaction into half-reactions - Determine the O.N.s for the species undergoing redox. +6 -1 +3 0 Cr is going from +6 to +3 I is going from -1 to 0 2. Balance atoms and charges in each half-reaction - 14H+(aq) + 2 + 7H2O(l) net: +6 Add 6e- to left. net: +12\n\n13. 2 + 2e- X 3 I-I2 I-I2 I-I2 6e- + 6e- + 14H+ + Cr2O72- Cr3+ 2 + 7H2O(l) 6 3 + 6e- 14H+(aq) + Cr2O72-(aq) + 6 I-(aq) 2Cr3+(aq) + 3I2(s) + 7H2O(l) 14H+(aq) + Cr2O72- Cr3+ + 7H2O(l) Balancing Redox Reactions in Acidic Solution continued 2 2 + 2e- Cr(+6) is the oxidizing agent and I(-1)is the reducing agent. 3. Multiply each half-reaction by an integer, if necessary - 4. Add the half-reactions together - Do a final check on atoms and charges.\n\n14. 14H2O + Cr2O72- + 6 I- 2Cr3+ + 3I2 + 7H2O + 14OH- 7H2O + Cr2O72- + 6 I- 2Cr3+ + 3I2 + 14OH- 14H+(aq) + Cr2O72-(aq) + 6 I-(aq) 2Cr3+(aq) + 3I2(s) + 7H2O(l) Balancing Redox Reactions in Basic Solution Balance the reaction in acid and then add OH- so as to neutralize the H+ ions. + 14OH-(aq) + 14OH-(aq) Reconcile the number of water molecules. Do a final check on atoms and charges.\n\n15. ELECTROCHEMISTRY Balancing Redox Reactions: MnO4- + C2O42- MnO2 + CO2 acidic: Cr2O72- + Fe2+ Cr3+ + Fe3+ As + ClO3- H3AsO3 + HClO Basic: CO2+ + H2O2 CO(OH)3 + H2O\n\n16. ELECTROCHEMICAL CELLS CHEMICALS AND EQUIPMENT NEEDED TO BUILD A SIMPLE CELL: The Cell: Voltmeter Two alligator clips Two beakers or glass jars The Electrodes: Metal electrode Metal salt solution The Salt Bridge: Glass or Plastic u-tube Na or K salt solution\n\n17. ELECTROCHEMISTRY A system consisting of electrodes that dip into an electrolyte and in which a chemical reaction uses or generates an electric current. Two Basic Types of Electrochemical cells: Galvanic (Voltaic) Cell: A spontaneous reaction generates an electric current. Chemical energy is converted into electrical energy Electrolytic Cell: An electric current drives a nonspontaneous reaction. Electrical energy is converted into chemical energy.\n\n18. Oxidation half-reaction X X+ + e- Oxidation half-reaction A- A + e- Reduction half-reaction Y++ e- Y Reduction half-reaction B++ e- B Overall (cell) reaction X + Y+ X+ + Y; DG < 0 Overall (cell) reaction A- + B+ A + B; DG > 0 General characteristics of voltaic and electrolytic cells. VOLTAIC CELL ELECTROLYTIC CELL Energy is released from spontaneous redox reaction Energy is absorbed to drive a nonspontaneous redox reaction\n\n19. ELECTROCHEMICAL CELLS A CHEMICAL CHANGE PRODUCES ELECTRICITY Theory: If a metal strip is placed in a solution of it’s metal ions, one of the following reactions may occur Mn+ + ne- M M Mn+ + ne- These reactions are called half-reactions or half cell reactions If different metal electrodes in their respective solutions were connected by a wire, and if the solutions were electrically connected by a porous membrane or a bridge that minimizes mixing of the solutions, a flow of electrons will move from one electrode, where the reaction is M1 M1n+ + ne- To the other electrode, where the reaction is M2n+ + ne- M2 The overall reaction would be M1 + M2n+ M2 + M1n+\n\n20. Electrochemical Cells • An electrochemical cell is a device in which an electric current (i.e. a flow of electrons through a circuit) is either produced by a spontaneous chemical reaction or used to bring about a nonspontaneous reaction. Moreover, a galvanic (or voltaic) cell is an electrochemical cell in which a spontaneous chemical reaction is used to generate an electric current. • Consider the generic example of a galvanic cell shown below:\n\n21. The cell consists of two electrodes, or metallic conductors, that make electrical contact with the contents of the cell, and an electrolyte, an ionically conducting medium, inside the cell. Oxidation takes place at one electrode as the species being oxidized releases electrons from the electrode. We can think of the overall chemical reaction as pushing electrons on to one electrode and pulling them off the other electrode. The electrode at which oxidation occurs is called the anode. The electrode at which reduction occurs is called the cathode. Finally, a salt bridge is a bridge-shaped tube containing a concentrated salt in a gel that acts as an electrolyte and provides a conducting path between the two compartments in the electrochemical circuit.\n\n22. Why Does a Voltaic Cell Work? The spontaneous reaction occurs as a result of the different abilities of materials (such as metals) to give up their electrons and the ability of the electrons to flow through the circuit. Ecell > 0 for a spontaneous reaction 1 Volt (V) = 1 Joule (J)/ Coulomb (C)\n\n23. More Positive Cathode(reduction) Eº Red (cathode) Eº cell Eºred (anode) Anode(oxidation) More Negative EºRed (V) A cell will always run spontaneous in the direction that produces a positive Eocell\n\n24. Zn(s) Zn2+(aq) + 2e- Cu2+(aq) + 2e- Cu(s) inert electrode Notation for a Voltaic Cell components of cathode compartment (reduction half-cell) components of anode compartment (oxidation half-cell) phase of lower oxidation state phase of lower oxidation state phase of higher oxidation state phase of higher oxidation state phase boundary between half-cells Examples: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s) graphite | I-(aq) | I2(s) || H+(aq), MnO4-(aq) | Mn2+(aq) | graphite\n\n25. NOTATION FOR VOLTAIC CELLS Zn + Cu2+ Zn2+ + Cu Zn(s)/Zn2+(aq) // Cu2+(aq)/Cu(s) Anode Cathode oxidation reduction salt bridge write the net ionic equation for: Al(s)/Al3+(aq)//Cu2+(aq)/Cu(s) Tl(s)/Tl+(aq)//Sn2+(aq)/Sn(s) Zn(s)/Zn2+(aq)//Fe3+(aq),Fe2+(aq)/Pt If given: Al(s)→Al3+(aq)+3e- and 2H+(aq)+2e-→H2(g) write the notation.\n\n26. The Hydrogen Electrode (Inactive Electrodes): At the hydrogen electrode, the half reaction involves a gas. 2 H+(aq) + 2e- H2(g) so an inert material must serve as the reaction site (Pt). Another inactive electode is C(graphite). H+(aq)/H2(g)/Pt cathode Pt/H2(g)/H+(aq) anode Therefore: Al(g)/Al3+(aq)//H+(aq)/H2(g)/Pt\n\n27. PROBLEM: Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO3)3 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode. e- Oxidation half-reaction Cr(s) Cr3+(aq) + 3e- K+ NO3- Reduction half-reaction Ag+(aq) + e- Ag(s) Overall (cell) reaction Cr(s) + Ag+(aq) Cr3+(aq) + Ag(s) Sample Problem: Diagramming Voltaic Cells PLAN: Identify the oxidation and reduction reactions and write each half-reaction. Associate the (-)(Cr) pole with the anode (oxidation) and the (+) pole with the cathode (reduction). SOLUTION: Voltmeter salt bridge Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)\n\n28. STANDARD REDUCTION POTENTIALS Individual potentials can not be measured so standard conditions: 1M H+ at 1 atm is arbitrarily measured as 0 V (Volts). Ecell = EoH+→H2 + EoZn→Zn2+ 0.76 V = (0 V) - (-0.76 V) cathode anode Ecell = Eocath – Eoanode The standard reduction potential is the Eo value for the reduction half reaction (cathode) and are found in tables.\n\n29. Standard Reduction Potentials in Water at 25°C Standard Potential (V)Reduction Half Reaction 2.87 F2(g) + 2e- 2F-(aq) 1.51 MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l) 1.36 Cl2(g) + 2e- 2Cl-(aq) 1.33 Cr2O72-(aq) + 14H+(aq) + 6e- 2Cr3+(aq) + H2O(l) 1.23 O2(g) + 4H+(aq) + 4e- 2H2O(l) 1.06 Br2(l) + 2e- 2Br-(aq) 0.96 NO3-(aq) + 4H+(aq) + 3e- NO(g) + H2O(l) 0.80 Ag+(aq) + e- Ag(s) 0.77 Fe3+(aq) + e- Fe2+(aq) 0.68 O2(g) + 2H+(aq) + 2e- H2O2(aq) 0.59 MnO4-(aq) + 2H2O(l) + 3e- MnO2(s) + 4OH-(aq) 0.54 I2(s) + 2e- 2I-(aq) 0.40 O2(g) + 2H2O(l) + 4e- 4OH-(aq) 0.34 Cu2+(aq) + 2e- Cu(s) 0 2H+(aq) + 2e- H2(g) -0.28 Ni2+(aq) + 2e- Ni(s) -0.44 Fe2+(aq) + 2e- Fe(s) -0.76 Zn2+(aq) + 2e- Zn(s) -0.83 2H2O(l) + 2e- H2(g) + 2OH-(aq) -1.66 Al3+(aq) + 3e- Al(s) -2.71 Na+(aq) + e- Na(s) -3.05 Li+(aq) + e- Li(s)\n\n30. The table of electrode potentials can be used to predict the direction of spontaneity. A spontaneous reaction has the strongest oxidizing agent as the reactant. Q1. Will dichromate ion oxidize Mn2+ to MnO4- in an acidic solution? Q2. Describe the galvanic cell based on Ag+ + e-→ Ag Eo = 0.80 V Fe3+ + e- → Fe2+Eo = 0.77V\n\n31. STANDARD REDUCTION POTENTIALS Intensive property 1. If the 1/2 reaction is reversed then the sign is reversed. 2. Electrons must balance so half-rx may be multiplied by a factor. The E° is unchanged. Q1. Consider the galvanic cell Al3+(aq) + Mg(s) ° Al(s) + Mg2+(aq) Give the balance cell reaction and calculate E° for the cell. Q2. MnO4- + 5e- + 8H+ Mn2+ + 4H2O ClO4- + 2H+ + 2e- ClO3- + H2O Give the balance cell reactions for the reduction of permanganate then calculate the E° cell.\n\n32. Electromotive Force The difference in electric potential between two points is called the POTENTIAL DIFFERENCE. Cell potential (Ecell) = electromotive force (emf). Electrical work = charge x potential difference J = C x V Joules = coulomb x Voltage The Faraday constant, F, describes the magnitude of charge of one mole of electrons. F = 9.65 x 104 C w = -F x Potential Difference wmax = -nFEcell Example : The emf of a particular cell is 0.500 V. Calculate the maxiumum electrical work of this cell for 1 g of aluminum. Al(s)/ Al3+(aq) // Cu2+(aq) / Cu(s)\n\n33. Galvanic cells differ in their abilities to generate an electrical current. The cell potential () is a measure of the ability of a cell reaction to force electrons through a circuit. A reaction with a lot of pushing-and-pulling power generates a high cell potential (and hence, a high voltage). This voltage can be read by a voltmeter. When taking both half reactions into account, for a reaction to be spontaneous, the overall cell potential (or emf, electromotive force) MUST BE POSITIVE. That is, is (+). Please note that the emf is generally measured when all the species taking part are in their standard states (i.e. pressure is 1 atm; all ions are at 1 M, and all liquids/solids are pure). Cell emf and reaction free energy (G) can be related via the following relationship: G = -n F E, where n=mol e-andF=Faraday’s Constant(96,500 C/mol e-)\n\n34. For a Voltaic Cell, the work done is electrical: DGo = wmax = -nFEocell Q1. Calculate the standard free energy change for the net reaction in a hydrogen-oxygen fuel cell. 2 H2 (g) + O2 (g) → 2 H2O (l) What is the emf for the cell? How does this compare to DGfo (H2O)l? Q2. A voltaic cell consists of Fe dipped in 1.0 M FeCl2 and the other cell is Ag dipped in 1.0 M AgNO3. Obtain the standard free energy change for this cell using DGfo. What is the emf for this cell?\n\n35. EXAMPLE 1: Consider the following unbalanced chemical equations: MnO4- + 5e- + 8H+ Mn2+ + 4H2O Fe2+(aq) + 2e-(aq) Fe(s) Use your table of standard reduction potentials in order to determine the following: A. Diagram the galvanic cell, indicating the direction of flow of electrons in the external circuit and the motion of the ions in the salt bridge. B. Write balanced chemical equations for the half-reactions at the anode, the cathode, and for the overall cell reaction. C. Calculate the standard cell potential for this galvanic cell. D. Calculate the standard free energy for this galvanic cell. E. Write the abbreviated notation to describe this cell.\n\n36. Example 2:A galvanic cell consists of a iron electrode immersed in a 1.0 M ferrous chloride solution and a silver electrode immersed in a 1.0 M silver nitrate solution. A salt bridge comprised of potassium nitrate connects the two half-cells. Use your table of standard reduction potentials in order to determine the following: A. Diagram the galvanic cell, indicating the direction of flow of electrons in the external circuit and the motion of the ions in the salt bridge. B. Write balanced chemical equations for the half-reactions at the anode, the cathode, and for the overall cell reaction. C. Calculate the standard cell potential for this galvanic cell. D. Calculate the standard free energy for this galvanic cell. E. Write the abbreviated notation to describe this cell.\n\n37. EXAMPLE 3: Consider the following unbalanced chemical equation: Cr2O72-(aq) + I-(aq) Cr+3(aq) + I2(s) Use your table of standard reduction potentials in order to determine the following: A. Diagram the galvanic cell, indicating the direction of flow of electrons in the external circuit and the motion of the ions in the salt bridge. B. Write balanced chemical equations for the half-reactions at the anode, the cathode, and for the overall cell reaction. C. Calculate the standard cell potential for this galvanic cell. D. Calculate the standard free energy for this galvanic cell. E. Write the abbreviated notation to describe this cell.\n\n38. Workshop on Galvanic/Voltaic Cells Use your table of standard reduction potentials in order to determine the following for questions 1 & 2 given below: A. Diagram the galvanic cell, indicating the direction of flow of electrons in the external circuit and the motion of the ions in the salt bridge. B. Write balanced chemical equations for the half-reactions at the anode, the cathode, and for the overall cell reaction. C. Calculate the standard cell potential for this galvanic cell. D. Calculate the standard free energy for this galvanic cell. E. Write the abbreviated notation to describe this cell. (1) A galvanic cell consists of a zinc electrode immersed in a zinc sulfate solution and a copper electrode immersed in a copper(II) sulfate solution. A salt bridge comprised of potassium nitrate connects the two half-cells. (2) An hydrogen-oxygen fuel cell follows the following overall reaction: 2H2 (g) + O2 (g) 2 H2O (l)\n\n39. Summary of Voltaic/Galvanic Cells 1. The cell potential should always be positive. 2. the electron flow is in the direction of a positive Eocell designate the anode (oxidation) & the cathode (reduction) RC & OA 4. be able to describe the nature of the electrodes (active vs. inactive)\n\n40. Cell Potential & Equilibrium One of the most useful applications of standard cell potentials is the calculation of equilibrium constants from electrochemical data. Recall, G = -nF and G = -RT ln Kc So: Eocell = RT/nF (ln K) = 2.303RT/nF (log K) The equilibrium constant of a reaction can be calculated from standard cell potentials by combining the equations for the half-reactions to give the cell reaction of interest and determining the standard cell potential of the corresponding cell. That is: Eocell = (0.0592/n) log (K) at 25oC\n\n41. Cell Potential & Equilibrium • Calculate the cell potential and equilibrium constant using the standard emf values for: • 1. Pb2+(aq) + Fe(s) → Pb(s) + Fe2+(aq) • S4O62- + Cr2+ → Cr3+ + S2O32-\n\n42. CONCENTRATION EFFECTS Finally, consider a galvanic cell where the concentrations of the solutions are NOT 1 M. As a reaction proceeds towards equilibrium, the concentrations of its reactants and products change, and Grxn approaches 0. Therefore, as reactants are consumed in an electrochemical cell, the cell potential decreases until finally it reaches 0. To understand this behavior quantitatively, we need to find how the cell emf varies with the concentrations of species in the cell. Recall: G = G + RT ln Q Because G = -nFE & G= -nFEo ஃ-nFE = -nFEo + RT ln Q\n\n43. CONCENTRATION EFFECTS When we divide through by -nF, we derive the Nernst Equation: = - (RT/nF) ln Q That is, the dependence of emf on composition is expressed via the Nernst equation, where Q is the reaction quotient for the cell reaction. Ecell = Eocell – (2.303RT/nF) log (Q)\n\n44. CONCENTRATION EFFECTS • Ecell = Eocell – (2.303RT/nF) log (Q) • 1. 2Al + 3Mn2+→ 2Al3+ + 3Mn Eocell =0.48V • Predict whether the cell potential is larger or smaller than the standard cell potential if: • [Al3+] = 2.0 M & [Mn2+] = 1.0 M • [Al3+] = 1.0M & [Mn2+] = 3.0M • Describe the cell based on: • VO2+ + 2H+ + e- → VO2+ + H2O Eocell = 1.0V • Zn2+ + 2e- → Zn Eocell = -0.76 • Where [VO2+]=2.0M, [H+]=0.5M, [VO2+]=0.01M & [Zn2+]=0.1M\n\n45. Workshop on Equilibrium & Cell Potential • Q1: Sn + Ag+ Sn+2 + Ag • A. Write the balanced net-ionic equation for this reaction. • B. Calculate the standard voltage of a cell involving the system above. • C. What is the equilibrium constant for the system above? • D.Calculate the voltage at 25 C of a cell involving the • system above when the concentration of Ag+ is 0.0010 • molar and that of Sn2+ is 0.20 molar. • Q2: Consider a galvanic cell in which a nickel electrode is immersed in a 1.0 molar nickel nitrate solution, and a zinc electrode is immersed in a 1.0 molar zinc nitrate solution. • Identify the anode of the cell and write the half reaction that occurs there. • Write the net ionic equation for the overall reaction that occurs as the cell operates and calculate the value of the standard cell potential. • C. Indicate how the value of the cell emf would be affected if the concentration of nickel nitrate was changed from 1.0 M to 0.10 M, and the concentration of zinc nitrate remained the same. Justify your answer. • D. Specify whether the value of the equilibrium constant for the cell reaction is less than 1, greater than 1, or equal to 1. Justify your answer.\n\n46. Workshop on Concentration Q1: Calculate the emf generated by the following cell at 298 K when [Al+3] = 4.0 x 10-3 M and [I-] = 0.010 M. Al(s) + I2(s) Al+3(aq) + I-(aq) Q2: Because cell potentials depend on concentration, one can construct galvanic cells where both compartments contain the same component but at different concentrations. These are known as concentration cells. Nature will try to equalize the concentrations of the respective ion in both compartments of the cell. Consider the schematic of a concentration cell shown below.\n\n47. Nernst Equation & pH Q1: A pH meter is constructed using hydrogen gas bubbling over an inert platinum electrode at a pressure of 1.2 atm. The other electrode is aluminum metal immersed in a 0.20M Al3+ solution. What is the cell emf when the hydrogen electrode is immersed in a sample of acid rain with pH of 4.0 at 25oC? If the electrode is placed in a sample of shampoo and the emf is 1.17 V, what is the pH of the shampoo? Q2: Calculate cell for the following: Pt(s) H2(g, 1 atm) H+(aq, pH = 4.0) H+(aq, pH = 3.0) H2(g, 1 atm) Pt(s)\n\n48. Workshop on pH Q1: What is the pH of a solution in the cathode compartment of a Zn-H+ cell when P(H2) = 1.0 atm, [Zn+2] = 0.10 M, and the cell emf is 0.542 V? Q2: A concentration cell is constructed with two Zn(s)-Zn+2(aq) half-cells. The first half-cell has [Zn2+] = 1.35 M, and the second half-cell has [Zn2+] = 3.75 x 10-4 M. Which half-cell is the anode? Determine the emf of the cell.\n\n49. Voltages of Some Voltaic Cells Voltaic Cell Voltage (V) 1.5 Common alkaline battery 2.0 Lead-acid car battery (6 cells = 12V) 1.3 Calculator battery (mercury) Electric eel (~5000 cells in 6-ft eel = 750V) 0.15 Nerve of giant squid (across cell membrane) 0.070\n\n50. Electrolysis • An electrolytic cell is an electrochemical cell in which electrolysis takes place. The arrangement of components in electrolytic cells is different from that in galvanic cells. Specifically, the two electrodes usually share the same compartment, there is usually only one electrolyte, and concentrations and pressures are usually far from standard. In an electrolytic cell, current supplied by an external source is used to drive the nonspontaneous reaction. • As in a galvanic cell, oxidation occurs at the anode and reduction occurs at the cathode, and electrons travel through the external wire from anode to cathode. But unlike the spontaneous current in a galvanic cell, a current MUST be supplied by an external electrical power source. The result is to force oxidation at one electrode and reduction at the other."
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https://www.datasciencecentral.com/bayesian-machine-learning-1/ | [
"# Bayesian Machine Learning\n\nBayesian Machine Learning (part – 3)\n\nBayesian Modelling\n\nIn this post we will see the methodology of building Bayesian models. In my previous post I used a Bayesian model for linear regression. The model looks like:",
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"So, let us first understand the construction of the above model:\n\n1. when there is an arrow pointing from one node to another, that implies start nodes causes end node. For example, in above case Target node depends on Weights node as well as Data node.\n2. Start node is known as Parent and the end node is known as Child\n3. Most importantly, cycles are avoided while building a Bayesian model.\n4. These structures normally are generated from the given data and experience\n5. Mathematical representation of the Bayesian model is done using Chain rule. For example, in the above diagram the chain rule is applied as follows:\n\nP(y,w,x) = P(y/w,x)P(w)P(x)\n\nGeneralized chain rule looks like:",
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"1. The Bayesian models are build based upon the subject matter expertise and experience of the developer.\n\nAn Example\n\nProblem Statement : Given are three variables : sprinkle, rain , wet grass, where sprinkle and rain are predictors and wet grass is a predicate variable. Design a Bayesian model over it.\n\nSolution:",
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"Theory behind above model:\n\n• Sprinkle is used to wet grass, therefore Sprinkle causes wet grass so Sprinkle node is parent to wet grass node\n• Rain also wet the grass, therefore Sprinkle causes wet grass so Rain node is parent to wet grass node\n• if there is rain, there is no need to sprinkle, therefore there Is a negative relation between Sprinkle and rain node. So, rain node is parent to Sprinkle node\n\nChain rule implementation is :\n\nP(S,R,W) = P(W/S,R)*P(S/R)*P(R)\n\nLatent Variable Introduction\n\nWiki definition: In statisticslatent variables (from Latinpresent participle of lateo (“lie hidden”), as opposed to observable variables), are variables that are not directly observed but are rather inferred (through a mathematical model) from other variables that are observed (directly measured)\n\nIn my words: Latent variables are hidden variables i.e. they are not observed. Latent variables are rather inferred and are been thought as the cause of the observed variables. Mostly in Bayesian models they are used when we end up in cycle generation in our model. Latent variables help us in simplifying the mathematical solution of our problem, but this is not always correct.\n\nLet us see with some examples\n\nSuppose we have a problem , hunger, eat and work . if we create a Bayesian model, the model looks like:",
null,
"The above model reads like, if we work, we feel hunger. If we feel hunger – we eat. If we eat, we have energy to work.\n\nNow this above model has a cycle in it and thus if chain rule is applied to it, the chain will become infinitely long. So, the above Bayesian model is not correct. Thus, we need to introduce the Latent variable here, let us call it as T",
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"Now the above mode states that, T is responsible for eat, hunger and work to happen. This variable T is not observed but can be inferred as the cause of happening of work, eat and hunger. This assumption seems to be correct also, as in a biological body – something resides in it that pushes it to eat and work, even though we cannot observe it physically.\n\nLet us write the chain rule equation for the above model:\n\nP(W,E,H,T) = P(W/T)*P(E/T)*P(H/T)*P(T)\n\nAnother Example\n\nLet us see another example. Suppose we have following variables: GPA, IQ, School. The model reads like if a person has good IQ, he/she will get good School and GPA, if he/she got good School, he/she will have good IQ and may get good GPA. If he/she gets good GPA, he/she may have good IQ, and he/she may be from good School. The model looks like:",
null,
"Now from above description of the model, all the nodes are connected to every other node. Thus chain rule cannot be applied to this model. So we need a Latent variable to be introduced here. Let us call the new Latent variable as I. The new model looks like:",
null,
"Now we read the above model as Latent variable I is responsible for all the other three variables. Now the chain rule can easily be applied. And it looks like:\n\nP(S,G,Q,I) = P(S/I)*P(G/I)*P(Q/I)*P(I)\n\nHence in this post we saw how to model and create Latent Variables. They mostly help in reducing the complexity of the problem.\n\nNow from then next post we will start much interesting part of Bayesian inferencing using the above Latent variables wherever required."
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https://office365-office.com/qa/do-you-round-up-if-its-5.html | [
"",
null,
"# Do You Round Up If It’S 5?\n\n## How do you round up to multiples of 5?\n\nSupposing that the number you want to round to closest 5 resides in cell A2, you can use on of the following formulas:To round a number down to nearest 5: =FLOOR(A2, 5)To round a number up to nearest 5: =CEILING(A2, 5)To round a number up or down to nearest 5: =MROUND(A2, 5).\n\n## How do you round to 2 decimal places?\n\nRounding to decimal placeslook at the first digit after the decimal point if rounding to one decimal place or the second digit for two decimal places.draw a vertical line to the right of the place value digit that is required.look at the next digit.if it’s 5 or more, increase the previous digit by one.More items…\n\n## How do you round off to the nearest 10000?\n\nThe rule for rounding to the nearest ten thousand is to look at the last four digits. If the last four digits are 5,000 or greater, then we round our ten thousands digit up, and if it is less than 5,000, then we keep our ten thousands digit the same. For example, 5,765 rounds up to 10,000. 43,567 rounds down to 40,000.\n\n## How do you round up a number?\n\nHere’s the general rule for rounding: If the number you are rounding is followed by 5, 6, 7, 8, or 9, round the number up. Example: 38 rounded to the nearest ten is 40. If the number you are rounding is followed by 0, 1, 2, 3, or 4, round the number down.\n\n## Why is Excel rounding my numbers?\n\nSome numbers having one or more decimal places may appear to rounded on your worksheet. If this isn’t the result of applying a rounding function, this can happen when the column isn’t wide enough for the entire number. … on the Home tab until you reach the number of decimal places you need to display.\n\n## Why is Excel not rounding up?\n\nThere are several possible reasons but the most common are these: Column width too narrow and doesn’t display the whole number. The number of decimal places is set to fewer digits than the actual decimal places. The number is too large and exceeds 15 digits; Excel is limited to display only 15 significant digits.\n\n## Why do you round up when it is a 5?\n\nThe reason is that 5 is directly in the middle of the digits we round, so we must round it up half the time, and down half the time. To make this more clear, look at the digits we round to another number: 1, 2, 3, 4 we round down.\n\n## How do you round to the nearest 5?\n\nRounding to the nearest 5 Rule: if the one’s digit is 1 or 2 round down to the nearest 5. if the one’s digit is 3 or 4 round up to the nearest 5. if the one’s digit is 6 or 7 round down to the nearest 5. if the one’s digit is 8 or 9 round up to the nearest 5.\n\n## What is the round off rule?\n\nRule 1: Determine what your rounding digit is and look at the digit to the right of it (highlighted digit). If the highlighted digit is 1, 2, 3, 4 simply drop all digits to the right of rounding digit. Example: 3.423 may be rounded off to 3.42 when rounded off to the nearest hundredths place.\n\n## What is 2.738 to 2 decimal places?\n\nSince it is larger than, you can round the 38 up to 40 . So now the number you have is 2.740 , but since the 0 does not need to be included, you have 2.74 , which is 2 decimal places.\n\n## How do you calculate round off?\n\nRounding off is a kind of estimating. To round off decimals: Find the place value you want (the “rounding digit”) and look at the digit just to the right of it. If that digit is less than 5, do not change the rounding digit but drop all digits to the right of it."
]
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null,
"https://mc.yandex.ru/watch/70836331",
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]
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https://corporatefinanceinstitute.com/resources/career-map/sell-side/capital-markets/kairi-relative-index-kri/ | [
"# Kairi Relative Index (KRI)\n\nA momentum oscillator indicator that measures the change of an asset price during a specific time period\n\n## What is the Kairi Relative Index (KRI)?\n\nThe Kairi Relative Index (KRI) is a type of oscillator indicator. It measures the deviation of an asset price from its daily average over a certain period of time, usually 10 to 20 days. The index was originally developed in Japan and now serves as a tool for technical analysis.\n\nTraders can time their transactions based on the KRI. However, the KRI is less commonly used nowadays, as other oscillators, such as the Relative Strength Index (RSI), are gaining more popularity.\n\n### Summary\n\n• The Kairi Relative Index (KRI) is a momentum oscillator indicator that measures the change of an asset price during a specific time period. The index compares the current price with its moving average over the look-back period.\n• The KRI swings up and down around zero. A value far above zero indicates overbought and gives a selling signal, while a value far below zero indicates oversold and gives a buying signal.\n• Other types of momentum oscillators include the Relative Strength Index (RSI), stochastic oscillator, etc.\n\n### Understanding the Kairi Relative Index\n\nThe Kairi Relative Index (KRI) is a momentum oscillator indicator that supports technical analysis by monitoring the rate of price change. A momentum oscillator measures the change of an asset price during a specific time period. It is usually calculated by dividing the current price by the price of a previous period. Based on the formula, the calculation of different oscillator indicators varies to serve different purposes better.\n\nTypically, the result is multiplied by 100, which gives a threshold of 100. A value below 100 indicates a negative momentum with decreasing price, and a value above 100 indicates a positive momentum with increasing price over the chosen time period.\n\nNo upper or lower boundaries exist for a momentum oscillator, but a high or low value supports an assumption that the trend will continue and thus gives a buy or sell signal. However, if the price increase or decrease is not confirmed by its momentum oscillator, a mean-reversal movement is signaled.",
null,
"The KRI is one of the many types of momentum oscillators. It compares the current price with the moving average of the price during the current chosen period is calculated as follows:",
null,
"Where:\n\n• Price – Current price\n• MVA – Simple moving average of price over the current period\n\nDue to a different method of calculation. The KRI can be either a positive or a negative number. When the current price is higher than the MVA, the KRI is positive; when the current price is lower than the MVA, the KRI is negative.\n\nFor example, if the price of a security is currently \\$50 and the current 10-day MVA is \\$60, the KRI will be -16.7 [(\\$50–\\$60) / \\$60*100]. If the price increases to \\$90 two weeks later, and the 10-day MVA also increases to \\$70, the KRI will be 28.6 [(\\$90–\\$70) / \\$70*100].\n\nLet’s assume a value of 28.6 is a high point for the KRI of the asset. It means the asset might be overbought, and a movement of mean-reversal might occur soon.\n\n### The Use of the Kari Relative Index\n\nThe KRI, as a tool for technical analysis, facilitates traders to make buy and sell decisions. The KRI is presented as a line moving up and down around zero. An upward sloping KRI line indicates an upward momentum of price, while a downward sloping KRI line indicates a downward momentum of price. The momentum is strong when the line is far away from zero.\n\nWhen the KRI is significantly high above zero, the asset is probably overbought (overvalued). This can be a selling signal to traders, expecting for mean-reversal or corrective pullback. A KRI significantly down below zero is a signal of oversold (undervalued), and traders typically make buy decisions based on the information.\n\n### Other Oscillators\n\nThe Relative Strength Index (RSI) is another type of momentum oscillator, which is now used more often than the KRI. It measures the speed of price movements. Unlike the KRI, which compares the current asset price and the moving average, the RSI compares the previous average gain and loss with the current gain and loss.\n\nThe RSI moves between 0 and 100 with a threshold at 70. Typically, an RSI above 70 indicates the asset may be oversold (overvalued), while a value below 30 indicates the asset is oversold (undervalued) and gives a selling signal.\n\nA stochastic oscillator compares the most recent closing price with the difference between the high and low of a previous period. The mean-reversal level of a stochastic oscillator depends on the price moving range over the look-back period, which is 14 days typically.\n\nThe use of a stochastic oscillator is based on the assumption that the most recent closing price should settle along with the current trend. Thus, it works most effectively within a consistent trading range, for example, in a choppy market."
]
| [
null,
"https://corporatefinanceinstitute.com/resources/career-map/sell-side/capital-markets/kairi-relative-index-kri/",
null,
"https://corporatefinanceinstitute.com/resources/career-map/sell-side/capital-markets/kairi-relative-index-kri/",
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]
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https://www.physicsforums.com/threads/problem-understand-pressure.709450/ | [
"# Problem understand Pressure\n\nPsychonautQQ\n\n## Homework Statement\n\nSo imagine a barrel with the area on top = ∏r^2\nThere is a tube of water 12 meters high ontop of the barrel.\nWhat is the net force exerted on the barrel?\n\nF = PA\nF = Density*Height*Gravity*(Area of barrel)\nHow come the volume of water in the tube isn't a factor in the net force? Only the area of the top of the barrel matter? I used this equation and I got the right answer. My teacher said the volume of water in the tube DOESN\"T matter, and I don't understand how that is possible. More total water = more total force?? Apparently not... Help me understand :)\n\nabrewmaster\nYour teacher is very misleading if that's what he said, the sheer volume of water does not always affect the pressure. The height of the water is in the equation, pressure changes with depth. As you get deeper underwater there becomes more and more pressure; however, if you make the volume of water larger but don't increase the depth (add more surface area of water) then the pressure at the bottom doesn't change. So for example you increase the height of the tube of water to 13 meters then your pressure and volume of water changes but if you simply add more water to the tube but don't change the height of the water (larger diameter tube) your volume changes but not your pressure.\n\nPsychonautQQ\nLets say the cross sectional area of the tube is 5 and it's 12 meters high and filled with water\nThat would produce more net force if the cross sectional area is 2 and it's 12 meters high,, right? Yet this variable is not represented in the equations and I still got the correct answer\n\nMentor\nYou might want to have a look at this diagram (from hyperphysics):",
null,
"Note that the shape of the container, and thus the total volume of fluid, has no bearing on the pressure at a given depth below the surface.\n\nSee: Static Fluid Pressure\n\nHomework Helper\nGold Member\nLets say the cross sectional area of the tube is 5 and it's 12 meters high and filled with water\nThat would produce more net force if the cross sectional area is 2 and it's 12 meters high,, right? Yet this variable is not represented in the equations and I still got the correct answer\nI think the problem is not worded correctly. If you were to add water to the barrel by say just filling it slowly from a water pitcher, the pressure at the top of the barrel would be zero. But if you fill it from the 12 m high thin tube, the pressure at the bot of the tube would be ρgh, where h = 12, and since the pressure would be uniformly distributed at that same value to the underside of the barrel top, the force on that top would be ρgh(A), where A is the area of the barrel top. Might be enough to cause the barrel to burst at its top or sides. Pressure at the bottom of the barrel would be even greater.\n\nPsychonautQQ\nI think the problem is not worded correctly. If you were to add water to the barrel by say just filling it slowly from a water pitcher, the pressure at the top of the barrel would be zero. But if you fill it from the 12 m high thin tube, the pressure at the bot of the tube would be ρgh, where h = 12, and since the pressure would be uniformly distributed at that same value to the underside of the barrel top, the force on that top would be ρgh(A), where A is the area of the barrel top. Might be enough to cause the barrel to burst at its top or sides. Pressure at the bottom of the barrel would be even greater.\n\nBut it's the force that the water EXERTS onto the barrel lid. So you're telling me if the barrel had an infinite surface area the force would be infinite as well even though the tube is still just that long slim volume?\n\nvoko\nI do not recommend going for an infinite or even \"very large\" volume here; it does not simplify anything and introduces instead a lot of complications. In a reasonably sized volume, pressure will depend only on the height of the water column above, and the force exerted onto the lid will be proportional to that pressure and the area of the lid.\n\nStaff Emeritus\nBut it's the force that the water EXERTS onto the barrel lid. So you're telling me if the barrel had an infinite surface area the force would be infinite as well even though the tube is still just that long slim volume?\nLet's get rid of that appeal to ridicule and replace \"infinite\" with \"large\":\nBut it's the force that the water EXERTS onto the barrel lid. So you're telling me if the barrel had an large surface area the force would be large as well even though the tube is still just that long slim volume?\nNow that's exactly right. It's the principle behind a hydraulic jack.\n\nHomework Helper\n\n## Homework Statement\n\nSo imagine a barrel with the area on top = ∏r^2\nThere is a tube of water 12 meters high ontop of the barrel.\nWhat is the net force exerted on the barrel?\n\nF = PA\nF = Density*Height*Gravity*(Area of barrel)\nHow come the volume of water in the tube isn't a factor in the net force? Only the area of the top of the barrel matter? I used this equation and I got the right answer. My teacher said the volume of water in the tube DOESN\"T matter, and I don't understand how that is possible. More total water = more total force?? Apparently not... Help me understand :)\n\nYour formula does involve volume. Height*(Area of barrel)=Volume of water. But if area of the barrel and height are fixed you can't change volume independently."
]
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"https://www.physicsforums.com/attachments/fp2-gif.163806/",
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https://www.studypug.com/au/au-year12/factorial-notation | [
"# Factorial notation\n\n### Factorial notation\n\n#### Lessons\n\n• 1.\nHow many ways are there to arrange 4 different books side by side on a bookshelf?\n\n• 2.\nfactorial notation: n! = n (n-1) (n-2) (n-3) (n-4) . . . . . (5) (4) (3) (2) (1)\nby definition : 0! = 1\na)\nEvaluate: 5!\n\nb)\nEvaluate: $\\frac{{7!}}{{5!}}$\n\nc)\nSimplify: $\\frac{{\\left( {n + 3} \\right)!}}{{n!}}$\n\nd)\nSimplify: $\\frac{{\\left( {n - 1} \\right)!}}{{\\left( {n + 2} \\right)!}}$\n\ne)\nSimplify: $\\frac{{\\left( {n + 1} \\right)!\\;\\;\\left( {n - 3} \\right)!}}{{{{(n!)}^2}}}$\n\nf)\nSolve: $\\frac{{n!}}{{\\left( {n - 2} \\right)!\\;\\;3!}} = 7$\n\n• 3.\narrangement of words \"without repititions\" = n!\nDetermine the number of different arrangements of all the letters in the following words:\na)\nDOG\n\nb)\nMATH\n\nc)\nCOMPUTER\n\n• 4.\narrangement of words \"with repititions\" = $\\frac{{n!}}{{\\left( {{1^{st}}\\;repetition} \\right)!\\;\\;\\;\\left( {{2^{nd}}\\;repetition} \\right)!\\;\\;\\;\\left( {{3^{rd}}\\;repetition} \\right)!\\; \\ldots ..}}$\nDetermine the number of different arrangements of all the letters in the following words:\na)\nABC vs ABB\n\nb)\n\nc)\nBANANA\n\nd)\nREPETITION\n\n• 5.\narrangement with restrictions: must deal with the restrictions first!\na)\n\nb)\nDetermine the number of different arrangements of all the letters in the word: COMPUTER\n(i) if there are no restrictions\n(ii) if the vowels must be together\n(iii) the vowels must not be together\n\nc)\ni) How many ways are there to arrange 3 Math books (Math 10, Math 11, Math 12), 2 physics books (Phys 11, Phys 12), and 5 English (Eng 8, Eng 9, Eng 10, Eng 11, Eng 12) on a bookshelf?\nii) What if the books on each subject must be kept together?\n\n• 6.\nseating arrangement\na)\nHow many ways can 4 girls and 4 boys sit in a row, if:\ni) they can sit anywhere?\nii) all the girls must sit together, and all the boys must sit together?\niii) all the girls must sit together, while the boys can pick their own seats?\niv) girls and boys alternate?\n\nb)\nThere are 3 couples, and they need to sit together. How many different ways can these 3 couples sit in a row?\n\nc)\nThere are 7 people A, B, C, D, E, F, and G sitting in a row. How many different seating arrangements are there, if:\ni) A must be to the left of B, but they do not need to sit together?\nii) A and B must sit together?\niii) A and B cannot sit together?\n\n• 7.\nSeating arrangements are considered to be different only when the positions of the people are different relative to each other.\nHow many seating arrangements are possible for 7 people sitting around a round table?"
]
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https://math.stackexchange.com/questions/271927/why-historically-do-we-multiply-matrices-as-we-do/271937 | [
"# Why, historically, do we multiply matrices as we do?\n\nMultiplication of matrices — taking the dot product of the $i$th row of the first matrix and the $j$th column of the second to yield the $ij$th entry of the product — is not a very intuitive operation: if you were to ask someone how to mutliply two matrices, he probably would not think of that method. Of course, it turns out to be very useful: matrix multiplication is precisely the operation that represents composition of transformations. But it's not intuitive. So my question is where it came from. Who thought of multiplying matrices in that way, and why? (Was it perhaps multiplication of a matrix and a vector first? If so, who thought of multiplying them in that way, and why?) My question is intact no matter whether matrix multiplication was done this way only after it was used as representation of composition of transformations, or whether, on the contrary, matrix multiplication came first. (Again, I'm not asking about the utility of multiplying matrices as we do: this is clear to me. I'm asking a question about history.)\n\n• As to the last, matrix multiplication definitely came first (centuries first), and I'm reasonably certain from a compact representation of systems of linear equations. Leibniz already had a determinant formula. As I have no historic sources for first use, this doesn't answer your question though. – gnometorule Jan 7 '13 at 4:19\n• Matrices are linear operators and have meaning only when it acts on vectors. Given matrices $A$ and $B$, what would we want the operator/matrix $BA$ to mean? Ideally, we would want $BA$ to mean the following. For all vectors $x$, we want $(BA)x = B(Ax)$ Once we have this i.e. $(BA)x = B(Ax)$ for all $x$, then we are forced to live with the way we currently multiply matrices. And as to why matrix-vector product is defined in the way it is, the primary reason for introducing matrices was to handle linear transformation in a notationally convenient way. – user17762 Jan 7 '13 at 4:21\n• There is another question, of course, which is not so much why matrix multiplication was defined like this, but why it stuck - why this apparently curious definition took off, and didn't die the death of so many putative definitions. And that was because it proved mathematically fruitful. – Mark Bennet Jan 7 '13 at 8:47\n• Possible duplicate of math.stackexchange.com/questions/192835/…. – lhf Jun 2 '14 at 12:45\n• Why is it \"not intuitive\"? If you ask someone how to multiply two matrices and they think about what that multiplication is supposed to mean, they absolutely will come up with the usual definition. – Matthew Towers Nov 30 '16 at 15:25\n\nMatrix multiplication is a symbolic way of substituting one linear change of variables into another one. If $x' = ax + by$ and $y' = cx+dy$, and $x'' = a'x' + b'y'$ and $y'' = c'x' + d'y'$ then we can plug the first pair of formulas into the second to express $x''$ and $y''$ in terms of $x$ and $y$: $$x'' = a'x' + b'y' = a'(ax + by) + b'(cx+dy) = (a'a + b'c)x + (a'b + b'd)y$$ and $$y'' = c'x' + d'y' = c'(ax+by) + d'(cx+dy) = (c'a+d'c)x + (c'b+d'd)y.$$ It can be tedious to keep writing the variables, so we use arrays to track the coefficients, with the formulas for $x'$ and $x''$ on the first row and for $y'$ and $y''$ on the second row. The above two linear substitutions coincide with the matrix product $$\\left( \\begin{array}{cc} a'&b'\\\\c'&d' \\end{array} \\right) \\left( \\begin{array}{cc} a&b\\\\c&d \\end{array} \\right) = \\left( \\begin{array}{cc} a'a+b'c&a'b+b'd\\\\c'a+d'c&c'b+d'd \\end{array} \\right).$$ So matrix multiplication is just a bookkeeping device for systems of linear substitutions plugged into one another (order matters). The formulas are not intuitive, but it's nothing other than the simple idea of combining two linear changes of variables in succession.\n\nMatrix multiplication was first defined explicitly in print by Cayley in 1858, in order to reflect the effect of composition of linear transformations. See paragraph 3 at http://darkwing.uoregon.edu/~vitulli/441.sp04/LinAlgHistory.html. However, the idea of tracking what happens to coefficients when one linear change of variables is substituted into another (which we view as matrix multiplication) goes back further. For instance, the work of number theorists in the early 19th century on binary quadratic forms $ax^2 + bxy + cy^2$ was full of linear changes of variables plugged into each other (especially linear changes of variable that we would recognize as coming from ${\\rm SL}_2({\\mathbf Z})$). For more on the background, see the paper by Thomas Hawkins on matrix theory in the 1974 ICM. Google \"ICM 1974 Thomas Hawkins\" and you'll find his paper among the top 3 hits.\n\nHere is an answer directly reflecting the historical perspective from the paper Memoir on the theory of matrices By Authur Cayley, 1857. This paper is available here.\n\nThis paper is credited with \"containing the first abstract definition of a matrix\" and \"a matrix algebra defining addition, multiplication, scalar multiplication and inverses\" (source).\n\nIn this paper a nonstandard notation is used. I will do my best to place it in a more \"modern\" (but still nonstandard) notation. The bulk of the contents of this post will come from pages 20-21.\n\nTo introduce notation, $$(X,Y,Z)= \\left( \\begin{array}{ccc} a & b & c \\\\ a' & b' & c' \\\\ a'' & b'' & c'' \\end{array} \\right)(x,y,z)$$\n\nwill represent the set of linear functions $(ax + by + cz, a'z + b'y + c'z, a''z + b''y + c''z)$ which are then called $(X,Y,Z)$.\n\nCayley defines addition and scalar multiplication and then moves to matrix multiplication or \"composition\". He specifically wants to deal with the issue of:\n\n$$(X,Y,Z)= \\left( \\begin{array}{ccc} a & b & c \\\\ a' & b' & c' \\\\ a'' & b'' & c'' \\end{array} \\right)(x,y,z) \\quad \\text{where} \\quad (x,y,z)= \\left( \\begin{array}{ccc} \\alpha & \\beta & \\gamma \\\\ \\alpha' & \\beta' & \\gamma' \\\\ \\alpha'' & \\beta'' & \\gamma'' \\\\ \\end{array} \\right)(\\xi,\\eta,\\zeta)$$\n\nHe now wants to represent $(X,Y,Z)$ in terms of $(\\xi,\\eta,\\zeta)$. He does this by creating another matrix that satisfies the equation:\n\n$$(X,Y,Z)= \\left( \\begin{array}{ccc} A & B & C \\\\ A' & B' & C' \\\\ A'' & B'' & C'' \\\\ \\end{array} \\right)(\\xi,\\eta,\\zeta)$$\n\nHe continues to write that the value we obtain is:\n\n\\begin{align}\\left( \\begin{array}{ccc} A & B & C \\\\ A' & B' & C' \\\\ A'' & B'' & C'' \\\\ \\end{array} \\right) &= \\left( \\begin{array}{ccc} a & b & c \\\\ a' & b' & c' \\\\ a'' & b'' & c'' \\end{array} \\right)\\left( \\begin{array}{ccc} \\alpha & \\beta & \\gamma \\\\ \\alpha' & \\beta' & \\gamma' \\\\ \\alpha'' & \\beta'' & \\gamma'' \\\\ \\end{array} \\right)\\\\[.25cm] &= \\left( \\begin{array}{ccc} a\\alpha+b\\alpha' + c\\alpha'' & a\\beta+b\\beta' + c\\beta'' & a\\gamma+b\\gamma' + c\\gamma'' \\\\ a'\\alpha+b'\\alpha' + c'\\alpha'' & a'\\beta+b'\\beta' + c'\\beta'' & a'\\gamma+b'\\gamma' + c'\\gamma'' \\\\ a''\\alpha+b''\\alpha' + c''\\alpha'' & a''\\beta+b''\\beta' + c''\\beta'' & a''\\gamma+b''\\gamma' + c''\\gamma''\\end{array} \\right)\\end{align}\n\nThis is the standard definition of matrix multiplication. I must believe that matrix multiplication was defined to deal with this specific problem. The paper continues to mention several properties of matrix multiplication such as non-commutativity, composition with unity and zero and exponentiation.\n\nHere is the written rule of composition:\n\nAny line of the compound matrix is obtained by combining the corresponding line of the first component matrix successively with the several columns of the second matrix (p. 21)\n\n• Should the set of linear functions $(ax + by + cz, a'z + b'y + c'z, a''z + b''y + c''z)$ be $(ax + by + cz, a'x + b'y + c'z, a''x + b''y + c''z)$? – Vilhelm Gray May 11 '17 at 18:11\n• Brad's is THE answer, I think. The eventual coefficients obtained from a double transformation of (x, y, z) require row elements of the left transform matrices to be multiplied by their corresponding column element of the right transform matrix. My own old idea was that if multiplication of a vector by a matrix is done like A (x, y, z) = (x', y', z'), i.e. using row vectors, then the resulting system will not \"read\" as clearly as if the (x, y, z) & (x', y', z') were written as a column vectors. And this is only for a 3-D system. Higher order systems harder still. But Brad's idea is dead-on. – Trunk Dec 28 '17 at 15:57\n\n\\begin{align} u & = 3x + 7y \\\\ v & = -2x + 11y \\\\ \\\\ \\\\ \\\\ p & =13u-20v \\\\ q & = 2u+6v \\end{align} Given $x$ and $y$, how do you find $p$ and $q$? How do you write: \\begin{align} p & = \\bullet\\, x + \\bullet\\, y \\\\ q & = \\bullet\\, x+\\bullet\\, y\\quad\\text{?} \\end{align} What numbers go where the four $\\bullet$'s are?\n\nThat is what matrix multiplication is. The rationale is mathematical, not historical."
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https://intl.siyavula.com/read/science/grade-11/geometrical-optics/05-geometrical-optics-07 | [
"We think you are located in United States. Is this correct?\n\n# 5.7 Critical angles and total internal reflection\n\n## 5.7 Critical angles and total internal reflection (ESBN9)\n\n### Total internal reflection (ESBNB)\n\nYou may have noticed when experimenting with ray boxes and glass blocks in the previous section that sometimes, when you changed the angle of incidence of the light, it was not refracted out into the air, but was reflected back through the block. When the entire incident light ray travelling through an optically denser medium is reflected back at the boundary between that medium and another of lower optical density, instead of passing through and being refracted, this is called total internal reflection.\n\nAs we increase the angle of incidence, we reach a point where the angle of refraction is $$\\text{90}$$$$\\text{°}$$ and the refracted ray travels along the boundary of the two media. This angle of incidence is called the critical angle.\n\nCritical angle\n\nThe critical angle is the angle of incidence where the angle of refraction is $$\\text{90}$$$$\\text{°}$$. The light must travel from an optically more dense medium to an optically less dense medium.",
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"Figure 5.15: When the angle of incidence is equal to the critical angle, the angle of refraction is equal to $$\\text{90}$$$$\\text{°}$$.\n\nIf the angle of incidence is bigger than this critical angle, the refracted ray will not emerge from the medium, but will be reflected back into the medium. This is called total internal reflection.\n\nThe conditions for total internal reflection are:\n\n1. light is travelling from an optically denser medium (higher refractive index) to an optically less dense medium (lower refractive index).\n2. the angle of incidence is greater than the critical angle.",
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"Figure 5.16: When the angle of incidence is greater than the critical angle, the light ray is reflected at the boundary of the two media and total internal reflection occurs.\n\nEach pair of media have their own unique critical angle. For example, the critical angle for light moving from glass to air is $$\\text{42}$$$$\\text{°}$$, and that of water to air is $$\\text{48,8}$$$$\\text{°}$$.\n\ntemp text\n\nA recommended experiment for informal assessment is also included. This covers determining the critical angle for light travelling through a rectangular glass block. Learners will need a rectangular glass block, a $$\\text{360}$$$$\\text{°}$$ protractor, pencil, paper, ruler and a ray box. Learners should all get similar results at the end of the experiment.\n\n#### Calculating the critical angle\n\nInstead of always having to measure the critical angles of different materials, it is possible to calculate the critical angle at the surface between two media using Snell's Law. To recap, Snell's Law states: $n_1 \\sin \\theta_1 = n_2 \\sin \\theta_2$ where $$n_1$$ is the refractive index of material $$\\text{1}$$, $$n_2$$ is the refractive index of material $$\\text{2}$$, $$\\theta_1$$ is the angle of incidence and $$\\theta_2$$ is the angle of refraction. For total internal reflection we know that the angle of incidence is the critical angle. So, $\\theta_1 = \\theta_c.$\n\nHowever, we also know that the angle of refraction at the critical angle is $$\\text{90}$$$$\\text{°}$$. So we have: $\\theta_2=90^{\\circ}.$\n\nWe can then write Snell's Law as: $n_1 \\sin{\\theta_c} = n_2 \\sin{90^{\\circ}}$\n\nSolving for $$\\theta_c$$ gives: \\begin{align*} n_1 \\sin \\theta_c &= n_2 \\sin \\text{90}\\text{°} \\\\ \\sin \\theta_c &= \\frac{n_2}{n_1}(1) \\end{align*} $\\boxed{\\therefore \\theta_c = \\sin^{-1}\\left(\\frac{n_2}{n_1}\\right)}$\n\nRemember that for total internal reflection the incident ray is always in the denser medium!\n\n## Worked example 5: Critical angle 1\n\nGiven that the refractive indices of air and water are $$\\text{1,00}$$ and $$\\text{1,33}$$ respectively, find the critical angle.\n\n### Determine how to approach the problem\n\nWe can use Snell's law to determine the critical angle since we know that when the angle of incidence equals the critical angle, the angle of refraction is $$\\text{90}$$$$\\text{°}$$.\n\n### Solve the problem\n\n\\begin{align*} n_1 \\sin{\\theta_c} &= n_2 \\sin{90^{\\circ}}\\\\ \\theta_c &=\\sin^{-1}\\left(\\frac{n_2}{n_1}\\right) \\\\ &=\\sin^{-1}\\left(\\frac{1}{\\text{1,33}}\\right) \\\\ &= \\text{48,8}\\text{°} \\end{align*}\n\nThe critical angle for light travelling from water to air is $$\\text{48,8}$$$$\\text{°}$$.\n\n## Worked example 6: Critical angle 2\n\nComplete the following ray diagrams to show the path of light in each situation.",
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"### Identify what is given and what is asked\n\nThe critical angle for water is $$\\text{48,8}$$$$\\text{°}$$\n\nWe are asked to complete the diagrams.\n\nFor incident angles smaller than $$\\text{48,8}$$$$\\text{°}$$ refraction will occur.\n\nFor incident angles greater than $$\\text{48,8}$$$$\\text{°}$$ total internal reflection will occur.\n\nFor incident angles equal to $$\\text{48,8}$$$$\\text{°}$$ refraction will occur at $$\\text{90}$$$$\\text{°}$$.\n\nThe light must travel from a medium with a higher refractive index (higher optical density) to a medium with lower refractive index (lower optical density).\n\n### Complete the diagrams",
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"Refraction occurs (ray is bent away from the normal).",
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"Total internal reflection occurs.",
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"$$\\theta_c = \\text{48,8}\\text{°}.$$",
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"Refraction towards the normal (air is less dense than water).\n\ntemp text\n\n#### Fibre optics\n\nTotal internal reflection is a very useful natural phenomenon since it can be used to confine light. One of the most common applications of total internal reflection is in fibre optics. An optical fibre is a thin, transparent fibre, usually made of glass or plastic, for transmitting light. Optical fibres are usually thinner than a human hair! The construction of a single optical fibre is shown in Figure 5.17.\n\nThe basic functional structure of an optical fibre consists of an outer protective cladding and an inner core through which light pulses travel. The overall diameter of the fibre is about $$\\text{125}$$ $$\\text{μm}$$ ($$\\text{125} \\times \\text{10}^{-\\text{6}}$$ $$\\text{m}$$) and that of the core is just about $$\\text{50}$$ $$\\text{μm}$$ ($$\\text{50} \\times \\text{10}^{-\\text{6}}$$ $$\\text{m}$$). The difference in refractive index of the cladding and the core allows total internal reflection to occur in the same way as happens at an air-water surface. If light is incident on a cable end with an angle of incidence greater than the critical angle then the light will remain trapped inside the glass strand. In this way, light travels very quickly down the length of the cable.\n\n##### Fibre Optics in Telecommunications\n\nOptical fibres are most common in telecommunications, because information can be transported over long distances, with minimal loss of data. This gives optical fibres an advantage over conventional cables.\n\nSignals are transmitted from one end of the fibre to another in the form of laser pulses. A single strand of fibre optic cable is capable of handling over $$\\text{3 000}$$ transmissions at the same time which is a huge improvement over the conventional co-axial cables. Multiple signal transmission is achieved by sending individual light pulses at slightly different angles. For example if one of the pulses makes a $$\\text{72,23}$$$$\\text{°}$$ angle of incidence then a separate pulse can be sent at an angle of $$\\text{72,26}$$$$\\text{°}$$! The transmitted signal is received almost instantaneously at the other end of the cable since the information coded onto the laser travels at the speed of light! During transmission over long distances repeater stations are used to amplify the signal which has weakened by the time it reaches the station. The amplified signals are then relayed towards their destination and may encounter several other repeater stations on the way.\n\n##### Fibre optics in medicine\n\nOptic fibres are used in medicine in endoscopes.\n\nEndoscopy means to look inside and refers to looking inside the human body for diagnosing medical conditions.\n\nThe main part of an endoscope is the optical fibre. Light is shone down the optical fibre and a medical doctor can use the endoscope to look inside the body of a patient. Endoscopes can be used to examine the inside of a patient's stomach, by inserting the endoscope down the patient's throat.\n\nEndoscopes also allow minimally invasive surgery. This means that a person can be diagnosed and treated through a small incision (cut). This has advantages over open surgery because endoscopy is quicker and cheaper and the patient recovers more quickly. The alternative is open surgery which is expensive, requires more time and is more traumatic for the patient.\n\n## Total internal reflection and fibre optics\n\nTextbook Exercise 5.5\n\nDescribe total internal reflection by using a diagram and referring to the conditions that must be satisfied for total internal reflection to occur.\n\nIf the angle of incidence is bigger than the critical angle, the refracted ray will not emerge from the medium, but will be reflected back into the medium. This is called total internal reflection.\n\nThe critical angle occurs when the angle of incidence where the angle of refraction is $$\\text{90}$$$$\\text{°}$$. The light must travel from an optically more dense medium to an optically less dense medium.\n\nThe conditions for total internal reflection are the the light is travelling from an optically denser medium (higher refractive index) to an optically less dense medium (lower refractive index) and that the angle of incidence is greater than the critical angle.\n\nRepresenting this on a diagram gives:",
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"Define what is meant by the critical angle when referring to total internal reflection. Include a ray diagram to explain the concept.\n\nThe critical angle occurs when the angle of incidence where the angle of refraction is $$\\text{90}$$$$\\text{°}$$. The light must travel from an optically more dense medium to an optically less dense medium.",
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"Will light travelling from diamond to silicon ever undergo total internal reflection?\n\nDiamond (index of refraction is about 3) is less optically dense than silicon (index of refraction is about 4) and so total internal reflection cannot occur.\n\nWill light travelling from sapphire to diamond undergo total internal reflection?\n\nSapphire (index of refraction is $$\\text{1,77}$$) is less optically dense than diamond (index of refraction is about 3) and so total internal reflection cannot occur.\n\nWhat is the critical angle for light travelling from air to acetone?\n\nThe critical angle is:\n\n\\begin{align*} n_1 \\sin{\\theta_c} &= n_2 \\sin{90^{\\circ}}\\\\ \\theta_c &=\\sin^{-1}\\left(\\frac{n_2}{n_1}\\right) \\\\ &=\\sin^{-1}\\left(\\frac{1}{\\text{1,36}}\\right) \\\\ &= \\text{47,33}\\text{°} \\end{align*}\n\nLight travelling from diamond to water strikes the interface with an angle of incidence of $$\\text{86}$$$$\\text{°}$$ as shown in the picture. Calculate the critical angle to determine whether the light be totally internally reflected and so be trapped within the diamond.",
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"The critical angle is:\n\n\\begin{align*} n_1 \\sin{\\theta_c} &= n_2 \\sin{90^{\\circ}}\\\\ \\theta_c &=\\sin^{-1}\\left(\\frac{n_2}{n_1}\\right) \\\\ &=\\sin^{-1}\\left(\\frac{\\text{1,33}}{\\text{2,419}}\\right) \\\\ &= \\text{33,35}\\text{°} \\end{align*}\n\nThe angle of incidence is greater than the critical angle and so the light will be trapped within the diamond.\n\nWhich of the following interfaces will have the largest critical angle?\n\n1. a glass to water interface\n2. a diamond to water interface\n3. a diamond to glass interface\na glass to water interface\n\nIf a fibre optic strand is made from glass, determine the critical angle of the light ray so that the ray stays within the fibre optic strand.\n\nThe critical angle is:\n\n\\begin{align*} n_1 \\sin{\\theta_c} &= n_2 \\sin{90^{\\circ}}\\\\ \\theta_c &=\\sin^{-1}\\left(\\frac{n_2}{n_1}\\right) \\\\ &=\\sin^{-1}\\left(\\frac{\\text{1}}{\\text{1,5}}\\right) \\\\ &= \\text{41,8}\\text{°} \\end{align*}\n\nA glass slab is inserted in a tank of water. If the refractive index of water is $$\\text{1,33}$$ and that of glass is $$\\text{1,5}$$, find the critical angle.\n\nThe critical angle is:\n\n\\begin{align*} n_1 \\sin{\\theta_c} &= n_2 \\sin{90^{\\circ}}\\\\ \\theta_c &=\\sin^{-1}\\left(\\frac{n_2}{n_1}\\right) \\\\ &=\\sin^{-1}\\left(\\frac{\\text{1,33}}{\\text{1,5}}\\right) \\\\ &= \\text{62,46}\\text{°} \\end{align*}\n\nNote that the light must be travelling from the glass into the water for total internal reflection to occur.\n\nA diamond ring is placed in a container full of glycerin. If the critical angle is found to be $$\\text{37,4}$$$$\\text{°}$$ and the refractive index of glycerin is given to be $$\\text{1,47}$$, find the refractive index of diamond.\n\nThe refractive index is:\n\n\\begin{align*} n_1 \\sin{\\theta_c} &= n_2 \\sin{90^{\\circ}}\\\\ n_{1} &= \\frac{n_1}{\\sin{\\theta_c}} \\\\ &= \\frac{\\text{1,47}}{ \\sin \\text{37,4}\\text{°}} \\\\ &= \\text{2,42}\\text{°} \\end{align*}\n\nAn optical fibre is made up of a core of refractive index $$\\text{1,9}$$, while the refractive index of the cladding is $$\\text{1,5}$$. Calculate the maximum angle which a light pulse can make with the wall of the core. NOTE: The question does not ask for the angle of incidence but for the angle made by the ray with the wall of the core, which will be equal to $$\\text{90}$$$$\\text{°}$$ − angle of incidence.",
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"The critical angle is:\n\n\\begin{align*} n_1 \\sin{\\theta_c} &= n_2 \\sin{90^{\\circ}}\\\\ \\theta_c &=\\sin^{-1}\\left(\\frac{n_2}{n_1}\\right) \\\\ &=\\sin^{-1}\\left(\\frac{\\text{1,5}}{\\text{1,9}}\\right) \\\\ &= \\text{52,14}\\text{°} \\end{align*}\n\nThe maximum angle is: $$\\text{90}\\text{°} - \\text{52,14}\\text{°} = \\text{37,86}\\text{°}$$"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.85980767,"math_prob":0.9981526,"size":9934,"snap":"2021-31-2021-39","text_gpt3_token_len":2393,"char_repetition_ratio":0.18700907,"word_repetition_ratio":0.14757282,"special_character_ratio":0.23525266,"punctuation_ratio":0.07918552,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999542,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-26T00:40:36Z\",\"WARC-Record-ID\":\"<urn:uuid:c81b7052-1e3f-4bd8-a4eb-b09e9cd5c443>\",\"Content-Length\":\"71939\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:179e2f6b-a130-4f57-a438-f7f6040c7111>\",\"WARC-Concurrent-To\":\"<urn:uuid:1d840738-310c-46b7-a7da-9bdf9d3f550c>\",\"WARC-IP-Address\":\"104.26.8.190\",\"WARC-Target-URI\":\"https://intl.siyavula.com/read/science/grade-11/geometrical-optics/05-geometrical-optics-07\",\"WARC-Payload-Digest\":\"sha1:6O7Q5F5YUOKARLGOPHVYJE2OMTV2CCDO\",\"WARC-Block-Digest\":\"sha1:D2WV47EI75NKINZLZELOXVJ2DCWDCOEM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057787.63_warc_CC-MAIN-20210925232725-20210926022725-00111.warc.gz\"}"} |
https://topfuturepoint.com/arctan-formula/ | [
"# arctan formula\n\narctan formula: The arc tangent to x is defined as the inverse tangent function of x when x is real (x∈ℝ). When tangent to y is equal to x: tan y = x. Then the arc tangent to x is equal to the inverse tangent function of x, which is equal to y: arctic x = tan – 1 x = y .\n\nAlso, how is Atan calculated?\n\nTan and atan are useful for converting between degree of slope and percentage of slope or “run over run”. Usually “25% slope” means that b/a = 0.25 in the figure above. The corresponding degree of slope will convert to atn(b/a) degrees, or atan(b/a)*180/pi . If b/a = 25%, then x = 14 degrees.\n\nHere, why is it called arcsin?\n\nIf you have a numerical value and you want the size of an angle that has this value of sine, you get something like this, where the value is a number and the arcsine is expressed in degrees of arc. This essentially reverses the process of the sine function. It’s called an “arcsin” because it gives you the measure of the arc.\n\nAlso to know what is the arcton of infinity? The arctangent is the inverse tangent function. The range of the arc tangent to x, when x is approaching infinity, is equal to pi/2 radians , or 90 degrees: the range of the arc tangent to x when x is approaching negative infinity, -pi/2 radians or -90 degree is equal to: arcton\n\nHow do you isolate arctan?\n\nHow do we differentiate y = arctic(x)? Step 1: Rearrange y = arctic(x) as tan(y) = x. Step 2: Use implicit differentiation to differentiate it with respect to x, which gives us: (dy/dx)*(sec(y))^2 = 1.\n\n## What is the arcton of infinity?\n\nThe arctangent is the inverse tangent function. The range of the arc tangent to x when x is approaching infinity is equal to pi/2 radians or 90 degrees: the range of the arc tangent to x when x is approaching negative infinity, -pi/2 radians or -90 degrees is equal to: arcton\n\n## What is arcsine equal to?\n\nThe arcsine function is\n\nthe inverse of the sine function\n\n. It returns the angle whose sine is the given number.\n\narcsine\n\n## What is arcsin equal to?\n\nThe arcsine function is\n\nthe inverse of the sine function\n\n. It returns the angle whose sine is the given number.\n\narcsine\n\n## Is Archos the same as SEC?\n\nThe secant, cosecant and cotangent, almost always written as sec, cosec and cot, are trigonometric functions such as sin, cos and tan. Note, sec x cos . is not the same as – 1 x (sometimes written as arccos x). Remember, you cannot divide by zero and so these definitions are only valid if the denominator is not zero.\n\n## Is Arctan the same as cot?\n\narctic (x)\n\nUsing tan – 1 x convention can lead to confusion about the difference between arctangent and cotangent. It turns out that arctan and cot are actually different things: cot(x) = 1/tan(x) , so the cotangent is basically the inverse of a tangent, or, in other words, the inverse of the multiplication.\n\n## What is the sin of infinity equal to?\n\nsin and cos infinity have only one finite value from 1 to -1 . between . But no one can say the exact value.\n\n## Is ln infinitesimal?\n\nThe answer to this question is . The natural log function is strictly increasing, so it is always increasing slowly. The derivative is y’=1x so it is never 0 and is always positive.\n\n## Is 1 infinity defined?\n\nInfinity is a concept, not a number; Therefore, the expression 1/infinity is actually undefined . In mathematics, the limit of a function is when x gets bigger and bigger as it approaches infinity, and 1/x gets smaller and smaller as it approaches zero.\n\n## Is arctan cos sin?\n\nThe functions are usually abbreviated: arcsine (arcSine), arccosine (arcos), arctic (arcton) arccosecant (rcscsc), arcsecond (arcsec), and arccotangent (arcot).\n\nMath2.org Math Tables:\n\n## What is sinx sin1?\n\nsin¯¹x is the inverse function of the sine, also known as the arcsine. It takes a ratio between -1 and +1 as input, and gives the angle measure as output. 1/ sinx is x . The reciprocal of the sine value of k, it is also called cosecant.\n\n## Is arcton always positive?\n\nThe arc tangent to a positive number is the angle of the first quadrant, tan-1(+) is in the quadrant I. The arc tangent to zero is zero, tan-1(0) 0.\nWhen you simplify an expression, make sure you use Arcsine.\n\n## What is the difference between arcsin and csc?\n\nArcsine is the inverse of the sine trigonometric function while the cosecant is the inverse of the sine . Since the sine is opposite the hypotenuse, the cosecant can be expressed as the hypotenuse on the opposite or 1/sine.\n\n## What is CSC equal to?\n\nThe secant of x 1 is divided by the cosine of x: sec x = 1 cos x, and the cosecant of x divided by the sine of x 1 is defined as: csc x = 1 sin x .\n\n## Is arcton equal to second?\n\nThe functions are usually abbreviated: arcsine (arcSine), arccosine (arcos), arctic (arcton) arccosecant (rcscsc), arcsecond (arcsec), and arccotangent (arcot).\n\nMath2.org Math Tables:\n\n## Is arcsec 1 the same as arccos?\n\n7 Answers. You can clearly see that it is not 1. In fact it is: arcsec(x)=arccos(1/x) .\n\n## Do tan and arctan cancel out?\n\nTan and arctan are two opposite operations. They cancel each other out .\n\n## Is cot the same as 1 body?\n\nTherefore – 1x = tan – 1 (x), sometimes interpreted as (tan(x)) – 1 = 1tan(x) = cot(x) or cotangent of x, trigonometric function tangent to product inverse (or inverse) (see above for ambiguity)\n\n## What is the value of cot pi by 2?\n\nThe correct value of cot(π2) cot ( 2 ) is 0 .\n\n## Is there a limit to sin?\n\nThe sine function oscillates from -1 to 1. Because of this the range does not converge to a single value. Which means the limit does not exist .\n\n## Why is sin not infinite?\n\nAs x approaches infinity, the y -values are 1 and −1 . oscillates between Therefore this limit does not exist.\n\n## What is the limit of sin infinity?\n\nThe range of y=sinx is R=[−1;+1]; The function oscillates between -1 and +1. Therefore, the limit when x approaches infinity is undefined ."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8768485,"math_prob":0.9956848,"size":6305,"snap":"2022-40-2023-06","text_gpt3_token_len":1772,"char_repetition_ratio":0.15473734,"word_repetition_ratio":0.17854077,"special_character_ratio":0.26804122,"punctuation_ratio":0.12950699,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994136,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-25T10:45:30Z\",\"WARC-Record-ID\":\"<urn:uuid:89bb519f-e704-4dea-8d57-bc049cc542db>\",\"Content-Length\":\"151806\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:af39635d-4c05-440b-a291-3287c1952837>\",\"WARC-Concurrent-To\":\"<urn:uuid:398cdbb2-8a54-4980-af1c-1abf52302f97>\",\"WARC-IP-Address\":\"172.67.174.157\",\"WARC-Target-URI\":\"https://topfuturepoint.com/arctan-formula/\",\"WARC-Payload-Digest\":\"sha1:V7K73EESGLISLQAKWMQFURAEXEBVQ5PN\",\"WARC-Block-Digest\":\"sha1:OIKUILVRPX4CMNMUVWHZVXQPA7TV7HKD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030334528.24_warc_CC-MAIN-20220925101046-20220925131046-00147.warc.gz\"}"} |
https://lucatrevisan.wordpress.com/tag/arv/ | [
"# ARV on Abelian Cayley Graphs\n\nContinuing from the previous post, we are going to prove the following result: let",
null,
"${G}$ be a",
null,
"${d}$-regular Cayley graph of an Abelian group,",
null,
"${\\phi(G)}$ be the normalized edge expansion of",
null,
"${G}$,",
null,
"${ARV(G)}$ be the value of the ARV semidefinite programming relaxation of sparsest cut on",
null,
"${G}$ (we will define it below), and",
null,
"${\\lambda_2(G)}$ be the second smallest normalized Laplacian eigenvalue of",
null,
"${G}$. Then we have",
null,
"$\\displaystyle \\lambda_2 (G) \\leq O(d) \\cdot (ARV (G))^2 \\ \\ \\ \\ \\ (1)$\n\nwhich, together with the fact that",
null,
"${ARV(G) \\leq 2 \\phi(G)}$ and",
null,
"${\\phi(G) \\leq \\sqrt{2 \\lambda_2}}$, implies the Buser inequality",
null,
"$\\displaystyle \\lambda_2 (G) \\leq O(d) \\cdot \\phi^2 (G) \\ \\ \\ \\ \\ (2)$\n\nand the approximation bound",
null,
"$\\displaystyle \\phi(G) \\leq O(\\sqrt d) \\cdot ARV(G) \\ \\ \\ \\ \\ (3)$\n\nThe proof of (1), due to Shayan Oveis Gharan and myself, is very similar to the proof by Bauer et al. of (2).\n\n# CS 294 Lecture 14: ARV Analysis, Part 3\n\nIn which we complete the analysis of the ARV rounding algorithm\n\nWe are finally going to complete the analysis of the Arora-Rao-Vazirani rounding algorithm, which rounds a Semidefinite Programming solution of a relaxation of sparsest cut into an actual cut, with an approximation ratio",
null,
"${O(\\sqrt {\\log |V|})}$.\n\nIn previous lectures, we reduced the analysis of the algorithm to the following claim.\n\n# CS294 Lecture 13: ARV Analysis, cont’d\n\nIn which we continue the analysis of the ARV rounding algorithm\n\nWe are continuing the analysis of the Arora-Rao-Vazirani rounding algorithm, which rounds a Semidefinite Programming solution of a relaxation of sparsest cut into an actual cut, with an approximation ratio",
null,
"${O(\\sqrt {\\log |V|})}$.\n\nIn previous lectures, we reduced the analysis of the algorithm to the following claim.\n\n# CS359G Lecture 14: ARV Rounding\n\nIn which we begin to discuss the Arora-Rao-Vazirani rounding procedure.\n\nRecall that, in a graph",
null,
"${G=(V,E)}$ with adjacency matrix",
null,
"${A}$, then ARV relaxation of the sparsest cut problem is the following semidefinite program.",
null,
"$\\displaystyle \\begin{array}{lll} {\\rm minimize} & \\frac 1 {2|E|} \\sum_{u,v} A_{u,v} || {\\bf x}_u - {\\bf x}_v ||^2\\\\ {\\rm subject\\ to}\\\\ & \\sum_{u,v} || {\\bf x}_u - {\\bf x}_v ||^2 = {|V|^2}\\\\ & || {\\bf x}_u - {\\bf x}_v||^2 \\leq || {\\bf x}_u - {\\bf x}_w||^2 + || {\\bf x}_w - {\\bf x}_v||^2 & \\forall u,v,w \\in V\\\\ & \\mbox{} {\\bf x}_u \\in {\\mathbb R}^d & \\forall u\\in V \\end{array}$\n\nIf we denote by",
null,
"${ARV(G)}$ the optimum of the relaxation, then we claimed that",
null,
"$\\displaystyle ARV(G) \\leq \\phi(G) \\leq O(\\sqrt{\\log |V|}) \\cdot ARV(G)$\n\nwhere the first inequality follows from the fact that",
null,
"${ARV(G)}$ is a relaxation of",
null,
"${\\phi(G)}$, and the second inequality is the result whose proof we begin to discuss today."
]
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"https://s0.wp.com/latex.php",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9030678,"math_prob":0.999574,"size":1940,"snap":"2023-14-2023-23","text_gpt3_token_len":418,"char_repetition_ratio":0.1373967,"word_repetition_ratio":0.32704404,"special_character_ratio":0.1927835,"punctuation_ratio":0.08882522,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998339,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-02T08:13:33Z\",\"WARC-Record-ID\":\"<urn:uuid:12b48052-1bb9-4dfb-926a-d7f169e53d64>\",\"Content-Length\":\"151570\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:40626116-de5c-4d0c-92e4-65a26bf9acf6>\",\"WARC-Concurrent-To\":\"<urn:uuid:da3b3f9a-205c-46f6-8906-455e704fb640>\",\"WARC-IP-Address\":\"192.0.78.12\",\"WARC-Target-URI\":\"https://lucatrevisan.wordpress.com/tag/arv/\",\"WARC-Payload-Digest\":\"sha1:E3EJY6QYFHWQM3NRBW75JGBHAXTYDSEB\",\"WARC-Block-Digest\":\"sha1:LV5HMP4FL7G2SDOW77SMB7J5BBLXVORC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224648465.70_warc_CC-MAIN-20230602072202-20230602102202-00499.warc.gz\"}"} |
https://file.scirp.org/Html/47788.html | [
" The Cross-Sectional Risk Premium of Decomposed Market Volatility in UK Stock Market",
null,
"Open Journal of Social Sciences, 2014, 2, 30-38 Published Online July 2014 in SciRes. http://www.scirp.org/journal/jss http://dx.doi.org/10.4236/jss.2014.27006 How to cite this paper: Yang, Y. and Copeland, L. (2014) The Cross-Sectional Risk Premium of Decomposed Market Volatili-ty in UK Stock Market. Open Journal of Social Sciences, 2, 30-38. http://dx.doi.org/10.4236/jss.2014.27006 The Cross-Sectional Risk Premium of Decomposed Market Volatility in UK Stock Market Yan Yang, Laurence Copeland Cardiff Business School, Cardiff University, Car diff, UK Email: [email protected], [email protected] Received May 2014 Abstract We decompose UK market volatility into short- and long-run components using EGARCH compo-nent model and examine the cross-sectional prices of the two components. Our empirical results suggest that these two components are significantly priced in the cross-section and the negative risk premia are consistent with the existing literature. The Fama-French three-factor model is improved by the inclusion of the two volatility components. However, our ICAPM model using market excess return and the decomposed volatility components as state variables compares infe-riorly to the traditional three-factor model. Keywords ICAPM, Decomposition of Stock Volatility 1. Introduction The most fundamental and best known model in asset-pricing theory is the Capital Asset Pricing Model (CAPM) which is essentially a “single factor” model of portfolio returns. However, the assumption of a single risk factor (market beta) limits the validity of this model and the effects of other risk factors have put this model under crit-icism. Specially, in the late 1970s, a research begins to uncover variables like size, various price ratios, and mo-mentum that add to the explanation of average returns associated with β. present evidence that beta almost has no explanatory. Their study demonstrates that size and book-to-mar- ket equity (BE/ME), combined to capture the cross-sectional variation in average stock returns together with the market β, leverage, and earnings-price (E/P) ratios. further suggest that the traditional CAPM does not ac-count for returns of size and book-to-market sorted portfolios. They show that there are common return factors related to size and BE/ME that help capture the cross-section of average stock returns in a way that is consistent with multifactor asset-pricing models. report the momentum effect that is left unexplained by the three-factor model. constructs a 4-factor model by including a momentum factor to three-factor model to capture the one-year momentum anomaly. FF",
null,
"Y. Yang, L. Copeland 31 three -factor model and Carhart’s four-factor model have been widely accepted and employed in empirical ana-lyses. In contrast to the lack of theoretical support of three (or four)-factor model, alternative responses to the poor performance of CAPM are to make modifications to the standard CAPM. Among all the developments, inter-temporal CAPM and conditional CAPM are most widely applied and various further extensions are proposed to better interpret the risk-return relation and portfolio structure. The advocates of conditional CAPM argue that the poor empirical performance of CAPM might be due to the failure to account for time-variation in conditional moments. Conditional CAPM tries to preserve the single fac-tor structure of the standard CAPM and assumes that all investors have the same conditional expectations for their asset returns. The ICAPM of suggests that when there is stochastic variation in investment opportunities, asset risk pre-mia are not only determined by covariation of returns with the market return, but also associated with innova-tions in the state variables that describe the investment opportunities. & points out, in cross-sectional as-set pricing studies, the factors in the model should be related to innovations in state variables that forecast future investment opportunities. There is no doubt that stock market volatility changes over time, but whether or not volatility represents a priced risk factor remains less certain. & tests the ICAPM and shows that investors care about risks both from the market return and from changes in forecasts of future market returns. Time-varying market volatility induces changes in the investment opportunity set by changing the expectation of future market returns, or by changing the risk-return trade off. develops the ICAPM in a framework in which the conditional means and variances of state variables are time varying and reflect changes in the investment opportunity set. Risk-averse investors tend to hedge against exposures to future market volatility. set up a standard two-factor pricing kernel with the market return and stochastic volatility as factors. They show that market volatility is a significant cross-sectional asset pricing factor. develop a new specification to model volatility process based on GARCH. They decompose volatility into permanent and transitory components. Following , the component GARCH model is applied to nu mer-ous economic areas and different countries1. [ 11] incorporate the component GARCH model of and the EGARCH model of to build up a new specification of volatility dynamics, the EGARCH component model. They present that the short- and long-run volatility components have negative, highly significant prices of risk which is robust across sets of portfolios, sub -periods, and volatility model specifications. The motivation for this paper stems from the fact that there are a growing number of papers dealing with the decomposition of the market volatility into components using Engle and Lee’s component GARCH Model. However, to my best knowledge, no studies examine the cross-sectional effect of the two decomposed compo-nents of market volatility, especially on UK stock market. Contrary to the existing empirical studies that simply employ Engle and Lee’s component GARCH model to explore the time-series effect of volatility, this thesis at-tempts to understand the cross-sectional effects of the transitory and permanent components of volatility. Fur-thermore, the EGARCH component framework is implemented together with the simple GARCH component model. Although Adrian and Rosenberg demonstrate that the short- and long-run volatility components are sig-nificantly priced in US stock market and their volatility components model compares favorably to the traditional CAPM, Fama-French model and several other model specifications, we are less confident that their volatility components are well priced in UK stock market and the superiority of their model. To test these, we apply the decomposition of market risk to UK stock market and investigate the pricing of short- and long-run volatility risk in the cross-section of stock returns. The object of this chapter is threefold. First, we attempt to both determine whether the two components of market volatility are priced risk factors and estimate the prices of these components. Second, we try to examine whether the short- and long-run component model remains superior in UK stock markets. Third, we examine the robustness of the volatility component model across sets of portfolios, sub-periods, and model specifications. The reminder of this article is organized as follows. Methodology and data description are presented in part II and part III. Empirical results are reported in part IV. The final section offers concluding remarks. 1Research include on futures market; on bond market; exploring US stock market; exploring middle-east stock market; exploring Malaysia stock market..",
null,
"Y. Yang, L. Copeland 32 2. Methodology 2.1. Theoretical Motivation of the Pricing of Volatility Risk The most parsimonious pricing ICAPM framework in which to study the relationship between innovations of state variables and expected return is given as: ( )( )( )2 ,,MERrra covRRcovRzλ≈ +∆ (1) R denotes return in excess of the risk-free rate and MR is stock market excess return. The state variables z are the variables that determine how well the investor can do in his maximization. rra is the elasticity of mar-ginal value with respect to wealth and is often referred as the coefficient of relative risk aversion. When investment opportunities vary over time, the multifactor models of and show that risk premia are associated with the conditional covariance between asset returns and innovations in state variables that de-scribe the time-variation of the investment opportunities. And hence, covariance with these innovations will therefore be priced. In the & ICAPM framework, investors care about risks both from the market return and from changes in forecasts of future market returns. extends Campbell’s model and shows that risk-averse investors tend to directly hedge against changes in future market volatility. Motivated by these multifactor models, express market volatility risk explicitly in Equation (1), ( )( )( )( )1112111,, 11 ,,KMt tttttktktkERrracov RRcov Rvcov Rfλλ++++++ +=≈+ ∆+∑ (2) where kf represent other factors other than aggregate volatility that induce changes in the investment opportu-nity set. For the convenience of empirical application, the above model can be written in terms of factor innovations. 1,mt mtRγ+−, 1,t vtvγ+− and ,1 ,kt ktfγ+− represent innovation in the market return, in aggregate volatility risk, and innovations to the other factors respectively. A true conditional multifactor representation of expected re-turns in the cross-section would take the following form: 1,1,,1,, ,1,1()( )()Ki iiMiittmttmtvtt vtktktktkRa Rvfβγβ γβγ++ ++==+− +−+−∑ (3) where 1itR+ is the excess return on stock i. ,imtβ, ,ivtβ and ,iktβ are the loadings on the excess market return, on market volatility risk, and on other risk factors, respectively. The conditional mean of the market return, ag-gregate volatility and other factors are denoted by ,mtγ, ,vtγ and ,ktγ, respectively. In equilibrium, the condi-tional mean of stock i is given by 1()Ki ii iimm vvkkKa ERβ λβλβλ=== ++∑ (4) where ,mtλ, ,vtλ and ,ktλ are the price of risk of the market factor, the price of aggregate volatility risk, and the prices of risk of the other factors, respectively. 2.2. Econometric Methodology—EGARCH Component Model As an extension of GARCH model, introduce a component GARCH model where the conditional variance is decomposed into transitory and permanent components. In this two-component model, transitory and perma-nent components are used to capture short- and long-term effects of shock, respectively. Many studies find that two-component volatility model is superior to one-component specification in ex-plaining equity market volatility and the log-normal model of EGARCH performs better than square-root or af-fine volatility specifications. Appealed to the merits of the component GARCH and the EGARCH models, incorporate the features of these two models and specify the dynamics of the market return in excess of risk-free rate MtR and the conditional volatility th as: Market return: 11MMt tttR hzµ++= + (1a) Market volatility: lnttthls= + (1b) Shor t-run component: ( )1451 612t ttts szzθθ θπ+ ++=++ − (1c)",
null,
"Y. Yang, L. Copeland 33 Long-run component: 1789 1101( 2)ttt tlsz zθθ θθπ+ ++=++ +− (1d) ( )... 0,1tziid N∼ and Mtµ is the one-period expected excess return. The log-volatility is the sum of two components tl and ts. Each component is an AR(1) processes with its own rate of mean reversion. Without loss of generality, let tl be the slowly mean-reverting, long-run component and ts be the quickly mean-re- verting, short-run component 48()θθ<. The unconditional mean of ts is normalized to zero. The terms 12tzπ+− in Equations (1c) and (1d) are the shocks to the volatility components. Their ex-pected values are equal to zero, given the normality of tz. For these error terms, equal-sized positive or nega-tive innovations results in the same volatility change, although the magnitude can be different for the short- and long -run components (6θ and 10θ). The asymmetric effect of returns on volatility is allowed by including the market innovation in Equations (1c) and (1d) with corresponding coefficients 5θ and 9θ. The market model defined by Equations (1a)-(1d) converges to a continuous-time, two-factor stochastic vola-tility process. An advantage of this specification is that it can be estimated in discrete time via maximum like-lihood. The daily log-likelihood function is: ( )( )( )212 13111 1()1;,|ln 2,22MMt ttttt ttttR slfs lRslhθθ θθπ−−−−−−− −=−− +− where t = 1,…, T is the daily time index, T is the total number of daily observations, and MtR is the daily mar-ket excess return. 3. Data We estimate the EGARCH-component volatility model using daily excess returns. The daily data are used in order to improve the estimation precision and then aggregated to a monthly frequency for the cross-sectional analysis. FTSE All Share Index with its dividend yield is used as the proxy for the market return, Mr, and one month return on Treasury Bills for the risk free rate, fr. The daily data range from 01/09/1980 to 31/12/2012 and are collected from LSPD (London Stock Price Database) and data stream. For the cross-sectional price test of the ICAPM, we apply the Fama and French 25 size and BE/ME portfolios and the portfolio returns and monthly factors are taken directly from website. In the spirit of French’s pro-vision of the data to the international academic community, construct the Fama and French size and B/M portfolio and the SMB (size factor), HML (value factor) and UMD (the momentum factor) of UK stock market and make all portfolios and factors downloadable from their website. 4. Empirical Results 4.1. Results of the Time Series Regression If the short- and long-run volatility components are also asset pricing factor, in the spirit of the ICAPM, the equilibrium pricing kernel thus depends on both short- and long-run volatility components as well as the excess market returns. Denote returns on asset i in excess of the risk free rate by itR. The equilibrium expected return for asset i is: ( )()()()11111 111, ,,iiMiittttsttlttERcov RRcov RscovRlλ λλ++++ +++= ++ (5) whe r e 1λ is the coefficient of relative risk aversion, and sλ andlλ are proportional to changes in the margin-al utility of wealth due to changes in the state variable ts and tl. Equation (5) shows that expected returns depend on three risk premia. The first risk premium arises from the covariance of the asset return with the excess market return, multiplied by relative risk a version 1λ. This is the risk-return tradeoff in a static CAPM. The second and third risk premia depend on the covariance of the asset return with the innovations in the short- and long-run factors. Market expected return Mtµ is defined as 12 3Mt ttslµ θθθ=++ (6) Examination of the risk-return relation is of fundamental importance to the asset pricing literature. Many au-thors either fail to detect a statistically significant intertemporal relation between risk and return of the market",
null,
"Y. Yang, L. Copeland 34 portfolio or find a negative relation2. The estimation results for the volatility model are shown in Table 1. In the expected return equation, short-run volatility has a significant positive coefficient 2θ, while 3θ, the coefficient of long-run volatility is signifi-cantly negative. The market excess return thus depends positively on short-run volatility but negatively on long -run volatility. identify a negative relationship between short-run volatility and market excess return but a positive relationship between long-run volatility and market excess return. Hence, short-run and long-run vola-tilities seem to have opposite effects on market excess return. This result might explain why previous research often have difficulty identifying a time-series relationship or mixed results of risk and return relation. We identify the short- and long-run components by their relative degrees of autocorrelation: the short-run vo-latility has an autoregressive coefficient of 0.807, and the long-run component has an autoregressive coefficient of 0.994. The long-run component is highly persistent. However, it’s not permanent, the hypothesis that 81θ= is rejected at the 1% significant level. The estimate of 4θ is smaller than that of 8θ, which indicates the short -run volatility is less persistent compared to long-run component. Because the short- and long-run compo-nents determine log-volatility additively, it’s impossible to identify the means of the two components separately, and only the mean the long-run component is estimated. 5θ and 9θ detect the asymmetric effect on volatility. Both the estimates of 5θ and 9θ are significantly negative and large than minus one. This suggests that a pos-itive surprise ( )tz increases both the short- and long-run volatility less than a negative surprise. Ljun g -box Q statistic suggests that there isn’t remaining serial correlation in the mean equation of the market excess return, while the ARCH-LM test reveals that there’s no additional ARCH effect exhibiting in the standar-dized residuals. 4.2. Results of the C ross -Sectional Tests Monthly data are employed to carry out the cross-sectional tests. The daily short- and long-run volatility com-ponents are obtained from the time-series regression. The 21-step out-of-sample forecasts of the short- and long -run components are made respectively. Daily innovations of the volatility components are calculated by subtracting the short- and long-run component from the forecasted values. The daily innovations in each month are then aggregated to a monthly frequency, and providing us with the monthly innovations of the short- and long -run component. Table 1. Time series estimation of the volatility components, daily, 01/09/1980 to 31/12/2012. Market excess return:112 31Mtt tttRsl hzθθ θ++=+ ++ 1θ 2θ 3θ Coef. −0.007 0.267 −0.499 Std.err. 0.010 0.106 0.031 p-value 0.475 0.012 0.000 Short -run component:1451 6 1( 2)t ttts szzθθ θπ+ ++=++ − 4θ 5θ 6θ Coef. 0.807 −0.046 −0.009 Std.err. 0.030 0.004 0.042 p-value 0.000 0.000 0.829 Long-run component:178 91101( 2)ttt tllz zθθ θθπ+ ++=++ +− 7θ 8θ 9θ 10θ Coef. 0.002 0.994 −0.032 0.028 Std.err. 0.000 0.001 0.001 0.002 p-value 0.000 0.000 0.000 0.000 2Examples include, , , and .",
null,
"Y. Yang, L. Copeland 35 211([ ]),Nmttttsress Es−== −∑ (7) 211([ ]),Nm ttttlresl El−== −∑ (8) where sres and lres denote innovations of short- and long-run volatility respectively. N is the trading days in month m. The market variance v is aggregated to a monthly frequency, and the time series follows an AR(2) process. Hence, variance innovations (vres) are estimated as residuals of a monthly autoregressive process with two lags. The statistics of the innovations and the other pricing factors are summarized in Table 2. Under the ICAPM described in Section 2.1, the pricing kernel is a linear function of the excess return on the market portfolio and the innovations in the state variables, so that the unconditional risk premium on asset i may be written as: ()iiiiimmssl la ERβ λβλβλ== ++ (9) where ,mtλ is the price of risk of the market factor, ,stλ is the price of the short-run volatility risk, and the ,ltλ is the price of risk of the long-run volatility. The implication of the ICAPM for stock returns can be tested directly by implementing the two-stage Fa-ma-MacBeth procedure. We use the correction procedure proposed by to account for the errors-in-variables problem. In the first stage, the 25 size and BE/ME portfolio returns are regressed on market excess returns, sres and lres. In the second stage, the portfolio returns are regressed on the estimated betas from the first stage to obtain the prices of market risk, short-run volatility risk and long-run volatility risk. Table 3 reports the pricing of volatility risk in the cross-section. The regressions of CAPM (column i), the FF three factor model (column ii), Carhart’s momentum model (column iii), and a model analogous to with in-novations to market variance as risk factor (column iv) are also presented in Table 3. In column (v), we can see that the short- and long-run volatility components are significant pricing factors at the 5% level. The prices of short- and long-run components are −0.334 and −0.842 respectively. This implies that an asset with a short-run volatility beta of unity requires a 0.334% lower returns than an asset with zero ex-posure to the short-run component. These results are consistent with hypothesis that the cro ss-section of stock returns reflects exposure to volatility risk. In column (vi) and (vii), the prices of risk when the short- and long -run volatility enters as separate factors are reported. Each of the components has a negative price of risk at 10% significant level. - show that investors intend to hedge against market volatility and they are willing to pay a premium for market downside risk. The hedge motive is indicative of a negative price of market volatil-ity. The negative prices of the decomposed components of market volatility would suggest that risk-averse in-vestors tempt to hedge the overall exposures to market risk, no matter whether the exposures are transitory or persistent. However, the short- and long-run volatility factor compares inferiorly to the FF three-factor model and Car-hart momentum model, in terms of pricing performance. The RMSPE (root-mean-squared pricing error) are re-ported to evaluate the pricing performance of the different models. The models with FF factors (column ii and viii) achieve the lowest pricing errors. It’s worth noting that adding the two volatility components to FF three-factor model and momentum model reduces the pricing errors. This suggests that the volatility components and the Fama and French factors (or Table 2. Summary statistics of the monthly pricing fact o r s . Pricing factors Mean Std. Dev. Skewness Kurtosis Short -run volatility(sres) 0.000 1.064 −0.309 3.192 Long-run volatility(lres ) 0.000 2.989 1.697 11.083 Market variance (vres) 0.000 14.520 4.429 47. 87 Value factor (HM L) 0.004 0.037 0.270 11.092 Size factor (SMB) 0.001 0.034 0.712 7.200 Momentum factor (UMD) 0.008 0.042 −1.258 10.844",
null,
"Y. Yang, L. Copeland 36 Table 3. Pricing the cross section of the 25 size and B/M sorted portfolios. (i) (ii) (iii) (iv) (v) (vi) (vii ) (viii) (ix) Excess market return Coef. 0.596 0.454 0.498 0.439 0.453 0.608 0.465 0.443 0.438 p-valu e 0.033 0.073 0.073 0.083 0.073 0.026 0.066 0.080 0.084 Short -run volatility (sres) Coef. −0.334 −0.265 −0.232 −0.287 p-valu e 0.047 0.097 0.091 0.062 Long-run volatility (lres) Coef. −0.842 −0.602 −0.623 −0.689 p-valu e 0.034 0.071 0.074 0.057 Market variance (vres) Coef. −3.251 p-valu e 0.049 Value factor (HML) Coef. 0.518 0.518 p-valu e 0.025 0.024 Size factor (SMB) Coef. 0.165 0.156 p-valu e 0.386 0.212 Momentum factor Coef. 0.890 0.801 (UMD) p-value 0.029 0.076 RMSPE 0.210 0.120 0.142 0.245 0.188 0.210 0.197 0.113 0.138 This table reports the two-stage cr oss-sectional regression results for the 25 size and B/M sorted portfolios under various model specifications, in-cluding ICAPM, FF three-factor model, and the CAPM. The t-ratios are calculated using Jagannathan and Wang (1998) and Newey and West (1987) procedures to account for the estimation errors in first-stage estimation and correct for the possible heteroskedasticity and autocorrelation. The cor-responding p-values are reported according to the adjusted t-ratios. Root mean square pricing errors (RMPSE) are reported. momentum factors) capture some orthogonal source of priced risk. The first-stage FM regression shows that the factor loadings have significant variation across the size di- mension, however but less significant varation across the B/M dimension3. Table 4 shows the risk premium on the 25 size and BE/ME portfolios. The price spread of market risk between small firms and large firms is −0.107, which translate into an annualized premium spread due to market risk of −1.3% per year. Similarly, the annualized premium spreads between small and large firms due to short -run volatility and long-run volatility are −1.76% and 4.14% respectively. Only the risk premium of the long -run volatility is positive, but with much bigger magnitude. Hence, combining the three risk-premium dif-ferences yields an average excess risk premium for small firm relative to large firms of 1.08%. The analysis suggests that the size effect of small cap firms earning higher risk adjusted returns may be attributed to the long -run volatility component. The difference of annualized risk premium between the high B/M firms and low B/M firms due to the risk of short -run volatility component is 0.50% per year, while the difference due to the long-run component is 0.33% per year. Hence, the short- and long-run volatility components have the same effect on the B/M firm portfolio. Combining the value spread due to market risk, the total effect that the high B/M firm portfolio earns 1.32% higher risk premium per year relative to low B/M firm portfolio. The analysis suggests that the B/M effect that high B/M firms earn higher returns may be explained by both the short- and long-run volatility components. 4.3. Robustness Tests4 In order to test the robustness of the results we find above, we proceed the following tests: 1) Estimate the models using three sub-periods of the whole sample periods. 2) The cross-sectional procedure is regressed on returns of different portfolios sorted by other criteria. 3) Different mean equations are specified to test the EGARCH component model. 4) We compare EGARCH component model with GARCH and GARCH component model by evaluate their 3For reasons of space, the results are available upon a request from the author. 4The test procedures and results are available upon a request.",
null,
"Y. Yang, L. Copeland 37 Table 4. Factor risk premia of the 25 size and B/M sorted portfolios. Market risk premium Small Size 2 Size 3 Size 4 Big Average Gr o wt h 0.392 0.421 0.4489 0.443 0.427 0.427 B/M 2 0.332 0.380 0.423 0.417 0.435 0.398 B/M 3 0.305 0.368 0.403 0.396 0.480 0.390 B/M 4 0.327 0.357 0.404 0.480 0.431 0.400 Value 0.313 0.378 0.403 0.462 0.429 0.397 Average 0.334 0.381 0.416 0.439 0.441 0.402 Short -run volatility risk premium Small Size 2 Size 3 Size 4 Big Average Gr o wt h −0.213 −0.102 −0.124 −0.072 −0.069 −0.106 B/M 2 −0.247 −0.184 −0.059 0.042 −0.037 −0.092 B/M 3 −0.139 −0.186 −0.109 0.079 −0.080 −0.056 B/M 4 −0.152 0.045 0 .0 11 −0.084 0.054 −0.039 Value −0.125 −0.050 0.022 −0. 113 0.000 −0.065 Average −0.175 −0.096 −0.052 −0.029 −0.027 −0.076 Long-run volatility risk premium Small Size 2 Size 3 Size 4 Big Average Gr o wt h 0.357 0.212 0.104 0.249 −0.035 0.177 B/M 2 0.393 0.341 0.074 0.184 0.032 0.205 B/M 3 0.339 0.286 0.105 0.225 0.045 0.200 B/M 4 0.332 0.286 0.069 0.215 0.109 0.202 Value 0.361 0.341 0.116 0.287 −0.079 0.205 Average 0.357 0.293 0.094 0.232 0.014 0.198 This table reports the risk premia of portfolio returns on the market excess return, short-run volatility innovations and long-run volatility innovations. The risk premia are computed by multiplying factor loadings from Fama-Macbeth first-stage regression and prices of risk of Table 3, column V, second stage Fama-Mac Bethregressoin. performance in the cross-sectional pricing. The magnitudes of the prices of risk for the volatility components are fairly similar across different sets of assets and sample periods. The benchmark specification is superior to the alternatives with lowest root-mean-squared error. 5. Conclusions ICAPM predicts that financial asset risk premia are not only due to covariation of returns with the market excess return, but also associated with innovations in the state variables that describe the investment opportunities. Multifactor models of risk already predict that aggregate volatility should be a cross-sectional risk factor. further decompose the aggregate volatility into a transitory and a permanent component. They conclude that the short - and long-run volatility components have negative, highly significant prices of risk using American stock market data. Applying decomposition of market risk to UK stock market, we find that the short- and long-run volatility components also have significantly negative prices of risk. The results are robust across sets of portfolios, sam-ple periods and model specifications. The short- and long-run volatility might provide an explanation of the size and value anomaly of financial market. Specifically, the size effect of small cap firms earning higher risk ad-",
null,
"Y. Yang, L. Copeland 38 justed returns may be attributed to the long-run volatility component. Whereas, the value effect that high B/M firms earn higher returns may be explained by both the short- and long-run volatility components. However, the performance of the decomposition model is superior to the Fama-French three factor model, and market va-riance model. This might suggest further investigation and improvement on Adrian and Rosenberg’s volatility decomposition model. References Fama, E.F. and French, K.R. (1992) The Cross-Section of Expected Stock Returns. The Journal of Finance, XLVII, 427-465. http://dx.doi.org/10.1111/j.1540-6261.1992.tb04398.x Fama, E.F. and French, K.R . (199 3) Common Risk Factors in the Returns on Stocks and Bonds. Journal of Financial Economics, 33, 3-56. htt p:// d x.do i.o rg/1 0.10 16 /03 04-405X(93)90023-5 Jegadeesh, N. and Titman, S. (199 3) Returns to Buying Winners and Selling Losers: Implications for Stock Market Ef-ficiency. The Journal of Finance, 48, 65-91. http://dx.doi.org/10.1111/j.1540-6261.1993.tb04702.x Carhart, M.M. (1997) On Persistence in Mutual Fund Performance. The Journal of Finance, 52, 57-82. http://dx.doi.org/10.1111/j.1540-6261.1997.tb03808.x Merton , R.C. (1973) An Intertemporal Capital Asset Pricing Model. Econometrica , 51, 867-887. http://dx.doi.org/10.2307/1913811 Campbell, J.Y. (1993) Intertemporal Asset Pricing without Consumption Data. American Economic Review, 83, 487 - 512. Campbell, J.Y. (1996) Understanding Risk and Return. Journal of Political Economy, 104, 298 -34 5. http://dx.doi.org/10.1086/262026 Chen, J. (2002) In tertempo ral, CAPM and the Cross-section of Stock Returns. Working P aper , University of Southern California, Los Angeles. Ang, A. , Hodrick, R. J., Xing, Y.H. and Zhang, X.Y. (2006) The Cross-Section of Volatility and Expected Returns. The Journal of Finance, LXI , 259-299. http://dx.doi.org/10.1111/j.1540-6261 .20 06 .00 836. x Engl e, R.F. and Lee, G. (1999) A Permanent and Transitory Component Model of Stock Return Volatility. Cointegra- tion, Causality and Forecasting: A Festschrift in Honor of Clive W.J. Granger. Oxford University Press, New York. Adr ian, T. and Rosenberg, J. (2008) Stock Returns and Volatility: Pricing the Short-Run and Long-Run Components of Market Risk. The Journal of Finance, 63, 2997 -30 30. http://dx.doi.org/10.1111/j.1540-6261.2008.01419.x Nelson, D.B. (19 91 ) Conditional Heteroskedasticity in Asset Returns: A New Approach. Economet rica, 59 , 347 -370. http://dx.doi.org/10.2307/2938260 Macmillan, H. and Tampoe, M. (2000) Strategic Management. Oxford University Press, Oxford. Simó n, S.R. and Amalia, M.Z. (2012) Volatility in EMU Sovereign Bond Yields: Permanent and Transitory Compo- nents. Applied Financial Economics, 22, 14 53-14 64. http://dx.doi.org/10.1080/09603107.2012.661397 Hertog, R.G.J.D. (1994) Pricing of Permanent and Transitory Volatility for US Stock Returns: A Composite GARCH Model. Economics Letters, 44, 421 -426. http://dx.doi.org/10.1016/0165-1765(94 )90 115 -5 Zarour, B . A. and Siriopoulos, C .P . (2008) Transitory and Permanent Volatility Components: The Case of the Middle East Stock Markets. Review of Middle East Economics and Finance, 4, 80-92. Mansor, H.I and Huson, J.A.A. (2014) Permanent and Transitory Oil Volatility and Aggregate Investment in Malaysia. Energy Po lic y, 67 , 552 -563. http://dx.doi.org/10.1016/j.enpol.2013.11.072 Ross, S.A. (1976) The Arbitrage Theory of Capital Asset Pricing. Journal of Economic Theory, 13, 341-360. http://dx.doi.org/10.1016/0022-0531(76)90046-6 Grego ry, A., Thar yan, R. and Huang, A. (2009) The Fama-French and Momentum Portfolios and Factors in the UK. University of Exeter Business School, XFI Centre for Finance and Investment Paper No. 09/05. Baillie, R.T. and Degennaro, R.P. (1990) Stock Returns and Volatility. Journal of Financial and Quantitative Analysis, 25, 203-214. http://dx.doi.org/10.2307/2330824 Whitelaw, R. (1994) Time Variations and Covariations in the Expectation and Volatility of Stock Returns. Journal of Finance, 49, 515-541. http://dx.doi.org/10.1111/j.1540-6261.1994.tb05150.x Harve y, C.R. (2001) The Specification of Conditional Expectations. Journal of Empirical Finance, 8, 573-638. http://dx.doi.org/10.1016/S0927-5398(01)00036-6 Jagannat han, R. and Wang, Z.Y. (1998) An Asymptotic Theory for Estimating Beta-Pricing Models Using Cross-Sec- tional Regression. Journal of Finance, 53, 1285-1309. ht tp :// dx. doi.o rg/ 10. 11 11/ 0022 -10 82.0 0053"
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https://whatisconvert.com/230-centimeters-in-feet | [
"## Convert 230 Centimeters to Feet\n\nTo calculate 230 Centimeters to the corresponding value in Feet, multiply the quantity in Centimeters by 0.032808398950131 (conversion factor). In this case we should multiply 230 Centimeters by 0.032808398950131 to get the equivalent result in Feet:\n\n230 Centimeters x 0.032808398950131 = 7.5459317585302 Feet\n\n230 Centimeters is equivalent to 7.5459317585302 Feet.\n\n## How to convert from Centimeters to Feet\n\nThe conversion factor from Centimeters to Feet is 0.032808398950131. To find out how many Centimeters in Feet, multiply by the conversion factor or use the Length converter above. Two hundred thirty Centimeters is equivalent to seven point five four six Feet.\n\n## Definition of Centimeter\n\nThe centimeter (symbol: cm) is a unit of length in the metric system. It is also the base unit in the centimeter-gram-second system of units. The centimeter practical unit of length for many everyday measurements. A centimeter is equal to 0.01(or 1E-2) meter.\n\n## Definition of Foot\n\nA foot (symbol: ft) is a unit of length. It is equal to 0.3048 m, and used in the imperial system of units and United States customary units. The unit of foot derived from the human foot. It is subdivided into 12 inches.\n\n## Using the Centimeters to Feet converter you can get answers to questions like the following:\n\n• How many Feet are in 230 Centimeters?\n• 230 Centimeters is equal to how many Feet?\n• How to convert 230 Centimeters to Feet?\n• How many is 230 Centimeters in Feet?\n• What is 230 Centimeters in Feet?\n• How much is 230 Centimeters in Feet?\n• How many ft are in 230 cm?\n• 230 cm is equal to how many ft?\n• How to convert 230 cm to ft?\n• How many is 230 cm in ft?\n• What is 230 cm in ft?\n• How much is 230 cm in ft?"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8667653,"math_prob":0.9870969,"size":1743,"snap":"2023-14-2023-23","text_gpt3_token_len":448,"char_repetition_ratio":0.21449108,"word_repetition_ratio":0.069078945,"special_character_ratio":0.30177855,"punctuation_ratio":0.11527377,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9992799,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-07T04:54:34Z\",\"WARC-Record-ID\":\"<urn:uuid:b187295f-d036-45a2-910c-7a5e3484a577>\",\"Content-Length\":\"28010\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fe8d90a9-7c50-4d92-9e88-ff26efbe52ba>\",\"WARC-Concurrent-To\":\"<urn:uuid:3fc9bd37-f1f9-4dd1-817c-7fb9e1822001>\",\"WARC-IP-Address\":\"104.21.13.210\",\"WARC-Target-URI\":\"https://whatisconvert.com/230-centimeters-in-feet\",\"WARC-Payload-Digest\":\"sha1:SYABC7ILMQLBLOW5ABA7AFY3NR6NPFUN\",\"WARC-Block-Digest\":\"sha1:FS2XFXH2XAQWITJBHU57PFX5JFBOV5NV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224653608.76_warc_CC-MAIN-20230607042751-20230607072751-00303.warc.gz\"}"} |
https://questioncove.com/updates/4d530fd1d04cb764c6b48481 | [
"Mathematics",
null,
"OpenStudy (anonymous):\n\nthe second of two numbers is 7 more than the first. their sum is 47. Find the numbers",
null,
"OpenStudy (anonymous):\n\nx+y=47, x-y = 7 means that if we add the two questions, 47 + 7 = x + y + x - y = 2x = 54. Then x = 27 and y = 20.",
null,
"OpenStudy (anonymous):\n\nI am still confused. I am sorry",
null,
"OpenStudy (sandra):\n\nok so you're trying to translate this word problem into two equations",
null,
"OpenStudy (sandra):\n\nthe point being, that two solve a problem with two variables (in this case the two numbers), you need two equations that have both variables in them",
null,
"OpenStudy (sandra):\n\nso for the first part of the word problem, it's saying that we have two numbers, and one of the numbers is 7 greater than the other",
null,
"OpenStudy (sandra):\n\nor in other words, y = x + 7, with y being the second number, and x being the first",
null,
"OpenStudy (sandra):\n\nnow the second piece of information tells you that when you add both together, they equal 47. so you can write that as an equation like this: x + y = 47",
null,
"OpenStudy (sandra):\n\nthe next step, once you have as many equations as you do variables to solve for, is to start substituting one into the other",
null,
"OpenStudy (sandra):\n\nso in this case, why don't we choose the first equation, that is, y = x + 7, and \"substitute\" y into the second equation. That is, we can replace all the \"y\" variables in the second equation (there's only one), with x+7",
null,
"OpenStudy (sandra):\n\nso x + y = 47, substituing x + 7 for y, we get x + x + 7 = 40",
null,
"OpenStudy (sandra):\n\nerrr sorry, x + (x + 7) = 47",
null,
"OpenStudy (sandra):\n\nif you keep solving this, then we see that 2x + 7 = 47 , 2x = 40, x = 20",
null,
"OpenStudy (sandra):\n\nok, so now we KNOW x=20",
null,
"OpenStudy (sandra):\n\nand then we substitute it back into either of our equations",
null,
"OpenStudy (sandra):\n\nso since x + y = 47, and we know x = 20, we see 20 + y = 47",
null,
"OpenStudy (sandra):\n\nand then y = 27",
null,
"OpenStudy (sandra):\n\ndoes that make more sense?",
null,
"OpenStudy (sandra):\n\nok so we had 2x + 7 = 47",
null,
"OpenStudy (sandra):\n\nwe need to subtract 7 from both sides",
null,
"OpenStudy (sandra):\n\nso 2x = 40",
null,
"OpenStudy (sandra):\n\nand now we divide each side by 2",
null,
"OpenStudy (sandra):\n\nso x=20",
null,
"OpenStudy (anonymous):\n\nok. so we sub. 7 to get X by itself?",
null,
"OpenStudy (sandra):\n\nexactly",
null,
"OpenStudy (sandra):\n\nthe reason we can do that is because we know if you subtract anything from two equal values (e.g. an eqation like 2x+7 = 47), we know that they are STILL equal - since we did the same operation to both values",
null,
"OpenStudy (sandra):\n\ni.e. if I have two baskets that have the same number of oranges in them, and I take out two oranges from each, regardless of how many they started with, they still have the same number",
null,
"OpenStudy (sandra):\n\nand this property holds true for any operation on equal values (the two sides of an equation)",
null,
"OpenStudy (sandra):\n\nas long as I do the same operation to both sides, I know they're still equal",
null,
"OpenStudy (anonymous):\n\ncan i post another one? I will try and work on it and see if i get it right?",
null,
"OpenStudy (sandra):\n\nok sure",
null,
"OpenStudy (anonymous):\n\ndo i post here or on the left?",
null,
"OpenStudy (sandra):\n\non the left =)\n\nLatest Questions",
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http://www.sanjuanciudadpatria.com/3xxihr2/how-to-prove-a-function-is-surjective-d00d9b | [
"Proving a Function is Injective Example 1. We say that is: f is injective iff: More useful in proofs is the contrapositive: f is surjective iff: . Thus, f : A ⟶ B is one-one. How to Prove Functions are Surjective(Onto) How to Prove a Function is a Bijection. 1. There are lots of ways one might go about doing it. Let f:ZxZ->Z be the function given by: f(m,n)=m2 - n2 a) show that f is not onto b) Find f-1 ({8}) I think -2 could be used to prove that f is not … Press J to jump to the feed. Create your account. Using math symbols, we can say that a function f: A → B is surjective if the range of f is B. Surjective (Also Called \"Onto\") A function f (from set A to B) is surjective if and only if for every y in B, there is at least one x in A such that f(x) = y, in other words f is surjective if and only if f(A) = B. Equivalently, for every b∈B, there exists some a∈A such that f(a)=b. Check the function using graphically method. this is what i did: y=x^3 and i said that that y belongs to Z and x^3 belong to Z so it is surjective Explain. i.e. Let f : A ⟶ B and g : X ⟶ Y be two functions represented by the following diagrams. Two simple properties that functions may have turn out to be exceptionally useful. Press question mark to learn the rest of the keyboard shortcuts JavaScript is disabled. Prove: f is surjective iff f has a right inverse. (Also, this function is not an injection.) Then: The image of f is defined to be: The graph of f can be thought of as the set . how do you prove that a function is surjective ? Then the rule f is called a function from A to B. ', Does there exist x in Z such that, for example, f(x)= x, Bringing atoms to a standstill: Researchers miniaturize laser cooling, Advances in modeling and sensors can help farmers and insurers manage risk, Squeezing a rock-star material could make it stable enough for solar cells. The typical method of showing that a function is surjective is to pick an arbitrary element in a given range and then find the element in the domain which maps to it. https://goo.gl/JQ8NysHow to Prove the Rational Function f(x) = 1/(x - 2) is Surjective(Onto) using the Definition f is surjective if for all b in B there is some a in A such that f(a) = b. f has a right inverse if there is a function h: B ---> A such that f(h(b)) = b for every b in B. i. when f(x 1 ) = f(x 2 ) ⇒ x 1 = x 2 Otherwise the function is many-one. {/eq} is said to be onto or surjective, if every element of {eq}Y how to prove that function is injective or surjective? How to check if function is one-one - Method 1 In this method, we check for each and every element manually if it has unique image It is not required that a is unique; The function f may map one or more elements of A to the same element of B. All rights reserved. f: X → Y Function f is one-one if every element has a unique image, i.e. {/eq} and read as f maps from A to B. A codomain is the space that solutions (output) of a function is … Sciences, Culinary Arts and Personal In simple terms: every B has some A. how can i prove if f(x)= x^3, where the domain and the codomain are both the set of all integers: Z, is surjective or otherwise...the thing is, when i do the prove it comes out to be surjective but my teacher said that it isn't. © copyright 2003-2021 Study.com. Please Subscribe here, thank you!!! Some of your past answers have not been well-received, and you're in danger of being blocked from answering. How to Write Proofs involving the Direct Image of a Set. Prove that an endomorphism is injective iff it is surjective, Proving that injectivity implies surjectivity, Prove that T is injective if and only if T* is surjective, Showing that a function is surjective onto a set, How can I prove it? Injective vs. Surjective: A function is injective if for every element in the domain there is a unique corresponding element in the codomain. Proving this with surjections isn't worth it, this is sufficent … answer! Do all bijections have inverses? a ≠ b ⇒ f(a) ≠ f(b) for all a, b ∈ A ⟺ f(a) = f(b) ⇒ a = b for all a, b ∈ A. e.g. On the left is a convex curve; the green lines, no matter where we draw them, will always be above the curve or lie on it. The identity function on a set X is the function for all Suppose is a function. Onto Function (surjective): If every element b in B has a corresponding element a in A such that f(a) = b. How to prove a function is surjective? Functions can be injections (one-to-one functions), surjections (onto functions) or bijections (both one-to-one and onto). Now, suppose the kernel contains only the zero vector. We already know that f(A) Bif fis a well-de ned function. Then, there can be no other element such that and Therefore, which proves the \"only if\" part of the proposition. Please Subscribe here, thank you!!! To prove a function, f: A!Bis surjective, or onto, we must show f(A) = B. The kernel of a linear map always includes the zero vector (see the lecture on kernels) because Suppose that is injective. So K is just a bijective function from N->E, namely the \"identity\" one, that just maps k->2k. How to prove that this function is a surjection? We can express that f is one-to-one using quantifiers as or equivalently , where the universe of discourse is the domain of the function.. Services, Working Scholars® Bringing Tuition-Free College to the Community. And I can write such that, like that. On the right, we are able to draw a number of lines between points on the graph which actually do dip below the graph. Earn Transferable Credit & Get your Degree, Get access to this video and our entire Q&A library. This means that for any y in B, there exists some x in A such that y=f(x). Therefore, d will be (c-2)/5. Putting f(x1) = f(x2) we have to prove x1 = x2 Since if f (x1) = f (x2) , then x1 = x2 ∴ It is one-one (injective) Check onto (surjective) f(x) = x3 Let f(x) = y , such that y ∈ N x3 = y x = ^(1/3) Here y is a natural number i.e. An onto function is also called a surjective function. A very simple scheduler implemented by the function random(0, number of processes - 1) expects this function to be surjective, otherwise some processes will never run. https://goo.gl/JQ8NysProof that if g o f is Surjective(Onto) then g is Surjective(Onto). how can i prove if f(x)= x^3, where the domain and the codomain are both the set of all integers: Z, is surjective or otherwise...the thing is, when i do the prove it comes out to be surjective but my teacher said that it isn't. Examples of Surjections. While most functions encountered in a course using algebraic functions are well-de … We note in passing that, according to the definitions, a function is surjective if and only if its codomain equals its range. If A and B are two non empty sets and f is a rule such that each element of A have image in B and no element of A have more than one image in B. (Two are shown, drawn in green and blue). Step 2: To prove that the given function is surjective. then f is an onto function. Proving a Function … How do you prove a Bijection between two sets? A function An injective (one-to-one) function A surjective (onto) function A bijective (one-to-one and onto) function A few words about notation: To de ne a speci c function one must de ne the domain, the codomain, and the rule of correspondence. A function f:A→B is surjective (onto) if the image of f equals its range. But g : X ⟶ Y is not one-one function because two distinct elements x1 and x3have the same image under function g. (i) Method to check the injectivity of a functi… Why do natural numbers and positive numbers have... How to determine if a function is surjective? It is not required that x be unique; the function f may map one … 06:02. In practice the scheduler has some sort of internal state that it modifies. When is a map locally injective jacobian? A function f : A ⟶ B is said to be a one-one function or an injection, if different elements of A have different images in B. (This is not the same as the restriction of a function … Please pay close attention to the following guidance: 02:13. To prove surjection, we have to show that for any point “c” in the range, there is a point “d” in the domain so that f (q) = p. Let, c = 5x+2. Clearly, f : A ⟶ B is a one-one function. (injection, bijection, surjection), Partial Differentiation -- If w=x+y and s=(x^3)+xy+(y^3), find w/s, Solving a second order differential equation. What that means is that if, for any and every b ∈ B, there is some a ∈ A such that f(a) = b, then the function is surjective. {/eq} is the... Our experts can answer your tough homework and study questions. This is written as {eq}f : A \\rightarrow B Vertical line test : A curve in the x-y plane is the graph of a function of iff no vertical line intersects the curve more than once. Become a Study.com member to unlock this Because, to repeat what I said, you need to show for every, 'Because, to repeat what I said, you need to show for every y, there exists an x such that f(x) = y! Why do injection and surjection give bijection... One-to-One Functions: Definitions and Examples, NMTA Elementary Education Subtest II (103): Practice & Study Guide, College Preparatory Mathematics: Help and Review, TECEP College Algebra: Study Guide & Test Prep, Business 104: Information Systems and Computer Applications, Biological and Biomedical Note: One can make a non-surjective function into a surjection by restricting its codomain to elements of its range. The easiest way to figure out if a graph is convex or not is by attempting to draw lines connecting random intervals. Any function can be made into a surjection by restricting the codomain to the range or image. And a function is surjective or onto, if for every element in your co-domain-- so let me write it this way, if for every, let's say y, that is a member of my co-domain, there exists-- that's the little shorthand notation for exists --there exists at least one x that's a member of x, such that. This curve is not convex at all on the interval being graphed. In mathematics, a function f from a set X to a set Y is surjective (also known as onto, or a surjection), if for every element y in the codomain Y of f, there is at least one element x in the domain X of f such that f (x) = y. Informally, an injection has each output mapped to by at most one input, a surjection includes the entire possible range in the output, and a bijection has both conditions be true. Suppose f has a right inverse h: B --> A such that f(h(b)) = b for every b … All other trademarks and copyrights are the property of their respective owners. Function: If A and B are two non empty sets and f is a rule such that each element of A have image in B and no element of A have more than one image in B. Onto or Surjective function: A function {eq}f: X \\rightarrow Y For a better experience, please enable JavaScript in your browser before proceeding. For example, the new function, f N (x):ℝ → [0,+∞) where f N (x) = x 2 is a surjective function. Does closure on a set mean the function is... How to prove that a function is onto Function? for a function $f:X \\to Y$, to show. Show that there exists an injective map f:R [41,42], i. e., f is defined for all non-negative real numbers x, and for all such x we have 41≤f(x)≤42. Functions in the first row are surjective, those in the second row are not. Proving a Function is Surjective Example 5. In other words, f: A!Bde ned by f: x7!f(x) is the full de nition of the function f. The most direct is to prove every element in the codomain has at least one preimage. Now, let's assume we have some bijection, f:N->F', where F' is all the functions in F that are bijective. A function f: X !Y is surjective (also called onto) if every element y 2Y is in the image of f, that is, if for any y 2Y, there is some x 2X with f(x) = y. One way to prove a function $f:A \\to B$ is surjective, is to define a function $g:B \\to A$ such that $f\\circ g = 1_B$, that is, show $f$ has a right-inverse. 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https://open.library.ubc.ca/cIRcle/collections/ubctheses/24/items/1.0378326 | [
"# Open Collections\n\n## UBC Theses and Dissertations",
null,
"## UBC Theses and Dissertations\n\n### Theoretical and numerical study of free-surface flow of viscoplastic fluids : 2D dambreaks, axisymmetrical… Liu, Ye 2019\n\nMedia\n24-ubc_2019_may_liu_ye.pdf [ 5.08MB ]\nJSON: 24-1.0378326.json\nJSON-LD: 24-1.0378326-ld.json\nRDF/XML (Pretty): 24-1.0378326-rdf.xml\nRDF/JSON: 24-1.0378326-rdf.json\nTurtle: 24-1.0378326-turtle.txt\nN-Triples: 24-1.0378326-rdf-ntriples.txt\nOriginal Record: 24-1.0378326-source.json\nFull Text\n24-1.0378326-fulltext.txt\nCitation\n24-1.0378326.ris\n\n#### Full Text\n\n`Theoretical and numerical study of free-surface flowof viscoplastic fluids: 2D dambreaks, axisymmetricalslumps and surges down an inclined slopebyYe LiuB.Sc, Mathematics, Peking University, 2012M.Sc, Mathematics, University of British Columbia, 2014A THESIS SUBMITTED IN PARTIAL FULFILLMENTOF THE REQUIREMENTS FOR THE DEGREE OFDoctor of PhilosophyinTHE FACULTY OF GRADUATE AND POSTDOCTORALSTUDIES(Mathematics)The University of British Columbia(Vancouver)May 2019c© Ye Liu, 2019The following individuals certify that they have read, and recommend to the Fac-ulty of Graduate and Postdoctoral Studies for acceptance, the thesis entitled:Theoretical and numerical study of free-surface flow of viscoplas-tic fluids: 2D dambreaks, axisymmetrical slumps and surgesdown an inclined slopesubmitted by Ye Liu in partial fulfillment of the requirements for the degree ofDoctor of Philosophy inMathematics.Examining Committee:Neil Balmforth, MathematicsSupervisorSarah Hormozi, Mechanical EngineeringCo-supervisorAnthony Wachs, Mathematics and Chemical & Biological EngineeringSupervisory Committee MemberJames J. Feng, Chemical and Biological EngineeringAdditional ExaminerChristian Schoof, Earth, Ocean and Atmospheric SciencesAdditional ExaminerAdditional Supervisory Committee Members:Anthony WachsSupervisory Committee MemberBrian WettonSupervisory Committee MemberiiAbstractThe dynamics of free surface flow of yield stress fluid under gravity has been anopen problem, both in theory and in computation. The contribution of this thesiscomes in three parts.First, we report the results of computations for two dimensional dambreaksof viscoplastic fluid, focusing on the phenomenology of the collapse, the modeof initial failure, and the final shape of the slump. The volume-of-fluid method(VOF) is used to evolve the surface of the viscoplastic fluid, and its rheology iscaptured by either regularizing the viscosity or using an augmented-Lagrangianscheme. The interface is tracked by the Piecewise-Linear-Interface-Calculation(PLIC) scheme, modified in order to avoid resolution issues associated with theover-ridden finger of ambient fluid that results from the no slip condition and theresulting inability to move the contact line. We establish that the regularized andaugmented-Lagrangian methods yield comparable results. The numerical resultsare compared with asymptotic theories valid for relatively shallow or verticallyslender flow, with a series of previously reported experiments, and with predic-tions based on plasticity theory.Second, we report computations of the axisymmetric slump of viscoplasticfluid, with the PLIC scheme improved for mass conservation. The critical yieldstress for failure is computed and bounded analytically using plasticity methods.The simulations are compared with experiments either taken from existing liter-ature or performed using Carbopol. The comparison is satisfying for lower yieldstresses; discrepancies for larger yield stresses suggest that the mechanism of re-iiilease may affect the experiments.Finally, we report asymptotic analyses and numerical computations for surgesof viscoplastic fluid down an incline with low inertia. The asymptotic theoryapplies for relatively shallow gravity currents. The anatomy of the surge consistsof an upstream region that converges to a uniform sheet flow, and over which atruly rigid plug sheaths the surge. The plug breaks further downstream due to thebuild up of the extensional stress acting upon it, leaving instead a weakly yieldedsuperficial layer, or pseudo-plug. Finally, the surge ends in a steep flow front thatlies beyond the validity of shallow asymptotics.ivLay SummaryYield-stress fluids are common in our daily life: toothpaste, ketchup, polymergels, fresh concrete, etc. The common feature of these fluids is that they all tendto form certain shapes, when poured on a plane, or squeezed out of a tube. Unlikewater, which behaves totally like liquid and deforms with even a tiny bit of force,yield-stress fluid behaves like solid, as long as the force applied on it is withincertain threshold, called the yield-stress.So what is the final shape of a can of ketchup when it is poured on a plane andspreads out? What is the stress state of that slump? In which condition will theketchup not deform at all? How it is related to practical application? In this study,we will answer these questions in a systematic manner.vPrefaceIn this section, we briefly explain the contents of the journal papers that are pub-lished or submitted for publication from this thesis and clarify the contributionsof co-authors in the papers.• [Liu, Y.], Balmforth, N.J. & Hormozi, S. & Hewitt D.R. (2016) Two-dimensional viscoplastic dambreaks. J. Non-Newtonian Fluid Mech.238, 65-79.This publication has mostly focused at trying to understand the flow dy-namics of yield stress fluid in 2D geometry. Chapter 2 includes the contentsof this publication. The author of this thesis was the principal contributorto this publication. Professor Sarah Hormozi assisted with code develop-ment. Dr. Hewitt assisted with the numerical computation. Professor NeilBalmforth supervised the research and assisted with writing the paper.• [Liu, Y.], Balmforth, N.J. &Hormozi, S. (2018) Axisymmetric viscoplas-tic dambreaks and the slump test. J. Non-Newtonian Fluid Mech. 258,45-57This publication has mostly focused at trying to understand the fluid dy-namics of yield stress fluid in axisymmetric geometry. Chapter 3 includesthe contents of this publication. The author of this thesis was the princi-pal contributor to this publication. Professor Sarah Hormozi assisted withcode development. Professor Neil Balmforth supervised the research andassisted with writing the paper.vi• [Liu, Y.], Balmforth, N.J. & Hormozi, S. (2018) Viscoplastic surgesdown an incline. J. Non-Newtonian Fluid Mech. Submitted for pub-lication, under review.This publication has mostly focused at trying to understand the flow dy-namics on a slope in 2D geometry. Chapter 4 includes the contents of thispublication. The author of this thesis was the principal contributor to thispublication. Professor Sarah Hormozi assisted with code development. Pro-fessor Neil Balmforth supervised the research and assisted with writing thepaper.viiTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiLay Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiList of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiList of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiiAcknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xixDedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xx1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Objectives of the thesis . . . . . . . . . . . . . . . . . . . . . . . 21.2 Viscoplastic fluids . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Mathematical Tools . . . . . . . . . . . . . . . . . . . . . . . . . 41.3.1 Lubrication theory . . . . . . . . . . . . . . . . . . . . . 41.3.2 Lubrication paradox . . . . . . . . . . . . . . . . . . . . . 51.3.3 Slipline model . . . . . . . . . . . . . . . . . . . . . . . . 61.3.4 Numerical tools . . . . . . . . . . . . . . . . . . . . . . . 7viii1.3.5 Volume of fluid . . . . . . . . . . . . . . . . . . . . . . . 81.3.6 Regularization method . . . . . . . . . . . . . . . . . . . 91.3.7 Augmented-Lagrangian method . . . . . . . . . . . . . . 121.3.8 PLIC advection scheme . . . . . . . . . . . . . . . . . . . 131.3.9 Treatment of moving contact lines . . . . . . . . . . . . . 151.4 Literature review . . . . . . . . . . . . . . . . . . . . . . . . . . 161.4.1 Two-dimensional viscoplastic dambreaks . . . . . . . . . 161.4.2 Axisymmetric viscoplastic dambreaks and the slump test . 181.4.3 Viscoplastic surges down an incline . . . . . . . . . . . . 181.5 Outline of the thesis . . . . . . . . . . . . . . . . . . . . . . . . . 192 Two dimensional viscoplastic dambreak . . . . . . . . . . . . . . . . 212.1 Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.1.1 Dambreak arrangement and solution strategy . . . . . . . 222.1.2 Model equations . . . . . . . . . . . . . . . . . . . . . . 232.2 Newtonian benchmark . . . . . . . . . . . . . . . . . . . . . . . 262.3 Bingham slumps . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.3.1 Slump and plug phenomenology . . . . . . . . . . . . . . 302.3.2 Shallow flow . . . . . . . . . . . . . . . . . . . . . . . . 352.3.3 Slender slumps . . . . . . . . . . . . . . . . . . . . . . . 402.3.4 Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.3.5 The final shape and slump statistics . . . . . . . . . . . . 482.4 Concluding remarks . . . . . . . . . . . . . . . . . . . . . . . . . 533 Axisymmetric viscoplastic dambreaks and the slump test . . . . . . 553.1 Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.1.1 Problem set-up and solution strategy . . . . . . . . . . . . 563.1.2 Dimensionless model equations . . . . . . . . . . . . . . 573.1.3 PLIC scheme with interface correction . . . . . . . . . . . 593.2 Bingham slumps . . . . . . . . . . . . . . . . . . . . . . . . . . . 603.2.1 Shallow flow . . . . . . . . . . . . . . . . . . . . . . . . 63ix3.2.2 Slender columns . . . . . . . . . . . . . . . . . . . . . . 673.2.3 Failure mode . . . . . . . . . . . . . . . . . . . . . . . . 693.3 Comparison with experiments . . . . . . . . . . . . . . . . . . . 743.3.1 Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . 743.3.2 Cylindrical dambreaks . . . . . . . . . . . . . . . . . . . 753.3.3 Extrusions . . . . . . . . . . . . . . . . . . . . . . . . . . 853.4 Concluding remarks . . . . . . . . . . . . . . . . . . . . . . . . . 894 Surges of viscoplastic fluids on slopes . . . . . . . . . . . . . . . . . 924.1 Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 934.1.1 Model equations . . . . . . . . . . . . . . . . . . . . . . 934.1.2 The sheet-flow solution . . . . . . . . . . . . . . . . . . . 954.1.3 Volume-of-fluid computations . . . . . . . . . . . . . . . 954.2 Asymptotic analysis . . . . . . . . . . . . . . . . . . . . . . . . . 974.2.1 Plugged Flow . . . . . . . . . . . . . . . . . . . . . . . . 984.2.2 Breaking the plug . . . . . . . . . . . . . . . . . . . . . . 1004.2.3 Lubrication Zone . . . . . . . . . . . . . . . . . . . . . . 1014.2.4 Improving the lubrication theory . . . . . . . . . . . . . . 1024.3 Numerical results . . . . . . . . . . . . . . . . . . . . . . . . . . 1054.3.1 Comparison with asymptotics . . . . . . . . . . . . . . . 1054.3.2 Comparison with experiments . . . . . . . . . . . . . . . 1114.4 Concluding remarks . . . . . . . . . . . . . . . . . . . . . . . . . 1165 Summary and future research directions . . . . . . . . . . . . . . . 1185.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1185.2 Limitations of the study . . . . . . . . . . . . . . . . . . . . . . . 1195.3 Summary of contributions . . . . . . . . . . . . . . . . . . . . . . 1205.4 Future research directions . . . . . . . . . . . . . . . . . . . . . . 121Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123xA Two-dimensional viscoplastic dam break . . . . . . . . . . . . . . . 131A.1 Further numerical notes . . . . . . . . . . . . . . . . . . . . . . . 131A.1.1 Parameter settings and other details . . . . . . . . . . . . 131A.1.2 The failure computation for Re = t = 0 . . . . . . . . . . . 132A.1.3 Thickness of the over-ridden finger . . . . . . . . . . . . . 133A.2 Shallow flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134A.3 Slender columns . . . . . . . . . . . . . . . . . . . . . . . . . . . 136A.4 Plasticity solutions . . . . . . . . . . . . . . . . . . . . . . . . . 138A.4.1 Slipline fields . . . . . . . . . . . . . . . . . . . . . . . . 138A.4.2 Simple failure modes . . . . . . . . . . . . . . . . . . . . 141B Axisymmetric viscoplastic dambreaks and the slump test . . . . . . 144B.1 Numerical convergence study . . . . . . . . . . . . . . . . . . . . 144B.2 Higher-order shallow asymptotics . . . . . . . . . . . . . . . . . 147B.3 Shallow slippy flows . . . . . . . . . . . . . . . . . . . . . . . . 149B.4 Tall slender cones . . . . . . . . . . . . . . . . . . . . . . . . . . 150C Surges of viscoplastic fluids on slopes . . . . . . . . . . . . . . . . . 152C.1 Additional numerical details . . . . . . . . . . . . . . . . . . . . 152C.1.1 Resolution study . . . . . . . . . . . . . . . . . . . . . . 153C.1.2 Inertial effects . . . . . . . . . . . . . . . . . . . . . . . . 154C.1.3 The effect of the back wall . . . . . . . . . . . . . . . . . 154C.2 Improved lubrication solution . . . . . . . . . . . . . . . . . . . . 157C.3 Improved slump profiles . . . . . . . . . . . . . . . . . . . . . . 160xiList of TablesTable 4.1 Experimental parameters from . . . . . . . . . . . . . . . . 110xiiList of FiguresFigure 1.1 A sketch of the slipline field . . . . . . . . . . . . . . . . . . 7Figure 2.1 A sketch of the geometry for the case of a rectangular initialblock. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23Figure 2.2 Snapshots of the evolving interface for a Newtonian fluid . . . 27Figure 2.3 Flow front X(t) plotted against time and interface shape forcomputations with Newtonian fluids . . . . . . . . . . . . . . 28Figure 2.4 Front position and central depth for Bingham dambreaks witha square initial block . . . . . . . . . . . . . . . . . . . . . . 31Figure 2.5 Snapshots of a collapsing square . . . . . . . . . . . . . . . . 32Figure 2.6 Snapshots of the evolving interface for a triangular initial con-dition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33Figure 2.7 Snapshots of the evolving interface for an incomplete slump . 35Figure 2.8 Slumps of slender columns . . . . . . . . . . . . . . . . . . . 36Figure 2.9 Comparison of shallow-layer theory with numerical results fora collapsing square block . . . . . . . . . . . . . . . . . . . . 38Figure 2.10 Critical yield stress, Bc, plotted against initial width X0 . . . . 42Figure 2.11 Trial velocity fields to compute lower bounds on Bc . . . . . . 44Figure 2.12 Strain-rate invariant plotted logarithmically as a density on the(x,z)−plane for blocks . . . . . . . . . . . . . . . . . . . . . 46Figure 2.13 Strain-rate invariant plotted logarithmically as a density on the(x,z)−plane for triangles . . . . . . . . . . . . . . . . . . . . 47xiiiFigure 2.14 Profiles of the final deposit . . . . . . . . . . . . . . . . . . . 49Figure 2.15 Final numerical solution for the slump of an initial square andsliplines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49Figure 2.16 Scaled final runout and central depth as a function of B/√X0for Bingham slumps from square initial conditions . . . . . . 51Figure 2.17 Scaled final runout and central depth as a function of B/√X0for Bingham slumps from rectangular and triangular initialconditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52Figure 3.1 A sketch of the geometry for the case of a cylindrical block . . 56Figure 3.2 Time series of flow height and front position for Binghamslumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61Figure 3.3 Profiles of the final deposit of cylindrical slumps . . . . . . . 62Figure 3.4 Evolution of the interface for a slumping cylinder with (R,B)=(1,0.1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63Figure 3.5 Final profile for a cylindrical dambreak with B = 0.0074 andR = 0.2546 . . . . . . . . . . . . . . . . . . . . . . . . . . . 66Figure 3.6 Collapse of a cylindrical column . . . . . . . . . . . . . . . . 69Figure 3.7 Critical yield stress for collapse Bc as a function of initial ra-dius R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70Figure 3.8 Strain-rate invariant plotted logarithmically as a density on the(r,z) plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71Figure 3.9 Comparison of experimental slump height with previously pub-lished data . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76Figure 3.10 Experimental and numerical slump heights against B . . . . . 77Figure 3.11 The central depth and radius of the final deposit against Bing-ham number . . . . . . . . . . . . . . . . . . . . . . . . . . . 80Figure 3.12 Side images from the slump experiments and theoretical finalprofiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82Figure 3.13 The dimensionless slump height for cylindrical dambreaks withR = 1, varying B and the Reynolds numbers indicated . . . . . 84xivFigure 3.14 Maximum radius versus depth for experimental extrusions overthe rough and smooth plexiglass . . . . . . . . . . . . . . . . 86Figure 3.15 Images and profiles from the extrusion experiments . . . . . . 88Figure 3.16 Slump heights for Bingham dambreaks with an initially coni-cal shape corresponding to the ASTM standard . . . . . . . . 91Figure 4.1 Sketch of the geometry of the surge in the frame of referencein which there is no net flux and the flow is steady. The freesurface is located at z = h; the level z = Y divides a fullyyield region underneath from either a true plug or a weaklyyielded pseudo-plug. The flow divides into three regions: aplugged flow region (PF) where the superficial layer of fluidis not yielded, a lubrication zone (LZ) where the pseudo-plugarises, and the flow front (FF) where the dynamics is not shallow. 93Figure 4.2 Asymptotic surge solutions for (a) the plastic limitY → 0, and(b) Y∞ = 1−S−1B = 0.2. The solid lines show the improvedlubrication solutions for h(x) plotted against S(x− x f ) with12εpiB = 0.2 (as given by either equations (4.41) and (4.42));the dotted lines indicate the predictions of the leading-ordertheory. In (b), the corresponding fake yield surfaces Y (x) arealso plotted. The insets show magnifications near the flow front.104Figure 4.3 Numerical solution for θ = 10◦ and Y∞ = 0.8 showing (a)τI, (b) τxz and (c) τxx. The darker (blue) lines show samplestreamlines and the dotted white line is the true yield surfacewhere τI = B. The dashed line shows the contour level whereτxz = B. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106xvFigure 4.4 Flow profiles and fake yield surfaces for (a) numerical simu-lations, (b) leading-order lubrication theory and (c) improvedasymptotic theory (using (4.42)), with θ = 10◦ and Y∞ = 0.4,0.6 and 0.8. The solid and dashed lines show h and Y , with Ydefined as the contour level where τxz = B for the simulations.The flow profiles deepen with increasing Y∞. The shaded re-gions in (a) show the true plugs (which extend up to the freesurface in each case, and with the shading darkening with in-creasing Y∞). . . . . . . . . . . . . . . . . . . . . . . . . . . 107Figure 4.5 Numerical solution for Y∞ = 0.8 and θ = 5◦ (B = 0.0175),showing a density plot of τI with z = h(x) and the fake yieldsurface superposed; the true plug is shaded black. The dashedlines show the prediction of the leading-order lubrication the-ory (with the flow front aligned). The dotted lines show thepredictions of the plugged-flow solution in (4.26), with 1−hmatched to the simulation at x = 5. . . . . . . . . . . . . . . . 109Figure 4.6 Experiment C5 of , showing (a) u(x,z) and (b) w(x,z) asdensities over the (x,z)−plane, for a qualitative comparisonwith their figure 6. A selection of streamlines is also shown(thinner blue lines). The dashed lines shows the levels whereu = 0 and z =Y (τxz = B). In leading-order lubrication theory,u = 0 along z =Y −Y [nY/h(2n+1)]n/(n+1); this prediction isalso drawn as the lighter (pink) solid line. . . . . . . . . . . . 110Figure 4.7 Experiment C2 of : shown are (a) log10(γ˙/0.26), (b) thesurface velocity, and (c)–(h) velocity profiles at the x−positionsindicated in (b), for a comparison with figure 10 in . The(red) dashed lines indicate the predictions of the leading-orderlubrication theory. In (a) the solid white line shows the con-tour τxz = B, whereas the dashed white line is the fake yieldsurface z = Y (x) of the leading-order lubrication theory. . . . 113xviFigure 4.8 A similar picture to figure 4.7, but for Experiment C5 of and for comparison with their figure 9 (with log10(γ˙/0.38)plotted in (a)) . . . . . . . . . . . . . . . . . . . . . . . . . . 114Figure 4.9 Flow profiles for a simulation with θ = 14.6◦, τY = 6 Pa, κ =6.65 Pa sn and n = 0.405 (the experiment C PIV of ), for(a) ub = 0.2, 0.4, 0.8, 1.2 and 1.4 m/s, in the steady state,and (b) ub = 1.6 m/s at a succession of times, starting fromthe initial profile shown by the dashed line (t = 0, 0.06, 0.08,0.11, 0.17 and 0.19s). In (a), the profile for ub = 1.4 m/s isshown by the solid line and several streamlines are also plotted(dotted lines). . . . . . . . . . . . . . . . . . . . . . . . . . . 116Figure A.1 Slender asymptotic solutions . . . . . . . . . . . . . . . . . . 139Figure A.2 Slipline solutions for a rectangle . . . . . . . . . . . . . . . . 141Figure B.1 Computations with varying grid size . . . . . . . . . . . . . . 145Figure B.2 Resolution studies for various numerical solvers . . . . . . . . 146Figure B.3 Scaled stress components and radial speed . . . . . . . . . . . 148xviiFigure C.1 Simulations of a Bingham surge profile with θ = 10◦ andB = 0.0522 (Y∞ = 0.7, fluid area of 24, ub = 0.041 and Re =1) with the grid sizes indicated. (a) shows h and Y ; (b) amagnification near the flow front; (c) contours of constantτxz/B, as indicated (with the surge profile of the finest res-olution case shown by the lighter grey line); (d) the shearstress along z = 0.025 (the lowest grid cell of the coarsestcomputation). In (a), the inset shows the flow front X f , meanshear stress, τxz/B =∫ ∫(τxz/B) c dx dz, and τxz at the point(x,z) = (18,0.4), all plotted against the vertical grid spacing∆z (∆x = 4∆z). The convergence of τxz is impeded by theneed to resolve the sharply localized region of high stress un-derneath the flow front (cf. figure 4.3 and panels (c) and (d)).. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155Figure C.2 Simulations of Bingham surges for θ = 10◦, ub = 0.041 andB = 0.0522 (Y∞ = 0.7) for the Reynolds numbers indicated(the total area of fluid is 24). Plotted are h, Y (τxz = B) and thestress levels τI/B = 1, 2 and 3. . . . . . . . . . . . . . . . . . 156Figure C.3 Simulations for θ = 10◦ and B = 0.0354 (Y∞ = 0.8) with dif-ferent left-hand boundary conditions: (a) (u,w) = (usheet ,0)and (b) (u,w) = (0,0). Shown is the second invariant τI asa density on the (x,z)−plane. The (true) yield surfaces areindicated by the solid (green) lines. The simulation in (a) cor-responds to the truncated solution shown in figure 4.3. Theprofiles and the true and fake yield surfaces are compared in(c). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162xviiiFigure C.4 Slump profiles for (a)-(b) S = 0 and (b)-(c) S 6= 0. For (a)-(b),scaled variables are plotted, which eliminates any free param-eters; in (c)-(d) the profiles are shown for 12εpiB = 0.2. Thedark solid lines show the solutions to (4.38), whereas the dot-ted lines show the solutions to the iterated version in(4.40); thedashed lines show the leading-order approximation. In (a)–(b)the lighter (red and blue) lines show the results of a series ofsimulations from with ε = 1, B = 0.02, 0.03, ..., 0.1 andthe flow fronts aligned. In (c)–(d) the lighter (red) lines showadditional simulations of the slump of a rectangular block onan incline with B = (0.1,0.2,0.3,0.4)/pi and θ = 5◦; the insetshows the aligned, but unscaled profiles. . . . . . . . . . . . 163xixAcknowledgmentsI would like to convey my gratitude to all people who gave me the possibility tocomplete this dissertation.In the first place, I would like to express my sincere gratitude to my supervi-sor, Professor Neil Balmforth for the continuous support of my Ph.D study andresearch, for his patience, motivation, enthusiasm, and immense knowledge. Hiswide knowledge and his logical way of thinking have been of great value to me.I am deeply grateful to my co-supervisor Sarah Hormozi for her guidance andassistance in my research. I would not be able to accomplish the development ofthe computational tool without her teaching and help.I gratefully acknowledge Professor Anthony Wachs, Brian Wetton and JamesFeng, for their supervision, advice, and guidance from the early stage of this re-search. Above all and the most needed, they provided me support and friendlyhelp in various ways.I would like to thank all my colleagues and schoolmates who have supportedme all along my study and research.Most importantly, I would like to thank my entire family. My immediate fam-ily to whom this dissertation is dedicated, has been a constant source of love,concern, support and strength all these years. I would like to express my heart-feltgratitude to all my friends who have supported and encouraged me throughout thisendeavor.xxDedicationTo my familyxxiChapter 1IntroductionViscoplastic fluids are commonly encountered in natural settings in geophysics(e.g. mud and and lava) and biology (mucus and blood clots), and feature inmany engineering processes in, for example, the food (fruit pulp, dairy productsand chocolate confections) and petroleum industries (drilling mud, cement andwaxy crude oil). These materials flow like viscous fluid once the stress exceedsa certain threshold (the yield stress), while remaining solid-like, with any defor-mation often assumed small and discarded. In that situation, which underpinspopular constitutive models such as the Bingham and Herschel-Bulkley laws, thestress state of the material becomes formally indeterminate . Together withthe need to track the boundaries of the yielded regions, this complicates signifi-cantly efforts aimed at theoretical modelling. Numerical strategies to overcomesuch difficulties have been developed in recent years and here we apply them tothe particular problem of the free-surface flow of viscoplastic fluid under gravity,including two dimensional dambreaks, axisymmetric slumps and surges onan inclined slope, which feature in a diverse range of problems from geophysicsto engineering. These flows can constitute natural or manmade hazards, as in thedisasters caused by mud surges and the collapse of mine tailing deposits. In anindustrial setting, the controlled release of a reservoir in a simple dambreak exper-iment forms the basis of a number of practical rheometers, including the slump1test for concrete [61, 68] and the Bostwick consistometer of food science .The Bostwick consistometer, in which materials such as ketchup are released in arectangular channel, is conveniently idealized by a two-dimensional flow (Chap-ter 2), while the slump test features the release of a cylinder of yield-stress fluid,which is widely used as a measurement of concrete rheology (Chapter 3). Finally,viscoplastic surges on an inclined plane (Chapter 4) have been studied experimen-tally, which provides useful results for theoretical comparisons.1.1 Objectives of the thesisThis thesis aims at the three topics described above. Overall our research seeks tofully understand the dynamics of fluid in each context. Different objectives of thisstudy include the following.1. We want to explore or improve the theoretical model in these problems.The lubrication model, widely used in simulating those processes, could beexpanded to higher order to provide more accurate results.2. We want to provide robust and accurate numerical simulations of theseproblems. The difficulties here mainly include the treatment of indeter-minate stress state in the plug and the advection of fluid interface.3. We want to understand if there is a true plug or a plug-like region held closeto, but above, the yield stress, in each case, which has been confusing andcontroversial in previous studies. We want to show it both asymptoticallyand numerically.4. By the theoretical and numerical tools we have developed, we want to per-form various numerical simulations of existing experiments in literature.We have also conducted some experiments ourselves, in order to supportthe improved model and numerical strategies.21.2 Viscoplastic fluidsBefore introducing the concept of viscoplastic fluid, we first review the rheologyof Newtonian fluid, where the shear stress tensor τi j(u) and shear rate tensorγ˙i j(u)are proportional,τi j(u) = µγ˙i j(u)where µ is the viscosity of the material(constant for Newtonian fluids). This equa-tion is called the constitutive law. Many fluids in reality, however, tend to followmore complicated constitutive laws, and are called non-Newtonian fluids.Viscoplastic fluid is a specific type of non-Newtonian fluid, which remainsundeformed until the stress exceeds certain value, called yield stress(τY ). Whenthe stress exceeds the yield stress, then the fluid deforms as viscous fluid, withcertain rheological models, as generated from varying materials, among whichtwo common ones are what we mainly discussed in this research.• Bingham modelThe Bingham model is the simplest viscoplastic model. Above the yieldstress, the constitutive law followsγ˙ jk = 0, τ < τY ,τ jk =(µ +τYγ˙)γ˙ jk, τ > τY ,(1.1)• Herschel-Bulkley modelThe Herschel-Bulkley model extends the simple power-law model to in-clude a yield stress as belowγ˙ jk = 0, τ < τY ,τ jk =(µγ˙n−1+τYγ˙)γ˙ jk, τ > τY ,(1.2)where τ =√12 ∑ j,k τ2jk and γ˙ =√12 ∑ j,k γ˙2jk. There are other models but we will3focus on the study of these two. The reason for studying the Bingham model isbecause of its simplicity, which reveals the yield stress behavior with minimumnumber of parameters considered. And the reason we use the Herschel-Bulkleymodel is because the most commonly used experiment material, Carbopol, is wellfitted by this model so that we can perform numerical simulations of existingexperiments with Carbopol. We believe that the extension to other yield stressmodels is straightforward, both theoretically and numerically.It is seen from the constitutive laws above that, below the yield stress, the rela-tion between the stress and strain rate remains unknown, which greatly increasesthe difficulty of the analysis and the computation. In the thesis, we will discuss indetail how we overcome this difficulty.1.3 Mathematical ToolsIn this section, we will introduce the main mathematical tools we have appliedin this thesis. For the theoretical part, we shall review the lubrication theory andthe slipline model in plasticity theory. For numerical part, we will describe thebasic idea of the regularization method, augmented-Lagrangian method and thePiecewise-Linear-Interface-Calculation (PLIC) method.1.3.1 Lubrication theoryIn fluid dynamics, lubrication theory describes the flow of fluids (liquids or gases)in a geometry in which one dimension is significantly smaller than the others. Thishelps simplify the momentum equation and allows for consistent asymptotic ex-pansions. Moreover, lubrication theory applies when the flow is slow so that onemay add the Stokes approximation. For free-surface viscoplastic flows, the appli-cation of lubrication theory dates back to Liu and Mei . Further theoreticalinsights and extensions were given by Balmforth et al. [6, 7, 10, 11, 15], Matsonand Hogg [43, 58], and Ancey and Cochard , based on which the leading order4solution for the final shape of a viscoplastic slump has been well known ash(x) =√2τYρg(x∞− x)in two-dimensional geometry, andh(r) =√2τYρg(r∞− r)in axisymmetric geometry, where h(x),h(r) refer to the flow height as a function ofthe horizontal coordinate, ρ is the fluid density, g is the gravitational acceleration,and x∞,r∞ are the maximum length of the slump.In this thesis, we will apply lubrication theory to all the three problems con-sidered, and improve those solutions to higher order.1.3.2 Lubrication paradoxThe so-called “lubrication paradox” is from Lipscomb & Denn , who appliedlubrication theory and predicted plug regions that were apparently below the yieldstress, while deformation rate is not zero. This inconsistency was later understood,as the paradox can be resolved by recognizing that the lubrication theory is theleading order term of an asymptotic expansion; an exploration of higher-orderterms demonstrates that the problematic plugs are actually slightly above the yieldstress, allowing for their deformation (Putz et al. , Walton & Bittleston ).The region has been termed a pseudo-plug to emphasize its nature, with the borderreferred to as a fake yield surface. Therefore there is no lubrication paradox.In the pseudo-plug, the stress equals the yield stress to the leading order, sothat it can be considered as a perfect plastic material, where the stress componentmust satisfyτ2xx + τ2zz +2τ2xz = 2τ2Y5in two-dimension, orτ2rr + τ2θθ + τ2zz +2τ2rz = 2τ2Yin axisymmetric geometry. This is the key assumption based on which the leadingorder asymptotic solution can be expanded to higher order. (And this is actuallytrue for slump problem in the steady state, even when it is not shallow. We willsee it in the following chapters.)1.3.3 Slipline modelSlipline field theory is used to model plastic deformation in plane strain only for asolid that can be represented as a rigid-plastic body. Elasticity is not included andthe loading has to be quasi-static. In terms of applications, the approach has beenapplied by Nye to study the “dynamics” of glaciers, and Dubash et al. to study the final shape of viscoplastic slump. In 2D geometry, the equilibriumequations for plane strain are:∂σxx∂x+∂τxz∂ z= 0∂τxz∂x+∂σzz∂ z= 0where σxx,σzz are the total stress components. There will always be two per-pendicular directions of maximum shear stress in a plane. These generate twoorthogonal families of slip lines called α lines and β lines, as shown in 1.1. Thevelocity distribution and stress state in the solid can always be determined fromthe geometry of these lines, and the Hencky relations givep+2kφ = constant along an α linep−2kφ = constant along a β line6Figure 1.1: A sketch of the slipline fieldwhere p is the hydrostatic pressure, k is the shear yield stress(τY in our case), andφ is the angle formed by the x-axis and the α line. Andσxx =−p− k sin2φσzz =−p+ k sin2φτxz = kcos2φBased on the relations above, we are able to construct a slipline field, given ap-propriate initial conditions. This technique has been applied in chapter 2 to solvethe final shape and stress state of two dimensional viscoplastic slump.1.3.4 Numerical toolsIn this thesis, I use the volume-of-fluid method (VOF) to solve the problem. Theadvantage of the VOF method is mass is well conserved. In our study, we usestructured rectangular grids for the simulation. The method works well with our7problem.In every time step, the code does two things:• Solving the Navier-Stokes equations of two fluids, to get the velocity field.• Solving the advection equation by given velocity field, to move the inter-face.There are two options for the Navier-Stokes solver: regularization method oraugmented-Lagrangian method, which are the most commonly used schemes forviscoplastic fluid. The advantage of the regularization method is that the compu-tation converges faster, while the disadvantage is that it is not quite accurate whenτ ∼ τY , so that it is not recommended in studying the incipient failure problem.The advantage of augmented-Lagrangian method is that it solves the exact consti-tutive equations. The disadvantage is that it converges much more slowly than theregularization method.All of the algorithms are implemented in C++ as an application of PELICANS,an object oriented platform developed at IRSN, France, to provide general frame-works and software components for the implementation of PDE solvers. PELI-CANS is distributed under the CeCILL license agreement.1.3.5 Volume of fluidWe use the concept of the volume-of-fluid method to simulate a two-phase fluidproblem. The volume fraction variable c = 1 for the lower layer, and c = 0 forthe upper layer. The problem is discretized into a Navier-Stokes equation and anadvection-diffusion equation. See Equation 1.3Re(c)[∂u∂ t+(u ·∇)u]=−∇p+∇ · τ(u)+Ri(c)g∇ ·u = 0∂c∂ t+∇ · (uc) = 0(1.3)8whereRe(c) :=(c+(1− c)ρ2ρ1)ρ1URµ1Ri(c) :=(c+(1− c)ρ2ρ1)ρ1gR2µ1U0τ(c) := cτ(u)+(1− c)µ2µ1γ(u)where τ(u) is defined in the following section.1.3.6 Regularization methodIn this section, we summarize the regularization method for single-phase fluid,but the generalization to VOF is straightforward. For a general Navier-Stokesequation in a computational domain Ω, with no-slip condition on the boundaryΓD and no-flux condition on the boundary ΓN , as belowρ[∂u∂ t+(u ·∇)u]=−∇p+∇ · τ(u)+ρg in Ω∇ ·u = 0 in Ωu = 0 in ΓD∇u ·n = 0 in ΓN(1.4)At a certain time tn, given the current velocity field un, time step ∆t, the discretescheme for solving un+1 isρ∆tun+1+(ρun ·∇)un+1−∇ · τ(un+1)+∇pn+1 = ρg+ ρ∆tun in Ω∇ ·un+1 = 0 in Ωun+1 = 0 in ΓD∇un+1 ·n = 0 in ΓN(1.5)9The variational form is: Find (u, p)∈ Su×V p so that for all trial functions (v,q)∈V u×V p:1∆t∫Ωρu · v+∫Ω(ρun ·∇)u · v+∫Ωτ(u) : ∇v−∫Ωp∇ · v =∫Ωρg · v+ 1∆t∫Ωρun · v∫Ωq∇ ·u = 0(1.6)where the function spaces Su,V u,V p are defined asSu = {v ∈ H1(Ω)|v = 0 in ΓD}V u = {v ∈ H1(Ω)|v = 0 in ΓD}V p = L2(Ω)(1.7)We can rewrite this in a simple way1∆tm(u,v)+ c(un;u,v)+ k(u,v)+b(v, p) = f (v)b(u,q) = 0(1.8)wherem(u,v) =∫Ωρu · vc(un;u,v) =∫Ωρ(un ·∇)u · vk(u,v) =∫Ωτ(u) : ∇vb(v, p) =−∫Ωp∇ · vf (v) =∫Ωρg · v+ 1∆tm(un,v)(1.9)10This variational equation is then solved using a Galerkin method. On a finitedimensional subspace V uh ⊂V u,V ph ⊂V p1∆tm(uh,vh)+ c(unh;uh,vh)+ k(uh,vh)+b(vh, ph) = f (vh)b(uh,qh) = 0(1.10)After a finite element basis is chosenXh = span{ψuk | 0≤ k < Nudo f }Mh = span{ψ pk | 0≤ k < N pdo f }The variational equation is transformed to a linear system{AU +BT P = FBU = 0(1.11)whereAIJ =1∆tm(ψuJ ,ψuI )+ c(unh;ψuJ ,ψuI )+ k(ψuJ ,ψuI ) 0≤ I,J < Nudo fBIJ = b(ψuJ ,ψpI ) 0≤ I < N pdo f ,0≤ J < Nudo fFI = f (ψuI ) 0≤ I,J < Nudo f(1.12)Notice that k(ψuJ ,ψuI ) is a nonlinear term becauseτ(u) =(µ +τy|∇u|)∇u |τ(u)|> τy∇u = 0 |τ(u)|< τy(1.13)Therefore we use a regularized model and an iteration scheme to overcome thisdifficulty.τ(u) =(µ +τy|∇u|+ ε)∇u (1.14)11is a simple regularized model of Eq 1.13. If we assume |∇u| on the denominatoris known, i.e.τ(u) =(µ +τy|∇uˆ|+ ε)∇u (1.15)where uˆ is from the previous iteration step, then the operator is linear. Thus wecan do the iteration on each time step t = tn. The implementation of this is inAlgorithm 1.Algorithm 1 Regularization methodRequire: GivenUn,Pn, Uˆ =Un, Imax, i = 0Generate matrix A,B and vector F from Un,Pn,Uˆwhile i < Imax doSolve the linear system(A BTB 0)(UP)=(F0)if |Uˆ−U |< tolerance thenreturn Un+1 =UelseUˆ =U Update A from Uˆend ifi = i+1end whilereturn Un+1 =U1.3.7 Augmented-Lagrangian methodA detailed description of the augmented-Lagrangian method can be found in .We just list the main ideas here.12Find (u, p,d ,λ ) so that for ∀(v, l,q)ρ∆t∫Ωu · v+∫Ω(ρun ·∇)u · v+∫Ωµ∇u ·∇v−∫Ωp∇ · v+∫Ωλ ·∇v+ r∫Ω(∇u−d) ·∇v =∫Ωf · v∫Ωl∇ ·u = 0d =1r(1− τy|λ + r∇u|)(λ + r∇u) if |λ + r∇u| ≥ τy0 if |λ + r∇u|< τy∫Ω(∇u−d) ·q = 0(1.16)where f = ρg+ ρ∆t un. The implementation of the scheme is in Algorithm 2.1.3.8 PLIC advection schemeWe used the PLIC (short for Piecewise Linear Interface Calculation) to solve theadvection of the fluid volumeC,∂C∂ t+u ·∇C = 0The scheme is designed to avoid numerical diffusion of the interface, and to in-crease accuracy in tracking the interface. It is 2nd order accurate in space. Theequation is discretized by a forward Euler schemeCn+1 =Cn−un∆t ·∇CnThe scheme is decomposed into x and y direction, so we have two sub-steps inevery time step. The key part of this scheme is to compute the line approximationgiven different circumstances. For the details of the this method, we refer toRenardy. An original implementation of PLIC scheme is in Algorithm 3.13Algorithm 2 Augmented-Lagrangian schemeRequire: Given un,r, Imaxi = 0 λ i = 0 d i = 0while i < Imax doi = i+1Solve the problem, find (ui, pi) so that ∀(v, l)ρ∆t∫Ωui · v+∫Ω(ρun ·∇)ui · v+∫Ω(µ + r)∇ui ·∇v−∫Ωpi∇ · v=−∫Ω(λ i−1− rd i−1) ·∇v+∫Ωf · v∫Ωl∇ ·ui = 0Update the velocity gradient dd i =1r(1− τy|λ i−1+ r∇ui|)(λ i−1+ r∇ui) if |λ i−1+ r∇ui| ≥ τy0 if |λ i−1+ r∇ui|< τyUpdate the Lagrangian multiplierλ i = λ i−1+ r(∇ui−d i)if max{|ui−ui−1|, |∇ui−d i|}< tolerance thenreturn un+1 = uiend ifend whilereturn un+1 = u14Algorithm 3 PLIC scheme 1.0Require: Given un,C,∆tEnsure: The CFL condition (udx+vdy)∆t <12Construct the linear approximation y = mx+b in each grid cell, where m = ∇Cand the volume fraction in the grid equals CCompute the new line y = mˆx+ bˆ after advection of un∆tCompute the flux from y = mx+b and y = mˆx+ bˆ and updateCRegularize C so that 0≤C ≤ 11.3.9 Treatment of moving contact linesA common problem in the computation of a two-phase fluid problem with a no-slip boundary condition is the movement of the contact line (the point where theinterface touches the static wall). Failure to address this problem will lead to anunresolved boundary layer (finger) of ambient fluid on the bottom. Commonly aslip boundary condition is used around the contact line to move it. However, theschemes always depend on the contact angle, or what parameterization is used toallow the contact line to move, while surface tension is neglected in our study.Currently we have implemented a simple correction in the PLIC scheme to avoidthe resolution issue associated with the finger. The algorithm is implemented inAlgorithm 4.We find that this scheme successfully removes the finger, leading to an appar-ent motion of the contact line, though none is in fact happening. Currently wehave tested that the solution is not affected by a slight change of the number from0.99 to 0.9. Moreover, the numerical results achieve much better spacial conver-gence using this correction than the original PLIC scheme. The main problem ofthis scheme is the violation of mass conservation, asC is always increasing duringadvection, which can affect the simulation at later times, particularly in axisym-metric geometry. To make the mass conserved, we make further modification of15Algorithm 4 PLIC scheme 1.1Require: Given un,C,∆tEnsure: The CFL condition (udx+vdy)∆t <12Construct the linear approximation y = mx+b in each grid, where m = ∇C andthe volume fraction in the grid equals CCompute the new line y = mˆx+ bˆ after advection of un∆tCompute the flux from y = mx+b and y = mˆx+ bˆ and updateCRegularize C so that 0≤C ≤ 1Check the value ofC on the bottom layer of the grid cell and the one just aboveit, we note them as C0 and C1, respectivelyIf C1 > 0.99, imposeC0 =C1our scheme, see Algorithm 5. In this way, the mass conservation is guaranteedand the solution converges even better.1.4 Literature review1.4.1 Two-dimensional viscoplastic dambreaksDespite wide-ranging practical applications, the theoretical modelling of viscoplas-tic dambreaks remains relatively unexplored. Asymptotic theories for shallow,slow flow have received previous attention and permit a degree of analytical in-sight into the problem (see [11, 52] and references therein). Numerical compu-tations of two-dimensional dambreaks have also been conducted to model flowsthat are not necessarily shallow . However, these simulations do not provide adetailed survey of the flow dynamics over a wide range of physical conditions andhave focused mainly on determining some of the more qualitative aspects of theend state of a slump, such as its final runout and maximum depth. Complement-ing both asymptotics and numerical simulation are cruder predictions of the final16Algorithm 5 PLIC scheme 1.2Require: Given un,C,∆tEnsure: The CFL condition (udx+vdy)∆t <12Construct the linear approximation y = mx+b in each grid, where m = ∇C andthe volume fraction in the grid equals CCompute the new line y = mˆx+ bˆ after advection of un∆tCompute the flux from y = mx+b and y = mˆx+ bˆ and updateCRegularize C so that 0≤C ≤ 1Check the value of C on the bottom layer of grid and its upper layer, we notethem as C0 and C1, respectivelyIf C1 > 0.99, imposeC0 =C1Compute the mass increment ∆C, and the total volume of the interface CIRescale the volume fraction in the interface by a factor of 1− ∆CCIshape based on solid mechanics and initial failure criteria derived from plasticitytheory [20, 21, 61].The key feature of a viscoplastic fluid that sets the problem apart from a clas-sical viscous dambreak is the yield stress. When sufficient, this stress can hold thefluid up against gravity, preventing any flow whatsoever. If collapse does occur,the yield stress brings the fluid to a final rest and can maintain localized rigid re-gions, or “plugs”, during the slump. The evolving plugs and their bordering yieldsurfaces present the main difficulty in theoretical models, particularly in numeri-cal approaches. Augmented-Lagrangian schemes that deal with the complicationsof the yield stress directly are often time-consuming to run, whereas regulariza-tions of the constitutive law that avoid true yield surfaces introduce their ownissues . For the dambreak problem, difficulties are compounded by the needto evolve the fluid surface and impose boundary conditions such as no-slip on thesubstrate underneath the fluid.171.4.2 Axisymmetric viscoplastic dambreaks and the slump testAxisymmetric collapses are exploited widely to gauge fluid rheology in the con-crete, mineral and food industries. The slump test, for example, is commonly usedto measure the yield stress of fresh concrete. In this test, a container filled withconcrete is lifted to release the material and allow it to spread under gravity; atstoppage, the vertical distance over which the concrete falls, the “slump height”,is measured as an indicator of yield stress. A number of experimental studies havebeen directed at establishing the precise relation of the slump height to the yieldstress for a variety of different types of viscoplastic fluids [25, 39, 61, 70].Theoretical studies of the slump test have been performed using either numer-ical computations, asymptotic analysis suitable for the limit of shallow flow, orestimates and bounds based on plasticity theory [11, 17, 21, 25, 31, 34, 61, 63,67, 67, 69, 71]. The numerical simulations have been conducted using a varietyof numerical techniques, although most of the algorithms employed were not spe-cially designed to capture yield-stress rheology or carefully track fluid interfaces.The current status of the modelling of the slump test is reviewed by .1.4.3 Viscoplastic surges down an inclineThe spreading of viscoplastic fluid over an inclined surface has been studied ex-perimentally in a number of previous studies, often with the goal of inferringthe yield stress from steady flows [28, 29, 32] or the shape of a final deposit[30, 33, 46, 60]. The most thorough and recent laboratory studies include thetransient dam-break-type experiments of Ancey and co-workers [3, 16, 27] and aseries of investigations on steady viscoplastic surges on inclined conveyor belts[22, 23, 36].Theoretically, a model for shallow viscoplastic flow based on Reynolds lu-brication theory has been widely used to complement such laboratory studies[10, 11, 51]. In this model, the long, thin flow is composed of a fully shearedregion adjacent to the underlying surface buffered from the free surface by a plug-like zone. Importantly, that zone is not truly rigid, but deforms weakly in the18direction of flow and is plug-like in that the transverse velocity profile is largely in-dependent of depth. This structure is common in many shallow viscoplastic flows[4, 81] and results from the separation of length scales in the directions aligned orperpendicular to flow. The border between the fully yield region and the plug-likezone is therefore not a true yield surface; instead, it is often referred to as a fakeyield surface, and the overlying zone as a pseudo-plug. Despite the fact that lubri-cation theory predicts the appearance of pseudo-plugs in shallow flows, it is alsoknown that genuine rigid plugs can appear within these zones surrounding pointsof symmetry [65, 81] or replace them in flow down almost uniform channels .This raises the question of whether the superficial regions of a free-surface flowcan also plug up in the far upstream extent of a steady surge flow, where the flowbecomes almost uniform, as discussed in a qualitative way by Piau . Indeed,for a truly steady surge that extends infinitely far upstream, one expects that theflow converges to a uniform sheet flow which, for a yield-stress fluid, is sheathedby a true plug.1.5 Outline of the thesisThe two dimensional viscoplastic dambreak problem is studied in chapter 2. Nu-merical schemes are validated with a Newtonian fluid problem first, after whicha thorough investigation of Bingham slump is illustrated, including the plug phe-nomenology, the final shape in shallow and slender geometry, and the criticalvalue for the slump to happen, etc. In comparison with the numerical result, animproved asymptotic analysis has been derived and presented. The theoreticaland numerical results are then compared with available experimental data fromliterature.In chapter 3, the axisymmetric slump of viscoplastic fluid is studied, withsimilar structure as in chapter 2. In addition, we conduct experiments of slumpand extrusion tests with Carbopol, and present the results, in comparison withnumerical simulations.In chapter 4, the surge problem of viscoplastic fluids on slopes is studied,19theoretically and numerically. The theory starts with a simple sheet flow solution,where a plug exists everywhere, then perturbation study has implied the existenceof a true plug and its breaking condition. The transition zone from a true plugto sheared flow, or the pseudo-plug is shown then, with an improved lubricationtheory. Finally, numerical simulations are conducted and compared with bothasymptotic results, and experimental results from .Chapter 5 of the thesis contains a summary of conclusions of the thesis andrecommendations for future work.Appendix A corresponds to Chapter 2. It provides more details of our parame-ter setting in the two dimensional viscoplastic dambreak problem. It also includesthe mathematical derivation of the improved asymptotic analysis, the main idea ofconstructing the slipline field, and ways to compute the critical value with simplefailure modes.Appendix B corresponds to Chapter 3. It includes a detailed resolution studyof the numerical simulation of axisymmetric dambreak problem. Moreover, asymp-totic analysis in axisymmetric geometry is also illustrated here.Appendix C corresponds to Chapter 4. It consists of a study of the effect ofinertia, or the back wall of the surge. Also a detailed asymptotic analysis is given.Most of the material in the appendices is original work; results arising fromprevious work are identified by citations in each section.20Chapter 2Two dimensional viscoplasticdambreak1In the chapter, we present numerical computations of viscoplastic dambreaksspanning a wide range of physical parameters. Our aim is to describe morefully the phenomenology of the collapse and its plugs, the form of the motionat initiation, and the detailed final shape. Our main interest is in the effect of theyield stress, so we consider a Bingham fluid, ignoring any rate-dependence of theplastic viscosity. We mathematically formulate the dambreak problem in section2.1 and outline the numerical strategies we use for its solution. We use both anaugmented-Lagrangian scheme and regularization of the constitutive law to ac-count for viscoplasticity; to deal with the free surface, we use the volume-of-fluidmethod. The latter method emplaces the viscoplastic fluid beneath a less denseand viscous fluid, then tracks the interface between the two using a concentrationfield. This effectively replaces the single-phase dambreak problem with that of atwo-phase miscible fluid displacement (we ignore surface tension), but introducesa significant complication when imposing a no-slip boundary condition: because1A version of chapter 2 has been published. [Liu, Y.], Balmforth, N.J. & Hormozi, S. & HewittD.R. (2016) Two-dimensional viscoplastic dambreaks. J. Non-Newtonian Fluid Mech. 238, 65-79. .21the lighter fluid cannot be displaced from the lower surface, the slumping heav-ier fluid over-rides a shallow finger of lighter fluid which lubricates the overlyingflow and thins continually, leading to difficulties with resolution. We expose thiscomplication for a viscous test case in section 2.2, and identify means to avoidit. We then move on to a discussion of Bingham dambreaks in section 2.3, beforeconcluding in section 2.4. The appendices contain additional technical details ofthe numerical schemes, asymptotic theories for shallow or slender flow, and somerelated plasticity solutions.2.1 Formulation2.1.1 Dambreak arrangement and solution strategyTo simulate the collapse of a Bingham fluid, we consider a two-fluid arrangement,with the yield-stress fluid emplaced underneath a lighter viscous fluid. We ig-nore any interfacial tension. The volume-of-fluid method is used to deal with theboundary between the two fluids: a concentration field c(x,y, t) smooths out andtracks the fluid-fluid interface; c = 1 represents the viscoplastic fluid and the over-lying Newtonian fluid has c = 0. The concentration field satisfies the advectionequation for a passive scalar; no explicit diffusion is included although some isunavoidable as a result of numerical imprecision. Figure 2.1 shows a sketch ofthe geometry; the initial block of viscoplastic fluid has a characteristic height Hand basal width 2L , but we assume that the flow remains symmetrical about theblock’s midline and consider only half of the spatial domain.To deal with the yield stress of the viscoplastic fluid, we use both an augmented-Lagrangian scheme and a regularization of the Bingham model. The numer-ical algorithm is implemented in C++ as an application of PELICANS2. We referthe reader to [44, 82] for a more detailed description of the numerical method and2 https://gforge.irsn.fr/gf/project/pelicans/; PELICANS is an object-oriented platform devel-oped at the French Institute for Radiological Protection and Nuclear Safety and is distributedunder the CeCILL license agreement (http://www.cecill.info/).22gxzLxLzLHρ1, µ1τYFluid 1c=1ρ2, µ2Fluid 2c=0u=w=0u=w=0u=0wx=0u=0wx=0Figure 2.1: A sketch of the geometry for the case of a rectangular initialblock.its implementation. We use the regularized scheme as the main computationaltool; the augmented-Lagragian algorithm is slower and was used more sparingly,specifically when looking at flow close to failure or during the final approach torest. In most situations, the agreement between the two computations is satisfac-tory (examples are given below in figure 2.4); only at the initiation or cessationof motion is there a noticeable difference, primarily in the stress field (discount-ing the solution for the plug, which is an artifact of the iteration algorithm in theaugmented-Lagrangian scheme).2.1.2 Model equationsWe quote conservation of mass, concentration and momentum for a two-dimensionalincompressible fluid in dimensionless form:∇ · u = 0, ∂c∂ t+(u ·∇)c = 0, (2.1)23ρRe[∂u∂ t+(u ·∇)u]=−∇p+∇ · τ −ρ zˆ, (2.2)In these equations, lengths x = (x,z) are scaled by the characteristic initial heightof the Bingham fluid,H , velocities u =(u,w) by the speed scaleU = ρ1gH2/µ1,and time t by H /U , where g is the gravitational acceleration; the stresses τand pressure p are scaled by ρ1gH . The Reynolds number is defined as Re =ρ1U H /µ1. Here, the subscript 1 or 2 on the (plastic) viscosity µ and density ρdistinguishes the two fluids, and linear interpolation with the concentration fieldc is used to reconstruct those quantities for the mixture; i.e. after scaling with thedenser fluid properties,ρ = c+(1− c)ρ2ρ1and µ = c+(1− c)µ2µ1. (2.3)In dimensionless form, the unregularized Bingham constitutive law isγ˙ jk = 0, τ < cB,τ jk =(µ +cBγ˙)γ˙ jk, τ > cB,τ =√12∑j,kτ2jk (2.4)whereB =τY Hµ1U≡ τYρ1gH(2.5)is a dimensionless parameter related to the yield stress τY , and the deformationrates are given byγ˙ jk =∂u j∂xk+∂uk∂x j, γ˙ =√12∑j,kγ˙2jk. (2.6)The regularized version that we employ isτ jk =(µ +cBγ˙ + ε)γ˙ jk, (2.7)24where ε is a small regularization parameter. We verified that the size of thisparameter had no discernible effect on the results presented below, as long as it issmall enough; we therefore consider irrelevant the precise form of the regulariza-tion (which is simple, but not necessarily optimal).We solve these equations over the domain 0 ≤ x ≤ ℓx = Lx/H and 0 ≤ z ≤ℓz = Lz/H , and subject to no-slip conditions, u = w = 0 on the top and bottomsurfaces (but see §2.2), and symmetry conditions on the left and right edges, u = 0and wx = 0. The computational domain is chosen sufficiently larger than the initialshape of Bingham fluid that the precise locations of the upper and right-handboundaries (i.e. ℓx and ℓz) exert little effect on the flow dynamics.Initially, both fluids are motionless, u(x,z,0) = w(x,z,0) = 0. We take the ini-tial shape of the viscoplastic fluid to be either a block or triangle; the concentrationfield then begins withc(x,z,0) = 1 for{0≤ x≤ X0 or X0(1− 12z),0≤ z≤ 1 or 2,and c(x,z,0) = 0 elsewhere, where X0 = L /H is the initial aspect ratio. Thedifferent maximum depths ensure that the initial conditions have the same areafor equal basal width X0.The main dimensionless parameters that we vary are the yield-stress parameterB (which we loosely refer to as a Bingham number) and initial aspect ratio X0.Unless otherwise stated, we set the other parameters to beℓx = 5, ℓz = 1.25,ρ2ρ1=µ2µ1= Re = 10−3.By fixing the density and viscosity ratios to be small, we attempted to minimizethe effect of the overlying viscous fluid (but see the discussion of the finger ofover-ridden fluid below). The relatively low Reynolds number reflects our inter-est in the limit of small inertia, although the PELICANS implementation requiresRe 6= 0, evolving the fluid from the motionless state; we established that adopt-25ing Re = 10−3 minimizes the effect of inertia beyond the initial transient. Someadditional technical details of the computations are summarized in A.1. In thisappendix, we also describe a second scheme that we used to study how the initialstate fails at t = 0; this scheme does not solve the initial-value problem, but calcu-lates the instantaneous velocity field at t = 0, assuming that Re = 0 and the initialstresses are in balance (the steady Stokes problem).2.2 Newtonian benchmarkThe collapse of the initial block of the heavier viscous fluid creates a slumpingcurrent that flows out above the bottom surface. However, because of the no-slip condition imposed there, the upper-layer fluid cannot be displaced from athin finger coating the base that is over-ridden by the advancing gravity current.Problematically, the finger becomes excessively thin and difficult to resolve withthe relatively small viscosity and density ratios that we used to minimize the effectof the lighter fluid. We illustrate the formation of the finger and its subsequentdevelopment in figure 2.2. A.1 features further discussion of the finger and itsevolution.The challenges associated with resolving the finger are illustrated in figure2.3, which plots the evolution in time of the flow front, X(t) (defined as the right-most position where c(X ,z, t) = 12), for computations with different grid spacing.The first panel in this figure shows the results using a relatively simple MUSCLscheme for tracking the interface , which was previously coded into PELI-CANS [44, 82]. This algorithm fails to track the interface well with the grid reso-lutions used: as the finger develops, it remains erroneously thick and the enhancedlubrication by the lighter viscous fluid causes the heavier current to advance tooquickly (cf. A.1.3).An interface-tracking scheme based on the PLIC algorithm proposed in performs better; see figure 2.3(b). The flow front now advances less quickly.However, the fine scale of the finger still leads to a relatively slow convergenceof the computations with grid spacing ∆x = ∆z (corresponding to finite elements26xz0 0.5 1 1.500.20.40.60.811 1.2 1.400.010.020.030.04Figure 2.2: Snapshots of the evolving interface for a Newtonian heavier fluidat the times t = 1,2,3,4,5; the inset shows a magnification of the fingerof over-ridden lighter fluid.in the PELICANS code with equal aspect ratio). Moreover, the resolution failureis again manifest as an enhancement in the runout of the current that results froma finger that does not thin sufficiently quickly. In A.1.3 we argue that this is anintrinsic feature of the volume-of-fluid algorithm when the interface is containedwithin the lowest grid cell of the numerical scheme.The resolution problems with the finger are compounded in computations withBingham fluid, for which the yield stress further decreases the effective viscosityratio. Although some sort of local mesh refinement and adaptation would be anatural way to help counter these problems, we elected to avoid them in a differ-ent fashion which was more easily incorporated into PELICANS. In particular, bymonitoring c(x,z, t) at z = ∆z for each time step, we determined when the fingerwas expected to be contained within the lowest grid cell. At this stage, to pre-270 50 100 150 200 25011.522.533.54tX(a) MUSCL 0 50 100 150 200 25011.522.533.54tX(b) PLIC 0 50 100 150 200 25011.522.533.54tX(c) PLIC with correction dz=1/40dz=1/80dz=1/160dz=1/320slipasympt.dz=1/40dz=1/80dz=1/160dz=1/320asympt.xz (d)0 0.5 1 1.5 2 2.5 3 3.500.2 PLIC PLIC with correctionPLIC with slipFigure 2.3: Flow front X(t) plotted against time for computations with New-tonian fluids using (a) the simple MUSCL scheme, (b) the PLIC im-provement, and (c) the PLIC scheme with the lower boundary con-dition on c(x,z, t) adjusted according to the algorithm outlined in themain text. In each case, runs with different resolution are shown. For(c), the (red) dashed-dotted line labelled slip shows a solution com-puted with the PLIC scheme, but with no slip imposed on the heavierfluid and free slip imposed on the lighter fluid at z = 0. The circlesshow the prediction of the leading-order, shallow-layer asymptoticsin A.2. In (d), we plot the interface shape at t = 250 for the highestresolution solutions computed with the PLIC scheme with the threedifferent lower boundary conditions.vent the resolution failure from artificially restricting the thinning of the finger inthe volume-of-fluid scheme (see A.1.3), we reset the concentration field to c = 1at z = 0. The finger was thereby truncated and the effective contact line moved.Practically, we reset c(x,0, t) when c(x,∆z, t) exceeded 0.99 (the results were in-sensitive to the exact choice for this value). As shown by figure 2.3(c), this led tocomputations that converged much more quickly with grid spacing and fell closeto both the most highly resolved computations with the original PLIC scheme and28the predictions of shallow-layer theory. Nevertheless, the adjustment destroys theability of the code to preserve the volumes of the two fluids. For the computa-tions we report here, less than about one percent of the volume of the lighter fluidwas lost as a result of the adjustment. But, as the velocity profile was then fullyresolved near the boundary and no other unexpected problems were found, weconsidered this flaw to be acceptable. Hereon, all reported computations use thisadjusted boundary condition.To provide a physical basis for the adjustment scheme, we would need todemonstrate that it corresponds to the addition of another physical effect, suchas van der Waals interactions. We did not do this here, but simply note that theadjustment acts like the numerical devices implemented in contact line problemswith surface tension to alleviate the stress singularity and allow the contact line tomove . Indeed, the scheme is much like limiting the dynamic contact angle tobe about 3pi/4 or less, by adjusting the interface over the scale of the bottom gridcell but without introducing explicitly any interfacial tension.An alternative strategy is to change the lower boundary condition so that thelighter fluid freely slips over the lower surface whilst the heavier fluid still satisfiesno slip. This strategy, which can be incorporated using a Navier-type slip law inwhich the slip length depends on c, leads to results that compare well with thescheme including the concentration correction (see figure 2.3(c)). However, forBingham fluid, the diffuse nature of the interface-tracking scheme and the PLICalgorithm eventually result in the recurrence of resolution problems over longertimes. By contrast, the adjustment scheme successfully survives the long timediffusion process.2.3 Bingham slumpsFor the collapse of a Bingham fluid, we first describe the general phenomenology,then explore the details of failure, and finally categorize the slumped end-states.Along the way, we indicate how the computations approach the asymptotic limitsof relatively shallow (low, squat) or slender (tall, thin) slumps.292.3.1 Slump and plug phenomenologyWhen the heavier fluid is viscoplastic, collapse only occurs provided the yieldstress does not exceed a critical value Bc that depends on initial geometry. ForB < Bc, the viscoplastic fluid collapses, but the yield stress eventually brings theflow to an almost complete halt (slumps with the regularized constitutive law nevertruly come to rest, and iteration errors in the augmented-Lagrangian scheme pre-vent the velocity field from vanishing identically). Figure 2.4 plots the position ofthe flow front X(t) and central depth H(t) = h(0, t) for computations with varyingB, beginning from a square (X0 = 1) initial block. For this shape, the critical valuebelow which collapse occurs is Bc ≈ 0.265, and unlike the inexorable advance ofthe Newtonian current (shown by a dashed line), X(t) and H(t) eventually con-verge to B−dependent constants (the case with the smallest B = 0.01 requires alonger computational time than is shown).Sample collapses from square blocks with B = 0.05 and 0.14 are illustratedin more detail in figure 2.5. The first example shows a slump with relatively lowyield stress, for which the fluid collapses into a shallow current. The case withhigher B collapses less far, with an obvious imprint left by the initial shape. Notethe stress concentration that arises for earlier times in the vicinity of the contactline (a feature of all the slumps, irrespective of initial condition and rheology).In general, for rectangular initial blocks with order one initial aspect ratio,we find that the flow features two different plug regions during the initial stagesof collapse (cf. ). First, at the centre of the fluid the stresses never becomesufficient to yield the fluid, and a rigid core persists throughout the evolution.Second, the top outer corner is not sufficiently stressed to move at the initiationof motion. This feature falls and rotates rigidly as fluid collapses underneath, buteventually liquifies and disappears for small B; at higher yield stress, the rigidcorner survives the fall and decorates the final deposit. As the fluid approachesits final shape and flow subsides, further plugs appear, particularly near the flowfront; for the deeper final deposits, these plugs appear to thicken and merge toleave relatively thin yielded zones.300 200 400 600 800 100011.522.533.544.5 B=0.1B=0.06B=0.03B=0.01X(a) X(t)B=0.20 200 400 600 800 10000.30.40.50.60.70.80.91tHB=0.1B=0.01(b) H(t)B=0.2Figure 2.4: (a) Front position X(t) and (b) central depth H(t) = h(0, t) forBingham dambreaks with a square initial block (X0= 1) and the valuesof B indicated. The dashed lines show the Newtonian results. For theviscoplastic cases, the circles show the result using the augmented-Lagrangian scheme and the lines indicate the result with a regularizedconstitutive law. The dotted lines show the prediction of the leading-order shallow-layer asymptotics for B = 0.01.When the initial shape is a triangle with X0 = O(1), only a few of the phe-nomenological details change (figure 2.6): the apex of the triangle now fallsrigidly as material spreads out underneath; this pinnacle decorates the final de-posit unless the yield stress is sufficiently small. Again, the slump features a rigid31Figure 2.5: Snapshots of a collapsing square with (a) B = 0.05 at t = 0, 2.5,5, 10, 20, 40, 70, 150, 450, 950, 4000, and (b) B = 0.14 at t = 0,10, 50, 100, 200, ..., 1000. The insets show density plots of the stressinvariant τ at the times indicated, with the yield surfaces drawn as solidlines (and common shading maps for the final two snapshots in eachcase).core and further plugs form near the nose over late times.With a relatively wide initial condition, the pattern of evolution is slightly32Figure 2.6: Snapshots of the evolving interface for a triangular initial condi-tion with X0 = 1 and (a) B = 0.05 and (b) B = 0.14, at t = 0, 2.5, 5, 10,20, 40, 100, 150, 450, 950, 4000. The insets show density plots of thestress invariant τ at the times indicated, with the yield surfaces drawnas solid lines (the two later time density plots have a common shadingscheme).33different: for the rectangle, the central rigid plug extends over the entire depthof the fluid layer and only the side of the block collapses. The final deposit thenfeatures a flat top at its centre, as illustrated in figure 2.7(a). Such incompleteslumps are predicted by shallow-layer theory to occur for 3BX0 > 1 . Thisestimate can be improved to 3BX0 > 1− 3piB/4 using the higher-order theoryoutlined in A.2 (and specifically the final-shape formula (2.9)), which adequatelyreproduces numerical results for B < 0.15; at higher B, the computations indicatethat BX0 must exceed 0.25±0.015 for the slump to preserve a rigid central blockIncomplete slumps of a different kind arise for an initial triangle. In this case,collapses begin over the central regions where the initial stresses are largest, andmay not reach the edge, where fluid is initially unyielded, if the triangle is toowide. The (leading-order) shallow-layer model predicts that collapse occurs butdoes not reach the fluid edge at x = X0 if 4> BX0 > 9/8. As illustrated in figure2.7(b), such incomplete slumps are also observed numerically, though again for aslightly different range of initial widths (the plugged toe of the triangle is relativelysmall in the example shown).Slender (i.e. tall, thin) initial blocks, with X0 ≪ 1, also collapse somewhatdifferently, with fluid yielding only at the base of the fluid and remaining rigid inan overlying solid cap; see figure 2.8, which shows computations for rectangles(slender triangular slumps are much the same). The lower section of the columnsubsequently spreads outwards with the rigid cap descending above it in a mannerreminiscent of the (axisymmetrical) slump test [61, 68]. Interestingly, the slenderslump also generates undulations in the thickness of the column. As illustrated byfigure 2.8(a), these undulations (which do not occur in the Newtonian problem)appear towards the end of the collapse and are linked to zigzag patterns in thestress invariant and yield surfaces. The features follow characteristic curves ofthe stress field (the “sliplines”) that intersect the side free surface, and which haveslopes close to forty-five degrees (see A.4.1); the wavelength of the pattern istherefore approximately the width of the column.34Figure 2.7: Snapshots of the evolving interface for an initial (a) rectangularwith (X0,B) = (3,0.11), and (b) triangle with (X0,B) = (8,0.14), atthe times t = 0, 10, 20, 30, 50, 100, 400, 700, 1000. The dotted line in(a) shows a modification of the prediction in (2.9) that incorporates thecentral plug (and which terminates at finite height). The insets showdensity plots of the stress invariant τ at the times indicated, with theyield surfaces drawn as solid lines.2.3.2 Shallow flowAs illustrated above, when B ≪ 1 the fluid collapses into a shallow current with|∂h/∂x| ≪ 1 whatever the initial condition. Asymptotic theory [11, 52] then pro-vides a complementary approach to the problem. As illustrated in figure 2.4, the35Figure 2.8: Slumps of slender columns: the four images on the left showcollapsing columns for B = 0.3 and X0 = 0.025 at the times indicatedin the top right corner. Shown are the interface, yield surfaces andcolormap of the stress invariant τ . The evolving side profile for thiscollapse is shown in panel (a) at the times t = 0, 4, 9, 16, 25, 36, 49,100, 400, 1000. Panels (b)–(e) show columns at t = 50 for the valuesof B indicated, all with X0 = 0.025. Panel (f) compares the final sideprofiles (blue) with the predictions of slender asymptotics (red). Theshading scheme for τ for all the colormaps is shown in (e).sample collapse with B = 0.01 is relatively shallow and the numerical solutionsfor the runout and central depth match the predictions of the shallow-layer asymp-totics.The asymptotics also predict that flow becomes plug-like over a region under-neath the interface (see and A.2). This “pseudo-plug” is not truly rigid but isweakly yielded and rides above a more strongly sheared lower layer. Horizontalvelocity profiles for a collapsing square block with B = 0.01 are shown in figure362.9(a) and compared with the predictions of the shallow-layer theory. Althoughthe pseudo-plugs are less obvious in the numerical profiles, the horizontal veloc-ity does become relatively uniform over the predicted pseudo-plug. Figure 2.9(b)illustrates how the superficial weak deformation rates associated with the pseudo-plug feature in a snapshot of log10 γ˙ , and how the transition to a more obviouslysheared layer underneath approximately follows asymptotic predictions.371 2 3 4 5 6x 10−300.050.10.150.20.25uzt=100200600 300(a) Profiles of uxh(c) Final profileB=0.01B=0.050 1 2 3 4 500.10.20.30.40.50.6Figure 2.9: Comparison of shallow-layer theory with numerical results for acollapsing square block (X0 = 1) with B = 0.01: (a) Horizontal veloc-ity profiles at x = 2.5 and the times indicated. (b) Logarithmic strain-rate invariant, A density map of log10 γ˙ on the (x,z)−plane, at t = 600.(c) Final shape. In (a) the crosses plot the numerical solution, and thepoints indicate the asymptotic profile (A.8); the star locates the bot-tom of the pseudo-plug. In (b), the solid (green) and dashed (blue)lines show the surface and fake yield surface (z = Y = h+B/hx) pre-dicted by asymptotics. In (c), the final profile for B = 0.05 is included.The solid lines show computed final states, the dotted lines denote theleading-order result (2.8) and the dashed line shows the higher-orderprediction (2.9). The dashed-dotted line is the asymptotic prediction(2.13) from .38Nevertheless, the numerical computations show notable disagreements withthe shallow-layer asymptotics. None of the true plugs appear in the asymptoticsolution, reflecting how they are associated with non-shallow flow dynamics: atthe midline, the asymptotics fail to incorporate properly the symmetry conditions,and at the flow front and the relic of the upper right corner, the surface alwaysremains too steep for a shallow approximation. The depth profiles predicted bythe asymptotics consequently do not show any of the finer secondary featuresimprinted by the true plugs, as illustrated by the profile for B = 0.05 also plottedin figure 2.9(c). Only when B is smaller are such features largely eliminated bythe fluid flow and the final shape well predicted by the asymptotics.Despite these details, the broad features of the numerical solutions are repro-duced by the shallow-layer asymptotics, particularly when first-order-correctionterms are included: for the final shape and if the initial block collapses completely,the leading-order solution ish(x) =√2B(X∞− x); (2.8)with the next-order corrections, we findh =√2B(X∞− x)+ pi2B (2.9)(A.2). The final runout X∞, or slump length, is dictated by matching the profile’sarea with that of the initial condition. This implies X∞ = (9X20/8B)1/3 for (2.8)and furnishes an algebraic problem to solve in the case of (2.9), with approximatesolution X∞ ≈ (9X20/8B)1/3[1− pi(B2/81X0)1/3]. As shown in figure 2.9(c), theprediction (2.9) agrees well with the shapes reached at the end of the numericalcomputations, even though the profile ends in a vertical cliff which violates theshallow-layer asymptotics.392.3.3 Slender slumpsFor a slender column with X0≪ 1, we can again make use of the small aspect ratioto construct an asymptotic solution. As summarized in A.3, this limit correspondsto theory of slender viscoplastic filaments and indicates that the final state isgiven byx = ξ (z) =X02Bexp(− z2B)for 0≤ z≤ Z, (2.10)whereZ =−2Bh(0,0) log(2B) (2.11)is the height dividing yielded fluid from the overlying rigid plug. The fluid adoptsits original shape over Z < z < Z + 2Bh(0,0), having fallen a vertical distance(1−2B)h(0,0)−Z. It follows that the column will not slump if B > 12,X∞ =X02Band H∞ = 2Bh(0,0)[1− log(2B)]. (2.12)The latter is equivalent to the “dimensionless slump” reported previously [61, 68],although it was not declared as an asymptotic result relying on the column beingslender. The profile (2.10) is compared with the final profiles of numerical com-putations in figure 2.8. Aside from the relatively short-wavelength undulationsin column thickness over the yielded base of the fluid (whose lengthscale vio-lates the slender approximation), the asymptotics are in broad agreement with thenumerical results.Note that overly slender configurations are likely susceptible to a symmetry-breaking instability in which the column topples over to one side . This is ruledout here in view of our imposition of symmetry conditions along the centrelinex = 0.402.3.4 FailureCritical yield stressThe critical yield stress, Bc, above which the fluid does not collapse is plottedagainst initial width, X0, in figure 2.10(a) for rectangular initial blocks. We cal-culate Bc in two ways: first, the final runout X∞ recorded in the slumps computedwith the PELICANS software (defined as in §2.3.5) converges linearly to the ini-tial width X0 as B → Bc. Second, in the inertialess problem, the initial stressesdictate the initial velocity field, and the maximum speed also falls linearly to zeroas B → Bc. Hence, we can determine Bc without performing any time steppingusing the scheme for Re = 0 described in A.1.2.410 0.2 0.4 0.6 0.8 1 1.20.250.30.350.40.450.5B c (a) Rectangles0 1 2 3 40.250.30.350.40.450.5X0B c(b) Triangles Figure 2.10: Critical yield stress, Bc, plotted against initial width X0 for (a)rectangular and (b) triangular blocks, found by from monitoring ei-ther the final runout X∞ (stars) or the initial maximum speed forRe = 0 (solid line). In (a), also shown are the bounds Bc = 0.2642and 0.2651 obtained from plastic limit-point analysis [55, 56, 62](dashed-dotted), results from slip-line theory (dashed line withpoints) and the simple lower bound Bc =12(√X20 +1−X0) (dot-ted line). The analogues of the latter two for the triangular blocks (seeA.4) are shown in (b). The dotted lines with open circles show im-provements of the simple lower bounds allowing for rotational failure(see A.4.2).42As illustrated in figure 2.10(a), Bc ≈ 0.2646 independently of X0 when X0 > 1.For such initial widths, collapses are incomplete and a solid core spans the fulldepth of the fluid, rendering the failure criterion independent of X0. The initialwidth matters for X0 < 1, leading to an increase of Bc. Eventually, Bc → 12 forX0→ 0, as expected from the slender column asymptotics in §2.3.3.Just below the critical yield stress, velocities are small and the Bingham prob-lem reduces to an analogous one in plasticity theory except over thin viscoplasticboundary layers. The incomplete slump is analogous to the classical geotechnicalproblem of the stability of a vertical embankment (e.g. ), provided no defor-mation occurs in z < 0. Classical arguments dating back to Coulomb, describe theform of failure in terms of the appearance of a slip surface dividing rigid blocks,allowing analytical estimations of Bc from balancing the plastic dissipation acrossthe slip surface with the potential energy release. In particular, assuming that thefailure occurs by the rotation of the top right corner above a circular arc, one ar-rives at an estimate Bc ≈ 0.261, after maximizing over all possible positions ofthe centre of rotation (cf. and figure 2.11(a)). However, this type of solu-tion strictly provides only a lower bound on Bc and may not be the actual modeof failure. Indeed, the bound has been optimized and improved to 0.2642, and acomplementary upper bound computed to be 0.2651 [55, 56, 62]; the optimizationsuggests that failure occurs over a relatively wide region of plastic deformation. The upper and lower bounds are included in figure 2.10(a), and are indistin-guishable on the scale of this picture, but bracket the value of Bc ≈ 0.2646 foundfor our Bingham slumps with X0 > 1.43−2 −1 0 100.511.52xz(a)−0.5 0 0.5 100.511.52zx(b)sα−0.5 0 0.5 1x(c)ssβFigure 2.11: Trial velocity fields to compute lower bounds on Bc for (a) thevertical embankment with a circular slip curve, and relatively slender(b) rectangular and (c) triangular initial blocks. In (b) and (c), thestraight (dashed) and circular (solid) failure surfaces for linear or ro-tational sliding motion are plotted; these surface can be parametrizedby the local slopes at the bottom corner, sα , and midline, sβ and s (forlinear sliding sα = sβ ).For a slender column, Chamberlain et al. provide an estimate of thecritical yield stress by assuming that two lines of failure occur: the lowest cutsoff a triangular basal section, whereas the upper cleaves off a second triangle thatslides away sideways, leaving the remaining overlying trapezoid to fall vertically;44see figure 2.11(b). Optimizing the slopes of the two cuts furnishes the lowerbound, Bc =12(√1+X20 −X0), which is included in figure 2.10. Chamberlain etal. also provide a numerical solution of the slipline field for a failure withthe form of an unconfined plastic deformation. This estimate converges towardstheir lower bound as X0 → 0, as indicated in figure 2.10(a). Our numericallydetermined values of Bc match Chamberlain et al.’s slipline solutions for X0 <0.5. For wider initial blocks, the slipline solutions deviate from the numericalresults and become inconsistent with the bounds for the vertical embankment forX0 > 0.8, highlighting how a different failure mechanism must operate.For triangles, no corresponding plasticity solutions exist in the literature. How-ever, Chamberlain et al.’s [20, 21] slipline solution and simple lower bound caneasily be generalized, as outlined in A.4 and illustrated in figure 2.11(c). Theslipline solution and bound are compared with numerical data in figure 2.10(b).Again, Bc → 12 for X0 → 0. Now, however, the slope of the initial free surfacecontinues to decline as X0 is increased, and so there is no convergence to a limitthat is independent of width. Instead, the shallow-layer theory of A.2 becomesrelevant and predicts that Bc → 4/X0 for X0 ≫ 1 (a limit lying well beyond thenumerical data in figure 2.10(b)).Note that Chamberlain et al.’s lower bounds can be improved by allowing thetriangle at the side to rotate out of position rather than slide linearly; see A.4.2.The failure surfaces then become circular arcs, as illustrated in figure 2.11(b,c).For the rectangle, the resulting improvement in the bound on Bc amounts to afew percent and is barely noticeable in figure 2.10(a). More significant is theimprovement of the bound for the triangle, which is brought much closer to thenumerical and plasticity results; see figure 2.10(b).Flow at failureThe failure modes of our rectangular viscoplastic solutions (for t = Re = 0) areillustrated in figure 2.12. The thinner initial columns yield only over a triangularshaped region at the base which closely matches that predicted by slipline theory45Figure 2.12: Strain-rate invariant plotted logarithmically as a density on the(x,z)−plane for solutions with B = 0.99BC and t = Re = 0 (A.1.2),at the values of X0 = 0.2,0.5,0.6,0.9,1,1.4 indicated by the x−axis.Also shown are a selection of streamlines. In (a)–(c), the solid bluelines indicate the border of the plastic region and expansion fan of thecorresponding slipline solutions (A.4). In (e) and (f), the blue linesindicate the circular failure surface of the lower bound solution. (see also A.4.1). The failure mode changes abruptly when X0 slightly exceedsabout 0.5. The failure zone then takes the form of a widening wedge extendingfrom the lower right corner of the initial block up to the centre of the top, with theentire side face remaining rigid. Evidently, this mode of failure is preferred overthe slipline solution at these values of X0, leading to the departure of the observedvalues of Bc from Chamberlain et al.’s curve in figure 2.10(a). The failure modechanges a second time for X0≈ 1: wider initial blocks fail chiefly over a relatively46Figure 2.13: A series of pictures similar to figure 2.12, but for initial trian-gles (with solid blue lines showing the yield surfaces of the sliplinesolution in (a), and the circular arcs of the bound of A.4.2 for rota-tional failure in (b) and (c)).narrow layer connecting the lower left corner to an off-centre location on the topsurface, which lies close to the circular failure surface of the simple lower boundin figure 2.11(a).For both the narrower and wider examples in figure 2.12, the failing deforma-tions are dominated by sharp viscoplastic boundary layers that likely correspondto yield surfaces. Spatially extensive regions (in comparison to the fluid depth orwidth) of plastic deformation do, nevertheless, occur, and the failure modes nevertake the form of a patchwork of sliding rigid blocks. Note that it is difficult tocleanly extract the yield surfaces as B→ Bc, which complicates the identificationof the failure mode. In figure 2.12, we have avoided showing these surfaces andinstead displayed the deformation rate and a selection of streamlines. Curiously,for X0 > 0.5, it is difficult to envision how one might construct corresponding sli-pline fields (there are no surfaces bounding the plastic zone with known stressesthat can be used to begin the slipline construction).47For a triangular initial condition, failure for smaller widths again occurs throughthe appearance of a lower plastic zone that compares well with slipline theory;see figure 2.13. This agreement is confirmed by the match of the observed criti-cal yield stress, Bc(X0), with the slipline predictions in figure 2.10(b). Unlike therectangle, however, there is no abrupt change in failure mode as X0 is increased,at least until the surface slope of the triangle becomes too shallow to apply Cham-berlain et al.’s construction (see A.4.1). At the largest widths, the trianglefails at the centre but not the edge, as already noted in section 2.3.1.2.3.5 The final shape and slump statisticsTo extract statistics of the final shape, we define a stopping criterion accordingto when the stress invariant first becomes equal or less than B at each point inthe domain. The resulting “final state” appears to be reached in a finite time(for both augmented-Lagrangian and regularized computations), in disagreementwith asymptotic theory , which predicts that flow halts only after an infinitetime (see also A.3). A selection of final profiles for varying Bingham number isdisplayed in figure 2.14 for both square and triangular initial conditions.Plasticity theory is again relevant in the limit that the slump approaches itsfinal state. This fact was used previously to construct the final profiles withslipline theory, following earlier work by Nye . A key assumption of thisconstruction is that the flow is under horizontal compression. The assumption canalso be used to continue the shallow-layer asymptotics to higher order to predictthe final profile,h =√2B(X∞− x)+ pi24B2− pi2B, (2.13)which agrees well with the slipline theory . Unfortunately, neither the sliplineprofiles or (2.13) compare well with laboratory experiments.48xzB ↑(a) Square0 0.5 1 1.5 2 2.5 3 3.5 4 4.500.20.40.60.81xz(b) TriangleB ↑0 0.5 1 1.5 2 2.5 3 3.5 4 4.500.20.40.60.811.21.41.61.82Figure 2.14: Profiles of the final deposit, starting from (a) a square blockand (b) a triangle, with X0 = 1, for B = 0.01, 0.02, 0.04, ..., 0.22 and0.24, together with the initial states.Figure 2.15: Final numerical solution for the slump of an initial square withB = 0.02, showing density maps of (a) pressure, (b) τxx and (c)τxz, and (d) the slipline field diagnosed from the numerical solution(solid) and built explicitly by integrating the slipline equations start-ing from the curve (2.9) (dotted). In (d), the plugs are shaded black,and no attempt has been made to match up the two sets of sliplines.49A sample final state from the numerical computations with a diagnosis of theassociated slipline field is displayed in figure 2.15. For the latter, we map outcurves of constant p− z±2Bθ , which are the invariants that are conserved alongthe two families of sliplines, where θ = −12tan−1(τxx/τxz) (see also A.4.1).As also shown by figure 2.15(d), the resulting curves compare well with an explicitcomputation of sliplines launched from the surface position predicted by (2.9),where p = 0 and the sliplines make an angle pi/4 with the local surface tangent.The sliplines in figure 2.15 follow a different pattern to Nye’s construction(see figures 5 and 6 in ). The reason for this discrepancy is that almost all ofour slumps comes to rest in a state of horizontal expansion, rather than compres-sion (we have observed regions under horizontal compression only in the slumpsof relatively wide triangles, as in figure 2.7(b)). Correcting this feature of the dy-namics leads to the revised higher-order asymptotics summarized in A.2, whichculminates in the prediction for the final profile in (2.9). As is evident from figure2.9(c), the asymptotic predictions for horizontal expansion compare much morefavourably with the numerical results than the slipline theory and asymptotics forhorizontal compression.The comparison is illustrated further in figure 2.16, which shows scaled finalrunouts and depths, X∞/√X0 and H∞/√X0, as functions of B/√X0 for the numer-ical computations, slipline theory and the various versions of the shallow-layerasymptotics. Scaling the runout and depth in this fashion corresponds to choosingthe square root of the initial area as the lengthscale in the non-dimensionalizationof the problem. The slipline theory and various versions of the shallow-layerasymptotics furnish curves of X∞/√X0 and H∞/√X0 against B/√X0 that are in-dependent of X0, implying an insensitivity to the initial condition. By contrast,the deeper final profiles of the numerical solutions with higher B, and their scaledfinal runout and depth, do depend on X0 and the initial shape. This dependence ishighlighted in figure 2.17, which compares data for initial triangles and rectangles.500.05 0.1 0.15 0.2 0.2511.522.533.544.55B /√X0X∞/√X0 (a)kaolin Carbopol 0.30 wt.%Carbopol 0.15 wt.%Joint Compound Cochard & Ancey0.05 0.1 0.15 0.2 0.250.20.40.60.811.2B /√X0H∞/√X0 (b)NumericsStaron et alSliplineH∞<<1H∞<<1, expansionH∞<<1, compressionX∞<<1 (Pashias et al)Figure 2.16: Scaled final (a) runout X∞/√X0 and (b) central depth H∞/√X0as a function of B/√X0 for Bingham slumps from square initial con-ditions (solid lines with dots). Also plotted using the symbols indi-cated are experimental data from and for slumps of aqueoussolutions of Carbopol, kaolin and “Joint Compound”. The leadingorder asymptotic prediction (2.8) is shown by the dotted line; thesolid lines with circles or squares plot the predictions in (2.9) and(2.13); the dashed lines shows the results of slipline theory . Theslender-column asymptotic prediction in (2.12) with X0 = 1 is shownby the dotted line and pentagrams. The dotted line and hexagramsshow the fit proposed by Staron et al. .510 0.1 0.2 0.3 0.411.522.533.544.5B/√X0X∞/√X0(a) 0 0.1 0.2 0.3 0.40.511.522.5B/√X0H∞/√X0(b)X0=1.5, blockX0=0.5, block X0=1, blockX0=1.5, triangleX0=0.5, triangleX0=1, triangleFigure 2.17: Scaled final (a) runout X∞/√X0 and (b) central depth H∞/√X0as a function of B/√X0 for Bingham slumps from rectangular andtriangular initial conditions with X0 = 0.5, 1 and 1.5. The solid linesshow the predictions of the higher-order asymptotics from (2.9) for acomplete slump.Figure 2.16 also includes data from laboratory experiments with Carbopol[26, 33] and some other fluids, which were conducted by Dubash et al. though notreported in their paper. None of these fluids are well fitted by the Bingham model,with a Herschel-Bulkley fit being superior. However, the final state is controlledby the yield stress and likely independent of the nonlinear viscosity of the mate-rial (at least provided inertia is not important), permitting a comparison betweenthe experiments and our Bingham computations. Although the numerical resultscompare more favourably with the experiments than the slipline theory, the com-parison with the leading-order asymptotic prediction is just as good. Thus, thediscrepancy between theory and experiment noted by is only partly due tothe assumption that the slump came to rest in a state of horizontal compression,but other factors must also be at work, such as stresses in the cross-stream di-rection and non-ideal material behaviour. Note that the range of dimensionless52yield stresses spanned by the experimental data lie well into the regime wherethere should be no significant dependence on the initial shape. This is comfort-ing in view of the fact that the experiments were conducted using different initialconditions, either by raising a vertical gate or tilting an inclined tank back to thehorizontal (which correspond roughly to our rectangular or triangular initial con-ditions).Finally, figure 2.16 includes the predictions of the slender column asymptoticsin (2.12) (see A.3) for X0 = 1, and a formula presented by Staron et al. basedon their volume-of-fluid computations with GERRIS and a regularized constitu-tive law. Given that the slumps from which the data in figure 2.16 are taken are notslender, it is not surprising that (2.12) compares poorly with the numerical results.We suspect that the disagreement between our results and the fit of Staron et al. originates either from an inadequate resolution of the over-ridden finger ofless dense fluid or the significance of inertial effects in their computations. Indeed,these authors quote a final shape that depends explicitly on the plastic viscosityof the heavier fluid, whereas this physical quantity is completely scaled out in ourcomputations when Re→ 0.2.4 Concluding remarksA yield stress introduces two key features in the collapse and spreading of a vis-coplastic fluid: failure occurs only provided the yield stress can be exceeded, and,when flow is initiated, the yield stress eventually brings motion to a halt. Here wehave provided numerical computations of the two-dimensional collapse of Bing-ham fluid, exploring the phenomenology of the flow for different initial shapes.We compared the results with asymptotic theory valid for relatively shallow (low,squat) or slender (tall, thin) slumps, and with solutions from plasticity theory ap-plying near the initiation and termination of flow. We verified that the compu-tations converge to the asymptotic solutions in the relevant limits and identifiedwhere the plasticity solutions apply. We studied both the initial form of failure,extracting criteria for a collapse to occur, and the shape of the final deposit, com-53paring its runout and depth with previous experiments and predictions.There are three key limitations of our study with regard to the collapses of vis-coplastic fluids in engineering or geophysical settings. First, our two-dimensionalgeometry is restrictive and an axisymmetric assumption prefereable for a range ofapplications such as the slump test. Such a generalization raises the interestingquestion of how incompressible viscoplastic flow avoids the inconsistency of thevon-Karman-Haar hypothesis (This will be further explained in Chapter 3).Second, the issues associated with the no-slip condition on the underlying sur-face are not merely numerical: viscoplastic fluids can suffer apparent slip ,demanding the inclusion of a slip law. Finally, inertia is important in many ap-plications, an effect that allows slumps to flow beyond the rest states we havecomputed. Other interesting generalizations include the incorporation of differentrheologies, such as thixotropy, and surface tension at small spatial scales.54Chapter 3Axisymmetric viscoplasticdambreaks and the slump test1One goal of this chapter is to provide a reliable solution of the benchmark problemproposed in , and to offer a more complete description of the slump behaviourover a wider range of physical conditions. For the task, we perform computationsbased on the VOF method to deal with the fluid interface and exploiting speciallydesigned codes to capture the yield-stress rheology. Our study follows on from thechapter 2 in which we considered two-dimensional dambreaks of viscoplas-tic fluid. We complement the computations with asymptotic theory for shallowgravity currents and slender vertical columns, and bounds from plasticity theoryto constrain the mode of failure for slumps near the critical yield stress whereatno collapse actually occurs.A second goal is to compare our theoretical modelling with experiments, col-lating some of the existing measurements from the literature [61, 70]. These ex-periments have not previously been performed sufficiently thoroughly to disentan-gle the effects of material rheology, the mechanism of release, and any interaction1A version of chapter 3 has been published. [Liu, Y.], Balmforth, N.J. & Hormozi, S. (2018)Axisymmetric viscoplastic dambreaks and the slump test. J. Non-Newtonian Fluid Mech. 258,45-57. .55with the underlying surface (effective slip). Therefore, we also perform our ownsuite of experiments using aqueous suspensions of Carbopol. This suspension isoften suggested to be well characterized by a Herschel-Bulkley rheology and po-tentially eliminates some of the confounding effects brought into experiments bynon-ideal material behaviour . We thereby provide a demanding test of thetheory whilst gauging the effects of the release mechanism and any effective slip.3.1 Formulation3.1.1 Problem set-up and solution strategyFigure 3.1: Sketch of the geometry.The geometry of the problem is shown in figure 3.1: we use an axisymmetriccylindrical polar coordinate system (r,z) to describe the sudden release of a cylin-der of incompressible Bingham fluid with radius Rˆ and height Hˆ. The fluid hasdensity ρ1, yield stress τY and plastic viscosity µ1 and is immersed in an ambientNewtonian fluid with density ρ2 and viscosity µ2. The density and viscosity ratiosare set at the small valuesρ2ρ1= µ2µ1 = 0.002 in order to minimize the effects ofthe ambient fluid (we have verified that the precise values of these ratios have nosignificant effect on the computations once one deals with the resolution issuesdescribed in §2.3). We use the VOF method to track the fluid interface, which56introduces an advected concentration field c(r,z, t) to distinguish the fluid phase:c = 1 represents the viscoplastic fluid, and c = 0 denotes the Newtonian ambient.Given c, bulk material parameters are computed using linear interpolation. Theinitial configuration isc(r,z,0) ={1 for 0≤ r ≤ Rˆ,0≤ z≤ Hˆ,0 elsewhere.The two fluids are miscible, eliminating interfacial tension.The computation domain extends to a height Lz and radius Lr, which are cho-sen to ensure that these boundaries remain remote and do not affect the evolutionof the slump. We impose no slip (u = 0,w = 0) on the bottom z = 0, regularityconditions on the axis r = 0 (i.e. u = ∂w/∂ r = 0), and no normal flow and freeslip conditions along r = Lr and z = Lz.We use two methods to deal with the yield-stress constitutive law: a regu-larization scheme that treats the unyielded region as a highly viscous fluid, andan augmented-Lagrangian scheme that explicitly treats the yield stress within aweak formulation of the problem [35, 80]. Both are implemented in C++ as anapplication of the PELICANS platform (e.g. ).3.1.2 Dimensionless model equationsWe scale lengths by the initial height Hˆ, velocities by the speed scaleU = ρ1gHˆ2/µ1,and time by Hˆ/U , where g is the gravitational acceleration; the stresses and pres-sure are scaled by ρ1gHˆ. The governing equations for the concentration field c,velocity u = (u,w), deviatoric stress tensor τ , and pressure p are then∇ · u = 0, ∂c∂ t+(u ·∇)c = 0,ρRe[∂u∂ t+(u ·∇)u]=−∇p+∇ · τ −ρ(01),(3.1)57whereρ = c+(1− c)ρ2ρ1and µ = c+(1− c)µ2µ1. (3.2)The unregularized Bingham constitutive law, used in the augmented-Lagrangianmethod, is γ˙ jk = 0, τ < cB,τ jk =(µ +cBγ˙)γ˙ jk, τ > cB,(3.3)whereas the regularization method uses the variant,τ jk =(µ +cBγ˙ + ε)γ˙ jk, (3.4)where τ =√12 ∑ j,k τ2jk and γ˙ =√12 ∑ j,k γ˙2jk denote second tensorial invariants, andthe deformation rates are given byγ˙rr γ˙rθ γ˙rzγ˙θ r γ˙θθ γ˙θ zγ˙zr γ˙zθ γ˙zz=2ur 0 uz +wr0 2u/r 0uz +wr 0 2wz , (3.5)with subscripts represent partial derivatives, except in the case of tensor compo-nents. The scalings introduce the dimensionless initial radius (or aspect ratio), andthe Reynolds and Bingham numbers,R =RˆHˆ, Re=ρ1UHˆµ1and B =τY Hˆµ1U. (3.6)The regularization parameter ε in (3.4) is taken to be 10−8, which was verified tobe sufficiently small that the modification of the constitutive law had an insignifi-cant effect on the results reported below (but see the comment at the end of §2.3).Given the concentration field, we define the instantaneous position of the surfaceof the slump to be given by c(r,z = h) = 12. For most of the computations, weselect parameters so that inertial effect is small, Re= 10−3, and vary B and R.583.1.3 PLIC scheme with interface correctionThe piecewise-linear-interface-construction (PLIC) [40, 66] is a contemporarystandard in the VOF method. The interface is represented by a line segment ineach grid cell, which is computed using the volume fraction c, as in . Then,given the velocity field, the line segment is advected to a new position, and c up-dated accordingly. In view of the boundary conditions, there is no flux of c into orout of the domain, which is incorporated into the scheme in the manner in whichthe solver advects c along the boundary. Importantly, the bottom boundary is noslip, which does not permit the contact line to move and introduces an awkwardresolution issue, as in the 2D problem in chapter 2. More specifically, as theviscoplastic fluid collapses and spreads out, a finger of ambient fluid adhering tothe bottom surface is over-ridden. For the relatively low density and viscosity ra-tios that we employ, this finger lubricates the slumping viscoplastic current andthins dramatically to introduce the resolution issue . A key problem is thatthe scheme fails to accurately evolve c when the interface is inside the lowest gridcell, leaving the finger artificially thick and lubricating.A common way of moving the contact line in problems with surface tensionand Newtonian fluid is to replace the boundary condition with another that permitsslip. However, numerical solutions may not converge with mesh refinement .Instead, in chapter 2 we suggested a correction scheme that eliminates thefinger in a different way, allowing the computations to remain well resolved overlong times. The main point is that, with no slip, a finger of ambient fluid must stillcoat the underlying surface. However, counter to the un-corrected VOF scheme,the finger actually becomes too thin to lubricate the slump and should instead beignored. Practically, the scheme implements this idea by removing all the ambientfluid from a grid cell adjacent to the base when c exceeds a threshold near unity(chosen to be 0.99). This procedure clips the interface when it invades the lowestgrid cells, thereby truncating the finger and rendering the computation convergentin grid spacing .Despite the success of the scheme for 2D dambreaks, the algorithm is not con-59servative, with the mass of viscoplastic fluid growing with time. This awkwardfeature does not impair computations in 2D, but it does become more problematicin axisymmetric geometry, for which the convergence of the solutions with meshrefinement is weakened. For the current computations we therefore modified thecorrection scheme so that it conserved mass. In particular, whenever a correctionto c was implemented, and some of the ambient fluid removed from the one of thelowest grid cells, the lost material was added back by uniformly redistributing itinto the grid cells containing the interface (where c = 12). Essentially, this redistri-bution incurs an error in the position of the interface that is of the order of a smallfraction of the grid spacing, but in such a way that mass is conserved. Thoughhard to justify from a physical perspective, the resulting conservative correctionscheme converges more satisfyingly with mesh refinement for slumps with New-tonian fluid than the original non-conservative scheme (see B.1), chiefly becausethe latter suffers a resolution-dependent mass loss.The conservative correction scheme also performs better for Bingham fluidusing either the regularized constitutive law or the augmented-Lagrangian solver.Moreover, both produce comparable results for the global properties of the slumps(see, for example, figure 3.2 below), and their interaction with the PLIC schemedoes not introduce any additional unexpected issues. In more detail, the regular-ization scheme has the drawback that the fluid can never truly come to rest andcan fail to correctly predict the positions of the yield surfaces . However, theregularization scheme is faster than the augmented-Lagrangian code. Therefore,we use regularization scheme for exploring global features of the slumps (definingthe flow to have come to rest when Max(|v|) < 10−5), while for the finer detailssuch as the yield surfaces, we compare both schemes.3.2 Bingham slumpsFigure 3.2 shows the slumps of cylinders of Bingham fluid for R = 1 and varyingyield stress B. Displayed is the central depth H (t) along with the radial positionof the flow front R(t) (defined as the largest radial extent of the interface, c = 12).600 100 200 300 400 500 600 700 800 900 10000.30.40.50.60.70.80.91(a)tH(t)0 200 400 600 800 100011.522.5tR(t)(b)Figure 3.2: Time series of flow height H (t) and front position R(t) forBingham slumps with R = 1 and B = 0.01, 0.03, 0.05, 0.08, 0.125, 0.2and 0.3. Blue curves are from regularization, and red circles from theaugmented-Lagrangian method.The computations suggest that flow is arrested in finite time and illustrate howcollapse only occurs when the yield stress is below a critical value Bc. A selectionof the final profiles is illustrated in figure 3.3(a).61(a)B ↑0 0.5 1 1.5 2 2.500.20.40.60.810 0.5 1 1.5 2 2.5 3 3.5 4 4.500.51 r(b)B=0.19B=0.1Figure 3.3: Profiles of the final deposit of cylindrical slumps with (a) R = 1,B= 0.01, 0.02, 0.04, 0.07, 0.125, 0.2 and 0.275, and (b) R= 4, B= 0.1and 0.19. The solid curves show numerical computations, the dashedcurves show the improved shallow-layer asymptotic result in (3.11)(for the lowest four values of B in (a)), and the dotted lines show theinitial cylinders.Further details of the phenomenology of a slump are shown in figure 3.4. Ini-tially, the stresses exceed the yield value throughout the cylinder except over asmall conical region at its core. Fluid subsequently slumps outwards, reducingthe stresses and allowing that plug to expand with time. Eventually, stresses de-cline towards B all the way to the flow front, bringing the fluid to rest. Notethat, there is only a single central plugged region; for equivalent two-dimensionaldambreaks plugs persist at the periphery of the initial configuration, leadingto sharp corners that decorate the final deposit. Rigid features of this sort cannotoccur in axisymmetric slumps because the fluid edge must expand in order to fall.However, analogous weakly yielded zones persist during the collapse, then plug62up to create sharp rings that disfigure some of the final shapes (see figure 3.3).Figure 3.4: Evolution of the interface for a slumping cylinder with (R,B) =(1,0.1). The main panel shows the interface at the times t = 0, 2, 4,6, 8, 20, 200 and 1000. The three insets show the stress invariant τ asa density on the (r,z)−plane for the times t = 2, 20 and 1000 (with acommon scale for the last two cases). The green curve shows the yieldsurface.As for 2D dambreaks , when the initial configuration is sufficiently wide,fluid only collapses near the edge, leaving an unyielded flat-topped central section.Such “incomplete slumps” arise when the central plug spans the entire fluid layerand are illustrated in figure 3.3(b).3.2.1 Shallow flowWhen flow is shallow and inertia is negligible, lubrication theory can be appliedto obtain analytical results . With our current scaling of the problem, thislimit is achieved when B ≪ 1. To account for the low aspect ratio, we rescalethe horizontal coordinate r = ε−1χ , deviatoric stress components τi j = ετˇi j and63Bingham number B = εBˇ using a small parameter ε ≪ 1, and then restate theforce balance equations for the viscoplastic layer:−pχ + εχ(χτˇrr)χ − εχτˇθθ + τˇrz,z = 0,−pz + ε2χ(χτˇrz)χ + ετˇzz,z = 1.(3.7)Neglecting the ambient fluid, the surface of the current can be located by theelevation z = h(χ , t), and is force free, demanding thatτˇrz+hχ(p− ετˇrr) = 0p− ετˇzz + ε2hχ τˇrz = 0}at z = h. (3.8)At leading order, we find that p ∼ h− z and τˇrz ∼ −hχ(h− z). The constitutivelaw and depth-integrated continuity equation can then be used to derive an evo-lution equation for h(χ , t) [11, 51]. However, the final profile arises when radialspreading speeds subside and τˇrz → Bˇ at the base of the fluid layer z = 0. Thenceτˇrz(χ ,0)∼−hhχ ∼ Bˇ, which givesh(r)∼√2B(r∞− r), (3.9)in terms of the original variables, where r∞ is the final radius.In chapter 2 a higher-order approximation for the final profile of a 2Dslump was developed by continuing the asymptotic solution to O(ε), assumingthat τˇ → Bˇ throughout the fluid layer. We follow suit here, although a signifi-cant complication emerges owing to the axisymmetric geometry. In particular, inaddition to the two force balance equations, we must also satisfyτˇrr + τˇθθ + τˇzz = 012(τˇ2rr + τˇ2θθ + τˇ2zz)+ τˇ2rz = Bˇ2,(3.10)leaving us one equation short for determining the full stress state (i.e. we have four64equations for the five unknowns p, τˇrr, τˇθθ , τˇrz and τˇzz). The origin of this indeter-minacy is the component τˇθθ , which does not arise in 2D and highlights how thestress field cannot, in general, be constructed independently of the velocity field.The situation is identical to classical plasticity theory where the so-called vonKarman-Haar hypothesis is often invoked to avoid this problem. The hypothesis,which states that τˇθθ must equal one of the principal stresses in the (r,z)−plane,implies that τˇ2θθ ≡ 13 Bˇ2 . However, τˇ2rz → Bˇ2 at z = 0 for the leading orderlubrication solution, indicating that τˇθθ must vanish at the base of the fluid layer.Therefore, the von Karman-Haar hypothesis contradicts the leading-order asymp-totic solution and cannot be invoked here.Instead, we add the approximation τˇrr ∼ τˇθθ , which is suggested by both thevelocity field of the leading-order asymptotic solution and the numerical compu-tations; see B.2. With this alternative hypothesis, the asymptotic analysis can becontinued to O(ε) in order to arrive at the higher-order asymptotic approximation,h(r)∼√2B(r∞− r)+√34piB (3.11)(again in terms of the original variables).The predictions in (3.9) and (3.11) are compared to a numerical simulation fora slump with B = 0.0074 and R = 0.2546 in figure 3.5. Figure 3.3 also comparesthe improved approximation (3.11) with computed final shapes over a wider rangeof B. Note that (3.11) predicts that the final profile ends in a vertical cliff, violatingthe shallow-layer asymptotics. Nevertheless, (3.11) provides a meaningful pre-diction along the flow body that furnishes a better approximation than the leadingorder result (3.9), even when the flow is not particularly shallow (B > 0.04).In the example of figure 3.5, the fluid yields significantly almost everywhere,removing any sign of the initial shape in the final profile. Indeed, computationsthat begin with the same amount of fluid but conical initial shape also lead tosimilar final profiles. Figure 3.5 includes a computation using an initial cone witha top radius of Rtop = 1/6 and bottom radius of Rbase = 1/3, which corresponds65rz−1 −0.500.050.1rˆ (m)zˆ(m)0 0.1 0.2 0.300.010.020.030.04Figure 3.5: Final profile for a cylindrical dambreak with B = 0.0074 andR = 0.2546. The thin black curve shows the numerical simulation.On the left, the data is plotted using dimensionless variables, and theleading-order and improved asymptotic predictions in (3.9) and(3.11)are included as the dotted (green) line and (red) points, respectively.On the right, the data is replotted in dimensional variables assumingHˆ = 0.3m. Also shown are three other final profiles: one from a sim-ulation beginning with a cone with the same volume, with a top radiusof Rtop = 1/6 and bottom radius of Rbase = 1/3 (the ASTM standardgeometry; thick light grey line); a second from a simulation with theregularized Bingham model in which the regularization parameter isincreased to 10−4 (dashed line); and a third from a simulation thatdoes not apply the interface correction scheme to remove the underly-ing finger of ambient fluid (thin red contour).to the ASTM standard geometry; the final state cannot be distinguished from thatof the cylindrical dambreak in the plot. Thus, the example shown in this figurecorresponds to the benchmark problem proposed in . Indeed, on the right-hand side of the figure, the profiles are replotted in dimensional variables usingHˆ = 30cm; the results can then be directly compared with figure 5 in .The agreement amongst the computations in is relatively poor, a discrep-ancy that may arise from different treatments of the contact line and the fluidrheology. By contrast, the current computations align satisfyingly with (3.11),have converged with respect to mesh refinement, and are independent of the nu-merical algorithm (augmented-Lagrangian or regularization). Our computationindicates that the final (dimensional) radius is 28cm, in comparison to the averageof 29.5cm quoted in . Figure 3.5 also tells the cautionary tale of the resultswhen we fail to resolve the finger of upper-layer fluid by applying the uncorrectedPLIC scheme, or overly regularize the constitutive model. Both deficiencies lead66to an enhancement in spreading. Note that there is essentially no effect of iner-tia on the profiles shown in figure 3.5: recomputing the results with a Reynoldsnumber (O(1)) based on the benchmark conditions, rather than the artificially lowvalue used for the bulk of our simulations, leads to no significant differences.When a complete slump does not occur but a central section of the initialcylinder survives, the shallow-layer solution is modified accordingly:h =1, 0< r < r∞− 12B√2B(r∞− r)+ pi√34B, r∞− 12B< r < r∞(3.12)This approximation is again compared with numerical final shapes in figure 3.3(b).3.2.2 Slender columnsFor a tall slender column (R ≪ 1), the asymptotic analysis of can be gen-eralized to determine the instantaneous radius r = r(a, t)≪ 1 in terms of a La-grangian coordinate a ∈ [0,1] corresponding to initial height: where the fluid islocally yielded, we findr2(a, t) = E(t)r20(a)+1−E(t)√3B∫ 1ar20(x)dxz(a, t) =∫ a0r20(x)r2(x, t)dx, E(t) = e− Bt√3 ,(3.13)where r0(a) represents the shape of the initial column. As t → ∞, we then obtainr2(a) =1√3B∫ 1ar20(x)dx,z(a) =√3B ln[∫ 10 r20(x)dx∫ 1a r20(x)dx].(3.14)67The solution in (3.13) and (3.14) must be matched to a plugged upper sectionof the column, which always remains unyielded. The yield surface is given byr0(aY ) = r(aY ). The dimensionless slump height s (the difference between theinitial and final heights) is therefores = aY +√3B ln[√3Br20(aY )∫ 10 r20(a)da]. (3.15)For an initial cylinder, r0(a) = R, ands = 1−√3B+√3B ln(√3B), (3.16)which is widely used as a prediction of the slump test . This immediatelyimplies that the column will not yield anywhere if B > Bc = 1/√3≈ 0.577.Figure 3.6 compares numerical computations with the slender-column asymp-totics for initial shape with r0(a) = R = 0.025. First, the dynamical evolution of acolumn with B= 0.15 is shown; second, the final shape is compared for cases withvarying yield stress B. Note that undulations in the surface profile do not appearnear the base of the column in the axisymmetric simulations, unlike in 2D .As a result, the computations and slender-column theory agree more satisfyingly,except at the very base of the column where the no slip condition is not correctlycaptured by the asymptotics.68rz(a) B=0.15, t=1, 2, 4, 8, 10000 0.02 0.0400.10.20.30.40.50.60.70.80.910 0.02 0.0400.10.20.30.40.50.60.70.80.91(b) Final shapes, varying BrFigure 3.6: (a) Collapse of a cylindrical column with B = 0.15, showingthe interface at the times indicated. (b) Final shapes of cylinder withR = 0.025 and B = 0.15, 0.20, 0.25, 0.30, 0.35, 0.40 and 0.45. Dashedlines indicate the slender asymptotic predictions (3.13) and (3.14);solid lines are from numerical simulations.3.2.3 Failure modeJust below the critical value Bc(R) for which no slump will occur, the collapse ischaracterized by a particular mode of failure that depends on the initial shape.To compute the critical yield stresses and find the failure modes, we use theaugmented-Lagrangian method and monitor the rate at which the iterations ofthe scheme converge during short-time computations ending at t = 1; iterationsconverge significantly faster when B > Bc. The results are plotted in figure 3.7.As R → 0, Bc converges to 1/√3, the limit for a slender column; as R → ∞, Bc69approaches the value 0.265, corresponding to the failure of a 2D vertical embank-ment [55, 56].0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.250.30.350.40.450.50.55RB c (a)Numerical velocity fieldChamberlain et al.(2004)Lower boundBc for R→∞Chamberlain et al. (2003)0 0.5 1 1.5 2 2.5 3 3.5 40.20.250.30.350.40.450.50.55RB c (b)Numerical velocity fieldChamberlain et al.(2004)Lower bound IBc for R→∞Lower bound IIFigure 3.7: Critical yield stress for collapse, Bc, as a function of initial radiusR. Blue stars indicate the results from numerical simulations; the solidline shows the lower bound assuming a mechanism involving basalcollapse (labelled I); the dotted line shows the result from . In (a)we show the range R < 1 and include experimental data from . In(b) we show the range R < 4 and an alternative lower bound assuminga peripheral collapse (labelled II).70Figure 3.8 collects together computational results for sample failure modes,extracted from the initial velocity field for simulations with B ≈ Bc. Much asexpected on physical grounds and found for 2D dambreaks, the fluid collapsesby failing first over a basal region for thinner initial cylinders, and only at theperiphery with a wider initial state; cylinders with radii around unity show mixedtypes of failure modes.Figure 3.8: Strain-rate invariant plotted logarithmically as a density on the(r,z) plane for solutions with B ≈ Bc,Re ≈ 0, t ≈ 0, at the values ofR = 0.1,0.25,0.5,1.4,4 indicated by the r-axis limits. Also shown area selection of streamlines. In (a)−(c), the solid blue lines indicate theborder of the plastic region predicted by limit analysis (3.17). In (e),the blue line indicates the circular failure surface of the lower boundsolution in 2D geometry .The critical yield stress can be bounded using methods from plasticity theory71. In particular, lower bounds on Bc can be established by maximizingB =∫∫wrdrdz∫∫Bγ˙rdrdz, (3.17)over families of trial velocity fields. Guided by the numerical failure modes infigure 3.8, we find a lower bound for slender initial cylinders by considering trialvelocity fields that are composed of a rigid central section in z < f (r), an overly-ing descending plug for z > g(r), and a plastically deforming region sandwichedinbetween. We take the “yield surfaces” f (r) and g(r) to have parabolic shapeand the (incompressible) plastic deformation to have a cubic horizontal velocityprofile. Thus, our trial isv =(0,−1) g(r)< z < 1(u,w) f (r)< z < g(r)(0,0) 0< z < f (r)(3.18)with u =6rg(r)− f (r)η(1−η)2w =−(6−8η +3η2)η2−6rη(1−η)2ηrη =z− f (r)g(r)− f (r)(3.19)Optimization of B can then be performed over the parameters c j of the parabolasdefining f = c1(R− r)+ c2(R− r)2 and g = c3+ c4(R− r)+ c5(R− r)2.Upper bounds on Bc can also be constructed using trial admissible stress fields. However, in axisymmetric geometry, without a velocity field to determineall the stress components, an additional assumption is needed such as the vonKarman-Haar hypothesis. This artifice permitted Chamberlain et al. to con-struct upper bounds on Bc. Unfortunately, this hypothesis is not appropriate forour axisymmetric slump and so their upper bound does not apply here.Note that the trial velocity field in (3.19) is continuous across the curves72z = f (r) and g(r), in line with the structure of the failure modes of figure 3.8.This contrasts sharply with the 2D problem in which failure can occur over dis-tinct curves that support velocity jumps and which become smoothed into viscousboundary layers in numerical solutions. Indeed, Chamberlain et al. havepreviously computed bounds for the failure of cylinders using trial velocity fieldsmore similar to the 2D failure modes. However, distinct failure lines and viscousboundary layers do not characterize our axisymmetric solutions, leading to thechoice of the cubic velocity field in (3.19). This choice complicates the optimiza-tion computation but significantly improves the results.The lower bound on Bc for basal failure is included in figure 3.7, and compareswell with numerical computations when R is small. Figure 3.8 also displays thepredictions of the optimization calculation for the surfaces z = f (r) and g(r),which have some correspondence with the computed yield surfaces. For largervalues of R, the bound diverges from the computations and the trial velocity fieldis less similar to the actual failure modes. Both occur because of the switch in theform of the failure mode, from a basal collapse to a peripheral one.For a lower bound on Bc for peripheral failure we require a different trialvelocity field. In 2D, a useful bound is found by assuming that a circular arc offailure connects the foot of the vertical face at the edge with some point on the topsurface; above this arc, material rotates rigidly out of initial position. For a simpleaxisymmetric generalization of this trial, we again assume that a circular arc offailure arises, but divide the 2D velocity field by r (with the horizontal coordinate xreplaced by r), which ensures that the trial is incompressible. The resulting boundis shown in figure 3.7, and always lies below the 2D bound (though converges toit for R→ ∞).733.3 Comparison with experiments3.3.1 MethodsTo complement the theory, we performed experiments using an aqueous suspen-sion of Carbopol Ultrez 21 (with a concentration of about 0.5 % by weight, andneutralized by sodium hydroxide). A Herschel-Bulkley fit to the flow curve mea-sured in a rheometer (MCR501, Anton Paar, with roughened parallel plates) gaveτY = 39Pa, n = 0.3 and K = 32Pa sn.Slump tests were conducted by filling a variety of cylinders with different ge-ometry with the suspension, smoothing the fluid’s upper surface with a sharp edge.The cylinders, with radii varying from 10mm to 95mm and heights over the rangeof 11mm to 562mm, were attached to a pivoted arm that raised each container ina relatively controlled and reproducible fashion. For most of the tests, the armwas raised relatively quickly, releasing the fluid over a time of about 1 second,and the surface over which the fluid slumped was a plexiglass plate roughenedwith sandpaper. Measurements of the final deposit were taken after waiting fora few minutes. However, we also conducted tests in which we varied the rate atwhich the cylinder was raised, or replaced the roughened plexiglass with either asmooth sheet or covered it with 60 grit sandpaper. Note that for the smaller aspectratio cylinders that we used (R < 12), there was a tendency for the slumps to toppleover sideways if the cylinder was not raised sufficiently slowly (which typicallylengthened the release time to a couple of seconds), highlighting an instability oftall thin columns . We abandoned any tests that showed substantial sidewaysmotion.To remove the effect of the manner in which the fluid was released, we con-ducted a second series of extrusion experiments. Here, the Carbopol was pumpedup onto the surface through a vent with a diameter of about 0.5mm. For boththe slump tests and extrusions, thickness profiles h(r, t) were extracted by takingphotographs from the side, allowing measurements of the final central height andradius (except for the extrusions over the smooth surface, as discussed below, the74tests were axisymmetric).Note that, for the slump tests, we continue to use the Bingham model to pro-vide theoretical solutions to compare against the experiments, despite the fact thatthe Carbopol is better fitted by the Herschel-Bulkley law. Thus, we implicitlyassume that the power-law viscosity plays a minor role in controlling the formof the final state of a slump. We explicitly confirmed this for isolated exam-ples of dambreak computations in which we implemented the Herschel-Bulkleymodel with a power-law index suggested from the flow curve of the Carbopol (i.e.n = 0.3, see the end of §4.2 and figure 3.12). In order to improve the comparisonwith experiments, we also used this alternative rheological model for our compu-tations of time-dependent extrusions. Specifically, the computations exploited theregularization,τ jk =[cγ˙n−1+µ2µ1(1− c)+ cBγ˙ + ε]γ˙ jk. (3.20)The characteristic velocity scale in this case is given by U = (ρgHˆn+1/K)1/n(ρ = 103 kg m−3, g = 9.81m s−2).3.3.2 Cylindrical dambreaksThe conventional slump test focuses on the distance fallen at the centre of thefluid, the so-called slump height S = 1−H∞, where H∞ is the final central depth.A summary of our results for this diagnostic is presented in figure 3.9. An impor-tant experimental parameter that we varied in the current suite of experiments isthe aspect ratio of the cylinder, R = Rˆ/Hˆ. Most previous experiments have con-ducted tests with aspect ratios close to a half, concluding that this parameter haslittle significant effect. However, the theoretical results summarized in §3 clearlyexpose how R plays an important role if varied over a sufficiently wide range.This is confirmed in our experiments, which clearly demonstrate how S decreaseswith R at fixed B; see the colour coding of the points in panel (a), and the plot forcylinders with similar initial height Hˆ in (c).The decrease of the slump height with increasing initial radius is a natural750 0.05 0.1 0.15 0.2 0.25 0.3 0.3500.10.20.30.40.50.60.70.80.91(a)R = Rˆ/HˆBS=1−H∞ 0.5 1 1.5 2GaoPashias−aPashias−bSaakClaytonCarbopol10−1 1000.250.30.350.40.45 (b) Fall vs rate of release (s−1)510Residual weight (g)0 0.5 1 1.5 2 2.5 300.10.20.30.4R (c) Fall against R1 − H∞1 − H cFigure 3.9: (a) Comparison of experimental slump height with previouslypublished data for R < 2. The points are colour coded by R, as indi-cated by the colour bar. The published data is taken from (Gao;R = 12), (Pashias-a: data at R≈ 12for differing materials; Pashias-b: data for red mud with varying R), (Saak; R = 0.52) and (Clayton; R = 12). On the right, slump height is plotted against (b) therate of release (i.e. the inverse of the time taken to lift up the cylinder)for Rˆ = 3cm and Hˆ = 3.8cm, and (c) R for Hˆ ≈ 3.1cm. Most slumpswere conducted by releasing the fluid over about a second, but notrecording the time precisely; several repeated slumps with this proto-col are plotted at release rate 1.25 s−1. In (b) and (c) the fall of thecorner of the initial cylinder 1−Hc is also recorded (open circles).Panel (b) also shows the residual mass attached to the cylinder afterthe release of the fluid, and data for slumps that were conducted over asmooth surface (red stars; in these cases the slumps were left to cometo rest for five minutes).consequence of the growth of the rigid core of the dambreak and the eventualemergence of an incomplete slump. Indeed, for the latter, H∞ = 1 and S = 0,rendering this diagnostic useless for sufficiently large R. Instead, a measure ofthe degree of slump at the edge of the initial cylinder is provided by the distance76Figure 3.10: (a) Experimental and (b) numerical slump heights against Bfor R < 2. Shown in (a) are both S (filled circles) and the distancefallen by the corner of the initial cylinder 1−Hc (open circles), asillustrated by the experimental image for (R,B) ≈ (1.25,0.13) insetabove (a) (and also shown in figure 3.12). Inset above (b) are thefinal profiles (red) of the computations with R = 1.25; that matchingthe value of B of the experiment is highlighted (blue). The colourcoding of the points by R is the same as in figure 3.9, and a collectionof previous data (with R = 12) from numerical computations are alsoplotted in (b) (crosses ; stars ; pentagrams ).77fallen by the upper circular corner, which clearly decorates the final deposit exceptfor very low yield stresses (see figures 3.3 and 3.10). This alternative diagnostic,denoted by 1−Hc where Hc is the final height of the corner, is compared with Sin figures 3.9 and 3.10. Evidently, 1−Hc is less sensitive to variations in R. Notethat, for some of the cylinders with relatively small aspect ratio, our side imagingof the thickness profile obscures the centre of the deposit when the circular cornerfalls less far; in this situation, the measurement of S is actually given by 1−Hc.Despite the reduced sensitivity of 1−Hc to R, this diagnostic does depend onthe rate at which the fluid is released (figure 3.9(b)). Evidently, the speed at whichthe cylinder is raised affects how much fluid yields at the fluid edge, which partlycontrols the fall there. Indeed, the release rate correlates closely with the amountof material left on the cylinder after it is raised (which we measured on a weighscale in this series of experiments; see figure 3.9(b)). The impact of the releasemechanism on the degree of slump has been reported previously , and includesthe possibility that both inertia and adhesion to the cylinder play important rolesin the dynamics. Here, our goal is to complement our inertialess axisymmetricdambreak computations with similar, if not identical experiments, and so the effectof the release mechanism is a distraction that precludes a quantitative comparisonof theory and experiment. In particular, the action of lifting the cylinder mustforce the fluid to yield even if B > Bc, which may well explain why the existingexperimental data at higher B does not progress to the unyielded slump value S= 0(0.265< Bc < 1/√3≈ 0.577 over the full range of R).Figure 3.10 compares in more detail the experimental slump heights with theresults of our simulations. The two agree qualitatively, if not quantitatively, withthe wider initial cylinders of the simulations slumping less far. The comparisonis worse for previously reported numerical data [31, 39, 63] for Bingham slumps,which mostly collapse even further. We suspect that this is due to resolution is-sues, as illustrated in figure 3.5.Figure 3.11 shows another comparison of the final central depth and radiusof the deposit against yield stress. Here, the data is scaled by R2/3, which cor-78responds to choosing a length scale based on the initial volume of the cylinderrather than its height, and constitutes a more natural choice in the shallow limitwhere much of the memory of the initial shape is lost. The rescaling collapsesboth the height and radius data close to a common curve matching the asymptoticprediction from (3.11) (at least for the aspect ratios used in the plot, with R < 4),in contrast with the slump height diagnostic of figures 3.9 and 3.10.7900.511.522.5H∞/R2/3(a)0 0.2 0.4 0.6 0.8 10.511.52(b)B /R 2/3R∞/R2/3Figure 3.11: The central depth and radius of the final deposit against Bing-ham number, all scaled by R2/3, for R < 4. Our results for Carbopolslumps are shown by the filled circles; the crosses show data from. The results from the numerical simulations also plotted in fig-ure 3.10 are shown by the solid lines; the improved asymptotic pre-diction from (3.11) is shown by the dotted line. The colour coding byR is the same as in figure 3.9.Previous experiments have also invariably been conducted over smooth sur-80faces. However, it is known that spreading drops of viscoplastic fluid cansuffer effective slip unless the surface is either chemically treated or roughened.Indeed, when we conduct slumps over the smooth plexiglass, the fluid collapsesnoticeably further, implying a degree of slip (the contact angle is also noticeablydifferent, exceeding 90◦ for the roughened surface, but not for the smooth one).By contrast, slumps over a surface covered with sandpaper are similar to thoseabove our roughened plexiglass, providing confidence that our surface roughen-ing significantly reduces slip in the bulk of our tests. Figure 3.12 compares thefinal shapes of a particular slump conducted over the different three surfaces, andfigure 3.9(b) includes data for the same slump over the smooth plexiglass.810 0.01 0.02 0.03 0.04 0.05 0.0600.0050.010.0150.020.025Free slipNo slipFigure 3.12: Side images from the slump experiments and theoretical fi-nal profiles: the photographs compare slumps over the roughened,smooth and (green) sandpaper surfaces (from top to bottom); ver-tical dashed lines indicate the final radii, and the scale of the pho-tographs is indicated in the first panel, for which the rectangle showsthe initial fluid cylinder, of height 3cm and radius 3.8cm. The plotshows computations matching the experiments (R = 1.29, B = 0.13and Re = 0.28) and using the surface boundary conditions indicated(blue line for no slip, red for free slip); the dashed lines indicate theshallow-layer asymptotic predictions in (3.11) and (B.14). For the noslip case, two other computations are shown: a computation with arti-ficially high Reynolds number (Re= 35; thick grey line) and one withHerschel-Bulkley model (n = 0.3; green dots). For the free slip case,the green dots again show a computation with the Herschel-Bulkleymodel (n = 0.3).82For a theoretical examination of the effect of surface slip, we conduct simula-tions in which the no slip condition is replaced by free slip (i.e. we impose τrz = 0at z = 0). Figure 3.12 compares the result with that using no slip for simulationsin which the geometrical parameters and yield stress are matched to the experi-ments. A free slip version of the asymptotic analysis for shallow flows can also beprovided, as summarized in B.3; the final shape is predicted to remain cylindrical,with a depth of 2B√3 and a radius of R/(12B2)1/4. Both this prediction and thatof the improved no slip theory in (3.11) are also plotted in figure 3.12. The fluidspreads noticeably further with free slip, mirroring the experiments. The spreadover the smooth plexiglass is, however, rather less significant than suggested bythe free slip computations, as would be the case if there was a residual surfaceinteraction.Figure 3.12 also includes results from a simulation in which inertial effects arepromoted by taking a higher Reynolds number than that matching the experimen-tal conditions (Re is about 125 times higher in this case). The two solutions canbarely be distinguished in the figure, highlighting how inertia plays little role incontrolling the final shape in the simulations. Indeed, our computations suggestthat inertial effects are relatively minor over most of the range of physical condi-tions of our experiments: figure 3.13, shows the slump height S and sample finalprofiles for computations with R = 1 and varying B and Re. Inertia plays no rolein controlling the final shape for Re < 1; the slump height increases for higherReynolds number, but the effect is modest for our experiments (with Re< 125).830 0.05 0.1 0.15 0.2 0.25 0.300.10.20.30.40.50.6BS Re=1125103B=0.2250.1250.05rz0.5 1 1.50.51Figure 3.13: The dimensionless slump height for cylindrical dambreaks withR = 1, varying B and the Reynolds numbers indicated. The insetshows three sample triplets of the final shape at the Bingham numbersindicated.A final feature of figure 3.12 is the inclusion of computations using the Herschel-Bulkley model in (3.20), rather than the Bingham law (for both no slip and freeslip surfaces). Evidently, as alluded to earlier, the power-law viscosity has littleeffect on the final shape. However, there do appear to be some minor quanti-tive differences, especially in the vicinity of the corner stemming from the upperedge of the initial cylinder. It is conceivable that these result from differences inthe time evolution of the plugs during the collapse, which then impacts the finaldeposit. Otherwise, the viscous stress is not expected to feature directly in theforce balance controlling the final shape. Nevertheless, any differences in the fi-nal radius and height that may be introduced in this manner from the power-lawrheology are unlikely to upset the comparison of experiments and Bingham theory84in figures 9-13.3.3.3 ExtrusionsThe impact of the release mechanism on the slump is removed in experimentsin which fluid is extruded slowly from a vent onto the underlying surface, al-lowing a clearer examination of the effect of surface slip and a more quantita-tive comparison with theory. Results of a sample extrusion are shown in fig-ure 3.14, which plots the instantaneous radius R(t)/B = ρgRˆ/τY against centraldepth H (t)/B = ρgHˆ/τY , both scaled by B. Rescaling the results in this wayremoves the scaling by Hˆ implicit in the non-dimensionalization of the problemin §2, which has no meaning for the extrusions. Instead, distance is scaled by thelength scale τY/ρg.850 5 10 15 20 25 30 35012345678910 R ( t )/BH(t)/B RoughSmooth√2H /B√2H /B+ pi√3/4No-sl i pFree -sl i pFigure 3.14: Maximum radius R(t)/B = ρgRˆ/τY versus depth H (t)/B =ρgHˆ/τY for experimental extrusions with a pump rate of 31.5 ml/minover the rough (stars) and smooth (squares) plexiglass. Also plot-ted are the leading-order and improved asymptotic predictions, andHerschel-Bulkley simulations matching the physical parameters ofthe experiment with either no slip or free slip boundary conditions.Figure 3.14 includes data from an extrusion with similar pump rate over thesmooth plexiglass. The radius-height relation,R/B versusH /B, is very differentand reflective of a shallower flow. This is illustrated further in figure 3.15, whichdisplays sample images and height profiles taken from the experiments. In fact,the extrusions over the smooth plexiglass became noticeably non-axisymmetricover late times, developing interesting lobe patterns towards the rim of the ex-truded dome (see the images shown at the top of figure 3.15). It is not clearwhether these patterns reflects an intrinsic instability of sliding extensional flow86[cf. 75] or a more pronounced sensitivity to surface imperfections at the contactline.When the extrusion flux is relatively small, the developing dome above thevent is expected to be close to a steady equilibrium state, allowing us to recyclethe shallow asymptotics of §3.1. In particular, the radius-height relation from(3.11) isHB∼√2RB+pi√34, (3.21)which is also included in figure 3.14, along with the leading-order result H /B =√2R/B. The improved model in (3.21) is unrealistic for R → 0, where theinvalid treatment of the edge predicts a finite central depth, which is the analogueof the vertical cliff predicted at the fluid edge by (3.11).For equivalent numerical computations, we perform simulations as outlined in§2, but using the Herschel-Bulkley law and replacing the no-penetration condi-tion w(r,0, t) = 0 with w(r,0, t) = 2pi−1(R2v − r2)/R4v for r < Rv, where Rv is thedimensionless radius of the vent. As initial condition, we take c(r,z,0) = 0 every-where except over a shallow prewetted film spanning the vent and of depth 0.05.In these computations, Hˆ no longer has any meaning as the characteristic heightof the initial configuration, but can be defined using the net dimensional flux Qˆ: inview of our prescription for the vertical velocity, the dimensionless flux is unity,so that Qˆ = Hˆ2U = Hˆ2(ρgHˆ1+n/K)1/n. Thus, Hˆ =(KQˆn/ρg)1/(1+3n).A numerical simulation designed to match the experimental conditions is in-cluded in figures 3.14 and 3.15. The numerical simulation over-predicts the cen-tral height in comparison to the experiments, but otherwise tracks the observedradius-height relation. The comparison of surface profiles illustrates how the the-ory reproduces the shape of the observed extruded dome, but again reveals theslight discrepancy between theory and experiment. We suspect that this originatesfrom the incomplete removal of slip over the underlying roughened plexiglasssurface (fluid in the experimental extrusion flows further than in the simulationand the side profiles are less steep). However, errors in the fluid rheology (theHerschel-Bulkley fit, or non-ideal properties of the Carbopol) might also be re-87−20 −15 −10 −5 0 5 10 15 20 25 300510No slip Free slipr/Bh/BFigure 3.15: Images and profiles from the extrusion experiments also shownin figure 3.14. The top images show top and side images of an ex-trusion over the rough (left) and smooth (right) plexiglass (a rulerstraddles the extrusions to add a scale for the side images). The firstrow of plots underneath show a sequence of height profiles at simi-lar times (extruded volumes; the side photographs correspond to thesixth profiles). The insets show magnifications of the contact line.The lowest row of plots shows height profiles from correspondingnumerical simulations applying either a no slip (left) or a free slip(right) boundary condition on z = 0 (less one profile for the free slipcase). The times of the snapshots are not perfectly matched owing toirregularities in pump rate in the experiments.88sponsible.Figures 3.14 and 3.15 also include simulations results for a computation inwhich the no slip condition on z = 0, r > Rv, is replaced by free slip. As forthe axisymmetric dambreaks, the enhanced spread of the fluid and the attendantmodification in the height profile are reminiscent of how the experiments on thesmooth plexiglass differ from those above the roughened surface, reinforcing ourconclusions regarding surface slip. Once more, however, the freely sliding com-putations spread even further than the extrusions on smooth plexiglass, suggestiveof residual surface traction.3.4 Concluding remarksIn this chapter, we have presented a theoretical analysis of the axisymmetricdambreak of a cylinder of viscoplastic fluid, and compared the results with ex-periments using a Carbopol gel. In the theory, we used computations with eithera regularized Bingham model or an augmented-Lagrangian scheme to study thefluid slump, complemented by asymptotic analyses relevant for shallow gravitycurrents or tall thin columns. We also examined the states at the brink of failureto determine the conditions under which the initial cylinder does not collapse.The computations, which are based on the VOFmethod to track the fluid inter-face, are complicated significantly by the need to resolve the flow adjacent to theunderlying no slip surface. Without special attention to this detail, computationsinevitably become unresolved and spread excessively far as a result. Both this andinaccurate treatments of the yield stress likely reduce the reliability of slump testcomputations. Here, we adopted a numerical device which ensures that our com-putations remain resolved, and acts by adjusting the fluid interface and allowingthe contact line to move over the surface.For dambreaks (cylindrical slump tests), theory and experiment agree qualita-tively and are consistent with previous experiments that measure the dimension-less “slump height” (the distance fallen by the centre of the cylinder divided bythe initial height), and use a variety of different kinds of (less ideal) viscoplastic89fluids. However, one must be careful to eliminate slip over the underlying surface,which can significantly enhance the collapse. Moreover, the mechanism by whichthe fluid is released introduces quantitative differences between theory and exper-iment, either through the interaction with the lifted container, or via the amountof inertia imparted at the moment of release. The effect of the release mechanismis avoided when fluid is pumped slowly through a vent onto the surface. In suchextrusions, the agreement between theory and experiment is improved, althoughthere are some remaining differences that are most likely due either to residualslip or an inaccurate treatment of fluid rheology.A main motivation of the current work was to shed further light on the fluiddynamics of the slump test. That practical device exploits a conical initial shaperather than a cylinder, begging the question of how the present results carry over tothis other geometry. Figure 3.16 plots simulation results for the slump height com-puted for Bingham fluid with an initial shape given by an ASTM standard cone(which is 30cm high, with a top radius of 5cm and a basal radius of 10cm). Asfor the cylinder, the theoretical slumps do not collapse as far as experiments (thistime taken from ) at higher yield stresses, a discrepancy that could arise fromeither the release mechanism or slip. However, although the threshold for failuredepends on the slip condition on the underlying surface (Bc ≈ 0.27 or 0.32 for thisgeometry with either no or free slip, respectively), the continued significant slumpof the experimental cones for higher yield stress suggests that the release mech-anism, not slip, is more important. Except for low yield stresses, the simulationsresults also disagree significantly with those reported by , especially for thefailure criterion.900 0.05 0.1 0.15 0.2 0.25 0.300.10.20.30.40.50.60.70.80.91Dimensionless yie ld stress B = τY/(ρgHˆ)DimensionlessslumpheightS Clayton − conesSlender coneNumerics − no−slipNumerics − free−slipRousselFigure 3.16: Slump heights for Bingham dambreaks with an initially con-ical shape corresponding to the ASTM standard, computed using ano slip (solid line with points) or free slip (dashed-dotted line withstars) condition on the underlying plane. Squares show experimentaldata from . The dashed line shows the prediction for a slenderconical column in B.4. The diamonds show the results of numericalcomputations of .91Chapter 4Surges of viscoplastic fluids onslopes1The purpose of this chapter is to explore models for steady, shallow viscoplasticsurges with a length that is sufficiently long that the flow converges to a steadysheet flow well upstream of the flow front; i.e. the theoretical analogue of theexperiments by Chambon et al. [22, 23, 36]. For this task, we reconsider the lu-brication analysis of Liu &Mei . First, we consider the upstream extent of thesurge to examine how the surge converges to uniform sheet flow, and how the cor-responding superficial plug breaks as one progresses downstream. This demandsa variation of the lubrication analysis that is designed for almost uniform flows,and parallels theory for flow down weakly varying channels . Second, oncethe upstream plugged flow gives way to a fully yielded surge with a pseudo-plug,standard lubrication theory applies; for this region, we improve that theory bycontinuing the analysis to higher order in order to account better for non-shalloweffects.We complement the shallow-flow analysis with computations using the Volume-of-Fluid (VOF) method and an augmented-Lagrangian scheme to deal with the1A version of chapter 4 has been submitted for publication (under review). [Liu, Y.], Balmforth,N.J. & Hormozi, S. Viscoplastic surges down an incline.92xzPF FFLZh(x)Y(x)ugUbFigure 4.1: Sketch of the geometry of the surge in the frame of referencein which there is no net flux and the flow is steady. The free surfaceis located at z = h; the level z = Y divides a fully yield region under-neath from either a true plug or a weakly yielded pseudo-plug. Theflow divides into three regions: a plugged flow region (PF) where thesuperficial layer of fluid is not yielded, a lubrication zone (LZ) wherethe pseudo-plug arises, and the flow front (FF) where the dynamics isnot shallow.yield stress. Such a combination of shallow-flow analysis and computation hasproven effective in our previous work studying dambreak flows and their finalshapes [53, 54], illustrated in chapter 2 and 3. Here, we examine the extent towhich the theoretical solutions match the observations of Chambon et al. and de-termine the conditions for which the pseudo-plug of lubrication theory locks upinto a true plug.4.1 Formulation4.1.1 Model equationsAs sketched in figure 4.1, we consider a two-dimensional surge of incompressibleviscoplastic fluid flowing steadily down a plane that is inclined at an angle θ to thehorizontal. We use Cartesian coordinates aligned with the plane to describe thegeometry; in the frame of reference of the surge, the inclined plane travels upslope93with a speed ub. We model the rheology of the fluid using the Herschel-Bulkleyconstitutive law. The governing equations for the velocity u = (u,w), deviatoricstress tensor τ , and pressure p are then∇ · u = 0,ρ[∂u∂ t+(u ·∇)u]=−∇p+∇ · τ +ρg(sinθ−cosθ),(4.1)and γ˙ jk = 0, τI < τY ,τ jk =(κγ˙n−1+τYγ˙)γ˙ jk, τI > τY ,(4.2)where ρ is the density, g is gravity, τY is the yield stress, the plastic viscosityµ = κγ˙n−1 introduces the consistency κ and power-law index n as two furtherrheological parameters, and τI =√12 ∑ j,k τ2jk and γ˙ =√12 ∑ j,k γ˙2jk denote secondtensorial invariants, withγ˙ =(2ux uz +wxuz +wx 2wz). (4.3)Here, subscripts on the velocity components represent partial derivatives.For the boundary conditions, we assume that there is no slip over the inclinedplane, u(x,0) = (−ub,0), and that the upper surface is stress free, so that∂h∂ t+u∂h∂x= w and (τ − pI) ·(−hx1)=(00), (4.4)on z = h(x, t). The surge ends at a flow front, x = X(t), where h→ 0, and extendsback upstream to where the flow converges to a uniform sheet flow.944.1.2 The sheet-flow solutionIn the frame of reference in which the net flux vanishes, the steady, uniform,sheet-flow solution is given by(p,τxz) = (1, tanθ)(1− zH)ρgcosθ (4.5)andu = usheet(z) =nU tan1/n θn+1××[n2n+1Y2+1/n∞ − (Y∞− zH )1+1/n], 0< zH<Y∞,n2n+1Y2+1/n∞ , Y∞ <zH< 1,(4.6)where H is the flow depth andU =(ρgH cosθκ)1/nH. (4.7)The level z = HY∞ corresponds to the yield surface above which the fluid isplugged, withY∞ = 1− τYρgH sinθ. (4.8)Note that, because there is no net flux along the plane in the frame of the surge,the profile in (4.6) demands that the speed of the inclined plane isub =nU tan1/n θn+1Y 1+1/n∞(1− nY∞2n+1). (4.9)The scales here can be used to non-dimensionalize the problem, as discussed later.4.1.3 Volume-of-fluid computationsIn our computations of the full problem, we use the VOF method to track the fluidinterface using an advected volume fraction c(x,z, t) (see [53, 54] or Chapter 2and 3). This scheme immerses the viscoplastic fluid beneath a miscible ambient95Newtonian fluid. The material properties of the bulk mixture are set by linearlyinterpolating between the two phases using c, with the density and viscosity of theNewtonian fluid taken to be relatively small in order to minimize the effect of theambient flow dynamics (cf. [53, 54] or Chapter 2 and 3).To solve the governing equations, we evolve the system as an initial-valueproblem until a steady state is reached. We use an augmented-Lagrangian schemeto deal with the yield stress within a weak formulation of the problem [35, 80],as implemented in C++ using the PELICANS platform (e.g. ). We refer thereader to our earlier work [53, 54] or Chapter 2 and 3) for further computationaldetails, including a discussion of how we avoid any resolution issues stemmingfrom the no-slip condition imposed on the underlying plane. Some additionaldetails relevant for the present computations (including a resolution study con-firming the fidelity of the computations) are provided in C.1.The computational domain is finite, extending up to a height Lz and to a lengthLx. The top and right-hand boundaries are chosen to be sufficiently distant thattheir positions do not affect the solution; free slip conditions are imposed to helpsuppress the ambient fluid dynamics. On the lower boundary we impose the fixedvelocity (u,w) = (−ub,0), where ub is chosen in a range that the flow adapts toreach a steady flow regime; for a long thin flow, the upstream current convergesto the sheet flow solution above, and so ub is given by (4.9).To minimize the influence of the left-hand boundary conditions and ensure thatthe surge is mostly long and thin, we select domain lengths Lx that are as large aspossible. To gauge the residual effect of the left-hand boundary conditions, wecompute solutions with two different conditions: an infinitely long flow can besimulated by imposing the velocity here as that given by the sheet-flow solution;i.e. (u,w) = (usheet ,0) at x = 0. Alternatively, the back wall of the conveyor-beltexperiment of [22, 23, 36] can be simulated by setting u = w = 0 at x = 0. C.1.3summarizes the role played by these boundary conditions.From c(x,z, t), we define the instantaneous position of the interface of theslump from the contour c(x,z = h) = 12. The surface z = h(x) plays a major role96in the asymptotic analysis of §4.2.For the computations with Bingham fluid that we compare with asymptotictheory, we select parameters to minimize inertial effects (see §4.3.1 and C.1.2).The Herschel-Bulkley simulations of §4.3.2 have parameter settings matched tocorresponding laboratory experiments; inertia may play a more significant role inthese examples.4.2 Asymptotic analysisThe anatomy of the surge is illustrated in figure 4.1: the flow body is divided intothree regions. A plugged flow (PF) region arises at the back, where the surgeconverges to the sheet flow and a true plug exist on top of the fluid. That plugthen breaks to leave a fully yielded flow with a more significant variation in thefree surface. Although there is no true plug, the flow remains relatively shallow;in this lubrication zone (LZ), standard shallow-layer analysis applies and there isa superficial pseudo-plug. Finally, at the flow front (FF), the free surface steepensup to terminate the surge and invalidate shallow-layer theory.To prepare the way for asymptotics, we rescale the equations to suit the shal-low geometry . We also focus on the special case of the Bingham model, withκ ≡ µ and n = 1 (the extension to Herschel-Bulkley model is straightforward andpartly described in C.2), and discard inertia. Hence, we introduce a characteristicdepth H and horizontal length L with ε = H/L≪ 1, and then setz = Hzˆ, x = Lxˆ, u =Uuˆ, w = εUwˆ,p = ρgH pˆcosθ , τ = ρgHε cosθ(σ ττ −σ),U =ρgH3 cosθκL, S =tanθε, B =τY LρgH2 cosθ.(4.10)Notice that S ∼ tanθ is assumed here. For steady flow, the dimensionless equa-97tions now become, after dropping the hat decoration,px = εσx + τz +S,pz = ε2τx− εσz−1,τ =(1+Bγ˙)γ˙ ,√τ2+σ2 > B,0=∫ h0u(x,z) dz,γ˙xx = 2εux, γ˙xz = uz+ ε2wx, γ˙ =√γ˙2xx + γ˙2xz,(4.11)with u(x,0) =−ub andτ +hx(p− εσ) = 0p+ εσ + ε2hxτ = 0}on z = h. (4.12)4.2.1 Plugged FlowIn the PF region, the flow is nearly uniform, so we introduce the asymptotic se-quences,h = 1+ εh1+ ..., p = 1− z+ ε p1+ ..., (4.13)τ = S(1− z)+ ετ1+ ..., u = u0+ εu1+ ... (4.14)and σ = σ0+ ... The leading-order terms from (4.11) recover the velocity profileof the sheet-flow solution, this time in dimensionless form:u0 =−ub + 12S×{z(2Y∞− z), 0< z <Y∞,Y 2∞, Y∞ < z < 1,(4.15)withub =16SY 2∞(3−Y∞) (4.16)98and Y∞ ≡ 1− BS . At O(ε) we now obtainp1x = τ1z +σ0x,p1y =−σ0z,(4.17)Expanding the free surface conditions about z = 1, we findp1+σ0 = h1,τ1 = Sh1,}at z = 1. (4.18)Hence, p1+σ0 = h1 throughout the fluid depth.Over the fully yielded region underneath the plug, the constitutive law impliesthatσ0 = 0 and τ1 = u1z. (4.19)It follows that p1 = h1 over this region, and soτ1 = h1xz+T and u1 =12h1xz2+T z, (4.20)where T (x) is not yet determined. However, in the overlying plug, σ0 cannot betaken to vanish and the stress state is indeterminate, as (4.17) do not determine allof p1, τ1 and σ0. Instead, the yield condition demands only that σ20 < B2−S2(1−z)2.The stress solution for the yielded region,τ ∼ S(1− z)+ ε(h1xz+T ) and σ = O(ε), (4.21)now implies that the yield surface is shifted to Y =Y∞ + εY1, whereY1 = S−1(h1xY∞ +T ). (4.22)However, the plug speed must remain equal to 12SY 2∞ as there is no deformation in99h > z > Y . This demands that u1(x,Y ) = 0, orT =−12Y∞h1x. (4.23)Finally, we impose the flux constraint,∫ h0udz = 0 (4.24)which gives, at O(ε),ubh1 =12SY 2∞h1+16h1xY3∞ +12TY 2∞. (4.25)Hence,h1x = 2Sh1 and Y1 =Y∞h1, (4.26)and so the departure from the uniform sheet solution grows exponentially in thedownslope direction, with an exponent given by 2S.4.2.2 Breaking the plugReturning to (4.17), we now observe that, over the plug,h1x = 2σ0x + τ1z, (4.27)which can be integrated from z =Y∞ upto z = 1 to give∂∂x∫ 1Y∞σ0 dz =12Sh1(1−Y∞). (4.28)Hence ∫ 1Y∞[σ0(x,z)]x−∞dz1−Y∞ =14h1, (4.29)which, given that |h1| grows exponentially, implies that the net jump in extensionalstress across the plug must also increase towards the flow front. Moreover, since100h1 must eventually become O(ε−1) to curve the surge, the plug must inevitablybreak.A further constraint is provided by the unyielded condition of the plug (giventhe leading-order shear stress τ ∼ S(1− z)),−√B2−S2(1− z)2 < σ0 <√B2−S2(1− z)2, (4.30)which bounds the integral on the left of (4.29). In particular, that integral cannotexceed 12piB in absolute size. Therefore, the plug must have broken when|h1|> 2piB. (4.31)4.2.3 Lubrication ZoneThe preceding analysis indicates that the plug breaks when h−1= O(ε). Down-stream, the departure from the sheet-flow solution grows further as we enter thelubrication zone and h− 1 becomes O(1). With h = h(x) 6= 1, the leading-ordersolution of (4.11) now furnishes the standard lubrication result for the velocityprofile,u∼−ub + 12(S−hx)×{z(2Y − z), 0< z <Y,Y 2, Y < z < h,(4.32)where Y = Y (x) is the position of the fake yield surface. The profile of the surgeitself follows from solvingY = h− BS−hx ,ubh = up(h− 13Y) (4.33)whereup =12(S−hx)Y 2 (4.34)101determines the speed of the overlying pseudo-plug. Over that region, the exten-sional stress is given byσ = sgn(upx)√B2− (S−hx)2(h− z)2, (4.35)which matches one of the limits in (4.30) for h→ 1+O(ε). Thus, the plugged flowis expected to match continuously to the lubrication zone once the plug breaks.Note that (4.33) predicts that h and Y converge exponentially to the sheet flowsolution as x→−∞. In particular,hx ∼ 12S(h−1)Y 2∞−3Y∞ +6Y 2∞−3Y∞ +3,Y −Y∞ ∼ 12Y∞(h−1)Y 2∞−2Y∞ +3Y 2∞−3Y∞ +3.(4.36)Because 0 < Y∞ < 1, this result implies that the fake yield surface lies above thetrue yield surface once we enter the plugged flow region where h− 1 = O(ε).Thus, one does not expect a pseudo-plug to intervene between the fully yieldedzone and the true plug (cf. ).4.2.4 Improving the lubrication theoryWith a little effort, as described in C.2, the lubrication theory can be continued tonext order to furnish the improved model,Y = h− BS−hx +12εpiB2hxx(S−hx)3 ,ubh = up(h− 13Y)+ 12εpiB2upx(S−hx)2 ,(4.37)in place of (4.33). In principle, this improved model better captures non-shalloweffects. In the limit of θ = 0 and ub = 0, the system in (4.37) reduces to theproblem for a collapsed two-dimensional slump (see Chapter 2). In the limitub = 0, the time dependence can be retained to furnish an improved lubrication102model for unsteady viscoplastic flow over an inclined surface.The plastic limit of (4.37) is achieved when Y → 0 throughout the surge, andpermits further analytical headway in constructing the surface profile. SettingY = 0 in the first equation in (4.37) furnishesh− BS−hx +12εpiB2hxx(S−hx)3 = 0. (4.38)Given thatY → 0 demands that B/S→ 1 at the back of the surge, the leading-ordersolution ish+ log(1−h)∼ S(x− x f ), (4.39)which corresponds to the final shape of an inclined dambreak [10, 51]. Continuingwith the improved model, (4.38) is equivalent to(S−hx)h−S+ 12εpiBhx = O(ε2), (4.40)orh− 12εpiB+(1− 12εpiB)log(1−h1− 12εpiB)∼ S(x− x f ), (4.41)if x = x f denotes the front position. Note that the front of the surge here has finiteheight, with h(x f ) =12εpiB, as found in (see Chapter 2). Further discussionof the slumped shapes predicted by (4.38) and (4.41) is provided in C.3.Away from the plastic limit, we can again iterate (4.37) to O(ε2) into thesecond-order system,B− (h−Y )(S−hx) = 12εpiB(hx−Yx),ubh−up(h− 13Y)= 14εpiBY [(2h−Y )Yx−Y hx].(4.42)Beginning from a position upstream where the surge is close to the sheet-flowsolution, this system can be integrated downstream to the flow front using ini-tial conditions based on (4.36). Again, the surge ends at finite depth where thefree surface becomes vertical and h and Y are O(ε). A sample solution with103Figure 4.2: Asymptotic surge solutions for (a) the plastic limit Y → 0, and(b) Y∞ = 1−S−1B = 0.2. The solid lines show the improved lubrica-tion solutions for h(x) plotted against S(x− x f ) with 12εpiB = 0.2 (asgiven by either equations (4.41) and (4.42)); the dotted lines indicatethe predictions of the leading-order theory. In (b), the correspondingfake yield surfaces Y (x) are also plotted. The insets show magnifica-tions near the flow front.Y∞ = 0.2 (B/S = 0.8) is shown in figure 4.2 and compared with the predictions ofthe leading-order lubrication model and the corresponding solutions in the plasticlimit.1044.3 Numerical results4.3.1 Comparison with asymptoticsIn this section, we report computations with the Bingham model, n = 1, for com-parison with the asymptotic analysis. Following along the lines of that theory, wealso place the problem into a dimensionless form by scaling variables using thedepth of the expected sheet flow H and its characteristic velocity U . The stressesare scaled by ρgH cosθ . In addition, in the computations, there is no significanceto the lengthscale L, so we set L = H which is equivalent to taking ε = 1. Shallowsurges then arise when the longitudinal length far exceeds the depth. We use Y∞and θ as the main parameters, which translate to a dimensionless belt speed andyield stress of ub =16Y 2∞(3−Y∞) tanθB = (1−Y∞) tanθ(4.43)In all the computations we report in this subsection, inertia is not sufficient to sig-nificantly affect the solution (see C.1.2), and we use left-hand boundary conditionsgiven by an upstream sheet-flow solution.105Figure 4.3: Numerical solution for θ = 10◦ and Y∞ = 0.8 showing (a) τI, (b)τxz and (c) τxx. The darker (blue) lines show sample streamlines andthe dotted white line is the true yield surface where τI = B. The dashedline shows the contour level where τxz = B.106Figure 4.4: Flow profiles and fake yield surfaces for (a) numerical simula-tions, (b) leading-order lubrication theory and (c) improved asymptotictheory (using (4.42)), with θ = 10◦ and Y∞ = 0.4, 0.6 and 0.8. Thesolid and dashed lines show h and Y , with Y defined as the contourlevel where τxz = B for the simulations. The flow profiles deepen withincreasing Y∞. The shaded regions in (a) show the true plugs (whichextend up to the free surface in each case, and with the shading dark-ening with increasing Y∞).107Figure 4.3 shows a sample solution with θ = 10◦ andY∞ = 0.8. The plots showthe deviatoric stress invariant and components as densities over the (x,z)−plane.Superposed are streamlines and the free surface, with an upstream section of thesolution not shown in order to remove the regions where the boundary conditionsat x = 0 play a role. The shear stress τxz matches the stress invariant τI throughoutthe lower part of the surge, but deviates over a superficial layer adjacent to thefree surface. There, the Augmented Lagrangian algorithm detects a genuine plug2for x . 15. This plug then breaks to leave, over 15 . x . 18, a weakly yieldedsuperficial region with τI ∼ B; i.e. a pseudo-plug. In the numerical simulations,the fake yield surface z = Y (x) can be picked out by determining the level whereτxy = B (see figure 4.3). The true and fake yield surfaces are continuous at thebreakage of the plug. All this anatomy of the surge was already anticipated by theasymptotic analysis in §4.2.Figure 4.4 shows sample flow profiles h(x) and fake yield surfaces Y (x), forθ = 10◦ and varyingY∞. The two panels compare the simulations with the predic-tions of the leading-order and improved asymptotic theories. The latter leads to amildly better comparison of h(x), correcting for a spread in the predicted profileswhich are found in the simulations to collapse closely to one another. However,the improved asymptotics theory does not lead to a substantially better agreementwith the simulations in the examples shown in figure 4.4 because the lubricationzone in these solutions is not particularly long, with the free surface steepening upquickly from the plugged flow upstream to the flow front.2Although the stress field is indeterminate here, the algorithm provides an admissible solutionfor the plug which is dictated by the iterative scheme. The borders of the true plug are a littlerough due to grid-dependent numerical errors. However, the plugs appear to be robustly detectedgiven that the stress invariant τI lies significantly below B over this part of the surge.108Figure 4.5: Numerical solution forY∞ = 0.8 and θ = 5◦ (B= 0.0175), show-ing a density plot of τI with z = h(x) and the fake yield surface su-perposed; the true plug is shaded black. The dashed lines show theprediction of the leading-order lubrication theory (with the flow frontaligned). The dotted lines show the predictions of the plugged-flowsolution in (4.26), with 1−h matched to the simulation at x = 5.We provide a more quantitative comparison of a numerical solution for Y∞ =0.8 and θ = 5◦ with the asymptotic theory in figure 4.5. This figure highlightsthe free surface, the true plug of the numerical soution, and the fake yield surfaceunderneath the pseudo-plug (again defined by τxz = B in the simulations). Theseare compared with the predictions of the plugged flow solution (4.26) and theleading-order lubrication theory (§4.2.3). The plug breaks in the simulation when1−h ≈ 0.12, in satisfying agreement with the prediction 1−h = 2piB = 0.11 of§4.2.2.109Table 4.1: Experimental parameters from .τY (Pa) κ (Pa sn) n θ (◦) ub (m/s)C2 7.2 5.1 0.41 11.9 0.26C5 7.2 5.0 0.43 15.3 0.148Figure 4.6: Experiment C5 of , showing (a) u(x,z) and (b) w(x,z) asdensities over the (x,z)−plane, for a qualitative comparison with theirfigure 6. A selection of streamlines is also shown (thinner blue lines).The dashed lines shows the levels where u = 0 and z = Y (τxz = B).In leading-order lubrication theory, u = 0 along z = Y −Y [nY/h(2n+1)]n/(n+1); this prediction is also drawn as the lighter (pink) solid line.1104.3.2 Comparison with experimentsChambon et al. [22, 23, 36] have presented a comprehensive experimental studyon viscoplastic surges on a conveyor belt. For comparison with their results, weperform numerical simulations matching some of their experimental parameters.More specifically, we choose the two sets of parameters denoted by C2 and C5 in. The rheology of the two samples and the inclination and speed of the beltare given in table 4.1. The fluid in these experiments is an aqueous suspension ofCarbopol with density ρ = 103 kg/m3. We employ a no-slip back wall to providethe left-hand boundary conditions.Figures 4.6–4.8 display the results of the computations. Figure 4.6 plots thevelocity field for experiment C5, and corresponds to figure 6 in . Figures 4.7and 4.8 show the strain-rate field, surface velocity and sample vertical profiles ofthe velocity components in the frame of the conveyor belt for both experiments(for comparison with their figures 9 and 10). Distances are scaled by the thick-ness of the expected uniform sheet-flow, H, and velocities by the mean downslopevelocity u (the scalings used by Chambon et al.). Although the identification ofthe true plug is more difficult (owing to the power-law viscosity of the Herschel-Bulkley law), the computations still detect that the stress invariant near the freesurface falls below the yield stress a scaled distance of about four units behindthe flow front for C2, and three units for C5. The figures also contrast the nu-merical results with the predictions of the leading-order lubrication theory, whichperforms well in reproducing the simulations. Evidently, the fine details of theplugged flow or the improved lubrication theory are minor, and non-shallow floweffects are insignificant away from the flow front. Importantly, the numerical so-lutions converge towards the uniform-flow state at the back of the surge (over alongitudinal distance of about 10H; see C.1.3), and there is no mismatch betweenH and the depth at the back of the surge. This supports the inference of Cham-bon et al. that some rheological effect is responsible for their observation that theexperimental surges are deeper than the expected sheet flow.The broad match between simulations and leading-order lubrication is some-111what better than Chambon et al.’s comparison of experiments and asymtptotics,although there is qualitative agreement between all three. For example, the differ-ences between the predictions of lubrication theory and simulations for the surgeprofile and surface velocity are very small in figures 4.7 and 4.8, unlike the cor-responding figures of , which reveal noticeable differences. The front of theexperimental surges are also rounded and overturn, as in the simulations but notthe asymptotics, although the nose occurs at dimensionless heights of about 0.25-0.3 whereas it lies well below 0.1 in the simulations. We interpret all this to implythat the shallow approximation is not responsible for the main quantitative dis-agreements between theory and experiment.112Figure 4.7: Experiment C2 of : shown are (a) log10(γ˙/0.26), (b) the sur-face velocity, and (c)–(h) velocity profiles at the x−positions indicatedin (b), for a comparison with figure 10 in . The (red) dashed linesindicate the predictions of the leading-order lubrication theory. In (a)the solid white line shows the contour τxz = B, whereas the dashedwhite line is the fake yield surface z =Y (x) of the leading-order lubri-cation theory.113Figure 4.8: A similar picture to figure 4.7, but for Experiment C5 of andfor comparison with their figure 9 (with log10(γ˙/0.38) plotted in (a))We quantify this further using two distance diagnostics defined by Chambonet al. The first of these records the distance from the flow front, x f c,u = |x− x f |,114where [∫ h0(uexp−uasy)2 dzh]1/2= 0.07uwhere the subscripts refer to PIV measurements and the leading-order asymptoticprediction, and u is again the mean donwslope speed. Chambon et al. quote valuesfor x f c,u from 0.5H to H for all their experiments. The same mean differencebetween the velocity profiles of our simulations and the asymptotics indicates thatx f c,u ≈ 0.34H for C2 and 0.18H for C5. In other words, the asymptotic predictionfor the velocity profile remains close to that of the simulation for distances muchnearer the flow front.The second diagnostic denotes the distance, x f c,h = |x− x f |, from the flowfront where |hsim−hasy| reaches 0.6 mm. Again, Chambon et al. find that x f c,h isorder of H; this time, they quote values between 1.5H and 2H for their tests. Bycontrast, in our C2 simulation, x f c,h = |hsim− hasy| ≈ 0.3H, whereas the differ-ence in flow depths never reaches such a threshold for C5, always being less than0.3 mm.All the experimental or asymptotic results described above relate to relativelyshallow surges. As the belt speed increases, however, the surge shortens and deep-ens, leading to profiles like those shown in figure 4.9(a). In these cases, the aspectratio of the flow profile is O(1), with the fluid beginning to climb up the (no-slip)back wall. Raising the belt speed still further leads to a sudden catastrophic over-turning event that interrupts the passage to a steady equilibrium, as illustrated infigure 4.9(b). In this case, the fluid climbs up the wall before collapsing down inthe manner of a breaking wave; bubbles of ambient fluid become entrained intothe surge and the reliability of the simulation is quickly lost, leading us to termi-nate the computation before any convergence to a steady state. At this stage it isnot clear whether the overturn heralds the loss of the steady state and the onset ofa continued cascading flow. As far as we are aware, this type of dynamics has notyet been observed experimentally.115Figure 4.9: Flow profiles for a simulation with θ = 14.6◦, τY = 6 Pa, κ =6.65 Pa sn and n = 0.405 (the experiment C PIV of ), for (a) ub =0.2, 0.4, 0.8, 1.2 and 1.4 m/s, in the steady state, and (b) ub = 1.6 m/sat a succession of times, starting from the initial profile shown by thedashed line (t = 0, 0.06, 0.08, 0.11, 0.17 and 0.19s). In (a), the profilefor ub = 1.4 m/s is shown by the solid line and several streamlines arealso plotted (dotted lines).4.4 Concluding remarksWe have conducted a theoretical study of viscoplastic surges down an inclinedsurface, combining asymptotic analysis with numerical simulations. We have fo-cussed on the steady states reached in frames moving at constant speed down the116slope, although we have also reported a situation in which such steady surges donot appear to be attainable and an unsteady cascading state is reached instead.The numerical computations of shallow surges mostly agree with the asymptoticpredictions, more so than the experiments by Chambon et al.[22, 23, 36], whichare less comparable with the asymptotics though still in broad agreement. Thatdiscrepancy between the experiments and asymptotics is not therefore the resultof non-shallow flow effects, but must originate elsewhere.The computations confirm the phenomenology expected for a surge, namelythat there is an upstream sheet flow with a rigid plug that breaks as one movesdownstream due to the build up of the extensional stress across the plug. Thisleaves a weakly yield zone atop the fluid, the pseudo-plug, as predicted by stan-dard lubrication theory. The surge eventually steepens and terminates at a rela-tively abrupt flow front.We have chiefly operated in the limit in which inertial effects play little rolein the surge, which is clearly a limitation with regard to many applications inthe geosciences. In particular, we have not catalogued any secondary instabilitiesof the steady surges (such as roll waves ) or found any multiple equilibria,both of which might well appear at higher Reynolds number. We leave suchconsiderations for future work.117Chapter 5Summary and future researchdirections5.1 SummaryThe work carried out in this thesis investigates the fluid dynamics of the free-surface flow of viscoplastic fluid in three geometric contexts: 2D dambreaks, ax-isymmetric slumps and surges on an inclined plane. This research is conductedfrom theoretical, computational and experimental perspectives.For 2D dambreaks, we have investigated the phenomenology of the collapse,the mode of initial failure, and the final shape of the slump. The volume-of-fluidmethod is used to evolve the surface of the viscoplastic fluid, and its rheology iscaptured by either regularizing the viscosity or using an augmented-Lagrangianscheme. We outline a modification to the volume-of-fluid scheme that eliminatesresolution problems associated with the no-slip condition applied on the underly-ing surface. We establish that the regularized and augmented-Lagrangian methodsyield comparable results, except for the stress field at the initiation or terminationof motion. The numerical results are compared with asymptotic theories valid forrelatively shallow or vertically slender flow, and good agreement is achieved.For axisymmetric slumps, we further modified the PLIC scheme in order to118achieve mass conservation. Numerical results are compared with asymptotic anal-yses for shallow gravity currents or slender vertical columns. The critical yieldstress for failure is computed and bounded analytically using plasticity methods.The simulations are compared with experiments either taken from existing liter-ature or performed using Carbopol. The comparison is satisfying for lower yieldstresses, while discrepancies for larger yield stresses suggest that the mechanismof release may affect the experiments.For surges on an inclined plane, through numerical simulation we have discov-ered that, the anatomy of the surge consists of an upstream region that convergesto a uniform sheet flow, and over which a truly rigid plug sheaths the surge. Theplug breaks further downstream due to the build up of the extensional stress act-ing upon it, leaving instead a weakly yielded superficial layer, or pseudo-plug.Finally, the surge ends in a steep flow front that lies beyond the validity of shallowasymptotics. Perturbation theory has been applied to prove the existence of a trueplug in the equilibrium state. The numerical solutions are compared with existingexperiments by , where good agreement is achieved.5.2 Limitations of the studyAlthough we have made a number of advances, as listed above, we must alsoacknowledge some limitations of our computational studies.One restriction is on not providing a physical background of the treatment ofthe over-ridden finger of ambient fluid, as illustrated in chapter 2 and 3. Cur-rently we can only show that the computational results converge better than otherschemes, and the flow profile and velocity agree with asymptotic solutions. How-ever, whether this treatment can be successfully applied to other problems remainsunknown.Another concern is that we only consider simple viscoplastic model, such asBingham or Herschel Bulkley model, while most viscoplastic fluids also showother rheological features, like thixotropy, which also affects the experimentalresults. This might explain why there is a discrepancy between numerical and119experimental results.Moreover, we have ignored the fact that effective slip happens during theslump process. In our study, we only consider either no-slip or free slip boundarycondition. However, our experiment with smooth and rough plates in chapter 3establishes that there must be effective slip. How this is related to the discrepancybetween numerical and experimental results is unknown as well.Finally, we are restricted by the computation ability of current scheme. In ourstudy, we use a fixed mesh for computation. Although our computation has beenproved to be converged, we are not able to deal with complex geometries, like thatcharacterizing the lifting of the container in the slump test and the release of thematerial inside.5.3 Summary of contributionsThe novel significant contributions of this thesis can be summarized as below.• Reliable numerical simulation of 2D or axisymmetric slump of viscoplasticfluid has been performed.• The unresolved layer of the upper-layer fluid is removed by a PLIC correc-tion scheme, which leads the computation to converge, while preserving theno-slip boundary condition, and conserving the mass.• Improved asymptotics of shallow/slender limit, and bounds of critical yieldstress have been derived.• Good agreement between asymptotic and numerical results has been achieved.• Qualitative agreement between numerical results and experiments usingCarbopol has been observed.1205.4 Future research directionsProbing deeper, the results in this thesis also provide a strong foundation for futurework. This section discusses several lines of research arising from this work whichcan be pursued.• The numerical scheme not only can solve slump problems, but it can alsosimulate other multi-layer fluid problems. Applications to multi-layer chan-nel flow , for example, is a good application of the numerical scheme.• Computational efficiency is extremely important for research progress. Cur-rently, the numerical scheme solves the system with a direct solver, which isgood for relatively lower number of grids, but unavailable for higher resolu-tions. The iterative solver, on the other hand, is suitable for high resolutionmesh, but converges slowly or never converge. How to improve the solver toguarantee the convergence of iterative solver is a promising topic to study.Related research has been conducted by .• The physical nature of moving contact lines has been investigated for manyyears [76, 78]. In this thesis, we have implemented artificial surgery toremove the finger of upper-layer fluid. However, the physical backgroundis lacking, especially if the contact line can actually move. Efforts in thisfield are needed.• Adaptive meshing has been proved to improve computing efficiency a lot.Our current numerical scheme supports domain with arbitrary shapes, withfixed rectangular grids. Whether an adaptive mesh can improve the com-putation of viscoplastic fluid is unknown. Related work can be found in.• In our numerical simulation of the slump or surge problem, only viscoplas-tic model is considered. In reality, viscoplastic fluid, like Carbopol, mayexhibit viscoelasticity or thixotropy as well. An elastoviscoplastic model121 could have been useful, and its application in our problem may helpexplain the discrepancy with experimental results.• Certain other effects have been considered negligible in our study, like sur-face tension, inertia, or the ambient fluid. It is worth continuing the studyand investigating their effect on the fluid dynamics when they become dom-inant.122Bibliography S. Afkhami, S. Zaleski, and M. Bussmann. A mesh-dependent model forapplying dynamic contact angles to vof simulations. Journal ofComputational Physics, 228:5370–5389, 2009. → page 59 C. Ancey and S. Cochard. The dam-break problem for herschelbulkleyviscoplastic fluids down steep flumes. Journal of Non-Newtonian FluidMechanics, 158:18–35, 2009. → page 4 N. Andreini, G. Epely-Chauvin, and C. Ancey. Internal dynamics ofnewtonian and viscoplastic fluid avalanches down a sloping bed. Physics ofFluids, 24:053101, 2012. → page 18 N. Balmforth. Viscoplastic asymptotics and other techniques. ViscoplasticFluids: From Theory to Application, CISM. Springer, 2018. → pages19, 149 N. Balmforth and J. Liu. Roll waves in mud. Journal of Fluid Mechanics,519:33–54, 2004. → page 117 N. J. Balmforth and R. V. Craster. A consistent thin-layer theory forBingham plastics. J. Non-Newtonian Fluid Mechanics, 84:65–81, 1999. →pages 4, 36, 135 N. J. Balmforth and R. V. Craster. A consistent thin-layer theory forBingham plastics. J. Non-Newtonian Fluid Mech., 84:65–81, 1999. →page 4 N. J. Balmforth and I. J. Hewitt. Viscoplastic sheets and threads. Journal ofNon-Newtonian Fluid Mechanics, 193:28–42, 2013. → pages 40, 74123 N. J. Balmforth, A. S. Burbidge, R. V. Craster, J. Salzig, and A. Shen.Visco-plastic models of isothermal lava domes. Journal of FluidMechanics, 403:37–65, 2000. → page 145 N. J. Balmforth, R. V. Craster, and R. Sassi. Shallow viscoplastic flow onan inclined plane. J. Fluid Mech., 470:1–29, 2002. → pages 4, 18, 103 N. J. Balmforth, R. V. Craster, A. C. Rust, and R. Sassi. Viscoplastic flowover an inclined surface. J.Non-Newtonian Fluid Mech., 139:103–127,2006. → pages 4, 16, 18, 35, 63, 64, 97, 134, 144, 147 N. J. Balmforth, R. V. Craster, P. Perona, A. C. Rust, and R. Sassi.Viscoplastic dam breaks and the Bostwick consistometer. J.Non-Newtonian Fluid Mechanics, 142:63–78, 2007. → pages 2, 34 N. J. Balmforth, N. Dubash, and A. C. Slim. Extensional dynamics ofviscoplastic filaments: II. Drips and bridges. J. Non-Newtonian FluidMech., 165:1147–1160, 2010. → pages 40, 134, 136, 137 N. J. Balmforth, I. Frigaard, and G. Ovarlez. Yielding to stress: Recentdevelopments in viscoplastic fluid mechanics. Ann. Rev. Fluid Mechanics,46:121–146, 2014. → pages 1, 56 N. Balmfoth, A. Burbidge, R. Craster, J. Salzig, and A. Shen. Visco-plasticmodels of isothermal lava domes. Journal of Fluid Mechanics, 403:37–65,2000. → page 4 B. Bates and C. Ancey. The dam-break problem for eroding viscoplasticfluids. Journal of Non-Newtonian Fluid Mechanics, 243:64–78, 2017. →page 18 A. Bouvet, E. Ghorbel, and R. Bennacer. The mini-conical slump flow test:Analysis and numerical study. Cement and Concrete Research, 40:1517–1523, 2010. → page 18 J. Chamberlain, J. Sader, K. Landman, D. Horrobin, and L. White.Incipient failure of a circular cylinder under gravity. International Journalof Mechanical Sciences, 44:1779–1800, 2002. → page 72 J. Chamberlain, S. Clayton, K. Landman, and J. Sader. Experimentalvalidation of incipient failure of yield stress materials under gravitationalloading. Journal of Rheology, 47(6):1317–1329, 2003. → page 70124 J. A. Chamberlain, J. E. Sader, K. A. Landman, and L. R. White. Incipientplane-strain failure of a rectangular block under gravity. Int. J. ofMechanical Sciences, 43:793–815, 2001. → pages17, 42, 45, 46, 48, 119, 138, 140 J. A. Chamberlain, D. J. Horrobin, K. A. Landman, and J. E. Sader. Upperand lower bounds for incipient failure in a body under gravitational loading.J. Applied Mechanics, 71:586–589, 2004. → pages17, 18, 20, 42, 44, 45, 70, 73, 138, 142 G. Chambon, A. Ghemmour, and D. Laigle. Gravity-driven surges of aviscoplastic fluid: An experimental study. Journal of Non-Newtonian FluidMechanics, 158:54–62, 2009. → pages 18, 92, 96, 111, 117 G. Chambon, A. Ghemmour, and M. Naaim. Experimental investigation ofviscoplastic free-surface flows in a steady uniform regime. Journal of FluidMechanics, 754:332–364, 2014. → pages xvi, 18, 92, 96, 111, 116, 117 W.-F. Chen, M. W. Giger, and H.-Y. Fang. On the limit analysis of stabilityof slopes. Soils and Foundations, 9(4):23–32, 1969. → pages 43, 141 S. Clayton, T. G. Grice, and D. V. Boger. Analysis of the slump test foron-site yield stress measurement of mineral suspensions. Int. J. Miner.Process, 70:3–21, 2003. → pages 18, 76, 90, 91, 150 S. Cochard. Measurements of time-dependent free-surface viscoplasticflows down steep slopes. PhD thesis, E´cole Polytechnique Fe´de´rale deLausanne, 2007. → pages 51, 52 S. Cochard and C. Ancey. Experimental investigation of the spreading ofviscoplastic fluids on inclined planes. Journal of Non-Newtonian FluidMechanics, 158:73–84, 2009. → page 18 P. Coussot and S. Boyer. Determination of yield stress fluid behaviour frominclined plane test. Rheologica Acta, 35(6):534–543, 1995. → page 18 P. Coussot and S. Proust. Slow, unconfined spreading of a mudflow. Journalof Geophysical Research: Solid Earth, 101(B11):25217–25229, 1996. →page 18125 P. Coussot, S. Proust, and C. Ancey. Rheological interpretation of depositsof yield stress fluids. Journal of Non-Newtonian Fluid Mechanics, 66(1):55–70, 1996. → page 18 M. R. Davidson, N. H. Khan, and Y. L. Yeow. Collapse of a cylinder ofbingham fluid. ANZIAM Journal, 42:499–517, 2000. → pages 18, 77, 78 D. De Kee, R. Chhabra, M. Powley, and S. Roy. Flow of viscoplastic fluidson an inclined plane: evaluation of yield stress. Chemical engineeringcommunications, 96(1):229–239, 1990. → page 18 N. Dubash, N. J. Balmforth, A. C. Slim, and S. Cochard. What is the finalshape of a viscoplastic slump? J. Non-Newtonian Fluid Mech., 158:91–100, 2009. → pages 6, 18, 38, 48, 50, 51, 52, 135 C. Famiglietti and J. Prevost. Solution of the slump test using a finitedeformation elasto-plastic drucker-prager model. International journal fornumerical methods in engineering, 37(22):3869–3903, 1994. → page 18 M. Fortin and R. Glowinski. Augmented Lagrangian Methods:Applications to the Numerical Solution of Boundary-Value Problems.North-Holland, 2006. → pages 12, 22, 57, 96, 131 P. Freydier, G. Chambon, and M. Naaim. Experimental characterization ofvelocity fields within the front of viscoplastic surges down an incline.Journal of Non-Newtonian Fluid Mechanics, 240:56–69, 2017. → pagesxi, xv, xvi, 18, 92, 96, 110, 111, 112, 113, 114, 117 I. Frigaard and D. Ryan. Flow of a visco-plastic fluid in a channel of slowlyvarying width. Journal of Non-Newtonian Fluid Mechanics, 123:67–83,2004. → pages 19, 92, 102 I. A. Frigaard and C. Nouar. On the usage of viscosity regularisationmethods for visco-plastic fluid flow computation. J. Non-Newtonian FluidMech., 127:1–26, 2005. → pages 17, 60 J. Gao and A. Fourie. Spread is better: An investigation of the mini-slumptest. Minerals Engineering, 71:120–132, 2014. → pages 18, 76, 77, 78, 80 D. Gueyffier, J. Li, A. Nadim, R. Scardovelli, and S. Zaleski.Volume-of-fluid interface tracking with smoothed surface stress methods126for three-dimensional flows. Journal of Computational Physics, 152:423–456, 1999. → page 59 J. Heyman. The stability of a vertical cut. International Journal ofMechanical Sciences, 15(10):845–854, 1973. → page 43 R. Hill. The mathematical theory of plasticity, volume 11. Oxforduniversity press, 1998. → page 65 A. Hogg and G. Matson. Slumps of viscoplastic fluids on slopes. Journal ofNon-Newtonian Fluid Mechanics, 158:101–112, 2009. → page 4 S. Hormozi, K. Wielage-Burchard, and I. Frigaard. Entry and start upeffects in visco-plastically lubricated viscous shear flow in pipe. J. FluidMechanics, 673:432–467, 2011. → pages 22, 26, 57, 96 S. Hormozi, K. Wielage-Burchard, and I. Frigaard. Multi-layer channelflows with yield stress fluids. Journal of Non-Newtonian Fluid Mechanics,166:262–278, 2011. → page 121 G. Hulme. The interpretation of lava flow morphology. GeophysicalJournal International, 39(2):361–383, 1974. → page 18 H. E. Huppert. The propagation of two-dimensional and axisymmetricviscous gravity currents over a rigid horizontal surface. J. Fluid Mech., 121:43–58, 1982. → pages 144, 145 M. Jalaal, N. J. Balmforth, and B. Stoeber. Slip of spreading viscoplasticdroplets. Langmuir, 31(44):12071–12075, 2015. → pages 54, 81 M. A. M. A. Khatib and S. D. R. Wilson. Slow dripping of yield-stressfluids. Journal of Fluids Engineering, 127:687–690, 2005. → page 137 G. G. Lipscomb and M. M. Denn. Flow of bingham fluids in complexgeometries. Journal of Non-Newtonian Fluid Mechanics, 14:337–346,1984. → page 5 K. F. Liu and C. C. Mei. Slow spreading of a sheet of Bingham fluid on aninclined plane. J. Fluid Mech., 207:505–529, 1989. → pages4, 18, 64, 92, 103, 149127 K. F. Liu and C. C. Mei. Approximate equations for the slow spreading of athin sheet of Bingham plastic fluid. Physics of Fluids A: Fluid Dynamics,2:30, 1990. → pages 16, 35, 134, 135 Y. Liu, N. J. Balmforth, S. Hormozi, and D. R. Hewitt. Two dimensionalviscoplastic dambreaks. Journal of Non-Newtonian Fluid Mechanics, 238:65–79, 2016. → pagesxviii, 21, 55, 59, 62, 63, 64, 67, 68, 71, 93, 95, 96, 102, 103, 144, 152, 153, 160, 163 Y. Liu, N. Balmforth, and S. Hormozi. Axisymmetric viscoplasticdambreaks and the slump test. Journal of Non-Newtonian Fluid Mechanics,258:45–57, 2018. → pages 55, 93, 95, 96, 152, 153, 160 A. V. Lyamin and S. W. Sloan. Lower bound limit analysis using non-linearprogramming. Int. J. Numerical Methods in Engineering, 55:573–611,2002. → pages 42, 43, 70 A. V. Lyamin and S. W. Sloan. Upper bound limit analysis using linearfinite elements and non-linear programming. Int. J. Numer. and Analyt.Methods in Geomechanics, 26:181–216, 2002. → pages 42, 43, 70 J. L. M. Renardy, Y. Renardy. Numerical simulation of moving contact lineproblems using a volume-of-fluid method. Journal of ComputationalPhysics, 171:243–263, 2001. → pages 13, 153 G. P. Matson and A. J. Hogg. Two-dimensional dam break flows ofHerschel-Bulkley fluids: The approach to the arrested state. J.Non-Newtonian Fluid Mech., 142:79–94, 2007. → pages 4, 48, 147 J. F. Nye. The flow of glaciers and ice-sheets as a problem in plasticity.Proc. R. Soc. Lond. Ser. A, 207:554–572, 1951. → pages 6, 48 D. I. Osmond and R. W. Griffiths. The static shape of yield strength fluidsslowly emplaced on slopes. Journal of Geophysical Research: Solid Earth,106(B8):16241–16250, 2001. → page 18 N. Pashias, D. V. Boger, J. Summers, and D. J. Glenister. A fifty centrheometer for yield stress measurement. Journal of Rheology, 40:1179,1996. → pages 2, 17, 18, 34, 40, 55, 68, 76128 J. Pastor, T.-H. Thai, and P. Francescato. Interior point optimization andlimit analysis: an application. Communications in Numerical Methods inEngineering, 19(10):779–785, 2003. → pages 42, 43 J.-M. Piau. Axisymmetric slump and spreading of cohesive plastic softmaterials: a yield stress measurement by consisto-rheometry. Journal ofRheology, 49(6):1253–1276, 2005. → pages 18, 19, 77, 78 W. Prager and P. G. Hodge. Theory of Perfectly Plastic Solids. Dover; NewEd edition, 1968. → pages 50, 54, 72, 140 A. Putz, I. Frigaard, and D. M. Martinez. The lubrication paradox and useof regularisation methods for lubrication flows. Journal of Non-NewtonianFluid Mechanics, 163:62–77, 2009. → pages 5, 19 M. Renardy, Y. Renardy, and J. Li. Numerical simulation of moving contactline problems using a volume-of-fluid method. Journal of ComputationalPhysics, 171:243–263, 2001. → pages 26, 59 N. Roussel and P. Coussot. Fifty-cent rheometer for yield stressmeasurements: From slump to spreading flow. Journal of Rheology, 49:705, 2005. → pages 18, 90, 91 N. Roussel and P. Coussot. Fifty-cent rheometer for yield stressmeasurements: From slump to spreading flow. Journal of Rheology, 49:705–718, 2005. → pages 2, 34, 40 N. Roussel, A. Gram, M. Cremonesi, L. Ferrara, K. Krenzer,V. Mechtcherine, S. Shyshko, J. Skocec, J. Spangenberg, O. Svec, L. N.Thrane, and K. Vasilic. Numerical simulations of concrete flow: Abenchmark comparison. Cement and Concrete Research, 79:265–271,2016. → pages 18, 55, 66 A. W. Saak, H. M. Jennings, and S. P. Shah. A generalized approach for thedetermination of yield stress by slump and slump flow. Cement andConcrete Research, 34:363–371, 2004. → pages 18, 55, 76 J. E. Sader and M. R. Davidson. Scaling behavior for gravity induced flowof a yield stress material. Journal of Rheology, 49(1):105–112, 2005. →page 18129 P. Saramito. A new elastoviscoplastic model based on the herschelbulkleyviscoplastic model. Journal of Non-Newtonian Fluid Mechanics, 158:154–161, 2009. → page 122 P. Saramito. A damped newton algorithm for computing viscoplastic fluidflows. Journal of Non-Newtonian Fluid Mechanics, 238:6–15, 2016. →page 121 P. Saramito and A. Wachs. Progress in numerical simulation of yield stressfluid flows. Rheologica Acta, 56:211–230, 2017. → page 121 R. Sayag, S. S. Pegler, and M. G. Worster. Floating extensional flows.Physics of Fluids, 24(9):091111, 2012. → page 87 J. Snoeijer and B. Andreotti. Moving contact lines: Scales, regimes, anddynamical transitions. Annual Review of Fluid Mechanics, 45:269–292,2013. → page 121 L. Staron, P. Lagree, P. Ray, and S. Popinet. Scaling laws for the slumpingof a Bingham plastic fluid. J. Rheol., 57:1265–1280, 2013. → pages16, 30, 51, 53 Y. Sui, H. Ding, and P. Spelt. Numerical simulations of flows with movingcontact lines. Ann. Rev. Fluid Mech., 46:97–119, 2014. → pages 29, 121 B. van Leer. Towards the ultimate conservative difference scheme. V. Asecond-order sequel to Godunov’s method. Journal of ComputationalPhysics, 142(1):101–136, 1979. → page 26 A. Wachs. Numerical simulation of steady bingham flow through aneccentric annular cross-section by distributed lagrange multiplier/fictitiousdomain and augmented lagrangian methods. Journal of Non-NewtonianFluid Mechanics, 142:183–198, 2007. → pages 57, 96 I. Walton and S. Bittleston. The axial flow of a bingham plastic in a narroweccentric annulus. Journal of fluid mechanics, 222:39–60, 1991. → pages5, 19 K. Wielage-Burchard and I. Frigaard. Static wall layers in plane channeldisplacement flows. J. Non-Newtonian Fluid Mechanics, 166:245–261,2011. → pages 22, 26130Appendix ATwo-dimensional viscoplastic dambreakA.1 Further numerical notesA.1.1 Parameter settings and other detailsPELICANS exploits a Galerkin finite element method to solve a weak form of theequations of motion using iteration. For the regularized constitutive model, we setε = 10−8 in (2.7). At each step, the regularized viscosity is computed using thevelocity field from the previous step, and iteration is continued until the L2−normof the velocity change over the entire spatial domain falls below 10−6/√n, wheren is the number of finite elements.As detailed in , the augmented-Lagrangian method introduces additionalvariables to solve iteratively the weak formulation of the equations of motionwhilst avoiding the singular viscosity arising at the yield surfaces and the stressindeterminacy of the plugs. The iterative scheme includes a relaxation parameterr which we select to be equal to B. Iteration is continued until the larger of theL2-norms of the change in the velocity field or γ˙ became less than 10−4/√n.131Based on resolution studies, we found that grids with ∆x = 0.01 and ∆z =0.005 were sufficient for mesh convergence in problems for which slumps wereorder one aspect ratio or shallow (once the no-slip boundary condition on thebase had been modified). For slender columns, we found it sufficient to take∆x = 5×10−4 with ℓx = 0.1 and distribute the vertical mesh non-uniformly suchthat the grid intervals formed a geometric series starting with 5× 10−4 at thebottom and ending at 0.005 at the top boundary z = ℓz = 1.1.A.1.2 The failure computation for Re = t = 0For the initial failure mode, we used an alternative numerical scheme that solvedthe equations at t = 0with Re≡ 0. The scheme employed an augmented-Lagrangianmethod to solve the Stokes equations over the domain shown in figure 1. In viewof our interest in yield stresses close to Bc, where the velocity field is small and theviscous stress of the outer fluid likely irrelevant, we simplified the computation bytaking a viscosity ratio of unity.At each step of the iteration procedure, the linear Stokes and continuity equa-tions were converted to a biharmonic equation which was solved using a Fouriersine transform in the x direction and second-order finite differences in z. Forrectangular slumps, the discontinuity in the yield stress between the two fluidswas imposed directly on the finite difference grid; for triangular shaped slumps,convergence was much improved by smoothing the discontinuity over a few gridpoints. In both cases, the forcing term in the biharmonic equation that arisesfrom the discontinuity in density between the two fluids was dealt with exactly inFourier space.As in our other scheme, the size of the domain was chosen to be sufficientlylarge so as to have a negligible effect on solutions (in most results, lz = 4/3 andlx = 4X0/3). We selected a vertical grid and truncation of the horizontal Fourierseries such that the resolution was ∆z = 1/999 and ∆x = X0/768; we establishedthat the solutions were independent of this choice. In the augmented-Lagrangianscheme, we again chose the relaxation parameter r = B, and the solutions were132considered to be converged when the maximum change in the strain-rate invariant,γ˙ , had fallen below 10−10.A.1.3 Thickness of the over-ridden fingerFor a relatively thin finger of depth ζ (x, t), flow is driven primarily by the shearstress on the interface,τζ ≈ µRuζζ, (A.1)where uζ is the horizontal velocity of the interface and µR ∼ µ2/µ1 is the viscosityratio. Conservation of mass implies∂ζ∂ t≈−12∂∂x(uζ ζ)≈− τζµRζ∂ζ∂x, (A.2)if τζ remains roughly independent of x. The shear stress τζ is of order one atthe beginning of a collapse and we use µ2/µ1 = 10−3. Hence uζ/ζ = O(103).Thus the finger effectively lubricates the overlying viscoplastic current until itsthickness becomes comparable to our grid spacing (∆z = 1/320). Moreover, thesolution of (A.2) then indicates that the finger thins like t−1. In other words, theeffective lower boundary condition on the viscoplastic current only reduces to noslip when the finger becomes difficult to resolve, and from then on, the resolutionproblem steadily worsens.The critical detail of the volume-of-fluid code regarding the finger is that ittreats each grid cell as a mixture with the rheology in (2.3) and (2.4) (or (2.7)). Inthe limit of low inertia, the stress state is dictated largely by the geometry of theslump and dominated by the shear stress at the base. Thus, over the bottom gridcell where c = c0,∂u∂ z=(τxz−Bc0)c0+(1− c0)µR . (A.3)The average horizontal speed over this cell is thenuM ∼ ∆z2c0(τxz−Bc0), (A.4)133if c0 lies well away from its limits and µR ≪ 1. On the other hand, when c0 takessuch a value, the interpretation is that the interface can be contained within thelowest grid cell. In that situation, a genuine sharp interface at 0 < z = ζ < ∆z,would imply∂u∂ z={τxz−B ζ < z < ∆z,µ−1R τxz 0< z < ζ .(A.5)It follows that the average horizontal speed should beuS ∼ τxzζ2µR∆z(2∆z−ζ ) (A.6)for µR ≪ 1. Evidently, uM/uS = O(µR). In other words, when the interface entersthe lowest grid cell, the treatment of the fluid as a mixture grossly underestimatesthe speed with which the finger will be advected horizontally; the finger is there-fore not swept away fast enough, remains too thick, and overly lubricates theviscoplastic gravity current. To cure this problematic feature, the finger must beremoved.A.2 Shallow flowIn this and the next appendix we ignore the upper viscous fluid and consider aspreading viscoplastic current with a stress-free surface. We summarize analysisand results that are based on existing literature [13, 52], highlighting any relevantnew developments; the reader is encouraged to consult the original references foradditional details of the basic theory.For relatively shallow flow (for which vertical gradients dominate horizon-tal gradients), the pressure becomes largely hydrostatic and only the shear stressfeatures in the main force balance [11, 52]. Thus,p = h− z and τxz =−hx(h− z), (A.7)134where z = h(x, t) is the position of the free surface. Here, and throughout thisand the following appendices, we use subscripts of x, z and t as shorthand for thecorresponding partial derivatives (except in the case of the stress components).The velocity field is now given byu =−12hx×{(2Y − z)z 0≤ z < Y,Y 2 Y ≤ z≤ h, (A.8)where z = Y = h+B/hx (for hx < 0) is where the leading-order shear stress fallsbelow B. This latter level is not a true yield surface because, although the overly-ing velocity field is plug-like, the fluid remains in extension and weakly yielded. Exploiting the depth-integrated expression of mass conservation, the problemthen reduces to solving the thin-layer equation ,ht =16∂∂x[(3h−Y )Y 2hx]. (A.9)Note that modifications are needed where Y = h+B/hx < 0, which signifies thatthe fluid is not sufficiently stressed to deform. This true yield criterion can beincorporated into the formulation simply by definingY (x, t) =Max(h+Bhx,0). (A.10)For our triangular initial condition, with h = 2(1− x/X0), the yield criterion in-dicates that the fluid will not collapse anywhere when B > 4/X0; by contrast, thevertical edge of the rectangular block ensures the fluid always collapses in theshallow limit. The final state is given by Y → 0, which leads to (2.8).The shallow-layer theory is the leading order of an asymptotic expansionwhich, the current dimensionless scalings, corresponds to the limit B ≪ 1. Forthe final shape one can go further with the analysis and compute higher-order cor-rections in the effort to extend the accuracy of the approximation. In particular,following the analysis in but bearing in mind that the slump comes to rest in135a state of horizontal expansion, we findp≈ h− z−√B2− v2, v =−hx(h− z), (A.11)τxx ≈√B2− v2− vτ1√B2− v2 , τxz ≈ v+ τ1, (A.12)whereτ1 =− ∂∂x[1hx(v√B2− v2+B2 sin−1 vB)]. (A.13)Imposing the lower boundary condition, τxz = B at z = 0, and integrating in x thengives12h2−h√B2−h2h2x−B2hxsin−1hhxB=C−Bx, (A.14)where C is an integration constant. Evaluating the higher-order corrections in(A.14) (i.e. the second and third terms on the left-hand side) using the leading-order approximation hhx =−B leads to (2.9) with C = BX∞−pi2B2/8.A.3 Slender columnsWhen the column of viscoplastic fluid remains slender throughout its collapse,we may use the thin-filament asymptotics outlined by . The key detail is thatthe horizontal gradients are much larger than the vertical ones and the verticalvelocity greatly exceeds the horizontal speed. Moreover, because the sides of thecolumn are stress free, shear stresses must remain much smaller than the exten-sional stresses and the vertical velocity cannot develop significant horizontal shearand remains largely plug-like. These consideration indicate that (see )w≈W (z, t), u≈−xWz, (A.15)andp≈ τxx =−τzz ≈ B−2Wz, (A.16)136the latter of which follows from the leading-order horizontal force balance (whichis (τxx− p)x ≈ 0) and constitutive law (given γ˙ ≈ 2|Wz|). The width-averagedmass conservation equation and vertical force balance then imply ξt +(ξW )z = 0 and ξ +2(ξ p)z = 0, (A.17)where x = ξ (z, t) is the local half-width.The equations in (A.17) can be solved analytically by transforming to La-grangian coordinates (a, t), where a denotes initial vertical position (i.e. themethod of characteristics; cf. ). For the rectangular or triangular blocks, wehave the initial condition ξ (a,0) = X0 or X0(1− 12a), respectively. The transfor-mation then indicates thatξ (a, t)∂ z∂a={X0X0(1− 12a)and ξt(a, t) =−ξWz. (A.18)Henceξ p = 12X0(1−a/a∗), (A.19)given that p = 0 at the top of the column where a = a∗ = 1 or 2. If the fluid isyielded, the constitutive law impliesp = B−2Wz = B+ 2ξtξ. (A.20)After a little algebra and the use of the bottom boundary condition z(a= 0, t)=0, we findξ (a, t)ξ (a,0)= E +1−E2B(1− aa∗), E = e−t/2B, (A.21)andz =2a∗B1−E log[1−E +2BE(1−E)(1−a/a∗)+2BE]. (A.22)The yield condition in this limit becomes p < −B, or a < a∗(1− 2B), which137translates toz < Z(t) =2a∗B(1− e−Bt/2) log[1− (1−2B)e−Bt/22B]. (A.23)The column does not therefore yield anywhere when B > 12. If fluid does yield,the base spreads out to a distance ξ (0, t)→ X∞ = X0/(2B), and the column fallsto a height H∞ = 2a∗B[1− log(2B)]. The yield condition and runout are the samefor both the rectangle and triangle because, in the slender limit, all that matters isthe weight of the overlying fluid. Sample solutions are shown in figure A.1.The main failing of the slender asymptotics is that the no-slip boundary con-dition is not imposed: the fluid slides freely over the base, leading to the collapsedcolumn being widest at z = 0, whereas the fluid actually rolls over the base in atank-treading motion (cf. figure 2.8). This failing must be remedied by adding aboundary layer at the bottom (with different asymptotic scalings). In any event,we attribute the lack of agreement between the slender asymptotics and the nu-merical results in figure 2.8 to this feature.A.4 Plasticity solutionsIn this appendix, we summarize slipline and bound computations based closely onexisting work in plasticity theory [20, 21], emphasizing some minor generaliza-tions for triangular initial shapes and circular failure curves. The reader is referredto [20, 21] for further details of the basic developments.A.4.1 Slipline fieldsThe slipline solutions of Chamberlain et al. begin from the side free surfacewhere the stress field is specified and are constructed as follows: settingP≡ pB+zB& (τxx,τxz) = B(−sin2θ ,cos2θ), (A.24)1380 0.05 0.100.511.52x(a) B=0.10.02 0.04x(b) 0.3Figure A.1: Slender asymptotic solutions for (a) B = 0.1 and (b) B = 0.3,for X0 = 0.025, at times t = 0, 4, 9, 16, 36, 100 and 1000. Collaps-ing rectangles (triangles) are shown by solid (dotted) lines; the dotsindicate the final profiles.139the side boundary conditions implyP = 1+ zˇ & θ =3pi4+φ (A.25)on xˇ = −Szˇ, where (xˇ, zˇ) = B−1(x,z), and (S,φ) = (0,0) for the rectangle and(X0/2, tan−1 12X0) for the triangle. On the α−characteristics,P+2θ = constant,dzdx= tanθ ; (A.26)for the β−characteristics,P−2θ = constant, dzdx=−cotθ . (A.27)Beginning from the section of the side, 0≤ zˇ≤ zˇP, with zˇP a parameter, the char-acteristics can be continued into the fluid interior using a standard finite differencescheme to solve the characteristics equations in (A.26)-(A.27) [20, 64]. Below theresulting web, an expansion fan is then added that spreads out from the base point(x,z)= (X0,0)with θP≤ θ ≤ 3pi/4+φ , where θP is a second parameter (cf. figureA.2). The combined slipline field is then continued to x= 0, or xˇ=−X0/B. At thispoint, the two characteristics that bound the complete slipline field must cross andterminate with θ = 3pi/4, in view of the symmetry conditions there. This selectsthe two parameters zˇP and θP. Finally, along the uppermost α−characteristic, thetotal vertical force must match the weight of the overlying plug, which translatesto imposing the condition,X0− 12SB2zˇ2P = B2∫ zˇP0cos2θ dzˇ+B2∫ 0X0/B(P− sin2θ)dxˇ, (A.28)and determines the relation B = Bcrit(X0) (as plotted in figure 2.10). Examples ofthe slipline field are shown in figure A.2. Note that the sliplines of the two familiesbegin to cross over one another near the top of the expansion fan if X0 is increasedpast some threshold . At that stage a curve of stress discontinuity must be1400 0.1 0.200.10.20.30.40.50.60.70.80.91yx(a)X0=0.20.1 0.2 0.3 0.4 0.5x(b) X0=0.50 0.5 100.20.40.60.811.21.41.61.82x(c) X0=1Figure A.2: Slipline solutions for a rectangle with (a) X0 = 0.2 and (b) X0 =0.5, and a triangle with X0 = 1. The dashed and dotted lines indicatethe lower bound and its improvement of A.4.2.introduced to render the slipline field single-valued. We avoid incorporatingthis detail here and only provide slipline solutions without a discontinuity, whichlimits the triangle data in figure 2.10(b) to X0 < 1. At still higher X0 > 2, theconstruction fails for the triangle altogether because the α−characteristic fromthe base of the free surface proceeds into z < 0.A.4.2 Simple failure modesFor the circular failure surface of a relatively wide initial rectangle, we refer thereader to existing literature (e.g. ). Here, we summarize the computation for141slender rectangles and triangles.We first consider the case where failure occurs on straight lines, as in .As illustrated in figure 2.11(b–c), we introduce two lines of failure, with slopestanγ and tanζ , that divide the initial block up into a lower stationary triangle, anintermediate triangle that slides out sideways, and the residual overlying materialthat falls vertically. When the downward speed of the top is W , the continuity ofthe normal velocity across each failure line demands that the intermediate triangleslides out parallel to the lower failure line with velocity U(cosγ,−sinγ), whereU = W secζ/(tanγ + tanζ ). Let AI and AII denote the areas of the intermediatetriangle and top block and LI and LII be the lengths of the lower and upper failurelines, both respectively. Equating the plastic dissipation across the failure lineswith the release of potential energy then furnishes (in our dimensionless notation)B(ULI +WLII secζtanγ + tanζ)= AIU sinγ +AIIW. (A.29)Geometric considerations allow us to express all the variables in terms of X0 andthe two angles γ and ζ . Equation (A.29) can therefore be formally written in thesuggestive form, B = B(γ,ζ ;X0). We then optimize the function B(γ,ζ ;X0) overall possible choices of the two angles (γ,ζ ) to arrive at the bound Bc(X0). It turnsout that Bc =12tanγ = 12tanζ = 12(√1+X20 −X0) for the rectangular block .In the case of the triangle, tanζ =√2(1+ tan2 γ)− tanγ , leaving a straightfor-ward algebraic problem to solve for the optimal γ and Bc, with solutions shownin figure 2.11(c) and 2.10(b). The failure lines of these bounds are compared tothe sample slipline fields in figure A.2, illustrating the manner in which the boundattempts to capture the actual plastic deformation.The streamlines of the numerically computed failure modes suggest that thepreceding bounds might be improved if the triangle at the side were allowed torotate out of position rather than slide linearly. In this situation, the failure sur-faces become circular arcs rather than straight lines which complicates the formof the power balance corresponding to (A.29) and the geometrical constraints.142Three optimization parameters are required to define the circular arcs; we use thelocal slopes at the bottom corner, sα , and midline, sβ and s, as illustrated in figure2.11(b,c). The optimization problem can then be continued through with the helpof the computer. We use the built-in function FMINSEARCH of Matlab to per-form the optimization of B(sα ,sβ ,s;X0) and improve the bounds on Bc(X0). Thecircular failure arcs corresponding to the three slipline solutions of figure A.2 areagain included in that picture.Note that the bounds for the triangle predict that sα falls to zero for X0 > 2.8with a straight failure surface and X0 > 1.2 for rotational failure. For wider initialstates, this parameter must then be removed from the optimization, which makesthe bounding procedure less effective. A more general and effective construction,that retains sα as a parameter, is to allow the lower circular failure arc to intersectthe base for x < X0, but not pass through that surface, and then continue beyond.That is, we allow the arc to proceed through a minimum at z= 0 and then intersectthe side surface at a finite height (cf. figure 2.13(b,c)). This extension permitscomputations of improved bounds for arbitrarily wide triangles and is plotted infigure 2.10(b). Figure 2.13(b,c) illustrates how the resulting arcs compare wellwith the computed failure modes for moderate width. Even for the widest trianglewith X0 = 8, the bound (Bc > 0.1635) is close to the computed value of Bc ≈0.1642. For X0≫ 1, the bound converges to Bc > 3/X0. By contrast, the shallow-layer asymptotics predict failure for Bc ∼ 4/X0 (see A.2), indicating that there isfurther room for improvement in this limit.143Appendix BAxisymmetric viscoplasticdambreaks and the slump testB.1 Numerical convergence studyFigure B.1 shows the results of a resolution study for the runout of a Newtonianslump with R = 1 and varying grid size. The first panel presents results for thePLIC scheme with the non-conservative correction of ; those with the con-servative correction are plotted in the second panel. Both are compared with theresults of lubrication theory [11, 47]. Also included in figure B.1(b)-(c) are corre-sponding results for the runout of a Bingham slump with B = 0.05 and a compar-ison of the fluid interfaces of the various solutions at t = 250, all for the case ofthe PLIC scheme with the conservative correction (and both values of B).Figure B.2 shows further details of the resolution studies with B = 0 and 0.05,plotting the runout and central depth of the slump at t = 250 for various compu-tations with differing resolution. Computations both with and without the correc-tion schemes are presented; the results for the original PLIC algorithm convergemuch more slowly than those for the corrected schemes, with the conservativecorrection scheme being superior. For B = 0.05, the three different treatments ofthe interface are shown for a computation using the regularization method to deal1441.52R(t) (a)1/401/801/1601/320Shallow0 50 100 150 200 25011.52tR(t)(b)rz(c)0 0.5 1 1.5 200.20.4Figure B.1: Computations with varying grid size (as indicated). Panels (a)and (b) show the flow front as a function of time for a Newtonianslump with R = 1 using the PLIC scheme with the non-conservativeand conservative corrections, respectively. Also inclued in (b) are cor-responding results for a Bingham slump with B = 0.05 (green curves).Panel (c) shows the interfaces at t = 250 of the computations in (b)(with inset showing magnifications near the flow fronts). The red cir-cles show leading-order shallow-layer solutions [9, 47].14540 80 160 3201.81.9Grid points pe r uni t l ength (a) R (250)B=0.0540 80 160 3200.460.48 (b) H (250)B=0.0540 80 160 3200.230.24 B=040 80 160 3202.42.6 B=0Figure B.2: (a) Runout and (b) central depth at t = 250 with varying gridsize for a Newtonian (B = 0; top) and Bingham slump (B = 0.05;bottom) using various interface-tracking schemes and Navier-Stokessolvers: regularization with the PLIC algorithm including the non-conservative (crosses), conservative (stars) or no correction (circles)scheme, or using the augmented-Lagrangian method with PLIC andthe conservative correction (squares). There is no difference betweenthe Navier-Stokes solvers for B = 0.146with the yield stress. Also shown is a series of computations using the augmented-Lagrangian method and the PLIC scheme with a conservative correction, whichillustrates how the two methods for the dealing with the yield stress yield similarresults.B.2 Higher-order shallow asymptoticsWhen flow becomes arrested, throughout the bulk of the fluid the constitutivemodel reduces to the plasticity law τi j = Bγ˙i j/γ˙ . In combination with the rescaledequations in (3.7) and (3.10), one may then deduce the leading-order stress com-ponents, τˇrz = Bˇ(1− z/h0),(τˇrr, τˇθθ ) = Bˇ(uχ ,χ−1u)√1− (1− z/h0)2√u2χ +u2/χ2+uuχ/χ,(B.1)where h0 =√1−χ/χ∞ and u are the leading-order profile and radial velocity,with χ∞ ≡ εr∞. If we instead neglect only the O(ε2) terms in (3.7)-(3.8), we findthat p+ ετˇzz = h− z+O(ε2). An integral of the radial force balance over thedepth of the fluid then furnishes the equationhhχ + Bˇ =3ε2ddχ[∫ h00(τˇrr + τˇθθ )dz]+ε2χ2ddχ[χ2∫ h00(τˇrr− τˇθθ )dz]. (B.2)To correct the profile, we evaluate the O(ε) terms on the right of (B.2) using theleading-order stress components.In order to accomplish this, however, we must first determine the leading-order radial velocity u in order to fix the indeterminacy of the axisymmetric stresscomponents. The convergence of the leading-order slump solution to rest hasbeen explored previously in , extending the work of Matson & Hogg for2D dambreaks. It was found that, for t → ∞, the radial velocity u is plug-like1470 0.2 0.4 0.6 0.8 100.20.40.60.81r /R(τrr,τθθ,τzz)/B τ z zτ r rτ θ θ(a)0 0.2 0.4 0.6 0.8 100.20.40.60.81Plug speed(b)r /R NumericsAsymptoticsFigure B.3: (a) Scaled stress components (τrr,τθθ ,τzz)/B and (b) radialspeed u/umax plotted against r/R along the level where τrz = 0 atthe end of the computation of a simulation with B = 0.01 and R = 1.throughout the fluid depth and given by u = Q2ξ/(4ξ ), whereddξ[(1−ξ 2)Q2ξ]= 2λ (1−ξ 2)(1−Q), (B.3)with ξ =√1−χ/χ∞ and λ ≈ 23.3855. With the solution of this problem in hand,we may then compute τˇrr and τˇθθ to leading order. The results are compared withcomponents extracted from the numerical simulations in figure B.3(a) for a sampleslump near the arrest of motion. Plotted are the stress components along the levelwhere τrz = 0 (which is z = h0 in the leading-order asymptotics), after recastingthe asymptotic predictions in terms of the original variables. Except near thecentre and edge of the collapsed cylinder (where the shallow-layer asymptoticsbreaks down), the two agree qualitatively. Evidently, τzz is almost constant inradius, whereas τrr and τθθ match one another up to a small correction.The approximation τˇrr ≈ τˇθθ (which corresponds to a plug speed u ∝ r; cf.figure B.3(b)), offers a simpler analytical pathway to a corrected final profile: theyield condition now implies thatτˇrr + τˇθθ ≈ 2Bˇ√3√1−(1− zh0)2, (B.4)148and so the corrected profile now follows from (B.2) ashhχ + Bˇ∼ pi√34εBˇhχ , (B.5)which provides the solution quoted in §3.2.1.B.3 Shallow slippy flowsWhen the fluid slides more freely over the underlying surface, the lubricationmodel of no longer captures the leading-order dynamics. In particular, theextensional stress components (τrr,τθθ ,τzz) become promoted to higher orderin comparison to the shear stress τrz because the reduced surface traction gen-erates little vertical shear and the horizontal flow becomes plug-like throughoutthe fluid layer. In this circumstance, the relevant asymptotic description is that ofa viscoplastic membrane model : rather than rescaling as in §3.1, we set onlyτrz = ετˇrz. Then, ignoring inertia, the leading-order momentum equations are0=−pχ + 1χ(χτrr)χ + τˇrz,z− τθθχ+O(ε), (B.6)0=−pz + τzz,z−1+O(ε), (B.7)With the rescaled horizontal velocity u∼ ε−1uˇ(χ , t)≫ w(χ ,z, t), the constitutivelaw predicts that(τrr,τθθ)∼ 2(1+Bγ˙)(uˇχ , uˇ/χ), (B.8)γ˙ ∼ 2√uˇ2χ + uˇuˇχ/χ + uˇ2/χ2, (B.9)which are independent of z. The surface boundary conditions in (3.8) are replacedbyτˇrz+hχ(τrr− p)p+ τzz}= O(ε) on z = h. (B.10)149The integral of (B.7) now furnishes p+ τzz = h− z to leading order. From thedepth integrated continuity equation and (B.6), it then follows thatht +1χ(χhuˇ)χ = 0, (B.11)0=−hχ − τbh+1h[h(2τrr + τθθ )]χ +τrr− τθθχ. (B.12)Here τb = τˇrz(r,0, t) is the basal shear stress, which must be specified by an addi-tional model of slip.For a cylindrical dambreak with free slip (τb = 0), the equations admit a simplesolution in which the fluid retains the shape of a cylinder, with h = H (t) anduˇ = χU (t) for χ < εR(t), and R˙ ≡RU . The extensional stresses become τrr =τθθ = 2U +B/√3, and mass conservation demands that R2H = R2. Although(B.12) is now satisfied, the combined gravitational and extensional stress must stillvanish at the edge of the cylinder, which implies that −12H 2+H (2τrr + τθθ ) =0, or 12H = 6U +B√3. Thus,H˙ =−2H R˙R= 16R√H (B√3− 12H ). (B.13)Provided B < 1/(2√3), the cylinder therefore collapses to a final state given byH = 2B√3 and R =R(12B2)1/4. (B.14)B.4 Tall slender conesMost practical slump tests take a truncated cone as the initial shape with the radiusat the top equal half of that at the base. The slender analysis in this case has beenderived by Clayton et al. using the Tresca yield condition. For von Mises,150the corresponding result iss = 1−h0−√3B ln( (1+α−1)3−1(1+α−1h0)3−1), (B.15)where h0 is the height of the unyielded region, given byB =α3√3(1+α−1h0)3−1(1+α−1h0)2and α = Rtop/(Rbase−Rtop).151Appendix CSurges of viscoplastic fluids onslopesC.1 Additional numerical detailsIn this appendix we provide further details of the numerical computations. We usethe dimensionless version of the problem in which lengths are scaled by H andvelocities by U as in §4.1.To find the steady surge states, we solve suites of initial-value problems inthe frame of reference of the surge, following the strategy outlined in [53, 54] orChapter 2 and 3. We begin from initial conditions in whichmotionless viscoplasticfluid is deposited on the inclined plane with a rectangular shape whose depthis set by the sheet-flow solution. The front of the rectangle is smoothed overa streamwise scale of order of a fraction of the fluid depth using a arc tangentfunction. The simulations do not appear to be sensitive to the initial condition,at least at the Reynolds numbers chosen for most of our simulations (§C.1.2),with no indication of multiple equilibria or unsteady states (but see the discussionsurrounding figure 4.9).The initial-value problem is solved exploiting the PLIC (Piecewise Linear In-terface Calculation) algorithm to evolve the interface within the volume-of-fluid152scheme. The essential details of the algorithm can be found in , although wemodify it slightly to surmount a numerical difficulty arising from an unresolvedlayer of the upper-layer fluid coating the inclined plane, as is described in [53, 54]or Chapter 2 and 3. In brief, the modification amounts to monitoring the vol-ume fraction adjacent to the inclined plane and adjusting the value of c there if itexceeds a threshold (set to 0.99). This replacement of ambient fluid with yield-stress material destroys mass conservation, which is restored by rescaling the flowheight uniformly over the length of the surge, incurring a further error, of order afraction of a grid spacing.The evolution observed in these initial-value problems suggest that the finalsteady states possess no contact line along the underlying plane. Rather, a contin-ually thinning finger of ambient fluid coats the plane as one progresses upstream.Therefore, although the adjustment to the PLIC algorithm applies an approxima-tion to allow the contact line of the initial condition to migrate with the plane tocreate this finger, once that feature is established, the scheme simply amounts toneglecting the small amount of ambient fluid within the finger once it becomesthinner than the lowest grid cell. This prevents interpolation errors in the veloc-ity field and shear stress of the finger from excessively lubricating the surge, andpermits the computation to otherwise remain resolved, as we now document.C.1.1 Resolution studyFigure C.1 shows the results of a resolution study for the profile of a Binghamsurge with θ = pi/18 and B = 0.0522 (Re = 1, ub = 0.041 and the total area offluid is 24). At the rear of the surge, a no-slip back wall is imposed, and the gridspacing is uniform with ∆x = 4∆z. Varying that grid spacing by a factor of 8 (asindicated) furnishes barely any discernible difference in the free surface profile.Indeed, the root-mean-square difference in the velocity field, defined as√∫∫ |u−u1|2c1 dxdy∫∫c1 dxdy,153between the coarsest and finest of these simulations is about 0.088ub, and de-creases to 0.025ub between the two finest simulations. Here, c1 and u1 denotethe reference solution, which is that for the finest simulation, interpolated ontothe grid of the coarser solution. Also shown are selected contours of constantinvariant τxz (which closely match those of τI below the plug), illustrating theconvergence with resolution, at least for stress levels sufficiently in excess of B.Contours closer to B show significantly less degree of convergence, as highlightedby the roughness of the yield surfaces plotted in figure 4.4, and the stress solutionremains sensitive to resolution for τI < B. These latter deficiencies are not prob-lematic as the solution is independent of the stress state over the plug as long asthe fluid there is not yielded. Note that the computations reported in §4 and theremainder of this appendix all use the finest grid of the resolution study.C.1.2 Inertial effectsFigure C.2 shows numerical simulations of Bingham surges with varying Reynoldsnumber Re= HU /κ from 0.1 to 10 (which span the range of all the simulationsreported in this study), with a back wall providing the left-hand boundary condi-tions. With the scalings of the problem outlined in §4.1, the dimensionless prob-lem retains the Reynolds number only as a factor in front of the inertial terms.Consequently, the flow profile becomes independent of Re in the inertialess limit.Indeed, the flow profiles and stress levels shown in figure C.2 closely collapse(save for the relatively rough yield surface) and the root-mean-square differencesin the velocity field and stress invariant are less than 4×10−3ub and 0.01B, respec-tively. Thus we conclude that inertial effects are not significant. The computationsreported in §4.1 are conducted with Re = 1.C.1.3 The effect of the back wallTo gauge the effect of the back wall on the computations, figure C.3 comparestwo simulations with different boundary conditions imposed along x = 0. In thefirst, the velocity field corresponding to the uniform sheet flow is imposed (so154Figure C.1: Simulations of a Bingham surge profile with θ = 10◦ and B =0.0522 (Y∞ = 0.7, fluid area of 24, ub = 0.041 and Re = 1) with thegrid sizes indicated. (a) shows h and Y ; (b) a magnification near theflow front; (c) contours of constant τxz/B, as indicated (with the surgeprofile of the finest resolution case shown by the lighter grey line);(d) the shear stress along z = 0.025 (the lowest grid cell of the coars-est computation). In (a), the inset shows the flow front X f , mean shearstress, τxz/B=∫ ∫(τxz/B) cdxdz, and τxz at the point (x,z)= (18,0.4),all plotted against the vertical grid spacing ∆z (∆x = 4∆z). The con-vergence of τxz is impeded by the need to resolve the sharply localizedregion of high stress underneath the flow front (cf. figure 4.3 and pan-els (c) and (d)).155Figure C.2: Simulations of Bingham surges for θ = 10◦, ub = 0.041 andB = 0.0522 (Y∞ = 0.7) for the Reynolds numbers indicated (the totalarea of fluid is 24). Plotted are h, Y (τxz = B) and the stress levelsτI/B = 1, 2 and 3.(u,w) = (usheet ,0)); for the second, we impose a no-slip condition, u = w = 0 Thefigure displays the stress invariant τI and plug regions; the solutions are much thesame except within a region near the back wall in which flow adjustments arisedue to the boundary condition there (and which changes the flow length slightly).In all our simulations, the extent of these flow adjustments was restricted to along-slope lengths of about 4H.156C.2 Improved lubrication solutionFor Herschel Bulkley fluid, the dimensionless system ispx = εσx + τz +S, pz =−εσz + ε2τx−1,0= τ(x,h, t)+ [p(x,h, t)− εσ(x,h, t)]hx,0= p(x,h, t)+ εσ(x,h, t)+O(ε2),(στ)=(γ˙n−1+Bγ˙)(2εuxuz + ε2wx)for 0< z < Y,B2 = σ2+ τ2 for Y < z < h,γ˙ =√4ε2u2x +(uz + ε2wx)2,ht =−(∫ h0udz)x.In the fully yielded region, uz ∼ O(1), σ ∼ O(ε) and γ˙ ∼ O(1), and droppingthe O(ε2) terms then givespz =−1 and px = τz+S (C.1)Hencep = h− z+Pτ = (S−hx−Px)(Y − z)+BY = h+T −BS−hx−Px(C.2)where P = P(x) and T = T (x), which further imply thatτ = B+unz ,uz = (S−hx−Px) 1n (Y − z) 1n ,u =nn+1(S−hx−Px)1n[Yn+1n − (Y − z) n+1n]−ub,(C.3)all to O(ε2).157In the pseudo-plug, u = up(x)−ub + εu1(x,z), and soγ˙ = ε√4u2px +u21z +O(ε) = εΓ.Hence, (στ)=BΓ(2upxu1z)+O(εn), (C.4)if n < 1. Thusσ =√B2− τ2+O(εn) (C.5)Force balance over this region demandspx = S+ τz+ εσx & pz + εσz =−1+O(ε2), (C.6)which, given the surface stress conditions, now providep = h− z− εσ +O(ε2),τ = (S−hx)(h− z)+2ε(∫ hzσdz)x+O(ε2),(C.7)or, given (C.5),τ = (S−hx)(h− z)+ 12εB2(2θ + sin2θS−hx)x+O(εn+1),whereθ = sin−1(S−hx)(h− z)B. (C.8)Next we observe that P ∼ O(ε2), because p = h− z+P and σ = O(ε) in thefully sheared region, but p = h− z− εσ +O(ε2) in the pseudo-plug. The matchof τ = (S−hx)(h− z)+T in z < Y with(S−hx)(h− z)+2ε(∫ hz√B2− τ2dz)x+O(εn+1)158for z >Y , then demands thatT = 2εB(∫ hY√1− (S−hx)2(h− z)2B2dz)x= 12εpiB2(1S−hx)x.(C.9)Now we match the velocity profile of the fully yielded region in (C.3) with that ofthe pseudo-plug u = up−ub + εu1, to find u1(x,Y, t) = 0 andup =nn+1(S−hx)1nYn+1n .Finally we compute u1 and the downslope flux: in the pseudo-plug,u1z2upx=τσ=(S−hx)(h− z)√B2− (S−hx)2(h− z)2+O(ε), (C.10)and so, given that (S−hx)(h−Y ) = B+O(ε),u1 = 2upx√B2− (S−hx)2(h− z)2S−hx +O(ε).The flux can then be computed as∫ h0udz =∫ Y0udz+∫ hY(up−ub + εu1)dz= up(h− n2n+1Y)−ubh+ 12εpiB2upx(S−hx)2 .The equations of the improved model quoted in the main text now follow, ontaking n = 1.159C.3 Improved slump profilesCuriously, the improved model for slumped shapes on slopes given by (4.38) im-plies thathhx ∼−12εpiB2(h−1x )x or h∼ [3εpiB2(x f − x)]1/3,for x → x f . This contrasts sharply with the solution of the iterated version ofthe model in (4.41) which has a finite depth at the edge. Evidently, the freedomafforded by the extra derivative allows us to reach the flow front with h → 0 andhx →−∞. This feature of the improved asymptotic theory was not appreciated inour earlier papers [53, 54] or Chapter 2 and 3, where the simpler iterated versionof the model was implemented.Figure C.4 compares the various asymptotic solutions, with or without a back-ground slope. The two versions of the improved model differ by O(ε2) away fromthe flow front; within a distance of O(ε2) of x = x f , however, the flow depths be-come O(ε) different, permitting the iterated version to terminate at finite depth.For comparison, figure C.4(a)-(b) also includes simulation data from figure 14 of for dambreaks on a horizontal surface with either square or triangular initialconditions. The (red) corners visible in panel (b) are relics of square initial condi-tions, whereas the (blue) sharp spikes at the back evident in panel (a) are remnantsfrom triangular initial conditions (in the scaled coordinates the slumps have dif-ferent lengths). Further computations for the slump of a rectangular block on anincline (with unit height and an upstream back wall) are included in figure C.4(c)-(d). Both comparisons indicate that the two versions of the improved model out-perform the leading-order theory away from the flow front. Near that steep fea-ture, the smooth decline to zero thickness of the non-iterated model provides aslightly more satisfying comparison with simulations. However, the asymptotictheory is not valid at the flow front where the slope of the free surface diverges.Moreover, the surface in the numerical simulations eventually overturns to createmulti-valued profiles with a finite elevation at the leading edge. Consequently, itis not clear which version of the improved model is superior; the iterated model160(which we have employed previously, and continue to use in the main text) hasthe advantage of being the simpler.161Figure C.3: Simulations for θ = 10◦ and B = 0.0354 (Y∞ = 0.8) with dif-ferent left-hand boundary conditions: (a) (u,w) = (usheet ,0) and (b)(u,w) = (0,0). Shown is the second invariant τI as a density onthe (x,z)−plane. The (true) yield surfaces are indicated by the solid(green) lines. The simulation in (a) corresponds to the truncated so-lution shown in figure 4.3. The profiles and the true and fake yieldsurfaces are compared in (c).162Figure C.4: Slump profiles for (a)-(b) S = 0 and (b)-(c) S 6= 0. For (a)-(b), scaled variables are plotted, which eliminates any free param-eters; in (c)-(d) the profiles are shown for 12εpiB = 0.2. The darksolid lines show the solutions to (4.38), whereas the dotted lines showthe solutions to the iterated version in(4.40); the dashed lines showthe leading-order approximation. In (a)–(b) the lighter (red and blue)lines show the results of a series of simulations from with ε = 1,B = 0.02, 0.03, ..., 0.1 and the flow fronts aligned. In (c)–(d) thelighter (red) lines show additional simulations of the slump of a rect-angular block on an incline with B = (0.1,0.2,0.3,0.4)/pi and θ = 5◦;the inset shows the aligned, but unscaled profiles.163`\n\n#### Cite\n\nCitation Scheme:\n\nCitations by CSL (citeproc-js)\n\n#### Embed\n\nCustomize your widget with the following options, then copy and paste the code below into the HTML of your page to embed this item in your website.\n``` ```\n<div id=\"ubcOpenCollectionsWidgetDisplay\">\n<script id=\"ubcOpenCollectionsWidget\"\nsrc=\"{[{embed.src}]}\"\ndata-item=\"{[{embed.item}]}\"\ndata-collection=\"{[{embed.collection}]}\"",
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"Our image viewer uses the IIIF 2.0 standard. To load this item in other compatible viewers, use this url:\n`https://iiif.library.ubc.ca/presentation/dsp.24.1-0378326/manifest`"
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https://stacks.math.columbia.edu/tag/0D0N | [
"Lemma 37.13.7. Let $f : X \\to Y$ be a morphism of schemes. The following are equivalent\n\n1. $f$ is smooth, and\n\n2. $f$ is locally of finite presentation, $H^{-1}(\\mathop{N\\! L}\\nolimits _{X/Y}) = 0$, and $H^0(\\mathop{N\\! L}\\nolimits _{X/Y}) = \\Omega _{X/Y}$ is finite locally free.\n\nProof. This follows from the definition of a smooth ring homomorphism (Algebra, Definition 10.137.1), Lemma 37.13.2, and the definition of a smooth morphism of schemes (Morphisms, Definition 29.34.1). We also use that finite locally free is the same as finite projective for modules over rings (Algebra, Lemma 10.78.2). $\\square$\n\nIn your comment you can use Markdown and LaTeX style mathematics (enclose it like $\\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar)."
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https://mathematica.stackexchange.com/questions/78429/using-map-sequentially | [
"Using Map sequentially\n\nI'm trying to simulate the probability of a balanced hand in Bridge (The Theory of Probability by Santosh S. Venkatesh, 2013) using Mathematica. The problem is solved in the textbook but I just wanted to use it to practice writing programs in Mathematica.\n\nThe idea is to shuffle the deck, partition it into 4 hands of 13 cards each, and check whether each of the 4 partitions contains an Ace. I'm designating the numbers {1,14,27,40} as the aces (they could have been any 4 different numbers).\n\nuno = Partition[RandomSample[Range, 52], 13];\nAnd[\nApply[Or, Map[MemberQ[uno[], #] &, {1, 14, 27, 40}]],\nApply[Or, Map[MemberQ[uno[], #] &, {1, 14, 27, 40}]],\nApply[Or, Map[MemberQ[uno[], #] &, {1, 14, 27, 40}]],\nApply[Or, Map[MemberQ[uno[], #] &, {1, 14, 27, 40}]]\n]\n\nI was proud of being able to use Map \"MemberQ\" to each of the four values {1,14,27,40}.\n\nWhere I got stucked is trying to Map it a second time to the four elements of uno.\n\n• Position[uno, Alternatives @@ {1, 14, 27, 40}][[All, 1]] == {1, 2, 3, 4} will give True/False for all 4 hands having an ace...\n– ciao\nMar 27 '15 at 22:46\n• FreeQ[#, Alternatives @@ {1, 14, 27, 40}, 1] & /@ uno Mar 27 '15 at 22:53\n• Thank you rasher & belisarius. I just added Alternatives to my toolkit. Mar 27 '15 at 23:09\n\nYour operation using the v10 operator form for MemberQ, along with Alternatives:\n\nuno = RandomSample[Range @ 52] ~Partition~ 13;\n\nAnd @@ MemberQ[1 | 14 | 27 | 40] /@ uno\n\nOr using modular arithmetic as already provided by others:\n\nx = RandomSample @ Range @ 52;\n\nSign[x ~Mod~ 13] ~Partition~ 4 // Total\n{12, 11, 13, 12}\n\nThis is the number of non-ace cards in each hand.\n\n• ...I need to user operator forms more... I also like labeling the \"aces\" 0 mod 13...and that it just happened to produce sample {1,2,0,1}...guess I am easily amused...+1 :) Mar 28 '15 at 11:44\n• Thank you Mr Wizard. Brief and elegant. This is the first time I see the shortcut ' ~ '. Mar 29 '15 at 18:01\n• @user23438 You're welcome. I'm kind of notorious for using ~infix~ syntax; personally I find it very useful to reduce the stacking of [[[brackets]]] which I find very hard to read; other people think the infix is hard to read. My advice it to experiment for yourself and see what you find more readable after experience with several styles. See also: (56504) Mar 29 '15 at 22:26\n\nUuuh, that's like a breeze....\n\nace= {1, 14, 27, 40};\nuno = Partition[RandomSample[Range, 52], 13];\nIntersection[#, ace] & /@ uno\n\nThis returns the Aces returned in each hand\n\nlet's go gambling....hi,hi,hi\n\n• Beautiful penguin77. Thank you very much. Gotta love Mathematica and its simplicity. Mar 27 '15 at 23:07\n\nJust write simpler code! Here's an example... first, the setup:\n\ncards = 52;\nuno = Partition[RandomSample[Range[cards], cards], cards/4];\naces = 1 + {0, 1, 2, 3} cards/4;\n\n(it's a bit easier to test if you set cards = 12).\n\nFreeQ[\nIntersection[aces, #] & /@ uno,\n{}]\n\nEnjoy.\n\n• Beautiful djp. Thank you very much. Gotta love Mathematica and its simplicity. Mar 27 '15 at 23:08\n\nJust for fun:\n\nYou can count number of \"aces\" (obviously arbitrary choice) 4 sets of 13 cards:\n\nTotal /@ Partition[\nBoole[Mod[#, 13] == 1] & /@ RandomSample[Range@52, 52], 13]\n\nFor simulation:\n\ntab[n_] :=\nTable[Total /@\nPartition[Boole[Mod[#, 13] == 1] & /@ RandomSample[Range@52, 52],\n13], {n}];\n\nNow cheating by just testing patterns and enumerating permutations for an approximation:\n\na = With[{res = tab},\nLength[Pick[res, # == {1, 1, 1, 1} & /@ res]]/1000000.];\nb = With[{res = tab},\nLength[Pick[res, # == {0, 0, 0, 4} & /@ res]]/1000000.];\nc = With[{res = tab},\nLength[Pick[res, # == {2, 2, 0, 0} & /@ res]]/1000000.];\nd = With[{res = tab},\nLength[Pick[res, # == {1, 1, 2, 0} & /@ res]]/1000000.];\ne = With[{res = tab},\nLength[Pick[res, # == {1, 3, 0, 0} & /@ res]]/1000000.];\nw = {1, 4, 6, 12, 12};\nsim = w {a, b, c, d, e};\n\nCalculating exactly:\n\ncalc = With[{den = Binomial[52, 4]},\nac = 13.^4/den;\nbc = 4. Binomial[13, 4]/den;\ncc = 6. Binomial[13, 2]^2/den;\ndc = 12. 13.^2 Binomial[13, 2]/den;\nec = 12. Binomial[13, 3] 13./den;\n{ac, bc, cc, dc, ec}\n];\n\nComparing probabilities:",
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https://la.mathworks.com/matlabcentral/cody/problems/463-looking-for-squares/solutions/2293991 | [
"Cody\n\n# Problem 463. Looking for Squares\n\nSolution 2293991\n\nSubmitted on 18 May 2020 by Karl Ezra Pilario\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\nfor n=4:6; sq=squares(n); assert(length(sq)==n) for k=1:n assert(sq(k)>0) assert(round(sqrt(sq(k)))==sqrt(sq(k))) end assert(round(sqrt(sum(sq)))==sqrt(sum(sq))) end"
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https://cs.nyu.edu/~yann/research/ebm/ | [
"",
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"# Energy-Based Models",
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"• Time Period: September 2003 - present.\n• Participants: Yann LeCun, Sumit Chopra, Raia Hadsell, Fu-Jie Huang, Marc'Aurelio Ranzato (Courant Institute/CBLL)\n\n Tutorials, Talks, and Videos\n\n Publications\n\n• [LeCun et al 2006]. A Tutorial on Energy-Based Learning, in Bakir et al. (eds) \"Predicting Structured Outputs\", MIT Press 2006: a 60-page tutorial on energy-based learning, with an emphasis on structured-output models. The tutorial includes an annotated bibliography of discriminative learning, with a simple view of CRF, maximum-margin Markov nets, and graph transformer networks.\n• [LeCun and Huang, 2005]. Loss Functions for Discriminative Training of Energy-Based Models. Proc. AI Stats 2005.\n• [Osadchy, Miller, and LeCun, 2004] Synergistic Face Detection and Pose Estimation Proc. NIPS 2004. This paper uses an energy-based model methodology and contrastive loss function to detect faces and simultaneously estimate their pose.\n\n Description\n\nProbabilistic graphical models associate a probability to each configuration of the relevant variables. Energy-based models (EBM) associate an energy to those configurations, eliminating the need for proper normalization of probability distributions. Making a decision (an inference) with an EBM consists in comparing the energies associated with various configurations of the variable to be predicted, and choosing the one with the smallest energy. Such systems must be trained discriminatively to associate low energies to the desired configurations and higher energies to undesired configurations. A wide variety of loss function can be used for this purpose. We give sufficient conditions that a loss function should satisfy so that its minimization will cause the system to approach to desired behavior.\n\nAn EBM can be seen as a scalar energy function of three (vector) variables E(Y,Z,X), parameterized by a set of trainable parameters W. X is the input, which is always observed (e.g. pixels from a camera), Y is the set of variables to be predicted (e.g. the label of the object in the input image), and Z is a set of latent variables that are never directly observed (e.g. the pose of the object).\n\nPerforming an inference consists in observing an X, and looking for the values of Z and Y that minimize the energy E(Y,Z,X).\n\nGiven a training set S={ (X1,Y1), (X2,Y2),....(Xp,Yp) }, training an EBM consists in \"digging holes\" in the energy surface at the training samples (Xi,Yi), and \"building hills\" everywhere else. This process will make the desired Yi become a minimum of the energy for the given Xi.\n\nThe main question is how to design a loss function so that minimizing this loss function with respect to the parameter vector W will have the effect of digging holes and building hills at the required places in the energy surface.\n\nIn the following, we demonstrate how various types of loss functions shape the energy surface. The task to learn is the square function Y=X^2. In the animations, the blue spheres indicate the location of training samples (a subset of the training samples). A good loss function will shape the energy surface so that the blue spheres are minima (over Y) of the energy for a given X.",
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"Here, we demonstrate how \"traditional\" supervised learning applied to a traditional neural net (viewed as an EBM) shapes the energy surface. The EBM architecture is a traditional 2-layer neural net with 20 hidden units, followed by a cost module that measures the L1 distance between the network output and the desired output Y. The value of Y that minimizes the energy is simply equal to the neural net output. The loss function used here is simply the average squared energy over the training set. This is the traditional square loss used in conventional neural nets. With this architecture, the shape of the energy as a function of Y is constant (it's a V-shaped function), only the location of its minimum can change. Therefore, digging a hole at the right Y will automatically make the other values of Y have higher energies.",
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"",
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"In this second example, we use a different architecture, shown at right, where both X and Y are fed to neural nets. The outputs of the nets are fed to an L1 distance module. The loss function used here is again the average squared energy over the training set. The shape of the energy as a function of Y is no longer constant, but depends on the neural net on the right. A catastrophic collapse happens: the two neural nets are perfectly happy to simply ignore X and Y and produce identical, constant outputs (the flat green surface is the solution, the orange-ish surface is the random initial condition). The loss function pulls down on the blue spheres, but nothing is pulled up, and nothing prevents the energy surface from becoming flat. It does become flat. Hence the collapse.",
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"",
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"In this third example, we use the same dual net architecture, but we use the so-called square-square generalized margin loss function which not only pulls down on the blue spheres, but also pulls up on points on the surface some distance away from the blue spheres. The point being pulled up for a given X is the Y with the lowest energy that is at least 0.3 away from the desired Y. This time, no collapse occurs, and the energy surface takes the right shape.",
null,
"",
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"In this fourth example, we use the same dual net architecture, trained with the negative-log-likelihood loss function. This loss pulls down on the blue spheres, and pulls up on all points of the surface, pulling harder on points with lower energy. This is the loss used to train traditional probabilistic models by maximizing the conditional likelihood of all the Yi given all the Xi. Again, no collapse occurs, but the computation is considerably more expensive than in the previous example, because we need to compute the integral over all values of Y of the derivative of the energy with respect to W (the derivative of the log partition function). Also, this loss function makes the energies of undesired Y grow without bounds, and it does not pull the energies of the blue spheres toward zero (only relative values matter).",
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https://articles.outlier.org/how-to-find-critical-value | [
"## If you could change one thing about college, what would it be?",
null,
"Outlier Articles Home\n\nStatistics\n\n# How To Find Critical Value In Statistics\n\n## Sarah Thomas\n\nSubject Matter Expert\n\nLearn how to find critical value, its importance, the different systems, and the steps to follow when calculating it.\n\n## Don't Overpay For College Statistics\n\nTake Intro to Statistics Online with Outlier.org\n\nFrom the co-founder of MasterClass, earn transferable college credits from the University of Pittsburgh (a top 50 global school). The world's best online college courses for 50% less than a traditional college.\n\nGet started",
null,
"In baseball, an ump cries “foul ball” any time a batter hits the ball into foul territory. In statistics, we have something similar to a foul zone. It’s called a rejection region. While foul lines, poles, and the stadium fence mark off the foul territory in baseball, in statistics numbers called critical values mark off rejection regions.\n\n## What Is a Critical Value?\n\nA critical value is a number that defines the rejection region of a hypothesis test. Critical values vary depending on the type of hypothesis test you run and the type of data you are working with.\n\nIn a hypothesis test called a two-tailed Z-test with a 95% confidence level, the critical values are 1.96 and -1.96. In this test, if the statistician’s results are greater than 1.96 or less than -1.96. We reject the null hypothesis in favor of the alternative hypothesis.\n\nIn Outlier's Intro to Statistics course, Dr. Gregory Matthews explains more about hypothesis testing and why to use it:\n\nThe figure below shows how the critical values mark the boundaries of two rejection regions (shaded in pink). Any test result greater than 1.96 falls into the rejection region in the distribution’s right tail, and any test result below -1.96 falls into the rejection region in the left tail of the distribution.",
null,
"A two-tailed Z-test with a 95% confidence level (or a significance level of ɑ = 0.05) has two critical values 1.96 and -1.96.\n\n## The Role of Critical Values in Hypothesis Tests\n\nBefore we dive deeper, let’s do a quick refresher on hypothesis testing. In statistics, a hypothesis test is a statistical test where you test an “alternative” hypothesis against a “null” hypothesis. The null hypothesis represents the default hypothesis or the status quo. It typically represents what the academic community or the general public believes to be true. The alternative hypothesis represents what you suspect could be true in place of the null hypothesis.\n\nFor example, I may hypothesize that as times have changed, the average age of first-time mothers in the U.S. has increased and that first-time mothers, on average, are now older than 25. Meanwhile, conventional wisdom or existing research may say that the average age of first-time mothers in the U.S. is 25 years old.\n\nIn this example, my hypothesis is the alternative hypothesis, and the conventional wisdom is the null hypothesis.\n\nAlternative Hypothesis $H_a$ = Average age of first-time mothers in the U.S. > 25\n\nNull Hypothesis $H_0$ = Average age of first-time mothers in the U.S. = 25\n\nIn a hypothesis test, the goal is to draw inferences about a population parameter (such as the population mean of first-time mothers in the U.S.) from sample data randomly drawn from the population.\n\nThe basic intuition behind hypothesis testing is this. If we assume that the null hypothesis is true, data collected from a random sample of first-time mothers should have a sample average that’s close to 25 years old. We don’t expect the sample to have the same average as the population, but we expect it to be pretty close. If we find this to be the case, we have evidence favoring the null hypothesis. If our sample average is far enough above 25, we have evidence that favors the alternative hypothesis.\n\nA major conundrum in hypothesis testing is deciding what counts as “close to 25” and what counts as being “far enough above 25”? If you randomly sample a thousand first-time mothers and the sample mean is 26 or 27 years old, should you favor the null hypothesis or the alternative?\n\nTo make this determination, you need to do the following:\n\n1. First, you convert your sample statistic into a test statistic.\n\nIn our first-time mother example, the sample statistic we have is the average age of the first-time mothers in our sample. Depending on the data we have, we might map this average to a Z-test statistic or a T-test statistic.\n\nA test statistic is just a number that maps a sample statistic to a value on a standardized distribution such as a normal distribution or a T-distribution. By converting our sample statistic to a test statistic, we can easily see how likely or unlikely it is to get our sample statistic under the assumption that the null hypothesis is true.\n\n2. Next, you select a significance level (also known as an alpha (ɑ) level) for your test.\n\nThe significance level is a measure of how confident you want to be in your decision to reject the null hypothesis in favor of the alternative. A commonly used significance level in hypothesis testing is 5% (or ɑ=0.05). An alpha-level of 0.05 means that you’ll only reject the null hypothesis if there is less than a 5% chance of wrongly favoring the alternative over the null.\n\n3. Third, you find the critical values that correspond to your test statistic and significance level.\n\nThe critical value(s) tell you how small or large your test statistic has to be in order to reject the null hypothesis at your chosen significance level.\n\n4. You check to see if your test statistic falls into the rejection region.\n\nCheck the value of the test statistic. Any test statistic that falls above a critical value in the right tail of the distribution is in the rejection region. Any test statistic located below a critical value in the left tail of the distribution is also in the rejection region. If your test statistic falls into the rejection region, you reject the null hypothesis in favor of the alternative hypothesis. If your test statistic does not fall into the rejection region, you fail to reject the null hypothesis.\n\nNotice that critical values play a crucial role in hypothesis testing. Without knowing what your critical values are, you cannot make the final determination of whether or not to reject the null hypothesis.\n\n## Factors That Influence Critical Values\n\nCritical values vary with the following traits of a hypothesis test.\n\n### What test statistic are you using?\n\nThis will depend on the type of research question you have and the type of data you are working with. In a first-year statistics course, you will often conduct hypothesis tests using Z-statistics (these correspond to a standard normal distribution), T-statistics (these correspond to a T-distribution), or chi-squared test statistics (these correspond to a chi-square distribution).\n\n### What significance level have you selected?\n\nThis is up to the person conducting the test. A significance level (or alpha level) is the probability of mistakenly rejecting the null hypothesis when it is actually true. By choosing a significance level, you are deciding how careful you want to be in avoiding such a mistake.\n\nYou might also hear a hypothesis test being described by a confidence level. Confidence levels are closely related to statistical significance. The confidence level of a test is equal to one minus the significance level or 1-ɑ.\n\n### Is it a one-tailed test or a two-tailed test?\n\nHypothesis tests can be one-tailed or two-tailed, depending on the alternative hypothesis. Null and alternative hypotheses are always mutually exclusive statements, but they can take different forms. If your alternative hypothesis is only concerned with positive effects or the right tail of the distribution, you will likely use a one-tailed upper-tail test.\n\nIf your alternative hypothesis is only concerned with negative effects or the left tail of the distribution, you will likely use a one-tailed lower-tail test. Finally, if your alternative hypothesis proposes a deviation in either direction from what the null hypothesis proposes, you’ll use a two-tailed test.\n\n## Critical Values for One-Tailed Tests & Two-Tailed Tests\n\nThe number of critical values in a hypothesis test depends on whether the test is a one-tailed test or a two-tailed test.\n\n### Critical Values for Two-Tailed Tests\n\nIn a two-tailed test, we divide the rejection region into two equal parts: one in the right tail of the distribution and one in the left tail of the distribution. Each of these rejection regions will contain an area of the distribution equal to ɑ/2. For example, in a two-tailed test with a significance level of 0.05, each rejection region will contain 0.05/2 = 0.025 = 2.5% of the area under the distribution. Because we split the rejection region, a two-tailed test has two critical values.\n\n### Critical Values for One-Tailed Tests\n\nA one-tailed test has one rejection region (either in the right tail or the left tail of the distribution) and one critical value. In a lower tail (or left-tailed) test, the critical value and rejection region will be in the left tail of the distribution. In an upper tail (or right-tailed) test, the critical value and rejection region will be in the right tail of the distribution.",
null,
"Two-tailed test",
null,
"One-tailed lower tail test",
null,
"One-tailed upper tail test\n\n## Commonly Used Critical Values\n\nThe tables below provide a list of critical values that are commonly used in hypothesis testing.\n\n### Z-Test Statistics (Using a Normal Distribution)\n\n CONFIDENCE LEVEL TAILS ALPHA (𝛂) CRITICAL VALUE(S) 90% Two-tailed 0.1 -1.64 and 1.64 Right-tailed 0.1 1.28 Left-tailed 0.1 -1.28 95% Two-tailed 0.05 -1.96 and 1.96 Right-tailed 0.05 1.65 Left-tailed 0.05 -1.65 99% Two-tailed 0.01 -2.58 and 2.58 Right-tailed 0.01 2.33 Left-tailed 0.01 -2.33\n\n### T-Test Statistics (Using a T Distribution)\n\n One-tailed test 𝛂 =______ 0.1 0.05 0.025 0.01 0.005 Two-tailed test 𝛂 =______ 0.2 0.1 0.05 0.02 0.01\n\n#### Degrees of Freedom (df)\n\n 1 3.078 6.314 12.71 31.82 63.66 2 1.886 2.92 4.303 6.965 9.925 3 1.638 2.353 3.182 4.541 5.841 4 1.533 2.132 2.776 3.747 4.604 5 1.476 2.015 2.571 3.365 4.032 6 1.44 1.943 2.447 3.143 3.707 7 1.415 1.895 2.365 2.998 3.499 8 1.397 1.86 2.306 2.896 3.355 9 1.383 1.833 2.262 2.821 3.25 10 1.372 1.812 2.228 2.764 3.169 11 1.363 1.796 2.201 2.718 3.106 12 1.356 1.782 2.179 2.681 3.055 13 1.35 1.771 2.16 2.65 3.012 14 1.345 1.761 2.145 2.624 2.977 15 1.341 1.753 2.131 2.602 2.947 16 1.337 1.746 2.12 2.583 2.921 17 1.333 1.74 2.11 2.567 2.898 18 1.33 1.734 2.101 2.552 2.878 19 1.328 1.729 2.093 2.539 2.861 20 1.325 1.725 2.086 2.528 2.845 21 1.323 1.721 2.08 2.518 2.831 22 1.321 1.717 2.074 2.508 2.819 23 1.319 1.714 2.069 2.5 2.807 24 1.318 1.711 2.064 2.492 2.797 25 1.316 1.708 2.06 2.485 2.787 26 1.315 1.706 2.056 2.479 2.779 27 1.314 1.703 2.052 2.473 2.771 28 1.313 1.701 2.048 2.467 2.763 29 1.311 1.699 2.045 2.462 2.756 30 1.31 1.697 2.042 2.457 2.75\n\n## How To Find the Critical Value in Statistics?\n\n### Finding a Critical Value for a Two-Tailed Z-Test\n\nSuppose you don’t remember what the critical values for a two-sided Z-test are. How would you go about finding them?\n\nTo find the critical value, you start with the significance level of your hypothesis test. Your significance level is equal to the total area of the rejection region. For example, with a 0.05 significance level, the entire rejection region will be equal to 5% of the area under the normal distribution.\n\nIn a two-tailed test Z-test, we split equally the rejection region into two parts. One rejection region is in the distribution’s right tail, and the other is in the left tail of the distribution. Each of these two parts will contain half of the total area of the rejection region. For a two-tailed Z-test with a significance level of ɑ=0.05, each rejection region will contain ɑ/2 = 0.025 or 2.5% of the distribution. This leaves a confidence interval of 0.95 (or 95%) between the two rejection regions.",
null,
"To find the critical values, you need to find the corresponding values (or Z-scores) in the Z-distribution. Make sure the percentage lying to the left of the first critical value is equal to ɑ/2. Also, check that the percentage of the distribution lying to the right of the second critical value is equal to ɑ/2. You can use a Z-table to look up these figures.\n\n### Solved Example: Two-Tailed Z-Test\n\nFor a two-tailed Z-test with a significance level of ɑ=0.05, we are looking for two critical values such that ɑ/2 or 2.5% of the normal distribution lies to the left of the first critical value and ɑ/2 or 0.025 of the normal distribution lies to the right of the second critical value.\n\nZ-tables will either show you probabilities to the LEFT or to the RIGHT of a particular value. We’ll stick to Z-tables showing probabilities to the LEFT.\n\nFor the first critical value, if the area to the left of the critical value is 0.025, we use the Z-table to find the number 0.025 in the table (we’ve shown this figure highlighted in an orange box). We then trace that value to the left to find the first two digits of the critical value (-1.9) and then up to the top to find the last digit (-0.06). If we put these together, we have the critical value -1.96. Z-tables provide Z-scores that are usually rounded to two decimal places.",
null,
"For the second critical value, 2.5% of the distribution will lie to the right, meaning 97.5% of the distribution will lie to the left of the critical value (1-0.025=0.0975). To find this critical value, we look for the number 0.0975 in the Z-table (we’ve shown this figure highlighted in a green box). We trace that value to the left to find the first two digits of the critical value (1.9) and then up to the top to find the last digit (0.06). Our second critical value is 1.96.",
null,
"Following similar steps, see if you can find the critical values for a Z-test with a significance level of ɑ=0.10. The critical values you find should be equal to -1.64 and 1.64.\n\n### Finding a Critical Value for a One-Tailed Z-Test\n\nIn a one-tailed test, there is just one rejection region, and the area of the rejection region is equal to the significance level.\n\nFor a one-tailed lower tail test, use the z-table to find a critical value where the total area to the left of the critical value is equal to alpha.\n\nFor a one-tailed upper tail test, use the z-table to find a critical value where the total area to the left of the critical value is equal to 1- alpha.\n\n### Solved Example: One-Tailed Z-Test\n\nLet’s see if we can use the Z-table to find the critical value for a lower tail Z-test with a significance level of 0.01.\n\nSince alpha equals 0.01, we are looking for this number in the Z-table. If you can’t find the exact number, you look for the closest number, which in this case is 0.0090. Once we’ve found this number, we trace the value to the first column to find the first two digits of the critical value and then up to the first row to find the last digit. The critical value is -2.33.",
null,
"Now let’s see if we can use the Z-table to find the critical value for an upper tail Z-test with a significance level of 0.10.\n\nSince this is an upper tail test, we need to use the Z-table to look for a critical value corresponding to 0.90 (1-ɑ = 1-0.10 = 0.90). The closest number to 0.90 we can find in the table is 0.89973. We trace this number to the left and then up to the top of the table to find a critical value of 1.28.",
null,
"## How To Find a Critical Value in R?\n\nTo find a critical value in R, you can use the qnorm() function for a Z-test or the qt() function for a T-test.\n\nHere are some examples of how you could use these functions in your critical value approach.\n\n### Z-Critical Values Using R\n\nFor a two-tailed Z-test with a 0.05 significance level, you would type:\n\nqnorm(p=0.05/2, lower.tail=FALSE)\n\nThis will give you one of your critical values. The second critical value is just the negative value of the first.\n\nFor a one-tailed lower tail Z-test with a 0.01 significance level, you would type:\n\nqnorm(p=0.01, lower.tail=TRUE)\n\nFor a one-tailed upper tail Z-test with a 0.01 significance level, you would type:\n\nqnorm(p=0.01, lower.tail=FALSE)\n\n### T-Critical Values Using R\n\nFor a two-tailed T-test with 15 degrees of freedom and a 0.1 significance level, you would type:\n\nqt(p=0.1/2, df=15, lower.tail=FALSE)\n\nThis will give you one of your critical values. The second critical value is just the negative value of the first.\n\nFor a one-tailed lower tail T-test with 10 degrees of freedom and a 0.05 significance level, you would type:\n\nqt(p=0.05, df=10, lower.tail=TRUE)\n\nFor a one-tailed upper tail T-test with 20 degrees of freedom and a 0.01 significance level, you would type:\n\nqt(p=0.01, df=20, lower.tail=FALSE)\n\nNow that you know the ins and outs of critical values, you’re one step closer to conducting hypothesis tests with ease!\n\n### Explore Outlier's Award-Winning For-Credit Courses\n\nOutlier (from the co-founder of MasterClass) has brought together some of the world's best instructors, game designers, and filmmakers to create the future of online college.\n\nCheck out these related courses:",
null,
"## Intro to Statistics\n\nHow data describes our world.\n\nExplore course",
null,
"## Intro to Microeconomics\n\nWhy small choices have big impact.\n\nExplore course",
null,
"## Intro to Macroeconomics\n\nHow money moves our world.\n\nExplore course",
null,
"## Intro to Psychology\n\nThe science of the mind.\n\nExplore course"
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https://libguides.cccua.edu/c.php?g=617810&p=4298781 | [
"# Copy of DIY Maths: Circumference of circles\n\n## CIRCUMFERENCE OF CIRCLES\n\nIn this module, you can study how to calculate the distance around a circle (its circumference) if you know its diameter.\n\nThe above video from Khan Academy provides an excellent introduction to understanding the radius, diameter and circumference measures of a circle.\n\n## Did you know?\n\nThe circumference of a circle is not equal to the length of two diameters.\n\nA circle's circumference divided by its diameter is the same for every circle. (It's pi, or 3.14)\n\nStudy and practice\n\n## Calculator",
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"https://libapps.s3.amazonaws.com/apps/libguides/images/google-logo.png",
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https://stage.geogebra.org/m/cpkeqyw9 | [
"# Properties of Rectangles, Rhombuses and Squares\n\nIn this activity, you will use what you have learned about parallelograms to discover properties about rectangles, rhombuses (rhombi), and squares. As you complete this activity, fill in the blanks on notes on the list of properties for each quadrilateral.\nA rectangle is a parallelogram with 4 right angles.\nRectangles also have properties that not all parallelograms possess. Let's discover these additional properties.\n\nMeasure the diagonals in the rectangle below by selecting the measurement tool (the button with \"cm\" and an arrow), then click on each endpoint of the segment you are measuring. The length will appear. What did you discover about diagonals AD and FB? Also, move one of the vertices to change the rectangle (Select the move tool first. The one with the white arrow). Does this change what you discovered? Finish the following sentence on your notes: The diagonals of a rectangle are __________________________.\n\n## Applet #1 RECTANGLE\n\nNow, let's explore rhombuses. A rhombus is a parallelgram with 4 congruent sides.\nRhombuses also have properties that not all parallelograms possess. Let's discover these additional properties.\n\nIn the rhombus below, measure the angles AHB, BHG, EHG, and AHE (Select the angle tool, then click on the points in order of the name of the angle.) What did you discover? Move one of the vertices to see if your findings remain the same for all rhombuses. Finish the following sentence on your notes: The diagonals of a rhombus are __________________________ to each other.\n\n## Applet #3 RHOMBUS\n\nUsing the rhombus applet above, measure angles ABH, GBH, GEH, and AEH. What did you discover about these angles? Manipulate the rhombus to see if your findings hold for all rhombuses.\n\nSo what does one diagonal do to a pair of opposite angles in a rhombus?\n\nDoes the other diagonal do the same thing?\n\nSo what do the diagonals of a rhombus do to both pair of opposite angles in a rhombus? Finish the following sentence on your notes: Each diagonal of a rhombus __________________________ opposite angles of the rhombus.\n\nFinally, a square is a parallelogram, rhombus, and a rectangle.\n\nSince a square possesses all the properties of a parallelogram, rectangle, and rhombus, Finish the following sentence on your notes: The diagonals of a square are __________________________. The diagonals of a square are __________________________ to each other. Each diagonal of a square __________________________ opposite angles of the square."
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http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1972.0 | [
"NTNUJAVA Virtual Physics LaboratoryEnjoy the fun of physics with simulations! Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/",
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"April 21, 2021, 09:17:54 am",
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"Welcome, Guest. Please login or register.Did you miss your activation email? 1 Hour 1 Day 1 Week 1 Month Forever Login with username, password and session length",
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"Home Help Search Login Register\nThere is a better way to do it; find it. ...\"Thomas Edison(1847-1931, American inventor, 1093 patients)\"\n Pages: Go Down",
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"Author Topic: Ejs Open Source Charge Particle in Magnetic Field B Java Applet in 3D (Read 29762 times) 0 Members and 1 Guest are viewing this topic. Click",
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"to toggle author information(expand message area).\nlookang\nModerator\nHero Member",
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"Offline\n\nPosts: 1796",
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"http://weelookang.blogspot.com",
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"",
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"« Embed this message on: October 04, 2010, 02:04:38 pm »",
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"Ejs Open Source Charge Particle in Magnetic Field B Java Applet in 3D\nreference:\nhttp://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1800.msg7327#msg7327 Created by prof Hwang Modified by Ahmed\nhttp://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1450msg5484;topicseen#msg5484 Created by prof Hwang\n\nCharge In B-Field\nThis 3D Ejs Charge particle In B-Field model allows the user to simulate a moving charged particle in two identical uniform magnetic fields separated by a zero magnetic field gap. A charge moving in a magnetic field experiences a magnetic force given by the Lorenz force law",
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"$\\vec{F}=\\vec{v}\\times\\vec{B}*q$ = v*B*q*sin(theta)\nwhere theta specifies the angle between the velocity vector v and the magnetic field B. In this simulation, the velocity and B-field are perpendicular (theta = 90 degrees) and the force is maximum and perpendicular to both v and B as predicted by",
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"$\\vec{v}\\times\\vec{B}$. You can adjust the magnitude of the magnetic field B, the mass m and charge q of the charged particles. The slider at the top controls the width of the field free region (it is a percentage of half the window width). The magnetic field is assumed to be uniform",
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"$Bz\\hat{z}$ inside the magnet region. and the field is zero when outside the boundary.\nYou can change the location and velocity of the charged particle with mouse drag and drop or with sliders.\n[/quote]\nEnjoy!\n\nEmbed a running copy of this simulation\n\nEmbed a running copy link(show simulation in a popuped window)\nFull screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list\nPress the Alt key and the left mouse button to drag the applet off the browser and onto the desktop. This work is licensed under a Creative Commons Attribution 2.5 Taiwan License\n• Please feel free to post your ideas about how to use the simulation for better teaching and learning.\n• Post questions to be asked to help students to think, to explore.\n• Upload worksheets as attached files to share with more users.\nLet's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!",
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"",
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"activities adapted from http://www.opensourcephysics.org/items/detail.cfm?ID=8984 Charge in Magnetic Field Model written by Fu-Kwun Hwang edited by Robert Mohr and Wolfgang Christian\n\n1 When",
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"$Bz\\hat{z}$ is positive and the charge particle is completely inside the",
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"$Bz\\hat{z}$ field region, which way do positively charged particle circle (clockwise or counter-clockwise as view from the top looking down). Use the Fleming left hand (thumb Force, second finger B field and middle finger current i ) or right-hand cross product rule",
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"$\\vec{F}=\\vec{v}\\times\\vec{B}*q$ to determine if the field points into or out of the screen?\n2 Explain why particle traveling in the region without B field, travel in a straight path.\n3 Charged particle that remain in the uniform B magnetic field (orange color field vectors) experience uniform circle motion. Why? What provides the centripetal force?\n4 design an experiment to investigate systematically, in a table the data and effects of varying the charge particle mass, m and charged q.\n5 Do they experience the same force F?\n6 What accounts for particle moving in circles of different radii (for the ones that stay in the uniform magnetic field)?\n7 How can you change the different parameters to decrease the radius? Explain why each change results in a smaller radius.\n8 challenging Optional: If you have EJS installed, modify this model to simulate a cyclotron. http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/cyclot.html It may be useful for you to know that Model ->Initialization page to see how the initial position and velocity of the particle(s) is(are) set as well as looking at the Model-> Custom page to see the equations of motion for a particle in the magnetic field as well as the gap region.\n « Last Edit: December 08, 2010, 01:34:54 pm by lookang »",
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"Logged\nlookang\nModerator\nHero Member",
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"Offline\n\nPosts: 1796",
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"http://weelookang.blogspot.com",
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"",
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"« Embed this message Reply #1 on: October 04, 2010, 02:07:56 pm »",
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"changes made\n\n1. color scheme\n2 magnet NS text and position\n3 B field visual start at edge of magnet for ease of associating the influence of F = q.v^B where ^ is cross product.\n4 rearrange the bottom panel\n5 add z and vz into the evolution page and values display for http://link.aip.org/link/?AJP/65/726/1 journal article has this student learning challenge (Bagno & Eylon, 1997)\n\nThe question is the paper is:\nThe velocity of a charged particle moving in a magnetic field is always perpendicular to the direction of the field.\n\nThe responses:\n37% think it is true,\nThe reasons and interviews analyzed indicated that the causes are:\na. Recitation of formula: v, B and F are always perpendicular according to left hand or right screw law 81\nb. No reason 19\nreference:\nBagno, E., & Eylon, B.-S. (1997). From problem solving to a knowledge structure: An example from the domain of electromagnetism. American Journal of Physics, 65( 8 ), 726-736. doi: 10.1119/1.18642\n\nMy thoughts:\nbut the answer is false. it could be a supposition of uniform velocity and circular motion, thus there is an angle =! 90o between v and B, much like a helix path\n\n6 add Force display value F = q.v^B\n7 activate the 3 axes coordinate system for ease of communicate and associating motion to x y z direction\n\nother good resources:\nhttp://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1431.0 Charged particle motion in static Electric/Magnetic field by Fu-Kwun Hwang\nhttp://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=36 Charged particle motion in E/B Field JDK version by Fu-Kwun Hwang\nhttp://www.opensourcephysics.org/items/detail.cfm?ID=8984 Charge in Magnetic Field Model written by Fu-Kwun Hwang\nedited by Robert Mohr and Wolfgang Christian\nhttp://www.opensourcephysics.org/items/detail.cfm?ID=9997 Charge Trajectories in 3D Electrostatic Fields Model written by Andrew Duffy\nhttp://www.opensourcephysics.org/items/detail.cfm?ID=8996 E x B Trajectory Model written by Anne Cox\nhttp://www.compadre.org/osp/items/detail.cfm?ID=8984 Charge in Magnetic Field Model written by Fu-Kwun Hwang edited by Robert Mohr and Wolfgang Christian\n *** There are 1 more attached files. You need to login to acces it! « Last Edit: December 07, 2010, 04:35:28 pm by lookang »",
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"Logged\nFu-Kwun Hwang\nAdministrator\nHero Member",
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"Offline\n\nPosts: 3086",
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"« Embed this message Reply #2 on: October 05, 2010, 10:08:36 am »",
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"What is the trajectory in your mind if there is an electron inside a vacuum with magnetic field?\n\nAssume the energy of the electron is 1keV: try to calculate it's velocity.\nAssume the magnetic field is 1kG(Gauss)=0.1T(Tesla) and velocity is perpendicular to the magnetic field.\n: try to calculate it radius and frequency.\n\nWith the above calculated data: try to think of the motion of the electron.\n\nIf the velocity of the electron is not perpendicular: assume 1% of velocity is parallel to the magnetic field.\nWhat do you think about the average motion of the electron will be look like?",
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"Logged\nlookang\nModerator\nHero Member",
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"Offline\n\nPosts: 1796",
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"http://weelookang.blogspot.com",
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"« Embed this message Reply #3 on: October 05, 2010, 10:38:01 am »",
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"What is the trajectory in your mind if there is an electron inside a vacuum with magnetic field?\na beam of electrons in helix?\n\nAssume the energy of the electron is 1keV: try to calculate it's velocity.\nAssume the magnetic field is 1kG(Gauss)=0.1T(Tesla) and velocity is perpendicular to the magnetic field.\n: try to calculate it radius and frequency.\n0.5*m*v^2 = E\n0.5*9.1*10-31*v^2 = 1*10^3*1.6*10^-19\nv = 1.88*10^6 m/s\n\nassume F = m*v^2/r is valid since perpendicular b and v\nB*v*q = m*v^2/r\n0.1*1.88*10^6*1.6*10^-19 = (9.1*10^-31*1.88*10^6)/r\nr = 1.06x10^-4 m\n\nassume circle path\n2*pi*r / T = v\n3.57x10^-10 s = T\n\nWith the above calculated data: try to think of the motion of the electron.\na beam of electrons in helix moving away from the circular axes?\n\nIf the velocity of the electron is not perpendicular: assume 1% of velocity is parallel to the magnetic field.\nWhat do you think about the average motion of the electron will be look like?\nlook like a beam in the vz direction? a beam of electrons in helix motion moving away from the circular axes x and y?\n\nAm i visualizing this.?",
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"http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?p=27760&sid=9a745d78c23d14d82acc194a8389c7a2\n\nnot sure whether this is what you mean",
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"« Last Edit: October 05, 2010, 10:44:48 am by lookang »",
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"Logged\nFu-Kwun Hwang\nAdministrator\nHero Member",
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"« Embed this message Reply #4 on: October 05, 2010, 10:17:50 pm »",
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"Think about the magnitude of the physics quantity is very important when we study physics.\n\nDid you notice that the radius is very small:",
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"$10^{-4}$ m\nAnd the period of one resolution is very small,too.",
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"$T=3.57\\times 10^{-10}$ s.\nSo the electron will make almost",
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"$3\\times 10^9$ turns in one second.\nThere is no way you can observed such trajectory with your eye or ordinary device.\nAnd the velocity is very large.",
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"$v=1.88\\times 10^6$ m/s.\nIf only 1% of velocity component is along the field line, the electron will move along the field line with speed",
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"$u\\approx 10^4$m/s",
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"$\\approx 10$ km/s.\n\nSo on average, the electron will move along the field line. That is also how electrons move between sourth pole and north pole-- following the field line of earth's magnetic field.\n\nWhen the magnetic field is very strong, it will force electron moving back -- that need more physics analysis.\n(or check out Helmholtz coil / particle trapped in magnetic mirror field)",
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"Logged\n Pages: Go Up\nThere is a better way to do it; find it. ...\"Thomas Edison(1847-1931, American inventor, 1093 patients)\"\nJump to:\n\n Related Topics Subject Started by Replies Views Last post",
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"Ejs open source Magnetic Field due to moving charges & current java applet Collaborative Community of EJS lookang 8 20778",
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"September 29, 2010, 07:43:38 pm by lookang",
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"August 15, 2013, 07:41:28 am by lookang",
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"Ejs Open Source Bar Magnetic Field Lines Model Java Applet Collaborative Community of EJS lookang 7 11627",
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"December 20, 2012, 06:55:06 pm by Fu-Kwun Hwang",
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"Ejs Open Source Magnetic Dipole Field Vector & Field Lines 3D Model Java Applet Collaborative Community of EJS lookang 2 9914",
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"December 13, 2012, 10:03:40 am by lookang\nPage created in 1.161 seconds with 23 queries.",
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https://code.tutsplus.com/tutorials/artificial-intelligence-series-part-1-path-finding--active-4439 | [
"Unlimited Plugins, WordPress themes, videos & courses! Unlimited asset downloads! From \\$16.50/m",
null,
"# Artificial Intelligence Series - Part 1: Path Finding\n\nDifficulty:IntermediateLength:LongLanguages:\n\nThis tutorial is the first of three which discuss how to give some Artificial Intelligence (AI) to games and apps you create. In this first tutorial we are going to learn about path finding.\n\n## Final Result Preview\n\nLet's take a look at the final result we will be working towards:\n\nClick a point, then another point and the AI will figure out a shortest possible path to get between them. You can use the drop-down list to select the AI algorithm to use.\n\n## Step 1: AI Series Overview\n\nThis tutorial will be the first of three which will discuss giving Artificial Intelligence (AI) to games and apps you create. You might think that this sounds just too hard, but it is actually pretty simple! I will explain two key aspects of the AI in games and then create a cool game using what we learn. I hope you enjoy following this short series!\n\n## Step 2: Introduction\n\nIn games, one of the most important aspects is to make it very efficient, to do what it has to do using minimal resources. There's a whole science just for that.\n\nIn this tutorial we are going to cover a key aspect in game development: pathfinding. We will discuss the 2 best methods, how they work, how we make them work in AS3, then last but not least, when to use them. Let's start..\n\n## Step 3: So What's Pathfinding About?\n\nSuppose you are in the middle of a forest with 50 different potential routes in front of you, you have almost no energy and need to find the best way out, how would you do it?\n\nOne option would be to follow each of the 50 routes; a trial and error method. You will certainly get out but that would take too much time. This isn't a new problem and many people have come up with ideas of how to solve it. One of them was Edsger W. Dijkstra who developed an algorithm to get the shortest path given a graph, a source and a target.\n\nBefore we can start solving the problem, we must create those elements, the graph which contains all the information, the nodes which are the waypoint of the graph, and the edges that connect the nodes. Think of it as the graph being a map: the nodes are the cities and the edges the highways that connect them.\n\n## Step 4: Setting up the Project\n\nBefore we start coding, let's first set up our workspace. Create a new folder and name it \"PathFinding\". Inside it create another folder and name it \"Graph\". Then in the PathFinding folder create a new Flash (AS3) FLA file and call it \"PathFinding\".\n\n## Step 5: Creating the Node Symbol\n\nIn the \"PathFinding.fla\" file, go to Insert > New Symbol and create a new MovieClip called Node. Select \"export for actionscript\" and as the class write \"Graph.Node\"\n\n## Step 6: Drawing the Node\n\nInside the Node Movieclip, create a new Circle with a radius of 25 and place it in the center. Then add a new text box, make it Dynamic and give it an instance name of \"idx_txt\" and position it at the center of the stage.\n\n## Step 7: Creating the Node\n\nEach node is composed of 2 main elements; an index in order to identify it, and its position on the graph. With this in mind create a new class inside the Graph folder/package and name it \"Node\". Make it extend the sprite class, then just add two variables: one int for the index, and a Vector3D for the position. Also, add their corresponding set and get methods:\n\n## Step 8: Creating the Edge\n\nThe edge contains three elements: the indices of the two nodes that it connects, and its \"cost\" (or distance between nodes in this case). Again in the Graph package/folder create a new class and call it \"Edge\"; make it extend the Sprite class again.\n\nNow add two ints for the nodes, and one Number for the cost, then add the corresponding set and get methods:\n\n## Step 9: Creating the Graph\n\nThe Graph is a more complex element since it must store all the information. In the Graph folder, create a new class and name it \"Graph\". Since it just contains data, there's no need to extend the Sprite class.\n\nThe graph will contain a Vector with all the nodes on the graph, another Vector with the edges that each node has and a static int in order to get the next index available for a node. This way, it will be easier to access all the elements in the graph, since the node index is the key of the edges vector, meaning that if you want to get the edges of a node with index 3, you access the edges vector at position 3 and you will get the edges for that node.\n\nNow let's create these variables:\n\nThe graph also needs methods to add and get nodes, and edges:\n\n## Step 10: Building a Graph\n\nSo now that we have all the elements needed, let's go ahead and build a graph.\n\nIn the PathFinding folder, create a new class and call it \"Main\", this will be our main class:\n\nNow we need to import the classes we just created since they are in the Graph folder. Then create two variables: one for the graph and another that is an array of the position of the nodes we want to add. After that just add those nodes and their edges:\n\n## Step 11: How to Choose the Right Path?\n\nAs I told you before there are many methods to obtain a path between two nodes, like the DFS(Deep First Search), or BFS(Broad First Search), but sometimes you don't just want to get a path that connect the nodes, you also want to get the \"best\" path, which most of the time will mean the one that takes less time or the closest, depending on what you want.\n\nDuring this tutorial I will just refer to it as the one that costs less to follow. (In our case, remember \"cost\" means distance. In a game, you may wish to find the least dangerous path, or the one that uses least fuel, or the least visible...)\n\nThere are algorithms that work depending on the cost, the most important ones being the Dijkstra Algorithm and the A* (A-Star) Algorithm. Lets talk first about the Dijkstra Algorithm\n\n## Step 12: Introduction to Dijkstra Algorithm\n\nThis algorithm was created by Edsger Dijkstra in 1959. It advances depending on the cost that each edge represents, the edges with lower cost will always be chosen first so when you reach the target, you will be sure that the path is the lowest costing path available.\n\n## Step 13: How the Dijkstra Algorithm Works\n\nBecause this algorithm works with costs, we must have a place where we will be storing the cost of moving to each node from the source (the starting node); call this a cost vector.\n\nFor example if the source is the node 0, when we access the cost vector at position 3, this will return the lowest cost until now of moving from 0 to 3. We will also need a way of storing the shortest path: I'll explain the Shortest Path Tree in the next step.\n\nThe basic steps of the algorithm are the following:\n\n1. Take the closest node not yet analyzed.\n2. Add its best edge to the Shortest Path Tree.\n3. If it is the target node, finish the search.\n4. Retrieve all the edges of this node.\n5. For each edge calculate the cost of moving from the source node to the arrival Node.\n6. If the total cost of this edge is less than the cost of the arrival node until now, then update the node cost with the new one.\n7. Take the next closest node not yet analyzed.\n8. Repeat this until the target is reached, or there are no more nodes available.\n\nAn animation will make more sense. Here we are trying to get from 1 to 6:\n\n## Step 14: Algorithm's Main Elements\n\nThis algorithm is composed of different vectors that will store all the information needed, for example the costs of the nodes, or the best path until now. I will explain this better:\n\n• The costs vector : Here we will store the best cost until now of reaching a certain node from the source. For example if the source node is 3 and we access element 6 of the cost vector, we will obtain the best cost found so far of moving from 3, the source node, to the node 6.\n• The Shortest Path Tree (SPT): This vector contains the lowest cost edge to get to a specific node. This means that if we access the element 7 of the SPT, it will return the best edge to access that node.\n• The Search Frontier (SF): This vector will store the best edge to get to a specific node, almost the same as the SPT, but this will have all the nodes that are not yet in the SPT. This means that the SF will work as a test area where we will test all the edges for one node, when all the edges are analyzed we will be sure that the SF contains the best one, so we can then add it to the SPT.\n• The Indexed Priority Queue (pq): A priority queue is a data structure which always keep its elements in an ordered way. As the algorithm needs to first get the lowest cost node, this structure will keep the nodes in a descending order depending on their cost, but because we want to retrieve the index of the node and not the cost itself, we use an indexed priority queue. For example, if the first element of the pq is node 4, this will mean that it is the one with the lowest cost.\n\nNote: AS3 doesn't contain many data structures including the Indexed Priority Queue, so I coded one to use in this tutorial. To get it just download the source files and import the `utils` folder to the PathFinding folder. The class is `utils.IndexPriorityQ`.\n\n## Step 15: Create an Algorithm Folder\n\nBefore we start coding, in the PathFinding folder, create a new Folder and call it \"Algorithms\".\n\nIn this new folder, create a new AS3 class named \"Dijkstra\":\n\n## Step 16: The Dijkstra Algorithm in AS3\n\nNow let's code the Algorithm. It will need three basic things: the graph itself, the source node and the target node; but we must also create the vectors we just talked about (Cost,SPT,SF). Remember to import the Graph classes:\n\n## Step 17: The Search Function\n\nOnce we have set up the class, we can start coding the algorithm which will be in the \"Search\" function. I will explain the code with comments, so If you still have some questions return to Steps 12 and 13 to remind yourself how this algorithm works:\n\n## Step 18: Getting the Path\n\nWhen the function finishes searching we will have the path all messes up on the SPT, so it is up to us to retrieve it. Since the SPT has the best edge to get to a node, we can work backwards to obtain the best path. Take as a reference the following image which comes from the previous animation:\n\nIf we access the SPT at the target node, which in this case is 6, it will return the best edge to get there. That would be the 6 - 5 edge. Now if we repeat this process with the new node that comes with the edge, the 5 we would obtain the best edge to get to that node, which is the 5 - 2 edge. Repeating this process once again with the node 2, will give us the edge 2 - 1, so we finally get to the beginning, now joining al those edges we get the best path: 6 > 5 > 2 >1.\n\nAs you see we have to work backwards starting on the target and moving to the source to get the best path.\n\n## Step 19: The getPath Function\n\nCreate a new function that will return an Array with the nodes of the path, call it \"getPath\":\n\n## Step 20: The Complete Dijkstra Class\n\nWe have finished almost everything, we just now have to fill the constructor and call the search function, so the class will look like this:\n\n## Step 21: Creating a New Graph\n\nIn order to see a better result of the algorithm, I wrote a class that helps with the creating of graphs. It's called Grapher and can be found inside the utils folder that comes with the source files. This class creates a grid of nodes from where we can observe how the algorithm is moving through the graph.\n\nWith this class, open again the \"Main.as\" file and modify it. We will now have the following code:\n\nSave and run the file and you will obtain this result:\n\n## Step 22: Using the Dijkstra Algorithm\n\nNow let's go ahead and perform a search with the new algorithm we have just done. Import the Dijkstra class, create an instance of it and call the getPath function:\n\nSave and run the file. You will see the edges that the algorithm analyzed as red lines, the best path found (from node 24 to node 35) appearing as a black line:\n\n## Step 23: Isn't There Something Better?\n\nAs we can see, the Dijkstra algorithm does find the shortest path between two nodes, but as the last image shows, there are a lot of red lines. This means that all those edges were analyzed, which isn't a big deal because we have a small graph, but imagine a bigger graph; that would be just too many edges to analyze. If we could just find a way to reduce this and make the algorithm even more efficient... well I present you the A* (A-Star) algorithm.\n\n## Step 24: The A* Algorithm\n\nThe A* algorithm is a modified version of Dijkstra. It takes into consideration a modified way of getting the cost of each node with an heuristic approach. This means that we provide a little \"help\" for the algorithm and tell him where to go, working as a compass and moving the algorithm directly to the target.\n\nInstead of calculating the cost of a node by summing the edge cost to the stored node cost, it is now calculated by summing the stored node cost and an Heuristic cost, which is an estimation of how close to the target the node we are analyzing is. This new cost is called the F cost\n\n## Step 25: The F Cost\n\nThe F cost is calculated using the following formula: F = G + H, where G is the cost of the node and H is the heuristic cost of that node to the target. In this tutorial the H cost will be calculated using the Euclidean distance between the node and the target, which is the straight line distance between the two nodes.\n\nWhat this does is that at the end, the nodes with a lower F cost will be the first ones tested and because the F cost will depend mostly on the H cost, at the end the algorithm will always prioritize the nodes closest to the target.\n\n## Step 26: The A* Algorithm in AS3\n\nIn the Algorithms folder, create a new class named \"Astar\". The variables inside the class will be almost the same as the Dijkstra class, but here we will have another vector to store the F cost of each node:\n\n## Step 27: The New Search Class\n\nThe only difference between the Dijkstra algorithm search function and this one will be that the nodes will be sorted (in the Indexed Priority Queue) depending on the F cost vector and that the F cost vector will be given by the calculated H cost and the stored G cost:\n\n## Step 28: The Complete A* Class\n\nThese are the only changes needed, since the getPath function is the same for both classes. At the end the class will be this one:\n\n## Step 29: Using the A* Algorithm\n\nOnce again open the \"Main.as\" file and import the Astar class, then erase the Dijkstra search we created, replacing it with an A* search:\n\nSave and run the file. As you can see, the result is very impressive, there are no red lines meaning that the search went directly from the source to the target without analyzing extra edges:\n\n## Step 30: Which One is better?\n\nWell, even though the A* algorithm is faster and better in getting a direct path from the source to the target, there will be some cases where Dijkstra will be the best option.\n\nFor example, imagine an RTS game, where you've instructed your villager to go and find some resources; with the A* algorithm you would have to do a search for every resource on the map and then analyze which one is closer. With a Dijkstra search, since it expands the same amount to all directions, the first resource it finds will be the best one to gather and you will have performed just one search versus the many searches of the A*.\n\nBasically, you will want to use the Dijkstra algorithm when you are doing a general search and the A* algorithm when you are looking for a specific item.\n\n## Conclusion\n\nThat's it for this tutorial, I hope you liked it and will use it for your projects. Keep track of the next tutorial of this AI series and let me know what you think about it."
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https://avesis.metu.edu.tr/yayin/2698af29-66b3-4680-8a4f-814fcdde77ff/a-decomposition-approach-for-undiscounted-two-person-zero-sum-stochastic-games | [
"## A decomposition approach for undiscounted two-person zero-sum stochastic games\n\nMATHEMATICAL METHODS OF OPERATIONS RESEARCH, cilt.49, sa.3, ss.483-500, 1999 (SCI İndekslerine Giren Dergi)",
null,
"",
null,
"• Yayın Türü: Makale / Tam Makale\n• Cilt numarası: 49 Konu: 3\n• Basım Tarihi: 1999\n• Doi Numarası: 10.1007/s001860050063\n• Dergi Adı: MATHEMATICAL METHODS OF OPERATIONS RESEARCH\n• Sayfa Sayıları: ss.483-500\n\n#### Özet\n\nTwo-person zero-sum stochastic games are considered under the long-run average expected payoff criterion. State and action spaces are assumed finite. By making use of the concept of maximal communicating classes, the following decomposition algorithm is introduced for solving two-person zero-sum stochastic games: First, the state space is decomposed into maximal communicating classes. Then, these classes are organized in an hierarchical order where each level may contain more than one maximal communicating class. Best stationary strategies for the states in a maximal communicating class at a level are determined by using the best stationary strategies of the states in the previous levels that are accessible from that class. At the initial level, a restricted game is defined for each closed maximal communicating class and these restricted games are solved independently. It is shown that the proposed decomposition algorithm is exact in the sense that the solution obtained from the decomposition procedure gives the best stationary strategies for the original stochastic game."
]
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null,
"https://avesis.metu.edu.tr/Content/images/integrations/small/integrationtype_2.png",
null,
"https://avesis.metu.edu.tr/Content/images/integrations/small/integrationtype_1.png",
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http://docs.optionscity.com/city_admin/reference/6/pages/199 | [
"# Margin Calculations\n\nCity API supports a modified version of live SPAN (standard portfolio analysis of risk) margin calculations for some exchanges. The exchanges currently supported are all Globex exchanges (CME, NYMEX, COMEX, CBOT, KCBT) and the ICE. If a position is held on instruments from other exchanges, they will not factor in to margin calculations.\n\nThe official documentation for SPAN can be found here.\n\nAt a high level, the margin requirement is the scan risk (also called portfolio risk), plus an intra-commodity spread charge (also called basis risk), minus an inter-commodity spread credit (not currently included), minus the net option value of the position. The greater of this value and the short option minimum is the margin requirement.\n\nThe inputs for scanning risk, basis risk and short option minimums is set by the exchange.\n\nThere are two differences between the margin calculated by City API and the official SPAN value. One, City API does not currently include the inter-commodity spread credit. Two, in some cases the exchange does not supply the scanning risk for a given strike. If this is the case, the scanning risk for the nearest available strike is used.\n\n### Accounting for Open Orders\n\nCity API needs to consider the highest possible margin requirement that could result at any instant to perform this check. That is, it assumes that any combination of open orders could conceivably trade at any instant. This calculation can become costly to compute, however if an account has many open orders (the number of calculations is exponential with the number of orders).\n\nTherefore, City API does not calculate the exact worst-case, but instead calculates an upper bound that is much cheaper to calculate (we always err on the side of requiring more, not less margin). Specifically, we make a modification to the calculation to address open orders: we do not apply any credits from risk arrays. That is, we assume the parts that add margin could be applicable, but not the parts that reduce margin."
]
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https://www.geeksforgeeks.org/greatest-odd-factor-of-an-even-number/?ref=rp | [
"# Greatest odd factor of an even number\n\nGiven an even number N, the task is to find the greatest possible odd factor of N.\nExamples:\n\nInput: N = 8642\nOutput: 4321\nExplanation:\nHere, factors of 8642 are {1, 8642, 2, 4321, 29, 298, 58, 149} in which odd factors are {1, 4321, 29, 149} and the greatest odd factor among all odd factors is 4321.\n\nInput: N = 100\nOutput: 25\nExplanation:\nHere, factors of 100 are {1, 100, 2, 50, 4, 25, 5, 20, 10} in which odd factors are {1, 25, 5} and the greatest odd factor among all odd factors is 25.\n\nNaive Approach: The naive approach is to find all the factors of N and then select the greatest odd factor from it.\nTime Complexity: O(sqrt(N))\n\nEfficient Approach: The efficient approach for this problem is to observe that every even number N can be represented as:\n\n```N = 2i*odd_number\n```\n\nTherefore to get the largest odd number we need to divide the given number N by 2 untill N becomes an odd number.\n\nBelow is the implementation of the above approach:\n\n## C++\n\n `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` ` `// Function to print greatest odd factor ` `int` `greatestOddFactor(``int` `n) ` `{ ` ` ``int` `pow_2 = (``int``)(``log``(n)); ` ` ` ` ``// Initialize i with 1 ` ` ``int` `i = 1; ` ` ` ` ``// Iterate till i <= pow_2 ` ` ``while` `(i <= pow_2) ` ` ``{ ` ` ` ` ``// Find the pow(2, i) ` ` ``int` `fac_2 = (2 * i); ` ` ``if` `(n % fac_2 == 0) ` ` ``{ ` ` ``// If factor is odd, then ` ` ``// print the number and break ` ` ``if` `((n / fac_2) % 2 == 1) ` ` ``{ ` ` ``return` `(n / fac_2); ` ` ``} ` ` ``} ` ` ` ` ``i += 1; ` ` ``} ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` ` ``// Given Number ` ` ``int` `N = 8642; ` ` ` ` ``// Function Call ` ` ``cout << greatestOddFactor(N); ` ` ``return` `0; ` `} ` ` ` `// This code is contributed by Amit Katiyar `\n\n## Java\n\n `// Java program for the above approach ` `class` `GFG{ ` ` ` `// Function to print greatest odd factor ` `public` `static` `int` `greatestOddFactor(``int` `n) ` `{ ` ` ``int` `pow_2 = (``int``)(Math.log(n)); ` ` ` ` ``// Initialize i with 1 ` ` ``int` `i = ``1``; ` ` ` ` ``// Iterate till i <= pow_2 ` ` ``while` `(i <= pow_2) ` ` ``{ ` ` ` ` ``// Find the pow(2, i) ` ` ``int` `fac_2 = (``2` `* i); ` ` ``if` `(n % fac_2 == ``0``) ` ` ``{ ` ` ` ` ``// If factor is odd, then ` ` ``// print the number and break ` ` ``if` `((n / fac_2) % ``2` `== ``1``) ` ` ``{ ` ` ``return` `(n / fac_2); ` ` ``} ` ` ``} ` ` ``i += ``1``; ` ` ``} ` ` ``return` `0``; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` ` ``// Given Number ` ` ``int` `N = ``8642``; ` ` ` ` ``// Function Call ` ` ``System.out.println(greatestOddFactor(N)); ` `} ` `} ` ` ` `// This code is contributed by divyeshrabadiya07 `\n\n## Python3\n\n `# Python3 program for the above approach ` ` ` `# importing Maths library ` `import` `math ` ` ` `# Function to print greatest odd factor ` `def` `greatestOddFactor(n): ` ` ` ` ``pow_2 ``=` `int``(math.log(n, ``2``)) ` ` ` `# Initialize i with 1 ` ` ``i ``=` `1` ` ` `# Iterate till i <= pow_2 ` ` ``while` `i <``=` `pow_2: ` ` ` `# find the pow(2, i) ` ` ``fac_2 ``=` `(``2``*``*``i) ` ` ` ` ``if` `(n ``%` `fac_2 ``=``=` `0``) : ` ` ` ` ``# If factor is odd, then print the ` ` ``# number and break ` ` ``if` `( (n ``/``/` `fac_2) ``%` `2` `=``=` `1``): ` ` ``print``(n ``/``/` `fac_2) ` ` ``break` ` ` ` ``i ``+``=` `1` ` ` `# Driver Code ` ` ` `# Given Number ` `N ``=` `8642` ` ` `# Function Call ` `greatestOddFactor(N) `\n\n## C#\n\n `// C# program for the above approach ` `using` `System; ` ` ` `class` `GFG{ ` ` ` `// Function to print greatest odd factor ` `public` `static` `int` `greatestOddFactor(``int` `n) ` `{ ` ` ``int` `pow_2 = (``int``)(Math.Log(n)); ` ` ` ` ``// Initialize i with 1 ` ` ``int` `i = 1; ` ` ` ` ``// Iterate till i <= pow_2 ` ` ``while` `(i <= pow_2) ` ` ``{ ` ` ` ` ``// Find the pow(2, i) ` ` ``int` `fac_2 = (2 * i); ` ` ``if` `(n % fac_2 == 0) ` ` ``{ ` ` ` ` ``// If factor is odd, then ` ` ``// print the number and break ` ` ``if` `((n / fac_2) % 2 == 1) ` ` ``{ ` ` ``return` `(n / fac_2); ` ` ``} ` ` ``} ` ` ``i += 1; ` ` ``} ` ` ``return` `0; ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ` ` ``// Given number ` ` ``int` `N = 8642; ` ` ` ` ``// Function call ` ` ``Console.WriteLine(greatestOddFactor(N)); ` `} ` `} ` ` ` `// This code is contributed by gauravrajput1 `\n\nOutput:\n\n```4321\n```\n\nTime Complexity: O(log2(N))\n\nAttention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.\n\nMy Personal Notes arrow_drop_up",
null,
"Check out this Author's contributed articles.\n\nIf you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.\n\nPlease Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below.\n\nArticle Tags :\nPractice Tags :\n\n2\n\nPlease write to us at [email protected] to report any issue with the above content."
]
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null,
"https://media.geeksforgeeks.org/auth/avatar.png",
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https://papers.nips.cc/paper/2019/file/3a20f62a0af1aa152670bab3c602feed-Reviews.html | [
"NeurIPS 2019\nSun Dec 8th through Sat the 14th, 2019 at Vancouver Convention Center\nPaper ID: 1560 PAC-Bayes under potentially heavy tails\n\n### Reviewer 1\n\nThe paper is quite pleasant to read and results are neatly introduced. The paper does not introduce radically new ideas, but rather elegantly combines existing concepts (the robust mean estimator from Catoni and Giulini, PAC-Bayes bounds from McAllester, Bégin et al., Alquier and Guedj) to obtain a new result: a PAC-Bayes bound with a slow regime ($\\mathcal{O}(1/\\sqrt{n})$) and with logarithmic confidence, holding for unbounded losses. Overall, the paper makes a significant contribution to the PAC-Bayes literature and I recommend acceptance. Major comments (no particular order): * it might be useful to novice readers to briefly explain why the strategy presented in Theorem 1 (McAllester's bound) fails when the loss is not bounded. * Section 3, the $x_i$s need to be independent for Lemma 3 to hold. The key idea to derive results from Section 4 is to pick $x_i:=\\ell(h;z_i)$ but this seems to prevent that hypotheses $h$ may be data-dependent (as this would no longer ensure that the $\\ell(h;z_i)$ would be independent). I see this as a limitation of the method, and would be interested in reading the authors' opinion. Lines 47-48 \"Denote by $\\mathcal{H}$ a model, from which the learner selects a candidate based on the $n$-sized sample\" are a bit awkward in that sense, as it is unclear whether the hypotheses are allowed to depend on data or not. * Practical value of $s$: as it is somewhat classical in the PAC-Bayes literature, a fine tuning of the parameters to the algorithm is crucial. While it is true that replacing $m_2$, $M_2$ and therefore $s$ with empirical counterparts yields a computable bound, this is a different bound. Perhaps an additional comment on the control of the approximation might be useful (see also the discussions in Catoni's, Audibert's papers about the tuning of the inverse temperature $\\lambda$ in the Gibbs posterior). * The shifting strategy used in Section 3 (line 141 and onwards) depends on a data-splitting scheme and the subsample size $k$ acts as an additional parameter. Why not optimize (9) with respect to $k$ and how does this choice of $k$ is reflected upon the main bound? * I am unsure that the numerical experiments add much value to the paper in their current form, as the setup appears quite limited ($n\\leq 100$, two data-generating distributions, actual variance used to compute the estimator, etc.) and does not shed much light on how the bound compares to other bounds in the literature. * The paper is lacking a comment on the computational complexity of the overall method. At the end of the story, it boils down to computing the mean of the (robustified) Gibbs posterior, which the authors' claim is reachable with Variational Bayes or MCMC methods. First, I am not sure VB is more tailored to the robustified Gibbs posterior than it is for the \"classical\" one, as the robust mean estimator makes the whole distribution likely to be further from, say, a Gaussian, than the empirical mean. In addition, VB particularly makes sense with a squared loss, which is not necessarily the case. Second, while it is true that the robust mean estimator \"is easy to compute [by design]\" (lines 4-5), it is not easier / cheaper than the empirical mean. I feel the paper should be more clear about where the proposed method stands regarding computational complexity. * The abstract reads: \"[we] demonstrate that the resulting optimal Gibbs posterior enjoys much stronger guarantees than are available for existing randomized learning algorithms\". I feel this might be a bit of an overclaim: the resulting algorithm (robustified Gibbs posterior) does enjoy a PAC-Bayes bound where the dependency on the confidence $\\delta$ is optimal -- but does that always make the bound tighter? As it is, no evidence is provided in the paper to support that claim. A (possibly numerical) more direct comparison between the bound between lines 202 and 203 (bound for the robustified Gibbs posterior) and e.g. the bound from Alquier and Guedj between lines 75 and 76 (which is a special case as their main bound holds in a more general setting), specialized to their optimal Gibbs posterior (Alquier and Guedj 2018, Section 4), would be a start. * The paper is also lacking a comparison with Tolstikhin and Seldin's bound, which also implies a variance-term. Tolstikhin and Seldin (2013), \"PAC-Bayesian-empirical-Bernstein inequality\", NIPS Minor comments: * $\\mathbf{K}(\\cdot;\\cdot)$ (the Kullback-Leibler divergence) is never defined. * Line 61, I suggest to rephrase \"Optimizing this upper bound\" as \"Optimizing this upper bound with respect to $\\rho$\" to make it even more clear to the reader. * Replace $l$ by $\\ell$ (more pleasant on the eye!) * Line 82, typo: the as --> as the * Line 89, $x_1,\\dots,x_n\\in\\mathbb{R}_+$ as negative values will not be considered in the sequel * Line 109, typo: the the --> the * Line 119, repeated \"assume\" * References: improper capitalization of Rényi === after rebuttal === I thank the authors for taking the time to address my concerns in their rebuttal, which I find satisfying. After reading other reviews, rebuttal and engaging in discussion, my score remains unchanged. I have a remark on the proof of Theorem 9 though. On page 17 of the supplementary material, right after equation (25) the authors invoke Proposition 4 to control the probabillity of deviation of X (lines 459--460 - I'll call this inequality (*)). In Proposition 4, the estimator $hat{x}$ depends on $s$ which depends on $\\delta$. So I suspect bounding $P(exp(cX)>exp(\\epsilon))$ uniformly in $\\epsilon$ might not be licit as it is. From my understanding, the bound should still be valid for any $\\epsilon$ large enough so that the right hand-side of inequality (*) is smaller than $\\delta$ -- but then integrating with respect to $\\epsilon$ might not be licit. I am very sorry for not spotting this in my initial review -- please correct if this is wrong.\n\n### Reviewer 2\n\nTo my knowledge, this is original research and the method is new. The authors make crystal clear comparison with previous work, and related papers are adequately cited (maybe except at the end of first paragraph, where only a review paper about variational approximations of the Gibbs posterior is cited, missing many papers that derive learning algorithms directly from minimizing PAC-Bayesian bounds, e.g. in Germain et al. (JMLR 2015), Roy et al. (AISTATS 2016)). All claims in the paper are thoroughly discussed demonstrated (in appendix). I did not check the detail of all the proofs in appendices, but as far as I checked I found no mistakes. The paper is very well written and organized. I appreciated the fact that the authors take the time (and space) to motivate the results, even by providing a so-called \"pre-theorem\" that is less general than their main one, but helps the reader understand the significance of the new bound. Finally, I have no doubt that the result is significant: the new PAC-Bayesian bound provided in this paper tackles a limitation of many PAC-Bayesian bounds from the luterature, contributing in opening the way to applying this theory to a broader family of problems.\n\n### Reviewer 3\n\nThe paper is in general well written. According to my reading, this study does bring some new insights into learning in the presence of heavy tails"
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https://socratic.org/questions/how-do-you-find-the-the-length-width-and-height-of-rectangular-prism-if-the-volu | [
"How do you find the the length, width and height of rectangular prism if the volume is h^3+h^2-20h cubic meters?\n\nIf l=length , w=width, h=height of the prism the volume is\n\n$V = l \\cdot w \\cdot h$\n\nBut we know that $V = {h}^{3} + {h}^{2} - 20 h$ hence\n\n$l \\cdot w \\cdot h = {h}^{3} + {h}^{2} - 20 h$\n\n$\\frac{l \\cdot w \\cdot h}{h} = \\frac{{h}^{3} + {h}^{2} - 20 h}{h}$\n\n$l \\cdot w = {h}^{2} + h - 20$\n\n$l \\cdot w = \\left(h + 5\\right) \\cdot \\left(h - 4\\right)$\n\nFrom the last equation we find that $l = \\left(h + 5\\right)$ m and $w = \\left(h - 4\\right)$ m and $h = h$ m"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7393364,"math_prob":1.0000094,"size":376,"snap":"2019-43-2019-47","text_gpt3_token_len":100,"char_repetition_ratio":0.11021505,"word_repetition_ratio":0.0,"special_character_ratio":0.2526596,"punctuation_ratio":0.06410257,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000074,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-16T00:00:55Z\",\"WARC-Record-ID\":\"<urn:uuid:af3104df-00d9-4b4a-a84f-a159b295485f>\",\"Content-Length\":\"32800\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eb43dc46-5a3d-41a7-88de-09d74419d935>\",\"WARC-Concurrent-To\":\"<urn:uuid:d30104d4-2bb2-465f-8979-5a7055b99abb>\",\"WARC-IP-Address\":\"54.221.217.175\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-find-the-the-length-width-and-height-of-rectangular-prism-if-the-volu\",\"WARC-Payload-Digest\":\"sha1:HN2FKROKTAQASNHFRNRW6BEU74VU7F6G\",\"WARC-Block-Digest\":\"sha1:N4QU3VOIUKAQMT5YEYSZKSIQGLAGVBCS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986660829.5_warc_CC-MAIN-20191015231925-20191016015425-00363.warc.gz\"}"} |
https://community.powerbi.com/t5/Desktop/Difference-between-MIN-and-MAX-dates/m-p/999835 | [
"cancel\nShowing results for\nDid you mean:\nHighlighted",
null,
"Helper III\n\n## \\$ Difference between MIN and MAX dates\n\nHello, I am trying to create a card showing the difference in ACV between the MAX date and the MIN date. This is tied to a date filter on a bar graph, so when the user switches years on the graph, the card will reflect the the \\$ change in ACV and % change (that's for another card later) for the new range of years. So, a simple subtraction calc between the first year and last year selected is what I'm looking for for the \\$ change.\n\nHowever, I cannot get it to work! Here is the sample data I am working with:\n\n Year Product Category ACV 2011 A \\$131,282.70 2012 A \\$1,397,633.70 2013 A \\$2,472,531.60 2014 A \\$4,010,372.20 2015 A \\$5,295,672.70 2016 A \\$6,553,524.60\n\nI have a date table in place. Tried several things posted before but I'm not getting anywhere. For example I'm attempting to subtract the ACV from 2012 from the ACV for 2014 to show the total change over time (so not counting 2013). I am not getting the correct value of roughly \\$2.6m\n\n1 ACCEPTED SOLUTION\n\nAccepted Solutions\nHighlighted",
null,
"Super User IV\n\n## Re: \\$ Difference between MIN and MAX dates\n\nTry like\n\n``````measure =\nvar _max = maxx('Date','Date'[Date])\nvar _min = Minx('Date','Date'[Date])\nreturn\nCALCULATE(SUM(Table[ACV]), FILTER(all('Date'), 'Date'[Date] =_max)) -CALCULATE(SUM(Table[ACV]), FILTER(all('Date'), 'Date'[Date] =_min))\n/////////////////////////////////Or\n\nmeasure =\nvar _max = maxx('Date','Date'[Date])\nvar _min = Minx('Date','Date'[Date])\nreturn\nCALCULATE(SUM(Table[ACV]), FILTER(all('Date'), year('Date'[Date]) =year(_max))) -CALCULATE(SUM(Table[ACV]), FILTER(all('Date'), year('Date'[Date]) =year(_min)))``````\n\nMy Recent Blog -Week is not so Weak Connect on Linkedin\n\nProud to be a Super User!\n\n4 REPLIES 4\nHighlighted",
null,
"Super User IV\n\n## Re: \\$ Difference between MIN and MAX dates\n\nTry like\n\n``````measure =\nvar _max = maxx('Date','Date'[Date])\nvar _min = Minx('Date','Date'[Date])\nreturn\nCALCULATE(SUM(Table[ACV]), FILTER(all('Date'), 'Date'[Date] =_max)) -CALCULATE(SUM(Table[ACV]), FILTER(all('Date'), 'Date'[Date] =_min))\n/////////////////////////////////Or\n\nmeasure =\nvar _max = maxx('Date','Date'[Date])\nvar _min = Minx('Date','Date'[Date])\nreturn\nCALCULATE(SUM(Table[ACV]), FILTER(all('Date'), year('Date'[Date]) =year(_max))) -CALCULATE(SUM(Table[ACV]), FILTER(all('Date'), year('Date'[Date]) =year(_min)))``````\n\nMy Recent Blog -Week is not so Weak Connect on Linkedin\n\nProud to be a Super User!\n\nHighlighted",
null,
"Super User IV\n\n## Re: \\$ Difference between MIN and MAX dates\n\n---------------------------------------\n\nPutting square pegs in round holes since 1972.\n\n##### I have a NEW book! DAX Cookbook from Packt\nOver 120 DAX Recipes!\n\nProud to be a Super User!\n\nHighlighted",
null,
"Community Support\n\n## Re: \\$ Difference between MIN and MAX dates\n\nYou could try the logic as below:\n\n``````Measure =\nvar _minyear=MIN('Table'[Year])\nvar _maxyear=MAX('Table'[Year])\nreturn\nCALCULATE(SUM('Table'[ACV]),FILTER(ALLEXCEPT('Table','Table'[Product Category]),'Table'[Year]=_maxyear))-CALCULATE(SUM('Table'[ACV]),FILTER(ALLEXCEPT('Table','Table'[Product Category]),'Table'[Year]=_minyear))``````\n``````Measure 2 =\nvar _minyear=MIN('Table'[Year])\nvar _maxyear=MAX('Table'[Year])\nreturn\nCALCULATE(SUM('Table'[ACV]),FILTER('Table','Table'[Year]=_maxyear))-CALCULATE(SUM('Table'[ACV]),FILTER('Table','Table'[Year]=_minyear))``````\n\nResult:",
null,
"Regards,\n\nLin\n\nCommunity Support Team _ Lin\nIf this post helps, then please consider Accept it as the solution to help the other members find it more quickly.\nHighlighted",
null,
"Helper III\n\n## Re: \\$ Difference between MIN and MAX dates\n\nAnnouncements",
null,
"#### Community Blog\n\nVisit our Community Blog for articles, guides, and information created by fellow community members.",
null,
"#### Using the Community\n\nNeed help with the Power BI Community? Our 'Using the Community' support articles are a great place to start.",
null,
"#### Galleries\n\nLooking for inspiration on how to present your data? Need instructional videos? Check out our Galleries!",
null,
"#### Community Summit North America\n\nInnovate, Collaborate, Grow. The top training and networking event across the globe for Microsoft Business Applications",
null,
"Top Solution Authors\nTop Kudoed Authors\nUsers online (2,543)"
]
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https://math.stackexchange.com/questions/1238667/is-2-aleph-0-aleph-1?noredirect=1 | [
"Is $2^{\\aleph_0} = \\aleph_1$?\n\nI was reading a thread on Examples of Common False Beliefs in Mathematics on MathOverflow, in which a user wrote:\n\n$$2^{\\aleph_0} = \\aleph_1$$\n\nThis is a pet peeve of mine, I'm always surprised at the number of people who think that $\\aleph_1$ is defined as $2^{\\aleph_0}$ or $|\\mathbb{R}|$.\n\nBut I thought that $\\aleph_0 = |\\Bbb{Z}|$ and $\\aleph_1 = 2^{\\aleph_0} = |\\Bbb{R}|$. I'm surprised to hear the opposite asserted. Who is right, and for what reasons?\n\nNote: I haven't had sufficient mathematical exposure to prove anything about cardinals, but I'd like to understand what's going on here.\n\n• you can read about the continuum hypothesis : en.wikipedia.org/wiki/Continuum_hypothesis – Tryss Apr 17 '15 at 7:04\n• $\\aleph_1$ is defined to be the cardinality of the set of countable ordinal numbers. We have $2^{\\aleph_0} = \\aleph_1$ iff the continuum hypothesis is true. – William Stagner Apr 17 '15 at 7:05\n• Thanks for the references -- I did not know about the continuum hypothesis. – Newb Apr 17 '15 at 7:16\n• – Martin Sleziak Apr 17 '15 at 7:18\n• It is truly a common false belief. (Unlike some of the others in that thread.) It has appeared in \"popular\" math books by non-mathematicians. – GEdgar Apr 17 '15 at 21:42\n\nThe cardinality of real numbers is indeed $2^{\\aleph_0}$. However $\\aleph_1$ is not defined as $2^{\\aleph_0}$ (the cardinality of the continuum, often denoted $\\mathfrak c$), but it is defined as the smallest cardinality strictly greater than $\\aleph_0$. Whether $2^{\\aleph_0}=\\aleph_1$ (termed the continuum hypothesis) can neither been proven nor disproven from ZFC. You can of course add it as additional axiom to ZFC, and in that extended set theory, it is then indeed true. However you can also add the axiom $2^{\\aleph_0}\\ne\\aleph_1$ (that is, the assumption that there exists a set whose cardinality lies strictly in between the natural and the real numbers) to ZFC and get another set theory in which the claim is false.\n\nNow usually mathematicians assume ZFC, not ZFC+continuum hypothesis (nor ZFC+negation of the continuum hypothesis), therefore it is a false belief that this relation must be true.\n\nIt is however not a false belief that this relation is true; that is only an unprovable belief. There's nothing inconsistent with assuming it to be true; you cannot disprove it; however you also cannot prove it. But that's true of many believes.\n\nHowever you cannot use it in a proof unless you explicitly specify it as precondition of what you want to prove (\"assuming the continuum hypothesis is true, …\").\n\n• If we took the continuum hypothesis as an additional axiom , what additional statement that were independent of zfc can be proven? And what if we took negation of continuum hypothesis as an axiom? – A Googler Apr 17 '15 at 7:35\n• math.stackexchange.com/q/178069/34930 – celtschk Apr 17 '15 at 7:49\n• Intuitively, can I imagine this is similar to the fifth postulate of Euclidean geometry? – MonkeyKing Apr 20 '15 at 2:18\n• @MonkeyKing: Indeed, there's a parallel (no pun intended): It was tried to prove it from the other axioms for some time before it was shown to be independent. However I don't think it was ever considered to be self-evident as the fifth postulate was for millennia. – celtschk Apr 20 '15 at 18:38\n\nEssentially, people are confusing $\\aleph_1$ with $\\beth_1$ (pronounced \"beth\", see beth numbers).\n\nMore precisely: People have heard of the $\\aleph$ numbers and cardinal exponentiation and think $\\aleph_1=2^{\\aleph_0}$ which is not true (unless we assume an extra axiom which \"by default\" we do not). However, people have not heard of the $\\beth$ numbers, and their definition:\n\n$$\\beth_{x+1}=2^{\\beth_x}$$\n\nseems to be just what some people think the $\\aleph$ numbers are.\n\nThere are good reasons that $2^{\\aleph_0}=\\aleph_1$ (the continuum hypothesis) is probably true, even if ZFC (+ large cardinals) alone cannot prove it. So people may not be too wrong when naively believing in this proposition.\n\nHowever, we cannot yet be sure, this is still ongoing research. You may want to look at Hugh Woodin's work on \"Ultimate-L\".\n\n• Probably true where? – Asaf Karagila Apr 17 '15 at 20:56\n• Out of curiosity, what are these \"good reasons\" you talk about? From what I've heard (I can't reference where I've heard that), the majority would rather believe CH to be false, not true. – Wojowu Apr 17 '15 at 21:27\n• True from a platonistic point of view, in the sense of \"really true\", just like the axioms of ZFC and large cardinal axioms (and projective determinacy and so on) are true facts about the \"real\" mathematical universe. – Gerald Apr 17 '15 at 21:30\n• @Wojowu: You are right that until about the end of the 20th century most logicians felt that CH should be false (Cohen: \"The power set operation is quite powerful\"). But this has changed. As mentioned, search for \"Woodin\" and \"V=Ultimate-L\". – Gerald Apr 17 '15 at 21:34"
]
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http://www.martinxia.me/2015/07/26/implementing-simple-hashtable/ | [
"Implementing Simple HashTable\n\nImplementing Simple HashTable",
null,
"Edit this problem on GitHub\n\nHash table is a very important and interesting structure. If I were you, I would love to have my own hash table implemented. So let’s get started with the most simple implementation. The following graph might help:",
null,
"As everyone already knows, hash table contains `(key, value)` pairs. For example, when you take “John Smith” as key of hash table, there will be a hash function to transform the `String` type key value to a `Integer` as index. With the index, you could easily get the value. In this case, the `(key, value)` pair will be called as entry of hash table.\n\nNow you are ready to implement it. Straight forward huh. Let’s get to the simplest part.\n\nThe code for hash table entry is:\n\n```public class HashEntry {\nprivate String key;\nprivate int value;\n\nHashEntry(String key, int value) {\nthis.key = key;\nthis.value = value;\n}\n\npublic String getKey() {\nreturn key;\n}\n\npublic int getValue() {\nreturn value;\n}\n}```\n\nThe code for hash table is:\n\n```public class SimpleHashTable {\nprivate int TABLE_SIZE;\nprivate int size;\nprivate HashEntry[] table;\n\n/* Constructor */\npublic SimpleHashTable(int ts) {\nsize = 0;\nTABLE_SIZE = ts;\ntable = new HashEntry[TABLE_SIZE];\nfor (int i = 0; i < TABLE_SIZE; i++)\ntable[i] = null;\n}\n/* Function to get number of key-value pairs */\npublic int getSize() {\nreturn size;\n}\n/* Function to clear hash table */\npublic void clean() {\nfor (int i = 0; i < TABLE_SIZE; i++)\ntable[i] = null;\nsize = 0;\n}\n\n/* function to retrieve value from the table according to key */\npublic int get(String key) {\nint index = findIndex(key);\nreturn table[index] == null ? -1 : table[index].getValue();\n}\n\n/* function to add value to the table */\npublic void put(String key, int value) {\ntable[findIndex(key)] = new HashEntry(key, value);\nsize++;\n}\n\n/* function to remove value to the table */\npublic void remove(String key) {\ntable[findIndex(key)] = null;\nsize--;\n}\n\n/* value to create the Hash code from he name entered */\nprivate int calculateHashCode(String key) {\nint mod = key.hashCode() % TABLE_SIZE;\nreturn mod < 0 ? mod + TABLE_SIZE : mod;\n}\n\nprivate int findIndex(String key) {\nint index = calculateHashCode(key);\nwhile (table[index] != null && !table[index].getKey().equals(key)) {\nindex = (index + 1) % TABLE_SIZE;\n}\nreturn index;\n}\n}```\n\nI guess I don’t have to explain everything, just the core, `findIndex()` and `calculateHashCode()`.\n\nExplanation On Important Functions\n\nThe responsibility of function `calculateHashCode()` is to transform the `String` type key value to a `Integer` as index. Java has a function `String.hashCode()` to similar things, we could exploit that.\n\n```Take s = key.toCharArray(),\nThen s.hashCode() = s*31^(n-1) + s*31^(n-2) + ... + s[n-1].```\n\nNow we have a 1 to 1 map from `String` to `Integer`, the only thing to do is to transform the integer to index of hash table. We can do this by modulo:\n\n```int mod = key.hashCode() % TABLE_SIZE;\nreturn mod < 0 ? mod + TABLE_SIZE : mod;```\n\nBut what if we got collisions on index of hash table? For example:\n\n```key1.hashCode() = 4;\nkey2.hashCode() = 8;\nTABLE_SIZE = 4;```\n\nThen we will need function `findIndex()`:\n\n```int index = calculateHashCode(key);\nwhile (table[index] != null && !table[index].getKey().equals(key)) {\nindex = (index + 1) % TABLE_SIZE;\n}```\n\nIf `index` has already been used in hash table, we use the larger index. Now we can get the efficiency of this simple hash table:\n\n``` Average\t Worst Case\nSpace\t O(n) O(n)\nSearch\t O(1)\t O(n)\nInsert\t O(1)\t O(n)\nDelete\t O(1)\t O(n)```\n\nNow if you pay enough attention, you will find out there is a flaw inside my simple hash table: I never handle hash-full situation. Basically, this hash table will enter a infinite loop in function `findIndex()` if the hash table is full. How do we handle that case?\n\nWe could fix that by using hash table chaining. Instead of having simple `(key, value)` pairs, we will have the `value` as\n\na linked list. If we are in a hash full situation, we just append it to the end of linked list.",
null,
"The code for hash table entry is:\n\n```public class LinkedHashEntry {\nString key;\nint value;\n\n/* Constructor */\n{\nthis.key = key;\nthis.value = value;\nthis.next = null;\n}\n}```\n\nThe code for hash table is:\n\n```public class NotSoSimpleHashTable {\nprivate int TABLE_SIZE;\nprivate int size;\n\n/* Constructor */\npublic NotSoSimpleHashTable(int ts) {\nsize = 0;\nTABLE_SIZE = ts;\nfor (int i = 0; i < TABLE_SIZE; i++)\ntable[i] = null;\n}\n/* Function to get number of key-value pairs */\npublic int getSize() {\nreturn size;\n}\n/* Function to clear hash table */\npublic void clean() {\nfor (int i = 0; i < TABLE_SIZE; i++)\ntable[i] = null;\nsize = 0;\n}\n/* Function to get value of a key */\npublic int get(String key) {\nint hash = (calculateHashCode( key ) % TABLE_SIZE);\nif (table[hash] == null)\nreturn -1;\nelse {\nwhile (entry != null && !entry.key.equals(key))\nentry = entry.next;\nif (entry == null)\nreturn -1;\nelse\nreturn entry.value;\n}\n}\n/* Function to insert a key value pair */\npublic void insert(String key, int value) {\nint hash = (calculateHashCode( key ) % TABLE_SIZE);\nif (table[hash] == null)\nelse {\nwhile (entry.next != null && !entry.key.equals(key))\nentry = entry.next;\nif (entry.key.equals(key))\nentry.value = value;\nelse\n}\nsize++;\n}\n\npublic void remove(String key) {\nint hash = (calculateHashCode( key ) % TABLE_SIZE);\nif (table[hash] != null) {\nwhile (entry.next != null && !entry.key.equals(key)) {\nprevEntry = entry;\nentry = entry.next;\n}\nif (entry.key.equals(key)) {\nif (prevEntry == null)\ntable[hash] = entry.next;\nelse\nprevEntry.next = entry.next;\nsize--;\n}\n}\n}\n/* Function calculateHashCode which gives a hash value for a given string */\nprivate int calculateHashCode(String key) {\nint mod = key.hashCode() % TABLE_SIZE;\nreturn mod < 0 ? mod + TABLE_SIZE : mod;\n}\n}```\n\nImprovements on Performance\n\nOf course, like everything else, there’s always a price for this improvement. The performance may be affected now that we have to go through a linked list instead of getting a single value. Besides, if some specific linked list is getting too long, we will need a rebalance function to improve performance. Those will not be included in this article 😀\n\nHowever, we could make some other improvement. We could use bit-masking instead of modulo.\n\n```private final int tableSize = Integer.highestOneBit(TABLE_SIZE) << 1;\nprivate final int tableMask = tableSize - 1;```\n\nThen we could get the remainder by `key.hashCode() & tableMask` instead of `key.hashCode() % TABLE_SIZE`.\n\nTake the following as example:\n\n``` decimal -> binary\n12345 -> 0011000000111001\n32 -> 0000000000100000\n32 - 1 -> 0000000000011111\n\n12345 & 31 -> 0000000000011001\n25 -> 0000000000011001```"
]
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null,
"http://www.martinxia.me/wp-content/uploads/2015/06/invertocat.png",
null,
"http://www.martinxia.me/wp-content/uploads/2015/07/simple-hashtable.png",
null,
"http://www.martinxia.me/wp-content/uploads/2015/07/hashtable-chaining.png",
null
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https://archive.softwareheritage.org/browse/content/sha1_git:d38934aca2d9b855290256dddf9a8d976f897d61/ | [
"\\name{best.first.search}\n\\alias{best.first.search}\n\\title{ Best-first search }\n\\description{\nThe algorithm for searching atrribute subset space.\n}\n\\usage{\nbest.first.search(attributes, eval.fun, max.backtracks = 5)\n}\n\\arguments{\n\\item{attributes}{ a character vector of all attributes to search in }\n\\item{eval.fun}{ a function taking as first parameter a character vector of all attributes and returning a numeric indicating how important a given subset is }\n\\item{max.backtracks}{ an integer indicating a maximum allowed number of backtracks, default is 5 }\n}\n\\details{\nThe algorithm is similar to \\code{\\link{forward.search}} besides the fact that is chooses the best node from all already evaluated ones and evaluates it. The selection of the best node is repeated approximately \\code{max.backtracks} times in case no better node found.\n}\n\\value{\nA character vector of selected attributes.\n}\n\\author{ Piotr Romanski }\n\\examples{\nlibrary(rpart)\ndata(iris)\n\nevaluator <- function(subset) {\n#k-fold cross validation\nk <- 5\nsplits <- runif(nrow(iris))\nresults = sapply(1:k, function(i) {\ntest.idx <- (splits >= (i - 1) / k) & (splits < i / k)\ntrain.idx <- !test.idx\ntest <- iris[test.idx, , drop=FALSE]\ntrain <- iris[train.idx, , drop=FALSE]\ntree <- rpart(as.simple.formula(subset, \"Species\"), train)\nerror.rate = sum(test\\$Species != predict(tree, test, type=\"c\")) / nrow(test)\nreturn(1 - error.rate)\n})\nprint(subset)\nprint(mean(results))\nreturn(mean(results))\n}\n\nsubset <- best.first.search(names(iris)[-5], evaluator)\nf <- as.simple.formula(subset, \"Species\")\nprint(f)\n\n}"
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http://www.nationaltrustcollections.org.uk/results?Collections=e70712bcfffffe0722480241528b9e78&Page=70 | [
"## You searched in “Baddesley Clinton, Warwickshire (Accredited Museum)”\n\nShow me:\nand\n\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 3,624 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8158451,"math_prob":0.9895909,"size":4696,"snap":"2020-34-2020-40","text_gpt3_token_len":1288,"char_repetition_ratio":0.37041774,"word_repetition_ratio":0.122349106,"special_character_ratio":0.2165673,"punctuation_ratio":0.18276763,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99962384,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-10T22:35:29Z\",\"WARC-Record-ID\":\"<urn:uuid:dcce11f0-43a9-4d63-9639-ff357e6d36dc>\",\"Content-Length\":\"227122\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6bff4f28-f398-4605-aaca-5ddb954e8986>\",\"WARC-Concurrent-To\":\"<urn:uuid:a62ec6d4-efee-4f3b-b596-71b8d23fcf09>\",\"WARC-IP-Address\":\"13.93.104.16\",\"WARC-Target-URI\":\"http://www.nationaltrustcollections.org.uk/results?Collections=e70712bcfffffe0722480241528b9e78&Page=70\",\"WARC-Payload-Digest\":\"sha1:KZ374MJIPBLLNCSQW37UCXTMMOLUJ57E\",\"WARC-Block-Digest\":\"sha1:DMK3KYWIHOAWROIN772HSFHRS6G65DHE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738699.68_warc_CC-MAIN-20200810205824-20200810235824-00298.warc.gz\"}"} |
https://www.daniweb.com/programming/software-development/threads/46375/simulation | [
"I am working on an assignment that involves simulating a volley ball game and I have NO idea. Can anyone help.\nHere is the question\nThanks\n\nOK, so i have worked on it a bit and have developed some code but am having problems with defining when the game should stop because the winning player needs to wins by 2 points and the highest score needs to be atleast 15. here is my code so far\n\n``````def main ():\nprint Intro ()\nprobA, probB, n = getInputs ()\nwinsA, winsB = simNGames (n, probA, probB)\nprintSummary (winsA, winsB)\ndef Intro ():\nprint \"This program simulates a gsme of volleyball between\"\nprint \"two players called A and B.\"\nprint \"player A always has the first serve\"\ndef getInputs ():\na = input (\"Player A wins a serve\")\nb = input (\"Player B wins a serve\")\nn = input (\"How many games do you want to simulate?\")\nreturn a, b, n\ndef simNGames(n, probA, probB) :\nwinsA= winsB = 0\nfor i in range(n):\nscoreA, scoreB = simOneGame (probA, probB)\nif scoreA > scoreB:\nwinsA =winsA + 1\nelse:\nwinsB = winB + 1\nreturn winsA, winsB\ndef simOneGame (probA, probB):\nserving= \"A\"\nscoreA = 0\nscoreB = 0\nwhile not gameOver (scoreA, scoreB):\nif serving == \"A\":\nif random () < probA:\nscoreA = scoreA + 1\nelse:\nserving = \"B\"\nelse:\nif random() < probB:\nscoreB = scoreB + 1\nelse:\nserving = \"A\"\n\nreturn scoreA, scoreB\ndef gameOver(a, b):\nif (a - b) > 2 or < -2\ndef printSummary(winsA, winsB):\nn= winsA + winsB\nprint \"\\nGames simulated\", n\nprint \"Wins for A: %d (%0.1f%%)\" % (winsA, float(winsA)/n*100)\nprint \"Wins for B: %d (%0.1f%%)\" % (winsB, float(winsA)/n*100)\nif __name__=='__main__':main()``````\nBe a part of the DaniWeb community\n\nWe're a friendly, industry-focused community of 1.18 million developers, IT pros, digital marketers, and technology enthusiasts learning and sharing knowledge."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9132043,"math_prob":0.9219842,"size":333,"snap":"2020-24-2020-29","text_gpt3_token_len":70,"char_repetition_ratio":0.06382979,"word_repetition_ratio":0.0,"special_character_ratio":0.1951952,"punctuation_ratio":0.12698413,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9773377,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-03T23:42:06Z\",\"WARC-Record-ID\":\"<urn:uuid:734de800-1bfe-4cb1-a865-26c981d85753>\",\"Content-Length\":\"49874\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:82622adf-5020-4f09-8a86-e7d6c2ad521d>\",\"WARC-Concurrent-To\":\"<urn:uuid:9c762148-a344-402f-ad94-40cdf88e05b0>\",\"WARC-IP-Address\":\"104.22.5.5\",\"WARC-Target-URI\":\"https://www.daniweb.com/programming/software-development/threads/46375/simulation\",\"WARC-Payload-Digest\":\"sha1:BQLEM5D3PG6HPGRRA7ENL6CUYQHJP6O3\",\"WARC-Block-Digest\":\"sha1:Y22BQELGTYRNPFATULLVYFRFU2GGXNWL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347436466.95_warc_CC-MAIN-20200603210112-20200604000112-00154.warc.gz\"}"} |
https://quantumcomputing.stackexchange.com/questions/16207/why-does-sum-n-langle-nm-m-rho-m-m-daggern-rangle-simplify-to-langle-p | [
"# Why does $\\sum_n \\langle n|M_m\\rho M_m^\\dagger|n\\rangle$ simplify to $\\langle \\psi|M_m^\\dagger M_m|\\psi\\rangle$?\n\nI was trying to derive the formula for $$p(m)$$ in exercise 8.2 on page 357 in Nielsen & Chuang. But I am wondering what rule I can apply to simplify this\n\n$$\\mathrm{tr}(\\mathcal{E}_m(\\rho) )= \\sum_n \\langle n | M_m \\rho M_m^{\\dagger} | n\\rangle = \\sum_n \\langle n | M_m | \\psi\\rangle \\langle \\psi | M_m^{\\dagger} | n\\rangle$$\n\nto $$\\langle\\psi|M_m^{\\dagger}M_m|\\psi\\rangle$$?\n\nBecause after this I can’t find any idea to boil down to this.\n\n• @glS.: Sometimes good titles just do not come to mind. It was not intentional. Feb 25 at 10:04\n\n$$\\sum_n \\langle n | M_m | \\psi\\rangle \\langle \\psi | M_m^{\\dagger} | n\\rangle = \\sum_n \\langle \\psi | M_m^{\\dagger} | n\\rangle \\langle n | M_m | \\psi\\rangle = \\langle \\psi | M_m^{\\dagger} I M_m | \\psi\\rangle = \\langle \\psi | M_m^{\\dagger} M_m | \\psi\\rangle$$\nnote that $$\\sum_n|n\\rangle\\langle n| = I$$"
]
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https://vdocuments.net/slope-stability-analysis-with-the-finite-element-slope-stability-analysis-with-the.html | [
"# slope stability analysis with the finite element...\n\nof 19 /19\n1 Slope Stability Analysis with the Finite Element Method Indra Noer Hamdhan 1, 2 Abstract In the evaluation of slopes, the factor of safety values still remain the main indexes for finding out how close or far slopes are from failure. The evaluation can be done my means of conventional slip circle analysis (the limit equilibrium methods) or by means of numerical methods such as the finite element method. This study presents the comparison between slope stability analysis with the finite element method and the limit equilibrium method. Keyword: slope stability analysis, limit equilibrium method, finite element method. Introduction The development of soil and rock mechanics will influence the development of slope stability analyses in geotechnical engineering. Assessing the stability of engineered and natural slopes is a common challenge to both theoreticians and practitioners. The balance of natural slopes may be interrupted by man or nature causing stability problems. Natural slopes that have been stable for many years may suddenly fail due to changes in topography, seismicity, groundwater flows, loss of shear strength, stress change, and weathering (Abramson et al., 2002). Duncan (1996) illustrated that the finite element method can be used to analyze the stability and deformations of slopes. Griffiths and Lane (1999) illustrated that the finite element method represents a powerful alternative method for slope stability analysis which is accurate, adaptable and requires less assumptions, especially concerning the failure mechanism. The failure mechanisms in the finite element method develop naturally through the regions wherein the shear strength of the soil is not sufficient to resist the shear stresses. The main objective of this section is to evaluate and to compare the methods of slope stability analysis between limit equilibrium and finite element method by assuming a Mohr-Coulomb failure criterion. 1 PhD. Student, Computational Geotechnics Group, Institute for Soil Mechanics and Foundation Engineering, Graz University of Technology, Austria. Email: 2 Lecturer, Department of Civil Engineering, National Institute of Technology, Bandung, Indonesia. Email:\n\nPost on 05-Mar-2018\n\n224 views\n\nCategory:\n\n## Documents\n\nEmbed Size (px)\n\nTRANSCRIPT",
null,
"1\n\nSlope Stability Analysis with the Finite Element Method Indra Noer Hamdhan\n\n1, 2\n\nAbstract\n\nIn the evaluation of slopes, the factor of safety values still remain the main indexes for\n\nfinding out how close or far slopes are from failure. The evaluation can be done my\n\nmeans of conventional slip circle analysis (the limit equilibrium methods) or by means\n\nof numerical methods such as the finite element method. This study presents the\n\ncomparison between slope stability analysis with the finite element method and the limit\n\nequilibrium method.\n\nKeyword: slope stability analysis, limit equilibrium method, finite element method.\n\nIntroduction\n\nThe development of soil and rock mechanics will influence the development of slope\n\nstability analyses in geotechnical engineering. Assessing the stability of engineered and\n\nnatural slopes is a common challenge to both theoreticians and practitioners. The\n\nbalance of natural slopes may be interrupted by man or nature causing stability\n\nproblems. Natural slopes that have been stable for many years may suddenly fail due to\n\nchanges in topography, seismicity, groundwater flows, loss of shear strength, stress\n\nchange, and weathering (Abramson et al., 2002).\n\nDuncan (1996) illustrated that the finite element method can be used to analyze the\n\nstability and deformations of slopes. Griffiths and Lane (1999) illustrated that the finite\n\nelement method represents a powerful alternative method for slope stability analysis\n\nwhich is accurate, adaptable and requires less assumptions, especially concerning the\n\nfailure mechanism. The failure mechanisms in the finite element method develop\n\nnaturally through the regions wherein the shear strength of the soil is not sufficient to\n\nresist the shear stresses.\n\nThe main objective of this section is to evaluate and to compare the methods of slope\n\nstability analysis between limit equilibrium and finite element method by assuming a\n\nMohr-Coulomb failure criterion.\n\n1 PhD. Student, Computational Geotechnics Group, Institute for Soil Mechanics and Foundation Engineering, Graz University of Technology, Austria. Email: [email protected] 2 Lecturer, Department of Civil Engineering, National Institute of Technology, Bandung, Indonesia. Email: [email protected]",
null,
"2\n\nLimit Equilibrium Methods\n\nLimit equilibrium methods are the most commonly used approaches in slope stability\n\nanalysis. The fundamental assumption in these methods is that failure occurs through\n\nsliding of a mass along a slip surface. The reputation of the limit equilibrium methods is\n\nprincipally due to their relative simplicity, the ability to evaluate the sensitivity of\n\nstability to various input parameters, and the experience geotechnical engineer have\n\nacquired over the years in calculating the factor of safety.\n\nThe assumptions in the limit equilibrium methods are that the failing soil mass can be\n\ndivided into slices and that forces act between the slices whereas different assumptions\n\nare made with respect to these forces in different methods. Some common features and\n\nlimitation for equilibrium methods in slope stability analysis are summarized in Table 1.\n\nAll methods use the same definition of the factor of safety:\n\nlibriumd for equiss requireShear stre\n\nilngth of soShear streFOS = (1)\n\nThe factor of safety is the factor by which the shear strength of the soil would have to be\n\ndivided to carry the slope into a state of barely stable equilibrium.\n\nThe findings related to the accuracy of the limit equilibrium methods can be reviewed as\n\nfollows:\n\n1) For effective stress analysis of flat slopes, the ordinary method of slices is highly\n\ninaccurate. The computed factor of safety is too low. This method is accurate for φ =\n\n0 analysis, and fairly accurate for any type of total stress analysis using circular slip\n\nsurfaces.\n\n2) For most conditions, the Bishop’s modified method is reasonably accurate. Because\n\nof numerical problems, sometimes encountered, the computed factor of safety using\n\nthe Bishop’s modified method is different from the factor of safety for the same\n\ncircle calculated using the ordinary method of slices.\n\n3) Computed factor of safety using force equilibrium methods are sensitive to the\n\nassumption of the inclination of side forces between slices. A bad assumption\n\nconcerning side force inclination will result in an inaccurate factor of safety.\n\n4) Janbu’s, Morgenstern and Prices’s and Spencer’s method that satisfy all conditions\n\nof equilibrium are accurate for any conditions. All of these methods have numerical\n\nproblems under some conditions.",
null,
"3\n\nTable 1: Features and Limitation for Traditional Equilibrium Methods in Slope\n\nStability Analysis (Duncan and Wright, 1980)\n\nMethod Features and Limitation\n\nSlope Stability Charts (Janbu,\n\n1968, Duncan et al, 1987)\n\n- Accurate enough for many purposes.\n\n- Faster than detailed computer analysis.\n\nOrdinary Method of Slices\n\n(Fellenius, 1927)\n\n- Only for circular slip surfaces.\n\n- Satisfies moment equilibrium.\n\n- Does not satisfy horizontal or vertical force\n\nequilibrium.\n\nBishop’s Modified Method\n\n(Bishop, 1955)\n\n- Only for circular slip surfaces.\n\n- Satisfies moment equilibrium.\n\n- Satisfies vertical force equilibrium.\n\n- Does not satisfy horizontal force equilibrium.\n\nForce Equilibrium Methods (e.g.\n\nLowe and Karafiath, 1960, Army\n\nCorps of Engineers, 1970)\n\n- Any shape of slip surfaces.\n\n- Does not satisfy moment equilibrium.\n\n- Satisfies both vertical and horizontal force\n\nequilibrium.\n\nJanbu’s Generalized Procedure of\n\nSlices (Janbu, 1968)\n\n- Any shape of slip surfaces.\n\n- Satisfies all conditions of equilibrium.\n\n- Permit side force locations to be varied.\n\n- More frequent numerical problems than some\n\nother methods.\n\nMorgenstern and Price’s Method\n\n(Morgenstern and Price, 1965)\n\n- Any shape of slip surfaces.\n\n- Satisfies all conditions of equilibrium.\n\n- Permit side force orientations to be varied.\n\nSpencer’s Method (Spencer,\n\n1967)\n\n- Any shape of slip surfaces.\n\n- Satisfies all conditions of equilibrium.\n\n- Side forces are assumed to be parallel.\n\nThe limitation of limit equilibrium method in slope stability analysis has been\n\ndemonstrated by Krahn (2003). This limitation is caused by the absence of a stress-\n\nstrain relationship in the method of analysis. The limit equilibrium method lacks a\n\nillustrated by Baker et al. (1993).\n\nFinite Element Method\n\nIn the finite element method, the latter analysis, the so-called shear strength reduction\n\n(SSR) technique (Matsui & San 1992, Dawson et al. 1999) can be applied. The angle of\n\ndilatancy, soil modulus or the solution domain size are not critical parameters in this",
null,
"4\n\ntechnique (Cheng, 1997). The safety factor can be obtained, assuming a Mohr-Coulomb\n\nfailure criterion, by reducing the strength parameters incrementally, starting from\n\nunfactored values ϕavailable and cavailable, until no equilibrium can be found in the\n\ncalculations. The corresponding strength parameters can be denoted as ϕfailure and cfailure\n\nand the safety factor ηfe is defined as:\n\nfailure\n\navailable\n\nfailure\n\navailablefe\n\nc\n\nc\n\n==\n\nϕ\n\nϕη\n\ntan\n\ntan (2)\n\nThere are two possibilities to arrive at the factor of safety as defined above.\n\nMethod 1: An analysis is performed with unfactored parameters modelling all\n\nconstruction stages required. The results represent the behaviour for working load\n\nconditions at the defined construction steps. This analysis is followed by an automatic\n\nreduction of strength parameters of the soil until equilibrium can be no longer achieved\n\nin the calculation. The procedure can be invoked at any construction step. This approach\n\nis sometimes referred to as ϕ/c-reduction technique.\n\nMethod 2: The analysis is performed with factored parameters from the outset, i.e.\n\nstrength values are reduced, again in increments, but a new analysis for all construction\n\nstages is performed for each set of parameters. If sufficiently small increments are used\n\nthe factor of safety is again obtained from the calculation where equilibrium could not\n\nbe achieved.\n\nBoth methods are straightforward to apply when using a standard Mohr-Coulomb\n\nfailure criterion. In the finite element method, failure occurs naturally through the zones\n\nwithin the soil mass wherein the shear strength of the soil is not capable to resist the\n\napplied shear stress, so there is no need to make assumption about the shape or location\n\nof the failure surface.\n\nMohr-Coulomb Failure Criterion\n\nThe Mohr-Coulomb failure criterion is commonly used to describe the strength of soil.\n\nThe relationship between shear strength and the principal stresses active on a mass of\n\nsoil can be represented in terms of the Mohr circle of stress where the limits on the\n\nprincipal stress axis represent the major and minor principal stresses, σ1 and σ3. Mohr-\n\nCoulomb's failure criterion (Figure 1) is a line forming a tangent to the circle at point a.\n\nThe slope of this line is the friction angle, φ, and the line intercepts the shear stress axis\n\nat the value of the soil's cohesion, c.",
null,
"5\n\nFigure 1: Mohr-Coulomb failure criterion\n\nSo, the Mohr-Coulomb failure envelope may be described by:\n\n'tan'' ϕστ += c (3)\n\nAlternatively the Mohr-Coulomb criterion can be formulated in terms of principal\n\nstresses as follows:\n\n( ) ( ) 0cossin''''322\n\n1\n\n322\n\n1\n\n1≤−++−= ϕϕσσσσ cf\n\na (4)\n\n( ) ( ) 0cossin''''232\n\n1\n\n232\n\n1\n\n1≤−++−= ϕϕσσσσ cf\n\nb (5)\n\n( ) ( ) 0cossin''''132\n\n1\n\n132\n\n1\n\n2≤−++−= ϕϕσσσσ cf\n\na (6)\n\n( ) ( ) 0cossin''''312\n\n1\n\n312\n\n1\n\n2≤−++−= ϕϕσσσσ cf\n\nb (7)\n\n( ) ( ) 0cossin''''212\n\n1\n\n212\n\n1\n\n3≤−++−= ϕϕσσσσ cf\n\na (8)\n\n( ) ( ) 0cossin''''122\n\n1\n\n122\n\n1\n\n3≤−++−= ϕϕσσσσ cf\n\nb (9)\n\nFigure 2 illustrates a fixed hexagonal cone in principal stress space with the condition fi\n\n= 0 for all yield function. The Mohr-Coulomb plastic potential functions that contain a\n\nthird plasticity parameter, the so-called dilatancy angle ψ are given by:\n\n( ) ( ) ψσσσσ sin''''322\n\n1\n\n322\n\n1\n\n1++−=\n\nag (10)\n\n( ) ( ) ψσσσσ sin''''232\n\n1\n\n232\n\n1\n\n1++−=\n\nbg (11)\n\n( ) ( ) ψσσσσ sin''''132\n\n1\n\n132\n\n1\n\n2++−=\n\nag (12)\n\n( ) ( ) ψσσσσ sin''''312\n\n1\n\n312\n\n1\n\n2++−=\n\nbg (13)\n\n( ) ( ) ψσσσσ sin''''212\n\n1\n\n212\n\n1\n\n3++−=\n\nag (14)\n\n( ) ( ) ψσσσσ sin''''122\n\n1\n\n122\n\n1\n\n3++−=\n\nbg (15)",
null,
"6\n\nFigure 2: The Mohr-Coulomb yield surface (c = 0) (Brinkgreve et al, 2010)\n\nGenerally, the linear perfectly-plastic Mohr-Coulomb model requires five parameter.\n\nThese parameters are the Young’s modulus (E), Poisson’s ratio (ν), cohesion (c),\n\nfriction angle (ϕ), and dilatancy angle (ψ).\n\nSlope Stability Examples\n\nIn this section, five examples of slope stability analysis from Griffiths and Lane (1999)\n\nwill be discussed. These examples will be analyzed by the finite element method and\n\nwill be compared with the limit equilibrium methods. The analysis was performed by\n\nutilizing PLAXIS for the finite element method and Slope/W for the limit equilibrium\n\nmethods. The soil model used in the analysis is the Mohr-Coulomb failure criterion.\n\nHomogeneous slope with no foundation layer\n\nThe height of the homogeneous slope is 10 m and the gradient (horizontal to vertical) is\n\n2:1. Figure 3 illustrates the geometry and the two dimensional finite element meshes\n\nconsisting of 390 15-noded elements.",
null,
"7\n\nFigure 3: Geometry and mesh for a homogeneous slope with no foundation layer\n\nThe soil parameters for this example are given in Table 2.\n\nTable 2: Soil Parameters for Example 1 with Mohr Coulomb Model\n\nDescription Symbol Unit Value\n\nUnit weight γ [kN/m3] 20\n\nEffective young's modulus E’ [kPa] 100000\n\nEffective poisson's ratio ν' [-] 0.3\n\nCohesion (effective shear strength) c' [kPa] 10\n\nFriction angle (effective shear strength) φ' [o] 20\n\nIn Figure 4 the failure mechanism for this example is presented for the finite element\n\nmethod. The result of the slope stability analysis using a limit equilibrium method is\n\npresented in Figure 5.\n\nFigure 4: Failure mechanism for a homogeneous slope with no foundation using the\n\nfinite element method\n\nFOS = 1.348",
null,
"8\n\n1\n\n1 2\n\n3 4\n\n1.386\n\nModel: Mohr-Coulomb\n\nUnit Weight: 20 kN/m³\n\nCohesion: 10 kPa\n\nPhi: 20 °\n\nDistance\n\n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34\n\nElevation\n\n0\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\nFigure 5: Morgenstern and Price Method for a homogeneous slope with no foundation\n\nThe difference of factor of safety between the finite element method and limit\n\nequilibrium methods is only 2.8% and the failure mechanism are similar.\n\nHomogeneous slope with a foundation layer\n\nThe homogeneous slope has a foundation layer with the thickness of half of the slope\n\nheight in this example. The height of the slope is 10 m and the gradient (horizontal to\n\nvertical) is 2:1. Figure 6 shows the geometry and the two dimensional finite element\n\nmesh consisting of 309 15-noded elements.\n\nFigure 6: Geometry and mesh for a homogeneous slope with a foundation layer\n\nThe soil parameters used for this example are the same as given in Table 2. In Figure 7\n\nthe failure mechanism is presented for the finite element method and the result of the\n\nanalysis using a limit equilibrium method is presented in Figure 8.",
null,
"9\n\nFigure 7: Failure mechanism for a homogeneous slope with a foundation layer using\n\nthe finite element method\n\n1\n\n1 2\n\n3 4\n\n5 6\n\n1.386\n\nModel: Mohr-Coulomb\n\nUnit Weight: 20 kN/m³\n\nCohesion: 10 kPa\n\nPhi: 20 °\n\nDistance\n\n0 10 20 30 40 50 60\n\nElevation\n\n0\n\n2\n\n4\n\n6\n\n8\n\n10\n\n12\n\n14\n\n16\n\nFigure 8: Morgenstern and Price Method for a homogeneous slope with a foundation\n\nlayer\n\nThe difference of factor of safety between the finite element method and limit\n\nequilibrium methods is only 3.5% and the computed failure mechanism are similar.\n\nAn undrained clay slope with a thin weak layer\n\nFigure 9 shows the geometry and the two dimensional finite element mesh of the\n\nexample of an undrained clay slope with a thin weak layer. The height of the slope is 10\n\nm and the slope is inclined at an angle of 26.57o (2:1) to the horizontal. The mesh\n\nconsist of 562 15-noded elements.\n\nFOS = 1.339",
null,
"10\n\nFigure 9: Geometry and mesh for an undrained clay slope with a thin weak layer\n\nThe soil parameters for this example are given in Table 3. The analyses are carried out\n\nusing a constant value of undrained shear strength of the soil (cu1) and five different\n\nvalues of undrained shear strength of the thin layer (cu2) with ratios cu2/cu1 equal to 1,\n\n0.8, 0.6, 0.4, and 0.2.\n\nTable 3: Soil Parameters for Example 3 with Mohr Coulomb Model\n\nDescription Symbol Unit Value\n\nUnit weight γ [kN/m3] 20\n\nEffective young's modulus E’ [kPa] 100000\n\nEffective poisson's ratio ν' [-] 0.3\n\nCohesion (undrained shear strength) cu1 [kPa] 50\n\nFriction angle (undrained shear strength) φu [o] 0\n\nIn Figure 10 computed failure mechanisms for this example are presented for the finite\n\nelement method with different ratios cu2/cu1 and the result of the analysis using a limit\n\nequilibrium method is presented in Figure 11.",
null,
"11\n\n(a) (b)\n\n(c) (d)\n\nFigure 10: Failure mechanism for an undrained clay slope with a thin weak layer using\n\nFinite Element Method; (a) cu2/cu1 = 0.8; (b) cu2/cu1 = 0.6; (c) cu2/cu1 = 0.4;\n\n(d) cu2/cu1 = 0.2\n\n1.446\n\nModel: Undrained (Phi=0)\n\nUnit Weight: 20 kN/m³\n\nCohesion: 50 kPa\n\nModel: Undrained (Phi=0)\n\nUnit Weight: 20 kN/m³\n\nCohesion: 40 kPa\n\nDistance\n\n0 10 20 30 40 50 60\n\nElevation\n\n0\n\n2\n\n4\n\n6\n\n8\n\n10\n\n12\n\n14\n\n16\n\n18\n\n20\n\n22\n\n1.400\n\nModel: Undrained (Phi=0)\n\nUnit Weight: 20 kN/m³\n\nCohesion: 50 kPa\n\nModel: Undrained (Phi=0)\n\nUnit Weight: 20 kN/m³\n\nCohesion: 30 kPa\n\nDistance\n\n0 10 20 30 40 50 60\n\nElevation\n\n0\n\n2\n\n4\n\n6\n\n8\n\n10\n\n12\n\n14\n\n16\n\n18\n\n20\n\n22\n\n(a) (b)\n\n0.902\n\nModel: Undrained (Phi=0)\n\nUnit Weight: 20 kN/m³\n\nCohesion: 50 kPa\n\nModel: Undrained (Phi=0)\n\nUnit Weight: 20 kN/m³\n\nCohesion: 20 kPa\n\nDistance\n\n0 10 20 30 40 50 60\n\nElevation\n\n0\n\n2\n\n4\n\n6\n\n8\n\n10\n\n12\n\n14\n\n16\n\n18\n\n20\n\n22\n\n0.451\n\nModel: Undrained (Phi=0) Unit Weight: 20 kN/m³\n\nCohesion: 50 kPa\n\nModel: Undrained (Phi=0)\n\nUnit Weight: 20 kN/m³\n\nCohesion: 10 kPa\n\nDistance\n\n0 10 20 30 40 50 60\n\nElevation\n\n0\n\n2\n\n4\n\n6\n\n8\n\n10\n\n12\n\n14\n\n16\n\n18\n\n20\n\n22\n\n(c) (d)\n\nFigure 11: Morgenstern and Price Method for an undrained clay slope with a thin weak\n\nlayer; (a) cu2/cu1 = 0.8; (b) cu2/cu1 = 0.6; (c) cu2/cu1 = 0.4; (d) cu2/cu1 = 0.2\n\nThe factor of safety obtained using the finite element and limit equilibrium methods for\n\nthis example are summarised in Table 4 and illustrated in Figure 12.\n\nFOS = 1.424 FOS = 1.366\n\nFOS = 0.954 FOS = 0.505",
null,
"12\n\nTable 4: Computed factor of safety for Example 3\n\ncu2/cu1\n\nFOS\n\nFinite Element Method\n\nMorgenstern and Price Method\n\n1.0 1.451 1.448\n\n0.8 1.424 1.446\n\n0.6 1.366 1.400\n\n0.4 0.954 0.902\n\n0.2 0.505 0.451\n\nThe computed factor of safety with the ratio cu2/cu1 > 0.6 using the finite element\n\nmethod are close to the Morgenstern and Price method and the failure mechanisms of\n\nthese methods are similar. With these ratios, the strength of the thin weak layer does not\n\naffect the safety factor of the slope and generate a circular (base) mechanism of failure.\n\nWhen the ratio cu2/cu1 reduced to 0.6, the finite element method produce two failure\n\nmechanisms. The first failure mechanism is a base mechanism combined with the weak\n\nlayer beyond the slope toe and the second failure mechanism is a non-circular\n\nmechanism closely following the geometry of the thin weak layer. This is shown in\n\nFigure 10. With this ratio, the Morgenstern and Price method only produce one failure\n\nmechanism, the so-called circular (base) mechanism. When the ratio cu2/cu1 is reduced\n\nto 0.4 and 0.2, the failure mechanism of the slope shows a non-circular mechanism\n\nclosely following the geometry of the thin weak layer. However, with the ratio cu2/cu1 ≤\n\n0.6, there is no significant difference of factor of safety between the finite element\n\nmethod and the Morgenstern and Price method.\n\ncu2/c\n\nu1\n\n0.0 0.2 0.4 0.6 0.8 1.0\n\nFOS\n\n0.2\n\n0.4\n\n0.6\n\n0.8\n\n1.0\n\n1.2\n\n1.4\n\n1.6\n\nFinite Element Method\n\nMorgenstern & Price Method\n\nFig. 12: Computed FOS for an undrained clay slope with a thin weak layer with\n\nvariations of cu2/cu1",
null,
"13\n\nAn undrained clay slope with a weak foundation layer\n\nIn this example analysis of an undrained clay slope of 10m height and a 10m thick\n\nfoundation layer is carried out. The slope is inclined at an angle of 26.57o (2:1) to the\n\nhorizontal. Figure 13 shows the geometry and the two dimensional finite element mesh\n\nconsisting of 562 15-noded elements.\n\nFigure 13: Geometry and mesh for an undrained clay slope with a weak foundation\n\nlayer\n\nThe soil parameters used for this example are given in Table 3. The analysis are carried\n\nout using a constant value of undrained shear strength of soil (cu1) and six different\n\nvalues of undrained shear strength of the foundation layer (cu2) with ratios cu2/cu1 equal\n\nto 0.5, 1.0, 1.5, 1.75, 2.0 and 2.5.\n\nIn Figure 14 computed failure mechanisms for this example are presented for the finite\n\nelement method and the results of slope stability analysis using limit equilibrium\n\nmethods are presented in Figure 15.\n\nThe factor of safety obtained using the finite element and limit equilibrium methods for\n\nthis example are summarised in Table 5 and illustrated in Figure 16. The average\n\ndifference of factor of safety between the finite element method and limit equilibrium\n\nmethods is only 2.2% and the failure mechanisms of these methods are similar except\n\nwhen the ratio cu2/cu1 = 1.5. When the ratio cu2/cu1 =1.5, the finite element method\n\ngenerates two failure mechanisms, namely a base mechanism and a toe mechanism. It\n\nrepresents the transition between these two fundamental mechanisms. However, with\n\nthis ratio, the Morgenstern and Price Method only generates one failure mechanism, the\n\nso-called toe mechanism.",
null,
"14\n\n(a) (b)\n\n(c) (d)\n\nFigure 14: Failure mechanism for an undrained clay slope with a weak foundation layer\n\nusing the finite element method; (a) cu2/cu1 = 0.5; (b) cu2/cu1 = 1.0; (c) cu2/cu1\n\n= 1.5; (d) cu2/cu1 = 1.75\n\n0.934\n\nModel: Undrained (Phi=0) Unit Weight: 20 kN/m³Cohesion: 50 kPa\n\nModel: Undrained (Phi=0) Unit Weight: 20 kN/m³Cohesion: 25 kPa\n\nDistance\n\n0 10 20 30 40 50 60\n\nElevation\n\n0\n\n2\n\n4\n\n6\n\n8\n\n10\n\n12\n\n14\n\n16\n\n18\n\n20\n\n22\n\n1.485\n\nModel: Undrained (Phi=0) Unit Weight: 20 kN/m³Cohesion: 50 kPa\n\nModel: Undrained (Phi=0) Unit Weight: 20 kN/m³Cohesion: 50 kPa\n\nDistance\n\n0 10 20 30 40 50 60\n\nElevation\n\n0\n\n2\n\n4\n\n6\n\n8\n\n10\n\n12\n\n14\n\n16\n\n18\n\n20\n\n22\n\n(a) (b)\n\n2.052\n\nModel: Undrained (Phi=0) Unit Weight: 20 kN/m³Cohesion: 50 kPa\n\nModel: Undrained (Phi=0) Unit Weight: 20 kN/m³Cohesion: 75 kPa\n\nDistance\n\n0 10 20 30 40 50 60\n\nElevation\n\n0\n\n2\n\n4\n\n6\n\n8\n\n10\n\n12\n\n14\n\n16\n\n18\n\n20\n\n22\n\n2.052\n\nModel: Undrained (Phi=0) Unit Weight: 20 kN/m³Cohesion: 50 kPa\n\nModel: Undrained (Phi=0) Unit Weight: 20 kN/m³Cohesion: 87.5 kPa\n\nDistance\n\n0 10 20 30 40 50 60\n\nElevation\n\n0\n\n2\n\n4\n\n6\n\n8\n\n10\n\n12\n\n14\n\n16\n\n18\n\n20\n\n22\n\n(c) (d)\n\nFigure 15: Morgenstern and Price Method for an undrained clay slope with a weak\n\nfoundation layer; (a) cu2/cu1 = 0.5; (b) cu2/cu1 = 1.0; (c) cu2/cu1 = 1.5; (d)\n\ncu2/cu1 = 1.75\n\nFOS = 0.852 FOS = 1.454\n\nFOS = 2.032 FOS = 2.069",
null,
"15\n\nTable 5: Computed factor of safety for Example 4\n\ncu2/cu1\n\nFOS\n\nFinite Element Method\n\nMorgenstern and Price Method\n\n0.50 0.852 0.934\n\n1.00 1.454 1.485\n\n1.50 2.032 2.052\n\n1.75 2.069 2.052\n\n2.00 2.076 2.064\n\n2.50 2.069 2.064\n\nFigure 16 shows that at cu2/cu1 = 1.5 the factor of safety remains constant for both\n\nmethods.\n\ncu2/c\n\nu1\n\n0.0 0.5 1.0 1.5 2.0 2.5\n\nFOS\n\n0.6\n\n0.8\n\n1.0\n\n1.2\n\n1.4\n\n1.6\n\n1.8\n\n2.0\n\n2.2\n\nFinite Element Method\n\nMorgenstern & Price Method\n\nFig. 16: Computed FOS for an undrained clay slope with a weak foundation layer\n\nwith variations of cu2/cu1\n\nHomogeneous slope with water level\n\nThe geometry, the two dimensional finite element mesh and the soil parameters of this\n\nexample are the same as the slope analysed in Example 1, combined with a water level\n\nat a depth L below the crest of the slope (Figure 17).",
null,
"16\n\nFigure 17: Geometry and mesh for homogeneous slope with water level\n\nIn this analysis, a slope with different drawdown ratios L/H, which has been varied\n\nfrom 0.0 (slope completely submerged with water level at the crest of the slope) to 1.0\n\n(water level at the toe of the slope) is considered. This example is the so-called slow\n\ndrawdown problem wherein a reservoir, initially at the crest of the slope, is slowly\n\nlowered to the base, with the water level within the slope maintaining the same level.\n\nFigure 18 shows the computed failure mechanisms for this example using the finite\n\nelement method and Figure 19 illustrates the results of slope stability analysis using\n\nlimit equilibrium methods.\n\n(a) (b)\n\n(c) (d)\n\n(e) (f)\n\nFigure 18: Failure mechanism for homogeneous slope with water level using the finite\n\nelement method; (a) L/H = 0.0; (b) L/H = 0.2; (c) L/H = 0.4; (d) L/H = 0.6;\n\n(e) L/H = 0.8; (f) L/H = 1.0\n\nFOS = 1.815 FOS = 1.552\n\nFOS = 1.366 FOS = 1.276\n\nFOS = 1.273 FOS = 1.349",
null,
"17\n\n1\n\n1 2\n\n3 4\n\n1.848\n\nModel: Mohr-Coulomb\n\nUnit Weight: 20 kN/m³\n\nCohesion: 10 kPa\n\nPhi: 20 °\n\nDistance\n\n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34\n\nElevation\n\n0\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\n1\n\n1 2\n\n3 4\n\n1.600\n\nModel: Mohr-Coulomb\n\nUnit Weight: 20 kN/m³\n\nCohesion: 10 kPa\n\nPhi: 20 °\n\nDistance\n\n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34\n\nElevation\n\n0\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\n(a) (b)\n\n1\n\n1 2\n\n3 4\n\n1.437\n\nModel: Mohr-Coulomb\n\nUnit Weight: 20 kN/m³\n\nCohesion: 10 kPa\n\nPhi: 20 °\n\nDistance\n\n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34\n\nElevation\n\n0\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\n1\n\n1 2\n\n3 4\n\n1.341\n\nModel: Mohr-Coulomb\n\nUnit Weight: 20 kN/m³\n\nCohesion: 10 kPa\n\nPhi: 20 °\n\nDistance\n\n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34\n\nElevation\n\n0\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\n(c) (d)\n\n1\n\n1 2\n\n3 4\n\n1.339\n\nModel: Mohr-Coulomb\n\nUnit Weight: 20 kN/m³\n\nCohesion: 10 kPa\n\nPhi: 20 °\n\nDistance\n\n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34\n\nElevation\n\n0\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\n1\n\n1 2\n\n3 4\n\n1.386\n\nModel: Mohr-Coulomb\n\nUnit Weight: 20 kN/m³\n\nCohesion: 10 kPa\n\nPhi: 20 °\n\nDistance\n\n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34\n\nElevation\n\n0\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\n(e) (f)\n\nFigure 19: Morgenstern and Price Method for homogeneous slope with horizontal\n\nwater level (L/H = 1.0); (a) L/H = 0.0; (b) L/H = 0.2; (c) L/H = 0.4; (d) L/H\n\n= 0.6; (e) L/H = 0.8; (f) L/H = 1.0\n\nThe factor of safety obtained using the finite element and limit equilibrium methods for\n\nthis example are summarised in Table 6 and illustrated in Figure 20. The average\n\ndifference of factor of safety between the finite element method and limit equilibrium\n\nmethods is only 3.7% and the failure mechanisms of these methods are similar.\n\nIn fully slow drawdown conditions, the factor of safety reaches a minimum when L/H =\n\n0.7 and the fully submerged slope (L/H = 0) is more stable than the dry slope (L/H = 1)\n\nas indicated by a higher factor of safety, which has been also demonstrated by Lane and\n\nGriffith (2000). The most severe condition is not when the water level was lowered to a\n\nminimum. It was observed that the movement near the toe was significantly upward and\n\nthe failure mechanism changed when the water level was lowered to the base.",
null,
"18\n\nTable 6: Computed factor of safety for Example 5\n\nL/H\n\nFOS\n\nFinite Element Method\n\nMorgenstern and Price Method\n\n-0.1 1.815 1.847\n\n0.0 1.815 1.858\n\n0.1 1.685 1.715\n\n0.2 1.552 1.600\n\n0.3 1.449 1.507\n\n0.4 1.366 1.437\n\n0.5 1.308 1.378\n\n0.6 1.276 1.341\n\n0.7 1.259 1.331\n\n0.8 1.273 1.339\n\n0.9 1.305 1.356\n\n1.0 1.349 1.386\n\nL/H\n\n-0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0\n\nFOS\n\n1.0\n\n1.1\n\n1.2\n\n1.3\n\n1.4\n\n1.5\n\n1.6\n\n1.7\n\n1.8\n\n1.9\n\n2.0\n\nFinite Element Method\n\nMorgenstern & Price Method\n\nFig. 20: Computed FOS for homogeneous slope with variations of L/H\n\nSummary\n\nThe two approaches of slope stability analyses, one based on limit equilibrium methods\n\nand the other based on the finite element method are widely used in geotechnical",
null,
"19\n\nengineering. The finite element method in combination with an elastic-perfectly plastic\n\n(Mohr-Coulomb) model has been shown to be suitable for slope stability analysis.\n\nIn simple cases similar factors of safety and failure mechanism are obtained as in limit\n\nequilibrium analysis, however under more complex conditions the finite element\n\nmethod is more versatile because no a priori assumptions on the shape of the failure\n\nReference\n\nAbramson, L., E.; Lee, T., S.; Sharma, S.; Boyce, G., M. (2002)\n\nSlope stability and stabilization methods. Second Edition, John Wiley & Sons,\n\nBaker, R.; Frydman, S.; Talesnick, M. (1993)\n\nNumerical and Analytical Methods in Geomechanics, Vol. 17 (1), 15-43.\n\nBrinkgreve, R.B.J.; Swolf, W. M.; and Engin, E. (2010)\n\nPlaxis, users manual. The Netherlands.\n\nCheng, Y., M.; Lansivaara, T.; Wei, W., B. (2007)\n\nTwo-dimensional slope stability analysis by limit equilibrium and strength\n\nreduction method. Computers and Geotechnics, Vol. 34 (3), 137-150.\n\nDawson, E., M.; Roth, W., H. ; Drescher, A. (1999)\n\nSlope stability analysis by strength reduction. Geotechnique, Vol. 49 (6), 835-840.\n\nDuncan, J., M. (1996)\n\nState of the art: Limit equilibrium and finite element analysis of slopes. Journal of\n\nGeotechnical Engineering, Vol. 122 (7), 577-596.\n\nGEO-SLOPE International (2008)\n\nStability modelling with Slope/W 2007, An Engineering Methodology, Fourth\n\nGriffiths, D., V.; Lane, P., A. (1999)\n\nSlope stability analysis by finite elements. Geotechnique, Vol. 49 (3), 387-403.\n\nKrahn, J. (2003)\n\nThe 2001 R.M. Hardy Lecture: The limits of limit equilibrium analyses. Canadian\n\nGeotechnical Journal, Vol. 40 (3), 643-660.\n\nMatsui, T.; San, K., C. (1992)\n\nFinite element slope stability analysis by shear strength reduction technique. Soils\n\nand Foundation, Vol. 32 (1), 59-70."
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null,
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null,
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null,
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null
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https://www.calculus-online.com/exercise/4556 | [
"# Calculating Triple Integrals – Fixed integration limits – Exercise 4556\n\nExercise\n\nCalculate the integral\n\n$$\\int\\int\\int_T z^2 e^{x+y} dxdydz$$\n\nWhere T is bounded by the surfaces\n\n$$x=0,x=1,y=0,y=1,z=0,z=1$$\n\n$$\\int\\int\\int_T z^2 e^{x+y} dxdydz=\\frac{1}{3}{(e-1)}^2$$"
]
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https://books.google.no/books?id=PToDAAAAQAAJ&q=fall&dq=editions:UOM39015067252117&lr=&hl=no&output=html_text&source=gbs_word_cloud_r&cad=5 | [
"# The Elements of Euclid, containing the first six books, with a selection of geometrical problems. To which is added the parts of the eleventh and twelfth books which are usually read at the universities. By J. Martin\n\n1874\n0 Anmeldelser\nAnmeldelsene blir ikke bekreftet, men Google ser etter falskt innhold og fjerner slikt innhold som avdekkes\n\n### Hva folk mener -Skriv en omtale\n\nVi har ikke funnet noen omtaler pĺ noen av de vanlige stedene.\n\n### Populćre avsnitt\n\nSide 1 - A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another.\nSide 6 - If a straight line meets two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles...\nSide 232 - If two triangles, which have two sides of the one proportional to two sides of the other, be joined at one angle, so as to have their homologous sides parallel to one another, the remaining sides shall be in a straight line. Let...\nSide 112 - The angle in a semicircle is a right angle; the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.\nSide 209 - ... triangles which have one angle in the one equal to one angle in the other, and their sides about the equal angles reciprocally proportional, are equal to one another.\nSide 269 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. D c A' D' Hyp. In triangles ABC and A'B'C', ZA = ZA'. To prove AABC = ABxAC. A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.\nSide 199 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.\nSide 23 - If two triangles have two angles of the one equal to two angles of the other, each to each, and also one side of the one equal to the corresponding side of the other, the triangles are congruent.\nSide 63 - If a straight line be divided into two equal, and also into two unequal parts, the squares on the two unequal parts are together double of the square on half the line and of the square on the line between the points of section. Let the straight line AB be divided into two equal parts...\nSide 32 - ... twice as many right angles as the figure has sides ; therefore all the angles of the figure together with four right angles, are equal to twice as many right angles as the figure has sides."
]
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https://www.studytonight.com/python/python-operator-overloading | [
"Dark Mode On/Off\n\nOperators are used in Python to perform specific operations on the given operands. The operation that any particular operator will perform on any predefined data type is already defined in Python.\n\nEach operator can be used in a different way for different types of operands. For example, `+` operator is used for adding two integers to give an integer as a result but when we use it with float operands, then the result is a float value and when `+` is used with string operands then it concatenates the two operands provided.\n\nThis different behaviour of a single operator for different types of operands is called Operator Overloading. The use of `+` operator with different types of operands is shown below:\n\n``````>>> x=10\n>>> y=20\n>>> x+y\n30\n\n>>> z=10.4\n>>> x+z\n20.4\n\n>>> s1 = 'hello '\n>>> s2 = 'world'\n>>> s1+s2\n'hello world'\n``````\n\n### Can `+` Operator Add anything?\n\nThe answer is No, it cannot. Can you use the `+` operator to add two objects of a class. The `+` operator can add two integer values, two float values or can be used to concatenate two strings only because these behaviours have been defined in python.\n\nSo if you want to use the same operator to add two objects of some user defined class then you will have to defined that behaviour yourself and inform python about that.\n\nIf you are still not clear, let's create a class and try to use the `+` operator to add two objects of that class,\n\n``````class Complex:\ndef __init__(self, r, i):\nself.real = r\nself.img = i\n\nc1 = Complex(5,3)\nc2 = Complex(2,4)\nprint(\"sum = \", c1+c2)\n``````\n\nTraceback (most recent call last): File \"/tmp/sessions/1dfbe78bb701d99d/main.py\", line 7, in print(\"sum = \", c1+c2) TypeError: unsupported operand type(s) for +: 'Complex' and 'Complex'\n\nSo we can see that the `+` operator is not supported in a user-defined class. But we can do the same by overloading the `+` operator for our class `Complex`. But how can we do that?\n\n## Special Functions in Python\n\nSpecial functions in python are the functions which are used to perform special tasks. These special functions have `__` as prefix and suffix to their name as we see in `__init__()` method which is also a special function. Some special functions used for overloading the operators are shown below:\n\n### Mathematical Operator\n\nBelow we have the names of the special functions to overload the mathematical operators in python.\n\nName Symbol Special Function\nAddition `+` ` __add__(self, other)`\nSubtraction `-` ` __sub__(self, other)`\nDivision `/` ` __truediv__(self, other)`\nFloor Division `//` ` __floordiv__(self, other)`\nModulus(or Remainder) `%` ` __mod__(self, other)`\nPower `**` ` __pow__(self, other)`\n\n### Assignment Operator\n\nBelow we have the names of the special functions to overload the assignment operators in python.\n\nName Symbol Special Function\nIncrement `+=` ` __iadd__(self, other)`\nDecrement `-=` ` __isub__(self, other)`\nProduct `*=` ` __imul__(self, other)`\nDivision `/=` ` __idiv__(self, other)`\nModulus `%=` `__imod__(self, other)`\nPower `**=` ` __ipow__(self, other)`\n\n### Relational Operator\n\nBelow we have the names of the special functions to overload the relational operators in python.\n\nName Symbol Special Function\nLess than `<` ` __lt__(self, other)`\nGreater than `>` ` __gt__(self, other)`\nEqual to `==` `__eq__(self, other)`\nNot equal `!=` ` __ne__(self, other)`\nLess than or equal to `<=` `__le__(self, other)`\nGreater than or equal to `> =` `__gt__(self, other)`\n\nIt's time to see a few code examples where we actually use the above specified special functions and overload some operators.\n\n### Overloading `+` operator\n\nIn the below code example we will overload the `+` operator for our class `Complex`,\n\n``````class Complex:\n# defining init method for class\ndef __init__(self, r, i):\nself.real = r\nself.img = i\n\nr = self.real + sec.real\ni = self.img + sec.img\nreturn complx(r,i)\n\n# string function to print object of Complex class\ndef __str__(self):\nreturn str(self.real)+' + '+str(self.img)+'i'\n\nc1 = Complex(5,3)\nc2 = Complex(2,4)\nprint(\"sum = \",c1+c2)``````\n\nsum = 7 + 7i\n\nIn the program above, ` __add__()` is used to overload the `+` operator i.e. when `+` operator is used with two `Complex` class objects then the function `__add__()` is called.\n\n` __str__()` is another special function which is used to provide a format of the object that is suitable for printing.\n\n### Overloading `<` operator\n\nNow let's overload the less than operator so that we can easily compare two `Complex` class object's values by using the less than operaton `<`.\n\nAs we know now, for doing so, we have to define the `__lt__` special function in our class.\n\nBased on your requirement of comparing the class object, you can define the logic for the special functions for overriding an operator. In the code above, we have given precedence to the real part of the complex number, if that is less then the whole complex number is less, if that is equal then we check for the imaginary part.\n\n### Conclusion\n\nOverloading operators is easy in python using the special functions and is less confusion too."
]
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https://www.litscape.com/word_analysis/tattles | [
"# Definition of tattles\n\n## \"tattles\" in the noun sense\n\n### 1. tattle, singing, telling\n\ndisclosing information or giving evidence about another\n\n## \"tattles\" in the verb sense\n\n### 1. chatter, piffle, palaver, prate, tittle-tattle, twaddle, clack, maunder, prattle, blab, gibber, tattle, blabber, gabble\n\nspeak (about unimportant matters) rapidly and incessantly\n\n### 2. spill the beans, let the cat out of the bag, talk, tattle, blab, peach, babble, sing, babble out, blab out\n\ndivulge confidential information or secrets\n\n\"Be careful\n\nSource: WordNet® (An amazing lexical database of English)\n\nWordNet®. Princeton University. 2010.\n\n# Quotations for tattles\n\nWine invents nothing; it only tattles. [ Schiller ]\n\n# tattles in Scrabble®\n\nThe word tattles is playable in Scrabble®, no blanks required.\n\nTATTLES\n(78 = 28 + 50)\n\ntattles\n\nTATTLES\n(78 = 28 + 50)\nTATTLES\n(74 = 24 + 50)\nTATTLES\n(74 = 24 + 50)\nTATTLES\n(74 = 24 + 50)\nTATTLES\n(74 = 24 + 50)\nTATTLES\n(74 = 24 + 50)\nTATTLES\n(74 = 24 + 50)\nTATTLES\n(74 = 24 + 50)\nTATTLES\n(74 = 24 + 50)\nTATTLES\n(71 = 21 + 50)\nTATTLES\n(68 = 18 + 50)\nTATTLES\n(68 = 18 + 50)\nTATTLES\n(68 = 18 + 50)\nTATTLES\n(68 = 18 + 50)\nTATTLES\n(68 = 18 + 50)\nTATTLES\n(68 = 18 + 50)\nTATTLES\n(66 = 16 + 50)\nTATTLES\n(66 = 16 + 50)\nTATTLES\n(66 = 16 + 50)\nTATTLES\n(66 = 16 + 50)\nTATTLES\n(66 = 16 + 50)\nTATTLES\n(66 = 16 + 50)\nTATTLES\n(66 = 16 + 50)\nTATTLES\n(66 = 16 + 50)\nTATTLES\n(64 = 14 + 50)\nTATTLES\n(64 = 14 + 50)\nTATTLES\n(64 = 14 + 50)\nTATTLES\n(64 = 14 + 50)\nTATTLES\n(64 = 14 + 50)\nTATTLES\n(61 = 11 + 50)\nTATTLES\n(61 = 11 + 50)\nTATTLES\n(61 = 11 + 50)\nTATTLES\n(60 = 10 + 50)\nTATTLES\n(60 = 10 + 50)\nTATTLES\n(59 = 9 + 50)\nTATTLES\n(59 = 9 + 50)\nTATTLES\n(59 = 9 + 50)\nTATTLES\n(59 = 9 + 50)\nTATTLES\n(59 = 9 + 50)\nTATTLES\n(59 = 9 + 50)\nTATTLES\n(59 = 9 + 50)\nTATTLES\n(58 = 8 + 50)\n\nTATTLES\n(78 = 28 + 50)\nTATTLES\n(74 = 24 + 50)\nTATTLES\n(74 = 24 + 50)\nTATTLES\n(74 = 24 + 50)\nTATTLES\n(74 = 24 + 50)\nTATTLES\n(74 = 24 + 50)\nTATTLES\n(74 = 24 + 50)\nTATTLES\n(74 = 24 + 50)\nTATTLES\n(74 = 24 + 50)\nTATTLES\n(71 = 21 + 50)\nTATTLES\n(68 = 18 + 50)\nTATTLES\n(68 = 18 + 50)\nTATTLES\n(68 = 18 + 50)\nTATTLES\n(68 = 18 + 50)\nTATTLES\n(68 = 18 + 50)\nTATTLES\n(68 = 18 + 50)\nTATTLES\n(66 = 16 + 50)\nTATTLES\n(66 = 16 + 50)\nTATTLES\n(66 = 16 + 50)\nTATTLES\n(66 = 16 + 50)\nTATTLES\n(66 = 16 + 50)\nTATTLES\n(66 = 16 + 50)\nTATTLES\n(66 = 16 + 50)\nTATTLES\n(66 = 16 + 50)\nTATTLES\n(64 = 14 + 50)\nTATTLES\n(64 = 14 + 50)\nTATTLES\n(64 = 14 + 50)\nTATTLES\n(64 = 14 + 50)\nTATTLES\n(64 = 14 + 50)\nTATTLES\n(61 = 11 + 50)\nTATTLES\n(61 = 11 + 50)\nTATTLES\n(61 = 11 + 50)\nTATTLES\n(60 = 10 + 50)\nTATTLES\n(60 = 10 + 50)\nTATTLES\n(59 = 9 + 50)\nTATTLES\n(59 = 9 + 50)\nTATTLES\n(59 = 9 + 50)\nTATTLES\n(59 = 9 + 50)\nTATTLES\n(59 = 9 + 50)\nTATTLES\n(59 = 9 + 50)\nTATTLES\n(59 = 9 + 50)\nTATTLES\n(58 = 8 + 50)\nLATTES\n(21)\nLATTES\n(21)\nTATTLE\n(21)\nTATTLE\n(21)\nATTEST\n(21)\nLATTES\n(21)\nATTEST\n(21)\nLATTES\n(21)\nATTEST\n(21)\nATTEST\n(21)\nTATTLE\n(21)\nLATEST\n(21)\nATTEST\n(21)\nATTEST\n(21)\nLATEST\n(21)\nLATTES\n(21)\nLATEST\n(21)\nLATEST\n(21)\nLATEST\n(21)\nLATEST\n(21)\nTATTLE\n(21)\nLATTES\n(21)\nTATTLE\n(21)\nTATTLE\n(21)\nSLATE\n(18)\nLATEST\n(18)\nLATTES\n(18)\nTEATS\n(18)\nTALES\n(18)\nTESLA\n(18)\nLEAST\n(18)\nSTEAL\n(18)\nSTALE\n(18)\nTEALS\n(18)\nSTALE\n(18)\nTASTE\n(18)\nSTALE\n(18)\nLATEST\n(18)\nTALES\n(18)\nTESLA\n(18)\nTESLA\n(18)\nSTATE\n(18)\nSTEAL\n(18)\nSTATE\n(18)\nTALES\n(18)\nSLATE\n(18)\nLEAST\n(18)\nLEAST\n(18)\nSTALE\n(18)\nLATTE\n(18)\nLATTE\n(18)\nSTATE\n(18)\nSTATE\n(18)\nLATTE\n(18)\nLATTES\n(18)\nTESLA\n(18)\nTEALS\n(18)\nTATTLE\n(18)\nATTEST\n(18)\nTEATS\n(18)\nSETAL\n(18)\nTEALS\n(18)\nTATTLE\n(18)\nTASTE\n(18)\nATTEST\n(18)\nSETAL\n(18)\nSETAL\n(18)\nTASTE\n(18)\nSLATE\n(18)\nTALES\n(18)\nTEALS\n(18)\nSTEAL\n(18)\nSTEAL\n(18)\nTEATS\n(18)\nSLATE\n(18)\nSETAL\n(18)\nTASTE\n(18)\nTEATS\n(18)\nLATTE\n(18)\nLEAST\n(18)\nTATTLE\n(16)\nTATTLE\n(16)\nTATTLE\n(16)\nTATTLE\n(16)\nLATEST\n(16)\nLATEST\n(16)\nLATTES\n(16)\nLATEST\n(16)\nLATTES\n(16)\nLATTES\n(16)\nLATTES\n(16)\nLATEST\n(16)\nATTEST\n(16)\nATTEST\n(16)\nATTEST\n(16)\nATTEST\n(16)\nLATTE\n(15)\nSTAT\n(15)\nTALES\n(15)\nSEAT\n(15)\nLAST\n(15)\nSTAT\n(15)\nTALES\n(15)\nLETS\n(15)\nTEATS\n(15)\nTASTE\n(15)\nSTALE\n(15)\nTASTE\n(15)\nETAS\n(15)\nSTALE\n(15)\nTASTE\n(15)\nTEATS\n(15)\nETAS\n(15)\nLEAS\n(15)\nSTALE\n(15)\nLEAS\n(15)\nLETS\n(15)\nTEATS\n(15)\nLEAST\n(15)\nLEAST\n(15)\nTEAS\n(15)\nLAST\n(15)\nTALES\n(15)\nTEAS\n(15)\nTESLA\n(15)\nSALT\n(15)\nLATTE\n(15)\nTEST\n(15)\nLATTE\n(15)\nSTEAL\n(15)\nSTEAL\n(15)\nTEST\n(15)\nSATE\n(15)\nSATE\n(15)\nSTATE\n(15)\nSALE\n(15)\nSTEAL\n(15)\nSEAL\n(15)\nTALE\n(15)\nSALE\n(15)\nSTATE\n(15)\nLATE\n(15)\nALES\n(15)\nSALT\n(15)\nTESLA\n(15)\nLATE\n(15)\nSEAL\n(15)\nTESLA\n(15)\nSTATE\n(15)\nTALE\n(15)\nALES\n(15)\nSETA\n(15)\nSEAT\n(15)\nSETAL\n(15)\nSETAL\n(15)\nTEAL\n(15)\nEAST\n(15)\nSLATE\n(15)\n\n# tattles in Words With Friends™\n\nThe word tattles is playable in Words With Friends™, no blanks required.\n\nTATTLES\n(77 = 42 + 35)\n\ntattles\n\nTATTLES\n(77 = 42 + 35)\nTATTLES\n(71 = 36 + 35)\nTATTLES\n(71 = 36 + 35)\nTATTLES\n(71 = 36 + 35)\nTATTLES\n(67 = 32 + 35)\nTATTLES\n(67 = 32 + 35)\nTATTLES\n(67 = 32 + 35)\nTATTLES\n(65 = 30 + 35)\nTATTLES\n(65 = 30 + 35)\nTATTLES\n(65 = 30 + 35)\nTATTLES\n(65 = 30 + 35)\nTATTLES\n(65 = 30 + 35)\nTATTLES\n(65 = 30 + 35)\nTATTLES\n(59 = 24 + 35)\nTATTLES\n(55 = 20 + 35)\nTATTLES\n(55 = 20 + 35)\nTATTLES\n(55 = 20 + 35)\nTATTLES\n(55 = 20 + 35)\nTATTLES\n(55 = 20 + 35)\nTATTLES\n(55 = 20 + 35)\nTATTLES\n(53 = 18 + 35)\nTATTLES\n(53 = 18 + 35)\nTATTLES\n(53 = 18 + 35)\nTATTLES\n(53 = 18 + 35)\nTATTLES\n(53 = 18 + 35)\nTATTLES\n(51 = 16 + 35)\nTATTLES\n(51 = 16 + 35)\nTATTLES\n(51 = 16 + 35)\nTATTLES\n(51 = 16 + 35)\nTATTLES\n(51 = 16 + 35)\nTATTLES\n(51 = 16 + 35)\nTATTLES\n(51 = 16 + 35)\nTATTLES\n(49 = 14 + 35)\nTATTLES\n(49 = 14 + 35)\nTATTLES\n(47 = 12 + 35)\nTATTLES\n(47 = 12 + 35)\nTATTLES\n(47 = 12 + 35)\nTATTLES\n(47 = 12 + 35)\nTATTLES\n(46 = 11 + 35)\nTATTLES\n(46 = 11 + 35)\nTATTLES\n(46 = 11 + 35)\nTATTLES\n(46 = 11 + 35)\nTATTLES\n(46 = 11 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(44 = 9 + 35)\nTATTLES\n(44 = 9 + 35)\nTATTLES\n(44 = 9 + 35)\nTATTLES\n(44 = 9 + 35)\nTATTLES\n(43 = 8 + 35)\n\nTATTLES\n(77 = 42 + 35)\nTATTLES\n(71 = 36 + 35)\nTATTLES\n(71 = 36 + 35)\nTATTLES\n(71 = 36 + 35)\nTATTLES\n(67 = 32 + 35)\nTATTLES\n(67 = 32 + 35)\nTATTLES\n(67 = 32 + 35)\nTATTLES\n(65 = 30 + 35)\nTATTLES\n(65 = 30 + 35)\nTATTLES\n(65 = 30 + 35)\nTATTLES\n(65 = 30 + 35)\nTATTLES\n(65 = 30 + 35)\nTATTLES\n(65 = 30 + 35)\nTATTLES\n(59 = 24 + 35)\nTATTLES\n(55 = 20 + 35)\nTATTLES\n(55 = 20 + 35)\nTATTLES\n(55 = 20 + 35)\nTATTLES\n(55 = 20 + 35)\nTATTLES\n(55 = 20 + 35)\nTATTLES\n(55 = 20 + 35)\nTATTLES\n(53 = 18 + 35)\nTATTLES\n(53 = 18 + 35)\nTATTLES\n(53 = 18 + 35)\nTATTLES\n(53 = 18 + 35)\nTATTLES\n(53 = 18 + 35)\nTATTLES\n(51 = 16 + 35)\nTATTLES\n(51 = 16 + 35)\nTATTLES\n(51 = 16 + 35)\nTATTLES\n(51 = 16 + 35)\nTATTLES\n(51 = 16 + 35)\nTATTLES\n(51 = 16 + 35)\nTATTLES\n(51 = 16 + 35)\nTATTLES\n(49 = 14 + 35)\nTATTLES\n(49 = 14 + 35)\nTATTLES\n(47 = 12 + 35)\nTATTLES\n(47 = 12 + 35)\nTATTLES\n(47 = 12 + 35)\nTATTLES\n(47 = 12 + 35)\nTATTLES\n(46 = 11 + 35)\nTATTLES\n(46 = 11 + 35)\nTATTLES\n(46 = 11 + 35)\nTATTLES\n(46 = 11 + 35)\nTATTLES\n(46 = 11 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(45 = 10 + 35)\nTATTLES\n(44 = 9 + 35)\nTATTLES\n(44 = 9 + 35)\nTATTLES\n(44 = 9 + 35)\nTATTLES\n(44 = 9 + 35)\nTATTLES\n(43 = 8 + 35)\nLATTES\n(39)\nLATEST\n(39)\nTATTLE\n(33)\nLATEST\n(33)\nTATTLE\n(33)\nLATEST\n(33)\nTATTLE\n(33)\nLATTES\n(33)\nLATTES\n(33)\nSETAL\n(30)\nATTEST\n(30)\nLATTE\n(30)\nATTEST\n(30)\nSTEAL\n(30)\nTEALS\n(30)\nLEAST\n(30)\nSTALE\n(30)\nTESLA\n(30)\nSLATE\n(30)\nLATTES\n(28)\nLATEST\n(28)\nTATTLE\n(28)\nLATTES\n(28)\nTATTLE\n(28)\nLATEST\n(28)\nLATEST\n(27)\nLATTES\n(27)\nTATTLE\n(27)\nLATE\n(27)\nSEAL\n(27)\nLATEST\n(27)\nTATTLE\n(27)\nLATTES\n(27)\nLATTES\n(27)\nLATEST\n(27)\nTEAL\n(27)\nLATEST\n(27)\nLATTES\n(27)\nTATTLE\n(27)\nTATTLE\n(27)\nLETS\n(27)\nLAST\n(27)\nLATTES\n(27)\nLEAS\n(27)\nLATEST\n(27)\nTATTLE\n(27)\nLEST\n(27)\nSTEAL\n(24)\nTEALS\n(24)\nSTEAL\n(24)\nTALES\n(24)\nTEALS\n(24)\nTALES\n(24)\nATTEST\n(24)\nTALES\n(24)\nTEALS\n(24)\nTALES\n(24)\nTALES\n(24)\nTEALS\n(24)\nLEAST\n(24)\nLEAST\n(24)\nLEAST\n(24)\nATTEST\n(24)\nATTEST\n(24)\nLEAST\n(24)\nATTEST\n(24)\nSTALE\n(24)\nTESLA\n(24)\nSTALE\n(24)\nTESLA\n(24)\nSTALE\n(24)\nTESLA\n(24)\nLATTE\n(24)\nSETAL\n(24)\nSETAL\n(24)\nATTEST\n(24)\nSETAL\n(24)\nSLATE\n(24)\nSLATE\n(24)\nSTEAL\n(24)\nSTEAL\n(24)\nATTEST\n(24)\nSLATE\n(24)\nLATTE\n(24)\nSTALE\n(24)\nATTEST\n(24)\nSETAL\n(24)\nLATTE\n(24)\nSLATE\n(24)\nATTEST\n(24)\nTESLA\n(24)\nLATTE\n(24)\nTATTLE\n(22)\nLATEST\n(22)\nLATTES\n(22)\nALES\n(21)\nLATTES\n(21)\nLEAS\n(21)\nALES\n(21)\nTASTE\n(21)\nSLAT\n(21)\nTASTE\n(21)\nLEST\n(21)\nSLAT\n(21)\nTASTE\n(21)\nTATTLE\n(21)\nTASTE\n(21)\nTEAL\n(21)\nLETS\n(21)\nLAST\n(21)\nTALE\n(21)\nSALT\n(21)\nTEATS\n(21)\nTEATS\n(21)\nLATE\n(21)\nSTATE\n(21)\nLATTES\n(21)\nSTATE\n(21)\nSALE\n(21)\nTEATS\n(21)\nSTATE\n(21)\nTATTLE\n(21)\nSALE\n(21)\nLATEST\n(21)\nSEAL\n(21)\nSTATE\n(21)\nTEATS\n(21)\nSALT\n(21)\nLATEST\n(21)\nTALE\n(21)\nLATTE\n(20)\nSTATE\n(20)\nTEATS\n(20)\nTASTE\n(20)\nLEAST\n(20)\nSETAL\n(20)\nSTEAL\n(20)\nLEAST\n(18)\nTATTLE\n(18)\nTATTLE\n(18)\nTATTLE\n(18)\nLEAST\n(18)\n\nat tattle\n\nta tattle\n\nretattles\n\ntittletattles"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.75430954,"math_prob":1.000009,"size":643,"snap":"2020-10-2020-16","text_gpt3_token_len":183,"char_repetition_ratio":0.12676056,"word_repetition_ratio":0.0,"special_character_ratio":0.22395024,"punctuation_ratio":0.2601626,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9973837,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-21T04:04:46Z\",\"WARC-Record-ID\":\"<urn:uuid:8d8bbac8-d418-4666-8d74-12474a5b706c>\",\"Content-Length\":\"135585\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:09acc77c-1a2b-488d-9eb8-446c6bf4307a>\",\"WARC-Concurrent-To\":\"<urn:uuid:6406752b-0445-419e-9da6-51583e203d82>\",\"WARC-IP-Address\":\"104.18.51.165\",\"WARC-Target-URI\":\"https://www.litscape.com/word_analysis/tattles\",\"WARC-Payload-Digest\":\"sha1:6JBE2272SAPMRX6UQQJBRYKKG7PEMV7A\",\"WARC-Block-Digest\":\"sha1:HVYGXULBDRRQBP7X3EXAIXKHXH3KGBH2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145438.12_warc_CC-MAIN-20200221014826-20200221044826-00521.warc.gz\"}"} |
https://pubs.nctm.org/browse?access=all&pageSize=10&sort=datedescending&t_0=number&t_1=geometry&t_2=probability&t_3=algebra_0&t_4=probability_5&t_5=algebra_5&t_6=algebra_2&t_7=math-topic&t_8=grades_6-8&t_9=measurement | [
"# Browse\n\n## You are looking at 1 - 1 of 1 items for :\n\n• Number\n• Geometry\n• Probability/Data Analysis and Statistics\n• Algebraic Thinking/Pre-algebra Concepts\n• Probability\n• Linear Equations and Inequalities\n• Expression/Equation\n• Math Topic"
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https://physicscalculatorpro.com/helical-coil-calculator/ | [
"# Helical Coil Calculator\n\nThe helical coil calculator aids in the calculation of a coil's various parameters. From aeroplanes to automobiles, and from electricity to heat exchangers, a helical coil is an essential component of any machine. It has a variety of qualities and is used in a variety of applications. For example, A coil is used as a spring because of its energy-storing and shock-absorbing capabilities, as well as in heat exchangers because of its large surface area.\n\nThe basic coil design formulas are the same regardless of the application. Continue reading to learn how to use the pipe coil length calculator to calculate parameters such as helical coil height and length.\n\n## What is a Helical Coil?\n\nA helical coil is a type of coil that has a helical shape. When a material is wound or twisted around a helix, a helical coil is created. For example, a helical coil is formed when a wire or tube is wrapped around a circular object, such as a pencil. Coil diameter, pitch or spacing, wire diameter, coil height, and other design parameters can be used to make a helical coil. The benefit is that these parameters can be used to customize this versatile part for a variety of applications, including a helical coil spring and a heat exchanger.\n\n### Designing Coils Formulas\n\nLet's take a look at some of the helical coil calculator's coil parameters.\n\nCoil diameter (Dc): The coil diameter is calculated by measuring the distance between the coil's center and the neutral circle.\n\nWire diameter (Dw): This measurement refers to the diameter of the coil wire. The diameters of the coil and wire are related using the equation: Dw = 2 x (Do - Dc)\n\nWhere Do = helical coil's outer diameter.\n\nTurns (N): The number of times the wire is wound on the helix axis is indicated by the number of turns (N).\n\nPitch (S): Pitch or spacing refers to the distance between successive coils.\n\nHeight of coil (H): The distance between the coil's top and bottommost points is defined as the coil's height (H). The equation can be used to calculate the coil's height: H = N x (S + Dw)\n\nLength of a helical coil (Lw): The length of a helical coil (Lw) refers to the total length of wire used to construct the coil. To put it another way, the circumference of the coil is multiplied by N. The formula for the length of a helical coil can be written as:L = π x Dc x N\n\nInductance (L): Inductance (L) is a material's capacity to modify the electric current that flows through it. It's counted in Henries. The inductance L is determined by coil characteristics such as: L = (Dc x N)2 / (18 x Dc + 40 x Lw)\n\nThe volume of wire used in the coil (V): The volume swept by the cross-section along the helix is called the volume of wire used in the coil (V).\n\nV = π x Dw2 x Lw / 4\n\nResonant frequency (Rf): The frequency at which the capacitance of an inductor resonates with the ideal inductance, resulting in high impedance, is known as the resonant frequency (Rf). The following equation can be used to find it: Frequency = 1 / (2 x π x √(L x C))\n\nWhere, C = capacitance of the coil.\n\n### How do you find out coil design parameters?\n\nThis tool is primarily used to calculate coil inductance and volume using coil parameters. You can also use it in reverse to find out coil parameters based on the inductance. To find out how much inductance a coil has, do the following:\n\n• Step 1: Enter the coil radius Rc or the coil diameter Dc (in Advanced mode).\n• Step 2: Fill in Dw for wire diameter.\n• Step 3: N is the total number of turns in the coil.\n• Step 4: Enter the coil pitch or spacing, S.\n• Step 5: The height, H, and wire length, Lw, of the coil spring will be returned by the coil spring calculator.\n• Step 6: The inductance, L in microHenries, and total volume of the coil V, are also calculated using the helical coil calculator.\n• Step 6: C in picofarads is the capacitance.\n• Step 7: The coil's resonant frequency will be returned by the calculator.\n\nGet similar concepts of physics all under one roof explained clearly with step by step process on Physicscalculatorpro.com a trusted portal for all your needs.\n\n### FAQs on Helical Coil Calculator\n\n1. What is a helical coil?\n\nHelical coils are metal tubes that have been bent into a spiral shape when referring to metal tubing. A helical coil can be as simple as one or two spiral turns or as complex as a series of spirals several feet long, depending on the finished product's requirements.\n\n2. What formula is used to determine the helical length?\n\nIt's calculated by multiplying the height and circumference by the square root of the sum of the squares. This helix coil formula can also be used to determine the length of a coil.\n\n3. What is a transformer's helical coil?\n\nA helical winding is made up of rectangular strips wound in a helix shape. The strips are wound radially in parallel, with each turn taking up the entire radial depth of winding. Helical coils are ideal for large transformer low voltage windings.\n\n4. What is a helical heat exchanger used for?\n\nHigh temperatures and extreme temperature differentials can be handled with helical geometry without causing high induced stresses or requiring expensive expansion joints. The exchanger's advantages are enhanced by its high-pressure capability and the ability to completely clean the service-fluid flow area.\n\n5. How do you find out the inductance of a coil spring?\n\nTo calculate the inductance of a helical coil, do the following:\n\n1. Multiply the coil's diameter by the number of turns.\n2. To get the numerator, square the resultant.\n3. To get the denominator, multiply the coil diameter by 18 and the wire length by 40.\n4. To find the inductance of the helical coil, divide the numerator by the denominator.\n\nL = (Dc x N)2 / (18 x Dc + 40 x Lw)"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9402467,"math_prob":0.980524,"size":4704,"snap":"2022-27-2022-33","text_gpt3_token_len":1084,"char_repetition_ratio":0.15808511,"word_repetition_ratio":0.03567182,"special_character_ratio":0.21981293,"punctuation_ratio":0.09120521,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99717695,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-03T06:10:23Z\",\"WARC-Record-ID\":\"<urn:uuid:a39b3f86-4fcc-48bd-8431-09c074ccfe9a>\",\"Content-Length\":\"26225\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9e2508e8-a7d9-4fb8-a3a6-a9f134b458da>\",\"WARC-Concurrent-To\":\"<urn:uuid:d9560705-0ec8-4c0d-a923-098cd712d025>\",\"WARC-IP-Address\":\"167.172.253.150\",\"WARC-Target-URI\":\"https://physicscalculatorpro.com/helical-coil-calculator/\",\"WARC-Payload-Digest\":\"sha1:PT3RF3H2JXSWWBPWMEYMYUPFMDUEHKT2\",\"WARC-Block-Digest\":\"sha1:LYIXCGDAFP7IWAKV3VWBQUEADFOIXDIX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104215790.65_warc_CC-MAIN-20220703043548-20220703073548-00752.warc.gz\"}"} |
https://socratic.org/questions/let-f-x-x-2-16-how-do-you-find-f-1-x | [
"# Let f(x) = x^2 - 16 how do you find f^-1(x)?\n\nMar 21, 2018\n\nThis is a way to express finding the inverse function of $f \\left(x\\right) = {x}^{2} - 16$\n\n#### Explanation:\n\nFirst, write the function as $y = {x}^{2} - 16$.\n\nNext, switch the $y$ and $x$ positions.\n\n$x = {y}^{2} - 16 \\rightarrow$ Solve for $y$ in terms of $x$\n\n$x + 16 = {y}^{2}$\n\n$y = \\sqrt{x + 16}$\n\nThe inverse function should be ${f}^{-} 1 \\left(x\\right) = \\sqrt{x + 16}$\n\nMar 21, 2018\n\n#### Explanation:\n\nSuppose that, $f : \\mathbb{R} \\to \\mathbb{R} : f \\left(x\\right) = {x}^{2} - 16$.\n\nObserve that, $f \\left(1\\right) = 1 - 16 = - 15 , \\mathmr{and} , f \\left(- 1\\right) = - 15$.\n\n$\\therefore f \\left(1\\right) = f \\left(- 1\\right)$.\n\n$\\therefore f \\text{ is not injective, or, } 1 - 1$.\n\n$\\therefore {f}^{-} 1$ does not exist.\n\nHowever, if $f$ is defined on a suitable domain, e.g.,\n\n${\\mathbb{R}}^{+}$, then ${f}^{-} 1$ exists as Respected Serena D. has shown."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7384094,"math_prob":1.0000085,"size":433,"snap":"2022-05-2022-21","text_gpt3_token_len":112,"char_repetition_ratio":0.121212125,"word_repetition_ratio":0.0,"special_character_ratio":0.25635105,"punctuation_ratio":0.11764706,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999995,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-18T06:38:04Z\",\"WARC-Record-ID\":\"<urn:uuid:60d93a15-658c-4fd1-9adf-24fc23b67bf7>\",\"Content-Length\":\"35445\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:025fed7a-f7ea-47c3-9cda-1883f6396199>\",\"WARC-Concurrent-To\":\"<urn:uuid:0bccdd5b-b7aa-40d5-b884-60b474ce4254>\",\"WARC-IP-Address\":\"216.239.34.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/let-f-x-x-2-16-how-do-you-find-f-1-x\",\"WARC-Payload-Digest\":\"sha1:BD7WXHKS5KYKS5I7NH5B7N7IW4SD2QTC\",\"WARC-Block-Digest\":\"sha1:3NPA3E5HRLT7SPSJOUYRCIZAWBDX2AM6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662521152.22_warc_CC-MAIN-20220518052503-20220518082503-00733.warc.gz\"}"} |
https://www.ncl.ucar.edu/Support/talk_archives/2013/2270.html | [
"# Rank correlation\n\nFrom: Guilherme Martins <jgmsantos_at_nyahnyahspammersnyahnyah>\nDate: Tue Aug 06 2013 - 17:51:44 MDT\n\nHi all,\n\nI'm trying to calculate the rank correlation, but appear a error like below:\n\n*(0) check_for_y_lat_coord: Warning: Data either does not contain a valid\nlatitude coordinate array or doesn't contain one at all.\n*\n*(0) A valid latitude coordinate array should have a 'units' attribute\nequal to one of the following values: *\n*(0) 'degrees_north' 'degrees-north' 'degree_north' 'degrees north'\n'degrees_N' 'Degrees_north' 'degree_N' 'degreeN' 'degreesN' 'deg north'*\n*(0) check_for_lon_coord: Warning: Data either does not contain a valid\nlongitude coordinate array or doesn't contain one at all.*\n*(0) A valid longitude coordinate array should have a 'units' attribute\nequal to one of the following values: *\n*(0) 'degrees_east' 'degrees-east' 'degree_east' 'degrees east'\n'degrees_E' 'Degrees_east' 'degree_E' 'degreeE' 'degreesE' 'deg east'*\n\nThen I put the:\n\n*spc!0 = \"lat\"*\n*spc!1 = \"lon\"*\n*spc&lat@units = \"degrees_north\"*\n*spc&lon@units = \"degrees_east\"*\n\nAnd the error continue, now appear:\n\n*fatal:No coordinate variable exists for dimension (lat) in variable (spc)*\n*fatal:(lat) is not coordinate variable in variable(spc).*\n*fatal:[\"Execute.c\":8128]:Execute: Error occurred at or near line 26 in\nfile fig13.correlacao.ncl*\n\nMy code:\n\nbegin\n\nprp_mod = f1->prec ; nt,ny,nx --> dimensão 0,1,2\nprp_obs = f2->precip ; nt,ny,nx --> dimensão 0,1,2\n\nspc = spcorr_n(prp_mod,prp_obs,0) ; correlation at dimension time\n\nspc!0 = \"lat\"\nspc!1 = \"lon\"\nspc&lat@units = \"degrees_north\"\nspc&lon@units = \"degrees_east\"\n\nprintVarSummary(spc)\n\nwks = gsn_open_wks(\"eps\",\"corr\")\n\nres = True\nres@cnFillOn = True\nres@cnLinesOn = False\n;res@mpMinLonF = -160.\n;res@mpMaxLonF = 0.\n;res@mpMinLatF = -60.\n;res@mpMaxLatF = 20.\n\nplot = gsn_csm_contour_map_ce(wks,spc,res)\n\nend\n\nThanks,\n\nGuilherme\n\n```--\n*Guilherme Martins*"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5717711,"math_prob":0.9315886,"size":2403,"snap":"2023-14-2023-23","text_gpt3_token_len":784,"char_repetition_ratio":0.13047104,"word_repetition_ratio":0.14184397,"special_character_ratio":0.32001665,"punctuation_ratio":0.19909503,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9608773,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-29T07:56:18Z\",\"WARC-Record-ID\":\"<urn:uuid:afd5bc54-a7fa-45f3-973d-800e5fa4a722>\",\"Content-Length\":\"8239\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b26a59a9-ab50-4c3b-928e-dc375fedd3a5>\",\"WARC-Concurrent-To\":\"<urn:uuid:ca036b66-3d1d-4a35-9814-9ca1d7f27322>\",\"WARC-IP-Address\":\"128.117.225.48\",\"WARC-Target-URI\":\"https://www.ncl.ucar.edu/Support/talk_archives/2013/2270.html\",\"WARC-Payload-Digest\":\"sha1:HD4FHB3ZCXQDVK7OG53IGGSLGIX4HY4M\",\"WARC-Block-Digest\":\"sha1:V6ZBTB2EFZFJ5X2VGDYK7DYJEC7NXHM6\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224644817.32_warc_CC-MAIN-20230529074001-20230529104001-00006.warc.gz\"}"} |
https://buddymantra.in/passive-components-capacitors-inductors/?amp=1 | [
"# Passive Components – Capacitors, Inductors\n\n### Capacitors:\n\nThe capacitor is the linear passive component which is used to store the electric charge. It will have the pair of electrodes between which have the insulating dielectric material.\n\nStored charge Q= CV (C= capacitive reactance, V= applied voltage)\n\nCurrent through capacitor I= C dv/dt\n\n### Types\n\nFixed capacitors(these offers the fixed reactance to flow the current)\n\n• Ceramic Capacitor",
null,
"(consists of the ceramic plates made by silver)\n\n• Electrolyte Capacitor",
null,
"• (These are polarized and used for the high value of capacitance)\n\nVariable Capacitors",
null,
"",
null,
"(They offer capacitance which can be varied by varying the distance between the plates. They can be air gap capacitors or vacuum capacitors.)\n\n### Inductors:\n\nAn inductor stores energy in form of a magnetic field. It generally consists of a conductor coil, which offers a resistance to the applied voltage. It works on the basic principle of Faraday’s law of inductance, according to which a magnetic field is created when current flows through the wire and the electromotive force developed opposes the applied voltage.",
null,
"stored energy E = LI^2. ( L is the inductance measured in Henries, I is the current flowing through it.)"
]
| [
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"data:image/svg+xml;base64,PHN2ZyBoZWlnaHQ9IjI1MCIgd2lkdGg9IjUzNCIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2ZXJzaW9uPSIxLjEiLz4=",
null,
"data:image/svg+xml;base64,PHN2ZyBoZWlnaHQ9IjM1MyIgd2lkdGg9IjUxNCIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2ZXJzaW9uPSIxLjEiLz4=",
null,
"data:image/svg+xml;base64,PHN2ZyBoZWlnaHQ9IjIzNyIgd2lkdGg9IjIzOCIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2ZXJzaW9uPSIxLjEiLz4=",
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"data:image/svg+xml;base64,PHN2ZyBoZWlnaHQ9IjMwMCIgd2lkdGg9IjIxNiIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2ZXJzaW9uPSIxLjEiLz4=",
null,
"data:image/svg+xml;base64,PHN2ZyBoZWlnaHQ9IjIzMyIgd2lkdGg9IjQwMyIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2ZXJzaW9uPSIxLjEiLz4=",
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]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9278502,"math_prob":0.9209235,"size":1171,"snap":"2023-14-2023-23","text_gpt3_token_len":256,"char_repetition_ratio":0.14310198,"word_repetition_ratio":0.0,"special_character_ratio":0.19299744,"punctuation_ratio":0.074257426,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96263736,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-08T01:15:06Z\",\"WARC-Record-ID\":\"<urn:uuid:3ee10d75-16bc-42ab-bf80-664706216d2e>\",\"Content-Length\":\"44192\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6b5b3f3b-f3a9-40f2-a20a-774370368580>\",\"WARC-Concurrent-To\":\"<urn:uuid:bcac74bd-563a-41e0-bc0a-5a9648cf9c76>\",\"WARC-IP-Address\":\"23.26.250.149\",\"WARC-Target-URI\":\"https://buddymantra.in/passive-components-capacitors-inductors/?amp=1\",\"WARC-Payload-Digest\":\"sha1:S6OVCFI2YBCIDFWA6SGPZ4YKCFANNFNH\",\"WARC-Block-Digest\":\"sha1:4ATLJ3AHREHRH3G5P2LFSU2ITK3ELG4G\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224654031.92_warc_CC-MAIN-20230608003500-20230608033500-00741.warc.gz\"}"} |
https://www.radzconsulting.com/quantitative-methods--and-analysis.php | [
"Quantitative Methods and Quantitative Analysis Training in Sacramento\n\nTopics covered in Basic and Advanced Training:\n\nBasics of Statistics Using MS Excel or SPSS\n\nBasic Statistics and Quantitative methods using Microsoft Excel is vital to develop analytical skills. Better decisions are outcome of such skills and it‘s for both researchers and business managers.\n\nWhat you will learn:\n1. Quantitative data collection,\n2. Data entry, data cleaning, analysis and decision making using Microsoft Excel / SPSS\n3. From problem statement to research questions and frameworks (models) and/or hypothesis testing using Microsoft Excel / SPSS\n4. Differentiate and use different statistical tests to solve problems using Microsoft Excel / SPSS\n5. Writing Interpretation and report.\n\nThe focus of this class is on using Microsoft Excel / SPSS to solve statistics problems.\n\nSubjects that we cover:\n- Basic Statistics (Microsoft Excel / SPSS)\n- Inferential Statistics (Microsoft Excel / SPSS)\n- Type of variables and measurement levels (Microsoft Excel / SPSS)\n- Sample and Population (Microsoft Excel / SPSS)\n- Data with Microsoft Excel / SPSS\n- Entering and Cleaning Data (Microsoft Excel / SPSS)\nDescriptive Statistics (Microsoft Excel / SPSS)\n- Frequency tables, Pie chart, Bar chart, Histogram (Microsoft Excel / SPSS)\n- Mean, Mode, Median (Microsoft Excel / SPSS)\n- Skewness, Standard Deviation, Variance (Microsoft Excel / SPSS)\nHypothesis Testing for differences (Microsoft Excel / SPSS)\n- t-tests ( 1-sample test , 2-smaple test and paired t-test) (Microsoft Excel / SPSS)\n- Analysis of Variance (ANOVA) (Microsoft Excel / SPSS)\nHypothesis Testing for Relations (Microsoft Excel / SPSS)\n- Graphing scatter diagram (Microsoft Excel / SPSS)\n- Correlations and Regression Analysis (Simple linear regression) (Microsoft Excel / SPSS)\n\n- graphing results\nDecision Analysis using the results from Microsoft Excel / SPSS\nInterpretation, sensitivity analysis, writing report, conclusion and more. (Microsoft Excel / SPSS)\n\n- Review of Descriptive Statistics\nScale of Measurement\n- Tests & Assumptions\n- Multivariate Statistics (Microsoft Excel / SPSS)\nWhy Study Multivariate Statistics\nUnivariate and Bivariate Statistics (Microsoft Excel / SPSS)\n- Reliability (Microsoft Excel / SPSS)\n- Data Appropriate for Multivariate Statistics\nCorrelation and Partial Correlation (Microsoft Excel / SPSS)\nHypothesis Testing (Microsoft Excel / SPSS)\n- Multiple Regression (1) (Microsoft Excel / SPSS)\nGeneral Purpose and Description\nTheoretical Issues: Assumptions\nData Analysis and Interpretation\n- Multiple Regression (2) (Microsoft Excel / SPSS)\nFactor Analysis (Exp. & Com.) SPSS\nBasics of Logistic Regression (LR) / SPSS\nInterpretation of LR / SPSS\n- Statistical technics to compare groups (1)\nNon Parametric statistics (Microsoft Excel / SPSS)\nT test MS Excel SPSS\nOne way ANOVA (Microsoft Excel / SPSS)\n- Statistical technics to compare groups (2)\nTwo ways ANOVA (Microsoft Excel / SPSS)\nMANOVA using SPSS\nAnalysis of Covariance (ANCOVA using SPSS)\nIntroduction to SEM (Structural Equation Modelling)"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6023938,"math_prob":0.47657394,"size":3118,"snap":"2021-31-2021-39","text_gpt3_token_len":697,"char_repetition_ratio":0.30378935,"word_repetition_ratio":0.089324616,"special_character_ratio":0.20654266,"punctuation_ratio":0.06849315,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99699956,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-05T22:03:02Z\",\"WARC-Record-ID\":\"<urn:uuid:4fd5d506-7748-4665-b2d8-c6bf284548a2>\",\"Content-Length\":\"51249\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eeca6c16-053e-4445-9801-01c4a32a16d3>\",\"WARC-Concurrent-To\":\"<urn:uuid:db0a1432-d53b-4379-ac28-9ad38669564f>\",\"WARC-IP-Address\":\"104.21.79.190\",\"WARC-Target-URI\":\"https://www.radzconsulting.com/quantitative-methods--and-analysis.php\",\"WARC-Payload-Digest\":\"sha1:XZ4SYVPEOOOJGTBAZWMLWPX2QE6UTNWN\",\"WARC-Block-Digest\":\"sha1:IUNJQREBRAXUU5YVLGT7XTWEGLZPSLJF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046157039.99_warc_CC-MAIN-20210805193327-20210805223327-00562.warc.gz\"}"} |
https://math.stackexchange.com/questions/619544/why-homology-with-coefficients | [
"# Why homology with coefficients?\n\nI am currently studying a bit of homology theory (on topological spaces). Let $H_n(X)$ denote the singular homology groups of the topological space $X$, then as you know we can define the singular homology with coefficients in the abelian group $G$ by $$H_n(X;G)=H_n(S_*(X)\\otimes G)$$ Now, I understand the definitions and computations, proofs etc., but I would like to understand the following:\n\n1. Why is this interesting to study? Does it make computations easier somewhere (for some groups $G$)? Does it allow us to prove interesting theorems we can't prove using $G=\\mathbb{Z}$?\n2. What are interesting examples of use of homology with coefficients?\n3. Is the theory of \"degree modulo $2$\" (treated e.g. in Milnor's book Topology from the differentiable viewpoint) in truth just degree theory on $H_n(X;\\mathbb{Z}_2)$?\n• First of all you want to be careful about what is meant by homology with coefficients. It is not true that $H_n(X,\\Bbb{Z}) \\otimes G$ is the same as first tensoring your chain complex and then calculating the homology there. There is an obstruction given by a Tor term in universal coefficients! – user38268 Dec 27 '13 at 13:23\n• @Benja Ah, I didn't know it. I removed the bracketed sentence. Thanks for the comment, would you care to give me a reference on the subject? – Daniel Robert-Nicoud Dec 27 '13 at 14:48\n• See my answer math.stackexchange.com/questions/47100/… for some examples of why $\\mathbb Z_2$ coefficients are useful. – Cheerful Parsnip Dec 27 '13 at 18:11\n• @DanielRobert-Nicoud Look at en.wikipedia.org/wiki/Universal_coefficient_theorem. If you're not sure about the Tor thing, you can learn about it by reading some posts on this site. It's just the left derived functor of the tensor product. – user38268 Dec 28 '13 at 10:02\n\nYou can find the construction of homology with general coefficients and the universal coefficient theorem in Hatcher's Algebraic Topology, which is available free from his website.\n\nThe answer to the second part of your first question is yes, especially in the case that we take $G$ to be a field, most often finite or $\\mathbb{Q}$, or $\\mathbb{R}$ in differential topology. Homology over a field is simple because $\\operatorname{Tor}$ always vanishes, so you get e.g. an exact duality between homology and cohomology. Homology with $\\mathbb{Z}_2$ coefficients is also the appropriate theory for many questions about non-orientable manifolds-their top $\\Bbb{Z}$-homology is zero, but their top $\\Bbb{Z}_2$ homology is $\\Bbb{Z}_2$, which leads to the degree theory in Milnor you were mentioning.\n\nCohomology with more general coefficients than $\\mathbb{Z}$ is even more useful than homology. For instance it leads to the result that if a manifold $M$ has any Betti number $b_i(M)<b_i(N)$, where $b_i$ is the rank of the free part of $H_i$, there's no map $M\\to N$ of non-zero degree. This has lots of quick corollaries-for instance, there's no surjection of $S^n$ onto any $n$-manifold with nontrivial lower homology! Edit: This is obviously false, and I no longer have any idea whether I meant anything true.\n\nBut in the end $H_*(X;G)$ is more of a stepping stone than anything else; it gets you thinking about how much variety there could be in theories satisfying the axioms of homology. It turns out there's almost none-singular homology with coefficients in $G$ is the only example-but if we rid ourselves of the \"dimension axiom\" $$H_*(\\star)=\\left\\{\\begin{matrix}\\mathbb{Z},*=0\\\\0,*>0\\end{matrix}\\right.$$ then we get a vast collection of \"generalized (co)homology theories,\" beginning with K-theory, cobordism, and stable homotopy, which really do contain new information. In some cases, so much new information that we can't actually compute them yet!\n\n• I don't understand your point about $S^n$ not surjecting onto an $n$-manifold with non-trivial lower homology. For example, here is a smooth surjection from $S^2$ to $S^1\\times S^1$: First, the projection map $S^2\\rightarrow D^2$ to a disc, given by $(x,y,z)\\mapsto(x,y)$ is definitely smooth. Now, a small modification of the usual square-with-edges-identified picture gives a smooth surjection from $D^2$ to $S^1\\times S^1$. (Of course, as you said, the degree will be $0$, but isn't this still a surjective smooth map?) (And sorry to be 3.5 years late!) – Jason DeVito Apr 25 '17 at 23:22\n• @JasonDeVito I also do not understand Kevin's point! – Kevin Arlin Apr 26 '17 at 5:49\n• I kept thinking about this a bit more, and I think the following is true: For given smooth manifolds $M$ and $N$ of dimension $m$ and $n$ respectively, assume $N$ is connected and $m\\geq n$. Then there is a smooth surjection $M\\rightarrow N$. Sketch: Map $M$ to $S^{m}$ by collapsing everything outside a small ball to a point, map $S^m$ to $D^m$ as in the previous comment. If $m > n$, project $D^m$ down to $D^n$ by just forgettnig coordinates. Equip $N$ with an auxillary Riemannian metric and imagine $D^n$ is a large ball in $T_p N$. Exponentiating gives a surjection by Hopf-Rinow. – Jason DeVito Apr 26 '17 at 17:06\n\nFirst of all, on the level of complexes $C_n(X;A)=C_n(X)\\otimes A$, but (in general) $H_n(X;A)\\ne H_n(X)\\otimes A$. Nevertheless, if you know ordinary homology you can compute homology with any coefficients — so these groups don't contain any really new information (see 'universal coefficient theorem' for details).\n\nMain reason to consider (co)homology with non-trivial coefficients, I believe, that sometimes these groups (and extra structure on these groups) are easier to compute.\n\nFor example, when you compute $H_n(X;\\mathbb Z/2)$ you can forget about all problems with signs etc (a concrete example: it's quite easy to compute (co)homology of $SO(n)$ with coefficients $\\mathbb Z/2$ but very hard even to describe answer for ordinary (co)homology group); when you compute homology with rational coefficients, you can forget about all torsion-related phenomena.\n\nSomewhat more advanced example: the algebra of stable cohomological operations mod p (see 'Steenrod squares' / 'Steenrod powers' ) is much easier to describe than the algebra of integral cohomological operations. (All this reminds slightly situation in number theory, where many problems are easier to solve mod p or maybe rationally than in integers.)\n\nAlso some constructions (obstructions, characteristic classes...) naturally 'live' in (co)homology with non-trivial coefficients (see e.g. 'Stiefel–Whitney classes').\n\nFinally, (co)homology group with coefficients in a field with char 0 have different analytical description ('de Rham cohomology' etc) and (at least in some situations) some extra structue (e.g. 'Hodge structures' in cohomology of projective complex manifolds).\n\nYou already probably know the advantages of thinking of homology as being a functor of $X$. By allowing an abelian group $G$ as coefficients, it becomes a functor of two variables, the space $X$ and the group $G$. Now you can exploit the $G$-functoriality as well.\n\nFor example, the short exact sequence of groups $$0 \\to \\mathbb Z \\to \\mathbb Z \\to \\mathbb Z/p\\mathbb Z \\to 0$$ (the second arrow is mult. by $p$) gives rise to the short exact sequence of cohomology groups $$0 \\to H^i(X,\\mathbb Z)/p H^i(X,\\mathbb Z) \\to H^i(X,\\mathbb Z/p\\mathbb Z) \\to p\\text{-torsion subgroup of } H^{i+1}(X,\\mathbb Z) \\to 0.$$ (Here I have switched from homology to cohomology, since I find that latter a little easier to think about, and easier to state these sorts of results about.)\n\nSo studying cohomology with coefficiens in $\\mathbb Z/p\\mathbb Z$ is closely related to studying torsion in cohomology with $\\mathbb Z$-coefficients. If $p$ is a prime (which is the case I was imagining, although the above is valid for any integer $p$), then $\\mathbb Z/p\\mathbb Z$ is a field, and so we can investigate torsion in cohomology by studying cohomology with coefficients in a (finite) field. The latter investigation has technical advantages, e.g. Poincare duality if $X$ is a manifold.\n\nThinking of cohomology as a functor of $G$ as well as of $X$ leads to the idea of sheaf cohomology (a sheaf of abelian groups is something like an abelian group that can vary from point to point over $X$). The possibility of taking $G$ to be a non-constant sheaf greatly increases the flexibility of cohomology as a tool.\n\nAnother example of utility of coefficients:\n\nIf $E \\to B$ is a fibre bundle with fibre $F$, and $B$ is connected and simply connected, then there is a spectral sequence (the Leray spectral sequence) $$E_2^{i,j} := H^i(B,H^j(F,G) ) \\implies H^{i+j}(E,G),$$ relating the cohomology of the base and fibre to that of the total space.\n\nNote that even if we take $G$ to be $\\mathbb Z$, the cohomology of the base has coefficients in the cohomology groups $H^j(F,\\mathbb Z)$, which are likely not just equal to $\\mathbb Z$.\n\nIf $B$ is not simply connected, then there is a similar spectral sequence, but the coefficients are a locally constant, but typically non-constant, sheaf on $B$.\n\n• Thanks for your answer. I an familiar with the basics of category theory, but (unfortunately?) the course of algebraic topology I took didn't delve too much into that point of view. Maybe in the second part of the course... Anyway, do you have a reference for a book looking at algebraic topology from a categorical setting? – Daniel Robert-Nicoud Dec 28 '13 at 0:30\n• @DanielRobert-Nicoud: Dear Daniel, Any alg. top. textbook should have some functorial/categorical discussion (e.g. because one of the important tools in applying (co)homology is not just the groups themselves, but the maps between them induced by maps between spaces). I wouldn't suggest going out of your way to learn more category theory; this is (in my opinion) a subject best learnt largely by osmosis as you see various categorical notions appearing in different contexts. What I would suggest is finding examples of spaces with torsion in their (co)homology, and then computing with ... – Matt E Dec 28 '13 at 1:59\n• ... various coefficients and seeing how the answers are related to one another. (E.g. see concretely how the $\\mathbb F_p$-Betti number differs from the $\\mathbb Q$-Betti number when there is $p$-torsion in the homology.) As you learn more topology/geometry, your understanding of the role of (co)homology will mature, as will your understanding of the role of different choices of coefficients. Regards, – Matt E Dec 28 '13 at 2:01"
]
| [
null
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https://www.gohonoring.com/2022/02/l5-binary-system-usage-of-binary-system.html | [
"### Ticker\n\n6/recent/ticker-posts\n\n# L5, Binary System, Usage of Binary System, package q10853\n\nThe binary number system is used both in mathematics and digital electronics.\n\nThe binary numeral system or base-2 numeral system represents numeric values using only two symbols - 0 (zero) and 1 (one).\n\nComputers have circuits (logic gates) which can either be in off or on state. These two states are represented by 0 (zero) and 1 (one).\n\nIt is for this reason computation in computers is performed using a binary number system (base-2) where all numbers are represented using 0's and 1's.\n\nEach binary digit i.e., a single 0 (zero) or 1 (one) is called a bit. A collection of 8 such bits is called a Byte.\n\nThe way we use grams and kilograms to measure, in computer terminology we have different names given to multiples of 210 (i.e., 1024 times existing value) .\n\nFor example:\n\n• 1 byte = 8 bits\n• 1 kilobyte = 1024 bytes\n• 1 megabyte = 1024 kilobytes\n• 1 gigabyte = 1024 megabytes\n• 1 terabyte = 1024 gigabytes\n• 1 petabyte = 1024 terabytes\n\nIn a computer when we type some text, store images, music, videos or any type of data, we should remember that they all are eventually stored in binary format on the disk.\n\nSelect all the correct statements given below.\n\npackage q10853 :-\n\nIn binary system the base is 2.\n\nA byte is composed of 10 bits.\n\n1MB (megabyte) = 8388608 bits.\n\nA decimal number cannot be represented as a binary number."
]
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null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8536976,"math_prob":0.9667066,"size":1339,"snap":"2023-40-2023-50","text_gpt3_token_len":349,"char_repetition_ratio":0.12734082,"word_repetition_ratio":0.008264462,"special_character_ratio":0.2785661,"punctuation_ratio":0.10344828,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99143064,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T20:26:34Z\",\"WARC-Record-ID\":\"<urn:uuid:f174e1fa-2b8f-482e-b8a8-ffaeeddc28f9>\",\"Content-Length\":\"469934\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b27b7f95-43db-432d-a9c3-7d5880df0172>\",\"WARC-Concurrent-To\":\"<urn:uuid:4c4fe3cc-788d-47c5-9b04-c1628c538a92>\",\"WARC-IP-Address\":\"142.251.163.121\",\"WARC-Target-URI\":\"https://www.gohonoring.com/2022/02/l5-binary-system-usage-of-binary-system.html\",\"WARC-Payload-Digest\":\"sha1:U7FCO7SH2SRDISCH73PFIIKLKQAJKY4P\",\"WARC-Block-Digest\":\"sha1:ZZ3TSX7W67RWQA53IM3O454ISTLOFBQ7\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100534.18_warc_CC-MAIN-20231204182901-20231204212901-00087.warc.gz\"}"} |
https://www.sdk.cn/details/Vqm1O8D0XRzW82RzGW | [
"# JS基础算法(正则表达式 &递归篇)\n\n• 前端\n• 面试\n• 算法",
null,
"原创\n\n### 1.重复的子字符串\n\n#### 示例 3:\n\nreturn reg.test(str)\n}\n\n\n### 2.正则表达式匹配\n\n‘.’ 匹配任意单个字符\n'\n’ 匹配零个或多个前面的那一个元素\n\ns 可能为空，且只包含从 a-z 的小写字母。\np 可能为空，且只包含从 a-z 的小写字母，以及字符 . 和 *。\n\ns = “aa”\np = “a”\n\ns = “aa”\np = “a*”\n\n#### 示例 3:\n\ns = “ab”\np = “.\"\n\n” 表示可匹配零个或多个（‘*’）任意字符（‘.’）。\n\ns = “aab”\np = “cab”\n\n#### 示例 5:\n\ns = “mississippi”\np = “misisp*.”\n\n• 一个字符一个字符对比，对比完一个扔掉一组，然后重复刚才的动作\n• 三种情况：无模式，有模式*,有模式.\nexport default (s,p)=>{\nlet isMatch=(s,p)=>{\n// 边界情况，如果s和p都为空，说明处理结束，返回true，否则返回false\nif(p.length<=0){\nreturn !s.length\n}\n// 判断p模式字符串的第一个字符和s字符串的第一个字符是不是匹配\nlet match=false\nif(s.length>0&&(p===s||p==='.')){\nmatch=true\n}\n// p有模式的，字符后面有*\nif(p.length>1&&p==='*'){\n// 第一种情况：s*匹配0个字符\n// 第二种情况：s*匹配1个字符，递归下去，用来表示s*匹配多个s\nreturn isMatch(s,p.slice(2))||(match && isMatch(s.slice(1),p))\n}else{// 字符后面没有*\nreturn match && isMatch(s.slice(1),p.slice(1))\n}\n}\nreturn isMatch(s,p)\n}\n\n\n### 3.复原IP地址\n\nP由4部分构成，每部分范围0~255（递归）\n\nexport default (str)=>{\n// 保存所有符合条件的ip\nlet r=[]\n// 递归函数\nlet search=(cur,sub)=>{\nif(cur.length===4&&cur.join('')===str){\nr.push(cur.join('.'))\n}else{\nfor(let i=0,len=Math.min(3,sub.length),tmp;i<len;i++){\ntmp=sub.substr(0,i+1)\nif(tmp<256){\nsearch(cur.concat([tmp]),sub.substr(i+1))\n}\n}\n}\n}\nsearch([],str)\nreturn r\n}\n\n\n• 递归的本质（就是一个while循环，但可以有多个条件或地方调用）\n• 每一个处理过程是相同的\n• 输入输出是相同的\n• 处理次数未知\n\n### 4.与所有单词相关联的字符串\n\n#### 示例 1：\n\ns = “barfoothefoobarman”,\nwords = [“foo”,“bar”]\n\n#### 示例 2：\n\ns = “wordgoodgoodgoodbestword”,\nwords = [“word”,“good”,“best”,“word”]\n\nexport default (str,words)=>{\n// 保存结果\nlet result=[]\n// 记录数组的长度，做边界条件计算\nlet num=words.length\n// 递归函数体\nlet range=(r,_arr)=>{\nif(r.length===num){\nresult.push(r)\n}else{\n_arr.forEach((item,idx)=>{\n// 当前元素踢出去，留下剩下的\nlet tmp=[].concat(_arr)\ntmp.splice(idx,1)\nrange(r.concat(item),tmp)\n})\n}\n}\nrange([], words)\n// [0,9,-1]\nreturn result.map(item=>{\nreturn str.indexOf(item.join(''))\n}).filter(item=>item!==-1).sort()\n}\n\n\n0\n\n0\n\n0"
]
| [
null,
"https://www.sdk.cn/_nuxt/img/original1.33619d8.svg",
null
]
| {"ft_lang_label":"__label__zh","ft_lang_prob":0.7632113,"math_prob":0.9969407,"size":2780,"snap":"2021-43-2021-49","text_gpt3_token_len":1618,"char_repetition_ratio":0.09978386,"word_repetition_ratio":0.018404908,"special_character_ratio":0.3852518,"punctuation_ratio":0.209622,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98783493,"pos_list":[0,1,2],"im_url_duplicate_count":[null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-27T02:42:14Z\",\"WARC-Record-ID\":\"<urn:uuid:091b152f-561d-474b-a26f-92709793addf>\",\"Content-Length\":\"264719\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3c4a38f8-563d-47a9-bc36-01c80407c54a>\",\"WARC-Concurrent-To\":\"<urn:uuid:00ab16be-f34d-4c4a-a80d-bd316a263fb1>\",\"WARC-IP-Address\":\"116.63.41.80\",\"WARC-Target-URI\":\"https://www.sdk.cn/details/Vqm1O8D0XRzW82RzGW\",\"WARC-Payload-Digest\":\"sha1:H4SVFNIENVI44JPVMVNE2VYG7OL32F7E\",\"WARC-Block-Digest\":\"sha1:RJHXNPCDR3CGEC6TMDVE2OFL52E5LQT3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358078.2_warc_CC-MAIN-20211127013935-20211127043935-00403.warc.gz\"}"} |
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