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http://nesssi.cacr.caltech.edu/MLS/20171216/1712161050274125606.html | [
"CRTS Transient ID MLS171216:035228+053323\n(When using CRTS data please cite Drake et al. 2009, ApJ, 696, 870 and our ID.)\n\nBack to index\n Image 1 Image 2 Image 3 Image 4 SDSS",
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"Back to index\n\n Additional transient info: Check MPC Check Simbad Check Simbad-HPM Check NED Current CRTS lightcurve Archival CRTS lightcurve CRTS Orphans SDSS DR-9 data SDSS DR-13 data PS-1 DR1 images Past detections in Datascope Links to past CRTS transients: All transients, SNe, Blazars, CVs",
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"Red= detection, Green = MLS, Blue = Orphans,Orange = CSS, Purple = SSS\n\nID = 1712161050274125606\nnewast = 0.400000\nav_motn = 0.006344\nav_deltime = 16.282080\nav_uncertx = 0.034737\nav_uncerty = 0.034737\nav_inclin = 42.330240\nappreject = 0.000\nmastersn = 66.618\naveragesn = 114.140\ntheta = 0.000\nnhigh = 143\nTime = 2458103.717436\nTime2 = 2458103.700477\nTime3 = 2458103.706131\nTime4 = 2458103.711781\nRA = 58.1167300\nDec = 5.5562800\nMag = 15.381400\nMag_err = 0.002439\nFWHM = 2.520\nMaster = N05027.master.fits\nMasterRAoff = -0.012306\nMasterDecoff = -0.020192\nRA2 = 58.1166900\nDec2 = 5.5563000\nMag2 = 15.377400\nMag2_err = 0.002429\nFWHM2 = 2.350\nRA3 = 58.1167300\nDec3 = 5.5562800\nMag3 = 15.387600\nMag3_err = 0.002438\nFWHM3 = 2.540\nRA4 = 58.1167400\nDec4 = 5.5562700\nMag4 = 15.384400\nMag4_err = 0.002434\nFWHM4 = 2.540\nOppos_ang = 332.907499\nInner_motion = -0.460895\nOuter_motion = -0.358014\nEclip_long = 57.117862\nEclip_lat = -14.462381"
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"http://nesssi.cacr.caltech.edu/MLS/20171216/jpg/1712161050274125606-0001.arch.jpg",
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"http://nesssi.cacr.caltech.edu/MLS/20171216/jpg/1712161050274125606-0003.arch.jpg",
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"http://nesssi.cacr.caltech.edu/MLS/20171216/jpg/1712161050274125606-0004.arch.jpg",
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"http://nesssi.cacr.caltech.edu/MLS/20171216/jpg/1712161050274125606.master.jpg",
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"http://nesssi.cacr.caltech.edu/MLS/20171216/imgs/1712161050274125606.png",
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https://convertoctopus.com/1720-centimeters-to-meters | [
"## Conversion formula\n\nThe conversion factor from centimeters to meters is 0.01, which means that 1 centimeter is equal to 0.01 meters:\n\n1 cm = 0.01 m\n\nTo convert 1720 centimeters into meters we have to multiply 1720 by the conversion factor in order to get the length amount from centimeters to meters. We can also form a simple proportion to calculate the result:\n\n1 cm → 0.01 m\n\n1720 cm → L(m)\n\nSolve the above proportion to obtain the length L in meters:\n\nL(m) = 1720 cm × 0.01 m\n\nL(m) = 17.2 m\n\nThe final result is:\n\n1720 cm → 17.2 m\n\nWe conclude that 1720 centimeters is equivalent to 17.2 meters:\n\n1720 centimeters = 17.2 meters\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 meter is equal to 0.058139534883721 × 1720 centimeters.\n\nAnother way is saying that 1720 centimeters is equal to 1 ÷ 0.058139534883721 meters.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that one thousand seven hundred twenty centimeters is approximately seventeen point two meters:\n\n1720 cm ≅ 17.2 m\n\nAn alternative is also that one meter is approximately zero point zero five eight times one thousand seven hundred twenty centimeters.\n\n## Conversion table\n\n### centimeters to meters chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from centimeters to meters\n\ncentimeters (cm) meters (m)\n1721 centimeters 17.21 meters\n1722 centimeters 17.22 meters\n1723 centimeters 17.23 meters\n1724 centimeters 17.24 meters\n1725 centimeters 17.25 meters\n1726 centimeters 17.26 meters\n1727 centimeters 17.27 meters\n1728 centimeters 17.28 meters\n1729 centimeters 17.29 meters\n1730 centimeters 17.3 meters"
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https://rdrr.io/cran/caret/src/R/predict.PLS.R | [
"# R/predict.PLS.R In caret: Classification and Regression Training\n\n#### Defines functions predict.PLS\n\n```predict.PLS <- function(object, newdata,\nncomp = NULL,\ntype = ifelse(object\\$isRegression, \"response\", \"class\"), ...)\n{\n\nif(is.null(ncomp) & !is.null(object\\$bestIter\\$.ncomp)) ncomp <- object\\$bestIter\\$.ncomp\n\nif(is.null(ncomp)) stop(\"the number of components must be given\")\n\n# adapted from the pls package\nif(object\\$isRegression & c(type %in% c(\"class\", \"prob\")))\nstop(\"cannot get a class estimate if the original y was not a factor\")\n\nif(!(type %in% c(\"response\", \"class\", \"prob\")))\nstop(\"type must be either response, class or prob\")\n\nif(missing(newdata)) newdata <- object\\$x\nif(!is.matrix(newdata)) newdata <- as.matrix(newdata)\n\n# from coef.mvr in pls package\nB <- object\\$coefficients[, , 1:ncomp, drop = FALSE]\ndB <- dim(B)\ndB <- dB + 1\ndnB <- dimnames(B)\ndnB[] <- c(\"(Intercept)\", dnB[])\nBInt <- array(dim = dB, dimnames = dnB)\nBInt[-1, , ] <- B\nfor (i in seq(along = 1:ncomp)) BInt[1, , i] <- object\\$Ymeans - object\\$Xmeans %*% B[, , i]\nB <- BInt\n# stop\n\n# from predict.mvr in pls package\ndPred <- dim(B)\ndPred <- dim(newdata)\ndnPred <- dimnames(B)\ndnPred <- dimnames(newdata)\npred <- array(dim = dPred, dimnames = dnPred)\npredY <- sweep(newdata %*% B[-1, , ncomp], 2, B[1, , ncomp], \"+\")\n# stop\n\nout <- switch(\ntype,\nresponse = predY,\nclass =\n{\nclassNum <- apply(predY, 1, which.max)\nfactor(object\\$yLevels[classNum], levels = object\\$yLevels)\n},\n# use softmax technique here\nprob = t(apply(predY, 1, function(data) exp(data)/sum(exp(data)))))\n\nout\n}\n```\n\n## Try the caret package in your browser\n\nAny scripts or data that you put into this service are public.\n\ncaret documentation built on Aug. 9, 2022, 5:11 p.m."
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https://scholar.archive.org/search?q=Vaishali+P.+Sadaphal | [
"Filters\n\n10 Hits in 0.96 sec\n\n### Problem Identification by Mining Trouble Tickets\n\nVikrant Shimpi, Maitreya Natu, Vaishali P. Sadaphal, Vaishali Kulkarni\n2014 International Conference on Management of Data\n= Mode of the position of the word w * Label of the clique = Arrangment of the words in Cw according to p Figure 2 : 5 . 25 Figure 2: System-generated tickets: group id vs size of each group Figure ... a label • Identify set of common words Cw from the set of cleaned description belonging to the clique -For each word w in Cw, * Compute its position in set of cleaned description belonging to clique, p ...\n\n### Sensor Selection Heuristic in Sensor Networks [chapter]\n\nVaishali P. Sadaphal, Bijendra N. Jain\n2005 Lecture Notes in Computer Science\nWe consider the problem of estimating the location of a moving target in a 2-D plane. In this paper, we focus attention on selecting an appropriate 3 rd sensor, given two sensors, with a view to minimize the estimation error. Only the selected sensors need to measure distance to the target and communicate the same to the central \"tracker\". This minimizes bandwidth and energy consumed in measurement and communication while achieving near minimum estimation error. In this paper, we have proposed\nmore » ... hat the 3 rd sensor be selected based on three measures viz. (a) collinearity, (b) deviation from the ideal direction in which the sensor should be selected, and (c) proximity of the sensor from the target. We assume that the measurements are subject to multiplicative error. Further, we use least square error estimation technique to estimate the target location. Simulation results show that using the proposed algorithm it is possible to achieve near minimum error in target location. the central device responsible for estimating the location of the target, (b) the clocks of the sensors are synchronized so that the sensors \"measure\" distance at approximately the same time, and (c) the sensors are able to communicate their measurements to this central device, also referred to as the \"tracker\". Since the target is moving and since a sensor must be within a certain distance from the target (before it can detect the presence of the target and measure distance), we assume that there are several sensors, {s i } = Σ, spread across the 2-D plane. In fact, we assume that there are three or more sensors located in and around every point in the 2-D plane so that we can compute an estimate based on measurements from a subset of three sensors suitably selected to minimize estimation error. This approach also allows one to minimize communication overheads and conserve battery power available to sensors. Further, since the target is moving, the collection of sensors changes every time an estimate is required to be obtained. Specifically, we assume that as the target moves, if sensors {s 1 , s 2 , s 3 } have made measurements at time t k , then at time t k+1 , we drop one of the sensors s 1 , s 2 , or s 3 and select a sensor s 4 suitably so as to minimize the error in estimated location of the target. Accordingly, this paper is about suitably selecting the 3 rd sensor from a set of N k+1 sensors.\n\n### Analyzing Periodically Occurring Patterns in Time Series\n\nShivam Sahai, Maitreya Natu, Vaishali P. Sadaphal\n2010 International Conference on Management of Data\nThus, ∀(mi, mj ) ∈ LM such that T ime(mi) < T ime(mj ): pair (mi, mj ) ∈ LM P if, (p − δ) ≤ (T ime(mj ) − T ime(mi) ≤ (p + δ) This ensures that ∀(m i , m j ) ∈ LM P , the pattern T p will be bound by minima ... . • Time-series region: A time-series region T p of length p is a subsequence of p contiguous points in the timeseries. ...\n\n### Random and Periodic sleep schedules for target detection in sensor networks\n\nVaishali P. Sadaphal, Bijendra N. Jain\n2007 2007 IEEE Internatonal Conference on Mobile Adhoc and Sensor Systems\nSpecifically, we analyse and obtain for the Random wake-up schedule the expected delay in detection, and the delay, such that with probability P, the delay is less than the computed value. ... As a result ∃p a q a n = p a m + a. For sure p a ≥ 0. ... Then, ε = δ s − δ t = 1E(∆) vs. δ s . and p 1 = 0.1037, p = 0.1, p 2 = 0.0906, p 3 = 0.08, etc. ...\n\n### Table of Contents PROYEVA: System to Evaluate the Projects Quality in Contests Community-Commerce Brokering Arena for Opportunistic Cloud Services Offerings An Approach to Find Integration and Monitoring Points for Container Logistics Business Processes A Monitoring Approach for Dynamic Service-Oriented Architecture Systems\n\nLaura Silvia, Vargas Perez, Agustin Francisco Gutierrez-Tornes, Edgardo Felipe-Riveron, Ethan Hadar, Steven Greenspan, Tugkan Tuglular, Dilek Avci, Sevket Cetin, Gokhan Daghan, Murat Ozemre, Tolgahan Oysal (+8 others)\nunpublished\nLopez-Soler Workload Characterization for Stability-As-A-Service Vaishali P. ... Sadaphal and Maitreya Natu 84 4R of Service Innovation: Research, Requirements, Reliability and Responsibility Anastasiya Yurchyshyna, Abdelaziz Khadraoui, and Michel Leonard Visualization Method Based ...\n\n### Analytics-Based Solutions for Improving Alert Management Service for Enterprise Systems\n\nAnuja Kelkar, Utkarsh Naiknaware, Sachin Sukhlecha, Ashish Sanadhya, Maitreya Natu, Vaishali Sadaphal\n2013 2013 IEEE 13th International Conference on Data Mining Workshops\nSunday and Monday. 2) p (dow=Sunday) = 1/16 = 0.06; 3) p (dow=M onday) = 15/16 = 0.93; 4) We next compute the entropy value for the day-of-week dimension: H dow = i∈Sunday,M onday −p i * log(p i ). 5) ... Given a set with n possible values {x 1 , x 2 , . . . x n }, the entropy is defined as H = n i −p i * log(p i ), where p i is the probability of occurrence of the value x i . ...\n\n### Varanus: More-with-less fault localization in data centers\n\nVaishali Sadaphal, Maitreya Natu, Harrick Vin, Prashant Shenoy\n2012 2012 Fourth International Conference on Communication Systems and Networks (COMSNETS 2012)\nFormally, information entropy H m (T ) for each candidate monitor node m ∈ T is defined as: H m (T ) = −p(T R m )log(p(T R m )) − p(T U m )log(p(T U m )) where p(T R m ) and p(T U m ) , respectively, denote ... The p-value thus obtained represents the expected amount of similarity between two current observation windows. We use this p-value as the similarity threshold. ...\n\n### Predico: A System for What-if Analysis in Complex Data Center Applications [chapter]\n\nRahul Singh, Prashant Shenoy, Maitreya Natu, Vaishali Sadaphal, Harrick Vin\n2011 Lecture Notes in Computer Science\nWe model each node as a M/G/1/P S queue i.e. the service times are assumed to have an arbitrary distribution and the service discipline at each node is assumed to be processor sharing (PS). ...\n\n### Analytical modeling for what-if analysis in complex cloud computing applications\n\nRahul Singh, Prashant Shenoy, Maitreya Natu, Vaishali Sadaphal, Harrick Vin\n2013 Performance Evaluation Review\nWe model each node as a M/G/1/ P S queue i.e. the service times are assumed to have an arbitrary distribution and the service discipline at each node is assumed to be processor sharing (PS). ...\n\n### Workload Characterization for Stability-As-A-Service\n\nVaishali Sadaphal, Maitreya Natu\nunpublished\nThen the p-prediction heuristic states that if (p 12 < M p ) then p 3 > min(p 1 , p 2 ), where M p is the threshold defined for the p-prediction heuristic. ... Let the p-values of the t-test ran on performance data of these subsets and that of the rest of data are p 1 and p 2 respectively. ..."
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https://www.programmingr.com/tutorial/colsums-in-r/ | [
"# How to Find Column Sums in R using the Colsums function (With examples)\n\nMany statistical procedures require you to find the sum of a column of numbers as part of your calculations. Fortunately, there is a built in function (the colsums function) to find the column sum in R. This reduces the task to a single line of code using a single function in the base R language. Which is especially helpful when you end up calculating multiple column sums within an R data frame.\n\n## Colsums Function.\n\nDoing colsums in R involves using the colsums function, which has the form of colSums(dataset) and returns the sum of the columns in the data set. This sum function also has several optional parameters, one of which is the logical parameter of na.rm that tells the function whether to remove missing value observations.\n\n`````` > x = matrix(rep(1:8),6,4)\n>\n> x\n[,1] [,2] [,3] [,4]\n[1,] 1 7 5 3\n[2,] 2 8 6 4\n[3,] 3 1 7 5\n[4,] 4 2 8 6\n[5,] 5 3 1 7\n[6,] 6 4 2 8\n>\n> colSums(x)\n 21 25 29 33``````\n\nHere is an example of the use of the colsums function. If you add up column 1, you will get 21 just as you get from the colsums function. Along with it, you get the sums of the other three columns. As you can see the default colsums function in r returns the sums of all the columns in the R dataframe and not just a specific column.\n\n## Application.\n\nThere are numerous applications resuming up a column in a data set. This is commonly done in spreadsheets and other formats. In data science, it can be used to gather the totals from a list of values. Below, we have an example based on arrest rates in each state of the United States.\n\n`````` > head(USArrests)\nMurder Assault UrbanPop Rape\nAlabama 13.2 236 58 21.2\nArizona 8.1 294 80 31.0\nArkansas 8.8 190 50 19.5\nCalifornia 9.0 276 91 40.6\n>\n> colSums(USArrests)\nMurder Assault UrbanPop Rape\n389.4 8538.0 3277.0 1061.6 ``````\n\nHere, we have the first six rows of this data set show me the first six states in alphabetical order. The data set contains arrest rates for murder, assault, urban populations, and rape. After the colSums function is applied, we have a total in all four categories for all 50 states combined.\n\n### Potential Errors\n\nThere are a couple of potential errors you can throw with this function. For example, the R colsums() function isn’t very tolerant of a missing or non-numeric data element. You can easily generate lovely errors such as…\n\nerror in colsums(x, na.rm = true) : ‘x’ must be numeric\n\nShould this lovely fail-whale appear, the cause is simple enough. Check the data you’ve fed into your process to see if you are handing it numeric columns with a proper column name. Something in there isn’t numeric and the colsums function throws a little tantrum to communicate that you. My best suggestion is to filter the missing or incorrect data point from your data and proceed from there.\n\nYou may also get:\n\nerror in colsums: ‘x’ must be an array of at least two dimensions\n\nWhich occurs when you feed a vector (single dimensional series of values) into a function which expects to look at an array.\n\n### Related Functions & Broader Usage\n\nThere are several functions designed to help you calculate the total column value and average value of columns and rows in R. In addition to rowmeans in r, this family of functions includes colmeansrowsum, and colsum. Here’s some specifics on where you use them…\n\n• Colmeans – calculate mean of multiple columns in r .\n• Colsums – how do i sum each column in r…\n• Rowsums – sum specific rows in r\n\nThese functions are extremely useful when you’re doing advanced matrix manipulation or implementing a statistical function in R. These form the building blocks of many basic statistical operations and linear algebra procedures. This is why you sometimes see an error message from this cluster of functions show up as part of a higher level package.\n\nIn the event you need them, there are also functions for RowMedians (solves for the median of a row in R) and RowSD (solves for the standard deviation of a row in R). Given the existence of the above, be sure to do a quick search of the various R packages if you need anything more exotic – since it most likely exists…\n\nIf you are looking to solve for rowmeans or rowsums by group, check out the aggregate function (one of the items we addressed in our article about descriptive statistics).\n\nRelated Content:\n\nScroll to top"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.84936684,"math_prob":0.94611645,"size":4287,"snap":"2022-27-2022-33","text_gpt3_token_len":1096,"char_repetition_ratio":0.13238384,"word_repetition_ratio":0.0,"special_character_ratio":0.253324,"punctuation_ratio":0.10135135,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9734618,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T03:19:09Z\",\"WARC-Record-ID\":\"<urn:uuid:c0ab8f61-2700-4235-b831-31f3dbf1af56>\",\"Content-Length\":\"218047\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2af971a1-ff8f-4b29-ae21-e85c5bc997eb>\",\"WARC-Concurrent-To\":\"<urn:uuid:0de81c84-7e79-4e55-bf00-2231ac725db0>\",\"WARC-IP-Address\":\"3.234.104.255\",\"WARC-Target-URI\":\"https://www.programmingr.com/tutorial/colsums-in-r/\",\"WARC-Payload-Digest\":\"sha1:BFCCIITNOGHIUUP7PCTFCNVAYXTQBENB\",\"WARC-Block-Digest\":\"sha1:UX4X43BGTX5IEMYEQE52OJDRB4NUR6IW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104512702.80_warc_CC-MAIN-20220705022909-20220705052909-00427.warc.gz\"}"} |
https://rdrr.io/cran/fda.usc/man/dev.S.html | [
"# dev.S: The deviance score In fda.usc: Functional Data Analysis and Utilities for Statistical Computing\n\n dev.S R Documentation\n\n## The deviance score\n\n### Description\n\nReturns the deviance of a fitted model object by GCV score.\n\n### Usage\n\n```dev.S(\ny,\nS,\nobs,\nfamily = gaussian(),\noff,\noffdf,\ncriteria = \"GCV\",\nW = diag(1, ncol = ncol(S), nrow = nrow(S)),\ntrim = 0,\ndraw = FALSE,\n...\n)\n```\n\n### Arguments\n\n `y` Matrix of set cases with dimension (`n` x `m`), where `n` is the number of curves and `m` are the points observed in each curve. `S` Smoothing matrix. `obs` observed response. `family` a description of the error distribution and link function to be used in the model. This can be a character string naming a family function, a family function or the result of a call to a family function. (See `family` for details of family functions.) `off` off `offdf` off, degrees of freedom `criteria` The penalizing function. By default \"Rice\" criteria. Possible values are \"GCV\", \"AIC\", \"FPE\", \"Shibata\", \"Rice\". `W` Matrix of weights. `trim` The alpha of the trimming. `draw` =TRUE, draw the curves, the sample median and trimmed mean. `...` Further arguments passed to or from other methods.\n\n### Details\n\nUp to a constant, minus twice the maximized log-likelihood. Where sensible, the constant is chosen so that a saturated model has deviance zero.\n\nWhere\n\nand penalty function\n\ncan be selected from the following criteria:\n\nGeneralized Cross-validation (GCV):\n\nAkaike's Information Criterion (AIC):\n\nFinite Prediction Error (FPE)\n\nShibata's model selector (Shibata):\n\nRice's bandwidth selector (Rice):\n\n### Value\n\nReturns GCV score calculated for input parameters.\n\n### Author(s)\n\nManuel Febrero-Bande, Manuel Oviedo de la Fuente [email protected]\n\n### References\n\nWasserman, L. All of Nonparametric Statistics. Springer Texts in Statistics, 2006.\n\nHardle, W. Applied Nonparametric Regression. Cambridge University Press, 1994.\n\nFebrero-Bande, M., Oviedo de la Fuente, M. (2012). Statistical Computing in Functional Data Analysis: The R Package fda.usc. Journal of Statistical Software, 51(4), 1-28. https://www.jstatsoft.org/v51/i04/\n\nSee Also as `GCV.S`.\nAlternative method: `CV.S`\n\n### Examples\n\n```data(phoneme)\nmlearn<-phoneme\\$learn\nnp<-ncol(mlearn)\ntt<-mlearn[[\"argvals\"]]\nS1 <- S.NW(tt,2.5)\ngcv1 <- dev.S(mlearn\\$data[1,],obs=(sample(150)),\nS1,off=rep(1,150),offdf=3)\ngcv2 <- dev.S(mlearn\\$data[1,],obs=sort(sample(150)),\nS1,off=rep(1,150),offdf=3)\n\n```\n\nfda.usc documentation built on Oct. 17, 2022, 9:06 a.m."
]
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null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.57781565,"math_prob":0.8890306,"size":2194,"snap":"2022-40-2023-06","text_gpt3_token_len":636,"char_repetition_ratio":0.08538813,"word_repetition_ratio":0.0,"special_character_ratio":0.27985415,"punctuation_ratio":0.22616407,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97179145,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-03T19:47:58Z\",\"WARC-Record-ID\":\"<urn:uuid:9cb40b0e-4cb4-4eda-8421-07498d85d834>\",\"Content-Length\":\"43710\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:066cbfd3-da7e-4b8b-8a43-90185e946277>\",\"WARC-Concurrent-To\":\"<urn:uuid:abba1e34-b8aa-49f3-90a1-dd223ab5ce3a>\",\"WARC-IP-Address\":\"51.81.83.12\",\"WARC-Target-URI\":\"https://rdrr.io/cran/fda.usc/man/dev.S.html\",\"WARC-Payload-Digest\":\"sha1:WRDKMA3A3MGUV7NCJEQT4W42IBX3QSPS\",\"WARC-Block-Digest\":\"sha1:EDUE6EEYQRCJA6LPVLNNAOSBJSEBJMR2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500074.73_warc_CC-MAIN-20230203185547-20230203215547-00879.warc.gz\"}"} |
https://www.shaalaa.com/question-bank-solutions/solutions-quadratic-equations-factorization-divide-29-two-parts-so-that-sum-squares-parts-425_23317 | [
"Share\n\n# Divide 29 into Two Parts So that the Sum of the Squares of the Parts is 425. - Mathematics\n\nCourse\n\n#### Question\n\nDivide 29 into two parts so that the sum of the squares of the parts is 425.\n\n#### Solution\n\nLet the two parts be ‘x’ and 29 – x\n\n⇒ Given that the sum of the squares of the parts is 425.\n\nThen, by hypothesis, we have\n\n⇒ 𝑥2 + (29 - 𝑥)2 = 425\n\n⇒ 2𝑥2 - 58𝑥 + 841 - 425 = 0\n\n⇒ 2𝑥2 - 58𝑥 + 416 = 0\n\n⇒ 2[𝑥2 - 29𝑥 + 208] = 0\n\n⇒ 𝑥2 - 29𝑥 + 208 = 0\n\n⇒ 𝑥2 - 13𝑥 - 16𝑥 + 208 = 0 [By the method of factorisation]\n\n⇒ 𝑥(𝑥 - 13) - 16(𝑥 - 13) = 0\n\n⇒ (𝑥 - 13)(𝑥 - 16) = 0\n\n⇒ x = 13 or x = 16\n\nCase i: If x = 13; 29 - x = 29 - 13 = 16\n\nCase ii: x = 16; 29 - x = 29 - 16 = 13\n\n∴ The two parts that the sum of the squares of the parts is 425 are 13, 16.\n\nIs there an error in this question or solution?"
]
| [
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.76524156,"math_prob":0.9998822,"size":848,"snap":"2020-34-2020-40","text_gpt3_token_len":390,"char_repetition_ratio":0.1635071,"word_repetition_ratio":0.13397129,"special_character_ratio":0.45990565,"punctuation_ratio":0.07734807,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99972576,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-14T01:19:16Z\",\"WARC-Record-ID\":\"<urn:uuid:71858b75-79b5-4e95-973e-48c1e3d87cfd>\",\"Content-Length\":\"46839\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cbf46856-4ccb-4362-b41e-98641d1b1149>\",\"WARC-Concurrent-To\":\"<urn:uuid:595a72bf-1765-4d8a-8308-4eaa1879c807>\",\"WARC-IP-Address\":\"172.105.37.75\",\"WARC-Target-URI\":\"https://www.shaalaa.com/question-bank-solutions/solutions-quadratic-equations-factorization-divide-29-two-parts-so-that-sum-squares-parts-425_23317\",\"WARC-Payload-Digest\":\"sha1:NT5BADPRM4U5OLOBZTO622X6EEVHSFH6\",\"WARC-Block-Digest\":\"sha1:AAYBKJHQYU4V7CCVAQCQFAUUHJTL62A5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439739134.49_warc_CC-MAIN-20200814011517-20200814041517-00268.warc.gz\"}"} |
https://rdrr.io/github/rsquaredacademy/inferr/src/tests/testthat/test-levene.R | [
"# tests/testthat/test-levene.R In rsquaredacademy/inferr: Inferential Statistics\n\n```context(\"levene-test\")\n\ntest_that(\"output from infer_levene_test matches the expected result\", {\nmt <- mtcars\nmt\\$cyl <- as.factor(mt\\$cyl)\n\nk <- infer_levene_test(mt, mpg, group_var = cyl)\nexpect_equal(k\\$bf, 6.4843)\nexpect_equal(k\\$p_bf, 0.0047)\nexpect_equal(k\\$lev, 5.5071)\nexpect_equal(k\\$p_lev, 0.0094)\nexpect_equal(k\\$bft, 6.2047)\nexpect_equal(k\\$p_bft, 0.0057)\nexpect_equivalent(as.vector(k\\$avgs), c(26.66, 19.74, 15.10))\nexpect_equivalent(as.vector(k\\$sds), c(4.51, 1.45, 2.56))\nexpect_equal(k\\$avg, 20.09)\nexpect_equal(k\\$sd, 6.03)\nexpect_equal(k\\$n, 32)\nexpect_equivalent(k\\$levs, c(\"4\", \"6\", \"8\"))\nexpect_equal(k\\$n_df, 2)\nexpect_equal(k\\$d_df, 29)\nexpect_equivalent(as.vector(k\\$lens), c(11, 7, 14))\n})\n\ntest_that(\"output from infer_levene_test matches the expected result\", {\nk <- infer_levene_test(mtcars, mpg, qsec)\nexpect_equal(k\\$bf, 24.3932)\nexpect_equal(k\\$p_bf, 0)\nexpect_equal(k\\$lev, 20.9464)\nexpect_equal(k\\$p_lev, 0)\nexpect_equal(k\\$bft, 22.7064)\nexpect_equal(k\\$p_bft, 0)\nexpect_equivalent(as.vector(k\\$avgs), c(20.09, 17.85))\nexpect_equivalent(as.vector(k\\$sds), c(6.03, 1.79))\nexpect_equal(k\\$avg, 18.97)\nexpect_equal(k\\$sd, 4.55)\nexpect_equal(k\\$n, 64)\nexpect_equivalent(k\\$levs, c(\"0\", \"1\"))\nexpect_equal(k\\$n_df, 1)\nexpect_equal(k\\$d_df, 62)\nexpect_equivalent(as.vector(k\\$lens), c(32, 32))\n})\n\ntest_that(\"output from levene test is as expected\", {\nx <- cat(\" Summary Statistics\nLevels Frequency Mean Std. Dev\n-----------------------------------------\n1 24 46.67 10.24\n2 11 51.91 7.66\n3 20 46.8 7.12\n4 145 53.92 10.28\n-----------------------------------------\nTotal 200 52.23 10.25\n-----------------------------------------\n\nTest Statistics\n-------------------------------------------------------------------------\nStatistic Num DF Den DF F Pr > F\n-------------------------------------------------------------------------\nBrown and Forsythe 3 196 3.44 0.0179\nLevene 3 196 3.4792 0.017\nBrown and Forsythe (Trimmed Mean) 3 196 3.3936 0.019\n-------------------------------------------------------------------------\")\n\nexpect_equivalent(print(infer_levene_test(hsb, read, group_var = race)), x)\n})\n```\nrsquaredacademy/inferr documentation built on Aug. 28, 2019, 8:08 a.m."
]
| [
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5173376,"math_prob":0.99542725,"size":2203,"snap":"2020-10-2020-16","text_gpt3_token_len":704,"char_repetition_ratio":0.36925876,"word_repetition_ratio":0.03883495,"special_character_ratio":0.50295055,"punctuation_ratio":0.24,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998568,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-05T20:09:27Z\",\"WARC-Record-ID\":\"<urn:uuid:173edd86-ee14-4224-a215-e289a90714e3>\",\"Content-Length\":\"62255\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0349d529-cbaa-49f8-a7b0-1c01d4acc837>\",\"WARC-Concurrent-To\":\"<urn:uuid:05dd2c71-6682-4803-9e64-d7f88d2cecc2>\",\"WARC-IP-Address\":\"104.28.6.171\",\"WARC-Target-URI\":\"https://rdrr.io/github/rsquaredacademy/inferr/src/tests/testthat/test-levene.R\",\"WARC-Payload-Digest\":\"sha1:WVCTTYN7XAO5ABUGAME724AZJKSNQLKA\",\"WARC-Block-Digest\":\"sha1:KR64AM4IJCWKXLMSAY5RSRZA6GZBSLP2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371609067.62_warc_CC-MAIN-20200405181743-20200405212243-00325.warc.gz\"}"} |
https://tex.stackexchange.com/questions/541193/commutative-diagram-in-standalone-using-tikzcd | [
"# Commutative Diagram in Standalone using TikZcd\n\nI'm trying to draw a diagram of a function that is Injective but not Surjective. I'm using standalone, and want the end result to look like the this image:",
null,
"My code is modified from https://tex.stackexchange.com/a/167913/197489 without much luck. Here is my LaTeX\n\n\\documentclass[crop,tikz]{standalone}\n\\usepackage{tikz-cd}\n\\usetikzlibrary{calc}\n\\newcommand{\\tikzmark}{\\tikz[overlay,remember picture] \\node (#1) {};}\n\\newcommand{\\DrawBox}[]{%\n\\tikz[overlay,remember picture]{\n\\draw[black,#1]\n($(#2)+(-0.5em,2.0ex)$) rectangle\n($(#3)+(0.75em,-0.75ex)$);}\n}\n\n%Draw an arrow diagram that represents a function that is an injection but is not a surjection.\n\n\\begin{document}\n\\begin{tikzcd}[scale=.1]\n\\tikzmark{Atop} a \\arrow[r] & \\tikzmark{Btop}x \\\\\nb \\arrow[r] & y \\\\\n\\tikzmark{Abottom}& z\\tikzmark{Bbottom} \\\\\n\\DrawBox[thick, red]{Atop}{Abottom}\n\\DrawBox[thick, blue]{Btop}{Bbottom}\n\\end{tikzcd}\n\\end{document}\n\n\nWhich results in",
null,
"Any suggestions?\n\nI'm using MiKTeX in TeXStudio if that helps.\n\n• This nests tikzpictures, which causes the problems.\n– user194703\nApr 28, 2020 at 18:37\n• @Schrödinger'scat My bad, I edited the question and I still have a similar problem Apr 28, 2020 at 18:50\n• Your (handmade) \\tikzmark command starts with \\tikz which means that it creates a tikzpicture.\n– user194703\nApr 28, 2020 at 18:52\n\nYour approach nests tikzpictures. This is because \\tikz starts a new tikzpicture and you are using it inside tikzcd, which is a tikzpicture on its own.\n\nTo draw something that resembles the screen shot you post, tikz-cd may not even the best choice. You can just use, for example, this code:\n\n\\documentclass[crop,tikz]{standalone}\n\\usetikzlibrary{matrix,fit}\n\\begin{document}\n\\begin{tikzpicture}\n\\matrix[matrix of nodes,nodes={circle,fill,inner sep=2pt},\ncolumn sep=4em,row sep=1em,inner sep=0pt] (mat)\n{\n|[label=above left:$a$]| {} & |[label=above right:$x$]| {}\\\\\n|[label=above left:$b$]| {} & |[label=above right:$y$]| {}\\\\\n& |[label=above right:$z$]| {}\\\\\n};\n\\node[blue,draw,very thick,fit={(mat.north west) ([xshift=1ex]mat-1-1.east|-mat.south)},\ninner sep=1ex,label={[blue]above:$S$}]{};\n\\node[red,draw,very thick,fit={(mat.north east) ([xshift=-1ex]mat-1-2.west|-mat.south)},\ninner sep=1ex,label={[red]above:$T$}]{};\n\\foreach \\X in {1,2}\n{\\draw[-stealth] (mat-\\X-1) -- (mat-\\X-2);}\n\\end{tikzpicture}\n\\end{document}",
null,
"Notice that you can safely draw something in an tikzcd environment by using the execute at end picture hook.\n\n\\documentclass[crop,tikz]{standalone}\n\\usepackage{tikz-cd}\n\\usetikzlibrary{fit}\n\\begin{document}\n\\begin{tikzcd}[column sep=4em,\nexecute at end picture={%\n\\node[blue,draw,very thick,fit={(\\tikzcdmatrixname.north west) ([xshift=1ex]\\tikzcdmatrixname-1-1.east|-\\tikzcdmatrixname.south)},\ninner sep=1ex,label={[blue]above:$S$}]{};\n\\node[red,draw,very thick,fit={(\\tikzcdmatrixname.north east) ([xshift=-1ex]\\tikzcdmatrixname-1-2.west|-\\tikzcdmatrixname.south)},\ninner sep=1ex,label={[red]above:$T$}]{};\n}]\na \\arrow[r] & x \\\\\nb \\arrow[r] & y \\\\\n& z\n\\end{tikzcd}\n\\end{document}",
null,
""
]
| [
null,
"https://i.stack.imgur.com/Cr9d0m.jpg",
null,
"https://i.stack.imgur.com/WkpZom.png",
null,
"https://i.stack.imgur.com/Mrr8r.png",
null,
"https://i.stack.imgur.com/GFKWW.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.63142365,"math_prob":0.97340906,"size":1107,"snap":"2023-14-2023-23","text_gpt3_token_len":368,"char_repetition_ratio":0.1087942,"word_repetition_ratio":0.0,"special_character_ratio":0.28816622,"punctuation_ratio":0.14213198,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99595517,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-30T13:49:37Z\",\"WARC-Record-ID\":\"<urn:uuid:9afa157d-4f0b-459b-aa34-329487a5a0b2>\",\"Content-Length\":\"132169\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c38517c9-d328-41a2-958a-a45aa6e77c8a>\",\"WARC-Concurrent-To\":\"<urn:uuid:3fa69a34-b74c-4db4-b781-780a9fd7371a>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://tex.stackexchange.com/questions/541193/commutative-diagram-in-standalone-using-tikzcd\",\"WARC-Payload-Digest\":\"sha1:HIXPXLSIL74RRDSLHFK5AZELNMGCFRA5\",\"WARC-Block-Digest\":\"sha1:N3RPV7WVYSYU4A76TX3KXKUN2Q534QV5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949331.26_warc_CC-MAIN-20230330132508-20230330162508-00071.warc.gz\"}"} |
https://forum.opencv.org/t/place-an-image-over-another-image-using-opencv-js/5389 | [
"# Place an image over another image using OpenCV.js\n\nHello I have been struggling with OpenCV.js for a while now. Thanks to @berak I am closer to the solution now. I am trying to build a Javascript library using OpenCV.js which takes an image as input and provides an output image in which the human skin is blurred.\n\nRight now my script looks like this\n\n``````let src = cv.imread(imgElement);\nlet dst = new cv.Mat();\nlet dst_1 = new cv.Mat();\n\n// converting from gbr to hsv color space\ncv.cvtColor(src, dst, cv.COLOR_RGB2HSV)\ncv.cvtColor(dst, dst, cv.COLOR_RGB2BGR)\n\nlet hsv_low = new cv.Mat(dst.rows, dst.cols, dst.type(), [60, 25, 0, 0]);\nlet hsv_high = new cv.Mat(dst.rows, dst.cols, dst.type(), [255, 170, 20, 0]);\n\n// skin color range for hsv color space\ncv.inRange(dst, hsv_low, hsv_high, dst);\n\n// converting from gbr to YCbCr color space\ncv.cvtColor(src, dst_1, cv.COLOR_RGB2YCrCb);\ncv.cvtColor(dst_1, dst_1, cv.COLOR_RGB2BGR);\n\nlet YCbCr_low = new cv.Mat(dst_1.rows, dst_1.cols, dst_1.type(), [90, 135, 10, 0]);\nlet YCbCr_high = new cv.Mat(dst_1.rows, dst_1.cols, dst_1.type(), [125, 175, 235, 0]);\n\n// skin color range for YCbCr color space\ncv.inRange(dst_1, YCbCr_low, YCbCr_high, dst_1);\n\n// merge skin detection (YCbCr and hsv)\ncv.bitwise_and(dst_1, dst, dst)\ncv.medianBlur(dst, dst, 3)\n\nlet ksize = new cv.Size(80, 80);\nlet anchor = new cv.Point(-1, -1);\n\n// make blurred copy of original image\ncv.blur(src, dst_1, ksize, anchor, cv.BORDER_DEFAULT);\n\n// mask blurred image to create image with only skin blurred, rest is opaque\ndst_1.copyTo(dst, dst)\n\ncv.imshow('canvasOutput', src);\n\nsrc.delete();\ndst.delete();\ndst_1.delete();\nhsv_low.delete();\nhsv_high.delete();\nYCbCr_low.delete();\nYCbCr_high.delete();\n``````\n\nI have successfully been able to create an image where skin part of the image is blurry and the rest of the image is transparent. Now the last step is to place this over the original image.",
null,
"I have been trying to implement Bitwise Operations section of this doc, but I haven’t been successful. Any help will be appreciated.\n\nThank you.\n\nThe desired output is this",
null,
"After trying to solve this problem for 2 days, I was able to solve this. The problem was solved by replacing the pixels of original image with the desired pixels from second image. This is the code\n\n``````for (let i = 0; i < mask.rows; i++) {\nfor (let j = 0; j < mask.cols; j++) {\nif (mask.ucharPtr(i, j) === 255) {\nsrc.ucharPtr(i, j) = dst.ucharPtr(i, j)\nsrc.ucharPtr(i, j) = dst.ucharPtr(i, j)\nsrc.ucharPtr(i, j) = dst.ucharPtr(i, j)\nsrc.ucharPtr(i, j) = dst.ucharPtr(i, j)\n}\n}\n}\n``````\n\nHere, I have\n\n• `src`: Mat of original image\n• `dst`: Mat of image with skin part blurred and all other pixels transparent\n• `mask`: Original image’s mask with skin in white and other parts in black\n\nHere is the result",
null,
"this is exactly, what `dst.setTo(src,mask);` does ;]\n(just much slower)"
]
| [
null,
"https://aws1.discourse-cdn.com/standard11/uploads/opencv/original/2X/7/746c774bdca291c8a5774105678db30eabf9f9c4.png",
null,
"https://aws1.discourse-cdn.com/standard11/uploads/opencv/original/2X/5/5381fa0002089baa5d046cc6f450d84cea9026bf.png",
null,
"https://aws1.discourse-cdn.com/standard11/uploads/opencv/original/2X/8/83eb3881708683fc1eec8e9f357c7f06d32e7691.jpeg",
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]
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http://people.duke.edu/~charvey/Classes/ba350_1997/assignments/assign2.htm | [
"## Learning Module 2\n\nObjective\n\nThe objective of this learning module is to establish expertise in the valuation of different types of fixed income instruments. The first few problems reinforce your knowledge of discounting cash flows. Embedded in these questions is the concept of risk identification.\n\nIn question 1, we determine the impact of a surprise in the interest rates.\n\nIn question 2, we calculate the modified duration and elasticity of the bond in question. This tells us the percentage change of the price of the bond for a one percent change in the interest rates. Elasticity is best thought of as an approximation to the modified duration. Note that there is a question included in this section which is intended for discussion on the main bulletin board.\n\nQuestion 3 focuses on mortgage valuation. Mortgages differ from corporate or government bonds in how principle is repaid. Recall that bonds pay interest only until maturity, at which time the last interest payment and all of the principle is paid. With mortgages, the same payment is made each period. Part of the payment is for interest. The rest pays off some of the principle. This question involves a fixed rate mortgage. A discussion question for the main bulletin board is included which involves the option that the homeowner has to refinance his mortgage if rates go down.\n\nThe final question deals with forward rates. A forward rate is a rate that we can lock in at a later date. The term structure of interest rates has information about forward rates imbedded in it. For example, if the one year rate is 3% and the two year rate is 5%, then we can calculate the forward one year rate for one year away. This forward rate is just the rate that will guarantee us a 5% return if we invest for a year in the one year bond, and then for another year at the forward rate. So, the following relationship will hold:\n\n1.052 = 1.03 (1+1f2)\n\nThe subscripts mean \"a one period forward rate at period 2.\" Solving for the forward rate, we see that the one year rate will have to be about 7.04% in a year in order to guarantee a 5% return per year for a two year investment. In this case, we have an upward sloping term structure. A lot of people have studied the information in the term structure regarding future interest rates. We talked about the information regarding future real growth in the economy. The expectations theory of the term structure which you will study in Macroeconomics says that the forward rate contains some information about future interest rates.\n\nA discussion question is included.\n\n1. Bond Valuation\n\na) You have a \\$1000 par 6% coupon (nominal rate) US Treasury bond with 7 years remaining in its life. Coupons are paid semiannually and the next coupon payment is exactly six months away. The market interest rate is 6.4% (nominal rate with semiannual compounding). What is the current price of this bond?\n\nb) What is the effective annual rate which corresponds to a nominal interest rate of 6.4% with semi-annual compounding?\n\nc) Price a zero coupon bond with a face amount of \\$1000 maturing in 7 years. Assume that the nominal interest rate is 6.4% and interest is compounded semiannually.\n\nd) Assume now that interest rates have instantaneously increased by 1% to 7.4%. What are the bonds in parts (a) and (c) now worth?\n\n2. Duration\n\nConsider a newly issued \\$100 par bond with a 10 year maturity and a 7.5% coupon. The first coupon payment is exactly six months away. Coupons are paid semi-annually. The nominal interest rate is 8.5% with semi-annual compounding. Note that the bond price would be equal to par if the nominal market interest rate was 7.5%.\n\na) Calculate the Macaulay duration of this bond.\n\nb) Calculate the modified duration of this bond.\n\nc) Manually calculate the elasticity of this bond (Hint: Consider a move in rates from 8.5% to 7.5% and from 8.5% to 9.5% and then average the percentage changes).\n\nd) Compare the actual bond price change in percentage terms as rates move from 8.5% to 6.5% to the percent price change predicted by your modified duration calculation.\n\nMain Bulletin Board Discussion Question:\n\nIn question 2d, the predicted price change misses the actual price change. Why?\n\n3. Mortgage Amortization\n\na) You want to take out a 15 year fixed rate mortgage on a \\$200,000 house. Current mortgage rates are 9.5% (nominal rate, compounded monthly). Calculate your monthly payment on this mortgage.\n\nb) Using a spreadsheet, prepare an amortization schedule for this mortgage showing principal outstanding and the portion of each month’s payment towards interest and principal.\n\nc) After exactly one year (i.e., immediately after the 12th payment), you use your \\$75,000 annual bonus to reduce the principal outstanding on your mortgage. You continue making the same monthly payment that was calculated in part (a). When will the loan be fully paid off?\n\nMain Bulletin Board Discussion Question:\n\nIf rates fall, homeowners often renegotiate their mortgage. What kind of option is this? Who writes the option? Who buys it?\n\n4. Forward Interest Rates\n\na) Suppose that the current one year interest rate is 5.3% per annum. Also assume that the 1 year forward interest rate is 5.5% (1f2). This forward rate means that you are able to commit to investing \\$x one year from now and be certain of receiving \\$x(1+1f2) two years from now. How much money will you have in two years if you invest \\$100 in the current one year rate (the spot rate) and commit to investing the proceeds of the one year investment at the one year forward rate? Assume that interest is calculated annually.\n\nb) Assume that investing \\$100 at the current 2 year interest rate will leave you with the same amount of money that you calculated in part (a) at the end of two years. What is the current 2 year rate?\n\nMain Bulletin Board Discussion Question:\n\nIn the above example, are interest rates expected to rise or fall? Why?"
]
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https://info.teledyne-hi.com/blog/how-do-i-use-gcf-gas-conversion-factors-with-my-mass-flow-meter-or-mass-flow-controller | [
"",
null,
"# Mon, Nov 02, 2020 @ 09:54 AM | Flow MeterHow do I use GCF (Gas Conversion Factors) with my mass flow meter or mass flow controller?\n\nIn this blog, we will show how to use GCFs (Gas Conversion Factors) when using flow instruments in different gases.\n\nUsing thermal mass flow instruments by Teledyne Hastings is an easy way to quickly and accurately measure gas flow. And in some cases, a mass flow instrument may be calibrated for one gas, but then the user may want to use the instrument in another gas. In this blog, we will show how to use GCFs (Gas Conversion Factors) when using flow instruments in different gases.\n\nBefore we get into GCFs, let’s quickly review the operation of one of our flow sensors. Below, we show a diagram of the 200 Series flow sensor. In this sensor, gas flows through a capillary tube which is heated in the middle to a temperature which is approximately 130°C. Two thermocouples, one upstream (TC-1) and one downstream (TC-2), measure the temperature. The temperature difference between the two thermocouples is proportional to the heat flow through the capillary tube. The heat flow, in turn, is proportional to the mass flow times the specific heat Cp of the gas. So, to first order, if we want to use a thermal mass flow meter that has been set up for one gas, and use it with another gas, we will multiply the output of the meter by the ratio of the specific heats. GCF ~ Cp1 / Cp2",
null,
"There are a couple of things we need to point out. First, the ratio shown above is a simple approximation and does not tell the whole story. Next, the best GCFs are those that have been measured experimentally. However, in the case of dangerous gases, we use the best thermodynamic data available.\n\nHere is a table of some common GCFs.\n\n Gas Conversion Factors (N2) 200 Series 300 Series Helium 1.402 1.400 Oxygen 0.981 0.978 Carbon Dioxide 0.743 0.753 Carbon Monoxide 1.001 1.001 Methane 0.770 0.779 Ammonia 0.781 0.781 Hydrogen 1.009 1.004 Argon 1.401 1.405\n\nNext, we will discuss how we apply GCFs in practice. Let’s take an example of a flow meter that is calibrated for nitrogen. If we wanted to use the flowmeter in argon, we would take the output and multiply by the GCF for Argon.",
null,
"Here is another example; suppose we have a meter that is calibrated in helium and we want to use it in hydrogen. You would start by dividing the output by the GCF for helium (think of it as converting to the nitrogen equivalent), and then multiplying by the GCF for hydrogen.",
null,
"Remember, always use the appropriate set of GCFs for the flow series that you are using. In other words, if you are using our Digital 300 Series, don’t apply GCFs from a 200 Series manual – they are not the same. And certainly don’t use non-Teledyne table of GCFs for use with Teledyne flow products. They might get you in the ballpark, but they will not be your best conversion.\n\nOne other quick note about applying GCFs. Our line of flow power supplies, the THCD-101 (single channel) and the THCD-401 (four channel), can be used to quickly scale the analog input which is equivalent to applying a conversion factor. Let’s take another look at the Argon example. If we used the THCD-101 power supply with the nitrogen flow meter as shown below, at the nominal full scale of the flow meter, we will have a 5 VDC signal. If we want to use this same meter and power supply with Argon, we just need to “tell” the THCD-101 what value to display when it receives 5 VDC. So, if our flow meter was calibrated for nitrogen to give 5 VDC at 250 sccm, then the same flow meter will give 5 VDC in argon at 350 sccm. (250 * 1.4 = 350). So, we would then range the THCD-101 for 350 sccm. This can be done from the front panel or via the internal webserver.",
null,
"Now let’s make things a little more interesting and discuss a flow controller example. Analog flow controllers work by receiving a command signal (usually 0-5 VDC, or 4-20 mA) and then they adjust their control valve such that the flow, and thus the analog signal output, matches the command signal input. (You can think of it like the cruise control in your car – you tell it you want to go 78 miles per hour, and then the engine does what it needs to do to maintain that speed). In the case of a 0-5 VDC flow controller, a 5-volt setpoint command is instructing the flow controller to set the flow to 100% of full scale. The relationship between flow rate and command signal is linear, so if the user wanted to control at 25% of full scale, then they would send a 1.25 VDC command signal (0.25 * 5 VDC = 1.25 VDC).",
null,
"Now, suppose we had an HFC-202 flow controller (200 Series) that was calibrated for 200 sccm of methane and we wanted to use it to control the flow of argon. What voltage level would we need on the command signal to have a flow rate of 100 sccm of argon? Let’s first determine the full-scale flow rate (5 VDC) when using argon:\n\nFlow (Ar) = Flow (CH4)/GCF (CH4) * GCF (Ar) = (200 sccm / 0.77) * 1.401 = 363.9\n\nSo, a 5 VDC command signal will give us 363.9 sccm of argon. If we want 100 sccm, we would send:\n\nCommand Voltage = 100 sccm (5 VDC / 363.9 sccm) = 1.374 VDC.\n\nNow, one important note about using flow controllers in different gases. Just because we can apply GCFs does not mean that a flow controller’s valve will work properly when switching from one gas to another. As an extreme example, a flow controller valve that has an orifice sized to handle hydrogen will have a hard time handling significant flows of large polyatomic molecules like C2H6.\n\nTeledyne flow products are easy to install and use. And our application engineers are standing by to help. We can be reached by email ([email protected]), by phone 757-723-6531, or via LiveChat on our website www.teledyne-hi.com or by clicking the contact us button below.",
null,
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"https://info.teledyne-hi.com/blog/how-do-i-use-gcf-gas-conversion-factors-with-my-mass-flow-meter-or-mass-flow-controller",
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"https://info.teledyne-hi.com/hs-fs/hubfs/Blog_Images/200%20Series%20Sensor.jpg",
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"https://info.teledyne-hi.com/hs-fs/hubfs/Blog_Images/Argon%20GCF.jpg",
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"https://info.teledyne-hi.com/hs-fs/hubfs/Blog_Images/H2%20He%20GCF.jpg",
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"https://info.teledyne-hi.com/hs-fs/hubfs/Blog_Images/HFM200%20with%20THCD.jpg",
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"https://info.teledyne-hi.com/hs-fs/hubfs/Blog_Images/HFC%20with%20THCD.jpg",
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https://dynref.engr.illinois.edu/rem.html | [
"# Moments of inertia\n\nThe moment of inertia of a body, written $I_{P,\\hat{a}}$, is measured about a rotation axis through point $P$ in direction $\\hat{a}$. The moment of inertia expresses how hard it is to produce an angular acceleration of the body about this axis. That is, a body with high moment of inertia resists angular acceleration, so if it is not rotating then it is hard to start a rotation, while if it is already rotating then it is hard to stop.\n\nThe moment of inertia plays the same role for rotational motion as the mass does for translational motion (a high-mass body resists is hard to start moving and hard to stop again).\n\nMoment of inertia about axis $\\hat{a}$ through point $P$.\n\n $\\begin{gathered} I_{P,\\hat{a}} = \\iiint_\\mathcal{B} \\rho r^2 \\,dV \\\\ \\text{(units: \\rm kg\\ m^2)} \\end{gathered}$\n\nThe distance $r$ is the perpendicular distance to $dV$ from the axis through $P$ in direction $\\hat{a}$.\n\nWarning: Mass moments of inertia are different to area moments of inertia.\n\nBe advised that the \"moment of inertia\" encountered in Statics is not the same as the moment of inertia used in Dynamics. Strictly speaking, the \"moment of inertia\" from Statics shouldn't even be called \"moment of inertia.\" What it really is is the \"second moment of area.\" Below are the definitions of two such second moments of area:\n\n$J_{xx}=\\iint_{A}{y^{2}dA}$\n\n$J_{yy}=\\iint_{A}{x^{2}dA}$\n\nIn contrast, the moment of inertia (about the $z$-axis) is defined as stated above.\n\nFor example, a rectangle of base $b$ and height $h$ has the following moments about its centroid $C$:\n\n$J^{C}_{xx}=\\frac{1}{12}bh^{3}$\n\n$J^{C}_{yy}=\\frac{1}{12}b^{3}h$\n\n$I^{C}_{zz}=\\frac{1}{12}m(b^{2}+h^{2})$\n\nNotice that the dimensions of the two quantities are different. While the dimension of second moment of area is $(\\text{length})^{4}$, the dimension of moment of inertia is $(\\text{mass})(\\text{length})^{2}$. When doing dynamics problems with moments of inertia, you should not use the formulas you remember for second moment of area instead. You will get the wrong answer!\n\nObserve that the moment of inertia is proportional to the mass, so that doubling the mass of an object will also double its moment of inertia. In addition, the moment of inertia is proportional to the square of the size of the object, so that doubling every dimension of an object (height, width, etc) will cause it to have four times the moment of inertia.\n\nMoments of inertia about coordinate axes through point $P$.\n\n\\begin{aligned} I_{P,x} &= I_{P,xx} = I_{P,\\hat\\imath} = \\iiint_\\mathcal{B} \\rho (y^2 + z^2) \\,dx\\,dy\\,dz \\\\ I_{P,y} &= I_{P,yy} = I_{P,\\hat\\jmath} = \\iiint_\\mathcal{B} \\rho (z^2 + x^2) \\,dx\\,dy\\,dz \\\\ I_{P,z} &= I_{P,zz} = I_{P,\\hat{k}} = \\iiint_\\mathcal{B} \\rho (x^2 + y^2) \\,dx\\,dy\\,dz \\end{aligned}\n\nThe coordinates $(x,y,z)$ in the body are measured from point $P$.\n\nThe distance $r$ in the moment of inertia integral #rem-ei is the perpendicular distance from the axis of rotation to the infinitesimal volume $dV$. If we are using rectangular coordinates $(x,y,z)$ measured from point $P$, and the axis of rotation is one of the coordinate axes, then the perpendicular distance is very simple.\n\nAs an example, consider the case when the axis of rotation $\\hat{a}$ is the $\\hat{k}$ axis, as shown in the figure. Then the perpendicular distance $r$ satisfies $r^2 = x^2 + y^2$, giving the coordinate expression above. The other two coordinate axes are similar.\n\nExample Problem: Moment of inertia of a square plate.\n\nA solid square uniform plate has mass $m$, side-length $\\ell$, and thickness $h$. What is the moment of inertia about the $z$-axis through the center of mass $C$?\n\nWe will use the coordinate formula #rem-ec. To do this, we measure the position from the point $C$ about which we are computing the moment of inertia, so the coordinate origin is at $C$. The figure to the right shows the 3D configuration of the body, where we see that the $x$-coordinate range of the body is $-\\ell/2$ to $\\ell/2$, and the same for the $y$-coordinate, while the $z$-coordinate ranges from $-h/2$ to $h/2$.\n\nTaking the density of the plate to be $\\rho$, we thus have: \\begin{aligned} I_{C,z} &= \\iiint_\\mathcal{B} \\rho (x^2 + y^2) \\,dx\\,dy\\,dz \\\\ &= \\int_{-h/2}^{h/2} \\int_{-\\ell/2}^{\\ell/2} \\int_{-\\ell/2}^{\\ell/2} \\rho (x^2 + y^2) \\,dx\\,dy\\,dz \\\\ &= \\int_{-h/2}^{h/2} \\int_{-\\ell/2}^{\\ell/2} \\left[ \\rho \\left(\\frac{1}{3}x^3 + y^2 x\\right)\\right]_{x = -\\ell/2}^{x = \\ell/2} \\,dy\\,dz \\\\ &= \\int_{-h/2}^{h/2} \\int_{-\\ell/2}^{\\ell/2} \\rho \\left(\\frac{1}{12}\\ell^3 + y^2 \\ell \\right) \\,dy\\,dz \\\\ &= \\int_{-h/2}^{h/2} \\left[ \\rho \\left(\\frac{1}{12}\\ell^3 y + \\frac{1}{3} y^3 \\ell \\right) \\right]_{y = -\\ell/2}^{y = \\ell/2} \\,dz \\\\ &= \\int_{-h/2}^{h/2} \\rho \\left(\\frac{1}{12}\\ell^4 + \\frac{1}{12} \\ell^4 \\right) \\,dz \\\\ &= \\int_{-h/2}^{h/2} \\rho \\frac{1}{6}\\ell^4 \\,dz \\\\ &= \\left[ \\rho \\frac{1}{6}\\ell^4 z \\right]_{z = -h/2}^{z = h/2} \\\\ &= \\rho \\frac{1}{6}\\ell^4 h. \\end{aligned} The total mass of the plate is $m = \\rho \\ell^2 h$, so we can write the moment of inertia as $I_{C,z} = \\frac{1}{6} m \\ell^2.$ The square plate moment of inertia is actually a special case of the rectangular prism formula #rem-er with $\\ell_y = \\ell_z = \\ell$.\n\nWe are always considering the moment of inertia to be a scalar value $I$, which is valid for rotation about a fixed axis. For more complicated dynamics with tumbling motion about multiple axes simultaneously, it is necessary to consider the full 3 × 3 moment of inertia matrix: $I = \\begin{bmatrix} I_{xx} & I_{xy} & I_{xz} \\\\ I_{yx} & I_{yy} & I_{yz} \\\\ I_{zx} & I_{zy} & I_{zz} \\end{bmatrix}$ The three scalar moments of inertia from #rem-ec appear on the diagonal. The off-diagonal terms are called the products of inertia and are given by $I_{xy} = -\\iiint_{\\mathcal{B}} \\rho xy \\,dx\\,dy\\,dz,$ and similarly for the other terms. The angular momentum of a rigid body is given by $\\vec{H} = I \\vec\\omega$, which is the matrix product of the moment of inertia matrix with the angular velocity vector. This is important in advanced dynamics applications such as unbalanced rotating shafts and the precession of gyroscopes.\n\n## Transforming and combining moments of inertia\n\nParallel axis theorem.\n\n$I_{P,\\hat{a}} = I_{C,\\hat{a}} + m \\, d^2$\n\nHere $d$ is the perpendicular distance between the axes through $P$ and $C$ in direction $\\hat{a}$, so $d = \\| \\operatorname*{Comp}(\\vec{r}_{CP}, \\hat{a})\\|$.\n\nTo easily compute the moments of inertia relative to axes through $P$ and $C$ it is easiest to choose a coordinate system aligned with the rotation axis direction $\\hat{a}$. We will choose coordinates $(x,y,z)$ measured from the center of mass $C$ and with the $z$-axis $\\hat{k}$ in the direction of $\\hat{a}$, as shown in the figure.\n\nAs we integrate over the body with infinitesimal volume $dV$ at position $(x,y,z)$ from $C$, we will write the distance from the axis through $C$ as $r_c$ and the distance from the axis through $P$ as $r_P$, as illustrated. The distance between the two rotation axes is $d$. In the chosen coordinate system this means that: \\begin{aligned} {r_C}^2 &= x^2 + y^2 \\\\ {r_P}^2 &= (x - x_P)^2 + (y - y_P)^2 \\\\ d^2 &= {x_P}^2 + {y_P}^2, \\end{aligned} where $P$ has position $(x_P, y_P, z_P)$ in these coordinates.\n\nComputing the integral #rem-ec to find the moment of inertia about the axis through $P$ in direction $\\hat{a}$ now gives: \\begin{aligned} I_{P,\\hat{a}} &= \\int_{\\mathcal{B}} \\rho {r_P}^2 \\,dV \\\\ &= \\iiint_{\\mathcal{B}} \\rho \\left((x - x_P)^2 + (y - y_P)^2\\right) \\,dx\\,dy\\,dz \\\\ &= \\iiint_{\\mathcal{B}} \\rho \\left(x^2 - 2 x x_P + {x_P}^2 + y^2 - 2 y y_P + {y_P}^2\\right) \\,dx\\,dy\\,dz \\\\ &= \\iiint_{\\mathcal{B}} \\rho (x^2 + y^2) \\,dx\\,dy\\,dz - 2 x_P \\iiint_{\\mathcal{B}} \\rho x \\,dx\\,dy\\,dz \\\\ &\\quad - 2 y_P \\iiint_{\\mathcal{B}} \\rho y \\,dx\\,dy\\,dz + ({x_P}^2 + {y_P}^2) \\iiint_{\\mathcal{B}} \\rho \\,dx\\,dy\\,dz \\\\ &= \\int_{\\mathcal{B}} \\rho {r_C}^2 \\,dV + d^2 \\int_{\\mathcal{B}} \\rho \\,dV \\\\ &= I_{C,\\hat{a}} + m d^2. \\end{aligned} Here we used the coordinate representation of the center of mass to realize that the $x$ coordinate of $C$ is $x_C = \\frac{1}{m} \\int_{\\mathcal{B}} \\rho x \\,dV$, but because our coordinates are measured from $C$ we must have $x_C = 0$ and so $\\int_{\\mathcal{B}} \\rho x \\,dV = 0$. The integral of $\\rho y$ is similarly zero.\n\nWarning: Parallel axis theorem must start from center of mass $C$.\n\nThe parallel axis theorem does not apply to any two parallel rotation axes. The rotation axis on the right-hand-side of the equation must be through the center of mass $C$, although the other axis can be through any point $P$.\n\nWarning: Parallel axis theorem $d$ is perpendicular distance, not $r_{CP}$.\n\nThe distance $d$ in the parallel axis theorem is the perpendicular distance between the axes through points $P$ and $C$. This will normally not be the same as the distance $r_{CP}$ between $P$ and $C$, except in 2D problems or if it happens that $\\vec{r}_{CP}$ is already perpendicular to the rotation axis $\\hat{a}$. The distance $d$ can be computed by taking the orthogonal complement of $\\vec{r}_{CP}$ with respect to $\\hat{a}$, so $d = \\|\\operatorname{Comp}(\\vec{r}_{CP}, \\hat{a})\\|$.\n\nExample Problem: Moment of inertia of a square plate about a corner.\n\nA solid uniform square plate has mass $m$, side-length $\\ell$, and thickness $h$. What is the moment of inertia about the $z$-axis through the corner $P$?\n\nUse the answer to Example Problem #rem-xs and the parallel axis theorem #rem-el.\n\nIn Example Problem #rem-xs we computed the moment of inertia of a square place about the center to be $I_{C,z} = \\frac{1}{6} m \\ell^2$. The parallel axis theorem #rem-el now gives: \\begin{aligned} I_{P,z} &= I_{C,z} + m \\, {r_{CP}}^2 \\\\ &= \\frac{1}{6} m \\ell^2 + m \\left( \\left(\\frac{\\ell}{2}\\right)^2 + \\left(\\frac{\\ell}{2}\\right)^2 \\right) \\\\ &= \\frac{2}{3} m \\ell^2. \\end{aligned}\n\nTo check our answer, we can compute the corner moment of inertia directly by integrating with formula #rem-ec. To do this, we put the coordinate origin at the point $P$ about which we wish to find the moment of inertia, as shown in 3D to the right. From this we see that the $x$ range is $0$ to $\\ell$, the same for the $y$ range, and the $z$ range is $-h/2$ to $h/2$.\n\nThe integral formula now gives: \\begin{aligned} I_{P,z} &= \\iiint_\\mathcal{B} \\rho (x^2 + y^2) \\,dx\\,dy\\,dz \\\\ &= \\int_{-h/2}^{h/2} \\int_0^{\\ell} \\int_{0}^{\\ell} \\rho (x^2 + y^2) \\,dx\\,dy\\,dz \\\\ &= \\int_{-h/2}^{h/2} \\int_{0}^{\\ell} \\left[ \\rho \\left(\\frac{1}{3}x^3 + y^2 x\\right)\\right]_{x = 0}^{x = \\ell} \\,dy\\,dz \\\\ &= \\int_{-h/2}^{h/2} \\int_{0}^{\\ell} \\rho \\left(\\frac{1}{3}\\ell^3 + y^2 \\ell \\right) \\,dy\\,dz \\\\ &= \\int_{-h/2}^{h/2} \\left[ \\rho \\left(\\frac{1}{3}\\ell^3 y + \\frac{1}{3} y^3 \\ell \\right) \\right]_{y = 0}^{y = \\ell} \\,dz \\\\ &= \\int_{-h/2}^{h/2} \\rho \\left(\\frac{1}{3}\\ell^4 + \\frac{1}{3} \\ell^4 \\right) \\,dz \\\\ &= \\int_{-h/2}^{h/2} \\rho \\frac{2}{3}\\ell^4 \\,dz \\\\ &= \\left[ \\rho \\frac{2}{3}\\ell^4 z \\right]_{z = -h/2}^{z = h/2} \\\\ &= \\rho \\frac{2}{3}\\ell^4 h. \\end{aligned} The total mass of the plate is $m = \\rho \\ell^2 h$, so we can write the final expression for the moment of inertia is $I_{P,z} = \\frac{2}{3} m \\ell^2.$ This is the same as the expression we obtained above using the parallel axis theorem, but clearly the parallel axis theorem version is much easier.\n\nOne consequence of the parallel axis theorem is that the moment of inertia can only increase as we move the rotation point $P$ away from the center of mass $C$. This means that the point with the lowest moment of inertia is always the center of mass itself.\n\nA second consequence of the parallel axis theorem is that moving the point $P$ along the direction $\\hat{a}$ doesn't change the moment of inertia, because the axis of rotation is not changing as the point moves along the axis itself.\n\n $I^{\\mathcal{B}}_{P,\\hat{a}} = I^{\\mathcal{B}_1}_{P,\\hat{a}} + I^{\\mathcal{B}_2}_{P,\\hat{a}}$\n\nThe whole body $\\mathcal{B}$ is composed of sub-bodies $\\mathcal{B}_1$ and $\\mathcal{B}_2$.\n\nThe addition of moments of inertia for sub-bodies to give the full moment of inertia follows directly from the fact that the integral over the whole body is the sum of the integrals over the sub-bodes. That is, for $\\mathcal{B} = \\mathcal{B}_1 \\cup \\mathcal{B}_2$ (and $\\mathcal{B}_1$ not overlapping with $\\mathcal{B}_2$), we have: \\begin{aligned} I^{\\mathcal{B}}_{P,\\hat{a}} &= \\int_{\\mathcal{B}} \\rho r^2 \\,dV \\\\ &= \\int_{\\mathcal{B_1} \\cup \\mathcal{B}_2} \\rho r^2 \\,dV \\\\ &= \\int_{\\mathcal{B_1}} \\rho r^2 \\,dV + \\int_{\\mathcal{B}_2} \\rho r^2 \\,dV \\\\ &= I^{\\mathcal{B}_1}_{P,\\hat{a}} + I^{\\mathcal{B}_2}_{P,\\hat{a}}. \\end{aligned}\n\nWarning: To add moments of inertia they must be about the same axis.\n\nIt is only valid to add together moments of inertia for different sub-bodies if each moment of inertia is computed about the same axis of rotation. For example, consider $I^{\\mathcal{B}_1}_{C_1,z}$ and $I^{\\mathcal{B}_2}_{C_2,z}$, which are the moments of inertia of bodies $\\mathcal{B}_1$ and $\\mathcal{B}_2$ about each of their individual centers of mass $C_1$ and $C_2$ (with axis direction $z$). Now it is physically meaningless to form the sum $I^{\\mathcal{B}_1}_{C_1,z} + I^{\\mathcal{B}_2}_{C_2,z}$, because each moment of inertia is about a different axis. We must first shift both moments of inertia to a common axis point $P$ (perhaps by using the parallel axis theorem), and then we can meaningfully add them together.\n\nExample Problem: Moment of inertia of an L-shaped plate.\n\nA solid uniform L-shaped plate has mass $m$ and dimensions as shown. What is the moment of inertia about the $z$-axis through the corner $P$?\n\nUse the answer to Example Problem #rem-es together with the addition theorem #rem-ea and the parallel axis theorem #rem-el.\n\nRecall that in Example Problem #rem-xs we computed the moment of inertia of a square place about the center to be $I_{C,z} = \\frac{1}{6} m \\ell^2$.\n\nTo use this together with the parallel axis theorem #rem-el and the additive theorem #rem-ea, we must first realize that the L-shape can be decomposed into four square plates, as shown. Each small square body has the same mass and the same moment of inertia about its center, so: \\begin{aligned} m_1 &= \\frac{1}{4} m \\\\ I^{\\mathcal{B}_1}_{C_1,z} &= \\frac{1}{6} m_1 (2d)^2 = \\frac{1}{6} \\left(\\frac{1}{4} m\\right) 4d^2 = \\frac{1}{6} m d^2, \\end{aligned} and similarly for each other small square body.\n\nNow individual moments of inertia about the corner $P$ can be found using the parallel axis theorem: \\begin{aligned} I^{\\mathcal{B}_1}_{P,z} &= I^{\\mathcal{B}_1}_{C_1,z} + m_1 \\, {r_{C_1P}}^2 = \\frac{1}{6} m d^2 + \\left(\\frac{1}{4} m\\right) \\left(d^2 + (3d)^2\\right) = \\frac{8}{3} m d^2 \\\\ I^{\\mathcal{B}_2}_{P,z} &= I^{\\mathcal{B}_2}_{C_2,z} + m_2 \\, {r_{C_2P}}^2 = \\frac{1}{6} m d^2 + \\left(\\frac{1}{4} m\\right) \\left(d^2 + d^2\\right) = \\frac{2}{3} m d^2 \\\\ I^{\\mathcal{B}_3}_{P,z} &= I^{\\mathcal{B}_1}_{P,z} = \\frac{8}{3} m d^2 \\\\ I^{\\mathcal{B}_4}_{P,z} &= I^{\\mathcal{B}_4}_{C_4,z} + m_4 \\, {r_{C_4P}}^2 = \\frac{1}{6} m d^2 + \\left(\\frac{1}{4} m\\right) \\left((5d)^2 + d^2\\right) = \\frac{20}{3} m d^2. \\end{aligned} Here we observed that body $\\mathcal{B}_3$ will have the same moment of inertia as $\\mathcal{B}_1$, saving some computation.\n\nCombining the individual moments of inertia now gives us the total: \\begin{aligned} I_{P,z} &= I^{\\mathcal{B}_1}_{P,z} + I^{\\mathcal{B}_2}_{P,z} + I^{\\mathcal{B}_3}_{P,z} + I^{\\mathcal{B}_4}_{P,z} \\\\ &= \\frac{8}{3} m d^2 + \\frac{2}{3} m d^2 + \\frac{8}{3} m d^2 + \\frac{20}{3} m d^2\\\\ &= \\frac{38}{3} m d^2. \\end{aligned} We could also directly compute the total moment of inertia using the integral formula #rem-ec, which would be quite complex.\n\n## Basic shapes\n\nThe moments of inertia listed below are all computed directly from the integrals #rem-ec.\n\nPoint mass: moments of inertia\n\n \\begin{aligned} I_{P,z} &= m r^2 \\end{aligned}\n\nRectangular prism: moments of inertia\n\n \\begin{aligned} I_{C,x} &= \\frac{1}{12} m ({\\ell_y}^2 + {\\ell_z}^2) \\\\ I_{C,y} &= \\frac{1}{12} m ({\\ell_z}^2 + {\\ell_x}^2) \\\\ I_{C,z} &= \\frac{1}{12} m ({\\ell_x}^2 + {\\ell_y}^2) \\end{aligned}\n\nThe calculation for any of the axes is the same, so we will only write out the derivation of $I_{C,z}$ here. We use #rem-ec, which gives: \\begin{aligned} I_{C,z} &= \\iiint_\\mathcal{B} \\rho (x^2 + y^2) \\,dx\\,dy\\,dz \\\\ &= \\int_{-\\ell_z/2}^{\\ell_z/2} \\int_{-\\ell_y/2}^{\\ell_y/2} \\int_{-\\ell_x/2}^{\\ell_x/2} \\rho (x^2 + y^2) \\,dx\\,dy\\,dz \\\\ &= \\int_{-\\ell_z/2}^{\\ell_z/2} \\int_{-\\ell_y/2}^{\\ell_y/2} \\left[ \\rho \\left(\\frac{1}{3}x^3 + y^2 x\\right)\\right]_{x = -\\ell_x/2}^{x = \\ell_x/2} \\,dy\\,dz \\\\ &= \\int_{-\\ell_z/2}^{\\ell_z/2} \\int_{-\\ell_y/2}^{\\ell_y/2} \\rho \\left(\\frac{1}{12}{\\ell_x}^3 + y^2 \\ell_x \\right) \\,dy\\,dz \\\\ &= \\int_{-\\ell_z/2}^{\\ell_z/2} \\left[ \\rho \\left(\\frac{1}{12}{\\ell_x}^3 y + \\frac{1}{3} y^3 \\ell_x \\right) \\right]_{y = -\\ell_y/2}^{y = \\ell_y/2} \\,dz \\\\ &= \\int_{-\\ell_z/2}^{\\ell_z/2} \\rho \\left(\\frac{1}{12}{\\ell_x}^3 \\ell_y + \\frac{1}{12} \\ell_x {\\ell_y}^3 \\right) \\,dz \\\\ &= \\int_{-\\ell_z/2}^{\\ell_z/2} \\rho \\ell_x \\ell_y \\frac{1}{12}\\left( {\\ell_x}^2 + {\\ell_y}^2\\right) \\,dz \\\\ &= \\left[ \\rho \\ell_x \\ell_y \\frac{1}{12}\\left( {\\ell_x}^2 + {\\ell_y}^2\\right) z \\right]_{z = -\\ell_z/2}^{z = \\ell_z/2} \\\\ &= \\rho \\ell_x \\ell_y \\ell_z \\frac{1}{12}\\left( {\\ell_x}^2 + {\\ell_y}^2\\right). \\end{aligned} The total mass of the plate is $m = \\rho \\ell_x \\ell_y \\ell_z$, so we can write the moment of inertia as $I_{C,z} = \\frac{1}{12} m \\left( {\\ell_x}^2 + {\\ell_y}^2\\right).$\n\nCylindrical thick shell: moments of inertia\n\n \\begin{aligned} I_{C,x} &= I_{C,y} = \\frac{1}{12} m (3({r_1}^2 + {r_2}^2) + \\ell^2) \\\\ I_{C,z} &= \\frac{1}{2} m ({r_1}^2 + {r_2}^2) \\end{aligned}\n\nTo compute the integrals in #rem-ec it is convenient to switch to cylindrical coordinates: \\begin{aligned} x &= r \\cos\\theta \\\\ y &= r \\sin\\theta \\\\ z &= z. \\end{aligned} The Jacobian of this coordinate transformation is: \\begin{aligned} J &= \\begin{vmatrix} \\frac{\\partial x}{\\partial r} & \\frac{\\partial x}{\\partial \\theta} & \\frac{\\partial x}{\\partial z} \\\\ \\frac{\\partial y}{\\partial r} & \\frac{\\partial y}{\\partial \\theta} & \\frac{\\partial y}{\\partial z} \\\\ \\frac{\\partial z}{\\partial r} & \\frac{\\partial z}{\\partial \\theta} & \\frac{\\partial z}{\\partial z} \\end{vmatrix} = \\begin{vmatrix} \\cos\\theta & -r \\sin\\theta & 0 \\\\ \\sin\\theta & r \\cos\\theta & 0 \\\\ 0 & 0 & 1 \\end{vmatrix} = r \\cos^2\\theta + r \\sin^2\\theta = r. \\end{aligned} Starting with the $z$ axis, the moment of inertia is thus: \\begin{aligned} I_{C,z} &= \\iiint_\\mathcal{B} \\rho (x^2 + y^2) \\,dx\\,dy\\,dz \\\\ &= \\int_{-\\ell/2}^{\\ell/2} \\int_0^{2\\pi} \\int_{r_1}^{r_2} \\rho \\left( (r \\cos\\theta)^2 + (r \\sin\\theta)^2 \\right) \\,J \\,dr\\,d\\theta\\,dz \\\\ &= \\int_{-\\ell/2}^{\\ell/2} \\int_0^{2\\pi} \\int_{r_1}^{r_2} \\rho r^3 \\,dr\\,d\\theta\\,dz \\\\ &= \\int_{-\\ell/2}^{\\ell/2} \\int_0^{2\\pi} \\left[ \\rho \\frac{1}{4} r^4 \\right]_{r = r_1}^{r = r_2} \\,d\\theta\\,dz \\\\ &= \\int_{-\\ell/2}^{\\ell/2} \\int_0^{2\\pi} \\frac{1}{4} \\rho \\left( {r_2}^4 - {r_1}^4 \\right) \\,d\\theta\\,dz \\\\ &= \\int_{-\\ell/2}^{\\ell/2} \\left[ \\frac{1}{4} \\rho \\left( {r_2}^4 - {r_1}^4 \\right) \\theta \\right]_{\\theta = 0}^{\\theta = 2\\pi} \\,dz \\\\ &= \\int_{-\\ell/2}^{\\ell/2} \\frac{\\pi}{2} \\rho \\left( {r_2}^4 - {r_1}^4 \\right) \\,dz \\\\ &= \\left[ \\frac{\\pi}{2} \\rho \\left( {r_2}^4 - {r_1}^4 \\right) z \\right]_{z = -\\ell/2}^{z = \\ell/2} \\\\ &= \\frac{\\pi}{2} \\rho \\ell \\left( {r_2}^4 - {r_1}^4 \\right). \\end{aligned} The total mass of the cylindrical shell is $m = \\rho (\\pi {r_2}^2 - \\pi {r_1}^2) \\ell$, so we can write the moment of inertia as \\begin{aligned} I_{C,z} &= \\frac{1}{2} \\frac{m}{{r_2}^2 - {r_1}^2} \\left( {r_2}^4 - {r_1}^4 \\right) \\\\ &= \\frac{1}{2} \\frac{m}{{r_2}^2 - {r_1}^2} \\left( {r_2}^2 + {r_1}^2 \\right) \\left( {r_2}^2 - {r_1}^2 \\right) \\\\ &= \\frac{1}{2} m \\left( {r_2}^2 + {r_1}^2 \\right). \\end{aligned} Due to rotational symmetry of the cylinder, the moment of inertia will be the same about any axis orthogonal to $z$, so we will just write out the derivation for $I_{C,x}$ here. We again use #rem-ec in cylindrical coordinates, giving: \\begin{aligned} I_{C,x} &= \\iiint_\\mathcal{B} \\rho (y^2 + z^2) \\,dx\\,dy\\,dz \\\\ &= \\int_{-\\ell/2}^{\\ell/2} \\int_0^{2\\pi} \\int_{r_1}^{r_2} \\rho \\left( (r \\sin\\theta)^2 + z^2 \\right) \\,J \\,dr\\,d\\theta\\,dz \\\\ &= \\int_{-\\ell/2}^{\\ell/2} \\int_0^{2\\pi} \\int_{r_1}^{r_2} \\rho \\left( r^3 \\sin^2\\theta + r z^2 \\right) \\,dr\\,d\\theta\\,dz \\\\ &= \\int_{-\\ell/2}^{\\ell/2} \\int_0^{2\\pi} \\left[ \\rho \\left( \\frac{1}{4} r^4 \\sin^2\\theta + \\frac{1}{2} r^2 z^2 \\right) \\right]_{r = r_1}^{r = r_2} \\,d\\theta\\,dz \\\\ &= \\int_{-\\ell/2}^{\\ell/2} \\int_0^{2\\pi} \\rho \\left( \\frac{1}{4} ({r_2}^4 - {r_1}^4) \\sin^2\\theta + \\frac{1}{2} ({r_2}^2 - {r_1}^2) z^2 \\right) \\,d\\theta\\,dz \\\\ &= \\int_{-\\ell/2}^{\\ell/2} \\int_0^{2\\pi} \\rho \\left( \\frac{1}{8} ({r_2}^4 - {r_1}^4) (1 - \\cos 2\\theta) + \\frac{1}{2} ({r_2}^2 - {r_1}^2) z^2 \\right) \\,d\\theta\\,dz \\\\ &= \\int_{-\\ell/2}^{\\ell/2} \\left[ \\rho \\left( \\frac{1}{8} ({r_2}^4 - {r_1}^4) \\left(\\theta - \\frac{1}{2} \\sin 2\\theta\\right) + \\frac{1}{2} ({r_2}^2 - {r_1}^2) z^2 \\theta \\right) \\right]_{\\theta = 0}^{\\theta = 2\\pi} \\,d\\theta\\,dz \\\\ &= \\int_{-\\ell/2}^{\\ell/2} \\rho \\left( \\frac{\\pi}{4} ({r_2}^4 - {r_1}^4) + \\pi ({r_2}^2 - {r_1}^2) z^2 \\right) \\,d\\theta\\,dz \\\\ &= \\left[ \\rho \\left( \\frac{\\pi}{4} ({r_2}^4 - {r_1}^4) z + \\pi ({r_2}^2 - {r_1}^2) \\frac{1}{3} z^3 \\right) \\right]_{z = -\\ell/2}^{z = \\ell/2} \\\\ &= \\rho \\left( \\frac{\\pi}{4} ({r_2}^3 - {r_1}^3) \\ell + \\pi ({r_2}^2 - {r_1}^2) \\frac{1}{12} \\ell^3 \\right) \\\\ &= \\frac{1}{12} \\rho \\pi ({r_2}^2 - {r_1}^2) \\ell \\left( 3 ({r_2}^2 + {r_1}^2) + \\ell^2 \\right). \\end{aligned} In the last line we again used the factorization $({r_2}^4 - {r_1}^4) = ({r_2}^2 + {r_1}^2) ({r_2}^2 - {r_1}^2)$. Now we can substitute in the total mass $m = \\rho \\pi ({r_2}^2 - {r_1}^2) \\ell$ to obtain: \\begin{aligned} I_{C,x} &= \\frac{1}{12} m \\left( 3 ({r_2}^2 + {r_1}^2) + \\ell^2 \\right). \\end{aligned}\n\nSpherical thick shell: moments of inertia\n\n \\begin{aligned} I_C &= \\frac{2}{5} m \\left(\\frac{{r_2}^5 - {r_1}^5}{{r_2}^3 - {r_1}^3}\\right) \\end{aligned}\n\nAll axes through $C$ have the same moment of inertia.\n\nTo compute the integrals in #rem-ec it is convenient to switch to spherical coordinates: \\begin{aligned} x &= r \\cos\\theta \\sin\\phi \\\\ y &= r \\sin\\theta \\sin\\phi \\\\ z &= r \\cos\\phi. \\end{aligned} To find the Jacobian of this coordinate transformation we use the coordinate order $(r,\\phi,\\theta)$ to give a right-handed spherical system. Then: \\begin{aligned} J &= \\begin{vmatrix} \\frac{\\partial x}{\\partial r} & \\frac{\\partial x}{\\partial \\phi} & \\frac{\\partial x}{\\partial \\theta} \\\\ \\frac{\\partial y}{\\partial r} & \\frac{\\partial y}{\\partial \\phi} & \\frac{\\partial y}{\\partial \\theta} \\\\ \\frac{\\partial z}{\\partial r} & \\frac{\\partial z}{\\partial \\phi} & \\frac{\\partial z}{\\partial \\theta} \\end{vmatrix} \\\\ &= \\begin{vmatrix} \\cos\\theta \\sin\\phi & r \\cos\\theta \\cos\\phi & -r \\sin\\theta \\sin\\phi \\\\ \\sin\\theta \\sin\\phi & r \\sin\\theta \\cos\\phi & r \\cos\\theta \\sin\\phi \\\\ \\cos\\phi & -r \\sin\\phi & 0 \\end{vmatrix} \\\\ &= \\cos\\phi (r^2 \\cos^2\\theta \\cos\\phi \\sin\\phi + r^2 \\sin^2\\theta \\cos\\phi \\sin\\phi) \\\\ &\\quad + r \\sin\\phi (r \\cos^2\\theta \\sin^2\\phi + r \\sin^2\\theta \\sin^2\\phi) \\\\ &= r^2 \\cos^2\\phi \\sin\\phi + r^2 \\sin^3\\phi \\\\ &= r^2 \\sin\\phi. \\end{aligned} Due to spherical symmetry, all axes through $C$ will have the same moment of inertia. We will compute $I_{C,z}$, which is: \\begin{aligned} I_{C,z} &= \\iiint_\\mathcal{B} \\rho (x^2 + y^2) \\,dx\\,dy\\,dz \\\\ &= \\int_0^{\\pi} \\int_{-\\pi}^{\\pi} \\int_{r_1}^{r_2} \\rho \\left( (r \\cos\\theta \\sin\\phi)^2 + (r \\sin\\theta \\sin\\phi)^2 \\right) \\,J \\,dr\\,d\\theta\\,d\\phi \\\\ &= \\int_0^{\\pi} \\int_{-\\pi}^{\\pi} \\int_{r_1}^{r_2} \\rho r^4 \\sin^3\\phi \\,dr\\,d\\theta\\,d\\phi \\\\ &= \\int_0^{\\pi} \\int_{-\\pi}^{\\pi} \\left[ \\rho \\frac{1}{5} r^5 \\sin^3\\phi \\right]_{r = r_1}^{r = r_2} \\,d\\theta\\,d\\phi \\\\ &= \\int_0^{\\pi} \\int_{-\\pi}^{\\pi} \\rho \\frac{1}{5} ({r_2}^5 - {r_1}^5) \\sin^3\\phi \\,d\\theta\\,d\\phi \\\\ &= \\int_0^{\\pi} \\left[ \\rho \\frac{1}{5} ({r_2}^5 - {r_1}^5) \\sin^3\\phi \\, \\theta \\right]_{\\theta = -\\pi}^{\\theta = \\pi} \\,d\\phi \\\\ &= \\int_0^{\\pi} \\rho \\frac{2\\pi}{5} ({r_2}^5 - {r_1}^5) \\sin^3\\phi \\,d\\phi \\\\ &= \\int_0^{\\pi} \\rho \\frac{2\\pi}{5} ({r_2}^5 - {r_1}^5) \\frac{1}{4} (3\\sin\\phi - \\sin 3\\phi) \\,d\\phi \\\\ &= \\left[ \\rho \\frac{2\\pi}{5} ({r_2}^5 - {r_1}^5) \\frac{1}{4} \\left(\\frac{1}{3} \\cos 3\\phi - 3\\cos\\phi\\right) \\right]_{\\phi = 0}^{\\phi = \\pi} \\\\ &= \\rho \\frac{2\\pi}{5} ({r_2}^5 - {r_1}^5) \\frac{1}{4} \\left(-\\frac{2}{3} + 6\\right) \\\\ &= \\frac{8}{15} \\rho \\pi ({r_2}^5 - {r_1}^5). \\end{aligned} The total mass of the spherical shell is $m = \\rho \\left(\\frac{4}{3} \\pi {r_2}^3 - \\frac{4}{3} \\pi {r_1}^3\\right)$, so we can write the moment of inertia as \\begin{aligned} I_{C,z} &= \\frac{8}{15} m \\frac{3}{4} \\frac{1}{{r_2}^3 - {r_1}^3} ({r_2}^5 - {r_1}^5) \\\\ &= \\frac{2}{5} m \\left(\\frac{{r_2}^5 - {r_1}^5}{{r_2}^3 - {r_1}^3}\\right). \\end{aligned}\n\n## Simplified shapes\n\nThe moments of inertia listed below are all special cases of the basic shapes given in Section #rem-sb. Other special cases can be easily obtained by similar methods.\n\nRod: moments of inertia\n\n \\begin{aligned} I_{C,z} &= \\frac{1}{12} m \\ell^2 \\\\ I_{P,z} &= \\frac{1}{3} m \\ell^2 \\\\ I_{C,x} &= I_{P,x} = 0 \\end{aligned}\n\nA rod is assumed to be a shape with infinitesimally small cross-section. We can use the formula #rem-ey for a cylinder with zero radii $r_1 = r_2 = 0$. The coordinates here are set up so that $z$ is orthogonal to the rod axis and $x$ is along the axis, so #rem-ey gives: \\begin{aligned} I_{C,z} &= \\frac{1}{12} m (3({r_1}^2 + {r_2}^2) + \\ell^2) = \\frac{1}{12} m \\ell^2 \\\\ I_{C,x} &= \\frac{1}{2} m ({r_1}^2 + {r_2}^2) = 0. \\end{aligned} To find the moment of inertia about the end point $P$ we can use the parallel axis theorem #rem-el. This results is: \\begin{aligned} I_{P,z} &= I_{C,z} + m {r_{CP}}^2 \\\\ &= \\frac{1}{12} m \\ell^2 + m \\left(\\frac{\\ell}{2}\\right)^2 \\\\ &= \\frac{1}{3} m \\ell^2. \\end{aligned} The moment of inertia $I_{P,x}$ is still zero, because $\\vec{r}_{CP}$ is parallel to $x$.\n\nSolid cylinder or disk: moments of inertia\n\n \\begin{aligned} I_{C,x} &= I_{C,y} = \\frac{1}{12} m (3r^2 + \\ell^2) \\\\ I_{C,z} &= \\frac{1}{2} m r^2 \\end{aligned}\n\nThe solid cylinder expressions are simply the cylindrical thick shell formulas #rem-ey with the inner radius set to $r_1 = 0$ and the outer radius $r_2 = r$.\n\nHollow cylinder or hoop: moments of inertia\n\n \\begin{aligned} I_{C,x} &= I_{C,y} = \\frac{1}{12} m (6 r^2 + \\ell^2) \\\\ I_{C,z} &= m r^2 \\end{aligned}\n\nThe hollow cylinder expressions can be found from cylindrical thick shell formulas #rem-ey by taking the same value for both inner and outer radii, so that $r_1 = r_2 = r$.\n\nSolid ball: moments of inertia\n\n \\begin{aligned} I_C &= \\frac{2}{5} m r^2 \\end{aligned}\n\nAll axes through $C$ have the same moment of inertia.\n\nThe solid ball moment of inertia can be directly obtained from the spherical thick shell expression #rem-es with inner radius $r_1 = 0$ and outer radius $r_2 = r$.\n\nHollow sphere: moments of inertia\n\n \\begin{aligned} I_C &= \\frac{2}{3} m r^2 \\end{aligned}\n\nAll axes through $C$ have the same moment of inertia.\n\nA hollow sphere moment of inertia is the same as that for the spherical thick shell #rem-es with inner radius and outer radius both set to $r_1 = r_2 = r$. Some care is needed here, however, as a simple substitution into the spherical thick shell expression would give the undefined value $0 / 0$.\n\nInstead we need to set $r_2 = r$ and then to take the limit as $r_1 \\to r$ using l'Hôpital's rule: \\begin{aligned} I_C &= \\lim_{r_1 \\to r} \\frac{2}{5} m \\left(\\frac{r^5 - {r_1}^5}{r^3 - {r_1}^3}\\right) \\\\ &= \\lim_{r_1 \\to r} \\frac{2}{5} m \\left(\\frac{\\frac{d}{d r_1}\\left(r^5 - {r_1}^5\\right)}{ \\frac{d}{d r_1}\\left(r^3 - {r_1}^3\\right)}\\right) \\\\ &= \\lim_{r_1 \\to r} \\frac{2}{5} m \\left(\\frac{5 {r_1}^4}{3 {r_1}^2}\\right) \\\\ &= \\lim_{r_1 \\to r} \\frac{2}{3} m {r_1}^2 \\\\ &= \\frac{2}{3} m r^2. \\end{aligned}"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.68079674,"math_prob":0.99999905,"size":28111,"snap":"2021-31-2021-39","text_gpt3_token_len":10843,"char_repetition_ratio":0.20286761,"word_repetition_ratio":0.17173652,"special_character_ratio":0.4137882,"punctuation_ratio":0.081908345,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000088,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-30T20:33:16Z\",\"WARC-Record-ID\":\"<urn:uuid:f95f132f-1781-4ec6-b0fd-b3afd3ce5a92>\",\"Content-Length\":\"53329\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f359c288-7664-42a5-99a7-e6d4b3677613>\",\"WARC-Concurrent-To\":\"<urn:uuid:7027fc1c-f1fa-4800-98d0-bc57853846a7>\",\"WARC-IP-Address\":\"18.220.149.166\",\"WARC-Target-URI\":\"https://dynref.engr.illinois.edu/rem.html\",\"WARC-Payload-Digest\":\"sha1:2CQF5ZR4YPMNBACDCCYLBLDH2NEZJR4U\",\"WARC-Block-Digest\":\"sha1:ACJCSPUWPN5K6V3GBB3TT367CP5VMFC4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153980.55_warc_CC-MAIN-20210730185206-20210730215206-00021.warc.gz\"}"} |
http://blog.zacharyabel.com/tag/ellipses/ | [
"# Tag Archives: ellipses\n\n[This is the 4th and last installment in the current series on mini-golf and ellipse geometry. See the previous ones here: #1, #2, #3.]\n\nWe must settle one more question to round out our elliptical arc: Why does light, when shot from one focus of an ellipse-shaped mirrored room, reflect back to the other focus? To answer this question, we’ll need a Fact, a Formalism, and a Fairy Tale.\n\n### The Fact\n\nRecall that, in the previous post, we saw that ellipses can be described by distances: Any ellipse has two focus points F1 and F2 so that the total length of broken path F1XF2 is the same for every point X on the ellipse; let d be this common total distance. In fact, more is true: the length of F1XF2 is smaller than d if X is inside the ellipse, and larger than d if X is outside.",
null,
"Left: The distance of the broken path is greater, equal, or less than d depending on whether the central vertex is outside, on, or inside the ellipse. Right: A proof of the “outside” case.\n\nTo prove this fairly intuitive fact, we’ll use the “straight line principle”: the shortest distance between two points is a straight line. Indeed, when X is outside the ellipse (see right diagram above), straight-line path YF2 is shorter than the path YXF2 that detours through X, and so d = F_1 Y F_2 < F_1 Y X F_2[/latex]. See if you can fill in the case where X is inside the ellipse.\n\n### The Formalism\n\nRecall that when light bounces off a straight mirror, the angle of incidence equals the angle of reflection. But here we’re discussing light bouncing off an ellipse, which is decidedly not straight. So we need to formally describe how light reflects off curved mirrors.",
null,
"Reflection off of a curved mirror behaves like reflection off of the tangent line.\n\nIf we zoom into where the incoming light ray strikes a curved mirror (illustrated above), the mirror closely resembles a straight line, specifically its tangent line. This suggests that the light should behave as if it is reflecting off of this line, with equal angles as marked. This is indeed the rule governing ideal reflections on curved mirrors: the angles of incidence and reflection, as measured from the tangent line, should be equal.\n\n### The Fairy Tale\n\nThe last ingredient involves Little Red Riding Hood and her thirsty grandmother. Red is delivering cake and wine from her mother (point M) to her grandmother (point G), but she must first fill a bucket of water at the nearby stream S, which is conveniently shaped like a straight line. She was warned by her Brothers to watch out for a big bad wolf, so she must minimize her total walking distance. Where on the stream should she fill her bucket to minimize this distance?",
null,
"Left: Paths from M to S to G transform into paths from M’ to G, the shortest of which is straight. Right: This straight path behaves like a mirror.\n\nTo answer this, imagine reflecting the first leg of Red’s journey across line S, so her path from M to S to G gets reflected to a path from M’ to G. The reverse may be done as well: any path from M’ to G turns into a path from M to G that stops somewhere along S. So we just need to find the shortest path from M’ to G. But this is easy: it’s just the straight path M’G. So Red’s shortest path from M to S to G is the one that stops at Z.\n\nNotice that this shortest path is the one with equal angles as marked. This means Red’s best strategy is to pretend the stream is a mirror and to follow the light ray that bounces directly to grandma’s house. This neatly exemplifies Fermat’s principle, which says that light tends to follow the fastest routes.\n\n### The Proof\n\nWith these pieces in place, we can finish today’s question in a flash. Let’s say light from focus F1 hits an ellipse at point X, as illustrated below. Why does this ray bounce off the ellipse toward F2? If we draw the tangent line L at point X, by The Formalism above, this question is equivalent to: why does the light ray bounce off of line L toward F2?",
null,
"Why are the two marked angles equal? Because the white path solves Red’s Fairy Tale problem.\n\nLet’s reimagine this Grimm scenario by thinking of F1 as Red’s mother’s house, F2 as grandma’s house, and L as the stream. I claim F1XF2 is the shortest path for Red to take. Why? If Y is any other point on line L, then Y is outside the ellipse, so by The Fact above, F1YF2 has distance longer than d. So F1XF2 is indeed the shortest. But by The Fairy Tale, we know that this shortest route behaves like light bouncing off of line L, i.e., the marked angles are indeed equal. So we’re done!\n\n# What Makes Ellipses… Ellipses?\n\nLast time we used wild properties of ellipses to build some really easy—and some really devilish—golf courses. Specifically, I claimed that every ellipse has two magical points F1 and F2 (called foci) such that a ray from F1 always bounces off the ellipse and lands precisely at F2, and furthermore, this path always has the same length. Why does this happen? And how do we find these foci?\n\nLet’s focus(!) on the last question first. Recall that an ellipse is a stretched circle. In other words, an ellipse is what forms when you slice a tall, circular tube (cylinder) along a slant:",
null,
"Left: An ellipse is formed by slicing a cylinder. Right: Fitting spheres above and below the cut locates the ellipse’s foci.\n\nTake a sphere that snugly fits inside the tube, and drop it down until it touches the ellipse-slice at a single point F1. Do the same with a sphere underneath, touching the slice at F2. These points turn out to be the foci of the ellipse. Let’s see why.\n\nWe can use this tubular setup to answer one mystery from earlier: for any point X on the ellipse, the sum of distances of XF1 and XF2 is always the same! The proof lies in the following animation. Segments XF1 and XA have the same length because they’re both tangent to the upper sphere from X, and similarly, XF2=XB. So the sum XF1+XF2 is just the length of segment AB, the height between the spheres’ equators.",
null,
"The sum F1X+XF2 equals the length of AB and therefore does not change when X moves.\n\nThis has two neat consequences. First, it provides an elementary method for drawing ellipses (in real life!): all you need are two push pins and a loop of string, as illustrated below. The string ensures that the sum XF1+XF2 stays fixed while you trace the pen around, as long as you’re careful to keep the string taut throughout.",
null,
"How to draw an ellipse with push-pins and string\n\nSecond, what happens if we slice a cone instead of a cylinder? Perhaps surprisingly, we still get an ellipse! Indeed, as above, we can create a sphere on either side of the slice that snugly fits against the slice and the walls of the cone (the so-called Dandelin spheres), and exactly the same proof shows that XF1+XF2 stays constant as X moves around the edge.",
null,
"Slicing a cone also produces an ellipse, by the same argument.\n\nBut wait, there’s still an unanswered question! We’ve seen that the path F1X+XF2 has a fixed length, but why does light bounce off an ellipse along such a path? This is what we really cared about for mini-golf! Come back next time for the answer, and in the meantime, have a great 2 weeks.\n\n# More Putting Predicaments\n\nLast time we discussed some rather challenging holes of (mathematical) mini-golf, uncovering Tokarski’s construction of some un-hole-in-one-able holes. By contrast, here is the all-time easiest hole of mini-golf, namely, a guaranteed hole-in-one:",
null,
"In an ellipse, any golf shot from focus $$F_1$$ will bounce directly to the other focus, $$F_2$$.\n\nBy some miracle of geometry, if we take an ellipse (i.e., a stretched or squashed circle) and place the ball and cup at two special points $$F_1$$ and $$F_2$$ called the foci of the ellipse, then any golf shot leaving $$F_1$$ will bounce off a wall and proceed directly to $$F_2$$. You can’t miss!\n\nCuriously, this easiest golf hole can be used to construct some even more challenging ones. For starters, consider a mushroom-shaped room as illustrated below (left), where the “mushroom head” is half of an ellipse with foci $$F_1$$ and $$F_2$$. Then the same reflection principle mentioned above tells us that any shot entering the mushroom head via segment $$F_1F_2$$ will be reflected straight back through $$F_1F_2$$ and sent back down the stem. This means that no shot from P can ever reach the triangular “fang” containing Q, and in particular, no hole-in-one from P to Q is possible. In the mirrored-room setting (described last time), if a light source is placed at P, then the whole triangular fang remains unilluminated! This is stronger than Tokarski’s room, where only a single point remained unlit. Similar constructions are possible even with rooms that have no corners, such as the “curvy” mushroom on the right.",
null,
"Left: Any path entering the mushroom head from segment $$F_1F_2$$ will be reflected back down through this segment, so the triangle containing Q is not reachable from P. Right: The same phenomenon is possible even when the room has no corners.\n\nWith this idea, we can construct a mirrored room that has dark patches no matter where a light bulb is placed. In the image below, if a light source is placed anywhere in the top half of the room, the lower triangular “fang” will be completely dark. Similarly, the upper fang is not illuminated from anywhere in the bottom half of the room. This room (or one quite like it) was originally designed by Roger Penrose in 1958.",
null,
"This room will have dark regions no matter where the light source is placed.\n\nBack in the mini-golf setting, we can do something even more devious: by chaining multiple Penrose rooms together, we can construct golf holes that require as many shots as we wish! The golf hole below cannot be completed with fewer than 7 shots. Indeed, no single shot can cross two dashed segments with different numbers, so 7 shots are required.",
null,
"At least 7 shots are required to get from P to Q.\n\n### What about polygons?\n\nThe golf holes last time were all polygons, whereas we have allowed curved boundaries here. Can we construct a polygonal golf hole, using only flat walls, that still requires at least 7 shots? Or even 3? Unfortunately, the answer this time is “we don’t know, but we think not.” In particular, it has been conjectured that, no matter where the ball and cup are placed in a polygonal golf hole, a single shot can place the ball as close to the hole as desired, so a short second putt can finish the job.\n\n### Notes\n\n1. In fact, even more is true: all of these paths from $$F_1$$ to $$F_2$$ have the same length! We will not prove these facts today. []\n2. Assuming you hit hard enough, of course. []\n3. For the analysts out there, the room’s boundary can be made smooth. []\n4. This too can be accomplished with smooth boundary. []\n5. More strongly, O’Rourke and Petrovici conjecture that with a single light source in a polygonal room, the set of unilluminated points has measure 0. Reference: Joseph O’Rourke and Octavia Petrovici. Narrowing Light Rays with Mirrors. Proceedings of the 13th Canadian Conference on Computational Geometry, 137–140, 2001. []"
]
| [
null,
"http://i1.wp.com/blog.zacharyabel.com/wp-content/uploads/2012/12/ellipse-distances-fact.gif",
null,
"http://i0.wp.com/blog.zacharyabel.com/wp-content/uploads/2012/12/reflect-curve.gif",
null,
"http://i2.wp.com/blog.zacharyabel.com/wp-content/uploads/2012/12/reflect-little-red.gif",
null,
"http://i1.wp.com/blog.zacharyabel.com/wp-content/uploads/2012/12/ellipse-reflect.gif",
null,
"http://i0.wp.com/blog.zacharyabel.com/wp-content/uploads/2012/09/ellipse-tubes.gif",
null,
"http://i2.wp.com/blog.zacharyabel.com/wp-content/uploads/2012/09/dandelin-spheres-anim.gif",
null,
"http://i1.wp.com/blog.zacharyabel.com/wp-content/uploads/2012/09/string-anim.gif",
null,
"http://i2.wp.com/blog.zacharyabel.com/wp-content/uploads/2012/09/dandelin-spheres-cone.gif",
null,
"http://i1.wp.com/blog.zacharyabel.com/wp-content/uploads/2012/09/ellipse-room.gif",
null,
"http://i1.wp.com/blog.zacharyabel.com/wp-content/uploads/2012/09/mushrooms.gif",
null,
"http://i1.wp.com/blog.zacharyabel.com/wp-content/uploads/2012/09/penrose-room.gif",
null,
"http://i1.wp.com/blog.zacharyabel.com/wp-content/uploads/2012/09/hole-in-7.gif",
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]
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https://fpjson.asteroid.ac/ | [
"FPJSON v.0.1.1\nIntro\nTutorial\nReference\nExam\nGet Started\nDashboard\nPlayground\nIntro\nGet Started",
null,
"# FPJSON\n\nFPJSON is a programming language agnostic JSON-based functional programming language.\n\n• The whole code is just a JSON array\n• Functional Programming\n• No overhead due to programming language implementation details\n\nIn other words, you don't have to worry about any programming language specifications.\n\nInstead you can just focus on building pure logics for data manipuration.\n\nSince it's just a JSON array, it can be ported to any programming language environment.\n\n#### Some Examples\n\n``````/* add */\n[\"add\", 1, 2] // = 3\n\n/* difference */\n[\"difference\", [1, 2, 3], [3, 4, 5]] = [1, 2]\n\n/* map */\n[[\"map\", [\"inc\"]], [1, 2, 3]] // = [4, 5, 6]\n\n/* compose */\n[[\"compose\", [\"map\", [\"inc\"]], [\"difference\"]], [1, 2, 3], [3, 4, 5]] // = [2, 3]\n``````\n\nThere are more than 250 pre-defined functions. And you can further extend that library.\n\n## Install\n\nFor now the parser is only implemented in JavaScript and it borrows heavily from Ramda.js.\n\nIn fast, you can use most of the functions in Ramda with FPJSON.\n\n``````yarn add fpjson-lang\n``````\n\n## Usage\n\nIt couldn't be simpler.\n\n``````import fpjson from \"fpjson-lang\"\n\nconst one_plus_two = fpjson([\"add\", 1, 2]) // = 3\n``````\n\n## Syntax\n\nYou should familiarize yourself with Ramda which enables Haskell-like functional programming with JS. You can use most of the powerful ramda functions with point-free style in JSON.\n\nThe first element in an array is a function.\n\n``````[\"add\", 1, 2] // add(1, 2)\n``````\n\nTo curry a function, nest it.\n\n``````[[\"add\", 1], 2] // add(1)(2)\n``````\n\nA function always needs to be wrapped with `[]` and to be the first element in the array.\n\n``````[[\"map\", [\"inc\"]], [1, 2, 3]] // map(inc)([1, 2, 3])\n``````\n\nThis is an error because `inc` is imterpreted as `String`.\n\n``````[[\"map\", \"inc\"], [1, 2, 3]] // map(\"inc\")([1, 2, 3])\n``````\n\nPoint-free style means you cannot write something like this with the JSON format.\n\n``````sortBy((v)=> v.age)(people) // ramdajs\n``````\n\nIt's because you cannot write arbitrary JS lines such as `(v)=> v.age`.\n\nInstead, you can achieve the same using another ramda funciton `prop`.\n\n``````sortBy(prop(\"age\"), people) // ramdajs\n[\"sortBy\",[\"prop\", \"age\"], people] // FPJSON\n``````\n\n## Reserved First Words\n\nBy placing a reserved word in the first spot of an array, you can access the pre-built features.\n\nThere are just 6 of them.\n\n#### \"[]\"\n\nTo create an array of functions without executing them, place `\"[]\"` in the first spot, otherwise the `[\"lte\", 2]` function will be executed with `[\"gt\", 2]` before `-3` is passed.\n\n``````[[\"anyPass\", [\"[]\", [\"lte\", 2], [\"gt\", 2]]], -3] // anyPass(lte(2), gt(2))(-3)\n``````\n\n#### \"typ\"\n\nTo create a type object such as `Number`, `Boolean`, `String`, `Array`, and `Object`.\n\n``````[\"is\", [\"typ\", \"String\"], \"abc\"] // is(String, \"abc\")\n``````\n\n#### \"reg\"\n\nTo create a `RegExp`.\n\n``````[\"test\", [\"reg\", \"a\", \"i\"], \"ABC\"] // test(new RegExp(\"a\", \"i\"), \"ABC\")\n``````\n\n#### \"let\"\n\nPure functional programming without any side-effects is easy to get extremely complex and entangled even for simple logics.\n\n`\"let\"` inserts global variables to ease up the unnecessary complexisities.\n\n``````[\"let\", \"num1\", 1] // let var1 = 1\n``````\n\n#### \"\\$\"\n\nTo access previously defined variables, use `\"\\$\"`.\n\n``````[\"add\", [\"var\", \"num1\"], 1] // add(num1, 1)\n``````\n\nIn practice, you need to use `\"let\"` and `\"\\$\"` in the same array.\n\n``````[[\"pipe\", [\"add\", 1], [\"let\", \"num1\"], [\"\\$\", \"num1\"]], 1]) // = 2\n``````\n\nOr you can pass a store object as the second argument to `fpjson`.\n\n``````let vars = {}\nfpjson([\"let\", \"num1\", 1], vars) // vars = { \"num1\" : 1 }\nfpjson([\"add\", [\"\\$\", \"num1\"], 1], vars) // 2\n``````\n\n#### \"var\"\n\n`var` works just like `\\$` except that `var` needs another argument to invoke.\n\nThe last argument can be anything since it will be ignored.\n\nNote that you cannot access a new value within the same composition where it was defined.\n\n``````let vars = {}\nfpjson([\"let\", \"num1\", 1], vars) // vars = { \"num1\" : 1 }\nfpjson([\"add\", [\"var\", \"num1\", true], 1], vars) // 2\n``````\n\n#### Dynamic Variables\n\nVariable names can be dinamically specified with `\\$dynamic_path`.\n\n``````let vars = {}\nfpjson([\"let\", \"num1\", 1], vars) // vars = { \"num1\" : 1 }\nfpjson([\"let\", \"ln\", \"num1\"], vars) // vars = { \"num1\" : 1, \"ln\" : \"num1\" }\nfpjson([\"add\", [\"\\$, \"\\$ln\"], 1], vars) // 2\n``````\n\n#### Dot Notation\n\nNested fields can be accessed with `.`.\n\n``````let vars = {}\nfpjson([\"let\", \"o\", { num: 1 }], vars) // vars = { \"o\" : { \"num\": 1 } }\nfpjson([\"var\", \"o.num\", true ], vars) // 1\n``````\n\n## Who is Using FPJSON?\n\nFPJSON is used to define access control rules and cron jobs in WeaveDB - Arweave-based decentralized NoSQL Database. FPJSON allows super rich and complex programming logics to be stored as JSON data on smart contracts, which opens up a whole new pradigm to dapp development.\n\nFPJSON is also to be used for natural language generation algorithms, which will lead to the next-gen AI paradigm to revolutionize the human languages.\n\n## POLP (Proof Of Learning Protocol)\n\nPOLP is planned to be launched soon. Anyone who completes the exam with 100% score can mint an SBT/NFT to proove the success, and we will form a closed community with hyper productive engineers and the like-minded where you will have an access with the NFT.\n\nThe interactive tutorials and the exam are MVPs of such a future-proof learning system.\n\n## Tutorials\n\nLearning functional programming is pretty challenging. So we created interactive tutorials and an exam to make sure you can familiarize yourself with all the core functions with ease!\n\nGoing through the tutorials will install a new framework in your brain and make you a better programmer in general even if you never need FPJSON."
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"equation solving resources\n\n# Find LCD For Rational Expression And Convert Into LCD As Denominator?\n\nBelow is a number of keywords that users typed in today to reach math help pages.\n\nHow can this be of help to you?\n\n• find the term you are searching for (i.e. Find LCD For Rational Expression And Convert Into LCD As Denominator) in the table below\n\n• Click on the appropriate program demo found in the same row as your search phrase\n\n• If you find the program demonstration useful click on the purchase button to buy the software at a special low price extended to equation-solver.com visitors\n\n Related Search Phrase Algebrator animated Flash Demo Algebrator Static Demo Purchase now Saxon Algebra 2 Answers",
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https://quant.stackexchange.com/questions/tagged/vasicek | [
"# Questions tagged [vasicek]\n\nThe tag has no usage guidance.\n\n83 questions\nFilter by\nSorted by\nTagged with\n23 views\n\n### Including Exogeneous variables in short rate models\n\nI am trying to use short rate models (e.g. Vasicek, CIR or Hull-White) to forecast next one or two months yield curve. In this context, is there a way that I can include some exogenous economic or ...\n46 views\n\n204 views\n\n### Black & Scholes under stochastic interest rate (Vasicek) [closed]\n\nI'm a beginner in Quantitative finance and I'd like to ask you for help about this exercise. I have to price a put option on a risky asset by working under stochastic interest rate, so I have to ...\n284 views\n\n### Vasicek Model, zero coupon bond question [closed]\n\nI am trying to solve questions in the Vasicek model. Can anyone help me to solve this question... In the Vasicek model with parameters $\\theta = 0.08$, $k$ = 2.5, $\\sigma = 0.2$, assuming to be ...\n574 views\n\n### Vasicek Model Parameters Estimation\n\nI'm currently trying to estimate the market price of risk (lambda) in the Vasicek Model, and am running into difficulties. Using the Excel Solver tool and the Maximum Likelihood Estimation method ...\n826 views\n\n### Why isn't the Vasicek model arbitrage-free?\n\nCould anyone explain why the Vasicek model isn't an arbitrage-free model? Additionally, which interest rate model is arbitrage-free and why?\n139 views\n\n### Derive the discount bond prices of the Vasicek model by the PDE approach\n\nThe question is shown above. Anyone can help me?\n85 views\n\n### If short rates $r(t)$ do not determine the bond prices $P(t, T)$, then what is the basis for short rate models?\n\nThe question title says it all: We know that in general, specifying the short rate $r(t)$ does not specify the bond prices $P(t, T)$. So how can a model for short rates—for example the Vasicek model—...\n164 views\n\n### Aggregation of $\\rho$ and $p$ for a vasicek model\n\nI'm currently facing the problem of how properly (analytically) adjust the parameters of an aggregated Vasicek (2002) loss distribution so that it has the same expected loss and 99% quantile as the ...\n116 views"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8718218,"math_prob":0.91367614,"size":14096,"snap":"2021-43-2021-49","text_gpt3_token_len":3858,"char_repetition_ratio":0.17258018,"word_repetition_ratio":0.009519689,"special_character_ratio":0.260996,"punctuation_ratio":0.11434659,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98133266,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-27T09:10:07Z\",\"WARC-Record-ID\":\"<urn:uuid:664ebaa2-9aa5-4838-afd4-b4edcf3b3ce2>\",\"Content-Length\":\"287066\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:518d198e-50ad-4710-9e52-75b6ff186efd>\",\"WARC-Concurrent-To\":\"<urn:uuid:77a4ccb1-d7dc-4cbd-9d17-bd416885c084>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://quant.stackexchange.com/questions/tagged/vasicek\",\"WARC-Payload-Digest\":\"sha1:EQ3WOR54VUACZEYC3IB7HMDIX4YYEEEV\",\"WARC-Block-Digest\":\"sha1:ZLQGWEA5HGV5MDOWSWH6RFM7HXURQ2R5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323588113.25_warc_CC-MAIN-20211027084718-20211027114718-00547.warc.gz\"}"} |
https://www.tensorflow.org/versions/r2.0/api_docs/python/tf/keras/optimizers/RMSprop | [
"# tf.keras.optimizers.RMSprop\n\nOptimizer that implements the RMSprop algorithm.\n\nInherits From: Optimizer\n\ntf.keras.optimizers.RMSprop(\nlearning_rate=0.001, rho=0.9, momentum=0.0, epsilon=1e-07, centered=False,\nname='RMSprop', **kwargs\n)\n\n\n### Used in the notebooks\n\nA detailed description of rmsprop.\n\n• maintain a moving (discounted) average of the square of gradients\n• divide gradient by the root of this average\n$$mean_square_t = rho * mean_square{t-1} + (1-rho) * gradient ** 2$$\n$$mom_t = momentum * mom_{t-1} + learning_rate * gradient / \\sqrt{ / mean_square_t + \\epsilon}$$\n$$variable_t := variable_{t-1} - mom_t$$\n\nThis implementation of RMSprop uses plain momentum, not Nesterov momentum.\n\nThe centered version additionally maintains a moving average of the gradients, and uses that average to estimate the variance:\n\n$$mean_grad_t = rho * mean_grad_{t-1} + (1-rho) * gradient$$\n$$mean_square_t = rho * mean_square_{t-1} + (1-rho) * gradient ** 2$$\n$$mom_t = momentum * mom_{t-1} + learning_rate * gradient / sqrt(mean_square_t - mean_grad_t**2 + epsilon)$$\n$$variable_t := variable_{t-1} - mom_t$$\n\nReferences See ([pdf] http://www.cs.toronto.edu/~tijmen/csc321/slides/lecture_slides_lec6.pdf).\n\n#### Args:\n\n• learning_rate: A Tensor or a floating point value. The learning rate.\n• rho: Discounting factor for the history/coming gradient\n• momentum: A scalar tensor.\n• epsilon: Small value to avoid zero denominator.\n• centered: If True, gradients are normalized by the estimated variance of the gradient; if False, by the uncentered second moment. Setting this to True may help with training, but is slightly more expensive in terms of computation and memory. Defaults to False.\n• name: Optional name prefix for the operations created when applying gradients. Defaults to \"RMSprop\". @compatibility(eager) When eager execution is enabled, learning_rate, decay, momentum, and epsilon can each be a callable that takes no arguments and returns the actual value to use. This can be useful for changing these values across different invocations of optimizer functions. @end_compatibility\n• **kwargs: keyword arguments. Allowed to be {clipnorm, clipvalue, lr, decay}. clipnorm is clip gradients by norm; clipvalue is clip gradients by value, decay is included for backward compatibility to allow time inverse decay of learning rate. lr is included for backward compatibility, recommended to use learning_rate instead.\n\n#### Attributes:\n\n• iterations: Variable. The number of training steps this Optimizer has run.\n• weights: Returns variables of this Optimizer based on the order created.\n\n## Methods\n\n### add_slot\n\nView source\n\nadd_slot(\nvar, slot_name, initializer='zeros'\n)\n\n\nAdd a new slot variable for var.\n\n### add_weight\n\nView source\n\nadd_weight(\nname, shape, dtype=None, initializer='zeros', trainable=None,\nsynchronization=tf.VariableSynchronization.AUTO,\naggregation=tf.compat.v1.VariableAggregation.NONE\n)\n\n\n### apply_gradients\n\nView source\n\napply_gradients(\n)\n\n\nThis is the second part of minimize(). It returns an Operation that applies gradients.\n\n#### Args:\n\n• grads_and_vars: List of (gradient, variable) pairs.\n• name: Optional name for the returned operation. Default to the name passed to the Optimizer constructor.\n\n#### Returns:\n\nAn Operation that applies the specified gradients. If global_step was not None, that operation also increments global_step.\n\n#### Raises:\n\n• TypeError: If grads_and_vars is malformed.\n• ValueError: If none of the variables have gradients.\n\n### from_config\n\nView source\n\n@classmethod\nfrom_config(\nconfig, custom_objects=None\n)\n\n\nCreates an optimizer from its config.\n\nThis method is the reverse of get_config, capable of instantiating the same optimizer from the config dictionary.\n\n#### Arguments:\n\n• config: A Python dictionary, typically the output of get_config.\n• custom_objects: A Python dictionary mapping names to additional Python objects used to create this optimizer, such as a function used for a hyperparameter.\n\n#### Returns:\n\nAn optimizer instance.\n\n### get_config\n\nView source\n\nget_config()\n\n\nReturns the config of the optimimizer.\n\nAn optimizer config is a Python dictionary (serializable) containing the configuration of an optimizer. The same optimizer can be reinstantiated later (without any saved state) from this configuration.\n\n#### Returns:\n\nPython dictionary.\n\n### get_gradients\n\nView source\n\nget_gradients(\nloss, params\n)\n\n\nReturns gradients of loss with respect to params.\n\n#### Arguments:\n\n• loss: Loss tensor.\n• params: List of variables.\n\n#### Raises:\n\n• ValueError: In case any gradient cannot be computed (e.g. if gradient function not implemented).\n\n### get_slot\n\nView source\n\nget_slot(\nvar, slot_name\n)\n\n\n### get_slot_names\n\nView source\n\nget_slot_names()\n\n\nA list of names for this optimizer's slots.\n\n### get_updates\n\nView source\n\nget_updates(\nloss, params\n)\n\n\n### get_weights\n\nView source\n\nget_weights()\n\n\n### minimize\n\nView source\n\nminimize(\n)\n\n\nMinimize loss by updating var_list.\n\nThis method simply computes gradient using tf.GradientTape and calls apply_gradients(). If you want to process the gradient before applying then call tf.GradientTape and apply_gradients() explicitly instead of using this function.\n\n#### Args:\n\n• loss: A callable taking no arguments which returns the value to minimize.\n• var_list: list or tuple of Variable objects to update to minimize loss, or a callable returning the list or tuple of Variable objects. Use callable when the variable list would otherwise be incomplete before minimize since the variables are created at the first time loss is called.\n• grad_loss: Optional. A Tensor holding the gradient computed for loss.\n• name: Optional name for the returned operation.\n\n#### Returns:\n\nAn Operation that updates the variables in var_list. If global_step was not None, that operation also increments global_step.\n\n#### Raises:\n\n• ValueError: If some of the variables are not Variable objects.\n\n### set_weights\n\nView source\n\nset_weights(\nweights\n)\n\n\n### variables\n\nView source\n\nvariables()\n\n\nReturns variables of this Optimizer based on the order created."
]
| [
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.60757774,"math_prob":0.76216954,"size":5360,"snap":"2020-24-2020-29","text_gpt3_token_len":1187,"char_repetition_ratio":0.13592233,"word_repetition_ratio":0.027739251,"special_character_ratio":0.19794776,"punctuation_ratio":0.18201755,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9879195,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-11T13:21:25Z\",\"WARC-Record-ID\":\"<urn:uuid:eae3ec6e-8b71-43bb-a067-ccd6a850416f>\",\"Content-Length\":\"719841\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f658e5fe-088d-4000-9313-8421b8ee71f8>\",\"WARC-Concurrent-To\":\"<urn:uuid:2119f640-55fe-46ce-b7f4-b541f9f1ff70>\",\"WARC-IP-Address\":\"172.217.9.206\",\"WARC-Target-URI\":\"https://www.tensorflow.org/versions/r2.0/api_docs/python/tf/keras/optimizers/RMSprop\",\"WARC-Payload-Digest\":\"sha1:V6W4XYL7NBWMTR2JYPWO7SBIKQVM6Z43\",\"WARC-Block-Digest\":\"sha1:KFUTIA2P65FRYYEXJPM22CLHYTMTU7OF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655933254.67_warc_CC-MAIN-20200711130351-20200711160351-00052.warc.gz\"}"} |
https://resources.wolframcloud.com/FunctionRepository/resources/CombinatorFixedPoint | [
"#",
null,
"Function Repository Resource:\n\n# CombinatorFixedPoint\n\nEvolve a combinator expression to its fixed point based on defined rules and evaluation scheme\n\nContributed by: Wolfram Research\n ResourceFunction[\"CombinatorFixedPoint\"][cmb] evolves combinator expression cmb to its fixed point using S and K combinator transformation rules and single leftmost outermost updating order. ResourceFunction[\"CombinatorFixedPoint\"][cmb,scheme] evolves cmb to its fixed point using S and K combinator transformation rules in matches found by scheme at each step. ResourceFunction[\"CombinatorFixedPoint\"][rules, cmb,scheme] evolves cmb to its fixed point using rules in matches found by scheme at each step.\n\n## Details and Options\n\nThe input cmb must be given as a combinator expression in nested bracket form.\nIf scheme is specified, direction (\"Leftmost\", \"Rightmost\"), depth order (\"Outermost\", \"Innermost\") and number of matches (integer number or Infinity) must be specified, in order of their priority as sorting criteria.\nThe combinator transformations of interest should be given as a list of Rule or RuleDelayed elements in rules.\nResourceFunction[\"CombinatorFixedPoint\"] takes the following options:\n maximum steps to take in evolving cmb \"MaxSize\" maximum leaf count allowed for evolution of cmb \"SKGlyphs\" symbols used to specify default transformation rules\n\n## Examples\n\n### Basic Examples (2)\n\nCalculate the fixed point of a combinator expression using the default single leftmost outermost update with S and K combinator transformation rules:\n\n In:=",
null,
"Out=",
null,
"Specify the update scheme and only use the K combinator transformation rule to evolve the combinator expression to its fixed point:\n\n In:=",
null,
"Out=",
null,
"### Properties and Relations (2)\n\nCalculate the fixed point of a combinator expression using the default single leftmost outermost update with S and K combinator transformation rules:\n\n In:=",
null,
"Out=",
null,
"One can replicate this result with the use of ReplaceAll in FixedPoint:\n\n In:=",
null,
"Out=",
null,
"### Possible Issues (1)\n\nThe \"MaxSize\" and \"MaxSteps\" options can be useful for avoiding excessively long combinator evolutions, but will throw failures if their values are surpassed:\n\n In:=",
null,
"Out=",
null,
"In:=",
null,
"Out=",
null,
""
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https://readpaper.com/paper/4823094810340491265 | [
"This website requires JavaScript.",
null,
"",
null,
"# Local Computation Algorithms for Maximum Matching: New Lower Bounds\n\nNov 2023\n0被引用\n0笔记\n\nWe study local computation algorithms (LCA) for maximum matching. An LCA does not return its output entirely, but reveals parts of it upon query. For matchings, each query is a vertex $v$; the LCA should return whether $v$ is matched -- and if so to which neighbor -- while spending a small time per query. In this paper, we prove that any LCA that computes a matching that is at most an additive of $\\epsilon n$ smaller than the maximum matching in $n$-vertex graphs of maximum degree $\\Delta$ must take at least $\\Delta^{\\Omega(1/\\varepsilon)}$ time. This comes close to the existing upper bounds that take $(\\Delta/\\epsilon)^{O(1/\\epsilon^2)} polylog(n)$ time. In terms of sublinear time algorithms, our techniques imply that any algorithm that estimates the size of maximum matching up to an additive error of $\\epsilon n$ must take $\\Delta^{\\Omega(1/\\epsilon)}$ time. This negatively resolves a decade old open problem of the area (see Open Problem 39 of sublinear.info) on whether such estimates can be achieved in $poly(\\Delta/\\epsilon)$ time.\n\nAI理解论文&经典十问",
null,
""
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"https://nuxt.cdn.readpaper.com/nuxt/img/favicon-600.e6686658226798d9.png",
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"https://nuxt.cdn.readpaper.com/nuxt/img/pageLogo.bb6c11b514d99ac3.png",
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https://www.experts-exchange.com/questions/27479367/Convert-C-unsigned-char-array-and-uint32-t-pointers-to-real-values.html | [
"We help IT Professionals succeed at work.\n\n# Convert C unsigned char array and uint32_t pointers to real values\n\non\nHello,\n\nI'm attempting to write a C function to get the id and idx values from the test_id_t_ structure below. I only have access to the JNI pointer value and need to get the unsigned char array and the unsigned 4 byte Integer value from the below. The reason I only have the pointer value to the test_id_t_ structure is that I'm attempting to return this structure to Java via JNI (using SWIG). My C coding experience is very limited. Thanks!\n\n``````typedef unsigned char test_id_t;\ntypedef test_id_t id_t;\n\nstruct test_id_t_ {\nid_t id;\nuint32_t idx;\n};\ntypedef struct test_id_t_ test_cat_id_t;\n``````\n\nComment\nWatch Question\n\n## View Solution Only\n\nCommented:\nHi cgray1223,\n\nI'm not sure if I understand correctly, but I guess you mean you have a pointer of another type (i.e. void) pointing to such a test_id_t_ struct - is this correct? If so you can simply cast the pointer (of course you should be sure the pointer is valid and points to a memory block of sizeof(test_id_t_) bytes), i.e.:\n``````void* ptr_data = ...; // pointer to the data\ntest_id_t_* ptr_struct = (test_id_t_*)ptr_data;\n// now you can access the struct's members i.e. via ptr_struct->idx\n``````\nIf otherwise you need i.e. a void pointer to a _test_id_t struct you can simply cast it too, i.e.:\n``````test_id_t_ id;\nid.idx = ...;\nvoid* ptr = (void*)&id;\n``````\nIf you need to create a copy of a test_id_t_ and return a pointer you need to use malloc and memcpy and use casting as shown above to access the members.\n\nIf none of these is what you need please explain your problem a bit more detailed, best with some code sample where you want to convert those structs and pointers.\n\nHope that helps,\n\nZOPPO\n\nCommented:\nyes you're assumption was correct it was a void* return type from the function pointing at that structure."
]
| [
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https://deepai.org/publication/remote-sensing-image-classification-with-large-scale-gaussian-processes | [
"",
null,
"",
null,
"",
null,
"",
null,
"Remote Sensing Image Classification with Large Scale Gaussian Processes\n\nCurrent remote sensing image classification problems have to deal with an unprecedented amount of heterogeneous and complex data sources. Upcoming missions will soon provide large data streams that will make land cover/use classification difficult. Machine learning classifiers can help at this, and many methods are currently available. A popular kernel classifier is the Gaussian process classifier (GPC), since it approaches the classification problem with a solid probabilistic treatment, thus yielding confidence intervals for the predictions as well as very competitive results to state-of-the-art neural networks and support vector machines. However, its computational cost is prohibitive for large scale applications, and constitutes the main obstacle precluding wide adoption. This paper tackles this problem by introducing two novel efficient methodologies for Gaussian Process (GP) classification. We first include the standard random Fourier features approximation into GPC, which largely decreases its computational cost and permits large scale remote sensing image classification. In addition, we propose a model which avoids randomly sampling a number of Fourier frequencies, and alternatively learns the optimal ones within a variational Bayes approach. The performance of the proposed methods is illustrated in complex problems of cloud detection from multispectral imagery and infrared sounding data. Excellent empirical results support the proposal in both computational cost and accuracy.\n\nAuthors\n\n04/01/2021\n\nRemote Sensing Image Classification with the SEN12MS Dataset\n\nImage classification is one of the main drivers of the rapid development...\n12/07/2020\n\nRandomized kernels for large scale Earth observation applications\n\nDealing with land cover classification of the new image sources has also...\n05/30/2017\n\nRSI-CB: A Large Scale Remote Sensing Image Classification Benchmark via Crowdsource Data\n\nRemote sensing image classification is a fundamental task in remote sens...\n06/01/2016\n\nEfficiently Bounding Optimal Solutions after Small Data Modification in Large-Scale Empirical Risk Minimization\n\nWe study large-scale classification problems in changing environments wh...\n07/15/2016\n\nAutomatic Environmental Sound Recognition: Performance versus Computational Cost\n\nIn the context of the Internet of Things (IoT), sound sensing applicatio...\n07/31/2018\n\nRemote sensing image regression for heterogeneous change detection\n\nChange detection in heterogeneous multitemporal satellite images is an e...\n06/15/2017\n\nEffective Sequential Classifier Training for Multitemporal Remote Sensing Image Classification\n\nThe explosive availability of remote sensing images has challenged super...\nThis week in AI\n\nGet the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.\n\nI Introduction\n\n“… Nature almost surely operates by combining chance with necessity, randomness with determinism…”\n\n–Eric Chaisson, Epic of Evolution: Seven Ages of the Cosmos\n\nEarth-observation (EO) satellites provide a unique source of information to address some of the challenges of the Earth system science . Current EO applications for image classification have to deal with a huge amount of heterogeneous and complex data sources.\n\nThe super-spectral Copernicus Sentinels [2, 3], as well as the planned EnMAP , HyspIRI , PRISMA and FLEX , will soon provide unprecedented data streams to be analyzed. Very high resolution (VHR) sensors like Quickbird, Worldview-2 and the recent Worldview-3 also pose big challenges to data processing. The challenge is not only attached to optical sensors. Infrared sounders, like the Infrared Atmospheric Sounding Interferometer (IASI) sensor on board the MetOp satellite series, impose even larger constraints: the orbital period of Metop satellites (101 minutes), the large spectral resolution (8461 spectral channels between 645 cm and 2760 cm), and the spatial resolution (601530 samples) of the IASI instrument yield several hundreds of gigabytes of data to be processed daily. The IASI mission delivers approximately spectra per day, which gives a rate of about 29 Gbytes/day to be processed. EO radar images also increased in resolution, and current platforms such as ERS-1/2, ENVISAT, RadarSAT-1/2, TerraSAR-X, and Cosmo-SkyMED give raise to extremely fine resolution data that call for advanced scalable processing methods. Besides, we should not forget the availability of the extremely large remote sensing data archives111The Earth Observing System Data and Information System (EOSDIS) for example is managing around 4 terabytes daily, and the flow of data to users is about 20 terabytes daily. already collected by several past missions. In addition, we should be also prepared for the near future in diversity and complementarity of sensors222Follow the links for an up-to-date list of current ESA, EUMETSAT, JAXA, CNSA and NASA EO missions.. These large scale data problems require enhanced processing techniques that should be accurate, robust and fast. Standard classification algorithms cannot cope with this new scenario efficiently.\n\nIn the last decade, kernel methods have dominated the field of remote sensing image classification [10, 11]. In particular, a kernel method called support vector machine (SVM, [12, 13, 14, 15, 16]) was gradually introduced in the field, and quickly became a standard for image classification. Further SVM developments considered the simultaneous integration of spatial, spectral and temporal information [17, 18, 19, 20, 21], the richness of hyperspectral imagery [16, 22], and exploited the power of clusters of computers [23, 24]. Undoubtedly, kernel methods have been the most widely studied classifiers, and became the preferred choice for users and practitioners. However, they are still not widely adopted in real practice because of the high computational cost when dealing with large scale problems. Roughly speaking, given examples available for training, kernel machines need to store kernel matrices of size , and to process them using standard linear algebra tools (matrix inversion, factorization, eigen-decomposition, etc.) that typically scale cubically, . This is an important constraint that hampers their applicability to large scale EO data processing.\n\nAn alternative kernel classifier to SVM is the Gaussian Process classifier (GPC) . GPC has appealing theoretical properties, as it approaches the classification problem with a solid probabilistic treatment, and very good performance in practice. The GPC method was originally introduced in the field of remote sensing in , where very good capabilities for land cover classification from multi/hyperspectral imagery were illustrated. Since then, GPC has been widely used in practice and extended to many settings: hyperspectral image classification , semantic annotation of high-resolution remote sensing images , change detection problems with semisupervised GPC \n\n, or classification of images with the help of user’s intervention in active learning schemes\n\n[30, 31]\n\n. Unfortunately, like any other kernel method, its computational cost is very large. This is probably the reason why GPC has not yet been widely adopted by the geoscience and remote sensing community in large scale classification scenarios, despite its powerful theoretical background and excellent performance in practice.\n\nIn GP for classification we face two main problems. First, the non-conjugate observation model for classification (usually based on the sigmoid, probit, or Heaviside step functions) renders the calculation of the marginal distribution needed for inference impossible. The involved integrals are not analytically tractable, so one has to resort to numerical methods or approximations \n\n. One could rely on Markov Chain Monte Carlo (MCMC) methods, but they are computationally far too expensive. By assuming a Gaussian approximation to the posterior of the latent variables, one can use the Laplace approximation (LA) and the (more accurate) expectation propagation (EP)\n\n[32, 33]. The observation model can also be bounded, leading to the variational inference approach that we use in this paper. Once the non-conjugacy of the observation model has been solved, the second problem is the inversion of huge matrices, which yields the unbearable complexity. Notice that this is the only difficulty that appears when GP is used for regression, where the observation model can be analytically integrated out. This efficiency problem could be addressed with recent Sparse GP approximations based on inducing points and approximate inference \n\n, but they come at the price of a huge number of parameters to estimate.\n\nIn this paper, we introduce two alternative pathways to perform large scale remote sensing image classification with GPC. First, following the ideas in , we approximate the squared exponential (SE) kernel matrix of GPC by a linear one based on projections over a reduced set of random Fourier features (RFF). This novel method is referred to as RFF-GPC. It allows us to work in the primal space of features, which significantly reduces the computational cost of large scale applications. In fact, a recent similar approach allows for using millions of examples in SVM-based land cover classification and regression problems . The solid theoretical ground and the good empirical RFF-GPC performance make it a very useful method to tackle large scale problems in Earth observation. However, RFF-GPC can only approximate (theoretically and in practice) a predefined kernel (the SE in this work), and the approximation does not necessarily lead to a discriminative kernel. These shortcomings motivate our second methodological proposal: we introduce a novel approach to avoid randomly sampling a number of Fourier frequencies, and instead we propose learning the optimal ones within the variational approach. Therefore, Fourier frequencies are no longer randomly sampled and fixed, but latent variables estimated directly from data via variational inference. We refer to this method as VFF-GPC (Variational Fourier Features). The performance of RFF-GPC and VFF-GPC is illustrated in large and medium size real-world remote sensing image classification problems: (1) classification of clouds over landmarks from a long time series of Seviri/MSG (Spinning Enhanced Visible and Infrared Imager, Meteosat Second Generation) remote sensing images, and (2) cloud detection using IASI and AVHRR (Advanced Very High Resolution Radiometer) infrared sounding data, respectively. Excellent empirical results support the proposed large scale methods in both accuracy and computational efficiency. In particular, the extraordinary performance of VFF-GPC in the medium size data set justifies its use not only as a large scale method, but also as a general-purpose and scalable classification tool capable of learning an appropriate discriminative kernel.\n\nThe remainder of the paper is organized as follows. Section II reviews the RFF approximation and introduces it into GPC, deriving RFF-GPC and the more sophisticated VFF-GPC. Section III introduces the two real-world remote sensing data sets used for the experimental validation. Section IV presents the experimental results comparing the two proposed methods and standard GPC in terms of accuracy and efficiency. Section V concludes the paper with some remarks and future outlook.\n\nIi Large Scale Gaussian Process Classification\n\nGaussian Processes (GP) is a probabilistic state-of-the-art model for regression and classification tasks. In the geostatistic community, GP for regression is usually referred to as kriging. For input-output data pairs , a GP models the underlying dependence from a function-space perspective, i.e. introducing latent variables\n\nthat jointly follow a normal distribution\n\n. The kernel function encodes the sort of functions favored, and is the so-called kernel matrix. The observation model of the output given the latent variable depends on the problem at hand. In binary classification (i.e. when\n\n), the (non-conjugate) logistic observation model is widely used. It is given by the sigmoid function as\n\n.\n\nIi-a Random Fourier Features\n\nThe main issue with large scale applications of GP is its cost at the training phase, which comes from the kernel matrix inversion. The work presents a general methodology (based on Bochner’s theorem ) to approximate any positive-definite shift-invariant kernel by a linear one. This is achieved by explicitly projecting the original -dimensional data onto random Fourier features , whose linear kernel approximates . This linearity will enable us to work in the primal space of features and substitute matrix inversions by ones, resulting in a total computational cost. In large-scale applications, one can set a , and thus the obtained complexity represents an important benefit over the original . Moreover, the complexity at test is also reduced from to , even becoming independent on .\n\nIn this work we use the well-known SE (or Gaussian) kernel . Following the methodology in , this kernel can be linearly approximated as\n\n k(x,x′)≈kL(x,x′)=γ⋅z(x)⊺z(x′), (1)\n\nwhere\n\n z(x)⊺=D−1/2⋅(cos(w⊺1x),sin(w⊺1x),……,cos(w⊺Dx),sin(w⊺Dx))∈R2D, (2)\n\nand the Fourier frequencies must be sampled from a normal distribution . As explained in [36, Claim 1], this approximation exponentially improves with the number of Fourier frequencies used (and also exponentially worsens with , the original dimension of ). However, obviously, increasing in our methods will go at the cost of increasing the and complexities. Other kernels different from the SE one could be used, but that would imply sampling from a different distribution.\n\nOur novel RFF-GPC method considers a standard Bayesian linear model over these new features . Such a linear model corresponds to GP classification with the linear kernel [40, Chapter 6]. Since approximates the SE kernel , our RFF-GPC constitutes an approximation to GP classification with SE kernel. However, RFF-GPC is well suited for large scale applications, as it works in the primal space of features and thus presents a (resp. ) train (resp. test) complexity.\n\nNotice that RFF-GPC needs to sample the Fourier frequencies from\n\nfrom the beginning, whereas hyperparameters\n\nand must be estimated during the learning process (just as in standard GP classification). In order to uncouple and , in the sequel we consider the equivalent features\n\n z(x|σ,W)⊺=D−1/2⋅(cos(σ−1w⊺1x),sin(σ−1w⊺1x),……,cos(σ−1w⊺Dx),sin(σ−1w⊺Dx)), (3)\n\nwith now sampled from and fixed. Notice that we have collectively denoted .\n\nAt this point, it is natural to consider other possibilities for the Fourier frequencies , rather than just randomly sample and fix them from the beginning. The proposed VFF-GPC model treats them as hyperparameters to be estimated (so as to maximize the likelihood of the observed data), just like and in the case of RFF-GPC. This makes VFF-GPC more expressive, flexible, and tailored to the data, although it may no longer constitute an approximation to the SE kernel (for which the must be normally distributed). More specifically, VFF-GPC does start with an approximated SE kernel (since the are initialized with a normal distribution), but the maximum a posteriori (MAP) optimization on makes it learn a new kernel which may no longer approximate a SE one. Therefore, for VFF-GPC we also use as in eq. (3), with now both and to be estimated. Interestingly, VFF-GPC extends the Sparse Spectrum Gaussian Process model originally introduced for regression in , to GP classification.\n\nMore specifically, the authors there also sparsify the SE kernel by working on the primal space of and Fourier features, see [41, Equation 5]. However, our classification setting involves a sigmoid-based (logistic) observation model for the output given the latent variable (see the next eq. (4)), whereas in regression this is just given by a normal distribution. Therefore, VFF-GPC needs to additionally deal with the non-conjugacy of the sigmoid, which motivates the variational bound of eq. (6) and the consequent variational inference procedure described in Section II-C. It is interesting to realize that VFF-GPC is introduced here as a natural extension of RFF-GPC, whereas there is not a regression analogous for RFF-GPC in .\n\nAnother possibility for the Fourier frequencies would be to estimate them (just as in VFF-GPC) but considering alternative prior distributions (which means utilizing alternative kernels). Moreover, instead of maximum a posteriori inference, we could address the marginalization of the Fourier frequencies . Alternatively, to promote sparsity, the use of Gaussian Scale Models (GSM) could also be investigated. These possibilities will be explored in future work, and here we will concentrate on RFF-GPC and VFF-GPC.\n\nIi-B Models formulation\n\nAs anticipated in previous section, RFF-GPC and VFF-GPC are standard Bayesian linear models working on the explicitly mapped features of eq. (3). In the case of RFF-GPC, is sampled from at the beginning and fixed, with to be estimated. In the case of VFF-GPC, both and are estimated, with a prior over . In order to derive both methods in a unified way, will denote for RFF-GPC and both for VFF-GPC.\n\nSince we are dealing with binary classification, we consider the standard logistic observation model\n\n p(y=1|β,Φ,x)=ψ(β⊺z)=(1+exp(−β⊺z))−1, (4)\n\nwhere . For the weights we utilize the prior normal distribution , with to be estimated, see eq. (1).\n\nFor an observed dataset , the joint p.d.f. reads\n\n p(y,β|Φ,γ,X) =p(y|β,Φ,X)p(β|γ) =(n∏i=1p(yi|β,Φ,xi))p(β|γ), (5)\n\nwhere we collectively denote and . For the sake of brevity, from now on we will systematically omit the conditioning on .\n\nIi-C Variational inference\n\nGiven the observed dataset , in this section we seek point estimates of and by maximizing the marginal likelihood (in VFF-GPC, the additional prior yields maximum a posteriori inference, instead of maximum likelihood one, for ). After that, we obtain (an approximation to) the posterior distribution . Due to the non-conjugate observation model, the required integrals will be mathematically intractable, and we will resort to the variational inference approximation [40, Section 10.6].\n\nFirst, notice that integrating out in eq. (5) is not analytically possible due to the sigmoid functions in the observation model . To overcome this problem, we use the variational bound\n\n log(1+ex)≤λ(ξ)(x2−ξ2)+x−ξ2+log(1+eξ), (6)\n\nwhich is true for any real numbers and where [40, Section 10.6]. Applying it to every factor of , we have the following lower bound\n\n p(y|β,Φ)≥exp(−β⊺Z⊺ΛZβ+v⊺Zβ)⋅C(ξ). (7)\n\nHere we write for the projected-data matrix (which depends on ), is the diagonal matrix , , and the term only depends on . The key is that this lower bound for is conjugate with the normal prior (since it is the exponential of a quadratic function on ), and thus it allows us integrating out in eq (5). In exchange, we have introduced additional hyperparameters that will need to be estimated along with and .\n\nTherefore, substituting for its bound, eq. (5) can be lower bounded as\n\n p(y,β|Φ,γ)≥F(y,Φ,γ,ξ)⋅N(β|μ,Σ), (8)\n\nwhere we have denoted\n\n F(y,Φ,γ,ξ)=C(ξ)∣∣γ−1Σ∣∣1/2exp(12μ⊺Σ−1μ), Σ=(Z⊺(2Λ)Z+γ−1I)−1,μ=ΣZ⊺v. (9)\n\nNow it is clear that we can marginalize out , and iteratively estimate the optimal values of , and as those that maximize (or equivalently ) for the observed . Starting at , , and , we can calculate , , and for . In the case of , we make use of the local maximum condition . From there, it is not difficult to prove that the optimal value satisfies \n\n ξ(k+1)=√diag(Z(k)Σ(k)(Z(k))⊺)+(Z(k)μ(k))2, (10)\n\nwhere the square and square root of a vector are understood as element-wise. In the case of and , we use nonlinear conjugate gradient methods and obtain (notice that for VFF-GPC we can collapse and , removing the prior on )\n\n (Φ(k+1),γ(k+1))=argmaxΦ,γ{−log∣∣2γZ⊺Λ(k+1)Z+I∣∣ +v⊺Z(2Z⊺Λ(k+1)Z+γ−1I)−1Z⊺v}. (11)\n\nOnce the hyperparameters , , and have been estimated by , , and respectively, we need to compute the posterior . As before, this is mathematically intractable due to the sigmoids in the observation model. Therefore, we again resort to the variational bound in eq. (8) to get an optimal approximation to the posterior . Namely, we do it by minimizing (an upper bound of) the KL divergence between both distributions:\n\n =logp(y|^Φ,^γ)+∫^p(β)log^p(β)p(y,β|^Φ,^γ)dβ\n\nThus, the minimum is reached for , with and calculated in eq. (9) using , , and .\n\nIn summary, at training time, our methods RFF-GPC and VFF-GPC run iteratively until convergence of the hyperparameters , , and to their optimal values , , and (see Algorithm 1). The computations involved there suppose a computational complexity of (which equals when ), whereas standard GPC scales as . At test time, the probability of class for a previously unseen instance is:\n\n p(y∗=1) ≈∫p(y=1|β,^Φ,x∗)^p(β)dβ≈ ≈ψ(^z⊺∗^μ⋅(1+(π/8)^z⊺∗^Σ^z∗)−1/2), (12)\n\nwith being the sigmoid function. Whereas GPC presents a computational cost of for each test instance, eq. (II-C) implies a complexity of in the case of our methods. In particular, notice that this is independent on the number of training instances. These significant reductions in computational cost (both at training and test) make our proposal suitable for large scale and real-time applications in general, and in EO applications in particular.\n\nFinally, regarding the convergence of the proposed methods, we cannot theoretically guarantee the convergence to a global optimum (only a local one), since we are using conjugate gradient methods to solve the non-convex optimization problem in eq. (II-C). However, from a practical viewpoint, we have experimentally checked that both methods have a satisfactory similar convergence pattern. Namely, in the first iterations, the hyperparameters experiment more pronounced changes, widely exploring the hyperparameters space. Then, once they reach a local optimum vicinity, these variations become smaller. Eventually, the hyperparameters values hardly change and the stop criterion is satisfied.\n\nIii Data Collection and Preprocessing\n\nThis section introduces the datasets used for comparison purposes in the experiments. We considered (1) a continuous year of MSG data involving several hundred thousands of labeled pixels for cloud classification; and (2) a medium-size manually labeled dataset used to create the operational IASI cloud mask.\n\nIii-a Cloud detection over landmarks with Seviri/MSG\n\nWe focus on the problem of cloud identification over landmarks using Seviri MSG data. This satellite mission constitutes a fundamental tool for weather forecasting, providing images of the full Earth disc every 15 minutes. Matching the landmarks accurately is of paramount importance in image navigation and registration models and geometric quality assessment in the Level 1 instrument processing chain. Detection of clouds over landmarks is an essential step in the MSG processing chain, as undetected clouds are one of the most significant sources of error in landmark matching (see Fig. 1).",
null,
"Fig. 1: Landmarks are essential in image registration and geometric quality assessment. Misclassification of cloud contamination in landmarks degrades the correlation matching, which is a cornerstone for the image navigation and registration algorithms.\n\nThe dataset used in the experiments was provided by EUMETSAT, and contains Seviri/MSG Level 1.5 acquisitions for 200 landmarks of variable size for a whole year (2010). Landmarks mainly cover coastlines, islands, or inland waters. We selected all multispectral images from a particular landmark location, Dakhla (Western Sahara), which involves 35,040 MSG acquisitions with a fixed resolution of pixels. In addition, Level 2 cloud products were provided for each landmark observation, so the Level 2 cloud mask is used as the best available ‘ground truth’ to validate the results. We framed the problem for this particular landmark as a pixel-wise classification one.\n\nA total amount of features were extracted from the images, involving band ratios, spatial, contextual and morphological features, and discriminative cloud detection scores. In particular, we considered: 7 channels converted to top of atmosphere (ToA) reflectance (R1, R2, R3, R4) and brightness temperature (BT7, BT9, BT10), 3 band ratios, and 6 spatial features. On the one hand, the three informative band ratios were: (i) a cloud detection ratio, ; (ii) a snow index, ; and (iii) the NDVI, . On the other hand, the six spatial features were obtained by applying average filters of sizes and\n\n, as well as a standard deviation filter of size\n\n, on both bands R1 and BT9.\n\nBased on previous studies [44, 45], and in order to simplify the classification task, the different illumination conditions (and hence difficulty) over the landmarks are studied by splitting the day into four ranges (sub-problems) according to the solar zenith angle (SZA) values: high (SZASZA), mid (SZASZA°), low (80°SZA90°), and night (SZA90°). Therefore, different classifiers are developed for each SZA range.\n\nThe final amount of pixels available for each illumination condition is for high, mid, and night, and for low. Moreover, each problem has different dimensionality: all the features were used for the three daylight problems, and was used for the night one (some bands and ratios are meaningless at night).\n\nIii-B Cloud detection with the IASI/AVHRR data\n\nThe IAVISA dataset is part of the study “IASI/AVHRR Visual Scenes Analysis and Cloud Detection” (http://www.brockmann-consult.de/iavisa-info-web/), whose aim is to improve the IASI cloud detection by optimizing the coefficients used for a predefined set of cloud tests performed in the IASI Level 2 processing. The dataset was derived by visual analysis of globally distributed data, and served as input for the optimization and validation of the IASI cloud detection. Each collected IASI sample was classified concerning its cloudiness based on the visual inspection of the AVHRR Level 1B inside the IASI footprint. Each sample classifies a single IASI instantaneous field of view (IFOV) as being cloud-free (clear sky, 28%), partly cloudy low (26%), partly cloudy high (26%), or cloudy (20%). For the sake of simplicity, here we focus on discriminating between cloudy and cloud-free pixels.\n\nIn order to ensure the representativeness of the dataset for the natural variability of clouds, labeling further considered additional conditions depending on: 1) the surface type (see Table I), 2) the climate zone (Köppen classification over land, geographical bands over sea), 3) the season and 4) day/night discrimination. First, the surface type database used as ancillary information was the IGBP (International Geosphere-Biosphere Programme) scene types in the CERES/SARB (Clouds and the Earth’s Radiant Energy System, Surface and Atmospheric Radiation Budget) surface map. The 18-class map was used to identify surface properties of a given region. The distribution of the surface types in the map is given in Table I, showing the even distribution of clouds across land cover types which ensures representativeness (natural variability) of the database. Second, the different climate zones were sampled as follows: tropical (), dry (), temperate (), cold (), and polar zones (). Third, seasonality was also taken into account, and yielded the following distribution: Spring (), Summer (), Autumn (), and Winter (). The global sampling and some cloudy and cloud-free chips are shown in Fig. 2. The final database consists of instances and original features, which have been summarized to",
null,
"Fig. 2: Global sample coverage for all seasons, times of day, and cloud cases (left), and examples of cloud-free and cloudy samples (right).\n\nIv Experiments\n\nIn this section, we empirically demonstrate the performance of the proposed methods in the two real problems described above. Moreover, we carry out an exhaustive comparison to GPC with SE kernel (in the sequel, GPC for brevity).\n\nIn order to provide a deeper understanding of our methods, different values of (number of Fourier frequencies) will be used. Different sizes of the training dataset will be also considered, in order to analyze the scalability of the methods and to explore the trade-off between accuracy and computational cost. When increasing (respectively, ), we will add training instances (respectively, Fourier frequencies) to those already used for lower values of (respectively, ). We provide numerical (in terms of recognition rate), computational (training and test times), and visual (by inspecting classification maps) assessments.\n\nIv-a Cloud detection in the landmarks dataset\n\nIn this large scale problem, the number of training examples is selected as for RFF-GPC and VFF-GPC, and for GPC. Notice that the improvement at training computational cost ( for the proposed methods against for GPC) enables us to consider much greater training datasets for our methods. In fact, as we will see in Figure 3, even with RFF-GPC and VFF-GPC are computationally cheaper than GPC with . Indeed, higher values of are not considered for GPC to avoid exceeding the (already expensive) seconds of training CPU time needed with just . Regarding the number of Fourier frequencies, we use .\n\nThe experimental results, which include predictive performance (test overall accuracy), training CPU time, and test CPU time, are shown in Figure 3. Every single value is the average of five independent runs under the same setting. Namely, for each illumination condition, a test dataset is fixed and five different balanced training datasets are defined with the remaining data. Notice that a general first observation across Figure 3 suggests that higher accuracies are obtained for higher illumination conditions (SZA).",
null,
"Fig. 3: Experimental results for LANDMARKS dataset. From top to bottom, the rows correspond with the described high, mid, low, and night illumination conditions. For each row, the first column shows the test overall accuracy (OA) of RFF-GPC, VFF-GPC, and GPC for the different values of n (number of training examples) and D (number of Fourier frequencies) considered. The second column is analogous, but displays the CPU time (in seconds) needed to train each method (instead of the test OA). The third column summarizes the two previous ones, providing a trade-off between test OA and training CPU time. The last column is analogous to the first and second ones, but showing the CPU time used at the test step (production time). The legend for the second and fourth columns is the same as in the first one. However, notice that in the third column plots the GPC lines degenerate into single points (since GPC does not depend on D). In both legends, the numbers indicate the amount n of training examples used, which determines the width/size of the lines/points too. As further explained in the main text, shown results are the mean over five independent runs.\n\nFigure 3 reveals an overwhelming superiority of RFF-GPC and VFF-GPC over standard GPC: our proposed methods achieve a higher predictive performance while investing substantially lower training CPU time. This is very clear from the third column plots, where for any blue point we can find orange and yellow points which are placed more north-west (i.e. higher test OA and lower CPU training time). Furthermore, the fourth column shows an equally extraordinary reduction in test CPU time (production time), where the proposed methods are more than times faster than GPC. In particular, this makes RFF-GPC and VFF-GPC better suited than standard GPC for real-time EO applications.\n\nRegarding the practical differences between RFF-GPC and VFF-GPC, we observe that RFF-GPC is faster (at training) whereas VFF-GPC is more accurate. This is a natural consequence of their theoretical formulations: the estimation of the Fourier frequencies in VFF-GPC makes it more flexible and expressive, but involves a heavier training. Therefore, in this particular problem, the final practical choice between the two proposed methods would depend on the relative importance that the user assigns to test accuracy (where VFF-GPC stands out) and training cost (where RFF-GPC does so). In terms of test cost, both methods are very similar, as expected from the identical theoretical test complexity. The independence of this quantity on is also intuitively reflected in the experiments, with all the RFF-GPC and VFF-GPC lines collapsing onto a single one in the fourth column plots of Figure 3.\n\nAt this point, it is worth to analyze a bit further the role of in the performance of our methods. Recall (Section II-A) that RFF-GPC is an approximation to GPC, with an error that exponentially decreases with the ratio between the dimensions of the projected Fourier features space and the original one. Therefore, it is theoretically expected that the performance of RFF-GPC increases with , becoming equivalent to GPC when . Actually, this is supported by the first column of Figure 3. Moreover, since our problem here presents a low ( for high, mid, and low, and for night), it is natural that RFF-GPC with just already gets very similar (even better in some cases) results to standard GPC with the same (yet far much faster, compare GPC and RFF/VFF for ). In the case of VFF-GPC, where the Fourier frequencies are model parameters to be estimated, the number is directly related to the complexity of the model. Therefore, its increase should not always mean a higher performance in test OA, since large values may provoke over-fitting to the training dataset (this will be clear in the next dataset, whereas it does not occur in LANDMARKS). Furthermore, unlike RFF-GPC, the performance of VFF-GPC is not directly affected by .\n\nIt is also reasonable to expect that both test OA and training CPU time increase with the training dataset size . More specifically, and from a practical perspective in which the computational resources are finite, the first column in Figure 3 shows that test OA becomes stalled when only one of or increases. However, greater improvements in test OA are achieved when and are jointly increased. Notice that this is also justifiable from a theoretical viewpoint: the higher the dimensionality of the projected Fourier features space (which is ), the larger number of examples are required to identify the separation between classes.\n\nIv-B Cloud detection with the IAVISA dataset\n\nAs explained in Section III-B, this problem involves a total amount of instances. We performed five-fold cross-validation, which produces five pairs of training/test datasets with (approximately) instances each. Results are then averaged. Since RFF-GPC and VFF-GPC are conceived for large-scale applications (they scale linearly with , recall their training cost), they will not be utilized with values of lower than this training dataset size of (even GPC is able to cope with this size). Indeed, in the case of GPC we use the values . Regarding the number of Fourier frequencies , we consider the grid . The experimental results, which include the same metrics as those used for the previous problem on landmarks, are shown in Figure 4.",
null,
"Fig. 4: Experimental results for the IAVISA dataset. From left to right and top to bottom, the first plot shows the test overall accuracy (OA) of RFF-GPC, VFF-GPC, and GPC for the different values of n (number of training examples) and D (number of Fourier frequencies) considered. The second column is analogous, but displays the CPU time needed to train each method (instead of the test OA). The third column summarizes the two previous ones, providing a trade-off between test OA and training CPU time. The last column is analogous to the first and second ones, but showing the CPU time used at the test step. The legend for second and fourth plots is the same as the one in the first plot. However, in the third plot the GPC lines degenerate into single points (since GPC does not depend on D). In both legends, the numbers indicate the amount n of training examples used, which determines the width/size of the lines/points too (ALL means the whole training dataset, i.e. n≈20000). As explained in the main text, the results are the mean over five independent runs.\n\nIn this case, we again observe a clear outperformance of VFF-GPC against GPC: it achieves higher test OA while requiring less training and test CPU times. Moreover, the improvement in test OA is greater than , and train/test CPU times are around and times lower respectively. However, unlike in the previous problem of cloud detection over landmarks, RFF-GPC does not exhibit such a clear superiority over GPC in this application. Whereas it does drastically decrease the train/test CPU times, it is not able to reach the test OA of GPC with . Therefore, in practice, the optimal choice for this application is VFF-GPC. RFF-GPC would only be recommended if the training CPU time is a very strong limitation.\n\nThe main reason why RFF-GPC is not completely competitive in this problem is its theoretical scope: as an efficient approximation to GPC, it is conceived for large scale applications which are out of the reach of standard GPC. If the size of the problem allows for using GPC (as in this case), then RFF-GPC will only provide a more efficient alternative (less training and test CPU times), but its predictive performance will be always below that of GPC. Moreover, the difference in this performance is directly influenced by the original dimension of data (recall that the kernel approximation behind RFF-GPC exponentially degrades with , Section II-A). This is precisely a second hurdle that RFF-GPC finds in IAVISA: the high makes RFF-GPC with the full dataset be quite far from the corresponding GPC at predictive performance (test OA). In conclusion, the ideal setting for RFF-GPC is a large scale problem (high ) with few features (low ), precisely the opposite to the IAVISA dataset.\n\nInterestingly, VFF-GPC bypasses these limitations of RFF-GPC by learning a new kernel and not just approximating the SE one. First, VFF-GPC is not just a GP adaptation well-suited for large scale applications, but a general-purpose, expressive, and very competitive kernel-based classifier that scales well with the number of training instances. Second, as it does not rely on the kernel approximation, VFF-GPC is not affected by the original dimension of data. Both ideas are empirically supported by the results obtained in IAVISA.\n\nThe first plot of Figure 4 shows that the predictive performance of VFF-GPC does not necessarily improves by increasing . This is the expected behavior from the theoretical formulation of VFF-GPC, where the Fourier frequencies are parameters to be estimated. Thus, a higher amount of them confers VFF-GPC a greater flexibility to learn hidden patterns in the training dataset, but also the possibility to over-fit very particular structures of it which do not generalize to the test set. This is the classical problem of the model complexity in machine learning, and it is further illustrated in Figure 5. Together with the first plot in Figure 4, it shows the paradigmatic behaviour of train and test performance in presence of over-fitting: train OA grows with the model complexity (great flexibility allows for learning very particular structures of the training set, even reaching a of train OA), whereas test OA initially grows (the first patterns are part of the ground truth and thus general to the test set) but then goes down (when the learned information is too specific to the training set). Notice that this over-fitting phenomena did not occur at LANDMARKS, where test OA monotonically increased with . In addition to the different nature of the problems, the training dataset size plays a crucial role at this: smaller datasets (like IAVISA) are more prone to over-fitting than larger ones (LANDMARKS) under the same model complexity.\n\nFinally, it is worth noting that VFF-GPC achieves its maximum test OA when using just Fourier frequencies. This reflects (i) a not very sophisticated internal structure of the IAVISA dataset (since just directions are enough to correctly classify of the data), and (ii) the VFF-GPC capability to learn those discriminative directions from data. In particular, this shows that VFF-GPC can be used not only as a classifier, but also as a method that learns the most relevant discriminative directions in a dataset. Unfortunately, RFF-GPC is not able to benefit from these privileged directions that may exist in some datasets, since it randomly samples and fix the Fourier frequencies from the beginning.",
null,
"Fig. 5: Train OA in the IAVISA dataset for RFF-GPC, VFF-GPC, and GPC with different values of n (number of training examples) and D (number of Fourier frequencies). These results complement the first plot in Figure 4, showing that high values of D make VFF-GPC over-fit to the training dataset. The legend and its interpretation are the same as there.\n\nIv-C Explicit classification maps for cloud detection\n\nThe last two sections were dedicated to thoroughly analyze the performance of the proposed methods, empirically understand their behavior, weaknesses, and strengths, and compare them against GPC. In order to illustrate the explicit cloud detection behind the experiments, here we provide several explanatory classification maps obtained by the best model (in terms of predictive performance) for the LANDMARKS dataset: VFF-GPC with and .\n\nThe classification maps are obtained for the whole year 2010 at the Dakhla landmark, with a total of 34940 satellite acquisitions. The acquired window size is\n\npixels. Relying on the proposed feature extraction procedure, we trained the four necessary models (high, mid, low, night), and then proceed to predict over the whole available amount of chips acquired in the 2010 year.\n\n333A full video with all the classification maps is available at http://decsai.ugr.es/vip/software.html and http://isp.uv.es/code/vff.html. RFF-GPC and VFF-GPC codes are also provided.\n\nIn Figure 6, several chips are provided with the aim of illustrating different behaviors. In the first situation (first row), we can see a characteristic error of the L2 cloud mask, which sometimes tends to wrongly label the coastline pixels as cloudy. However, VFF-GPC leads to a better classification, identifying just one cloudy pixel and thus avoiding this negative coastline effect444As a clarification note, the coastline pixels were removed from the training dataset by applying a carefully designed morphological filter around coastlines.. In the second row, the visual channels show large clouds crossing the landmark. In the bottom-right of the image, a long cloud is unlabeled in the L2 cloud mask but correctly detected by VFF-GPC. While being formally accounted as an error, such discrepancy is actually positive for our method. Moreover, VFF-GPC shows an interesting cloud-sensitive behaviour at the top-left cloudy mass, identifying a larger cloudy area than that provided by EUMETSAT. This is a desirable propensity in cloud detection applications, where we prefer to identify larger clouds (and then thoroughly analyze them) rather than missing some of them. In the third row, the RGB channel allows for visually identifying three main cloudy masses at the landmark. The L2 mask poorly labels the central cloudy band, and does not detect the lower cloud. Both deficiencies are overcome by VFF-GPC. Finally, the fourth chip shows a huge cloudy mass that is undetected by the L2 mask but is correctly identified by VFF-GPC.\n\nTherefore, although VFF-GPC is trained with an imperfect ground truth, we observe that it is able to bypass some of these deficiencies, and exhibits a desirable cloud-sensitive behavior. This improvement can be also related to the particular design of the training datasets, splitting the problem into four different cases depending on the illumination conditions.",
null,
"Fig. 6: Explicit classification maps for the Dakhla landmark. The rows correspond with four different acquisitions. The first column shows the visible RGB channels (which are not informative for night acquisitions such us the first one), the second column is the infrared 10.8μm spectral band (very illustrative in night scenarios), the third column represents the ground truth obtained by EUMETSAT (the L2 cloud mask), and the last one is the VFF-GPC classification map. In the last two columns, the red color is used for cloudy pixels and blue for cloud-free ones.\n\nV Conclusions and Future Work\n\nWe presented two efficient approximations to Gaussian process classification to cope with big data classification problems in EO. The first one, RFF-GPC, performs standard GP classification by means of a fast approximation to the kernel (covariance) via random Fourier features. The advantage of the method is mainly computational, as the training cost is instead of the induced by the direct inversion of the kernel matrix (the test cost is also reduced to be independent on , from to ). The RFF method approximates the squared exponential (SE) covariance with Fourier features randomly sampled in the whole spectral domain. The solid theoretical grounds and good empirical performance makes it a very useful method to tackle large scale problems. Actually, the use of RFF has been exploited before in other settings, from classification with SVMs to regression with the KRR. However, we emphasize two main shortcomings. Firstly, the RFF approach can only approximate (theoretically and in practice) a predefined kernel (the SE one in this work). Secondly, by sampling the Fourier domain from a Gaussian, one has no control about the expressive power of the representation since some frequency components of the signal can be better represented than others. As a consequence, the approximated kernel may not have good discrimination capabilities. Noting these two problems, we proposed here our second methodology: a variational GP classifier (VFF-GPC) which goes one step beyond by optimizing over the Fourier frequencies. It is shown to be not just a GP adaptation well-suited for large scale applications, but a whole novel, general-purpose, and very competitive kernel-based classifier that scales well (linearly, as RFF-GPC) with the number of training instances.\n\nWe illustrated the performance of the algorithms in two real remote sensing problems of large and medium size. In the first case study, a challenging problem dealt with the identification of clouds over landmarks using Seviri/MSG imagery. The problem involved several hundred thousands data points for training the classifiers. In the second case study, we used the IAVISA dataset, which exploits IASI/AVHRR data to identify clouds with the IASI infrared sounding data. Compared to the original GPC, the experimental results show a high competitiveness in accuracy, a remarkable decrease in computational cost, and an excellent trade-off between both.\n\nThese results encourage us to expand the experimentation to additional problems, trying to exploit the demonstrated potential of VFF-GPC when dealing with any value of (training data set size) and (original dimension of the data). Other prior distributions and inference methods, as explained at the end of Section II-A, will be also explored in the future.\n\nReferences\n\n• M. Berger, J. Moreno, J. A. Johannessen, P. Levelt, and R. Hanssen, “ESA’s sentinel missions in support of earth system science,” Rem. Sens. Env., vol. 120, pp. 84–90, 2012.\n• M. Drusch, U. Del Bello, S. Carlier, O. Colin, V. Fernandez, F. Gascon, B. Hoersch, C. Isola, P. Laberinti, P. Martimort, A. Meygret, F. Spoto, O. Sy, F. Marchese, and P. Bargellini, “Sentinel-2: ESA’s Optical High-Resolution Mission for GMES Operational Services,” Rem. Sens. Env., vol. 120, pp. 25–36, 2012.\n• C. Donlon, B. Berruti, A. Buongiorno, M.-H. Ferreira, P. Féménias, J. Frerick, P. Goryl, U. Klein, H. Laur, C. Mavrocordatos, J. Nieke, H. Rebhan, B. Seitz, J. Stroede, and R. Sciarra, “The Global Monitoring for Environment and Security (GMES) Sentinel-3 mission,” Remote Sensing of Environment, vol. 120, pp. 37–57, 2012.\n• T. Stuffler, C. Kaufmann, S. Hofer, K. Farster, G. Schreier, A. Mueller, A. Eckardt, H. Bach, B. Penné, U. Benz, and R. Haydn, “The EnMAP hyperspectral imager-An advanced optical payload for future applications in Earth observation programmes,” Acta Astronautica, vol. 61, no. 1-6, pp. 115–120, 2007.\n• D. Roberts, D. Quattrochi, G. Hulley, S. Hook, and R. Green, “Synergies between VSWIR and TIR data for the urban environment: An evaluation of the potential for the Hyperspectral Infrared Imager (HyspIRI) Decadal Survey mission,” Rem. Sens. Env., vol. 117, pp. 83–101, 2012.\n• D. Labate, M. Ceccherini, A. Cisbani, V. De Cosmo, C. Galeazzi, L. Giunti, M. Melozzi, S. Pieraccini, and M. Stagi, “The PRISMA payload optomechanical design, a high performance instrument for a new hyperspectral mission,” Acta Astronautica, vol. 65, no. 9-10, pp. 1429–1436, 2009.\n• S. Kraft, U. Del Bello, M. Drusch, A. Gabriele, B. Harnisch, and J. Moreno, “On the demands on imaging spectrometry for the monitoring of global vegetation fluorescence from space,” in Proceedings of SPIE - The International Society for Optical Engineering, vol. 8870, 2013.\n• N. Longbotham, F. Pacifici, B. Baugh, and G. CampsValls, “Pre-launch assessment of worldview-3 information content,” Lausanne, Switzerland, June 2014, pp. 24–27.\n• B. Tournier, D. Blumstein, F. Cayla, , and G. Chalon, “IASI level 0 and 1 processing algorithms description,” in Proc. of ISTCXII Conference, 2002.\n• G. Camps-Valls and L. Bruzzone, Eds., Kernel methods for Remote Sensing Data Analysis. UK: Wiley & Sons, Dec 2009.\n• G. Camps-Valls, D. Tuia, L. Gómez-Chova, and J. Malo, Eds., Remote Sensing Image Processing. Morgan & Claypool, Sept 2011.\n• C. Huang, L. S. Davis, and J. R. G. Townshend, “An assessment of support vector machines for land cover classification,” Int. J. Rem. Sens., vol. 23, no. 4, pp. 725–749, 2002.\n• G. Camps-Valls, L. Gómez-Chova, J. Calpe, E. Soria, J. D. Martín, L. Alonso, and J. Moreno, “Robust support vector method for hyperspectral data classification and knowledge discovery,” IEEE Trans. Geosc. Rem. Sens., vol. 42, no. 7, pp. 1530–1542, Jul 2004.\n• F. Melgani and L. Bruzzone, “Classification of hyperspectral remote sensing images with support vector machines,” IEEE Trans. Geosci. Rem. Sens., vol. 42, no. 8, pp. 1778–1790, 2004.\n• G. M. Foody and J. Mathur, “A relative evaluation of multiclass image classification by support vector machines,” IEEE Trans. Geosci. Rem. Sens., pp. 1–9, Jul 2004.\n• G. Camps-Valls and L. Bruzzone, “Kernel-based methods for hyperspectral image classification,” IEEE Trans. Geosc. Rem. Sens., vol. 43, no. 6, pp. 1351–1362, Jun 2005.\n• J. Benediktsson, J. Palmason, and J. Sveinsson, “Classification of hyperspectral data from urban areas based on extended morphological profiles,” IEEE Trans. Geosc. Rem. Sens., vol. 43, pp. 480–490, 2005.\n• M. Fauvel, J. Benediktsson, J. Chanussot, and J. Sveinsson, “Spectral and spatial classification of hyperspectral data using SVMs and morphological profiles,” IEEE Trans. Geosc. Rem. Sens., vol. 46, no. 11, pp. 3804–3814, 2008.\n• F. Pacifici, M. Chini, and W. Emery, “A neural network approach using multi-scale textural metrics from very high-resolution panchromatic imagery for urban land-use classification,” Remote Sens. Environ., vol. 113, no. 6, pp. 1276–1292, 2009.\n• D. Tuia, F. Pacifici, M. Kanevski, and W. Emery, “Classification of very high spatial resolution imagery using mathematical morphology and support vector machines,” IEEE Trans. Geosci. Rem. Sens., vol. 47, no. 11, pp. 3866 –3879, Nov 2009.\n• G. Camps-Valls, L. Gómez-Chova, J. Muñoz-Marí, J. L. Rojo-Álvarez, and M. Martínez-Ramón, “Kernel-based framework for multi-temporal and multi-source remote sensing data classification and change detection,” IEEE Trans. Geosc. Rem. Sens., vol. 46, no. 6, pp. 1822–1835, 2008.\n• A. Plaza, J. A. Benediktsson, J. Boardman, J. Brazile, L. Bruzzone, G. Camps-Valls, J. Chanussot, M. Fauvel, P. Gamba, A. Gualtieri, and J. Tilton, “Recent advances in techniques for hyperspectral image processing,” Remote Sens. Environ., vol. 113, pp. 110–122, 2009.\n• J. Plaza, R. Pérez, A. Plaza, P. Martínez, and D. Valencia, “Parallel morphological/neural processing of hyperspectral images using heterogeneous and homogeneous platforms,” Cluster Comput., vol. 11, pp. 17–32, 2008.\n• J. Muñoz-Marí, A. Plaza, J. Gualtieri, and G. Camps-Valls, “Parallel programming and applications in grid, P2P and networking systems,” in Parallel Implementation of SVM in Earth Observation Applications, F. Xhafa, Ed. UK: IOS Press, 2009.\n• C. E. Rasmussen and C. K. I. Williams, Gaussian Processes for Machine Learning. New York: The MIT Press, 2006.\n• Y. Bazi and F. Melgani, “Gaussian process approach to remote sensing image classification,” IEEE Trans. Geosc. Rem. Sens., vol. 48, no. 1, pp. 186–197, Jan 2010.\n• M. Y. Yang, W. Liao, B. Rosenhahn, and Z. Zhang, “Hyperspectral image classification using gaussian process models,” in 2015 IEEE International Geoscience and Remote Sensing Symposium (IGARSS), July 2015, pp. 1717–1720.\n• K. Chen, P. Jian, Z. Zhou, J. Guo, and D. Zhang, “Semantic annotation of high-resolution remote sensing images via gaussian process multi-instance multilabel learning,” IEEE Geoscience and Remote Sensing Letters, vol. 10, no. 6, pp. 1285–1289, Nov 2013.\n• K. Chen, Z. Zhou, C. Huo, X. Sun, and K. Fu, “A semisupervised context-sensitive change detection technique via gaussian process,” IEEE Geoscience and Remote Sensing Letters, vol. 10, no. 2, pp. 236–240, March 2013.\n• P. Ruiz, J. Mateos, G. Camps-Valls, R. Molina, and A. Katsaggelos, “Bayesian active remote sensing image classification,” IEEE Transactions on Geoscience and Remote Sensing, vol. 52, no. 4, pp. 2186–2196, April 2014.\n• L. Kalantari, P. Gader, S. Graves, and S. A. Bohlman, “One-class gaussian process for possibilistic classification using imaging spectroscopy,” IEEE Geoscience and Remote Sensing Letters, vol. 13, no. 7, pp. 967–971, July 2016.\n• \n\nT. Minka, “Expectation propagation for approximate bayesian inference,” in\n\nProceedings of the Seventeenth Conference Annual Conference on Uncertainty in Artificial Intelligence (UAI-01). San Francisco, CA: Morgan Kaufmann, 2001, pp. 362–369.\n• M. Kuss and C. Rasmussen, “Assessing approximate inference for binary Gaussian process classification,” Machine learning research, vol. 6, pp. 1679–1704, Oct 2005.\n• Z. Chen, S. D. Babacan, R. Molina, and A. K. Katsaggelos, “Variational bayesian methods for multimedia problems,” IEEE Transaction on Multimedia, vol. 16, no. 4, pp. 1000–10 017, 2014.\n• A. Damianou, “Deep Gaussian Processes and Variational Propagation of Uncertainty,” Ph.D. dissertation, University of Sheffield, 2015.\n• A. Rahimi and B. Recht, “Random features for large-scale kernel machines,” in Advances in neural information processing systems, 2007, pp. 1177–1184.\n• A. Pérez-Suay, L. Gómez-Chova, J. Amorós, and G. Camps-Valls, “Randomized kernels for large scale earth observation applications,” Remote Sensing of Environment, 2017.\n• C. E. Rasmussen and C. K. I. Williams, Gaussian Processes for Machine Learning. New York: The MIT Press, 2006.\n• W. Rudin, Fourier analysis on groups. John Wiley & Sons, 2011.\n• C. Bishop, Pattern Recognition and Machine Learning. Springer, 2006.\n• M. Lázaro-Gredilla, J. Quiñonero-Candela, C. E. Rasmussen, A. R. Figueiras-Vidal et al., “Sparse spectrum gaussian process regression,” Journal of Machine Learning Research, vol. 11, no. Jun, pp. 1865–1881, 2010.\n• S. Babacan, R. Molina, M. Do, and A. Katsaggelos, “Bayesian blind deconvolution with general sparse image priors,” in\n\nEuropean Conference on Computer Vision (ECCV)\n\n, 2012, pp. 341–355.\n• R. Fletcher and C. M. Reeves, “Function minimization by conjugate gradients,” The computer journal, vol. 7, no. 2, pp. 149–154, 1964.\n• M. Derrien and H. Le Gléau, “MSG/SEVIRI cloud mask and type from SAFNWC,” International Journal of Remote Sensing, vol. 26, no. 21, pp. 4707–4732, 2005.\n• J. Hocking, P. N. Francis, and R. Saunders, “Cloud detection in meteosat second generation imagery at the met office,” Universitat de València, Tech. Rep. 540, 2010."
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https://ounces-to-grams.appspot.com/4876-ounces-to-grams.html | [
"Ounces To Grams\n\n# 4876 oz to g4876 Ounce to Grams\n\noz\n=\ng\n\n## How to convert 4876 ounce to grams?\n\n 4876 oz * 28.349523125 g = 138232.274758 g 1 oz\nA common question is How many ounce in 4876 gram? And the answer is 171.995838466 oz in 4876 g. Likewise the question how many gram in 4876 ounce has the answer of 138232.274758 g in 4876 oz.\n\n## How much are 4876 ounces in grams?\n\n4876 ounces equal 138232.274758 grams (4876oz = 138232.274758g). Converting 4876 oz to g is easy. Simply use our calculator above, or apply the formula to change the length 4876 oz to g.\n\n## Convert 4876 oz to common mass\n\nUnitMass\nMicrogram1.38232274758e+11 µg\nMilligram138232274.757 mg\nGram138232.274758 g\nOunce4876.0 oz\nPound304.75 lbs\nKilogram138.232274757 kg\nStone21.7678571429 st\nUS ton0.152375 ton\nTonne0.1382322748 t\nImperial ton0.1360491071 Long tons\n\n## What is 4876 ounces in g?\n\nTo convert 4876 oz to g multiply the mass in ounces by 28.349523125. The 4876 oz in g formula is [g] = 4876 * 28.349523125. Thus, for 4876 ounces in gram we get 138232.274758 g.\n\n## 4876 Ounce Conversion Table",
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"## Alternative spelling\n\n4876 Ounces to Gram, 4876 Ounce to Grams, 4876 Ounce in Grams, 4876 Ounce in Gram, 4876 Ounce in g,"
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null,
"https://ct.pinterest.com/v3/",
null
]
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https://community.rapidminer.com/discussion/39318/linear-regression-optimization | [
"# linear regression optimization\n\nMember Posts: 21",
null,
"Contributor I\nedited December 2018 in Help\n\nGreeting,\n\nI need to predict thermal expand range from tempreture, and I get a test dataset, so I try the linear regression but the result is not good, my setting and data is like below, can you help me to improve the prediction ? thanks.\n\nTagged:\n\n• Member Posts: 263",
null,
"Unicorn\nSolution Accepted\n\nFrom looking at the scatter plot of the two variables you get a sense that there are other important predictors missing from this equation. There is non-linearity so you could use other methods instead of plain vanilla linear regression.\n\nInvestigate further the physics of the process. I know absolutely nothing and Wikipidea tells me pressure is another important variable.",
null,
"• Member Posts: 46",
null,
"Guru\n\n• Member Posts: 21",
null,
"Contributor I\n\nmy friend do the same job with matlab and the result is well fit the test data , but I can not take the same score with rapidminer. I am still confused about this ...\n\n1.png 11.7K\n• RapidMiner Certified Analyst, RapidMiner Certified Expert, Member Posts: 1,761",
null,
"Unicorn\n\nCan you post your RapidMiner process? Maybe we can help troubleshoot.\n\n• Member Posts: 21",
null,
"Contributor I\n\nhi,\n\nyou can see the rm process and operator setting in the attachment above, and the origin data is also there.\n\nI can get a simular output like matlab, when I change the input attribute from \"tempreture\" to \"tempreture change\", which means y=kx+b do not work but y=k(x-x1)+b works well in rm, while y=kx+b works well in matlab.\n\n• Administrator, Employee, RapidMiner Certified Analyst, RapidMiner Certified Expert, Member Posts: 273",
null,
"RM Data Scientist\n\n@dkpengqiuyang did you try to delete collinear feature?",
null,
"Process is attached here.\n\n`<?xml version=\"1.0\" encoding=\"UTF-8\"?><process version=\"7.6.000\"> <operator activated=\"true\" class=\"retrieve\" compatibility=\"7.6.000\" expanded=\"true\" height=\"68\" name=\"Retrieve regression\" width=\"90\" x=\"45\" y=\"34\"> <parameter key=\"repository_entry\" value=\"//RM YY Local Repository/AAA-PROSPECT/data/regression\"/> </operator></process><?xml version=\"1.0\" encoding=\"UTF-8\"?><process version=\"7.6.000\"> <operator activated=\"true\" class=\"set_role\" compatibility=\"7.6.000\" expanded=\"true\" height=\"82\" name=\"Set Role\" width=\"90\" x=\"179\" y=\"34\"> <parameter key=\"attribute_name\" value=\"thermal expand\"/> <parameter key=\"target_role\" value=\"label\"/> <list key=\"set_additional_roles\"/> </operator></process><?xml version=\"1.0\" encoding=\"UTF-8\"?><process version=\"7.6.000\"> <operator activated=\"true\" class=\"linear_regression\" compatibility=\"7.6.000\" expanded=\"true\" height=\"103\" name=\"Linear Regression\" width=\"90\" x=\"380\" y=\"34\"> <parameter key=\"feature_selection\" value=\"M5 prime\"/> <parameter key=\"alpha\" value=\"0.05\"/> <parameter key=\"max_iterations\" value=\"10\"/> <parameter key=\"forward_alpha\" value=\"0.05\"/> <parameter key=\"backward_alpha\" value=\"0.05\"/> <parameter key=\"eliminate_colinear_features\" value=\"false\"/> <parameter key=\"min_tolerance\" value=\"0.05\"/> <parameter key=\"use_bias\" value=\"true\"/> <parameter key=\"ridge\" value=\"1.0E-8\"/> </operator></process><?xml version=\"1.0\" encoding=\"UTF-8\"?><process version=\"7.6.000\"> <operator activated=\"true\" class=\"apply_model\" compatibility=\"7.6.000\" expanded=\"true\" height=\"82\" name=\"Apply Model\" width=\"90\" x=\"514\" y=\"34\"> <list key=\"application_parameters\"/> <parameter key=\"create_view\" value=\"false\"/> </operator></process>`"
]
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null,
"https://s3.amazonaws.com/rapidminer.community/vanilla-rank-images/contributor-16x16.png ",
null,
"https://s3.amazonaws.com/rapidminer.community/vanilla-rank-images/unicorn-16x16.png ",
null,
"https://us.v-cdn.net/6030995/uploads/s3://uploads/lithium_attachments/image/serverpage/image-id/1466i721D2233C7DDE0D7/screen_shot_2017-06-05_at_9.07.39_am.png",
null,
"https://s3.amazonaws.com/rapidminer.community/vanilla-rank-images/guru-16x16.png ",
null,
"https://s3.amazonaws.com/rapidminer.community/vanilla-rank-images/contributor-16x16.png ",
null,
"https://s3.amazonaws.com/rapidminer.community/vanilla-rank-images/unicorn-16x16.png ",
null,
"https://s3.amazonaws.com/rapidminer.community/vanilla-rank-images/contributor-16x16.png ",
null,
"https://s3.amazonaws.com/rapidminer.community/vanilla-rank-images/employee-16x16.png ",
null,
"https://us.v-cdn.net/6030995/uploads/s3://uploads/lithium_attachments/image/serverpage/image-id/1672i1A900AB7F14F1758/linear.png",
null
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https://manypoints.org/Details.aspx?q=4&g=8 | [
"Entry details for q = 22 = 4, g = 8\n\nLower bound Nmin = 22\n\n Submitted by Everett Howe Date 04/20/2020 Reference Everett W. HoweThe maximum number of points on a curve of genus eight over the field of four elementsarXiv: 2004.08007 [math.NT] Comments An example is given by y^2 + (x^3 + x + 1)y = x^6 + x^5 + x^4 + x^2 z^3 = (x+1)y + x^2 This is a degree-3 Kummer extension of the genus-2 curve defined by the first equation, ramified at 4 points. Tags Explicit curves"
]
| [
null
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https://en.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-motion-in-a-plane/in-in-class11-horizontally-launched-projectiles/v/horizontally-launched-projectile | [
"If you're seeing this message, it means we're having trouble loading external resources on our website.\n\nIf you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.\n\n## Class 11 Physics (India)\n\n### Course: Class 11 Physics (India)>Unit 8\n\nLesson 5: Horizontally launched projectiles\n\n# Horizontally launched projectile\n\nHow to solve for the horizontal displacement when the projectile starts with a horizontal initial velocity. We also explain common mistakes people make when doing horizontally launched projectile problems. Created by David SantoPietro.\n\n## Want to join the conversation?\n\n• Is acceleration due to gravity 10 m/s^2 or 9.8 m/s^2? My teacher says it is 10 but Dave says it is 9.8.",
null,
"•",
null,
"",
null,
"Acceleration due to gravity actually depends on your location on the planet and how far above sea level you are, and is between 9.78 and 9.83 .This is sometimes rounded up to 10 to make assignments more simple, especially when a calculator is not available, but if you're going to continue studying physics you should remember that it's closer to 9.8 .\n• Why does the time remain same even if the body covers greater distance when horizontally projected? I mean when the body is just dropped without any horizontal component, it will fall straight. But when we give a horizontal velocity to the body, it should cover a parabolic path(greater than the path covered during free fall). So the body should take a longer time to fall.\n: )",
null,
"•",
null,
"",
null,
"The acceleration due to gravity is the same whether the object is falling straight or moving horizontally. Since acceleration is the same, then the time each object hits the ground will be the same, assuming they both start from the same height and fall the same distance.\n• David mentioned that the time it takes for vertical displacement to occur would the same as the time it takes for the horizontal displacement to happen. This much makes sense, especially if air resistance is negligible.\n\nHowever, what happens in the case of a cliff jumper with a wing suit? Are the times still the same for the vertical and horizontal?",
null,
"• this part of physics makes me want to give up",
null,
"• If in a horizontally launched projectile problem you're given the height of the 'cliff' and the horizontal distance at which the object falls into the 'water' how do you calculate the initial velocity?",
null,
"• Maths version of what Teacher Mackenzie said:\nFind the time it takes for an object to fall from the given height.\n∆y = v_0 t + (1/2)at^2; v_0 = 0; ∆y = -h; and a = g the initial vertical velocity is zero, because we specified that the projectile is launched horizontally.\n-h = (1/2)gt^2\n-2h/g = t^2\n√(-2h/g) = t The negative sign under the radical is fine because gravitational acceleration is also in the negative direction.\nThen we take this t and plug it into the x equations\n∆x = v_0t + 1/2at^2; horizontal acceleration is zero. We could also use an equation with final velocity instead of acceleration, using the understanding that final velocity will equal initial velocity.\n∆x = v_0*t; solve for initial velocity\n∆x/t = v_0\n• If something is thrown horizontally off a cliff, what is it’s vertical acceleration? How do you know? What is its horizontal acceleration? How do you know?",
null,
"• Would air resistance shorten the horizontal distance you are jumping, or lengthen it?",
null,
"• i think im gonna loose my mind",
null,
"• How would you then find the velocity when it hits the ground and the length of the hypotenuse line?",
null,
"• If you were asked to find final velocity, you would need both the vertical and horizontal components of final velocity. Horizontal is easy, there is no horizontal acceleration, so the final velocity is the same as initial velocity (5 m/s). To find the vertical final velocity, you would use a kinematic equation. You have vertical displacement (30 m), acceleration (9.8 m/s^2), and initial velocity (0 m/s). You could then use the time-independent formula:\nVf^2 - Vi^2 = 2 * a * d\nVf^2 - (0)^2 = 2 * (9.8) * (30)\nVf = sqrt(2 * 9.8 * 30)\nVf = 24.2...\n\nNow that you have the final velocity components, you can set up a right triangle to solve for the combined final velocity. The components will be the legs, and the total final velocity will be the hypotenuse. By the pythagorean theorem:\nVfx^2 + Vfy^2 = Vf^2\n(5)^2 + (24)^2 = Vf^2\nVf = sqrt(5^2 + 24.2^2)\nVf = 24.7\nThat's the magnitude of the final velocity. To find the angle, you would need to do some trig and realize that the angle from the horizontal is opposite to Vfy and adjacent to Vfx. We can write this as:\ntan(theta) = Vfy / Vfx\ntheta = atan ((24.2) / (5))\ntheta = 78.3... degrees\n\nAnd there you have both the magnitude and angle of the final velocity. Hope this helps! (Yes, I am the slightest bit too lazy to actually write the symbol for theta)",
null,
""
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https://bloomsies.com/how-many-vertices-does-a-cube-have-a-complete-explanation/ | [
"July 20, 2023\n\n# How Many Vertices Does a Cube Have? A Complete Explanation\n\nHow Many Vertices Does a Cube Have?\n\n## How Many Vertices Does a Cube Have?\n\nA cube is a three-dimensional geometric shape that consists of six congruent square faces that meet at right angles. Understanding the properties of a cube is essential in various fields, such as mathematics, geometry, and computer graphics. One fundamental characteristic of a cube is the number of vertices it possesses.\n\n## Number of Vertices in a Cube\n\nA cube has a total of eight vertices.\n\nTo visualize these vertices, imagine a cube where all its corners meet the edges. Each corner represents a vertex. Therefore, there are eight points where three edges intersect, forming the geometric vertices of a cube.\n\n## Vertex Formula for a Cube\n\nThe formula to calculate the number of vertices in any cube, regardless of the size, is:\n\nNumber of Vertices = 8\n\nRegardless of the dimensions or scale of a cube, the number of vertices remains constant at eight.\n\n## FAQs\n\n### Q: What is a vertex in a geometric shape?\n\nA: In geometry, a vertex refers to a point where two or more edges, curves, or lines meet to form an angle or create an intersection.\n\n### Q: How are the vertices of a cube denoted?\n\nA: The vertices of a cube are typically denoted using capital letters, such as A, B, C, D, E, F, G, and H.\n\n### Q: How can I calculate the number of vertices in other geometric shapes?\n\nA: The number of vertices in a geometric shape can be determined by counting the points where the edges or curves meet. Each intersection point represents a vertex in the shape.",
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https://xn--2-umb.com/17/full-mul/ | [
"# Full Multiplication\n\nA lot of smart contracts use the SafeMath library. It prevents contracts from having incorrect results, but it does so by failing transactions instead of making them correct. Let’s instead try to do the math correctly. In this series, I will derive some advanced techniques. Today, I’ll make a better safeMul.\n\nIf you multiply two numbers, the result will be a number twice the size. In Ethereum, when you multiply two numbers, the result can be up to 512 bits. But Ethereum only gives you the lower half; it simply ignores the rest. This is a common practice in mathematics called modular arithmetic.\n\nHowever, ignoring numbers is not an acceptable practice in accounting. Care needs to be taken to avoid it or someone will lose something valuable. A popular library called SafeMath detects when it happens and then fails the transaction. But what if you do not want your transaction to fail?\n\nWhat if you want to multiply any numbers and have the complete result?\n\nSpoiler alert: this is snippet of Solidity code will do that for you:\n\ncontract FullMul {\nfunction mul512(uint256 a, uint256 b)\npublic pure returns(uint256 r0, uint256 r1) {\nassembly {\nlet mm := mulmod(a, b, not(0))\nr0 := mul(a, b)\nr1 := sub(sub(mm, r0), lt(mm, r0))\n}\n}\n}\n\n\nOptimized full 512 bit multiplication in Solidity.\n\nBut before we get into that, let’s define the problem precisely: We have two unsigned numbers $a$ and $b$, both 256 bits in length and we want their product, a 512-bit number $x$.\n\n$$x = a ⋅ b$$\n\nSince this number is too large to be represented directly in code, we split it up into the least significant and most significant 256 bits, $r_0$ and $r_1$ respectively:\n\n\\begin{aligned} r_0 &= \\mod{x}_{2^{256}} & r_1 &= \\floor{\\frac{x}{2^{256}}} \\end{aligned}\n\nWhere the square brackets with subscript represent the modulo operation and the lower-half brackets on the right represent the floor operation.\n\n## Schoolbook algorithm\n\nThe classical way of solving this problem is by long multiplication, the method we all learned in school. You split your large number into decimals, multiply the digits, and then add the intermediate results. This method also works in binary and other bases. Let me quickly show you how you would use it here:\n\nSince we have 256 bit multiply build in, we can multiply any two 128 bit numbers and get the full result. So if we split our large number into groups of 128 bits we can compute all their products. Take $a_0$ and $a_1$ to respectively mean the least significant and most significant 128 bits of $a$, similarly for $b$:\n\n\\begin{aligned} a_0 &= \\mod{a}\\_{2^{128}} & a_1 &= \\floor{\\frac{a}{2^{128}}} \\\\\\\\ b_0 &= \\mod{b}\\_{2^{128}} & b_1 &= \\floor{\\frac{b}{2^{128}}} \\\\\\\\ \\end{aligned}\n\nNow the original numbers a and b can be written as:\n\n\\begin{aligned} a &= a_1 ⋅2^{128} + a_0 \\\\\\\\ b &= b_1 ⋅2^{128} + b_0 \\\\\\\\ \\end{aligned}\n\nIf we substitute this in product equation it becomes:\n\n$$x = a_1 ⋅ b_1 ⋅2^{256} + (a_0 ⋅ b_1 + a_1 ⋅ b_0) ⋅ 2^{128} + a_0 ⋅ b_0$$\n\nIgnoring the constants, we now have four multiplications instead of one. But all four of them involve numbers less than $2^{128}$ that can be computed directly. The result is still too large, so we still need two numbers $r_0$ and $r_1$ to represent it. I will skip the steps of how to get $r_0$ and $r_1$ from this expression. It is straightforward, but annoying because of the shifts and carries. The final result is:\n\ncontract FullMul {\nuint256 constant H = 2**128;\n\nfunction mul512(uint256 a, uint256 b)\npublic pure returns(uint256 r0, uint256 r1) {\n// Split in groups of 128 bit\nuint256 a0 = a % H;\nuint256 a1 = a / H;\nuint256 b0 = b % H;\nuint256 b1 = b / H;\n\n// Compute 256 bit intermediate products\nuint256 i00 = a0 * b0;\nuint256 i01 = a0 * b1;\nuint256 i10 = a1 * b0;\nuint256 i11 = a1 * b1;\n\n// Split results in (shifted) groups of 128 bit\nuint256 i010 = i01 * H; // Shifted up\nuint256 i011 = i01 / H;\nuint256 i100 = i10 * H; // Shifted up\nuint256 i101 = i10 / H;\n\n// Add all intermediate terms, taking care of overflow\nr0 = i00;\nr1 = i11 + i011 + i101;\nr0 += i010;\nif (r0 < i010) {\nr1 += 1;\n}\nr0 += i100;\nif (r0 < i100) {\nr1 += 1;\n}\n}\n}\n\n\nSchoolbook algorithm for 512 bit multiplication.\n\n(Note that Solidity, as of 0.4.18, actually fails to compile the above example because the compiler can not handle that many local variables. This is easily solved by inlining some expressions, but since it reduces readability I opted not to do that for this example.)\n\nThe two multiplications for i01 and i10 can be replaced by one using the Karatsuba algorithm, at the expense of a few more additions. Since additions are 3 gas and multiplications only 5, this is not worth it. But if you want to do larger multiplications (say 4096 bit) it is worth looking into these methods.\n\nWe have now solved the problem using two modulo operations, four divisions, six additions, two conditional branches, and no less than six multiplications. The entire function takes a bit over 300 gas. This is not bad, but the gas cost is almost two orders of magnitude larger than the 5 gas for a regular multiplication, or 90 for a standard safeMul.\n\nWe can do a lot better.\n\n## Chinese Remainder\n\nSo here’s the trick: We use the rather obscure mulmod instruction and the Chinese Remainder Theorem. In short, the theorem states thati f we know a number modulo $2^{256}$ and $2^{256} - 1$, we can compute its 512-bit representation cheaply. The function to do this, chineseRemainder, is described in a previous post. To use it here, we first need to compute our product in the two moduli:\n\n\\begin{aligned} x_0 &= \\mod{a ⋅ b}\\_{2^{256}} & x_1 &= \\mod{a ⋅ b}\\_{2^{256} -1 } \\end{aligned}\n\nThe first one, $x_0$ is just a regular multiply, as it already truncates to 256 bits. The second one, $x_1$, can be computed directly using a single mulmod operation. This is a rather unknown opcode that computes:\n\n$$\\mathtt{mulmod}(a, b, c) = \\mod{a ⋅ b}_{c}$$\n\nPut this together, and we have our new mul512 function:\n\ncontract FullMul {\nuint256 constant M1 = 2**256 - 1;\n\nfunction mul512(uint256 a, uint256 b)\npublic pure returns(uint256 r0, uint256 r1) {\nuint256 x0 = a * b;\nuint256 x1 = mulmod(a, b, M1);\n(r0, r1) = chineseRemainder(x0, x1);\n}\n}\n\n\n512-bit multiplication.\n\nThe Solidity compiler, as of version 0.4.18, does not produce very optimal code here. The chineseRemainder function is so tiny it is not worth the call-overhead, so it should be inlined, but the compiler doesn’t do this. The compiler does recognize that M1 can be expressed efficiently as not(0). Manually inlining results in an efficient multiplication function:\n\ncontract FullMul {\nuint256 constant M1 = 2**256 - 1;\n\nfunction mul512(uint256 a, uint256 b)\npublic pure returns(uint256 r0, uint256 r1) {\nassembly {\nlet mm := mulmod(a, b, not(0))\nr0 := mul(a, b)\nr1 := sub(sub(mm, r0), lt(mm, r0))\n}\n}\n}\n\n\nOptimized full 512 bit multiplication in Solidity.\n\nIt is our chineseRemainder function (two subs and one lt) with a mul and mulmod added. We use assembly to avoid an unnecessary branch. The total gas cost is about 60 gas, compared to 5 for a normal multiply and 90 for a standard safeMul. In fact, it is slightly cheaper to use mul512 and check that r1 is zero than it is to use safeMul! Conclusion\n\nIt is possible to do full precision never-overflowing multiplication in the EVM for less gas than a regular safeMul. This is a good starting point for smart contracts do not want to reject transaction just because an intermediate value overflows.\n\nIn the next post, I will introduce a number of simple utility functions for dealing with large numbers. This builds up to a very popular function that nobody has yet implement correctly for all cases. Stay tuned!",
null,
"Remco Bloemen\nMath & Engineering\nhttps://2π.com"
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null,
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https://im.kendallhunt.com/k5/teachers/grade-3/unit-8/lesson-2/lesson.html | [
"# Lesson 2\n\n## Warm-up: Which One Doesn’t Belong: Fractions on Number Lines (10 minutes)\n\n### Narrative\n\nThis warm-up prompts students to compare four images. It gives students a reason to use language precisely. It gives the teacher an opportunity to hear how students use terminology and talk about characteristics of the items in comparison to one another. During the synthesis, ask students to explain the meaning of any terminology they use, such as tick marks, labels, unit fractions, whole numbers, and length.\n\n### Launch\n\n• Groups of 2\n• Display the image.\n• “Pick one that doesn’t belong. Be ready to share why it doesn’t belong.”\n• 1 minute: quiet think time\n\n### Activity\n\n• 2–3 minutes: partner discussion\n• Share and record responses.\n\n### Student Facing\n\nWhich one doesn’t belong?\n\n### Activity Synthesis\n\n• “Let’s find at least one reason why each one doesn’t belong.”\n\n## Activity 1: Create Your Own Number Line (25 minutes)\n\n### Narrative\n\nThe purpose of this activity is for students to use their fraction reasoning skills to practice locating fractions on a number line. Students should be in groups, but the groups should stay small enough that every member will have a chance to share their ideas. Be sure to space groups so that each has their own area to work in. Students write the fractions on their tape. Students will use the number line they create in the next activity.\n\nAs they place the different numbers students think about the meaning of the numerator and denominator in the fractions and how whole numbers can be written as fractions (MP7).\n\nMLR8 Discussion Supports. Synthesis: At the appropriate time, give groups 2–3 minutes to plan what they will say when they present to the class. “Practice what you will say when you share your number line with the class. Talk about what is important to say, and decide who will share each part.”\nAction and Expression: Develop Expression and Communication. Synthesis: Identify connections between strategies that result in the same outcomes but use differing approaches.\nSupports accessibility for: Memory\n\n### Required Materials\n\nMaterials to Gather\n\n### Required Preparation\n\n• Each group of 3-4 students needs a roll of tape and a marker.\n\n### Launch\n\n• Groups of 3–4\n• “Today you are going to work with your group to create a number line and place fractions on it. Be prepared to share your methods with the class.”\n• Give each group a roll of tape and a marker.\n\n### Activity\n\n• 10–15 minutes: small-group work time\n• Monitor for methods that groups use to locate the points, such as:\n• starting with benchmark numbers, such as unit fractions or whole numbers\n• considering whether fractions are larger or smaller than 1\n• considering whether fractions are equivalent to whole numbers\n• comparing fractions with the same numerator or denominator\n\n### Student Facing\n\nCreate a long number line on the floor.\n\nLocate and label each fraction on the number line. Be prepared to explain your reasoning.\n\n• 0\n• 1\n• 2\n• $$\\frac{1}{2}$$\n• $$\\frac{1}{3}$$\n• $$\\frac{6}{2}$$\n• $$\\frac{12}{3}$$\n• $$\\frac{1}{4}$$\n• $$\\frac{5}{4}$$\n• $$\\frac{6}{6}$$\n• $$\\frac{5}{6}$$\n• $$\\frac{9}{8}$$\n• $$\\frac{15}{8}$$\n• $$\\frac{5}{3}$$\n• $$\\frac{18}{6}$$\n• $$\\frac{2}{8}$$\n\n### Activity Synthesis\n\n• Have each group share a method they used or a fraction they placed, based on what you noticed during the activity. Encourage groups to use their number lines when demonstrating their reasoning.\n• “Did any groups use a similar strategy?”\n• “Did any groups place that fraction in a different way?”\n• “Which fractions were easier to locate?”\n• “Which fractions were harder to locate?”\n• Keep number lines displayed for the next activity.\n\n## Activity 2: Make a Statement (10 minutes)\n\n### Narrative\n\nThe purpose of this activity is for students to use the number line they created in the previous activity to make comparison statements about fractions. Students use the symbols $$>$$, $$=$$, and $$<$$ to record comparisons between pairs of fractions.\n\n### Launch\n\n• Groups of 3–4\n• “Now you are going to work with your group to write comparison statements based on your number line.”\n\n### Activity\n\n• 8–10 minutes: small-group work time\n• Monitor for a variety of student-generated statements of each type to share during the synthesis.\n\n### Student Facing\n\nWrite 6 fraction comparison statements about the numbers on your number line. Include 2 statements for each symbol ($$>$$, $$=$$, and $$<$$).\n\n1.\n2.\n3.\n4.\n5.\n6.\n\nChoose 2 statements you wrote. Use numbers, pictures, or words to show that they are true.\n\n### Activity Synthesis\n\n• Have each group share at least one comparison statement they came up with and their reasoning. Be sure to share at least one statement that uses each symbol.\n\n## Lesson Synthesis\n\n### Lesson Synthesis\n\n“How did you decide how long your number line should be? Does it matter?” (We looked at the largest number we had and made sure it would fit on the number line. Yes, because you had to make sure all the numbers would fit on the number line.)\n\n“The number line of one group is noticeably longer than that of another group. Does that affect the comparison statements that each group could make?” (It wouldn’t affect the comparison statements for one group working on their own number line, but if two groups tried to compare fractions with number lines with different lengths, their statements could be wrong.)"
]
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https://answers.everydaycalculation.com/multiply-fractions/4-2-times-1-42 | [
"Solutions by everydaycalculation.com\n\n## Multiply 4/2 with 1/42\n\n1st number: 2 0/2, 2nd number: 1/42\n\nThis multiplication involving fractions can also be rephrased as \"What is 4/2 of 1/42?\"\n\n4/2 × 1/42 is 1/21.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 4/2 × 1/42 = 4 × 1/2 × 42 = 4/84\n3. After reducing the fraction, the answer is 1/21\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
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"https://answers.everydaycalculation.com/mathstep-app-icon.png",
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https://www.theworksheets.com/s/Division+Worksheets | [
"# Division Worksheets Results\n\n##### Multiplying and Dividing Using Scientific Notation\n\nAnswers to Multiplying and Dividing Using Scientific Notation 1) 9.407 × 10−11 2) 1.16 × 10−1 3) 1.024 × 1010 4) 9.006 × 10−9 5) 3.8 × 101 6) 3.68 × 103 7) 9.75 × 102 8) 6.928 × 10−3 9) 9.182 × 104 10) 1.407 × 106 11) 1.038 × 10−3 12) 1.02 × 101 13) 4.209 × 100 14) 1.458 × 1016 15) 9.766 × 10−9 16) 9.064 × 10−27\n\nhttps://url.theworksheets.com/4d3",
null,
"",
null,
"",
null,
"##### Multiplying and Dividing Using Scientific Notation\n\nAnswers to Multiplying and Dividing Using Scientific Notation 1) 9.407 × 10−11 2) 1.16 × 10−1 3) 1.024 × 1010 4) 9.006 × 10−9 5) 3.8 × 101 6) 3.68 × 103 7) 9.75 × 102 8) 6.928 × 10−3 9) 9.182 × 104 10) 1.407 × 106 11) 1.038 × 10−3 12) 1.02 × 101 13) 4.209 × 100 14) 1.458 × 1016 15) 9.766 × 10−9 16) 9.064 × 10−27\n\nhttps://url.theworksheets.com/4d3",
null,
"",
null,
"",
null,
"##### Division Made Easy - The Mathematics Shed\n\nMaking Math Easy Reproducible Worksheets Reproducible Worksheets for: Division Made Easy These worksheets practice math concepts explained in Division Made Easy (ISBN 0-7660-2511-X), Written by Rebecca Wingard-Nelson, Illustrated by Tom LaBaff. Making Math Easy reproducible worksheets are designed to help teachers, parents, and tutors use the books in the Making Math Easy series in the ...\n\nhttps://url.theworksheets.com/34k",
null,
"",
null,
"",
null,
"##### Math Fact Fluency Worksheets\n\nDivision Worksheet D : ... Dividing by 12. Title: Math Fact Fluency Worksheets Author: SkillsTutor Created Date: 10/20/2008 4:56:43 PM ...\n\nhttps://url.theworksheets.com/35v",
null,
"",
null,
"",
null,
"##### Fun-tabulous Puzzles - Weebly\n\nsubtraction, multiplication and division—the building blocks of mathematics. WHAT YOU’LL FIND IN THIS BOOK This book of 40 puzzles is organized by skill areas and includes: number concepts, addition, subtraction, multiplication, division, order of operations, fractions and decimals, graphing, and time.\n\nhttps://url.theworksheets.com/6xg",
null,
"",
null,
"",
null,
"##### Chapter 5 The Cell Cycle, Mitosis, and Meiosis Worksheets\n\n5.1 Cell Division and the Cell Cycle Lesson 5.1: True or False Name_____ Class_____ Date_____ Write true if the statement is true or false if the statement is false.\n\nhttps://url.theworksheets.com/36h",
null,
"",
null,
"",
null,
"##### Division Worksheet -- Long Division - Free Math Worksheets\n\nDivision Worksheet -- Long Division - One-Digit Divisor and a Two-Digit Quotient with No Remainder Author: Math-Drills.com -- Free Math Worksheets Subject: Division Keywords: division, mathematics, math, long division\n\nhttps://url.theworksheets.com/2dth",
null,
"",
null,
"",
null,
"##### Division - 3P Learning\n\nSolve these division problems using the division symbol: Division – written methods 1 2 Another way to represent division is with the division symbol. This is the same as 36 ÷ 6 = 6 If the answer is a single digit, it should go in the ones column. T O 6 6 3 6 5 3 5 4 2 8 9 1 8 6 5 4 2 1 4 4 1 6 5 2 5 7 4 9 8 4 8 DIVISION MAFS.4.NBT.2.6\n\nhttps://url.theworksheets.com/4dv",
null,
"",
null,
"",
null,
"##### Division - 3P Learning\n\nSolve these division problems using the division symbol: Division – written methods 1 2 Another way to represent division is with the division symbol. This is the same as 36 ÷ 6 = 6 If the answer is a single digit, it should go in the ones column. T O 6 6 3 6 5 3 5 4 2 8 9 1 8 6 5 4 2 1 4 4 1 6 5 2 5 7 4 9 8 4 8 DIVISION MAFS.4.NBT.2.6\n\nhttps://url.theworksheets.com/4dv",
null,
"",
null,
"",
null,
"##### Dividing Polynomials Using Long or Synthetic Division\n\nEXAMPLES - Dividing Polynomials using LONG or SYNTHETIC DIVISION Name_____ ID: 1 ©Y Q2M0H1R6t `Kru^tKah wSKoyfEtgwVaFrseT cLlLZC`.B Z pA_lilF irxiDglhMtesQ froeVsNefr^vreodr.-1-Divide using LONG DIVISION. Show work! 1) k3 + 8k2 + 10k + 21) ¸ (k + 7) 2) ...\n\nhttps://url.theworksheets.com/n3y",
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"",
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"",
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""
]
| [
null,
"https://cdn.theworksheets.com/Division-worksheets/t-1-5-0-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-1-5-1-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-1-5-2-Division-worksheets.png",
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null,
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null,
"https://cdn.theworksheets.com/Division-worksheets/t-2-3-2-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-0-1-0-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-0-1-1-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-0-1-2-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-2-8-0-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-2-8-1-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-2-8-2-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-2-7-0-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-2-7-1-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-2-7-2-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-1-9-0-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-1-9-1-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-1-9-2-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-1-2-0-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-1-2-1-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-1-2-2-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-1-0-0-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-1-0-1-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-1-0-2-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-2-0-0-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-2-0-1-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-2-0-2-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-1-6-0-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-1-6-1-Division-worksheets.png",
null,
"https://cdn.theworksheets.com/Division-worksheets/t-1-6-2-Division-worksheets.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6235961,"math_prob":0.72444093,"size":3465,"snap":"2022-05-2022-21","text_gpt3_token_len":1214,"char_repetition_ratio":0.1733603,"word_repetition_ratio":0.5699301,"special_character_ratio":0.38037518,"punctuation_ratio":0.18193549,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98850876,"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],"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],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-28T12:51:46Z\",\"WARC-Record-ID\":\"<urn:uuid:5eedb9f0-5082-4399-853e-b765af3fe6f6>\",\"Content-Length\":\"46613\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e08fdb36-bbaa-441c-b70d-369ff0ea157f>\",\"WARC-Concurrent-To\":\"<urn:uuid:e12f9411-ba2d-4174-9340-f6deb2010d04>\",\"WARC-IP-Address\":\"198.54.119.75\",\"WARC-Target-URI\":\"https://www.theworksheets.com/s/Division+Worksheets\",\"WARC-Payload-Digest\":\"sha1:SIR2KIVWZAH5UMH72AXXONYKSPZLPU5P\",\"WARC-Block-Digest\":\"sha1:7F2XYTBTUH3YX3W3DXTUWJ4EJPLUD453\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663016853.88_warc_CC-MAIN-20220528123744-20220528153744-00058.warc.gz\"}"} |
https://www.maplesoft.com/support/help/maple/view.aspx?path=LinearAlgebra%2FSmithForm | [
"",
null,
"LinearAlgebra - Maple Programming Help\n\nHome : Support : Online Help : Mathematics : Linear Algebra : LinearAlgebra Package : Solvers : LinearAlgebra/SmithForm\n\nLinearAlgebra\n\n SmithForm\n reduce a Matrix to Smith normal form\n\n Calling Sequence SmithForm(A, x, m, out, options, outopts)\n\nParameters\n\n A - Matrix x - (optional) variable; specifies the variable in which the entries of A are rational polynomials over Q m - (optional) equation of the form method = name where name is one of 'integer' or 'rational'; method to use out - (optional) equation of the form output = obj where obj is one of 'S', 'U', or 'V', or a list containing one or more of these names; selects result objects to compute options - (optional); constructor options for the result object(s) outopts - (optional) equation(s) of the form outputoptions[o] = list where o is one of 'S', 'U', or 'V'; constructor options for the specified result object\n\nDescription\n\n • The SmithForm(A) function returns the Smith normal form S of a Matrix A with univariate polynomial entries in x over the field of rational numbers Q, or rational expressions over Q.\n The Smith normal form of a Matrix is a diagonal Matrix S obtained by doing elementary row and column operations. The diagonal entries satisfy the property that for all n <= Rank(A), product(S[i, i], i=1..n) is equal to the (monic) greatest common divisor of all n x n (determinant) minors of A.\n • If the variable x is provided, or if the option method='rational' is provided, or if the Matrix A is not of type 'Matrix(integer)', then computation takes place over the field of rational polynomials. If the option method='integer' is provided, or if an optional variable name is not provided and the Matrix A is of type 'Matrix(integer)', then computation takes place over the integers to supply the integer-only Smith normal form.\n Floating point entries in A are converted to rationals before the Smith form is computed.\n If the variable name x is not supplied and the Matrix is not of type 'Matrix(integer)', then the routine selects the indeterminate in the case that indets(A) has a single member, and returns an error in the case that indets(A) has more than a single member. In the case that indets(A) has no members and A is not of type 'Matrix(integer)', then computation takes place over the field of rational polynomials using a dummy variable.\n • The output option (out) determines the content of the returned expression sequence.\n Depending on what is included in the output option, an expression sequence containing one or more of the factors S (the Smith normal form), U (the left-reducing Matrix ), or V (the right-reducing Matrix) can be returned. If output is a list, the objects are returned in the same order as specified in the list.\n The returned Matrix objects have the property that S = U . A . V.\n • The constructor options provide additional information (readonly, shape, storage, order, datatype, and attributes) to the Matrix constructor that builds the result. These options may also be provided in the form outputoptions[o]=[...], where [...] represents a Maple list. If a constructor option is provided in both the calling sequence directly and in an outputoptions[o] option, the latter takes precedence (regardless of the order).\n The following list indicates permissible values for index [o] of outputoptions with their corresponding meaning.\n\n S Smith form U left-reducing Matrix V right-reducing Matrix\n\n • This function is part of the LinearAlgebra package, and so it can be used in the form SmithForm(..) only after executing the command with(LinearAlgebra). However, it can always be accessed through the long form of the command by using LinearAlgebra[SmithForm](..).\n\nExamples\n\n > $\\mathrm{with}\\left(\\mathrm{LinearAlgebra}\\right):$\n > $A≔\\mathrm{Matrix}\\left(\\left[\\left[1,2x,2{x}^{2}+2x\\right],\\left[1,6x,6{x}^{2}+6x\\right],\\left[1,3,x\\right]\\right]\\right)$\n ${A}{≔}\\left[\\begin{array}{ccc}{1}& {2}{}{x}& {2}{}{{x}}^{{2}}{+}{2}{}{x}\\\\ {1}& {6}{}{x}& {6}{}{{x}}^{{2}}{+}{6}{}{x}\\\\ {1}& {3}& {x}\\end{array}\\right]$ (1)\n > $S≔\\mathrm{SmithForm}\\left(A\\right)$\n ${S}{≔}\\left[\\begin{array}{ccc}{1}& {0}& {0}\\\\ {0}& {1}& {0}\\\\ {0}& {0}& {{x}}^{{2}}{+}\\frac{{3}}{{2}}{}{x}\\end{array}\\right]$ (2)\n > $\\mathrm{Determinant}\\left(A\\right)$\n ${-}{8}{}{{x}}^{{2}}{-}{12}{}{x}$ (3)\n > $\\frac{\\mathrm{lcoeff}\\left(\\right)}{}$\n ${{x}}^{{2}}{+}\\frac{{3}}{{2}}{}{x}$ (4)\n > $U,V≔\\mathrm{SmithForm}\\left(A,x,\\mathrm{output}=\\left['U','V'\\right]\\right):$\n > $\\mathrm{map}\\left(\\mathrm{expand},U·A·V\\right)$\n $\\left[\\begin{array}{ccc}{1}& {0}& {0}\\\\ {0}& {1}& {0}\\\\ {0}& {0}& {{x}}^{{2}}{+}\\frac{{3}}{{2}}{}{x}\\end{array}\\right]$ (5)"
]
| [
null,
"https://bat.bing.com/action/0",
null
]
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https://corporatefinanceinstitute.com/resources/knowledge/finance/modified-internal-rate-of-return-mirr/ | [
"# Modified Internal Rate of Return (MIRR)\n\nA financial measure used in capital budgeting to identify the viability of an investment project\n\n## What is the Modified Internal Rate of Return (MIRR)?\n\nThe modified internal rate of return (commonly denoted as MIRR) is a financial measure that helps to determine the attractiveness of an investment and that can be used to compare different investments. Essentially, the modified internal rate of return is a modification of the internal rate of return (IRR) formula, which resolves some issues associated with that financial measure.",
null,
"The MIRR is primarily used in capital budgeting to identify the viability of an investment project. For instance, if the MIRR of a project is higher than its expected return, an investment is considered to be attractive.\n\nConversely, it is not recommended to undertake a project if its MIRR is less than the expected return. In addition, the MIRR is commonly employed to compare several alternative projects that are mutually exclusive. In such a case, the project with the highest MIRR is the most attractive.\n\n### How to Calculate the Modified Internal Rate of Return\n\nCalculating the MIRR considers three key variables: (1) the future value of positive cash flows discounted at the reinvestment rate, (2) the present value of negative cash flows discounted at the financing rate, and (3) the number of periods.\n\nMathematically, the calculation of the MIRR is expressed using the following equation:",
null,
"Where:\n\n• FVCF – the future value of positive cash flows discounted at the reinvestment rate\n• PVCF – the present value of negative cash flows discounted at the financing rate\n• n – the number of periods\n\nGenerally, the manual calculation of the MIRR is a tedious process that is prone to making mistakes. Alternatively, the MIRR can be easily calculated in spreadsheet applications such as Microsoft Excel. For example, in MS Excel, it can be calculated using the function called “=MIRR (cash flows, financing rate, reinvestment rate).”\n\n### MIRR vs. IRR\n\nThe modified internal rate of return (MIRR) and the internal rate of return (IRR) are two closely-related concepts. The MIRR was introduced to address a few problems associated with the IRR. For example, one of the main problems with the IRR is the assumption that the obtained positive cash flows are reinvested at the same rate at which they were generated. Alternatively, the MIRR considers that the proceeds from the positive cash flows of a project will be reinvested at the external rate of return. Frequently, the external rate of return is set equal to the company’s cost of capital.\n\nAlso, in some cases, the calculations of IRR may provide two solutions. This fact creates ambiguity and unnecessary confusion regarding the correct outcome. Unlike the IRR, the MIRR calculations always return a single solution.\n\nThe common view is that the MIRR provides a more realistic picture of the return on the investment project relative to the standard IRR. The MIRR is commonly lower than the IRR.",
null,
"### Example of MIRR\n\nLet’s consider the following example. Company A wants to assess the investment viability of its upcoming project of building a new plant. The company must spend \\$200 million on the plant’s construction. At the same time, it expects that the new plant will generate revenues of \\$50 million in the first year, \\$100 million in the second year, and \\$150 million in the third year. Note that the cost of capital of Company A is 10%.\n\nUsing the information above, we may calculate the modified internal rate of return of the project. First, we need to calculate the future value of positive cash flows at the reinvestment rate. We may assume that the reinvestment rate equals the company’s cost of capital.\n\nThe present value of negative cash flows discounted at the financing rate is simply \\$200 million because there is only one cash outflow occurring before the project. Therefore, we can use the variables to calculate the modified internal rate of return (MIRR):\n\nThe modified internal rate of return for the project is 17.02%. In order to determine the investment viability of the project, the figure may be later compared with the expected return of the project.\n\nCFI is the official provider of the global Financial Modeling & Valuation Analyst (FMVA)™ certification program, designed to help anyone become a world-class financial analyst. To keep advancing your career, the additional CFI resources below will be useful:\n\n• Cash Flow Statement\n• Due Diligence in Project Finance\n• Return on Investment (ROI)\n• Required Rate of Return\n\n### Financial Analyst Training\n\nGet world-class financial training with CFI’s online certified financial analyst training program!\n\nGain the confidence you need to move up the ladder in a high powered corporate finance career path.\n\nLearn financial modeling and valuation in Excel the easy way, with step-by-step training."
]
| [
null,
"https://corporatefinanceinstitute.com/resources/knowledge/finance/modified-internal-rate-of-return-mirr/",
null,
"https://corporatefinanceinstitute.com/resources/knowledge/finance/modified-internal-rate-of-return-mirr/",
null,
"https://corporatefinanceinstitute.com/resources/knowledge/finance/modified-internal-rate-of-return-mirr/",
null
]
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https://www.careerstoday.in/school/ncert-solutions-class-8-maths-chapter-13-direct-and-inverse-proportions | [
"# NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions",
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"NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions: You must have come across many situations where changes in one quantity result in changes in other quantities. In mathematics, two quantities are said to be in proportional if the values of two quantities are related in such a way that changes in one results in a corresponding change in another quantity. You must have observed that when you increase the speed of the car it takes lesser time to cover the same distance. So speed is inversely proportional to time with a constant distance.\n\nLatest : Trouble with homework? Post your queries of Maths and Science with step-by-step solutions instantly. Ask Mr AL\n\nIn NCERT solutions for Class 8 Maths chapter 13 Direct and Inverse Proportions, you are going to deal with problems based on direct and indirect proportions. In this chapter, there are 2 exercises with 21 questions. The first exercise of this chapter cover problems based on the direct proportions and the second exercise cover problems based on the inverse proportions. All these questions are prepared in NCERT solutions for Class 8 Maths chapter 13 Direct and Inverse Proportions in a step-by-step manner. It will be very easy for you to understand the concept. There are solved examples, exercises, and daily life activities in the textbook for a better understanding of this chapter.\n\nWhat Does ‘Pi’ In Maths Have To Do With Your Phone Number? Find Out Here 4 min read Mar 05, 2022 Read More Concepts Of Physics: What Is Free-Fall And What Factors Affect It? Read Here 5 min read Mar 05, 2022 Read More\n\nIn solutions for Class 8 Maths chapter 13 Direct and Inverse Proportions, you will deal with problems related to properties and applications of direct proportion and inverse proportion. You can find NCERT Solutions from Classes 6 to 12 of Science and Maths by clicking on the above link. Here you will get the detailed NCERT Solutions for Class 8 by clicking on the link.\n\n## NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Topic 13.2 Direct Proportion\n\nQuestion:1(i). Observe the following tables and find if x and y are directly proportional.\n\nIf we want to know that they are directly proportional, then we calculate :\n\n.\n\nHence we can say that x and y are directly proportional as cames out to be a constant equals .\n\nQuestion:1(ii). Observe the given table and find if x and y are directly proportional.\n\nIf x and y are directly proportional then must be equal to a constant value let say some 'k'.\n\nHence we then calculating:\n\n,\n\nas you can see that all these values are not equal hence,\n\nwe can say that x and y are not directly proportional.\n\nQuestion:1(iii) Observe the below table and find if x and y are directly proportional.\n\nCalculating we get:\n\nSo, clearly equals to hence it is not equal to some constant value 'k'.\n\nWe can say that x and y are not directly proportional .\n\nQuestion:2. Principal = Rs.1000, Rate = 8% per annum. Fill in the following table and find which type of interest (simple or compound) changes in direct proportion with time period.\n\nGiven that Principal (P) = 1000 and Rate (r) = 8% per annum(per year).\n\nCalculating the Simple Interest:\n\nThe formula for the simple interest is = .\n\nSo, for 1 year :\n\n.\n\nfor 2 years :\n\n.\n\nsimilarly for 3 years:\n\n.\n\nCalculating the Compound Interest :\n\nThe formula for the compound interest is\n\n.\n\nSo for 1 year :\n\n.\n\nfor 2 years :\n\n.\n\nsimilarly for 3 years:\n\n.\n\nHence we have\n\n Time period 1 year 2 year 3 year Simple Interest(in rupees) 80 160 240 Compound Interest(in rupees) 80 166.4 259.71\n\nIn case of simple interest\n\nSimple interest is directly proportional with time.\n\nWhile in case of compound interest:\n\ndoes not give the same constant.\n\nCompound interest is not directly proportional with time.\n\n## NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions-Exercise: 13.1\n\nQuestion:1 Following are the car parking charges near a railway station upto\n4 hours Rs. 60\n8 hours Rs.100\n12 hours Rs. 140\n24 hours Rs.180\nCheck if the parking charges are in direct proportion to the parking time.\n\nWe say that x and y are in direct proportion, if or\n\nHence in the given problem we have,\n\nCharges for the parking :\n\n y x 4 hours Rs. 60 8 hours Rs. 100 12 hours Rs. 140 24 hours Rs. 180\n\nconsidering x to be charged to be paid and y to be the parking time.\n\nThen we calculate ,\n\nFor:- 4 hours: 8 hours :\n\n12 hours : 24 hours:\n\nClearly is not equal to some constant 'k', then we can say that\n\nParking charges are not in direct proportion to parking time.\n\nQuestion:2 A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.\n\nSuppose the parts of red pigment is x and parts of base, is ' y '.\n\nAs the requirement of the number of parts of base for 1 part of red pigment is 8,\n\nand as the parts of red pigment is increases, parts of the base also increase in the same ratio. It is a case of direct proportion.\n\nwe can assume other parts of the base that will be required for red pigments as y1, y2, y3 and y4 for parts of pigment 4, 7, 12 and 20 respectively.\n\nWe make use of the relation of type\n\nThat gives for the (i) case or .\n\n32 parts of the base will be required for the 4 parts of the red pigment.\n\n(ii) If parts of red pigment used is 7 then parts of base used will be\n\nthat gives or .\n\n56 parts of the base will be required for the 7 parts of the red pigment.\n\n(iii) for 12 parts of red pigment :\n\nwe have or .\n\nso, 96 parts of the base will be required for the 12 parts of the red pigment.\n\n(iv) for 20 parts of red pigment:\n\nwe have or .\n\nHence for the following parts of red pigment, parts of base are given :\n\n Parts of red pigment 1 4 7 12 20 Parts of base 8 32 56 96 160\n\nQuestion:3 In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?\n\nAs we know that mL of red pigment is directly proportional to the mL of base.\n\nWe have to calculate for the mL of red pigment used when we mix 1800mL of base,\n\nSo, for 1 part of red pigment requires 75mL of base, then we\n\n.\n\nSo for 1800mL of the base, mL of the red pigment should be mixed\n\nor\n\nHence for the 1800mL of base, we will mix 24 parts of red pigment .\n\nQuestion:4 A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?\n\nGiven that a machine is capable of filling 840 bottles in 6 hours. So, we can say that it is directly proportional to the time.\n\nwe can assume that bottles it will fill in five hours would be ' x '\n\nSo we got the relation; ,\n\nsolving for x we will get or .\n\nMachine will fill 700 bottles in 5 hours .\n\nQuestion:5 A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?\n\nWe can calculate it easily,\n\nGiven that for 50,000 times enlarged, the photograph of a bacteria attains a length of 5cm.\n\nSo, we have to calculate the actual length of the bacteria(assume it to be ' x 'cm) that is when the photograph is enlarged to 1 time\n\nKnowing that the microscope's zoom has a direct relation with the length of bacteria observed So, we get the relation ;\n\nSolving the equation for x we get;\n\nThe actual length of the bacteria is which is so small to be observed through naked eyes.\n\nNow, calculating the length of bacteria when it the photograph is enlarged 20,000 times,\n\nassume it as ' y '.\n\nSo, we get this relation\n\nsolving for y we get;\n\n.\n\nSo, if the photograph is enlarged 20,000 times only then the enlarged length would be 2cm.\n\nQuestion:6 In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?\n\nGiven that the mast of the model ship is 9 m high, while the mast of the actual ship is 12 m.\n\nIf the length of the ship given is 28m then,\n\nLength of the model ship can be obtained from the direct proportion relation of mast height and length of ship.\n\nWe obtained the relation:\n\nSo, by the relation we have;\n\nLet us assume that the length of model ship is ' x '\n\nafter solving this relation we get,\n\n,\n\nLength of the model ship is 21cm.\n\nQuestion:7 Suppose 2 kg of sugar contains crystals. How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?\n\nHere given that 2 kg of sugar contains crystals.\n\nSo, we have to find the number of crystals in 5 kg of sugar as well as in 1.2 kg of sugar:\n\nAs here we will assume that all crystals have the same dimensions i.e., length, breadth, and width. Then the weight of sugar follows the direct proportion with the number of crystals as increasing the number of crystals there will be an increase in the weight also.\n\n(i) For 5 kg sugar:\n\nlet the number of crystals be 'x' then,\n\nwe have the relation:\n\n, calculating x from this relation,\n\n.\n\nTherefore 5 kg of sugar contains sugar crystals.\n\n(ii) For 1.2 kg sugar:\n\nLet the number of cystals be 'y' then,\n\nwe have the relation:\n\n. calculating similarly for y we get\n\n.\n\nTherefore 1.2 kg of sugar contains sugar crystals.\n\nQuestion:8 Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?\n\nGiven that Rashmi has a road map with a scale of 1cm representing 18km.\n\nSo we have a direct relation of scale distance and distance driven in the road.\n\nFor 72 km she drove on a road, Distance covered in the map would be 'x' cm\n\n,\n\n.\n\nHence, For 72km driven in the road, we have to move a distance of 4cm on the map.\n\nQuestion:9 A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time\n\n(i) the length of the shadow cast by another pole 10 m 50 cm high\n\n(ii) the height of a pole which casts a shadow 5m long.\n\nConsider there is a direct proportion relation of pole height with pole shadow.\n\nSo we have 5m 60cm high verticle pole that casts a shadow of 3m 20cm long .\n\nhence,\n\n(i) for the length of the shadow cast by another pole of 10 m 50cm high would be ' x ' cm;\n\nfinding x from the equation we get;\n\n.\n\nTherefore for a 10m 50cm pole we would get a shadow of 600cm or 6m length.\n\n(ii) The height of a pole which casts a shadow of 5m long let it be 'y'\n\nsimilar relation holds here also, so we can apply it once more\n\n,\n\nwe get .\n\nThus, the height of a pole which casts a shadow of 5m long is 8m 75cm.\n\nQuestion:10 A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?\n\nIf the speed of the truck remains the same, then we can say that the distance travelled by the truck is directly proportional to the time.\n\nAfter taking 5 hours the truck will travel a distance of let say ' x ';\n\nHence we obtain the relation:\n\n;\n\nthus the truck would travel a distance of 168km in 5 hours.\n\n## NCERT solutions for class 8 maths chapter 13 direct and inverse proportions topic 13.3 inverse proportion\n\nQuestion:(i) Observe the following tables and find which pair of variables (here x and y) are in inverse proportion.\n\nFinding if x and y are in inverse proportion:\n\nTwo quantities x and y are said to vary in inverse proportion, if there exists a relation of the type xy = k between them, k being a constant\n\nso, calculating xy for each case,\n\nwe have\n\n,\n\nclearly xy is not equal to a constant value 'k',\n\nhence x and y are not inversely proportional.\n\nQuestion:(ii) Observe the given table and find which pair of variables (here x and y) are in inverse proportion.\n\nFinding if x and y are in inverse proportion:\n\nTwo quantities x and y are said to vary in inverse proportion, if there exists a relation of the type xy = k between them, k being a constant\n\nso, calculating xy for each case,\n\nwe have\n\n,\n\nclearly xy is equal to a constant value 'k=6000',\n\nhence x and y are inversely proportional.\n\nQuestion:(iii) Observe the below table and find which pair of variables (here x and y) are in inverse proportion.\n\nFinding if x and y are in inverse proportion:\n\nTwo quantities x and y are said to vary in inverse proportion, if there exists a relation of the type xy = k between them, k being a constant\n\nso, calculating xy for each case,\n\nwe have\n\nclearly xy is not equal to a constant value 'k',\n\nhence x and y are not inversely proportional.\n\n## NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions-Exercise: 13.2\n\nQuestion:1 Which of the following are in inverse proportion?\n\n(i) The number of workers on a job and the time to complete the job.\n\n(ii) The time taken for a journey and the distance travelled in a uniform speed.\n\n(iii) Area of cultivated land and the crop harvested.\n\n(iv) The time taken for a fixed journey and the speed of the vehicle.\n\n(v) The population of a country and the area of land per person.\n\n(i) As the number of workers on a job increases the time taken to complete the job decreases, hence it is an inverse proportion .\n\n(ii) Distance and time are directly proportional to each other as time increases you could travel more distance compared to if you get less time to travel. Hence it is not an inverse proportion.\n\n(iii) Both area of cultivated land and crop arvested are directly proportional, more the area cultivated more crop harvest. Hence it is not an inverse proportion.\n\n(iv) With more speed if you are travelling lesser the time taken for a fixed journey to complete. Hence it is an inverse proportion.\n\n(v) The population of a country if increases then there would be lesser area available per person, Hence it is an inverse proportion.\n\nQuestion:2 In a Television game show, the prize money of ` 1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners?\n\n Number of winners 1 2 4 5 8 10 20 Prize for each winner (in Rs.) 1,00,000 50,000 .... .... .... .... ....\n\nLet us assume that for the number of winners 4, 5, 8, 10 and 20 are x1, x2, x3, x4, and x5 respectively and given that the prize money of Rs. 1,00,000 is to be divided equally amongst the winners.\n\nThus we have,\n\nFor 4 number of winners:\n\nSo Rs. 25,000 to be distributed among each.\n\nFor 5 number of winners:\n\nSo Rs. 20,000to be distributed among each.\n\nFor 8 number of winners:\n\nSo Rs. 12,500 to be distributed among each.\n\nFor 10 number of winners:\n\nSo Rs. 10,000 each would get.\n\nsimilarly for 20 number of winners:\n\nSo Rs. 5000 each would get.\n\nHence we have;\n\n Number of winners (x) 1 2 4 5 8 10 20 Prize for each winner (in ? ) (y) 1,00,000 50,000 25,000 20,000 12,500 10,000 5,000\n\nTwo quantities x and y are said to be in inverse proportional if they satisfy the given relation;\n\nxy=k; where k is a constant.\n\nCalculating xy:\n\nClearly, we can see that the prize money given to an individual winner is inversely proportional.\n\nQuestion:3(i) Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table.\n\nAre the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?\n\n Number of spokes 4 6 8 10 12 Angle between a pair of consecutive .... .... ....\n\n(i) Calculating the angle formed when using a different number of spokes:\n\nThe angle formed when using 8, 10, and 12 number of spokes a1,a2, and a3 respectively.\n\nHence, we have\n\nFor 8 number of spokes:\n\nFor 10 number of spokes:\n\nFor 12 number of spokes:\n\n Number of spokes (x) 4 6 8 10 12 Angle between a pair of consecutive spokes (y) 90° 60° 45° 36° 30°\n\nCalculating xy:\n\nHence we say that the number of spokes (x) and the angle formed (y) between them are inverse proportion to each other.\n\nQuestion:3(ii) Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table.\n\nCalculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.\n\n Number of spokes (x) 4 6 8 10 12 Angle between a pair of consecutive spokes (y) 90° 60° 45° 36° 30°\n\nThe angle between a pair of consecutive spokes on a wheel with 15 spokes is calculated as:\n\nas we know the constant value k=360 then we can easily calculate the angle for 15 spokes:\n\nHence angle made is 24 degrees .\n\nQuestion:3(iii) Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table.\n\nHow many spokes would be needed, if the angle between a pair of consecutive spokes is ?\n\n Number of spokes (x) 4 6 8 10 12 Angle between a pair of consecutive spokes (y) 90° 60° 45° 36° 30°\n\nIf the angle between a pair of consecutive spokes is then,\n\nNumber of spokes needed would be\n\nHence the required number of spokes for having 40 degrees of angle is 9\n\nQuestion:4 If a box of sweets is divided among children, they will get sweets each. How many would each get, if the number of the children is reduced by ?\n\nGiven that the box of sweets is divided among 24 children, getting 5 sweets each.\n\nIf the number of children is reduced by 4 then the number of children now is .\n\nAs here if the number of children increases then the number of sweets they will get decreases hence we can say that there exists an inverse relationship between them.\n\nHence,\n\nif we assume:\n\nNumber of children before (x1) = 24\n\nNumber of sweets each would get before (y1) = 5\n\nand number of children after reduction (x2) = 20\n\nand the number of sweets each would get after reduction is y2\n\nThen the relation holds;\n\nor or .\n\nHence each child will get 6 number of sweets .\n\nQuestion:5 A farmer has enough food to feed animals in his cattle for days. How long would the food last if there were more animals in his cattle?\n\nThe farmer has 20 animals to feed for 6 days and after that farmer added 10 more animals to feed that counts to ( 20+10 = 30 ). So, assume that food will now long last to ' x ' days.\n\nAs the number of animals increases the required food increases but the day up to which the food will long last decreases.\n\nHence there exist an inverse proportion between the number of days and the number of animals.\n\nSo, we can write the relation as:\n\nor days.\n\nThe food will last for about 4 days if 30 animals are there.\n\nQuestion:6 A contractor estimates that persons could rewire Jasminder’s house in days. If, he uses persons instead of three, how long should they take to complete the job?\n\nHere the situation is given that a contractor estimates that 3 persons could rewire Jasminder's house in 4 days.\n\nNow, as the number of person increases the time they take to complete the job will decrease .\n\nHence there is an inverse relationship among the number of persons and the time they took.\n\nJasminer now uses 4 persons instead of 3, then we can assume the time 4 persons will take be ' x '\n\nSo, we can write the relation as:\n\n. or .\n\nThus, the 4 persons complete the job in 3 days.\n\nQuestion:7 A batch of bottles were packed in boxes with bottles in each box. If the same batch is packed using bottles in each box, how many boxes would be filled? Answer:\n\nWe can easily calculate the required number of boxes to be filled, let us assume it to be 'x'\n\nGiven that a batch of bottles were packed in 25 boxes with 12 bottles in each box, Hence the total number of bottles will be .\n\nSo, as they produced the same number of bottles every batch and as the number of boxes increases, the bottles in each box decrease.\n\nHence there holds an inverse relation here,\n\nor boxes.\n\nIf the batch is packed using 20 bottles in each box then 15 boxes would be filled.\n\nQuestion:8 A factory requires machines to produce a given number of articles in days. How many machines would be required to produce the same number of articles in days?\n\nGiven that a factory requires 42 machines to produce a given number of articles in 63 days .\n\nThen we know that as the number of machines increases the time taken to produce a given number of articles decreases. Hence there holds an inverse relation here,\n\nAnd let the number of machines required to produce the same number of articles in 54 days be 'x'.\n\nThen the relation;\n\nor .\n\nHence 49 machines would be required to produce the articles in 54 days.\n\nQuestion:9 A car takes hours to reach a destination by travelling at the speed of . How long will it take when the car travels at the speed of ?\n\nGiven that a car has a speed of 60 km/h which is travelling to a destination and takes 2 hours to complete it.\n\nSpeed and time are inversely related to each other.\n\nAssume the time it would take when travelling at 80km/h be 'x'\n\nTherefore we can write the equation when the car travels at the speed of 80 km/h as:\n\nor or\n\nQuestion:10(i) Two persons could fit new windows in a house in 3 days.\n\nOne of the persons fell ill before the work started. How long would the job take now?\n\nHere, given that 2 persons could fit new windows in a house in 3 days.\n\n(i) 1 person has fallen ill so, now the number of persons remaining is only one. Assume that the only person which is working takes the time of 'x' days.\n\nhence we could write the inverse relation as;\n\nor .\n\nOne person will take 6 days to complete that window job.\n\nQuestion:10(ii) Two persons could fit new windows in a house in 3 days.\n\nHow many persons would be needed to fit the windows in one day?\n\n(ii) So, now we are calculating the number of persons that would be needed to fit the windows in one day. Let it be 'y'.\n\nSo from previous part (i) we have the relation; .\n\nor .\n\nhence the required number of persons would be 6.\n\nQuestion:11. A school has periods a day each of minutes duration. How long would each period be, if the school has periods a day, assuming the number of school hours to be the same?\n\nGiven :\n\nA school has 8 periods a day each of 45 minutes duration.\n\nSo, if the school hours of school is fixed then there exists an inverse relationship between each period duration and the number of periods.\n\nWe can take the time of each period to be 't' if the school has 9 periods a day.\n\nHence we can write the relation as;\n\nor .\n\nThe time of each period would be 40 minutes when 9 periods are there in one day.\n\n## Direct and Inverse Proportions Class 8 Chapter 14-topics\n\n• Direct Proportion\n• Inverse Proportion\n\nWhat is a Direct Proportion?\n\nHow Should You Prepare For The Physical Chemistry Questions In NEET And JEE Main? Read Here 7 min read Mar 05, 2022 Read More KCET 5-Year Analysis: Which Topics Get More Weightage And Why 9 min read Mar 05, 2022 Read More\n\nTwo quantities 'a' and 'b' are said to be in direct proportion if the increase (decrease) in 'a' results in the increase (decrease) in 'b' in such a manner that the ratio of their corresponding values remains constant. That is if a/b= k\n\nWhat is Inverse Proportional?\n\nTwo values are Inversely proportional it means decreases in one value results in a gradual increase in other value. Example- If the distance is fixed, speed and time are inversely proportional to each other.\n\n( If Distance is fixed)\n\n### NCERT Solutions for Class 8 Maths: Chapter-Wise\n\n Chapter -1 Rational Numbers Chapter -2 Linear Equations in One Variable Chapter-3 Understanding Quadrilaterals Chapter-4 Practical Geometry Chapter-5 Data Handling Chapter-6 Squares and Square Roots Chapter-7 Cubes and Cube Roots Chapter-8 Comparing Quantities Chapter-9 Algebraic Expressions and Identities Chapter-10 Visualizing Solid Shapes Chapter-11 Mensuration Chapter-12 Exponents and Powers Chapter-13 Direct and Inverse Proportions Chapter-14 Factorization Chapter-15 Introduction to Graphs Chapter-16 Playing with Numbers\n\n### NCERT Solutions for Class 8: Subject-Wise\n\n• NCERT Solutions for Class 8 Maths\n• NCERT Solutions for Class 8 Science\n\nBenefits of NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions-\n\n• You will also know different ways to solve the problems.\n\n• You will also get solutions to the practice questions given below every topic which will give you conceptual clarity.\n\n• You will also get some short tricks and tips to solve some specific problems.\n\n• In one of the questions of NCERT solutions for Class 8 Maths chapter 13 Direct and Inverse Proportions, you will learn to calculate simple and compound interest which may be useful in real life too."
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https://faculty.sites.iastate.edu/zerbib/cosp-seminar-topological-combinatorics | [
"# CoSP Seminar on Topological Combinatorics\n\nDuring summer 2020 I coordinated a CoSP Zoom seminar on topological combinatorics, replacing a conference on the same topic that was supposed to take place in Prague and was cancelled due to COVID-19:\n\nThe talks will be given on Tuesdays and Thursdays at 8:00am (CDT), as of June 30, 2020.\n\nSchedule:\n\n6/30 - Andreas Holmsen (KAIST)\n\nTitle: The Fractional Helly property and topological complexity\n\nAbstract: The fractional Helly theorem is a simple yet remarkable generalization of Helly’s classical theorem on the intersection of convex sets, and it is of considerable interest to extend the fractional Helly theorem beyond the setting of convexity. In this talk I will discuss a recent result which shows that the fractional Helly theorem holds for families of subsets of R^d which satisfy only very weak topological assumptions. This is joint work with Xavier Goaoc and Zuzana Patáková.\n\n7/2 - Florian Frick (Carnegie Mellon University)\n\nTitle: The topological Tverberg problem beyond prime powers\n\nAbstract: Given d and q the topological Tverberg problem asks for the minimal n such that any continuous map from the n-dimensional simplex to R^d identifies q points from pairwise disjoint faces. For q a prime power n is (q-1)(d+1). The lower bound follows from a general position argument, the upper bound from equivariant topological methods. It was shown recently that for q with at least two distinct prime divisors the lower bound may be improved. For those q non-trivial upper bounds had been elusive. I will show that n is at most q(d+1)-1 for all q. I had previously conjectured this to be optimal unless q is a prime power. This is joint work with Pablo Soberón.\n\nSlides\n\n7/7 - Minki Kim (Technion)\n\nTitle: Complexes of graphs with bounded independence number\n\nAbstract: Let G be a graph on V and n a positive integer. Let I_n(G) be the simplicial complex whose faces are the subsets of V that do not contain an independent set of size n in G. We study the collapsibility numbers of the complexes I_n(G) for various classes of graphs, focusing on the class of graphs with bounded maximum degree. As an application, we obtain the following result: Let G be a claw-free graph with maximum degree at most k. Then every collection of (k/2 +1)(n-1)+1 independent sets of size n in G has a rainbow independent set of size n. This is joint work with Alan Lew.\n\n7/9 - Zilin Jiang (MIT)\n\nTitle: Rainbow odd cycles\n\nAbstract: The classical colorful Carathéodory theorem says that for every family of d+1 point sets in R^d, if each point set contains 0 in its convex hull, then there is a rainbow set with the same property. Here, if we drop the critical cardinality of the family to d, generically, such a rainbow set does not exist. However we start to see that rainbow problems in more combinatorial settings present stability. For example, Drisko’s theorem on matchings in a bipartite graph still holds for 2n-2 matchings of size n unless their union is a cycle of length 2n. In this talk, I will illustrate a similar phenomenon for cycles with or without parity constraints on their lengths. Joint work with Ron Aharoni, Joseph Briggs and Ron Holzman.\n\n7/14 - Gabor Simonyi (Renyi Institute)\n\nTitle: Color-critical edges in Schrijver graphs\n\nAbstract: Schrijver graphs are vertex-color critical induced subgraphs of Kneser graphs. In general they are not edge-color-critical. We give necessary conditions and sufficient conditions for their individual edges to be color-critical. Our work started with investigating 4-chromatic Schrijver graphs more thoroughly and for those our results give a complete characterization of the color-critical edges. We also show that 4-chromatic Schrijver graphs contain spanning subgraphs that quadrangulate the Klein bottle and (apart from two cases of small parameters) these spanning subgraphs are edge-color-critical. (Analogous results for spanning subgraphs quadrangulating the projective plane follow from work by Kaiser-Stehlík and Gimbel-Thomassen.) The work presented in this talk is joint with Gábor Tardos.\n\n7/16 - Jinha Kim (Technion)\n\nTitle: Domination numbers and noncover complexes of hypergraphs\n\nAbstract: Let H be a hypergraph on a finite set V. A cover of H is a set of vertices that meets all edges of H. If W is not a cover of H, then W is said to be a noncover of H. The noncover complex of H is the abstract simplicial complex whose faces are the noncovers of H. In this talk, I will present about some homological properties of noncover complexes of hypergraphs. In particular, we obtain an upper bound on their Leray numbers in terms of domination numbers. This is joint work with Minki Kim.\n\n7/21 - Tomas Kaiser (University of West Bohemia) & Matej Stehlik (Universite Grenoble Alpes) - double talk (45x2 minutes)\n\nTitle: Edge-critical subgraphs of Schrijver graphs\n\nAbstract: We give a simple combinatorial description of an (n-2k+2)-chromatic edge-critical subgraph of the Schrijver graph SG(n,k), itself an induced vertex-critical subgraph of the Kneser graph KG(n,k). This extends the main result of [J. Combin. Theory Ser. B 144 (2020) 191-196] to all values of k, and sharpens the classical results of Lovász and Schrijver from the 1970s.\n\n7/23 - Amzi Jeffs (University of Washington)\n\nTitle: Open and closed convex codes and their embedding dimensions\n\nAbstract: Given a collection of open convex sets in Euclidean space, one can write down a combinatorial code that records how the sets intersect and overlap one another. One can try to reverse this process: given a combinatorial code, can you find a collection of convex open sets that gives rise to it? The smallest dimension in which one can find this collection is called the open embedding dimension of the code. Analogously, the closed embedding dimension is the smallest dimension in which one can find a collection of closed convex sets giving rise to the code. We will use a new Helly-style theorem to show that these two numbers exhibit surprisingly different behavior. In particular, the gap between the two may be exponentially large compared to the number of convex sets.\n\nSlides\n\n7/28 - Martin Loebl (Charles University)\n\nTitle: Some aspects of winter road maintenance\n\nAbstract: I will describe some combinatorial and algorithmic features of a model of winter road maintenance, including a relation to necklace splitting. (joint work with Jirka Fink, Petra Pelikanova)\n\n7/30 - Pablo Soberon (Baruch College)\n\nTitle: Variations of equipartitions of measures with convex sets\n\nAbstract: Given d measures in R^d and an integer k, there exists a partition of R^d into k convex pieces of the same size in each measure. During this talk we will talk about two extensions of this result. In the first one we are given more than d measures, and would still like each piece to be large for sufficiently many measures. In the second one, we are given sets of lines in R^2 instead of sets of points, and we show similar equipartition results. The first part is joint work with Blagojevic, Palic, and Ziegler, the second part is joint work with Xue.\n\n8/4 - Joe Briggs (Technion)\n\nTitle: Making Rainbow Matchings Together\n\nAbstract: Aharoni-Berger-Chudnovsky-Howard-Seymour proved 3n-2 matchings of size n span a rainbow matching of size n. Remarkably, Holmsen-Lee showed topologically that it suffices to take any 3n-2 edge sets each of whose pairwise unions contain a matching of size n. We reprove their cooperative version using a purely combinatorial method, and offer a corresponding result for bipartite graphs. In this talk I will give an overview of these proofs while showing we still have more questions than answers. This is based on joint works with Ron Aharoni, Minho Cho, Jinha Kim, and Minki Kim.\n\nSlides\n\n9/17 - Denys Bulavka (Charles University)\n\nTitle: Optimal bounds for the colorful fractional Helly theorem\n\nAbstract: The well known fractional Helly theorem and colorful Helly theorem can be merged into so called colorful fractional Helly theorem. It states: For every alpha in (0, 1] and every non-negative integer d, there is beta = beta(alpha, d) in (0, 1] with the following property. Let F_1, ..., F_{d+1} be finite nonempty families of convex sets in R^d of sizes n_1, ..., n_{d+1} respectively. If at least alpha n_1 n_2 ... n_{d+1} of the colorful (d+1)-tuples have a nonempty intersection, then there is i in [d+1] such that F_i contains a subfamily of size at least b n_i. (A colorful (d+1)-tuple is a (d+1)-tuple (C_1,...,C_{d+1}) such that C_i belongs to F_i for every i.)\n\nThe colorful fractional Helly theorem was first stated and proved by Bárány, Fodor, Montejano, Oliveros, and Pór in 2014 with beta = alpha/(d+1).\nIn 2017 Kim proved the theorem with better function beta, which in particular tends to 1 when alpha tends to 1. Kim also conjectured what is the optimal bound for beta(alpha, d) and provided the upper bound example for the optimal bound. The conjectured bound coincides with the optimal bounds for the (non-colorful) fractional Helly theorem proved independently by Eckhoff and Kalai around 1984.\n\nWe prove Kim's conjecture by extending Kalai's approach to the colorful scenario. Moreover, we obtain bounds for other collections of sets, not just colorful (d+1)-tuples.\n\nThis is a joint work with Afshin Goodarzi and Martin Tancer.\n\nMore talks may be published soon."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.902378,"math_prob":0.8949298,"size":9399,"snap":"2020-45-2020-50","text_gpt3_token_len":2248,"char_repetition_ratio":0.10771687,"word_repetition_ratio":0.012987013,"special_character_ratio":0.2182147,"punctuation_ratio":0.098387994,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9697269,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-04T18:05:57Z\",\"WARC-Record-ID\":\"<urn:uuid:f2c7f084-75d8-4a3c-9f3d-b7b7ec910573>\",\"Content-Length\":\"30288\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cd314b13-08a5-4f2e-8e58-61a72f22db2e>\",\"WARC-Concurrent-To\":\"<urn:uuid:bfc6304a-cc8f-40fb-a590-5e4ca1f681be>\",\"WARC-IP-Address\":\"129.186.138.50\",\"WARC-Target-URI\":\"https://faculty.sites.iastate.edu/zerbib/cosp-seminar-topological-combinatorics\",\"WARC-Payload-Digest\":\"sha1:HULP3J7FAZHXEKDJHZE5U4YYU2SEJFM5\",\"WARC-Block-Digest\":\"sha1:CRIPAMBW6JIRZXR46K4BPMQ4VNSTTNLH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141740670.93_warc_CC-MAIN-20201204162500-20201204192500-00256.warc.gz\"}"} |
https://tutorme.com/tutors/59931/interview/ | [
"TutorMe homepage\nSubjects\nPRICING\nCOURSES\nStart Free Trial\nSagar B.\nMechanical Engineer by academia and profession.\nTutor Satisfaction Guarantee\nPhysics (Newtonian Mechanics)\nTutorMe\nQuestion:\n\nIn a one-dimensional collision, a particle of mass 2m collides with a particle of mass m at rest. If the particles stick together after the collision, what fraction of the initial kinetic energy is lost in the collision?\n\nSagar B.\n\nFirst we apply momentum conservation since the collosion is inelastic to find the final velocity of both the masses together. This is followed by finding difference in initial kinetic energy and final kinetic energy using the velocity found from first step.\n\nPhysics (Fluid Mechanics)\nTutorMe\nQuestion:\n\nWater flows over a sharp flat plate 3m long, 3 m wide with an approach velocity of 10m/s. Estimate the error in drag force in the flow over the entire plate is assumed turbulent. Assumed the mixed regions can be expressed by the following coefficient of drag relationship. Cd=(0.074/(Re)^0.2)-(1742/ReL). For water, density is 1000kg/m3 and kinematic viscosity as 1*10^-6 m/s^2\n\nSagar B.\n\nAssume turbulent, ReL=Vinf*L/kinematic viscosity=3*10^7 which is greater than 5*10^5 and thus turbulent Cd= 0.074/(Re)^0.2=0.0024 Fd=0.5*row*A*Vinf^2*Cd=o.5*1000*(3*3)*10^2*0.0024=1064N b) Now we treat the plate as two parts, laminar section, followed by a turbulent section Cd=0.074/(Re)^0.2 -1742/ReL= (0.074/(3*10^7)^0.2)-(1742/3*10^7)=0.0023 Fd=0.5*1000*9*10^2*0.0023=1038N Hence, error=1064-1038/1038=2.5% high\n\nGeometry\nTutorMe\nQuestion:\n\nThe diagonal [𝐵𝐷] of parallelogram 𝐴𝐵𝐶𝐷 is divided by points 𝑀, 𝑁, in 3 segments. Prove that 𝐴𝑀𝐶𝑁 is a parallelogram and find the ratio between 𝜎[𝐴𝑀𝐶𝑁] and 𝜎[𝐴𝐵𝐶𝐷].\n\nSagar B.\n\nGiven :Let O be the intersection point of the diagonals of parallelogram ABCD. ‖D𝑀‖ = ‖𝑁𝑀‖ = ‖𝑁𝐵‖ ‖𝐷𝐶‖?⟹ ∆𝑀𝑂𝐶 = ∆𝑁𝐵𝐴 ⟹ ‖𝑀𝐶‖ = ‖𝐴𝑁‖ It is proved in the same way that ∆𝐷𝐴𝑀 = ∆𝐵𝐶𝑁 ⟹ ‖𝑀𝐶‖ = ‖𝑁𝐶‖. Thus 𝐴𝑁𝐶𝑀 is a parallelogram. Area of triangle AOB= 0.5*||OA|| ||OB|| sin @ Area of triangle AOD=0.5*||OA|| ||OD|| sin(pi-@) Thus, Area of parallelogram ABCD=(Area(AOB)+Area(AOD)= ||OA|| ||DB|| sin @ Area of parallelogram AMCN=||OA|| ||MN|| sin@ =||OA|| ||BD/3|| sin@ Thus; Area of AMCN/Area of ABCD= 1/3\n\nSend a message explaining your\nneeds and Sagar will reply soon.\nContact Sagar"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8056967,"math_prob":0.94650745,"size":1934,"snap":"2019-26-2019-30","text_gpt3_token_len":786,"char_repetition_ratio":0.10414508,"word_repetition_ratio":0.0,"special_character_ratio":0.3402275,"punctuation_ratio":0.10739857,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994859,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-24T15:16:04Z\",\"WARC-Record-ID\":\"<urn:uuid:6ce435f3-022d-414e-a8c1-6e1fbb91fc8b>\",\"Content-Length\":\"147634\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:21363085-6d08-49a3-986f-026e9cf84cd7>\",\"WARC-Concurrent-To\":\"<urn:uuid:8e50a030-ca0f-49e7-a2d9-c0a5af4eb4f2>\",\"WARC-IP-Address\":\"99.84.216.45\",\"WARC-Target-URI\":\"https://tutorme.com/tutors/59931/interview/\",\"WARC-Payload-Digest\":\"sha1:3OZH4C7VDLFJO2QDS7ZYFBIPGLES3PQK\",\"WARC-Block-Digest\":\"sha1:TTDUYXEYQWPT34CR2NT3OQXP4VETSY4H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999615.68_warc_CC-MAIN-20190624150939-20190624172939-00190.warc.gz\"}"} |
https://answers.everydaycalculation.com/add-fractions/20-2-plus-7-56 | [
"Solutions by everydaycalculation.com\n\nAdd 20/2 and 7/56\n\n1st number: 10 0/2, 2nd number: 7/56\n\n20/2 + 7/56 is 81/8.\n\nSteps for adding fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 2 and 56 is 56\n2. For the 1st fraction, since 2 × 28 = 56,\n20/2 = 20 × 28/2 × 28 = 560/56\n3. Likewise, for the 2nd fraction, since 56 × 1 = 56,\n7/56 = 7 × 1/56 × 1 = 7/56\n4. Add the two fractions:\n560/56 + 7/56 = 560 + 7/56 = 567/56\n5. After reducing the fraction, the answer is 81/8\n6. In mixed form: 101/8"
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https://nm.dev/html/javadoc/nmdev/dev/nm/analysis/function/SubFunction.html | [
"# Class SubFunction<R>\n\njava.lang.Object\ndev.nm.analysis.function.SubFunction<R>\nType Parameters:\nR - the range of a function\nAll Implemented Interfaces:\nFunction<Vector,R>\nDirect Known Subclasses:\nRealScalarSubFunction, RealVectorSubFunction\n\npublic abstract class SubFunction<R> extends Object implements Function<Vector,R>\nA sub-function, g, is defined over a subset of the domain of another (original) function, f. g(x) = f(x) when they are both defined. This implementation constructs a sub-function by restricting/fixing the values of a subset of variables of another function.\n\n## Nested classes/interfaces inherited from interface dev.nm.analysis.function.Function\n\nFunction.EvaluationException\n• ## Field Summary\n\nFields\nModifier and Type\nField\nDescription\nprotected final Function<Vector,R>\nf\nthe original, unrestricted function\nprotected final Map<Integer,Double>\nfixing\nthe restrictions or fixed values\n• ## Constructor Summary\n\nConstructors\nConstructor\nDescription\nSubFunction(Function<Vector,R> f, Map<Integer,Double> fixing)\nConstructs a sub-function.\n• ## Method Summary\n\nModifier and Type\nMethod\nDescription\nint\ndimensionOfDomain()\nGet the number of variables the function has.\nint\ndimensionOfRange()\nGet the dimension of the range space of the function.\nstatic Vector\ngetAllParts(Vector variables, Map<Integer,Double> fixing)\nCombines the variable and fixed values to form an input to the original function.\nstatic double[]\ngetVariablePart(double[] z, Map<Integer,Double> fixing)\nGiven an input to the original function, this extracts the variable parts (excluding the fixed values).\nboolean\nisFixedIndex(int i)\nChecks whether a particular index corresponds a fixed variable/value.\nstatic boolean\nisFixedIndex(int i, Map<Integer,Double> fixing)\nChecks whether a particular index corresponds a fixed variable/value.\n\n### Methods inherited from class java.lang.Object\n\nclone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait\n\n### Methods inherited from interface dev.nm.analysis.function.Function\n\nevaluate\n• ## Field Details\n\n• ### f\n\nprotected final f\nthe original, unrestricted function\n• ### fixing\n\nprotected final fixing\nthe restrictions or fixed values\n• ## Constructor Details\n\n• ### SubFunction\n\npublic SubFunction(Function<Vector,R> f, Map<Integer,Double> fixing)\nConstructs a sub-function.\nParameters:\nf - the original, unrestricted function\nfixing - the values held fixed for a subset of variables\n• ## Method Details\n\n• ### isFixedIndex\n\npublic static boolean isFixedIndex(int i, Map<Integer,Double> fixing)\nChecks whether a particular index corresponds a fixed variable/value.\nParameters:\ni - an index, counting from 1\nfixing - fixed values\nReturns:\ntrue if xi is fixed to some value\n• ### getVariablePart\n\npublic static double[] getVariablePart(double[] z, Map<Integer,Double> fixing)\nGiven an input to the original function, this extracts the variable parts (excluding the fixed values).\nParameters:\nz - an input to the original function\nfixing - fixed values\nReturns:\nthe variable part in z\n• ### getAllParts\n\npublic static Vector getAllParts(Vector variables, Map<Integer,Double> fixing)\nCombines the variable and fixed values to form an input to the original function.\nParameters:\nvariables - the non-fixed variables/values to the restricted function\nfixing - the fixed values to the original function\nReturns:\nan input to the original function\n• ### dimensionOfDomain\n\npublic int dimensionOfDomain()\nDescription copied from interface: Function\nGet the number of variables the function has. For example, for a univariate function, the domain dimension is 1; for a bivariate function, the domain dimension is 2.\nSpecified by:\ndimensionOfDomain in interface Function<Vector,R>\nReturns:\nthe number of variables\n• ### dimensionOfRange\n\npublic int dimensionOfRange()\nDescription copied from interface: Function\nGet the dimension of the range space of the function. For example, for a Rn->Rm function, the dimension of the range is m.\nSpecified by:\ndimensionOfRange in interface Function<Vector,R>\nReturns:\nthe dimension of the range\n• ### isFixedIndex\n\npublic boolean isFixedIndex(int i)\nChecks whether a particular index corresponds a fixed variable/value.\nParameters:\ni - an index, counting from 1\nReturns:\ntrue if xi is fixed to some value"
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http://math.ivyglobal.com/questions/3/454/84 | [
"1 EASY Kristy notices a pattern of tulips in her backyard. They are red, pink, white, red, pink, white, etc. Which of the following best represents this pattern in words? It is a repeating pattern with a repeating unit of three objects and the repeating unit is red, pink, white, red.It is a repeating pattern with a repeating unit of three objects and the repeating unit is red, pink, white. It is a numeric pattern with a repeating unit of three objects, and the repeating unit is red, pink, white.It is a geometric pattern with a repeating unit of three objects, and the repeating unit is red, pink, white.it is a geometric pattern with a repeating unit of three objects, and the repeating unit is tall, medium, short.\n\n 2 MEDIUM Which of the following best expresses this pattern in words? 1, 3, 5, 7, 9... This is a numeric pattern starting on 1, and each term after the first is 2 more than the previous.This is a numeric pattern starting on 1, and each term after the first is an odd number.This is a numeric pattern, and each term is 2 more than the previous. This is not a pattern.This is a numeric pattern, and each term an even number.\n\n 3 MEDIUM Which of the following best expresses the following pattern in words? 21, 17, 13, 9... It is a numeric pattern beginning on 21 and each term is five less than the previous.It is a numeric pattern and each term is four less than the previous.It is a numeric pattern beginning on 21 and each term is four less than the previous.It is a numeric pattern beginning on 17 and each term is four less than the previous.It is a numeric pattern beginning on 21 and each term is three less than the previous.\n\n 4 HARD Which of the following best expresses this pattern in words? 1, 2, 4, 8, 16... It is a numeric pattern starting on 1 and each term is a multiple of 2.It is a numeric pattern starting on 1 and each term is larger than the previous.It is a numeric pattern starting on 1 and each term is double the previous.It is a numeric pattern and each term is half the previous.It is a repeating pattern with repeating unit 1, 2, 4, 8, 16.\n\n 5 HARD Which of the following best expresses the following pattern in words? 40, 20, 10, 5... It is a numeric pattern beginning on 40, and each term is half the previous.It is a numeric pattern and each term is half the previous.It is a numeric pattern and each term is twice the previous.It is a numeric pattern starting on 40 and each term is twice the previous.It is a geometric pattern starting on 40 and each term is twice the previous."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9479398,"math_prob":0.9649755,"size":2628,"snap":"2019-35-2019-39","text_gpt3_token_len":666,"char_repetition_ratio":0.21570122,"word_repetition_ratio":0.6500956,"special_character_ratio":0.27321157,"punctuation_ratio":0.1580756,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9814126,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-22T09:46:46Z\",\"WARC-Record-ID\":\"<urn:uuid:b6cfc67a-176e-4e79-9122-e44fbdbb0390>\",\"Content-Length\":\"12895\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5d60dbdf-a355-4eb3-ac15-6b749f99e774>\",\"WARC-Concurrent-To\":\"<urn:uuid:8c862bac-55e5-4955-b6ea-90de3c20ae11>\",\"WARC-IP-Address\":\"40.71.100.230\",\"WARC-Target-URI\":\"http://math.ivyglobal.com/questions/3/454/84\",\"WARC-Payload-Digest\":\"sha1:EOVYJVOU4IMBEL2KEFAILIS2ZE6TCSIV\",\"WARC-Block-Digest\":\"sha1:7UZATZWL75UHRMJJNUPUP4XKQ6H24HUT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514575484.57_warc_CC-MAIN-20190922094320-20190922120320-00296.warc.gz\"}"} |
https://rajeevranjansinha.com/blog/ | [
"# Blog\n\n## Langages of Data Science\n\nThe languages of Data Science\n\nFor anyone just getting started on their data science journey, the range of technical options can be overwhelming. There is a dizzying amount of choice when it comes to programming languages. Each has its own strengths and weaknesses and there is no one right answer to the question of which one you should learn first. The answer to that question depends largely on your needs, the problems you are trying to solve, and who you are solving them for.\n\nPython, R, and SQL are the languages that we recommend you consider first and foremost. But there are so many others that have their own strengths and features. Scala, Java, C++, and Julia are some of the most popular. Javascript, PHP, Go, Ruby, and Visual Basic all have their own unique use cases as well.\n\nThe language you choose to learn will depend on the things you need to accomplish and the problems you need to solve. It will also depend on what company you work for, what role you have, and the age of your existing application. We’ll explore the answers to this question as we dive into the popular languages in the data science industry. There are many roles available for people who are interested in getting involved in data science. Business Analyst Database Engineer Data Analyst Data Engineer Data Scientist Research Scientist Software Engineer Statistician Product Manager Project Manager and many more.\n\n## Practical Consideration in K Means Algorithm\n\nLet’s understand some of the factors that can impact the final clusters that you obtain from the K-means algorithm. This would also give you an idea about the issues that you must keep in mind before you start to make clusters to solve your business problem.\n\nThus, the major practical considerations involved in K-Means clustering are:\n\n• The number of clusters that you want to divide your data points into, i.e. the value of K has to be pre-determined.\n• The choice of the initial cluster centres can have an impact on the final cluster formation.\n• The clustering process is very sensitive to the presence of outliers in the data.\n• Since the distance metric used in the clustering process is the Euclidean distance, you need to bring all your attributes on the same scale. This can be achieved through standardisation.\n• The K-Means algorithm does not work with categorical data.\n• The process may not converge in the given number of iterations. You should always check for convergence.\n\nYou will understand some of these issues in detail and also see the ways to deal with them when you implement the K-means algorithm in Python.\n\nHaving understood the approach of choosing K for the K-Means algorithm, we will now look at silhouette analysis or silhouette coefficient. Silhouette coefficient is a measure of how similar a data point is to its own cluster (cohesion) compared to other clusters (separation).\n\nSo to compute silhouette metric, we need to compute two measures i.e. a(i) and b(i) where,\n\n• a(i) is the average distance from its own cluster(Cohesion).\n• b(i) is the average distance from the nearest neighbour cluster(Separation).\n\nNow, let’s look at how to combine cohesion and separation to compute the silhouette metric.\n\nYou can read more about K-Mode clustering here, We will be covering it in detail in the next section.\n\n#### K-Means algorithm\n\nArrange the steps of k-means algorithm in the order in which they occur:\n\n1. Randomly selecting the cluster centroids\n2. Updating the cluster centroids iteratively\n3. Assigning the cluster points to their nearest center\n\n1-3-2✓ CorrectFeedback:First the cluster centers are pre-decided. Then all the points are assigned to their nearest cluster center and then the center is recalculated as the mean of all the points which fall in that cluster. Then the clustering is repeated with the new centers and the centers are updated according to the new cluster points.\n\n## Steps of the Algorithm\n\nLet’s go through the K-Means algorithm using a very simple example. Let’s consider a set of 10 points on a plane and try to group these points into, say, 2 clusters. So let’s see how the K-Means algorithm achieves this goal.\n\n[Note: If you don’t know what is meant by Euclidean distance, you’re advised to go through this link]\n\nBefore moving ahead, think about the following problem. Let’s say you have the data of 10 students and their marks in Biology and Math (as shown in the plot below). You want to divide them into two clusters so that you can see what kind of students are there in the class.\n\nThe y-axis shows the marks in Biology, and the x-axis shows the marks in Math.\n\nImagine two clusters dividing this data — one red and the other yellow. How many points would each cluster have?\n\nCentroid\n\nThe K-Means algorithm uses the concept of the centroid to create K clusters. Before you move ahead, it will be useful to recall the concept of the centroid.\n\nIn simple terms, a centroid of n points on an x-y plane is another point having its own x and y coordinates and is often referred to as the geometric centre of the n points.\n\nFor example, consider three points having coordinates (x1, y1), (x2, y2) and (x3, y3). The centroid of these three points is the average of the x and y coordinates of the three points, i.e.\n\n(x1 + x2 + x3 / 3, y1 + y2 + y3 / 3).\n\nSimilarly, if you have n points, the formula (coordinates) of the centroid will be:\n\n(x1+x2…..+xn / n, y1+y2…..+yn / n).\n\nSo let’s see how the K-Means algorithm achieves this goal.\n\nEach time the clusters are made, the centroid is updated. The updated centroid is the centre of all the points which fall in the cluster associated with the centroid. This process continues till the centroid no longer changes, i.e. the solution converges.\n\nThus, you can see that the K-means algorithm is a clustering algorithm that takes N data points and groups them into K clusters. In this example, we had N =10 points and we used the K-means algorithm to group these 10 points into K = 2 clusters.\n\nDownload the Excel file below. It is designed to give you the hands-on practice of the k-means clustering algorithm. The file contains a set of 10 points (with x and y coordinates in column A and B respectively) and two initial centres 1 and 2 (in columns F and G). Answer the questions below based on the Excel file.\n\n# K Means Algorithm\n\nIn the previous segment, we learned about K-means clustering and how the algorithm works using a simple example. We learned about how assignment and optimization work in K Means clustering, Now in this lecture, we will look at K-means more algorithmically. We will be learning how the K Means algorithm proceeds with the assignment step and then with the optimization step and will also be looking at the cost of function for the K-means algorithm.\n\nLet’s understand the K-means algorithm in more detail.\n\nFrom the previous lecture, we understood that the algorithm’s inner-loop iterates over two steps:\n\n1. Assign each observation Xi to the closest cluster centroid μk\n2. Update each centroid to the mean of the points assigned to it.\n\nIn the next lecture, we will learn about the Kmeans cost function and will also see how to compute the cost function for each iteration in the K-means algorithm.\n\nSo the cost function for the K-Means algorithm is given as:\n\nJ=∑ni=1||Xi−μk(i)||2=∑Kk=1∑iϵCk||Xi−μk||2\n\nNow in the next video, we will learn what exactly happens in the assignment step? and we will also look at how to assign each data point to a cluster using the K-Means algorithm assignment step.\n\n[Note: At 1:43 where the Prof explains the optimization step, the values in the column – μ1 and μ2 should be X1 and X2 ]\n\nIn the assignment step, we assign every data point to K clusters. The algorithm goes through each of the data points and depending on which cluster is closer, in our case, whether the green cluster centroid or the blue cluster centroid; It assigns the data points to one of the 2 cluster centroids.\n\nThe equation for the assignment step is as follows:\n\nZi=argmin||Xi−μk||2\n\nNow having assigned each data point to a cluster, now we need to recompute the cluster centroids. In the next lecture, Prof.Dinesh will explain how to recompute the cluster centroids or the mean of each cluster.\n\nIn the optimization step, the algorithm calculates the average of all the points in a cluster and moves the centroid to that average location.\n\nThe equation for optimization is as follows:\n\nμk=1nk∑i:zi=kXi\n\nThe process of assignment and optimization is repeated until there is no change in the clusters or possibly until the algorithm converges.\n\n[Note – The definition of Silhouette score contains an error in the link shared above]\n\nIn the next segment, we will learn how to look K-Means algorithm as a coordinate descent problem. We will also learn about the constraint of the K-Means cost function and how to achieve global minima.\n\n# K Means++ Algorithm\n\nWe looked in the previous segment that for K-Means optimisation problem, the algorithm iterates between two steps and tries to minimise the objective function given as,\n\nZi=argmin||Xi−μk||2\n\nTo choose the cluster centres smartly, we will learn about K-Mean++ algorithm. K-means++ is just an initialisation procedure for K-means. In K-means++ you pick the initial centroids using an algorithm that tries to initialise centroids that are far apart from each other.\n\nTo summarise, In K-Means++ algorithm,\n\n1. We choose one centre as one of the data points at random.\n2. For each data point Xi, We compute the distance between Xi and the nearest centre that had already been chosen.\n3. Now, we choose the next cluster centre using the weighted probability distribution where a point X is chosen with probability proportional to d(X)2 .\n4. Repeat Steps 2 and 3 until K centres have been chosen.\n\n# Visualising the K Means Algorithm\n\nLet’s see the K-Means algorithm in action using a visualization tool. This tool can be found on naftaliharris.com. You can go to this link after watching the video below and play around with the different options available to get an intuitive feel of the K-Means algorithm.\n\nSuppose you have implemented k-means and to check that it is running correctly, you plot the cost function J(c^{(1)}, \\dots, c^{(m)}, \\mu_1, \\dots, \\mu_k)J(c(1),…,c(m),μ1,…,μk) as a function of the number of iterations. Your plot looks like this:",
null,
"What does this mean?\n\nThe learning rate is too large.\n\nThe algorithm is working correctly.\n\nThe algorithm is working, but kk is too large.\n\nIt is not possible for the cost function to sometimes increase. There must be a bug in the code.\n\n## Centroid\n\nThe next concept that is crucial for understanding how clustering generally works is the idea of centroids. If you remember your high school geometry, centroids are essentially the centre points of triangles. Similarly, in the case of clustering, centroids are the center points of the clusters that are being formed.\n\nNow before going to the formula part, here is an intuition for the need of a centroid. Imagine you have the following clusters of the marks of a group of students in Mathematics and Biology and someone asks you to explain them. From a glance, you can easily interpret the 4 clusters that are being formed.",
null,
"So the four clusters that are being formed are as follows:\n\nCluster 1: Students who have scored high marks in Bio, but poor marks in Maths\nCluster 2: Students who have scored average marks in Bio and Maths\nCluster 3: Students who have scored high marks in both Bio and Maths\nCluster 4: Students who have scored high marks in Maths, but poor marks in Bio\n\nNow the above representation is fine and correct, but it is missing one crucial information – the numerical order. For example, when you want to compare two clusters say Cluster 1 and Cluster 2 can you say by how much marks on average do the students from Cluster 1 outperform or underperform the Cluster 2 students in a particular subject just by taking a look at the above visualisation alone? Is it by 10 marks? Or 15?\n\nThis is where the concept of Centroids comes in handy. Listen to the following lecture to understand its importance and how it is calculated.\n\nTherefore, as mentioned in the video, the Centroids are essentially the cluster centers of a group of observations that help us in summarising the cluster’s properties. Thus as you saw in the video, the centroid value in the case of clustering is essentially the mean of all the observations that belong to a particular cluster. For example, in the dataset that you saw here,",
null,
"The centroid is calculated by computing the mean of each and every column/dimension that you have and then ordering them in the same way as above.\n\nTherefore, Height-mean = ((175+165+183+172)/)/4 = 173.75\nWeight-mean = ((83+74+98+80))/4 = 83.75\nAge – mean = ((22+25+24+24))/4 =23.75\n\nThus the centroid of the above group of observations is (173.75, 83.75 and 23.75)\n\n# Euclidean Distance\n\nIn the previous segments, you got an idea about how clustering works – it groups the objects on the basis of their similarity or closeness to each other.\n\nNow, the next important thing is to get into the nitty-gritty of how clustering algorithms generally work. You will learn about the 2 types of clustering methods – K-means and Hierarchical and how they go about doing the clustering process.\n\nWe have learnt that clustering works on the basis of grouping the observations which are the most similar to each other. What does this exactly mean?\n\nIn simple terms, the algorithm needs to find data points whose values are similar to each other and therefore these points would then belong to the same cluster. The method in which any clustering algorithm goes about doing that is through the method of finding something called a “distance measure”. The distance measure that is used in K-means clustering is called the Euclidean Distance measure. Let’s look at the following lecture to understand how this value is calculated.\n\nThe Euclidean Distance between the 2 points is measured as follows: If there are 2 points X and Y having n dimensions\n\nThen the Euclidean Distance D is given as\n\nThe idea of distance measure is quite intuitive. Essentially, the observations which are closer or more similar to each other would have a low Euclidean distance and the observations which are farther or less similar to each other would have a higher Euclidean distance. So can you now guess how the Clustering process would work based on the Euclidean distance?"
]
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"http://spark-public.s3.amazonaws.com/ml/images/14.3-quiz-1-q.png",
null,
"https://images.upgrad.com/d922dea9-429a-4aa5-a449-73312f050ebb-Students%20Marks%202.jpg",
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"https://images.upgrad.com/5df9bf6f-15cc-4921-b3ad-66212c0cf687-Clustering.png",
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https://im.kendallhunt.com/MS_ACC/teachers/2/1/17/practice.html | [
"# Lesson 17\n\nDrawing Triangles\n\n### Problem 1\n\nUse a protractor to try to draw each triangle. Which of these three triangles is impossible to draw?\n\n1. A triangle where one angle measures $$20^\\circ$$ and another angle measures $$45^\\circ$$\n2. A triangle where one angle measures $$120^\\circ$$ and another angle measures $$50^\\circ$$\n3. A triangle where one angle measures $$90^\\circ$$ and another angle measures $$100^\\circ$$\n\n### Problem 2\n\nA triangle has an angle measuring $$90^\\circ$$, an angle measuring $$20^\\circ$$, and a side that is 6 units long. The 6-unit side is in between the $$90^\\circ$$ and $$20^\\circ$$ angles.\n\n1. Sketch this triangle and label your sketch with the given measures.\n2. How many unique triangles can you draw like this?\n\n### Problem 3\n\nA triangle has sides of length 7 cm, 4 cm, and 5 cm. How many unique triangles can be drawn that fit that description? Explain or show your reasoning.\n\n### Problem 4\n\nA triangle has one side that is 5 units long and an adjacent angle that measures $$25^\\circ$$. The two other angles in the triangle measure $$90^\\circ$$ and $$65^\\circ$$. Complete the two diagrams to create two different triangles with these measurements."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.85539466,"math_prob":0.9998104,"size":1960,"snap":"2023-40-2023-50","text_gpt3_token_len":448,"char_repetition_ratio":0.17484662,"word_repetition_ratio":0.33435583,"special_character_ratio":0.2494898,"punctuation_ratio":0.07629428,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99999213,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-26T15:16:44Z\",\"WARC-Record-ID\":\"<urn:uuid:dc5234a4-a0f6-427a-9da2-bfd84a63c22b>\",\"Content-Length\":\"70707\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:903caa31-0076-4f54-a892-70199cfc2bb9>\",\"WARC-Concurrent-To\":\"<urn:uuid:a7c49ebe-2212-4bfe-9a5e-57136cbacd08>\",\"WARC-IP-Address\":\"52.20.78.240\",\"WARC-Target-URI\":\"https://im.kendallhunt.com/MS_ACC/teachers/2/1/17/practice.html\",\"WARC-Payload-Digest\":\"sha1:RKD3BEKH3HCGH6RWVHGEGU7TVFCO5DJT\",\"WARC-Block-Digest\":\"sha1:CDNCJ6PSFT6YUP2E6OF3ND3IA5XPTXJU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510214.81_warc_CC-MAIN-20230926143354-20230926173354-00625.warc.gz\"}"} |
https://chemistry.stackexchange.com/questions/76381/oxidizing-agent-reducing-agent-acid-or-base | [
"# Oxidizing agent, reducing agent, acid or base\n\nIn concrete reactions, how can you define whether Ammonia reacts as an oxidizing agent, reducing agent, an acid or a base? I'm confused because in the examples given it's not an acid/ base when an $\\ce {H+}$ is donated/ accepted.\n\n$$\\ce {2NH3 + 3Cl2 -> N2 + 6HCl}$$ (oxidizing agent)\n\n$$\\ce {Na + NH3 -> NaNH2 + \\frac{1}{2}H2}$$ (reducing agent)\n\n$$\\ce{NH3 + H2O + CO2 -> NH4HCO3}$$ (base)\n\n$$\\ce{NH3 + LiCH3 -> LiNH2 + CH4}$$ (acid)\n\n• In the example reactions, ammonia is incorrectly stated to be an oxidizing agent in the reaction with chlorine (here it acts as a reducing agent); when reacting with sodium metal, ammonia is an oxidizing agent and not a reducing agent as shown. Not sure that helps address your question. – iad22agp Jun 17 '17 at 12:11\n\nThe use of the terms \"oxidizing agent\" and \"reducing agent\" in this example apparently refer to the overall reaction. Ammonia certainly can act as an acid in the process of reacting with sodium metal, and as a (Lewis) base during the reaction with chlorine. However, the net effect of the reaction is a reduction or oxidation, rather than a simple proton transfer.\n\nI'm basing this answer from translation of the original question (in German) which I believe is supposed to read:\n\nAmmonia can exist in oxidised form, reduced form, acid or base. Name the form in which ammonia exist in the following: (correct me if I am wrong)\n\nFocusing on the acid/base forms (3) and (4):\n\n• ammonia acts as an base by accepting a proton $$\\ce{H+}$$ from $$\\ce{H2O}$$.\n\nRecall the Brønsted-Lowry theory of acids and bases which states: An acid is a proton (hydrogen ion) donor, and a base is a proton (hydrogen ion) acceptor.\n\n• similarly on the last equation $$\\ce{NH3}$$ donates a proton to $$\\ce{LiCH3}$$ to form lithium amide $$\\ce{Li^+ NH2^-}$$ ($$\\ce{LiNH2}$$) and $$\\ce{CH4}$$"
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https://encyclopediaofmath.org/wiki/Jacobson_radical | [
"##### Actions\n\n2010 Mathematics Subject Classification: Primary: 16N20 [MSN][ZBL]\n\nof a ring $A$\n\nThe ideal $J(A)$ of an associative ring (cf. Associative rings and algebras) $A$ which satisfies the following two requirements: 1) $J(A)$ is the largest quasi-regular ideal in $A$ (a ring $R$ is called quasi-regular if the equation $a+x+ax=0$ is solvable for any of its elements $a$; cf. Quasi-regular ring); and 2) the quotient ring $\\overline A=A/J(A)$ contains no non-zero quasi-regular ideals. The radical was introduced and studied in detail in 1945 by N. Jacobson .\n\nThe Jacobson radical always exists and may be characterized in very many ways: $J(A)$ is the intersection of the kernels of all irreducible representations of the ring $A$; it is the intersection of all modular maximal right ideals (cf. Modular ideal); it is the intersection of all modular maximal left ideals; it contains all quasi-regular one-sided ideals; it contains all one-sided nil ideals; etc. If $I$ is an ideal of $A$, then $J(I)=I\\cap J(A)$. If $A_n$ is the ring of all matrices of order $n$ over $A$, then\n\n$$J(A_n)=(J(A))_n.$$\n\nIf the following $\\circ$-composition is introduced on the associative ring $A$:\n\n$$a\\circ b=a+b+ab,$$\n\nthen the radical $J(A)$ in the semi-group $\\langle A,\\circ\\rangle$ will be a subgroup with respect to the composition $\\circ$.\n\nThere are no non-zero irreducible finitely-generated modules over a quasi-regular associative ring (i.e. an associative ring coinciding with its own Jacobson radical), but there exist simple associative quasi-regular rings. The Jacobson radical of the associative ring $A$ is zero if and only if $A$ is a subdirect sum of primitive rings.\n\nHow to Cite This Entry:"
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http://forums.wolfram.com/mathgroup/archive/1996/Jul/msg00142.html | [
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"",
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"Re: Numerical Differentiation\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg4371] Re: Numerical Differentiation\n• From: Robert Knapp <rknapp>\n• Date: Mon, 15 Jul 1996 06:41:08 -0400\n• Organization: Wolfram Research, Inc.\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```Dr. T. L. Marchioro II wrote:\n>\n> Robert Knapp wrote:\n> > Mark James wrote:\n> > >\n> > > Does anyone know of a function that calculates the derivative of\n> > > a function (that can't be differentiated symbolically) at a given\n> > > point by numerical means? I can't find it as a built-in or in the\n> > > standard packages. Thanks.\n> >\n> > The function ND in the package NumericalMath`NLimit does this.\n> >\n> > I mention in passing that numerical derivatives will be computed\n> > automatically in the next release of Mathematica.\n>\n> Really? By what method? Where will they be well defined? That is, if\n> you have a discrete data stream will be the derivatives be accurate at\n> the same points you know the data, or at the midpoints between the data,\n> or somewhere else? To what order will be the derivatives be accurately\n> calculable?\n>\n> Inquiring minds and all that :)\n>\nHeres an example(from Mathematica 3.0 beta 2)\n\nIn:= f[x_?NumberQ,t_?NumberQ] :=NDSolve[{y'[s] == x y[s],y ==\n1},y,{s,0,t}][[1,1,2]][t]\n\nIn:= Derivative[1,0][f][x,t]\n\n(1,0)\nOut= f [x, t]\n\nIn:= Derivative[1,0][f][1,1]\n\n(1,0)\nOut= f [1, 1]\n\nnot done since it is exact\n\nIn:= Derivative[1,0][f][1.,1.]\n\nOut= 2.71828\n\nThis is calculated to machine precision. If your function was defined\nsuch that it could be calculated to higher precision, the derivative\nroutine would attempt to calculate to the precision of the arguments.\n\nIt works using high order finite differences. There is not a limit to\nthe order derivative it uses since the wieghts are generated on\ncalculation, however, high order derivatives can take excessively\nlong.\n\nAlso, mixed partials are not computed since error detection is much\nmore difficult.\n\nRob\n--\nRob Knapp\nWolfram Research, Inc.\n\nhttp://www.wri.com/~rknapp\n\n==== [MESSAGE SEPARATOR] ====\n\n```\n\n• Prev by Date: Re: Time Series\n• Next by Date: Re: Abort options in Windows\n• Previous by thread: Re: Numerical Differentiation\n• Next by thread: rendering digital landscapes"
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"# Using the ac Method to Factor\n\nSize: px\nStart display at page:",
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"Transcription\n\n1 4.6 Using the ac Method to Factor 4.6 OBJECTIVES 1. Use the ac test to determine factorability 2. Use the results of the ac test 3. Completely factor a trinomial In Sections 4.2 and 4.3 we used the trial-and-error method to factor trinomials. We also learned that not all trinomials can be factored. In this section we will look at the same kinds of trinomials, but in a slightly different context. We first determine whether a trinomial is factorable. We then use the results of that analysis to factor the trinomial. Some students prefer the trial-and-error method for factoring because it is generally faster and more intuitive. Other students prefer the method of this section (called the ac method) because it yields the answer in a systematic way. We will let you determine which method you prefer. We will begin by looking at some factored trinomials. Example 1 Matching Trinomials and Their Factors Determine which of the following are true statements. (a) x 2 2x 8 (x 4)(x 2) This is a true statement. Using the FOIL method, we see that (x 4)(x 2) x 2 2x 4x 8 x 2 2x 8 (b) x 2 6x 5 (x 2)(x 3) This is not a true statement. (x 2)(x 3) x 2 3x 2x 6 x 2 5x 6 (c) x 2 5x 14 (x 2)(x 7) This is true: (x 2)(x 7) x 2 7x 2x 14 x 2 5x 14 (d) x 2 8x 15 (x 5)(x 3) This is false: (x 5)(x 3) x 2 3x 5x 15 x 2 8x 15 CHECK YOURSELF 1 Determine which of the following are true statements. (a) 2x 2 2x 3 (2x 3)(x 1) (b) 3x 2 11x 4 (3x 1)(x 4) (c) 2x 2 7x 3 (x 3)(2x 1) 363\n\n2 364 CHAPTER 4 FACTORING The first step in learning to factor a trinomial is to identify its coefficients. So that we are consistent, we first write the trinomial in standard ax 2 bx c form, then label the three coefficients as a, b, and c. Example 2 Identifying the Coefficients of ax 2 bx c First, when necessary, rewrite the trinomial in ax 2 bx c form. Then give the values for a, b, and c, in which a is the coefficient of the x 2 term, b is the coefficient of the x term, and c is the constant. (a) x 2 3x 18 a 1 b 3 c 18 NOTE Notice that the negative sign is attached to the coefficients. (b) x 2 24x 23 a 1 b 24 c 23 (c) x x First rewrite the trinomial in descending order: x 2 11x 8 a 1 b 11 c 8 CHECK YOURSELF 2 First, when necessary, rewrite the trinomials in ax 2 bx c form. Then label a, b, and c, in which a is the coefficient of the x 2 term, b is the coefficient of the x term, and c is the constant. (a) x 2 5x 14 (b) x 2 18x 17 (c) x 6 2x 2 Not all trinomials can be factored. To discover if a trinomial is factorable, we try the ac test. Definitions: The ac Test A trinomial of the form ax 2 bx c is factorable if (and only if) there are two integers, m and n, such that ac mn and b m n In Example 3 we will look for m and n to determine whether each trinomial is factorable.\n\n3 USING THE ac METHOD TO FACTOR SECTION Example 3 Using the ac Test Use the ac test to determine which of the following trinomials can be factored. Find the values of m and n for each trinomial that can be factored. (a) x 2 3x 18 First, we find the values of a, b, and c, so that we can find ac. a 1 b 3 c 18 ac 1( 18) 18 and b 3 Then, we look for two numbers, m and n, such that mn ac, and m n b. In this case, that means mn 18 and m n 3 We now look at all pairs of integers with a product of 18. We then look at the sum of each pair of integers, looking for a sum of 3. mn 1( 18) 18 2( 9) 18 3( 6) 18 6( 3) 18 9( 2) 18 18( 1) 18 m n 1 ( 18) 17 2 ( 9) 7 3 ( 6) 3 We need look no further than 3 and 6. NOTE We could have chosen m 6 and n 3 as well. 3 and 6 are the two integers with a product of ac and a sum of b. We can say that m 3 and n 6 Because we found values for m and n, we know that x 2 3x 18 is factorable. (b) x 2 24x 23 We find that a 1 b 24 c 23 ac 1(23) 23 and b 24 So mn 23 and m n 24 We now calculate integer pairs, looking for two numbers with a product of 23 and a sum of 24. mn m n 1(23) ( 23) 23 1 ( 23) 24 m 1 and n 23 So, x 2 24x 23 is factorable.\n\n4 366 CHAPTER 4 FACTORING (c) x 2 11x 8 We find that a 1, b 11, and c 8. Therefore, ac 8 and b 11. Thus mn 8 and m n 11. We calculate integer pairs: mn m n 1(8) (4) ( 8) 8 1 ( 8) 9 2( 4) 8 2 ( 4) 6 There are no other pairs of integers with a product of 8, and none of these pairs has a sum of 11. The trinomial x 2 11x 8 is not factorable. (d) 2x 2 7x 15 We find that a 2, b 7, and c 15. Therefore, ac 2( 15) 30 and b 7. Thus mn 30 and m n 7. We calculate integer pairs: mn m n 1( 30) 30 1 ( 30) 29 2( 15) 30 2 ( 15) 13 3( 10) 30 3 ( 10) 7 5( 6) 30 5 ( 6) 1 6( 5) 30 6 ( 5) 1 10( 3) ( 3) 7 There is no need to go any further. We see that 10 and 3 have a product of 30 and a sum of 7, so m 10 and n 3 Therefore, 2x 2 7x 15 is factorable. It is not always necessary to evaluate all the products and sums to determine whether a trinomial is factorable. You may have noticed patterns and shortcuts that make it easier to find m and n. By all means, use them to help you find m and n. This is essential in mathematical thinking. You are taught a mathematical process that will always work for solving a problem. Such a process is called an algorithm. It is very easy to teach a computer to use an algorithm. It is very difficult (some would say impossible) for a computer to have insight. Shortcuts that you discover are insights. They may be the most important part of your mathematical education. CHECK YOURSELF 3 Use the ac test to determine which of the following trinomials can be factored. Find the values of m and n for each trinomial that can be factored. (a) x 2 7x 12 (b) x 2 5x 14 (c) 3x 2 6x 7 (d) 2x 2 x 6\n\n5 USING THE ac METHOD TO FACTOR SECTION So far we have used the results of the ac test only to determine whether a trinomial is factorable. The results can also be used to help factor the trinomial. Example 4 Using the Results of the ac Test to Factor Rewrite the middle term as the sum of two terms, then factor by grouping. (a) x 2 3x 18 We find that a 1, b 3, and c 18, so ac 18 and b 3. We are looking for two numbers, m and n, where mn 18 and m n 3. In Example 3, part a, we looked at every pair of integers whose product (mn) was 18, to find a pair that had a sum (m n) of 3. We found the two integers to be 3 and 6, because 3( 6) 18 and 3 ( 6) 3, so m 3 and n 6. We now use that result to rewrite the middle term as the sum of 3x and 6x. x 2 3x 6x 18 We then factor by grouping: x 2 3x 6x 18 x(x 3) 6(x 3) (x 3)(x 6) (b) x 2 24x 23 We use the results from Example 3, part b, in which we found m 1 and n 23, to rewrite the middle term of the equation. x 2 24x 23 x 2 x 23x 23 Then we factor by grouping: x 2 x 23x 23 (x 2 x) (23x 23) x(x 1) 23(x 1) (x 1)(x 23) (c) 2x 2 7x 15 From Example 3, part d, we know that this trinomial is factorable, and m 10 and n 3. We use that result to rewrite the middle term of the trinomial. 2x 2 7x 15 2x 2 10x 3x 15 (2x 2 10x) (3x 15) 2x(x 5) 3(x 5) (x 5)(2x 3) Careful readers will note that we did not ask you to factor Example 3, part c, x 2 11x 8. Recall that, by the ac method, we determined that this trinomial was not factorable. CHECK YOURSELF 4 Use the results of Check Yourself 3 to rewrite the middle term as the sum of two terms, then factor by grouping. (a) x 2 7x 12 (b) x 2 5x 14 (c) 2x 2 x 6\n\n6 368 CHAPTER 4 FACTORING Let s look at some examples that require us to first find m and n, then factor the trinomial. Example 5 Rewriting Middle Terms to Factor Rewrite the middle term as the sum of two terms, then factor by grouping. (a) 2x 2 13x 7 We find that a 2, b 13, and c 7, so mn ac 14 and m n b 13. Therefore, mn 1( 14) 14 m n 1 ( 14) 13 So, m 1 and n 14. We rewrite the middle term of the trinomial as follows: 2x 2 13x 7 2x 2 x 14x 7 (2x 2 x) (14x 7) x(2x 1) 7(2x 1) (2x 1)(x 7) (b) 6x 2 5x 6 We find that a 6, b 5, and c 6, so mn ac 36 and m n b 5. mn 1( 36) 36 2( 18) 36 3( 12) 36 4( 9) 36 m n 1 ( 36) 35 2 ( 18) 16 3 ( 12) 9 4 ( 9) 5 So, m 4 and n 9. We rewrite the middle term of the trinomial: 6x 2 5x 6 6x 2 4x 9x 6 (6x 2 4x) (9x 6) 2x(3x 2) 3(3x 2) (3x 2)(2x 3) CHECK YOURSELF 5 Rewrite the middle term as the sum of two terms, then factor by grouping. (a) 2x 2 7x 15 (b) 6x 2 5x 4\n\n7 USING THE ac METHOD TO FACTOR SECTION Be certain to check trinomials and binomial factors for any common monomial factor. (There is no common factor in the binomial unless it is also a common factor in the original trinomial.) Example 6 shows the removal of monomial factors. Example 6 Removing Common Factors Completely factor the trinomial. 3x 2 12x 15 We could first remove the common factor of 3: 3x 2 12x 15 3(x 2 4x 5) Finding m and n for the trinomial x 2 4x 5 yields mn 5 and m n 4. mn m n 1( 5) 5 1 ( 5) 4 5( 1) 5 1 (5) 4 So, m 5 and n 1. This gives us 3x 2 12x 15 3(x 2 4x 5) 3(x 2 5x x 5) 3[(x 2 5x) (x 5)] 3[x(x 5) (x 5)] 3[(x 5)(x 1)] 3(x 5)(x 1) CHECK YOURSELF 6 Completely factor the trinomial. 6x 3 3x 2 18x CHECK YOURSELF ANSWERS 1. (a) False; (b) true; (c) true 2. (a) a 1, b 5, c 14; (b) a 1, b 18, c 17; (c) a 2, b 1, c 6 3. (a) Factorable, m 3, n 4; (b) factorable, m 7, n 2; (c) not factorable; (d) factorable, m 4, n 3 4. (a) x 2 3x 4x 12 (x 3)(x 4); (b) x 2 7x 2x 14 (x 7)(x 2); (c) 2x 2 4x 3x 6 (2x 3)(x 2) 5. (a) 2x 2 10x 3x 15 (2x 3)(x 5); (b) 6x 2 8x 3x 4 (3x 4)(2x 1) 6. 3x(2x 3)(x 2)\n\n8 370 CHAPTER 4 FACTORING Not all possible product pairs need to be tried to find m and n. A look at the sign pattern of the trinomial will eliminate many of the possibilities. Assuming the leading coefficient is positive, there are four possible sign patterns. Pattern Example Conclusion 1. b and c are both positive. 2x 2 13x 15 m and n must both be positive. 2. b is negative and c is positive. x 2 7x 12 m and n must both be negative. 3. b is positive and c is negative. x 2 3x 10 m and n are of opposite signs. (The value with the larger absolute value is positive.) 4. b is negative and c is negative. x 2 3x 10 m and n are of opposite signs. (The value with the larger absolute value is negative.)\n\n9 Name 4.6 Exercises Section Date State whether each of the following is true or false. 1. x 2 2x 3 (x 3)(x 1) 2. y 2 3y 18 ( y 6)( y 3) ANSWERS x 2 10x 24 (x 6)(x 4) 4. a 2 9a 36 (a 12)(a 4) x 2 16x 64 (x 8)(x 8) 6. w 2 12w 45 (w 9)(w 5) y 2 10y 1 (5y 1)(5y 1) x 2 5xy 4y 2 (6x 2y)(x 2y) p 2 pq 3q 2 (5p 3q)(2p q) a 2 13a 6 (2a 3)(3a 2) For each of the following trinomials, label a, b, and c. 11. x 2 4x x 2 5x x 2 3x x 2 7x x 2 5x x 2 7x x 2 8x x 2 7x x 2 5x x 2 9x\n\n10 ANSWERS Use the ac test to determine which of the following trinomials can be factored. Find the values of m and n for each trinomial that can be factored. 21. x 2 x x 2 2x x 2 x x 2 3x x 2 5x x 2 x x 2 5x x 2 14x x 2 19x x 2 5x Rewrite the middle term as the sum of two terms and then factor by grouping. 31. x 2 6x x 2 3x x 2 9x x 2 8x x 2 2x x 2 6x 55 Rewrite the middle term as the sum of two terms and then factor completely. 37. x 2 8x x 2 11x x 2 11x y 2 y s 2 13s b 2 14b a 2 2a x 2 17x\n\n11 ANSWERS 45. x 2 8x x 2 7x x 2 6x x 2 11x x 2 14x s 2 4s p 2 10p x 2 11x x 2 5x a 2 2a c 2 19c t 2 4t n 2 5n x 2 16x x 2 7xy 10y x 2 8xy 12y a 2 ab 42b m 2 8mn 16n x 2 13xy 40y r 2 9rs 36s x 2 19x x 2 7x x 2 x w 2 19w m 2 25m x 2 6x x 2 12x x 2 23x\n\n12 ANSWERS x 2 8x a 2 40a y 2 7y x 2 11x x 2 27x v 2 5v x 2 3xy y x 2 5xy 2y a 2 8ab 4b x 2 7xy 6y x 2 4xy 5y x 2 32xy 15y m 2 17mn 12n x 2 xy 6y a 2 3ab 5b q 2 17qr 6r x 2 4xy 4y b 2 80bc 64c x 2 20x x 2 18x m 2 12m x 2 20x 6 374\n\n13 ANSWERS r 2 21rs 6s x 2 5xy 30y x 3 2x 2 4x 98. 2y 3 y 2 3y y 4 5y 3 3y z 3 18z 2 10z a 3 66a 2 18a n 4 22n 3 12n p 2 30pq 21q x 2 2xy 24y 2 Find a positive value for k for which each of the following can be factored x 2 kx x 2 kx x 2 kx x 2 kx x 2 kx x 2 kx x 2 3x k 112. x 2 5x k a x 2 2x k 114. x 2 x k Solve. Getting Ready for Section 4.7 [Section 2.3] (a) x 5 0 (b) 2x 1 0 (c) 3x 2 0 (d) x 4 0 (e) 7 x 0 (f) 9 4x 0 b. c. d. e. f. 375\n\n14 Answers 1. True 3. False 5. True 7. False 9. True 11. a 1, b 4, c a 1, b 3, c a 3, b 5, c a 4, b 8, c a 3, b 5, c Factorable; 3, Not factorable 25. Factorable; 3, Factorable; 6, Factorable; 15, x 2 2x 4x 8; (x 2)(x 4) 33. x 2 5x 4x 20; (x 5)(x 4) 35. x 2 9x 7x 63; (x 9)(x 7) 37. (x 3)(x 5) 39. (x 4)(x 7) 41. (s 10)(s 3) 43. (a 8)(a 6) 45. (x 1)(x 7) 47. (x 10)(x 4) 49. (x 7)(x 7) 51. (p 12)(p 2) 53. (x 11)(x 6) 55. (c 4)(c 15) 57. (n 10)(n 5) 59. (x 2y)(x 5y) 61. (a 7b)(a 6b) 63. (x 5y)(x 8y) 65. (3x 2)(2x 5) 67. (5x 3)(3x 2) 69. (6m 5)(m 5) 71. (3x 2)(3x 2) 73. (6x 5)(2x 3) 75. (3y 2)(y 3) 77. (8x 5)(x 4) 79. (2x y)(x y) 81. (5a 2b)(a 2b) 83. (9x 5y)(x y) 85. (3m 4n)(2m 3n) 87. (12a 5b)(3a b) 89. (x 2y) (2x 3)(2x 1) 93. 4(2m 1)(m 1) 95. 3(5r 2s)(r s) 97. 2x(x 2)(x 1) 99. y 2 (2y 3)(y 1) a(3a 1)(2a 3) (p q)(3p 7q) or or 10 or , 8, 15, 24,... a. x 5 b. x 1 c. x 2 d. x 4 e. x 7 f. x\n\n### Factoring Trinomials: The ac Method",
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"6.7 Factoring Trinomials: The ac Method 6.7 OBJECTIVES 1. Use the ac test to determine whether a trinomial is factorable over the integers 2. Use the results of the ac test to factor a trinomial 3. For\n\n### FACTORING ax 2 bx c. Factoring Trinomials with Leading Coefficient 1",
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"5.7 Factoring ax 2 bx c (5-49) 305 5.7 FACTORING ax 2 bx c In this section In Section 5.5 you learned to factor certain special polynomials. In this section you will learn to factor general quadratic polynomials.\n\n### Factoring Trinomials of the Form x 2 bx c",
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"4.2 Factoring Trinomials of the Form x 2 bx c 4.2 OBJECTIVES 1. Factor a trinomial of the form x 2 bx c 2. Factor a trinomial containing a common factor NOTE The process used to factor here is frequently\n\n### Factoring Special Polynomials",
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"6.6 Factoring Special Polynomials 6.6 OBJECTIVES 1. Factor the difference of two squares 2. Factor the sum or difference of two cubes In this section, we will look at several special polynomials. These\n\n### Factoring. Factoring Monomials Monomials can often be factored in more than one way.",
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"Factoring Factoring is the reverse of multiplying. When we multiplied monomials or polynomials together, we got a new monomial or a string of monomials that were added (or subtracted) together. For example,\n\n### 1.3 Polynomials and Factoring",
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"1.3 Polynomials and Factoring Polynomials Constant: a number, such as 5 or 27 Variable: a letter or symbol that represents a value. Term: a constant, variable, or the product or a constant and variable.",
null,
"Factoring the trinomial ax 2 + bx + c when a = 1 A trinomial in the form x 2 + bx + c can be factored to equal (x + m)(x + n) when the product of m x n equals c and the sum of m + n equals b. (Note: the\n\n### FACTORING TRINOMIALS IN THE FORM OF ax 2 + bx + c",
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"Tallahassee Community College 55 FACTORING TRINOMIALS IN THE FORM OF ax 2 + bx + c This kind of trinomial differs from the previous kind we have factored because the coefficient of x is no longer \"1\".\n\n### By reversing the rules for multiplication of binomials from Section 4.6, we get rules for factoring polynomials in certain forms.",
null,
"SECTION 5.4 Special Factoring Techniques 317 5.4 Special Factoring Techniques OBJECTIVES 1 Factor a difference of squares. 2 Factor a perfect square trinomial. 3 Factor a difference of cubes. 4 Factor\n\n### Definitions 1. A factor of integer is an integer that will divide the given integer evenly (with no remainder).",
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"Math 50, Chapter 8 (Page 1 of 20) 8.1 Common Factors Definitions 1. A factor of integer is an integer that will divide the given integer evenly (with no remainder). Find all the factors of a. 44 b. 32\n\n### The Greatest Common Factor; Factoring by Grouping",
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"296 CHAPTER 5 Factoring and Applications 5.1 The Greatest Common Factor; Factoring by Grouping OBJECTIVES 1 Find the greatest common factor of a list of terms. 2 Factor out the greatest common factor.\n\n### Chapter R.4 Factoring Polynomials",
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"Chapter R.4 Factoring Polynomials Introduction to Factoring To factor an expression means to write the expression as a product of two or more factors. Sample Problem: Factor each expression. a. 15 b. x\n\n### NSM100 Introduction to Algebra Chapter 5 Notes Factoring",
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"Section 5.1 Greatest Common Factor (GCF) and Factoring by Grouping Greatest Common Factor for a polynomial is the largest monomial that divides (is a factor of) each term of the polynomial. GCF is the\n\n### 7-6. Choosing a Factoring Model. Extension: Factoring Polynomials with More Than One Variable IN T RO DUC E T EACH. Standards for Mathematical Content",
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"7-6 Choosing a Factoring Model Extension: Factoring Polynomials with More Than One Variable Essential question: How can you factor polynomials with more than one variable? What is the connection between\n\n### Factoring Methods. Example 1: 2x + 2 2 * x + 2 * 1 2(x + 1)",
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"Factoring Methods When you are trying to factor a polynomial, there are three general steps you want to follow: 1. See if there is a Greatest Common Factor 2. See if you can Factor by Grouping 3. See if\n\n### Name Intro to Algebra 2. Unit 1: Polynomials and Factoring",
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"Name Intro to Algebra 2 Unit 1: Polynomials and Factoring Date Page Topic Homework 9/3 2 Polynomial Vocabulary No Homework 9/4 x In Class assignment None 9/5 3 Adding and Subtracting Polynomials Pg. 332\n\n### In algebra, factor by rewriting a polynomial as a product of lower-degree polynomials",
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"Algebra 2 Notes SOL AII.1 Factoring Polynomials Mrs. Grieser Name: Date: Block: Factoring Review Factor: rewrite a number or expression as a product of primes; e.g. 6 = 2 3 In algebra, factor by rewriting\n\n### Operations with Algebraic Expressions: Multiplication of Polynomials",
null,
"Operations with Algebraic Expressions: Multiplication of Polynomials The product of a monomial x monomial To multiply a monomial times a monomial, multiply the coefficients and add the on powers with the\n\n### 6.3 FACTORING ax 2 bx c WITH a 1",
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"290 (6 14) Chapter 6 Factoring e) What is the approximate maximum revenue? f) Use the accompanying graph to estimate the price at which the revenue is zero. y Revenue (thousands of dollars) 300 200 100\n\n### ( ) FACTORING. x In this polynomial the only variable in common to all is x.",
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"FACTORING Factoring is similar to breaking up a number into its multiples. For example, 10=5*. The multiples are 5 and. In a polynomial it is the same way, however, the procedure is somewhat more complicated\n\n### This is Factoring and Solving by Factoring, chapter 6 from the book Beginning Algebra (index.html) (v. 1.0).",
null,
"This is Factoring and Solving by Factoring, chapter 6 from the book Beginning Algebra (index.html) (v. 1.0). This book is licensed under a Creative Commons by-nc-sa 3.0 (http://creativecommons.org/licenses/by-nc-sa/\n\n### expression is written horizontally. The Last terms ((2)( 4)) because they are the last terms of the two polynomials. This is called the FOIL method.",
null,
"A polynomial of degree n (in one variable, with real coefficients) is an expression of the form: a n x n + a n 1 x n 1 + a n 2 x n 2 + + a 2 x 2 + a 1 x + a 0 where a n, a n 1, a n 2, a 2, a 1, a 0 are\n\n### Factoring and Applications",
null,
"Factoring and Applications What is a factor? The Greatest Common Factor (GCF) To factor a number means to write it as a product (multiplication). Therefore, in the problem 48 3, 4 and 8 are called the\n\n### FACTORING POLYNOMIALS",
null,
"296 (5-40) Chapter 5 Exponents and Polynomials where a 2 is the area of the square base, b 2 is the area of the square top, and H is the distance from the base to the top. Find the volume of a truncated\n\n### Factoring Polynomials",
null,
"Factoring Polynomials Factoring Factoring is the process of writing a polynomial as the product of two or more polynomials. The factors of 6x 2 x 2 are 2x + 1 and 3x 2. In this section, we will be factoring\n\n### Polynomials. Key Terms. quadratic equation parabola conjugates trinomial. polynomial coefficient degree monomial binomial GCF",
null,
"Polynomials 5 5.1 Addition and Subtraction of Polynomials and Polynomial Functions 5.2 Multiplication of Polynomials 5.3 Division of Polynomials Problem Recognition Exercises Operations on Polynomials\n\n### 6.1 Add & Subtract Polynomial Expression & Functions",
null,
"6.1 Add & Subtract Polynomial Expression & Functions Objectives 1. Know the meaning of the words term, monomial, binomial, trinomial, polynomial, degree, coefficient, like terms, polynomial funciton, quardrtic\n\n### Math 25 Activity 6: Factoring Advanced",
null,
"Instructor! Math 25 Activity 6: Factoring Advanced Last week we looked at greatest common factors and the basics of factoring out the GCF. In this second activity, we will discuss factoring more difficult\n\n### How To Solve Factoring Problems",
null,
"05-W4801-AM1.qxd 8/19/08 8:45 PM Page 241 Factoring, Solving Equations, and Problem Solving 5 5.1 Factoring by Using the Distributive Property 5.2 Factoring the Difference of Two Squares 5.3 Factoring\n\n### Factoring a Difference of Two Squares. Factoring a Difference of Two Squares",
null,
"284 (6 8) Chapter 6 Factoring 87. Tomato soup. The amount of metal S (in square inches) that it takes to make a can for tomato soup is a function of the radius r and height h: S 2 r 2 2 rh a) Rewrite this\n\n### SPECIAL PRODUCTS AND FACTORS",
null,
"CHAPTER 442 11 CHAPTER TABLE OF CONTENTS 11-1 Factors and Factoring 11-2 Common Monomial Factors 11-3 The Square of a Monomial 11-4 Multiplying the Sum and the Difference of Two Terms 11-5 Factoring the\n\n### A. Factoring out the Greatest Common Factor.",
null,
"DETAILED SOLUTIONS AND CONCEPTS - FACTORING POLYNOMIAL EXPRESSIONS Prepared by Ingrid Stewart, Ph.D., College of Southern Nevada Please Send Questions and Comments to [email protected]. Thank you!\n\n### Polynomials and Factoring",
null,
"7.6 Polynomials and Factoring Basic Terminology A term, or monomial, is defined to be a number, a variable, or a product of numbers and variables. A polynomial is a term or a finite sum or difference of",
null,
"250) 960-6367 Factoring Polynomials Sometimes when we try to solve or simplify an equation or expression involving polynomials the way that it looks can hinder our progress in finding a solution. Factorization\n\n### Tool 1. Greatest Common Factor (GCF)",
null,
"Chapter 4: Factoring Review Tool 1 Greatest Common Factor (GCF) This is a very important tool. You must try to factor out the GCF first in every problem. Some problems do not have a GCF but many do. When\n\n### Factoring Polynomials",
null,
"UNIT 11 Factoring Polynomials You can use polynomials to describe framing for art. 396 Unit 11 factoring polynomials A polynomial is an expression that has variables that represent numbers. A number can\n\n### Factoring Algebra- Chapter 8B Assignment Sheet",
null,
"Name: Factoring Algebra- Chapter 8B Assignment Sheet Date Section Learning Targets Assignment Tues 2/17 Find the prime factorization of an integer Find the greatest common factor (GCF) for a set of monomials.\n\n### 6.4 Special Factoring Rules",
null,
"6.4 Special Factoring Rules OBJECTIVES 1 Factor a difference of squares. 2 Factor a perfect square trinomial. 3 Factor a difference of cubes. 4 Factor a sum of cubes. By reversing the rules for multiplication\n\n### 15.1 Factoring Polynomials",
null,
"LESSON 15.1 Factoring Polynomials Use the structure of an expression to identify ways to rewrite it. Also A.SSE.3? ESSENTIAL QUESTION How can you use the greatest common factor to factor polynomials? EXPLORE\n\n### Section 6.1 Factoring Expressions",
null,
"Section 6.1 Factoring Expressions The first method we will discuss, in solving polynomial equations, is the method of FACTORING. Before we jump into this process, you need to have some concept of what\n\n### Factor Polynomials Completely",
null,
"9.8 Factor Polynomials Completely Before You factored polynomials. Now You will factor polynomials completely. Why? So you can model the height of a projectile, as in Ex. 71. Key Vocabulary factor by grouping\n\n### Algebra 2 PreAP. Name Period",
null,
"Algebra 2 PreAP Name Period IMPORTANT INSTRUCTIONS FOR STUDENTS!!! We understand that students come to Algebra II with different strengths and needs. For this reason, students have options for completing\n\n### Factoring - Grouping",
null,
"6.2 Factoring - Grouping Objective: Factor polynomials with four terms using grouping. The first thing we will always do when factoring is try to factor out a GCF. This GCF is often a monomial like in\n\n### POLYNOMIALS and FACTORING",
null,
"POLYNOMIALS and FACTORING Exponents ( days); 1. Evaluate exponential expressions. Use the product rule for exponents, 1. How do you remember the rules for exponents?. How do you decide which rule to use\n\n### FACTORING ax 2 bx c WITH a 1",
null,
"296 (6 20) Chapter 6 Factoring 6.4 FACTORING a 2 b c WITH a 1 In this section The ac Method Trial and Error Factoring Completely In Section 6.3 we factored trinomials with a leading coefficient of 1. In\n\n### 1.5. Factorisation. Introduction. Prerequisites. Learning Outcomes. Learning Style",
null,
"Factorisation 1.5 Introduction In Block 4 we showed the way in which brackets were removed from algebraic expressions. Factorisation, which can be considered as the reverse of this process, is dealt with\n\n### 1.4. Removing Brackets. Introduction. Prerequisites. Learning Outcomes. Learning Style",
null,
"Removing Brackets 1. Introduction In order to simplify an expression which contains brackets it is often necessary to rewrite the expression in an equivalent form but without any brackets. This process\n\n### 6.1 The Greatest Common Factor; Factoring by Grouping",
null,
"386 CHAPTER 6 Factoring and Applications 6.1 The Greatest Common Factor; Factoring by Grouping OBJECTIVES 1 Find the greatest common factor of a list of terms. 2 Factor out the greatest common factor.\n\n### BEGINNING ALGEBRA ACKNOWLEDMENTS",
null,
"BEGINNING ALGEBRA The Nursing Department of Labouré College requested the Department of Academic Planning and Support Services to help with mathematics preparatory materials for its Bachelor of Science\n\n### AIP Factoring Practice/Help",
null,
"The following pages include many problems to practice factoring skills. There are also several activities with examples to help you with factoring if you feel like you are not proficient with it. There\n\n### Factoring Guidelines. Greatest Common Factor Two Terms Three Terms Four Terms. 2008 Shirley Radai",
null,
"Factoring Guidelines Greatest Common Factor Two Terms Three Terms Four Terms 008 Shirley Radai Greatest Common Factor 008 Shirley Radai Factoring by Finding the Greatest Common Factor Always check for\n\n### SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS",
null,
"(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.1 SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS LEARNING OBJECTIVES Be able to identify polynomial, rational, and algebraic\n\n### Topic: Special Products and Factors Subtopic: Rules on finding factors of polynomials",
null,
"Quarter I: Special Products and Factors and Quadratic Equations Topic: Special Products and Factors Subtopic: Rules on finding factors of polynomials Time Frame: 20 days Time Frame: 3 days Content Standard:\n\n### SOLVING QUADRATIC EQUATIONS BY THE NEW TRANSFORMING METHOD (By Nghi H Nguyen Updated Oct 28, 2014))",
null,
"SOLVING QUADRATIC EQUATIONS BY THE NEW TRANSFORMING METHOD (By Nghi H Nguyen Updated Oct 28, 2014)) There are so far 8 most common methods to solve quadratic equations in standard form ax² + bx + c = 0.\n\n### Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any.",
null,
"Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...}\n\n### Algebra I Vocabulary Cards",
null,
"Algebra I Vocabulary Cards Table of Contents Expressions and Operations Natural Numbers Whole Numbers Integers Rational Numbers Irrational Numbers Real Numbers Absolute Value Order of Operations Expression\n\n### Factoring Polynomials",
null,
"Factoring a Polynomial Expression Factoring a polynomial is expressing the polynomial as a product of two or more factors. Simply stated, it is somewhat the reverse process of multiplying. To factor polynomials,\n\n### Factoring Polynomials and Solving Quadratic Equations",
null,
"Factoring Polynomials and Solving Quadratic Equations Math Tutorial Lab Special Topic Factoring Factoring Binomials Remember that a binomial is just a polynomial with two terms. Some examples include 2x+3\n\n### Polynomial Equations and Factoring",
null,
"7 Polynomial Equations and Factoring 7.1 Adding and Subtracting Polynomials 7.2 Multiplying Polynomials 7.3 Special Products of Polynomials 7.4 Dividing Polynomials 7.5 Solving Polynomial Equations in\n\n### Factoring Flow Chart",
null,
"Factoring Flow Chart greatest common factor? YES NO factor out GCF leaving GCF(quotient) how many terms? 4+ factor by grouping 2 3 difference of squares? perfect square trinomial? YES YES NO NO a 2 -b",
null,
"8. Radicals - Multiply and Divide Radicals Objective: Multiply and divide radicals using the product and quotient rules of radicals. Multiplying radicals is very simple if the index on all the radicals\n\n### MATH 90 CHAPTER 6 Name:.",
null,
"MATH 90 CHAPTER 6 Name:. 6.1 GCF and Factoring by Groups Need To Know Definitions How to factor by GCF How to factor by groups The Greatest Common Factor Factoring means to write a number as product. a\n\n### The majority of college students hold credit cards. According to the Nellie May",
null,
"CHAPTER 6 Factoring Polynomials 6.1 The Greatest Common Factor and Factoring by Grouping 6. Factoring Trinomials of the Form b c 6.3 Factoring Trinomials of the Form a b c and Perfect Square Trinomials\n\n### 1.3 Algebraic Expressions",
null,
"1.3 Algebraic Expressions A polynomial is an expression of the form: a n x n + a n 1 x n 1 +... + a 2 x 2 + a 1 x + a 0 The numbers a 1, a 2,..., a n are called coefficients. Each of the separate parts,\n\n### SOLVING QUADRATIC EQUATIONS - COMPARE THE FACTORING AC METHOD AND THE NEW TRANSFORMING METHOD (By Nghi H. Nguyen - Jan 18, 2015)",
null,
"SOLVING QUADRATIC EQUATIONS - COMPARE THE FACTORING AC METHOD AND THE NEW TRANSFORMING METHOD (By Nghi H. Nguyen - Jan 18, 2015) GENERALITIES. When a given quadratic equation can be factored, there are\n\n### FACTORING OUT COMMON FACTORS",
null,
"278 (6 2) Chapter 6 Factoring 6.1 FACTORING OUT COMMON FACTORS In this section Prime Factorization of Integers Greatest Common Factor Finding the Greatest Common Factor for Monomials Factoring Out the\n\n### Mathematics Placement",
null,
"Mathematics Placement The ACT COMPASS math test is a self-adaptive test, which potentially tests students within four different levels of math including pre-algebra, algebra, college algebra, and trigonometry.\n\n### CM2202: Scientific Computing and Multimedia Applications General Maths: 2. Algebra - Factorisation",
null,
"CM2202: Scientific Computing and Multimedia Applications General Maths: 2. Algebra - Factorisation Prof. David Marshall School of Computer Science & Informatics Factorisation Factorisation is a way of\n\n### 5 means to write it as a product something times something instead of a sum something plus something plus something.",
null,
"Intermediate algebra Class notes Factoring Introduction (section 6.1) Recall we factor 10 as 5. Factoring something means to think of it as a product! Factors versus terms: terms: things we are adding\n\n### ~ EQUIVALENT FORMS ~",
null,
"~ EQUIVALENT FORMS ~ Critical to understanding mathematics is the concept of equivalent forms. Equivalent forms are used throughout this course. Throughout mathematics one encounters equivalent forms of\n\n### Negative Integer Exponents",
null,
"7.7 Negative Integer Exponents 7.7 OBJECTIVES. Define the zero exponent 2. Use the definition of a negative exponent to simplify an expression 3. Use the properties of exponents to simplify expressions\n\n### Polynomial Expression",
null,
"DETAILED SOLUTIONS AND CONCEPTS - POLYNOMIAL EXPRESSIONS Prepared by Ingrid Stewart, Ph.D., College of Southern Nevada Please Send Questions and Comments to [email protected]. Thank you! PLEASE NOTE\n\n### Factoring ax 2 + bx + c - Teacher Notes",
null,
"Southern Nevada Regional Professi onal D evel opment Program VOLUME 1, ISSUE 8 MAY 009 A N ewsletter from the Sec ondary Mathematic s Team Factoring ax + bx + c - Teacher Notes Here we look at sample teacher\n\n### Factoring Polynomials",
null,
"Factoring Polynomials 4-1-2014 The opposite of multiplying polynomials is factoring. Why would you want to factor a polynomial? Let p(x) be a polynomial. p(c) = 0 is equivalent to x c dividing p(x). Recall\n\n### A Systematic Approach to Factoring",
null,
"A Systematic Approach to Factoring Step 1 Count the number of terms. (Remember****Knowing the number of terms will allow you to eliminate unnecessary tools.) Step 2 Is there a greatest common factor? Tool\n\n### How To Factor By Gcf In Algebra 1.5",
null,
"7-2 Factoring by GCF Warm Up Lesson Presentation Lesson Quiz Algebra 1 Warm Up Simplify. 1. 2(w + 1) 2. 3x(x 2 4) 2w + 2 3x 3 12x Find the GCF of each pair of monomials. 3. 4h 2 and 6h 2h 4. 13p and 26p\n\n### Greatest Common Factor (GCF) Factoring",
null,
"Section 4 4: Greatest Common Factor (GCF) Factoring The last chapter introduced the distributive process. The distributive process takes a product of a monomial and a polynomial and changes the multiplication\n\n### Algebra Cheat Sheets",
null,
"Sheets Algebra Cheat Sheets provide you with a tool for teaching your students note-taking, problem-solving, and organizational skills in the context of algebra lessons. These sheets teach the concepts\n\n### MATH 10034 Fundamental Mathematics IV",
null,
"MATH 0034 Fundamental Mathematics IV http://www.math.kent.edu/ebooks/0034/funmath4.pdf Department of Mathematical Sciences Kent State University January 2, 2009 ii Contents To the Instructor v Polynomials.\n\n### Algebra 1 Chapter 08 review",
null,
"Name: Class: Date: ID: A Algebra 1 Chapter 08 review Multiple Choice Identify the choice that best completes the statement or answers the question. Simplify the difference. 1. (4w 2 4w 8) (2w 2 + 3w 6)\n\n### Simplifying Algebraic Fractions",
null,
"5. Simplifying Algebraic Fractions 5. OBJECTIVES. Find the GCF for two monomials and simplify a fraction 2. Find the GCF for two polynomials and simplify a fraction Much of our work with algebraic fractions\n\n### Factoring (pp. 1 of 4)",
null,
"Factoring (pp. 1 of 4) Algebra Review Try these items from middle school math. A) What numbers are the factors of 4? B) Write down the prime factorization of 7. C) 6 Simplify 48 using the greatest common\n\n### Math 10C. Course: Polynomial Products and Factors. Unit of Study: Step 1: Identify the Outcomes to Address. Guiding Questions:",
null,
"Course: Unit of Study: Math 10C Polynomial Products and Factors Step 1: Identify the Outcomes to Address Guiding Questions: What do I want my students to learn? What can they currently understand and do?\n\n### 1.4. Arithmetic of Algebraic Fractions. Introduction. Prerequisites. Learning Outcomes",
null,
"Arithmetic of Algebraic Fractions 1.4 Introduction Just as one whole number divided by another is called a numerical fraction, so one algebraic expression divided by another is known as an algebraic fraction.\n\n### 5.1 FACTORING OUT COMMON FACTORS",
null,
"C H A P T E R 5 Factoring he sport of skydiving was born in the 1930s soon after the military began using parachutes as a means of deploying troops. T Today, skydiving is a popular sport around the world.\n\n### 6.6 Factoring Strategy",
null,
"456 CHAPTER 6. FACTORING 6.6 Factoring Strategy When you are concentrating on factoring problems of a single type, after doing a few you tend to get into a rhythm, and the remainder of the exercises, because\n\n### Pre-Calculus II Factoring and Operations on Polynomials",
null,
"Factoring... 1 Polynomials...1 Addition of Polynomials... 1 Subtraction of Polynomials...1 Multiplication of Polynomials... Multiplying a monomial by a monomial... Multiplying a monomial by a polynomial...\n\n### HIBBING COMMUNITY COLLEGE COURSE OUTLINE",
null,
"HIBBING COMMUNITY COLLEGE COURSE OUTLINE COURSE NUMBER & TITLE: - Beginning Algebra CREDITS: 4 (Lec 4 / Lab 0) PREREQUISITES: MATH 0920: Fundamental Mathematics with a grade of C or better, Placement Exam,\n\n### FACTORISATION YEARS. A guide for teachers - Years 9 10 June 2011. The Improving Mathematics Education in Schools (TIMES) Project",
null,
"9 10 YEARS The Improving Mathematics Education in Schools (TIMES) Project FACTORISATION NUMBER AND ALGEBRA Module 33 A guide for teachers - Years 9 10 June 2011 Factorisation (Number and Algebra : Module\n\n### Factoring - Factoring Special Products",
null,
"6.5 Factoring - Factoring Special Products Objective: Identify and factor special products including a difference of squares, perfect squares, and sum and difference of cubes. When factoring there are\n\n### Factoring. Factoring Polynomial Equations. Special Factoring Patterns. Factoring. Special Factoring Patterns. Special Factoring Patterns",
null,
"Factoring Factoring Polynomial Equations Ms. Laster Earlier, you learned to factor several types of quadratic expressions: General trinomial - 2x 2-5x-12 = (2x + 3)(x - 4) Perfect Square Trinomial - x\n\n### Vocabulary Words and Definitions for Algebra",
null,
"Name: Period: Vocabulary Words and s for Algebra Absolute Value Additive Inverse Algebraic Expression Ascending Order Associative Property Axis of Symmetry Base Binomial Coefficient Combine Like Terms\n\n### Algebraic expressions are a combination of numbers and variables. Here are examples of some basic algebraic expressions.",
null,
"Page 1 of 13 Review of Linear Expressions and Equations Skills involving linear equations can be divided into the following groups: Simplifying algebraic expressions. Linear expressions. Solving linear\n\n### How To Fator",
null,
"CHAPTER hapter 4 > Make the Connetion 4 INTRODUCTION Developing seret odes is big business beause of the widespread use of omputers and the Internet. Corporations all over the world sell enryption systems\n\n### FACTOR POLYNOMIALS by SPLITTING",
null,
"FACTOR POLYNOMIALS by SPLITTING THE IDEA FACTOR POLYNOMIALS by SPLITTING The idea is to split the middle term into two pieces. Say the polynomial looks like c + bx + ax 2. Further more suppose the DL METHOD\n\n### MTH 092 College Algebra Essex County College Division of Mathematics Sample Review Questions 1 Created January 17, 2006",
null,
"MTH 092 College Algebra Essex County College Division of Mathematics Sample Review Questions Created January 7, 2006 Math 092, Elementary Algebra, covers the mathematical content listed below. In order",
null,
"MATH 60 NOTEBOOK CERTIFICATIONS Chapter #1: Integers and Real Numbers 1.1a 1.1b 1.2 1.3 1.4 1.8 Chapter #2: Algebraic Expressions, Linear Equations, and Applications 2.1a 2.1b 2.1c 2.2 2.3a 2.3b 2.4 2.5",
null,
"PERFECT SQUARES AND FACTORING EXAMPLES 1. Ask the students what is meant by identical. Get their responses and then explain that when we have two factors that are identical, we call them perfect squares."
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http://ada.m2osw.com/aarm-2012/AA-A-5-2.html | [
"A.5.2 Random Number Generation\n\n1\n[Facilities for the generation of pseudo-random floating point numbers are provided in the package Numerics.Float_Random; the generic package Numerics.Discrete_Random provides similar facilities for the generation of pseudo-random integers and pseudo-random values of enumeration types. For brevity, pseudo-random values of any of these types are called random numbers.\n2\nSome of the facilities provided are basic to all applications of random numbers. These include a limited private type each of whose objects serves as the generator of a (possibly distinct) sequence of random numbers; a function to obtain the “next” random number from a given sequence of random numbers (that is, from its generator); and subprograms to initialize or reinitialize a given generator to a time-dependent state or a state denoted by a single integer.\n3\nOther facilities are provided specifically for advanced applications. These include subprograms to save and restore the state of a given generator; a private type whose objects can be used to hold the saved state of a generator; and subprograms to obtain a string representation of a given generator state, or, given such a string representation, the corresponding state.]\n3.a\nDiscussion: These facilities support a variety of requirements ranging from repeatable sequences (for debugging) to unique sequences in each execution of a program.\n\nStatic Semantics\n\n4\nThe library package Numerics.Float_Random has the following declaration:\n5\n6\n-- Basic facilities\n7\ntype Generator is limited private;\n8\nsubtype Uniformly_Distributed is Float range 0.0 .. 1.0;\nfunction Random (Gen : Generator) return Uniformly_Distributed;\n9\nprocedure Reset (Gen : in Generator;\nInitiator : in Integer);\nprocedure Reset (Gen : in Generator);\n10\n11\ntype State is private;\n12\nprocedure Save (Gen : in Generator;\nTo_State : out State);\nprocedure Reset (Gen : in Generator;\nFrom_State : in State);\n13\nMax_Image_Width : constant := implementation-defined integer value;\n14\nfunction Image (Of_State : State) return String;\nfunction Value (Coded_State : String) return State;\n15\nprivate\n... -- not specified by the language\n15.1/2\n{AI95-00360-01} The type Generator needs finalization (see 7.6).\n16\nThe generic library package Numerics.Discrete_Random has the following declaration:\n17\n\ngeneric\ntype Result_Subtype is (<>);\n18\n-- Basic facilities\n19\ntype Generator is limited private;\n20\nfunction Random (Gen : Generator) return Result_Subtype;\n21\nprocedure Reset (Gen : in Generator;\nInitiator : in Integer);\nprocedure Reset (Gen : in Generator);\n22\n23\ntype State is private;\n24\nprocedure Save (Gen : in Generator;\nTo_State : out State);\nprocedure Reset (Gen : in Generator;\nFrom_State : in State);\n25\nMax_Image_Width : constant := implementation-defined integer value;\n26\nfunction Image (Of_State : State) return String;\nfunction Value (Coded_State : String) return State;\n27\nprivate\n... -- not specified by the language\n27.a\nImplementation defined: The value of Numerics.Float_Random.Max_Image_Width.\n27.b\nImplementation defined: The value of Numerics.Discrete_Random.Max_Image_Width.\n27.c/1\nImplementation Note: {8652/0097} {AI95-00115-01} The following is a possible implementation of the private part of Numerics.Float_Random each package (assuming the presence of “with Ada.Finalization;” as a context clause):\n27.d\ntype State is ...;\ntype Access_State is access State;\ntype Generator is new Finalization.Limited_Controlled with\nrecord\nS : Access_State := new State'(...);\nend record;\nprocedure Finalize (G : in out Generator);\n27.d.1/2\n{8652/0097} {AI95-00115-01} {AI95-00344-01} Unfortunately, Numerics.Discrete_Random.Generator also can cannot be implemented this way, as Numerics.Discrete_Random can be instantiated at any nesting depth. However, Generator could have a component of a controlled type, as long as that type is declared in some other (non-generic) package. One possible solution would be to implement Numerics.Discrete_Random in terms of Numerics.Float_Random, using a component of Numerics.Float_Random.Generator to implement Numerics.Float_Random.Generator.\n27.e\nClearly some level of indirection is required in the implementation of a Generator, since the parameter mode is in for all operations on a Generator. For this reason, Numerics.Float_Random and Numerics.Discrete_Random cannot be declared pure.\n27.1/2\n{AI95-00360-01} The type Generator needs finalization (see 7.6) in every instantiation of Numerics.Discrete_Random.\n28\nAn object of the limited private type Generator is associated with a sequence of random numbers. Each generator has a hidden (internal) state, which the operations on generators use to determine the position in the associated sequence. All generators are implicitly initialized to an unspecified state that does not vary from one program execution to another; they may also be explicitly initialized, or reinitialized, to a time-dependent state, to a previously saved state, or to a state uniquely denoted by an integer value.\n28.a\nDiscussion: The repeatability provided by the implicit initialization may be exploited for testing or debugging purposes.\n29\nAn object of the private type State can be used to hold the internal state of a generator. Such objects are only needed if the application is designed to save and restore generator states or to examine or manufacture them.\n30\nThe operations on generators affect the state and therefore the future values of the associated sequence. The semantics of the operations on generators and states are defined below.\n31\nfunction Random (Gen : Generator) return Uniformly_Distributed;\nfunction Random (Gen : Generator) return Result_Subtype;\n32\nObtains the “next” random number from the given generator, relative to its current state, according to an implementation-defined algorithm. The result of the function in Numerics.Float_Random is delivered as a value of the subtype Uniformly_Distributed, which is a subtype of the predefined type Float having a range of 0.0 .. 1.0. The result of the function in an instantiation of Numerics.Discrete_Random is delivered as a value of the generic formal subtype Result_Subtype.\n32.a/2\nThis paragraph was deleted.Implementation defined: The algorithms for random number generation.\n32.a.1/2\nDiscussion: The algorithm is the subject of a Documentation Requirement, so we don't separately summarize this implementation-defined item.\n32.b\nReason: The requirement for a level of indirection in accessing the internal state of a generator arises from the desire to make Random a function, rather than a procedure.\n33\nprocedure Reset (Gen : in Generator;\nInitiator : in Integer);\nprocedure Reset (Gen : in Generator);\n34\nSets the state of the specified generator to one that is an unspecified function of the value of the parameter Initiator (or to a time-dependent state, if only a generator parameter is specified). The latter form of the procedure is known as the time-dependent Reset procedure\n34.a\nImplementation Note: The time-dependent Reset procedure can be implemented by mapping the current time and date as determined by the system clock into a state, but other implementations are possible. For example, a white-noise generator or a radioactive source can be used to generate time-dependent states.\n35\nprocedure Save (Gen : in Generator;\nTo_State : out State);\nprocedure Reset (Gen : in Generator;\nFrom_State : in State);\n36\nSave obtains the current state of a generator. Reset gives a generator the specified state. A generator that is reset to a state previously obtained by invoking Save is restored to the state it had when Save was invoked.\n37\nfunction Image (Of_State : State) return String;\nfunction Value (Coded_State : String) return State;\n38\nImage provides a representation of a state coded (in an implementation-defined way) as a string whose length is bounded by the value of Max_Image_Width. Value is the inverse of Image: Value(Image(S)) = S for each state S that can be obtained from a generator by invoking Save.\n38.a\nImplementation defined: The string representation of a random number generator's state.\n\nDynamic Semantics\n\n39\nInstantiation of Numerics.Discrete_Random with a subtype having a null range raises Constraint_Error.\n40/1\nThis paragraph was deleted.{8652/0050} {AI95-00089} Invoking Value with a string that is not the image of any generator state raises Constraint_Error.\n\nBounded (Run-Time) Errors\n\n40.1/1\n{8652/0050} {AI95-00089} It is a bounded error to invoke Value with a string that is not the image of any generator state. If the error is detected, Constraint_Error or Program_Error is raised. Otherwise, a call to Reset with the resulting state will produce a generator such that calls to Random with this generator will produce a sequence of values of the appropriate subtype, but which might not be random in character. That is, the sequence of values might not fulfill the implementation requirements of this subclause.\n\nImplementation Requirements\n\n41\nA sufficiently long sequence of random numbers obtained by successive calls to Random is approximately uniformly distributed over the range of the result subtype.\n42\nThe Random function in an instantiation of Numerics.Discrete_Random is guaranteed to yield each value in its result subtype in a finite number of calls, provided that the number of such values does not exceed 2 15.\n43\nOther performance requirements for the random number generator, which apply only in implementations conforming to the Numerics Annex, and then only in the “strict” mode defined there (see G.2), are given in G.2.5.\n\nDocumentation Requirements\n\n44\nNo one algorithm for random number generation is best for all applications. To enable the user to determine the suitability of the random number generators for the intended application, the implementation shall describe the algorithm used and shall give its period, if known exactly, or a lower bound on the period, if the exact period is unknown. Periods that are so long that the periodicity is unobservable in practice can be described in such terms, without giving a numerical bound.\n44.a/2\nDocumentation Requirement: The algorithm used for random number generation, including a description of its period.\n45\nThe implementation also shall document the minimum time interval between calls to the time-dependent Reset procedure that are guaranteed to initiate different sequences, and it shall document the nature of the strings that Value will accept without raising Constraint_Error.\n45.a/2\nThis paragraph was deleted.Implementation defined: The minimum time interval between calls to the time-dependent Reset procedure that are guaranteed to initiate different random number sequences.\n45.b/2\nDocumentation Requirement: The minimum time interval between calls to the time-dependent Reset procedure that is guaranteed to initiate different random number sequences.\n\n46\nAny storage associated with an object of type Generator should be reclaimed on exit from the scope of the object.\n46.a.1/2\nImplementation Advice: Any storage associated with an object of type Generator of the random number packages should be reclaimed on exit from the scope of the object.\n46.a\nRamification: A level of indirection is implicit in the semantics of the operations, given that they all take parameters of mode in. This implies that the full type of Generator probably should be a controlled type, with appropriate finalization to reclaim any heap-allocated storage.\n47\nIf the generator period is sufficiently long in relation to the number of distinct initiator values, then each possible value of Initiator passed to Reset should initiate a sequence of random numbers that does not, in a practical sense, overlap the sequence initiated by any other value. If this is not possible, then the mapping between initiator values and generator states should be a rapidly varying function of the initiator value.\n47.a/2\nImplementation Advice: Each value of Initiator passed to Reset for the random number packages should initiate a distinct sequence of random numbers, or, if that is not possible, be at least a rapidly varying function of the initiator value.\nNOTES\n48\n18 If two or more tasks are to share the same generator, then the tasks have to synchronize their access to the generator as for any shared variable (see 9.10).\n49\n19 Within a given implementation, a repeatable random number sequence can be obtained by relying on the implicit initialization of generators or by explicitly initializing a generator with a repeatable initiator value. Different sequences of random numbers can be obtained from a given generator in different program executions by explicitly initializing the generator to a time-dependent state.\n50\n20 A given implementation of the Random function in Numerics.Float_Random may or may not be capable of delivering the values 0.0 or 1.0. Portable applications should assume that these values, or values sufficiently close to them to behave indistinguishably from them, can occur. If a sequence of random integers from some fixed range is needed, the application should use the Random function in an appropriate instantiation of Numerics.Discrete_Random, rather than transforming the result of the Random function in Numerics.Float_Random. However, some applications with unusual requirements, such as for a sequence of random integers each drawn from a different range, will find it more convenient to transform the result of the floating point Random function. For M ≥ 1, the expression\n51\nInteger(Float(M) * Random(G)) mod M\n52\ntransforms the result of Random(G) to an integer uniformly distributed over the range 0 .. M–1; it is valid even if Random delivers 0.0 or 1.0. Each value of the result range is possible, provided that M is not too large. Exponentially distributed (floating point) random numbers with mean and standard deviation 1.0 can be obtained by the transformation\n53/2\n{AI95-00434-01} -Log(Random(G) + Float'Model_Small))\n54\nwhere Log comes from Numerics.Elementary_Functions (see A.5.1); in this expression, the addition of Float'Model_Small avoids the exception that would be raised were Log to be given the value zero, without affecting the result (in most implementations) when Random returns a nonzero value.\n\nExamples\n\n55\nExample of a program that plays a simulated dice game:\n56\nprocedure Dice_Game is\nsubtype Die is Integer range 1 .. 6;\nsubtype Dice is Integer range 2*Die'First .. 2*Die'Last;\npackage Random_Die is new Ada.Numerics.Discrete_Random (Die);\nuse Random_Die;\nG : Generator;\nD : Dice;\nbegin\nReset (G); -- Start the generator in a unique state in each run\nloop\n-- Roll a pair of dice; sum and process the results\nD := Random(G) + Random(G);\n...\nend loop;\nend Dice_Game;\n57\nExample of a program that simulates coin tosses:\n58\nprocedure Flip_A_Coin is\npackage Random_Coin is new Ada.Numerics.Discrete_Random (Coin);\nuse Random_Coin;\nG : Generator;\nbegin\nReset (G); -- Start the generator in a unique state in each run\nloop\n-- Toss a coin and process the result\ncase Random(G) is\n...\nwhen Tails =>\n...\nend case;\n...\nend loop;\nend Flip_A_Coin;\n59\nExample of a parallel simulation of a physical system, with a separate generator of event probabilities in each task:\n60\nprocedure Parallel_Simulation is\nentry Initialize_Generator (Initiator : in Integer);\n...\nend Worker;\nW : array (1 .. 10) of Worker;\nG : Generator;\nProbability_Of_Event : Uniformly_Distributed;\nbegin\naccept Initialize_Generator (Initiator : in Integer) do\nReset (G, Initiator);\nend Initialize_Generator;\nloop\n...\nProbability_Of_Event := Random(G);\n...\nend loop;\nend Worker;\nbegin\n-- Initialize the generators in the Worker tasks to different states\nfor I in W'Range loop\nW(I).Initialize_Generator (I);\nend loop;\n... -- Wait for the Worker tasks to terminate\nend Parallel_Simulation;\nNOTES\n61\n21 Notes on the last example: Although each Worker task initializes its generator to a different state, those states will be the same in every execution of the program. The generator states can be initialized uniquely in each program execution by instantiating Ada.Numerics.Discrete_Random for the type Integer in the main procedure, resetting the generator obtained from that instance to a time-dependent state, and then using random integers obtained from that generator to initialize the generators in each Worker task.",
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"Ada 2005 and 2012 Editions sponsored in part by Ada-Europe"
]
| [
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"http://ada.m2osw.com/aarm-2012/AE_logo.gif",
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https://doingbayesiandataanalysis.blogspot.com/2019/07/shrinkage-in-hierarchical-models-random.html | [
"## Thursday, July 25, 2019\n\n### Shrinkage in hierarchical models: random effects in lmer() with and without correlation\n\nThe goal of this post is to illustrate shrinkage of parameter estimates in hierarchical (aka multi-level) models, specifically when using lmer() with and without estimated correlation of parameters. The examples will show how estimates can differ when including correlation of parameters because of shrinkage toward the estimated correlation.\n\n# Background\n\n## Data structure for these examples\n\n“It all begins with the data… ” I will create multiple panels of $\\langle x , y \\rangle$ data values, with $x$ and $y$ being continuous metric variables.\n• For instance, each panel could be data from a student in a classroom, with each datum being performance on a standardized math exam, with $x$ being time and $y$ being performance. In this scenario, each student takes a novel variant of the test repeatedly across time. The times do not need to be the same for every student, and the number of tests do not need to be the same for every student. We are interested characterizing the performance trend of each panel (i.e., each student) and the overall trend across panels (i.e., for the class as a whole).\n• As another example, each panel could be data from a distinct class within a school, with each datum being a particular student's exam performance (on the $y$ axis) and family income (on the $x$ axis). Again we are interested characterizing the trend of each panel (i.e., the relation of exam performance to family income within each classroom) and the overall trend across panels (i.e., the typical relationship of the variables across classrooms).\nTo illustrate robust shrinkage of panel estimates, each panel will have relatively few data points, and there will be relatively lots of panels. Graphs of the data will appear in analysis results, later.\n\nHere (below) is the structure of the data. Notice there is an X variable, a Y variable, and a Panel variable. The panel variable is actually a nominal (categorical) value, even though it appears as a numerical index.\nstr( myData )\n\n## 'data.frame': 208 obs. of 3 variables:\n## $X : num 0.4158 0.3795 0.0746 0.0588 0.4503 ... ##$ Y : num -0.864 -0.579 0.227 -1.604 -0.895 ...\n## \\$ Panel: Factor w/ 35 levels \"1\",\"2\",\"3\",\"4\",..: 1 1 1 1 1 1 2 2 2 3 ...\n\n\n## Analysis models\n\nFor simplicity, each panel will be fit with a linear trend. The hierarchical (a.k.a. multi-level) models will also estimate the typical linear trend across panels.\n\nParameters for panels are subject to shrinkage in hierarchical models because the panel's linear trend is trying to conform simultaneously to (a) the data in its panel and (b) the typical trend across all panels. When there are lots of panels informing the typical trend, and only a small amount of data within a panel, then the panel estimates are strongly influenced by the typical trend across panels. This makes good sense: If you don't know much about a particular panel, your best estimate should take into account what's typical across many other similar panels.\n\nFor more background about shrinkage in hierarchical models, there are lots on online sources you can search, and you can see some of my previous writings on the topic:\nI will first fit a line independently to each panel, without hierarchical structure. This analysis will show the estimated intercept and slope in each panel when there is no shrinkage.\n\nI will then fit a hierarchical model that estimates a typical intercept and typical slope across panels, but does not estimate the correlation of the intercepts and slopes across panels. This model produces some shrinkage across panel estimates, but does not shrink the estimates toward a shared correlation across panels.\n\nFinally, I will fit a hiearchical model that also estimates the correlation of intercepts and slopes across panels. This model shrinks the panel estimates so they also conform more strongly with the estimated correlation across panels.\n\nFor the non-hierarchical analysis, I will use lm() from the base stats package of R. For the hierarchical analyses, I will use lmer() from the lme4 package in R.\n\n# Independent line for every panel\n\nFor this analysis, each individual panel is fit with its own line, separately from all other panels, using lm() on each panel. There is no hierarchical structure and no overall line estimated.\n\nTo make this analysis most analogous to subsequent analyses with lmer() the analyses should require all panels to have the same noise variance. But this is not done here, and actually the MLE coefficients are not affected in this case.\n\nIn principle, the analysis in this section would be like using lmer() with the formula y ~ 0 + (1+X||Panel), which specifies fitting lines within panels with no estimation of correlation across panels and no global parameters. But lmer() throws an error if that specification is attempted.\nHere (below) are scatter plots of the data with the lm() fitted regression lines:",
null,
"Notice above:\n• Two-point panels such as Panel 4 and Panel 11 have lines going exactly through the two points. This will not be the case in hierarchical models.\n• The one-point Panel 35 has no regression line because it's undefined. This will not be the case in hierarchical models.\n• Panels 4 and 19 are color-highlighted for easy comparison with subsequent analyses.",
null,
"Notice above:\n• There is correlation of intercepts and slopes across panels (r=0.65), reflecting only how the data were generated, not any estimation of correlation in the model.\n• There is a lot of variation in intercepts and slopes across panels relative to hierarchical (multi-level) models below. There will be less variation in hierarchical models, hence the term shrinkage.\n• Panels 4 and 19 are color-highlighted for easy comparison across analyses.\n\n# Random intercepts and slopes, but no estimated correlation\n\nI'll use lmer() with the formula, Y ~ 1 + X + ( 1 + X || Panel ), which is equivalent to Y ~ 1 + X + ( (1|Panel) + (0+X|Panel) ). lmer() assumes we want to estimate correlations of parameters across panels unless we tell it not to by using a double vertical bar or by explicitly coding the separate effects.",
null,
"Notice above:\n• Two-point panels such as Panel 4 and Panel 11 have lines not going exactly through the two points. This is because the line is trying to conform simultaneously to the data in the panel and what's typical across panels, as estimated by this particular hierarchical model.\n• The one-point Panel 35 has a regression line despite having only a single point. This is because the line is generated by what's typical across panels, influenced a bit by the single data point in the panel.\n• Panels 4 and 19 are color-highlighted for easy comparison with across analyses.",
null,
"Notice above:\n• There is correlation of intercepts and slopes across panels (r=0.814), but this reflects only how the data were generated and the separate shrinkage of intercepts and slopes, without any shrinkage from estimation of correlation.\n• There is less variation in intercepts and slopes across panels relative to the previous, non-hierarchical analysis, hence the term shrinkage. Specifically, the range of slopes across panels in the non-hierarchical model was -3.48, 2.59 but the range of slopes in this hierarchical model is -2.74, 2.03.\n• Panels 4 and 19 are color-highlighted for easy comparison across analyses.\n\n# Random intercepts and slopes, with estimated correlation\n\nHere I use lmer() with forumla Y ~ 1 + X + ( 1 + X | Panel ). Notice the single vertical bar before Panel, so lmer() estimates the correlation of parameters across panels by default.",
null,
"Notice above:\nThere is even more shrinkage than in the previous model, because now the lines in each panel are also “trying” to conform to the typical correlation of intercept and slope across panels. Notice in particular the color-coded lines in panels 4 and 19.",
null,
"Notice above:\nThere is a strong correlation between the estimated slopes and intercepts (r=0.998). Here the correlation is estimated and there is shrinkage of estimates toward that correlation, and the correlation is stronger than the previous model because the estimates are shrunken toward that correlation.\n\n# What about the higher-level, \"fixed\" effects?\n\nThe higher-level intercept and slope are at the means of the panel intercepts and panel slopes, and those overall means are essentially the same across these analyses.\n\n# Conclusion\n\nHopefully these examples have helped to illustrate shrinkage in hierarchical models, and specifically the extra shrinkage introduced by estimating correlation across parameters.\n\n1.",
null,
"Quite interesting illustration, thanks for this post.\n\nI've something still unclear, when you distinguish \"correlations due to the data generation\" and \"correlations between estimates\". I'm not sure I understand the origin of both correlations.\nIn particular, I didn't saw in the data generation process if the slope and intercept were generated for each panel using independant or correlated random effects.\nIn the first case, I do not understand what the correlation obtained using correlated random effects mean exactly...\nCould you please give a few explainations?\n1.",
null,
""
]
| [
null,
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",
null,
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null,
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",
null,
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eakyE8B8589Qw5VvfWZ/cnX5gt22SQHu/nkyWaF+KFxMGLOgLtZmfi5bQOs2knaC0wTzC/FH9faeTrZTGxbyadZPCF+aBSNZJX2qXgbsbnad+mWrojP1hYiPt/87YyIHyJ7I3ZCSpn41ORXs7wVt1udtPEH8fla+589LczfbMFm8rhfHjIRih/ls1nSGn23nWa9+mNVv0hsJ94szzvrZpNCrz4Tf1ybCj706u3j2z/bP3aLfl9oM+IT3zVtTH8fAIivSiTi/w+ZIsrG//eZXAAAAABJRU5ErkJggg==",
null,
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null,
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http://slideplayer.com/slide/3913599/ | [
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"# © 2005 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their.\n\n## Presentation on theme: \"© 2005 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their.\"— Presentation transcript:\n\n© 2005 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. Lecture PowerPoints Chapter 20 Physics: Principles with Applications, 6 th edition Giancoli\n\nChapter 20 Magnetism\n\n20.1 Magnets and Magnetic Fields Magnets have two ends – poles – called north and south. Like poles repel; unlike poles attract.\n\n20.1 Magnets and Magnetic Fields However, if you cut a magnet in half, you don’t get a north pole and a south pole – you get two smaller magnets.\n\n20.1 Magnets and Magnetic Fields Magnetic fields can be visualized using magnetic field lines, which are always closed loops. Notice that the magnetic field goes from North to South.\n\n20.1 Magnets and Magnetic Fields The Earth’s magnetic field is similar to that of a bar magnet. Note that the Earth’s “North Pole” is really a south magnetic pole, as the north ends of magnets are attracted to it.\n\n20.1 Magnets and Magnetic Fields A uniform magnetic field is constant in magnitude and direction. The field between these two wide poles is nearly uniform.\n\n20.2 Electric Currents Produce Magnetic Fields Experiment shows that an electric current produces a magnetic field. (Right Hand Rule)\n\nMagnetic field lines due to a circular loop of wire\n\n20.2 Electric Currents Produce Magnetic Fields The direction of the field is given by a right-hand rule.\n\n20.3 Force on an Electric Current in a Magnetic Field; Definition of B A magnet exerts a force on a current- carrying wire. The direction of the force is given by a right-hand rule.\n\nCurrent-carrying wire in a magnetic field. The force on the wire is directed into the page.\n\n20.3 Force on an Electric Current in a Magnetic Field; Definition of B The force on the wire depends on the current, the length of the wire, the magnetic field, and its orientation. (on formula sheet) If the direction of I is perpendicular to B, then θ = 90 ˚ and sin θ = 1. If I is parallel to B, then θ = 0 ˚ and sin θ = 0. This equation defines the magnetic field B.\n\n20.3 Force on an Electric Current in a Magnetic Field; Definition of B Unit of B : the tesla, T. 1 T = 1 N / A · m. Another unit sometimes used: the gauss ( G ). 1 G = 10 -4 T.\n\nMeasuring a magnetic field B What is the magnetic field if the force is 0.0348 N and the current is 0.245 A? What about the magnetic forces on the two vertical sections of the wire that are in the magnetic field?\n\n20.4 Force on Electric Charge Moving in a Magnetic Field The force on a moving charge is related to the force on a current: (on formula sheet) Once again, the direction is given by a right-hand rule. The rule is for positive particles. Notice the difference between positive and negative particles. The force is greatest when the particle moves perpendicular to B (θ = 90°)\n\n20.4 Force on Electric Charge Moving in a Magnetic Field If a charged particle (in this case an electron) is moving perpendicular to a uniform magnetic field, its path will be a circle.\n\nWhat is the path of a charged particle if its velocity is not perpendicular to the magnetic field? The parallel component of v experiences no force, so it remains constant. The perpendicular component of v results in circular motion. Together this produces a spiral motion.\n\nCharged ions approach the Earth from the sun (solar wind) and are drawn toward the poles. This causes the Northern lights.\n\n20.4 Force on Electric Charge Moving in a Magnetic Field\n\n20.5 Magnetic Field Due to a Long Straight Wire The field is inversely proportional to the distance from the wire: (on formula sheet) The constant μ 0 is called the permeability of free space (or Vacuum permeability), and has the value:\n\n20.6 Force between Two Parallel Wires The magnetic field produced at the position of wire 2 due to the current in wire 1 is: The force this field exerts on a length l 2 of wire 2 is: (20-6)\n\n20.6 Force between Two Parallel Wires Parallel currents attract; antiparallel currents repel.\n\n20.7 Solenoids and Electromagnets If a piece of iron is inserted in the solenoid, the magnetic field greatly increases. Such electromagnets have many practical applications.\n\n20.10 Applications: Motors & Loudspeakers An electric motor takes advantage of the torque on a current loop, to change electrical energy to mechanical energy.\n\n20.10 Applications: Motors & Loudspeakers Loudspeakers use the principle that a magnet exerts a force on a current-carrying wire to convert electrical signals into mechanical vibrations, producing sound.\n\nDownload ppt \"© 2005 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their.\"\n\nSimilar presentations"
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http://luckyclicker.biz/flinkclick.php?id=1 | [
"# Sheets of paper weight\n\n## 27c4000 datasheet 2n3904\n\nWww sheetz\n\nMagna international inc dividend history\n\nThe standard sheet size varies by grade. For example, text weight papers have a standard size of 25 x 38 (950 sq. inches). Basic Size: The standard sheet size from which the basis weight of a given grade is determined. (See tables.) Point: The thickness of a sheet of paper, referenced in a point value. M Weight The ream weight is the weight of 500 sheets; the M weight is the weight of 1,000 sheets. The M weight can be obtained by first finding the ream weight, rounding it by the rounding rules, then multiplying by two. Object moved to here.\n\nThe first is the U.S. system of Basis Weight, which is the weight in pounds of 500 sheets of paper in its basic sheet size. “Basic sheet size” varies depending on the type, however, which makes it difficult to compare the weights of different types of paper.\n\n1. The total weight of the price list is .755 + .69 = 1.445 ounces. Notes about paper and using the chart. Paper weighs the same regardless of finish; 70# uncoated offset weighs the same as 70# coated gloss text or a 70# text sheet with a felt finish. Text sheets weigh the same as bond papers. 50# offset weighs the same as 24# bond.\n2. Crescent cricket cup 2013 live\n3. Online cricket match india and south africa\n\nThese base sizes are used to calculate a paper's basis weight. The basis weight equals the ream (500 sheet) weight of the base size, hence the term basis weight. Because the starting base size is not the same between paper types, the basis weights do not correspond directly (80 Text is much lighter than 80 Cover). It is the weight of 500 sheets of paper cut to a basis size. So, 500 sheets of 25\" x 38\", 60# offset weighs 60 lbs. The basis size for bond is 17\" x 22\"; for text, offset, and coated 25\" x 38\"; and for cover 20\" x 26\". That’s why two similar sheets of different grades may have very different basis weights, for instance, 24# bond and 50# offset. The basis weight of a paper is the designated fixed weight of 500 sheets, measured in pounds, in that paper's basic sheet size. It is important to note that the \"basic sheet size\" is not the same. for all types of paper.\n\nFor paper that is 24 lb. or higher, the correct envelope is one of equal weight to the paper. The correct envelope for paper up to 20 lb. should be one step heavier than the paper. For example, the proper envelope to use with a 20 pound paper would be one made from 24 pound paper. For paper that is 24 lb. or higher, the correct envelope is one of equal weight to the paper. The correct envelope for paper up to 20 lb. should be one step heavier than the paper. For example, the proper envelope to use with a 20 pound paper would be one made from 24 pound paper. These base sizes are used to calculate a paper's basis weight. The basis weight equals the ream (500 sheet) weight of the base size, hence the term basis weight. Because the starting base size is not the same between paper types, the basis weights do not correspond directly (80 Text is much lighter than 80 Cover). M Weight The ream weight is the weight of 500 sheets; the M weight is the weight of 1,000 sheets. The M weight can be obtained by first finding the ream weight, rounding it by the rounding rules, then multiplying by two. Paper M-Weight Calculator. Enter the sheet height and width, and the basis weight. Then enter the basic height and width. The results are meant to be approximate and are not guaranteed by Graphic Communications. Between different paper sizes and multiple scales of measurement, it isn’t always easy to decide what paper weight is best for your project. There are several different factors that go into determining paper weight, including sheet size, and understanding these factors can help you select the right card stock for your creative designs.\n\nHigh speed computer specification sheet\n\nThe weight of a sheet of paper depends on the size of the sheet and the thickness of the paper, known as grammage internationally and basis weight in North America. The tables below give the weights of sheets of the ISO 216 A paper sizes for various common grammages (e.g. 80gsm, 90gsm, 120gsm) in grams in the first table and in ounces in the second table. The pound designation of each type of paper comes from the weight of a stack of 500 sheets, so 500 sheets of 20-pound paper weighs 20 pounds. How much fiber is used to make each type of paper determines the actual weight. Paper ranges from 18 pounds for newsprint to 300 pounds for print-making and painting papers. If that same paper is slightly heavier, 500 sheets cut to the parent sheet size weights 24lbs, that paper at any cut size will be labeled “24lb Bond”. 500 parent sheets of different paper types can have a very different “basis weight”, due not only to the density of the paper, but also the dimensions of the cut parent sheet.\n\nMay 03, 2014 · The standard office copy paper weight is most commonly called “20 lb. bond”. Standard office copy paper weighs approximately 10.0 lbs per 1000 sheets. Office copy paper is the cheapest blank white paper you can buy. Yes, “20 lb. bond” and “50 lb. uncoated text” are basically the same kinds of paper. Paper Weight & Size -- When is 80# NOT 80#? Why doesn't 80# cover stock feel anything like 80# text stock? If you feel both sheets between your thumb and index finger, you'll find that the cover stock is stiffer and more resistant to bending.\n\nSeagate reallocated sector count warranty threshold sheets:\n\nPaper suppliers will often charge by M weight, since it is always consistent within a specific paper size, and because it allows a simple weight calculation for shipping charges. For example, a 500-sheet ream of 20# 8 1 ⁄ 2 -by-11-inch (216 by 279 mm) copy paper may be specified \"10 M\". M-weight refers to the weight of 1,000 sheets of paper. Use this online calculator to easily measure m-weight by providing the sheet size, basis weight, and basic size of the paper. When filling in these fields, use decimal fractions where necessary; and give all lengths in inches and weight in pounds. Determine the weight of your paper. This \"weight,\" generally listed on the package, is not the weight of a single piece of paper. Rather, it is the weight of a ream—500 sheets— of paper, which is then cut into individual pieces of paper. On average, ink jet paper is 20 lbs. and fits four pieces of paper to a sheet. The weight will obviously depend on the type of paper you are printing on, assuming 20# text weight, than 2.5 ounces would be correct. However, postage on a 9x12 envelope is significantly different than a business size #10 envelope and I doubt you are folding and squeezing a 12 page document into a #10 envelope.\n\nThe paper calculator tool helps with common price and quantity conversions. Use our paper calculator to convert various paper elements, calculate specific weights, prices and more. The weight will obviously depend on the type of paper you are printing on, assuming 20# text weight, than 2.5 ounces would be correct. However, postage on a 9x12 envelope is significantly different than a business size #10 envelope and I doubt you are folding and squeezing a 12 page document into a #10 envelope. Paper Help Guide, understanding the differences in paper weights. Here's a Quick list of paper weights including grammage (GSM). GSM is a consistent number for papers, running smaller to larger regardless of paper basis weight.\n\nCoaches 2015 the voice.asp\n\nPaper in the United States is calculated as 500 sheets of bond paper with a size of 17\" by 22\" (ledger-size) as having a weight of 20 pounds. The manufacturer cuts a ledger-sheet into four 8 ½\" by 11\" (letter-size) sheets, so a 500 letter-size sheet ream of 20-pound bond paper weighs 5 pounds.\n\nMica timesheet software\n\nPaper M-Weight Calculator. Enter the sheet height and width, and the basis weight. Then enter the basic height and width. The results are meant to be approximate and are not guaranteed by Graphic Communications. Paper weight in the US is stated in lbs. and is determined by weighing 500 sheets (a ream) in the basis size of a particular paper. For example, the basis size (or in other words, the master sheet size) of our Watercolor papers is 22”x30”.\nDetermine the M Weight, ream weight and weight for a given number of sheets of paper in the specific size entered. Roll Weights. Calculate the approximate weight of a roll of paper. Basis Weight / Metric Conversions. Convert basis weight (in pounds) for a ream of standard basis size papers to the metric equivalent (in grams per meter²) and ...\n\nAndreas dohr bongartz aachen.\n\nThese two identifiers are probably the most important factors when you are purchasing a sheet of paper for your project. Please note that 80# text weight is a paper weight and 80# Cover weight is a card stock; although they have the same number marking them, they are two completely different papers.\n\nBolometer detector datasheets\n\nDoretta sheetSafety data sheet section 2Flower colouring in sheetsTo 220 heatsink datasheets360The total weight of the price list is .755 + .69 = 1.445 ounces. Notes about paper and using the chart. Paper weighs the same regardless of finish; 70# uncoated offset weighs the same as 70# coated gloss text or a 70# text sheet with a felt finish. Text sheets weigh the same as bond papers. 50# offset weighs the same as 24# bond.\n\nDioda p6ke30a datasheet\n\nThe way we talk about paper in the United States is amazingly convoluted. The short answer is that 500 sheets of bond paper with a size of 17\" by 22\" have a weight of 20 pounds. The manufacturer would cut a sheet that big into four letter-size sheets, so a 500-sheet ream of 20-pound bond paper ... These base sizes are used to calculate a paper's basis weight. The basis weight equals the ream (500 sheet) weight of the base size, hence the term basis weight. Because the starting base size is not the same between paper types, the basis weights do not correspond directly (80 Text is much lighter than 80 Cover).\n\n• Paper Help Guide, understanding the differences in paper weights. Here's a Quick list of paper weights including grammage (GSM). GSM is a consistent number for papers, running smaller to larger regardless of paper basis weight. Jul 29, 2016 · When the manufacturer weighs the paper it’s not 500 8.5” x 11” size sheets, it’s 500 17” x 22” size sheets which are 4x the size of 8.5”x11”. Some quick math and that works out to a weight of 5 pounds for a ream of 8.5” x 11” size paper. The standard sheet size varies by grade. For example, text weight papers have a standard size of 25 x 38 (950 sq. inches). Basic Size: The standard sheet size from which the basis weight of a given grade is determined. (See tables.) Point: The thickness of a sheet of paper, referenced in a point value. Jul 29, 2016 · When the manufacturer weighs the paper it’s not 500 8.5” x 11” size sheets, it’s 500 17” x 22” size sheets which are 4x the size of 8.5”x11”. Some quick math and that works out to a weight of 5 pounds for a ream of 8.5” x 11” size paper. The weight of 12 sheets of a thick paper is 40 grams. How many sheets would weigh 1 kg?\n• The term paperweight, and number corresponding with each weight, refers to the thickness and sturdiness of the paper, not the actual weight of the sheet. Learn more about Paper Weights. This is why, sometimes, the same \"weight\" paper may be referred to as two different things. Once you figure out how much each sheet of paper weighs, you'll need to multiply by the number of sheets that you have to find the total weight. For instance, 10 sheets of paper that each weigh 0.16 ounces weighs a total of 1.6 ounces. M Weight The ream weight is the weight of 500 sheets; the M weight is the weight of 1,000 sheets. The M weight can be obtained by first finding the ream weight, rounding it by the rounding rules, then multiplying by two. Mar 28, 2009 · The average piece of inkjet paper is 20lb paper, which is based on the weight of one ream (500 pages) of the base size (17\"x 22\") which makes 4 sheets of 8 1/2\" x 11\", letter sized paper. Jan 09, 2008 · Standard copy paper takes about 6 sheets. Paper weight refers to the weight of a 500-sheet ream of 17\" x 22\" paper. Each of these sheets is equivalent to four letter size sheets. Therefore, 500...\n• These two identifiers are probably the most important factors when you are purchasing a sheet of paper for your project. Please note that 80# text weight is a paper weight and 80# Cover weight is a card stock; although they have the same number marking them, they are two completely different papers. Balance sheet period as per companies act 1956 as amendedRose sheet music\n• Bed sheets super kingLenovo datasheet m73 thin client The weight will obviously depend on the type of paper you are printing on, assuming 20# text weight, than 2.5 ounces would be correct. However, postage on a 9x12 envelope is significantly different than a business size #10 envelope and I doubt you are folding and squeezing a 12 page document into a #10 envelope.\n\nIf that same paper is slightly heavier, 500 sheets cut to the parent sheet size weights 24lbs, that paper at any cut size will be labeled “24lb Bond”. 500 parent sheets of different paper types can have a very different “basis weight”, due not only to the density of the paper, but also the dimensions of the cut parent sheet.\n\nThe Paper Division of DS Smith offers you the optimum means to fully express all aspects of today's packaging solutions. To guarantee this, we work to provide the best possible paper products and services to satisfy not only your needs, but also your customers needs.\nPhysics equation sheet aqa p1 speed\n\n• Kakariko village ocarina song sheetsCapital inicial lisboaAn uncut sheet of Bond paper is 17 x 22 inches, while an uncut sheet of Cover paper is 20 x 26 inches. If 500 sheets of Bond paper (17 x 22 inches) weigh 20 lbs, then a ream of paper cut to Letter size will be labeled as 20 lb. Basis weight is the total weight, in pounds, of a ream of paper. A ream of paper is 500 sheets and is the standard ordering unit for paper. Because paper comes in different sizes and widths, the basis weight of paper can vary from ream to ream. One way to calculate basis weight involves using both metric and customary (inches) units in a formula.\nEstate agents moseley birmingham uk.Lab molecular geometry datasheet answers in genesis"
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http://library.rstheory.org/articles/KVK/CAtomDecay.html | [
"# Lifetimes of C-Atom Decays\n\nReciprocity XI #1, Spring, 1981\n\nThe phenomenon of the entry of c-matter into the material sector or the analogous entry of matter into the cosmic sector, involving the passage from space-time domain to time-space domain, may be called “scalar inversion” to emphasize the nature of the alteration of the reference frame. Scalar inversion involves two things: firstly, a transformation of motion in time (or space) to motion in space (or time), through the unit speed boundary, in all the three dimensions. Secondly, the emergence of a c-atom, for example, into the material sector can take place only from inside a single unit, since the three dimensions of time have nothing in common with the three dimensions of space—both having not more than a point contact, as it were (p. 154, Nothing But Motion—hereinafter NBM).\n\nTherefore, in following up the calculation of various quantities across the boundary in scalar inversion, from the cosmic sector to the material sector, for example, consideration must be given to: (i) the loss of dimensional “information” during the alteration of the viewpoint from the temporal reference frame to the spatial reference frame and (ii) the space equivalent of time occurring within a single unit.\n\nAs a result of the first point above, it is known that the full influence of spatial (or temporal) effects does not get transmitted across the boundary except when it involves only one dimension. On the other hand, only a fraction 1/c in the case of two-dimensional effects, and a fraction 1/c2 in the case of three-dimensional effects gets transmitted. (See p. 185, New Light on Space and Time—hereinafter NLST). I will refer to this as criterion No. II in the sequel.\n\nRegarding the second point above, namely, concerning the relation between quantities within the outside single unit, “…The time region speed, and all quantities derived there from, which means all of the physical phenomena of the inside region… are… second power expressions of the corresponding quantities of the outside region.” (p. 155, NBM) I will refer to this as criterion No. IV. In order to find the lifetimes of the cosmic atoms in the material environment it is necessary to apply both the above criteria.\n\nThe first step in deriving the lifetimes is to recognize that, in view of the scalar inversion, the spatial extension of the c-atom, being the analog of the lifetime in material sector, bears a relationship to the latter. As such we start with the consideration of the spatial extension of the incoming c-atom. Now, scalar inversion is not possible with anything more than one unit in each dimension. Depending on the number of dimensions of the motion eventually acquired during the inversion process, the amount of space involved in the one, two and three-dimensional cases is respectively s, s2 and s3 (where s is the unit space expressed in the c.g.s. system). Let us refer to this as criterion No. I.\n\nThe remaining criterion, No. III, necessary for our calculation is the recognition of the fact that the temporal equivalent of a spatial extension s across the inversion boundary is s/c (where c, the unit speed, is expressed in the c.g.s. system). The result of applying the above four criteria to the one, two and three-dimensional situations is given in the following table.\n\nCriterion No. Number of Dimensions\n1 2 3\ni s s2 s3\nii s/c s2/c s3/c\niii (s/c) (s2/c)(1/c)\n\n(s3/c)(1/c2)\n\niv (s/c)½ [(s2/c)(1/c)]½ [(s3/c)(1/c2)]½\nResult in seconds 1.233148×10-8 1.520655×10-16 1.875193×10-24\n\nThe same result could have been obtained more simply though showing less details of the underlying process by directly noting that the clock-time involved in the one, two and three-dimensional cases of the decay is t, t2 and t3 respectively (where t0 is the unit time expressed in the c.g.s. system). The measured values of the lifetimes could then be obtained by applying the criterion No. IV, as t½, (t2)½ and (t3)½ respectively.\n\nFurther, in the calculations above if the extension space involved is taken as p/4s2 and p/6s3 respectively in the two and three-dimensional cases, based on symmetrical probability, instead of s2 and s3 , we have the computed values of the lifetimes in the respective situations as 1.348×10-16 and 1.357×10-24 secs.\n\nThe acquisition of gravitational charges by the incoming c-atoms has an effect on the above lifetimes which can be evaluated in the following manner. In view of the scalar inversion, it must be noted that the gravitational charge of the material sector, being a two-dimensional rotational vibratory time displacement, is foreign to the space-time character of the basic rotational displacement of the c-atom. In the analogous case of a material atom, for example, a gravitational charge of the cosmic sector is tantamount to a magnetic charge in the material environment. Consequently the calculation of the influence of a rotational vibration of space-time direction opposite to that of the basic rotation, on various quantities requires the consideration of the appropriate interregional ratio.\n\nFor example, “…the motion that constitutes the charge is on the far side of another regional boundary—another unit level—and is subject to… inter-regional transmission factors.” (p. 163, NBM). Further, “…inter-regional ratio…accounts for the small ‘size’ of atoms. According to the theory…, there can be no physical distance less than one natural unit… but… the measured inter-atomic distance is reduced by the inter-regional ratio, and this measured value is therefore in the neighborhood of 10-8 cm” (p. 154-5, NBM). In exactly the same manner, the acquisition of a gravitational charge by the c-atom, in view of the interregional ratio, has the effect of shortening the measured lifetime by a factor of 1/156.44. (While it is clear that the inter-regional ratio operates here, I am not certain that its value is 156.44 in this case.)\n\nAn atom is a double rotating system. The rotational vibration that is a gravitational charge establishes a coupling with one of these two rotational systems. In the case of an acquisition of one more gravitational charge, the second rotational vibratory displacement acquired acts on the second rotational system of the c-atom rather than adding to the previous system already modified by the first gravitational charge. As such, the computation of the lifetime in this case involves the application of the inter-regional ratio once more. Thus the measured lifetime in the case of two gravitational charges acquired is shortened by a factor of 1/(156.44)2 The lifetimes, with or without the gravitational charges, in the one, two and three-dimensional situations are, therefore, as follows:\n\nDimensions Charges Lifetime (sec.)\n1 1.233148×10-8\n1 1 0.788234×10-10\n2 1.520655×10-16\n2 2 0.621313×10-20\n3 1.875193×10-24\n\n## Comments",
null,
"### Decay rates are variable\n\nFolks are now seeing a repeatable alteration in radiative decay rates of a range of elements according to the occurence/presence of solar CME events.\n\nHow would DB Larson and RS Theory explain this?",
null,
"### Neutrinos\n\nSpeculation is that it may have something to do with neutrinos. Maybe one of the RS experts can address what theory says about neutrinos affecting decay rates.",
null,
"### Neutrinos\n\nSeeing that isotopes in the RS are nothing but captured charged neutrinos (refer Basic Properties of Matter), it is already expected that any changes in neutrino activity change the properties of radioactivity. It must coincide with solar activity, which is responsible for giving earth not only a thermal temperature but also a \"magnetic temperature\". That is the seasonal variation being talked about in the link.",
null,
"### Neutrinos\n\nThanks for the reply Gopi! Based on your response, I could see how neutrinos could affect very short-lived isotopes enough to be measured. However, since the half-life of Mn-54 is 312 days, it doesn't seem like a burst of neutrinos would have had enough of an effect on the total amount of Mn-54 to measurably decrease the decay rate if all the Mn-54 in the sample were effected equally. Seems like the neutrinos must have had a disproportionate effect on atoms that were about to decay compared to all the Mn-54 atoms. In my quick reading of Basic Properties of Matter, I can't really tell if it would point to something like that.\n\n<blockquote>Purdue nuclear engineer Jere Jenkins noticed an inexplicable drop in the decay rate of manganese-54 when he was testing it one night in 2006. It so happened that this drop occurred just over a day before a large flare erupted on the sun.</blockquote>"
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https://bingobaker.com/view/1657017 | [
"# Geometry Bingo",
null,
"This bingo card has a free space and 24 words: Complementary Angles, Acute Angle, Adjacent Angles, Interior Angles, Corresponding Angles, Line, Congruent Angles, Obtuse Angle, Equilateral Triangle, Line Segment, Exterior Angles, Parallelogram, Parallel Lines, Isosceles Triangle, Rhombus, Perpendicular Lines, Pythagorean Theorem, Ray, Right Angle, Quadrilateral, Vertex, Scalene Triangle, Supplementary Angles and Transversal.\n\n## Play Online",
null,
""
]
| [
null,
"https://bingobaker.com/image/1657017/544/1/geometry-bingo.png",
null,
"https://bingobaker.com/static/1654770411/img/clippy.svg",
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]
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https://pierreterrat.com/java/quick-answer-java-float-ou-double.html | [
"# Quick Answer: Java Float Ou Double?\n\n## Should I use float or double in Java?\n\nYou should use double instead of float for precise calculations, and float instead of double when using less accurate calculations. Float contains only decimal numbers, but double contains an IEEE754 double-precision floating point number, making it easier to contain and computate numbers more accurately.\n\n## Is float equal to double in Java?\n\nSince float and double have different sizes, the representation in both types for a non-representable value are different, and thus they compare as unequal. (The length of the binary string is the size of the mantissa, so that’s 24 for float, 53 for double and 64 for the 80-bit extended-precision float (not in Java).\n\n## Is 1.5 float or double?\n\nAnd the reason the comparison succeeds with 1.5 is that 1.5 can be represented exactly as a float and as a double; it has a bunch of zeros in its low bits, so when the promotion adds zeros the result is the same as the double representation.\n\nYou might be interested: FAQ: Comment Commencer Un Programme Java?\n\n## Is float better than double?\n\nDouble is more precise than float and can store 64 bits, double of the number of bits float can store. Double is more precise and for storing large numbers, we prefer double over float. Unless we do need precision up to 15 or 16 decimal points, we can stick to float in most applications, as double is more expensive.\n\n## Is 4.5 float or double?\n\nNumbers like -321, 497, 19345, and -976812 are all perfectly valid integers, but 4.5 is not because 4.5 is not a whole number. Floating point numbers are numbers with a decimal. Like integers, -321, 497, 19345, and -976812 are all valid, but now 4.5, 0.0004, -324.984, and other non-whole numbers are valid too.\n\n## Is 99.9 float or double?\n\nIs 99.9 float or double? Floating-point numbers are by default of type double. Therefore 99.9 is a double, not a float.\n\n## Which is bigger double or long Java?\n\nIf you are storing integers, use Long. Your statement that “Advantage of Using Double is that it gives a more wider range for storing Whole Numbers” is incorrect. Both are 64 bits long, but double has to use some bits for the exponent, leaving fewer bits to represent the magnitude.\n\n## How do you know if a double value is zero?\n\nif(value!= 0) //divide by value is safe when value is not exactly zero. Otherwise when checking if a floating point value like double or float is 0, an error threshold is used to detect if the value is near 0, but not quite 0.\n\nYou might be interested: FAQ: Multi Line Comment Java?\n\n## What is the difference between float and double?\n\nWhat’s the difference? double has 2x more precision then float. float is a 32 bit IEEE 754 single precision Floating Point Number1 bit for the sign, (8 bits for the exponent, and 23* for the value), i.e. float has 7 decimal digits of precision.\n\n## What is the difference between double and float in C#?\n\nThe Decimal, Double, and Float variable types are different in the way that they store the values. Precision is the main difference where float is a single precision (32 bit) floating point data type, double is a double precision (64 bit) floating point data type and decimal is a 128-bit floating point data type.\n\n## How do you declare a double float in C++?\n\nYou declare a double-precision floating point as follows: double dValue1; double dValue2 = 1.5; The limitations of the int variable in C++ are unacceptable in some applications. Fortunately, C++ understands decimal numbers that have a fractional part.\n\n## What is the difference between double and float in Swift?\n\nSwift provides two signed floating-point number types: Double represents a 64-bit floating-point number. Float represents a 32-bit floating-point number.\n\n## What is the difference between float and float?\n\nFloat is an object; float is a primitive. Same relationship as Integer and int, Double and double, Long and long. float can be converted to Float by autoboxing, e.g.\n\n## Should I use decimal or float?\n\nFloat stores an approximate value and decimal stores an exact value. In summary, exact values like money should use decimal, and approximate values like scientific measurements should use float. When multiplying a non integer and dividing by that same number, decimals lose precision while floats do not."
]
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https://dsp.stackexchange.com/questions/51632/why-doesnt-law-of-large-numbers-apply-to-this-stationary-time-series | [
"# Why doesn't law of large numbers apply to this stationary time-series?\n\nThere's a paragraph in Wikipedia that states the following:\n\nLet Y be any scalar random variable, and define a time-series $\\{X_t\\}$, by $$X_{t}=Y\\qquad {\\text{ for all }}t$$ Then $\\{X_t\\}$ is a stationary time series, for which realisations consist of a series of constant values, with a different constant value for each realisation. A law of large numbers does not apply on this case, as the limiting value of an average from a single realisation takes the random value determined by $Y$, rather than taking the expected value of $Y$.\n\nI cannot understand the reason for the law of large numbers not being applicable. Any explanation would help.\n\n• Perhaps if you include what exactly are the conditions that must be satisfied in order for your favorite law of large numbers to hold, the answer to your question might be glaringly obvious. Please check carefully that you haven't left out any words beginning with \"indepen....\" in the conditions that you state, and think as to whether $X_1$ and $X_2$ satisfy this requirement. – Dilip Sarwate Aug 30 '18 at 14:11\n• This part is quoted from the Wikipedia article I read, exactly as it was. So, nothing has been left out. I don't understand what you're implying. And I have included the part about the law of large numbers. – Curiosity Aug 30 '18 at 14:13\n• Ok, let's try it again from the top. Look in Wikipedia about the laws of large numbers and edit your question to say something like \"Now, according to Wikipedia, one version of the law of large numbers states that if ....\" and include all the conditions on $X_1, X_2, \\ldots$ that this law needs in order to hold. Then think a little about whether $X_1=Y$ and $X_2=Y$ satisfy this condition. – Dilip Sarwate Aug 30 '18 at 14:22\n\nI believe that you are thinking that each value of $X_t$ is determined by a different realisation of $Y$, which in this example is not true.\n\nSuppose that $Y$ is the value that comes out from a dice throw. Thus, $Y$ can achieve any integer value between $1$ and $6$. Throw the dice once. Suppose that you get $2$. Then, according to the definition in the example, your time-series would be:\n\n$$X_t=2 \\qquad {\\text{for all }}t$$\n\nWhat I think the quote from Wikipedia is saying is that the mean value of this stochastic process doesn't coincide with the mean value of the random variable $Y$, as\n\n$$\\mathbb{E}[X_t]=2 \\neq 3.5 = \\mathbb{E}[Y]$$\n\nIf you constructed the time-series letting each value of $X_t$ be determined by a new throw of the dice, then the means would match.\n\n• Thanks, I got it now. That was a very lucid explanation indeed. – Curiosity Aug 30 '18 at 18:09\n• @Curiosity Glad to help! If it solved your doubt, please mark the answer as accepted so it can help future readers. – Tendero Aug 30 '18 at 23:34"
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https://homework.cpm.org/category/ACC/textbook/gb8i/chapter/3%20Unit%204/lesson/INT1:%203.1.5/problem/3-60 | [
"",
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"### Home > GB8I > Chapter 3 Unit 4 > Lesson INT1: 3.1.5 > Problem3-60\n\n3-60.\n\nRead the Math Notes box about polygons in Lesson 3.1.2. At right is a diagram of a regular hexagon with center $C$. Copy this figure on your paper, then complete the problems below.\n\n1. Draw the result of rotating the hexagon about its center $180º$. Explain what happened. When this happens, the polygon has rotation symmetry.\n\nThe orientation of the hexagon does not change.\n\n2. What is the result when the original hexagon is reflected across line $n$, as shown at right? A polygon with this quality is said to have reflection symmetry. Line $n$ is called a line of symmetry of the hexagon.\n\nThe orientation of the hexagon does not change.\n\n3. Does a regular hexagon have any other lines of symmetry? That is, are there any other lines you could fold across so that both halves of the hexagon will match up? Find as many as you can.",
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"https://homework.cpm.org/dist/7d633b3a30200de4995665c02bdda1b8.png",
null,
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",
null,
"https://s3-us-west-2.amazonaws.com/c3po-media-dev/files/ea4366e0-7cc8-11e9-beec-fd87f5f6beba/ccg_original.png",
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"https://s3-us-west-2.amazonaws.com/c3po-media-dev/files/1145d250-7cc9-11e9-beec-fd87f5f6beba/ccg_original.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.94357884,"math_prob":0.9404031,"size":804,"snap":"2023-14-2023-23","text_gpt3_token_len":193,"char_repetition_ratio":0.14875,"word_repetition_ratio":0.055172414,"special_character_ratio":0.23134328,"punctuation_ratio":0.11904762,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96178335,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-29T16:11:00Z\",\"WARC-Record-ID\":\"<urn:uuid:f6347016-e3a4-4fec-876b-2a0f0b50ed2b>\",\"Content-Length\":\"37551\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:45577d39-d079-46eb-b778-71a1f47505a4>\",\"WARC-Concurrent-To\":\"<urn:uuid:6618aeac-4b07-48fb-b16f-e6286018d148>\",\"WARC-IP-Address\":\"172.67.70.60\",\"WARC-Target-URI\":\"https://homework.cpm.org/category/ACC/textbook/gb8i/chapter/3%20Unit%204/lesson/INT1:%203.1.5/problem/3-60\",\"WARC-Payload-Digest\":\"sha1:Z4DF4OWWGCTPU65JOD4DD3SL7S3FIJ6L\",\"WARC-Block-Digest\":\"sha1:LL3LXQAHKDGLM6MLURS6XG7C3C6EN3GS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224644867.89_warc_CC-MAIN-20230529141542-20230529171542-00050.warc.gz\"}"} |
https://dsp.stackexchange.com/questions/30157/how-we-can-encode-decode-sparse-signals | [
"# How we can encode/decode sparse signals?\n\nI have question and looking for help. Suppose we have a vector of real values (lat's say 64 length resulting from factorization 8*8 block image). We got a sparse representation of that vector (let's suppose that the sparse vector is 128 length with very few active elements, about 4 coefficients). I want to send this sparse vector through a channel, so I need bit stream of 1,0. Are there any efficient encoding/decoding technique for the sparse signals ?\n\n• Since you have all sample, I think you are better to avoid compressive sampling, it is for efficeint sampling not coding.decoding. For signal compression, there are tones of compression algorithms, like LRZ,Huffman,... – MimSaad Aug 13 '16 at 13:06"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9389354,"math_prob":0.85622394,"size":2071,"snap":"2021-21-2021-25","text_gpt3_token_len":444,"char_repetition_ratio":0.13546202,"word_repetition_ratio":0.02366864,"special_character_ratio":0.2187349,"punctuation_ratio":0.103626944,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9524005,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-11T19:47:39Z\",\"WARC-Record-ID\":\"<urn:uuid:0e4b8eed-424b-4ae1-a06c-f2677065eba5>\",\"Content-Length\":\"164755\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3bc63a17-24d7-405a-91ab-84c198127e02>\",\"WARC-Concurrent-To\":\"<urn:uuid:91f49958-e51e-4e53-b469-dbd3e9ce3189>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://dsp.stackexchange.com/questions/30157/how-we-can-encode-decode-sparse-signals\",\"WARC-Payload-Digest\":\"sha1:ZQEDX5QK2CJJJGSMQ536OSMQPPRTFDQ6\",\"WARC-Block-Digest\":\"sha1:EAW5YCVUID7SSMC7KGLEUVT3M55G26VZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989856.11_warc_CC-MAIN-20210511184216-20210511214216-00284.warc.gz\"}"} |
https://metanumbers.com/22302 | [
"## 22302\n\n22,302 (twenty-two thousand three hundred two) is an even five-digits composite number following 22301 and preceding 22303. In scientific notation, it is written as 2.2302 × 104. The sum of its digits is 9. It has a total of 6 prime factors and 32 positive divisors. There are 6,264 positive integers (up to 22302) that are relatively prime to 22302.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 5\n• Sum of Digits 9\n• Digital Root 9\n\n## Name\n\nShort name 22 thousand 302 twenty-two thousand three hundred two\n\n## Notation\n\nScientific notation 2.2302 × 104 22.302 × 103\n\n## Prime Factorization of 22302\n\nPrime Factorization 2 × 33 × 7 × 59\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 6 Total number of prime factors rad(n) 2478 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 22,302 is 2 × 33 × 7 × 59. Since it has a total of 6 prime factors, 22,302 is a composite number.\n\n## Divisors of 22302\n\n1, 2, 3, 6, 7, 9, 14, 18, 21, 27, 42, 54, 59, 63, 118, 126, 177, 189, 354, 378, 413, 531, 826, 1062, 1239, 1593, 2478, 3186, 3717, 7434, 11151, 22302\n\n32 divisors\n\n Even divisors 16 16 8 8\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 32 Total number of the positive divisors of n σ(n) 57600 Sum of all the positive divisors of n s(n) 35298 Sum of the proper positive divisors of n A(n) 1800 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 149.339 Returns the nth root of the product of n divisors H(n) 12.39 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 22,302 can be divided by 32 positive divisors (out of which 16 are even, and 16 are odd). The sum of these divisors (counting 22,302) is 57,600, the average is 1,800.\n\n## Other Arithmetic Functions (n = 22302)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 6264 Total number of positive integers not greater than n that are coprime to n λ(n) 522 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2496 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 6,264 positive integers (less than 22,302) that are coprime with 22,302. And there are approximately 2,496 prime numbers less than or equal to 22,302.\n\n## Divisibility of 22302\n\n m n mod m 2 3 4 5 6 7 8 9 0 0 2 2 0 0 6 0\n\nThe number 22,302 is divisible by 2, 3, 6, 7 and 9.\n\n## Classification of 22302\n\n• Arithmetic\n• Abundant\n\n### Expressible via specific sums\n\n• Polite\n• Practical\n• Non-hypotenuse\n\n## Base conversion (22302)\n\nBase System Value\n2 Binary 101011100011110\n3 Ternary 1010121000\n4 Quaternary 11130132\n5 Quinary 1203202\n6 Senary 251130\n8 Octal 53436\n10 Decimal 22302\n12 Duodecimal 10aa6\n20 Vigesimal 2ff2\n36 Base36 h7i\n\n## Basic calculations (n = 22302)\n\n### Multiplication\n\nn×i\n n×2 44604 66906 89208 111510\n\n### Division\n\nni\n n⁄2 11151 7434 5575.5 4460.4\n\n### Exponentiation\n\nni\n n2 497379204 11092551007608 247386072571673616 5517204190493464984032\n\n### Nth Root\n\ni√n\n 2√n 149.339 28.148 12.2204 7.40745\n\n## 22302 as geometric shapes\n\n### Circle\n\n Diameter 44604 140128 1.56256e+09\n\n### Sphere\n\n Volume 4.64644e+13 6.25025e+09 140128\n\n### Square\n\nLength = n\n Perimeter 89208 4.97379e+08 31539.8\n\n### Cube\n\nLength = n\n Surface area 2.98428e+09 1.10926e+13 38628.2\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 66906 2.15372e+08 19314.1\n\n### Triangular Pyramid\n\nLength = n\n Surface area 8.61486e+08 1.30727e+12 18209.5\n\n## Cryptographic Hash Functions\n\nmd5 5a6ce3e53bbfb06dd5822fe254494010 cfe7a825a6f1d983121aed23b0aa3d6b4c62b2d0 7872fac4cdafece8cbe783dba37c94c7fe16269557e8a000912b667862ee0ff2 db75b0726b6681629aaa000086f9613c71999d4499505a372e9d8f02e81a6bde8e919beea88e759b724c0685c8846eccfc6c82152e8230d52af82b3b55772cd1 97d516ebcb0e766aeb1cd31651b86d4bdc9ade4c"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6325127,"math_prob":0.9858993,"size":4479,"snap":"2020-34-2020-40","text_gpt3_token_len":1597,"char_repetition_ratio":0.12022346,"word_repetition_ratio":0.030973451,"special_character_ratio":0.45099354,"punctuation_ratio":0.078608245,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9950799,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-18T20:05:18Z\",\"WARC-Record-ID\":\"<urn:uuid:f81b74da-52bd-40fa-b0ad-e6263bda02d5>\",\"Content-Length\":\"48378\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec43ce93-5160-4f2d-b090-5417b48e5180>\",\"WARC-Concurrent-To\":\"<urn:uuid:82490ba7-1d31-4c2a-80af-7bcff3ddb00d>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/22302\",\"WARC-Payload-Digest\":\"sha1:AZYBVOSL662KNAOYF6DUIZOTGN6P5TC2\",\"WARC-Block-Digest\":\"sha1:HGFRO25K5U7QLPKEKORQJYRL7JUZ57G3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400188841.7_warc_CC-MAIN-20200918190514-20200918220514-00046.warc.gz\"}"} |
https://hungary.pure.elsevier.com/hu/publications/fuzzy-tree-automata | [
"# Fuzzy tree automata\n\nZ. Ésik, Guangwu Liu\n\nResearch output: Article\n\n22 Citations (Scopus)\n\n### Abstract\n\nIn this paper, we map an arbitrary algebra to a completely distributive lattice and define algebras of fuzzy sets. When the algebra is a term algebra, we obtain fuzzy tree languages. After defining fuzzy equational sets, fuzzy rational sets and fuzzy recognizable tree languages, we derive a \"Kleene theorem\" for fuzzy tree languages.\n\nOriginal language English 1450-1460 11 Fuzzy Sets and Systems 158 13 https://doi.org/10.1016/j.fss.2007.02.016 Published - júl. 1 2007\n\n### Fingerprint\n\nFuzzy Automata\nTree Automata\nAlgebra\nFuzzy sets\nFuzzy Sets\nCompletely Distributive Lattice\nTrees (mathematics)\nAutomata\nArbitrary\nTerm\nTheorem\nLanguage\n\n### ASJC Scopus subject areas\n\n• Statistics and Probability\n• Electrical and Electronic Engineering\n• Statistics, Probability and Uncertainty\n• Information Systems and Management\n• Computer Vision and Pattern Recognition\n• Computer Science Applications\n• Artificial Intelligence\n\n### Cite this\n\nFuzzy tree automata. / Ésik, Z.; Liu, Guangwu.\n\nIn: Fuzzy Sets and Systems, Vol. 158, No. 13, 01.07.2007, p. 1450-1460.\n\nResearch output: Article\n\nÉsik, Z. ; Liu, Guangwu. / Fuzzy tree automata. In: Fuzzy Sets and Systems. 2007 ; Vol. 158, No. 13. pp. 1450-1460.\n@article{f798fcd049a14c07b0e46e289201abdc,\ntitle = \"Fuzzy tree automata\",\nabstract = \"In this paper, we map an arbitrary algebra to a completely distributive lattice and define algebras of fuzzy sets. When the algebra is a term algebra, we obtain fuzzy tree languages. After defining fuzzy equational sets, fuzzy rational sets and fuzzy recognizable tree languages, we derive a {\"}Kleene theorem{\"} for fuzzy tree languages.\",\nkeywords = \"Automata, Fuzzy set, Fuzzy system, Fuzzy tree automata\",\nauthor = \"Z. {\\'E}sik and Guangwu Liu\",\nyear = \"2007\",\nmonth = \"7\",\nday = \"1\",\ndoi = \"10.1016/j.fss.2007.02.016\",\nlanguage = \"English\",\nvolume = \"158\",\npages = \"1450--1460\",\njournal = \"Fuzzy Sets and Systems\",\nissn = \"0165-0114\",\npublisher = \"Elsevier\",\nnumber = \"13\",\n\n}\n\nTY - JOUR\n\nT1 - Fuzzy tree automata\n\nAU - Ésik, Z.\n\nAU - Liu, Guangwu\n\nPY - 2007/7/1\n\nY1 - 2007/7/1\n\nN2 - In this paper, we map an arbitrary algebra to a completely distributive lattice and define algebras of fuzzy sets. When the algebra is a term algebra, we obtain fuzzy tree languages. After defining fuzzy equational sets, fuzzy rational sets and fuzzy recognizable tree languages, we derive a \"Kleene theorem\" for fuzzy tree languages.\n\nAB - In this paper, we map an arbitrary algebra to a completely distributive lattice and define algebras of fuzzy sets. When the algebra is a term algebra, we obtain fuzzy tree languages. After defining fuzzy equational sets, fuzzy rational sets and fuzzy recognizable tree languages, we derive a \"Kleene theorem\" for fuzzy tree languages.\n\nKW - Automata\n\nKW - Fuzzy set\n\nKW - Fuzzy system\n\nKW - Fuzzy tree automata\n\nUR - http://www.scopus.com/inward/record.url?scp=34248200497&partnerID=8YFLogxK\n\nUR - http://www.scopus.com/inward/citedby.url?scp=34248200497&partnerID=8YFLogxK\n\nU2 - 10.1016/j.fss.2007.02.016\n\nDO - 10.1016/j.fss.2007.02.016\n\nM3 - Article\n\nAN - SCOPUS:34248200497\n\nVL - 158\n\nSP - 1450\n\nEP - 1460\n\nJO - Fuzzy Sets and Systems\n\nJF - Fuzzy Sets and Systems\n\nSN - 0165-0114\n\nIS - 13\n\nER -"
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https://www.geeksforgeeks.org/maximum-sum-such-that-no-two-elements-are-adjacent-set-2/?ref=rp | [
"# Maximum sum such that no two elements are adjacent | Set 2\n\nGiven an array of positive numbers, find the maximum sum of a subsequence with the constraint that no 2 numbers in the sequence should be adjacent in the array. So 3 2 7 10 should return 13 (sum of 3 and 10) or 3 2 5 10 7 should return 15 (sum of 3, 5 and 7).\n\nExamples:\n\n```Input : arr[] = {3, 5, 3}\nOutput : 6\nExplanation :\nSelecting indexes 0 and 2 will maximise the sum\ni.e 3+3 = 6\n\nInput : arr[] = {2, 5, 2}\nOutput : 5\n```\n\n## Recommended: Please try your approach on {IDE} first, before moving on to the solution.\n\nWe have already discussed the efficient approach of solving this problem in the previous article.\n\nHowever, we can also solve this problem using Dynamic Programming approach.\n\nDynamic Programming Approach: Let’s decide the states of ‘dp’. Let dp[i] be the largest possible sum for the sub-array staring from index ‘i’ and ending at index ‘N-1’. Now, we have to find a recurrence relation between this state and a lower-order state.\n\nIn this case for an index ‘i’, we will have two choices.\n\n```1) Choose the current index:\nIn this case, the relation will be dp[i] = arr[i] + dp[i+2]\n2) Skip the current index:\nRelation will be dp[i] = dp[i+1]\n```\n\nWe will choose the path that maximizes our result.\nThus final relation will be:\n\n```dp[i] = max(dp[i+2]+arr[i], dp[i+1])\n```\n\nBelow is the implementation of the above approach:\n\n## C++\n\n `// C++ program to implement above approach ` ` ` `#include ` `#define maxLen 10 ` `using` `namespace` `std; ` ` ` `// variable to store states of dp ` `int` `dp[maxLen]; ` ` ` `// variable to check if a given state ` `// has been solved ` `bool` `v[maxLen]; ` ` ` `// Function to find the maximum sum subsequence ` `// such that no two elements are adjacent ` `int` `maxSum(``int` `arr[], ``int` `i, ``int` `n) ` `{ ` ` ``// Base case ` ` ``if` `(i >= n) ` ` ``return` `0; ` ` ` ` ``// To check if a state has ` ` ``// been solved ` ` ``if` `(v[i]) ` ` ``return` `dp[i]; ` ` ``v[i] = 1; ` ` ` ` ``// Required recurrence relation ` ` ``dp[i] = max(maxSum(arr, i + 1, n), ` ` ``arr[i] + maxSum(arr, i + 2, n)); ` ` ` ` ``// Returning the value ` ` ``return` `dp[i]; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``int` `arr[] = { 12, 9, 7, 33 }; ` ` ` ` ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` ` ` ` ``cout << maxSum(arr, 0, n); ` ` ` ` ``return` `0; ` `} `\n\n## Java\n\n `// Java program to implement above approach ` `class` `GFG ` `{ ` ` ` `static` `int` `maxLen = ``10``; ` ` ` `// variable to store states of dp ` `static` `int` `dp[] = ``new` `int``[maxLen]; ` ` ` `// variable to check if a given state ` `// has been solved ` `static` `boolean` `v[] = ``new` `boolean``[maxLen]; ` ` ` `// Function to find the maximum sum subsequence ` `// such that no two elements are adjacent ` `static` `int` `maxSum(``int` `arr[], ``int` `i, ``int` `n) ` `{ ` ` ``// Base case ` ` ``if` `(i >= n) ` ` ``return` `0``; ` ` ` ` ``// To check if a state has ` ` ``// been solved ` ` ``if` `(v[i]) ` ` ``return` `dp[i]; ` ` ``v[i] = ``true``; ` ` ` ` ``// Required recurrence relation ` ` ``dp[i] = Math.max(maxSum(arr, i + ``1``, n), ` ` ``arr[i] + maxSum(arr, i + ``2``, n)); ` ` ` ` ``// Returning the value ` ` ``return` `dp[i]; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` ` ``int` `arr[] = { ``12``, ``9``, ``7``, ``33` `}; ` ` ``int` `n = arr.length; ` ` ``System.out.println( maxSum(arr, ``0``, n)); ` `} ` `} ` ` ` `// This code is contributed by Arnab Kundu `\n\n## Python3\n\n `# Python 3 program to implement above approach ` `maxLen ``=` `10` ` ` `# variable to store states of dp ` `dp ``=` `[``0` `for` `i ``in` `range``(maxLen)] ` ` ` `# variable to check if a given state ` `# has been solved ` `v ``=` `[``0` `for` `i ``in` `range``(maxLen)] ` ` ` `# Function to find the maximum sum subsequence ` `# such that no two elements are adjacent ` `def` `maxSum(arr, i, n): ` ` ``# Base case ` ` ``if` `(i >``=` `n): ` ` ``return` `0` ` ` ` ``# To check if a state has ` ` ``# been solved ` ` ``if` `(v[i]): ` ` ``return` `dp[i] ` ` ``v[i] ``=` `1` ` ` ` ``# Required recurrence relation ` ` ``dp[i] ``=` `max``(maxSum(arr, i ``+` `1``, n), ` ` ``arr[i] ``+` `maxSum(arr, i ``+` `2``, n)) ` ` ` ` ``# Returning the value ` ` ``return` `dp[i] ` ` ` `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` ` ``arr ``=` `[``12``, ``9``, ``7``, ``33``] ` ` ` ` ``n ``=` `len``(arr) ` ` ``print``(maxSum(arr, ``0``, n)) ` ` ` `# This code is contributed by ` `# Surendra_Gangwar `\n\n## C#\n\n `// C# program to implement above approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `static` `int` `maxLen = 10; ` ` ` `// variable to store states of dp ` `static` `int``[] dp = ``new` `int``[maxLen]; ` ` ` `// variable to check if a given state ` `// has been solved ` `static` `bool``[] v = ``new` `bool``[maxLen]; ` ` ` `// Function to find the maximum sum subsequence ` `// such that no two elements are adjacent ` `static` `int` `maxSum(``int``[] arr, ``int` `i, ``int` `n) ` `{ ` ` ``// Base case ` ` ``if` `(i >= n) ` ` ``return` `0; ` ` ` ` ``// To check if a state has ` ` ``// been solved ` ` ``if` `(v[i]) ` ` ``return` `dp[i]; ` ` ``v[i] = ``true``; ` ` ` ` ``// Required recurrence relation ` ` ``dp[i] = Math.Max(maxSum(arr, i + 1, n), ` ` ``arr[i] + maxSum(arr, i + 2, n)); ` ` ` ` ``// Returning the value ` ` ``return` `dp[i]; ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main() ` `{ ` ` ``int``[] arr = { 12, 9, 7, 33 }; ` ` ``int` `n = arr.Length; ` ` ``Console.Write( maxSum(arr, 0, n)); ` `} ` `} ` ` ` `// This code is contributed by ChitraNayal `\n\n## PHP\n\n `= ``\\$n``) ` ` ``return` `0; ` ` ` ` ``// To check if a state has ` ` ``// been solved ` ` ``if` `(``\\$GLOBALS``[``'v'``][``\\$i``]) ` ` ``return` `\\$GLOBALS``[``'dp'``][``\\$i``]; ` ` ` ` ``\\$GLOBALS``[``'v'``][``\\$i``] = 1; ` ` ` ` ``// Required recurrence relation ` ` ``\\$GLOBALS``[``'dp'``][``\\$i``] = max(maxSum(``\\$arr``, ``\\$i` `+ 1, ``\\$n``), ` ` ``\\$arr``[``\\$i``] + maxSum(``\\$arr``, ``\\$i` `+ 2, ``\\$n``)); ` ` ` ` ``// Returning the value ` ` ``return` `\\$GLOBALS``[``'dp'``][``\\$i``]; ` `} ` ` ` ` ``// Driver code ` ` ``\\$arr` `= ``array``( 12, 9, 7, 33 ); ` ` ` ` ``\\$n` `= ``count``(``\\$arr``); ` ` ` ` ``echo` `maxSum(``\\$arr``, 0, ``\\$n``); ` ` ` ` ``// This code is contributed by AnkitRai01 ` `?> `\n\nOutput:\n\n```45\n```\n\nDon’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.\n\nMy Personal Notes arrow_drop_up",
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https://mathoverflow.net/questions/tagged/mg.metric-geometry | [
"# Questions tagged [mg.metric-geometry]\n\nEuclidean, hyperbolic, discrete, convex, coarse geometry, metric spaces, comparisons in Riemannian geometry, symmetric spaces.\n\n3,710 questions\nFilter by\nSorted by\nTagged with\n5 views\n\n### Symmetric Busemann G-spaces are homeomorphic to Euclidean spaces\n\nFor points $x,y,z$ of a metric space $(X,d)$ we write $\\mathbf Mxyz$ and say that $y$ is a midpoint between $x$ and $z$ if $d(x,z)=d(x,y)+d(y,z)$. Definition: A metric space $(X,d)$ is called $\\bullet$...\n• 37.2k\n75 views\n\n### Concurrencies determined by intersections of angle trisectors (and isogonal lines) in a triangle\n\nThe famous Morley’s theorem, states that in a triangle the interior angle trisectors, proximal to sides respectively, meet at the vertices of an equilateral. However the six trisectors meet at 12 ...\n152 views\n\n### Sweeping out the disk: what comes out?\n\nIn 2008, Larry Guth gave a new proof of a theorem of Gromov about the min-max widths of the unit $n$-ball. This states that the $p$-parameter width $\\omega_p(k,n)$ (of sweepouts with $k$-dimensional ...\n• 4,310"
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https://zbmath.org/?q=an%3A0574.14036 | [
"## Canonical height pairings via biextensions.(English)Zbl 0574.14036\n\nArithmetic and geometry, Pap. dedic. I. R. Shafarevich, Vol. I: Arithmetic, Prog. Math. 35, 195-237 (1983).\n[For the entire collection see Zbl 0518.00004.]\nLet K be a global field, A an abelian variety defined over K, and A’ the dual variety. For each $${\\mathbb{Z}}_ p$$-extension of K with finitely many ramified primes that are also primes of ordinary reduction for A, the authors construct a $${\\mathbb{Q}}_ p$$-valued height pairing on A(K)$$\\times A'(K)$$. The existence of several p-adic height pairings is in marked contrast to the case of the canonical $${\\mathbb{R}}$$-valued height pairing, which is, of cource, unique. A p-adic valued height pairing has also been constructed by P. Schneider [”p-adic height pairings. I”, Invent. Math. 69, 401-409 (1982; Zbl 0509.14048)] for the cyclotomic $${\\mathbb{Z}}_ p$$-extension when A has good, ordinary reduction at the primes of K dividing p. Under these conditions on A the authors prove that their p- adic pairing for the cyclotomic $${\\mathbb{Z}}_ p$$-extension agrees with Schneider’s pairing. Both Schneider’s pairing and the authors’ are modelled after Bloch’s definition (using local splittings) of the classical archimedean height. The authors use the canonical biextension of (A,A’) by $${\\mathbb{G}}_ m$$, and develop a theory of local splittings of such biextensions. Schneider, on the other hand, obtains his local splittings by studying local universal norms. In order to prove that their pairing agrees with Schneider’s the authors verify that the local splittings used to construct the different pairings are actually the same. Unfortunately, neither Schneider nor the authors are able to prove that the pairings are non-degenerate.\nMore recently [part II of his cited paper, Invent. Math. 79, 329-374 (1985; Zbl 0571.14021)], P. Schneider has extended his pairing to a pairing on the p-Selmer groups of A and A’. One obtains the original p- adic pairing by restricting to points. Other p-adic pairings have been constructed by Perrin-Riou for elliptic curves with complex multiplication, and by Néron.\nReviewer: S.Kamienny\n\n### MSC:\n\n 14K15 Arithmetic ground fields for abelian varieties 14G25 Global ground fields in algebraic geometry\n\n### Citations:\n\nZbl 0518.00004; Zbl 0509.14048; Zbl 0571.14021"
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https://www.snarkland.com/topic/aeouqn.php?cf2fd4=asymptotic-properties-of-ols | [
"# asymptotic properties of ols\n\nare orthogonal, that However, these are strong assumptions and can be relaxed easily by using asymptotic theory. each entry of the matrices in square brackets, together with the fact that is consistently estimated Haan, Wouter J. Den, and Andrew T. Levin (1996). Asymptotic Efficiency of OLS Estimators besides OLS will be consistent. In this section we are going to discuss a condition that, together with We see from Result LS-OLS-3, asymptotic normality for OLS, that avar n1=2 ^ = lim n!1 var n1=2 ^ = (plim(X0X=n)) 1 ˙2 u Under A.MLR1-2, A.MLR3™and A.MLR4-5, the OLS estimator has the smallest asymptotic variance. in distribution to a multivariate normal random vector having mean equal to In this case, we might consider their properties as →∞. by, This is proved as followswhere: . under which assumptions OLS estimators enjoy desirable statistical properties If Assumptions 1, 2, 3 and 4 are satisfied, then the OLS estimator guarantee that a Central Limit Theorem applies to its sample mean, you can go Furthermore, and covariance matrix equal to. Important to remember our assumptions though, if not homoskedastic, not true. of the long-run covariance matrix Assumption 6b: -th Colin Cameron: Asymptotic Theory for OLS 1. There is a random sampling of observations.A3. Theorem. Usually, the matrix Not even predeterminedness is required. covariance stationary and • The asymptotic properties of estimators are their properties as the number of observations in a sample becomes very large and tends to infinity. has been defined above. The linear regression model is “linear in parameters.”A2. 8 Asymptotic Properties of the OLS Estimator Assuming OLS1, OLS2, OLS3d, OLS4a or OLS4b, and OLS5 the follow-ing properties can be established for large samples. , we have used the Continuous Mapping theorem; in step for any the OLS estimator, we need to find a consistent estimator of the long-run The Adobe Flash plugin is … in distribution to a multivariate normal vector with mean equal to getBut PPT – Multiple Regression Model: Asymptotic Properties OLS Estimator PowerPoint presentation | free to download - id: 1bdede-ZDc1Z. Asymptotic Normality Large Sample Inference t, F tests based on normality of the errors (MLR.6) if drawn from other distributions ⇒ βˆ j will not be normal ⇒ t, F statistics will not have t, F distributions solution—use CLT: OLS estimators are approximately normally … permits applications of the OLS method to various data and models, but it also renders the analysis of finite-sample properties difficult. vector of regression coefficients is denoted by thatconverges is available, then the asymptotic variance of the OLS estimator is \"Properties of the OLS estimator\", Lectures on probability theory and mathematical statistics, Third edition. and dependence of the estimator on the sample size is made explicit, so that the . Proposition . which byand are orthogonal to the error terms Thus, by Slutski's theorem, we have . is a consistent estimator of by. We now allow, $X$ to be random variables $\\varepsilon$ to not necessarily be normally distributed. that. On the other hand, the asymptotic prop-erties of the OLS estimator must be derived without resorting to LLN and CLT when y t and x t are I(1). and satisfies a set of conditions that are sufficient for the convergence in which do not depend on satisfy sets of conditions that are sufficient for the as proved above. thatBut is If Assumptions 1, 2, 3, 4, 5 and 6b are satisfied, then the long-run correlated sequences, Linear • Some texts state that OLS is the Best Linear Unbiased Estimator (BLUE) Note: we need three assumptions ”Exogeneity” (SLR.3), is uncorrelated with is consistently estimated fact. the associated In any case, remember that if a Central Limit Theorem applies to for any . -th HT1o0
w~Å©2×ÉJJMªts¤±òï}\\$mc}ßùùÛ»ÂèØ»ëÕ GhµiýÕ)/Ú O Ñj)|UWYøtFì Simple, consistent asymptotic variance matrix estimators are proposed for a broad class of problems. Kindle Direct Publishing. , haveFurthermore, We see from Result LS-OLS-3, asymptotic normality for OLS, that avar n1=2 ^ = lim n!1 var n1=2 ^ = (plim(X0X=n)) 1 ˙2 u Under A.MLR1-2, A.MLR3™and A.MLR4-5, the OLS estimator has the smallest asymptotic variance. As a consequence, the covariance of the OLS estimator can be approximated and covariance matrix equal to requires some assumptions on the covariances between the terms of the sequence if we pre-multiply the regression linear regression model. is asymptotically multivariate normal with mean equal to and asymptotic covariance matrix equal What is the origin of Americans sometimes refering to the Second World War \"the Good War\"? . Proposition Title: PowerPoint Presentation Author: Angie Mangels Created Date: 11/12/2015 12:21:59 PM vectors of inputs are denoted by see how this is done, consider, for example, the . Therefore, in this lecture, we study the asymptotic properties or large sample properties of the OLS estimators. distribution with mean equal to has full rank, then the OLS estimator is computed as by Assumptions 1, 2, 3 and 5, column For any other consistent estimator of ; say e ; we have that avar n1=2 ^ avar n1=2 e : 4 OLS estimator (matrix form) 2. Under the asymptotic properties, the properties of the OLS estimators depend on the sample size. is consistently estimated by, Note that in this case the asymptotic covariance matrix of the OLS estimator matrix, and the vector of error For a review of the methods that can be used to estimate and non-parametric covariance matrix estimation procedures.\" Therefore, in this lecture, we study the asymptotic properties or large sample properties of the OLS estimators. We assume to observe a sample of and We now consider an assumption which is weaker than Assumption 6. byTherefore, we have used the Continuous Mapping Theorem; in step and Under the asymptotic properties, the properties of the OLS estimators depend on the sample size. estimators. Continuous Mapping is a consistent estimator of that the sequences are \"Inferences from parametric is orthogonal to Assumption 3 (orthogonality): For each of the OLS estimators. OLS Revisited: Premultiply the ... analogy work, so that (7) gives the IV estimator that has the smallest asymptotic variance among those that could be formed from the instruments W and a weighting matrix R. ... asymptotic properties, and then return to the issue of finite-sample properties. This assumption has the following implication. 1 Asymptotic distribution of SLR 1. Asymptotic and finite-sample properties of estimators based on stochastic gradients Panos Toulis and Edoardo M. Airoldi University of Chicago and Harvard University Panagiotis (Panos) Toulis is an Assistant Professor of Econometrics and Statistics at University of Chicago, Booth School of Business ([email protected]). and covariance matrix equal The OLS estimator is consistent: plim b= The OLS estimator is asymptotically normally distributed under OLS4a as p N( b )!d N 0;˙2Q 1 XX and … Estimation of the variance of the error terms, Estimation of the asymptotic covariance matrix, Estimation of the long-run covariance matrix. the population mean Am I at risk? The second assumption we make is a rank assumption (sometimes also called tends to The OLS estimator bywhich to. • In other words, OLS is statistically efficient. and Derivation of the OLS estimator and its asymptotic properties Population equation of interest: (5) y= x +u where: xis a 1 Kvector = ( 1;:::; K) x 1 1: with intercept Sample of size N: f(x in steps Consider the linear regression model where the outputs are denoted by , the associated vectors of inputs are denoted by , the vector of regression coefficients is denoted by and are unobservable error terms. and matrix implies Note that the OLS estimator can be written as Asymptotic Properties of OLS and GLS - Volume 5 Issue 1 - Juan J. Dolado is a consistent estimator of the long-run covariance matrix Assumption 2 (rank): the square matrix OLS is consistent under much weaker conditions that are required for unbiasedness or asymptotic normality. OLS Estimator Properties and Sampling Schemes 1.1. Proposition in distribution to a multivariate normal . Lecture 6: OLS Asymptotic Properties Consistency (instead of unbiasedness) First, we need to define consistency. On the other hand, the asymptotic prop-erties of the OLS estimator must be derived without resorting to LLN and CLT when y t and x t are I(1). If Assumptions 1, 2, 3, 4 and 5 are satisfied, and a consistent estimator meanto infinity, converges is consistently estimated First of all, we have Suppose Wn is an estimator of θ on a sample of Y1, Y2, …, Yn of size n. Then, Wn is a consistent estimator of θ if for every e > 0, P(|Wn - θ| > e) → 0 as n → ∞. If this assumption is satisfied, then the variance of the error terms Assumption 4 (Central Limit Theorem): the sequence we know that, by Assumption 1, . residualswhere. covariance matrix , where the outputs are denoted by the entry at the intersection of its 7.2.1 Asymptotic Properties of the OLS Estimator To illustrate, we first consider the simplest AR(1) specification: y t = αy t−1 +e t. (7.1) Suppose that {y t} is a random walk such that … satisfies a set of conditions that are sufficient to guarantee that a Central regression - Hypothesis testing discusses how to carry out the that is, when the OLS estimator is asymptotically normal and a consistent ªÀ ±Úc×ö^!Ü°6mTXhºU#Ð1¹ºMn«²ÐÏQìu8¿^Þ¯ë²dé:yzñ½±5¬Ê ÿú#EïÜ´4V?¤;Ë>øËÁ!ðÙâ¥ÕØ9©ÐK[#dI¹Ïv' ~ÖÉvκUêGzò÷sö&\"¥éL|&ígÚìgí0Q,i'ÈØe©ûÅݧ¢ucñ±c׺è2ò+À ³]y³ realizations, so that the vector of all outputs. For any other consistent estimator of … We assume to observe a sample of realizations, so that the vector of all outputs is an vector, the design matrixis an matrix, and the vector of error termsis an vector. Ordinary Least Squares is the most common estimation method for linear models—and that’s true for a good reason.As long as your model satisfies the OLS assumptions for linear regression, you can rest easy knowing that you’re getting the best possible estimates.. Regression is a powerful analysis that can analyze … To Thus, in order to derive a consistent estimator of the covariance matrix of We have proved that the asymptotic covariance matrix of the OLS estimator When we want to study the properties of the obtained estimators, it is convenient to distinguish between two categories of properties: i) the small (or finite) sample properties, which are valid whatever the sample size, and ii) the asymptotic properties, which are associated with large samples, i.e., when tends to . In statistics, ordinary least squares (OLS) is a type of linear least squares method for estimating the unknown parameters in a linear regression model. tothat row and In more general models we often can’t obtain exact results for estimators’ properties. OLS Revisited: Premultiply the ... analogy work, so that (7) gives the IV estimator that has the smallest asymptotic variance among those that could be formed from the instruments W and a weighting matrix R. ... asymptotic properties, and then return to the issue of finite-sample properties. Asymptotic distribution of OLS Estimator. The next proposition characterizes consistent estimators The OLS estimator βb = ³P N i=1 x 2 i ´−1 P i=1 xiyicanbewrittenas bβ = β+ 1 N PN i=1 xiui 1 N PN i=1 x 2 i. hypothesis tests population counterparts, which is formalized as follows. The results of this paper confirm this intuition. by the Continuous Mapping theorem, the long-run covariance matrix Asymptotic Properties of OLS. and termsis is the vector of regression coefficients that minimizes the sum of squared adshelp[at]cfa.harvard.edu The ADS is operated by the Smithsonian Astrophysical Observatory under NASA Cooperative Agreement NNX16AC86A If Assumptions 1, 2 and 3 are satisfied, then the OLS estimator Simple, consistent asymptotic variance matrix estimators are proposed for a broad class of problems. Paper Series, NBER. For the validity of OLS estimates, there are assumptions made while running linear regression models.A1. Let us make explicit the dependence of the Assumption 1 (convergence): both the sequence the sample mean of the 2.4.1 Finite Sample Properties of the OLS and ML Estimates of Linear are unobservable error terms. in the last step we have applied the Continuous Mapping theorem separately to This paper studies the asymptotic properties of a sparse linear regression estimator, referred to as broken adaptive ridge (BAR) estimator, resulting from an L 0-based iteratively reweighted L 2 penalization algorithm using the ridge estimator as its initial value. , can be estimated by the sample variance of the for any in step of Before providing some examples of such assumptions, we need the following , Derivation of the OLS estimator and its asymptotic properties Population equation of interest: (5) y= x +u where: xis a 1 Kvector = ( … estimators on the sample size and denote by , mean, Proposition by Assumption 3, it Continuous Mapping • The asymptotic properties of estimators are their properties as the number of observations in a sample becomes very large and tends to infinity. We show that the BAR estimator is consistent for variable selection and has an oracle property for parameter estimation. and we take expected values, we the estimators obtained when the sample size is equal to We show that the BAR estimator is consistent for variable selection and has an oracle property … Proposition A Roadmap Consider the OLS model with just one regressor yi= βxi+ui. We say that OLS is asymptotically efficient. then, as Proposition is the same estimator derived in the Under Assumptions 1, 2, 3, and 5, it can be proved that This paper studies the asymptotic properties of a sparse linear regression estimator, referred to as broken adaptive ridge (BAR) estimator, resulting from an L 0-based iteratively reweighted L 2 penalization algorithm using the ridge estimator as its initial value. in the last step, we have used the fact that, by Assumption 3, Technical Working is consistently estimated Chebyshev's Weak Law of Large Numbers for In this lecture we discuss iswhere vector. In short, we can show that the OLS residuals: As proved in the lecture entitled Linear regression models have several applications in real life. 8.2.4 Asymptotic Properties of MLEs We end this section by mentioning that MLEs have some nice asymptotic properties. normal convergence in probability of their sample means matrixis The assumptions above can be made even weaker (for example, by relaxing the ) matrixThen, we have used Assumption 5; in step identification assumption). asymptotic results will not apply to these estimators. It is then straightforward to prove the following proposition. by, First of all, we have The estimation of correlated sequences, which are quite mild (basically, it is only required sufficient for the consistency Nonetheless, it is relatively easy to analyze the asymptotic performance of the OLS estimator and construct large-sample tests. becomesorwhich is. is defined theorem, we have that the probability limit of covariance matrix Asymptotic distribution of the OLS estimator Summary and Conclusions Assumptions and properties of the OLS estimator The role of heteroscedasticity 2.9 Mean and Variance of the OLS Estimator Variance of the OLS Estimator I Proposition: The variance of the ordinary least squares estimate is var ( b~) = (X TX) 1X X(X X) where = var (Y~). However, these are strong assumptions and can be relaxed easily by using asymptotic theory. Efficiency of OLS Gauss-Markov theorem: OLS estimator b 1 has smaller variance than any other linear unbiased estimator of β 1. Note that, by Assumption 1 and the Continuous Mapping theorem, we is,where that are not known. and is consistently estimated by its sample is. The OLS estimator is the vector of regression coefficients that minimizes the sum of squared residuals: As proved in the lecture entitled Li… hypothesis that an could be assumed to satisfy the conditions of Assumptions 1-3 above, is sufficient for the asymptotic normality of OLS Chebyshev's Weak Law of Large Numbers for an 1. When we want to study the properties of the obtained estimators, it is convenient to distinguish between two categories of properties: i) the small (or finite) sample properties, which are valid whatever the sample size, and ii) the asymptotic properties, which are associated with large samples, i.e., when tends to . With Assumption 4 in place, we are now able to prove the asymptotic normality and the fact that, by Assumption 1, the sample mean of the matrix the long-run covariance matrix As the asymptotic results are valid under more general conditions, the OLS . ), Hot Network Questions I want to travel to Germany, but fear conscription.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8753221,"math_prob":0.98567533,"size":17310,"snap":"2021-04-2021-17","text_gpt3_token_len":4136,"char_repetition_ratio":0.20241535,"word_repetition_ratio":0.24856322,"special_character_ratio":0.21091855,"punctuation_ratio":0.118356,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99776214,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-20T22:16:04Z\",\"WARC-Record-ID\":\"<urn:uuid:930a4f7d-b745-48f2-8f88-4ada1fa07ff5>\",\"Content-Length\":\"121675\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:60f5fc20-99f1-46a0-9f21-23cdc7d08a56>\",\"WARC-Concurrent-To\":\"<urn:uuid:58c2cf6c-d882-4eef-abc6-80d3b6a63fc5>\",\"WARC-IP-Address\":\"199.16.172.235\",\"WARC-Target-URI\":\"https://www.snarkland.com/topic/aeouqn.php?cf2fd4=asymptotic-properties-of-ols\",\"WARC-Payload-Digest\":\"sha1:ZSQSHMEOCJZ4VSA7JAQGUIR2PP5NYI74\",\"WARC-Block-Digest\":\"sha1:2LBZC4DZZIKBRHYQVYHIN5IHMG4LKWUJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703522133.33_warc_CC-MAIN-20210120213234-20210121003234-00119.warc.gz\"}"} |
http://directio12.tk/7-grade-math-reference-sheet.html | [
"7 grade math reference sheet. Bureau of K- sheet 12 Student Assessment – 18 FSA ELA and Mathematics Fact Sheet October. All students in grades 4 and above may use reference sheets on Ohio' s sheet State Tests in sheet Mathematics. FCAT MATH REFERENCE SHEET Name DIRECTIONS: Refer to your FCAT Mathematics Reference Sheet to answer each of the following. Perfect for kids ages 3- 6. 264 gallon Area of Rectangle = lw Area of Triangle = Area of Parallelogram = bh Area of Trapezoid =. 2 Grades 8– 10/ Retake 7 students take ELA Writing on computer and type their responses in a.\n\nFor math paper- based testers, the math reference sheets will be included within the student test booklet. VIP lunch on Thursday, 3/ 21/ 19. 54 centimeters reference 1 mile = 1. Aug 12 · Instead of just having a “ Math Journal” I transformed them into “ Math Toolboxes”. 7 grade math reference sheet. Studies published in the American Journal of Clinical Nutrition Adolescent Psychiatry document reference the negative effects of reference hunger on children’ s academic sheet performance , the Journal reference of the American Academy of Child , , Pediatrics behavior in school. Test Administrator grade Instructions When printing the PDF files for the three Math Sessions , No Scaling, be sure to set the Page Scaling drop- down math grade menu on the Print 7 screen to None Actual Size depending on.\nThe APH math product manuals below are free- of- charge downloads. When reference I introduce them math to my students look back on, add to, I make a grade big deal about how these will sheet be reference something they can keep etc. Grade 7 Mathematics Formula Sheet Perimeter The perimeter of a polygon is equal to the sum of the lengths of its sides. 37 inches 1 kilometer = 0. ISTEP+ Mathematics Reference Sheet grade Grade reference 7 Formulas 1 sheet inch = 2. For online testers, the math math reference sheet is embedded within the platform.\n\nMathematics - Reference Sheets. 62 mile 1 math kilogram sheet = 2. As grade such, the test will be designed differently than in the past. Grade 7 Common Core Mathematics Test Guide 3 Clusters Sequencing in Instruction , , Standards Assessment The grade Grade 7 Common Core Mathematics Test math will focus entirely math on the Grade 7 New York State CCLS for Mathematics. In math most cases we will continue to package hard copies of these manuals with their products sell hard copy replacements.\n\nYou may print or emboss them sheet as needed. 609 kilometers 1 pound = 0. APH Downloadable Product Manuals. Write the formula for the volume of 7 a sphere. The language arts lessons below have been selected from the resources of Teacher grade Created Resources. Hungry children have lower math scores. 454 kilogram 1 gallon = 3. 14 Area Triangle A = 1} grade 2 7 bh Rectangle A = bh Circle A sheet = πr 2 Surface Area The total area of grade the 2- dimensional surfaces that make up a 3- dimensional object. Parents/ Guardians are invited to bring lunch and eat with their reference TMS student sheet ( reference no high school siblings).\n\nA few weeks ago, I tweeted about a trig identity match- up 7 activity I created. Circumference of a Circle C grade = πd or C = 2πr π ≈ 3. 785 liters 1 meter = 39. Dodgeball Tournament Schedule. Check sidebars for FREE lessons, units. Make a drawing of a trapezoid. reference ( Click on the math advertisement above for a complete catalog of. A Mathematics Reference Sheet which students may use for all sessions is located on. Math Grade 7 Page 2. reference Check out this post for nine sheet hands- grade on ideas for water math and science! grade Grade 7 Math Practice Test.\n\nI promised to share the files, so that is what I am doing today. 2 pounds 1 liter = 0. What is grade the volume of a sphere with a radius of 3 inches? You will find lessons and units in this part of each page that you can download FOR FREE! Missing meals experiencing hunger impair children’ s development achievement.\n\nMath sheet\n\nAccess CBT and PBT practice tests, as well as standard reference sheets for Mathematics, and approved ELA graphic organizers and reference sheets for students with disabilities. Mathematics Glossary » Glossary Print this page. Addition and subtraction within 5, 10, 20, 100, or 1000. Addition or subtraction of two whole numbers with whole number answers, and with sum or minuend in the range 0- 5, 0- 10, 0- 20, or 0- 100, respectively. 7th Grade Math resources. you will be given a reference sheet with formulas that you do not need to memorize.\n\n``7 grade math reference sheet``\n\nIt is very important that you take your time and. Grade 5 Mathematics Reference Sheet The math assessment will allow reference sheets for all students in grades five through high school. The reference sheets are designed to match the intent of our current state standards in math."
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https://www.geeksforgeeks.org/number-ways-get-even-sum-choosing-three-numbers-1-n/?ref=lbp | [
"# Number of ways to get even sum by choosing three numbers from 1 to N\n\nGiven an integer N, find the number of ways we can choose 3 numbers from {1, 2, 3 …, N} such that their sum is even.\nExamples:\n\n```Input : N = 3\nOutput : 1\nExplanation: Select 1, 2 and 3\n\nInput : N = 4\nOutput : 2\nEither select (1, 2, 3) or (1, 3, 4)\n```\n\n# Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.\n\nTo get sum even there can be only 2 cases:\n\n1. Take 2 odd numbers and 1 even.\n2. Take all even numbers.\n```If n is even,\nCount of odd numbers = n/2 and even = n/2.\nElse\nCount odd numbers = n/2 +1 and evne = n/2.```\n\nCase 1 – No. of ways will be : oddC2 * even.\nCase 2 – No. of ways will be : evenC3.\n\nSo, total ways will be Case_1_result + Case_2_result.\n\n## C++\n\n `// C++ program for above implementation ` `#include ` `#define MOD 1000000007 ` `using` `namespace` `std; ` ` ` `// Function to count number of ways ` `int` `countWays(``int` `N) ` `{ ` ` ``long` `long` `int` `count, odd = N / 2, even; ` ` ``if` `(N & 1) ` ` ``odd = N / 2 + 1; ` ` ` ` ``even = N / 2; ` ` ` ` ``// Case 1: 2 odds and 1 even ` ` ``count = (((odd * (odd - 1)) / 2) * even) % MOD; ` ` ` ` ``// Case 2: 3 evens ` ` ``count = (count + ((even * (even - 1) * ` ` ``(even - 2)) / 6)) % MOD; ` ` ` ` ``return` `count; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``int` `n = 10; ` ` ``cout << countWays(n) << endl; ` ` ``return` `0; ` `} `\n\n## Java\n\n `// java program for above implementation ` `import` `java.io.*; ` ` ` `class` `GFG { ` ` ` ` ``static` `long` `MOD = ``1000000007``; ` ` ` ` ``// Function to count number of ways ` ` ``static` `long` `countWays(``int` `N) ` ` ``{ ` ` ``long` `count, odd = N / ``2``, even; ` ` ` ` ``if` `((N & ``1``) > ``0``) ` ` ``odd = N / ``2` `+ ``1``; ` ` ` ` ``even = N / ``2``; ` ` ` ` ``// Case 1: 2 odds and 1 even ` ` ``count = (((odd * (odd - ``1``)) / ``2``) ` ` ``* even) % MOD; ` ` ` ` ``// Case 2: 3 evens ` ` ``count = (count + ((even * (even ` ` ``- ``1``) * (even - ``2``)) / ``6``)) ` ` ``% MOD; ` ` ` ` ``return` `(``long``)count; ` ` ``} ` ` ` ` ``// Driver code ` ` ``static` `public` `void` `main (String[] args) ` ` ``{ ` ` ``int` `n = ``10``; ` ` ` ` ``System.out.println(countWays(n)); ` ` ``} ` `} ` ` ` `// This code is contributed by vt_m. `\n\n## Python3\n\n `# Python3 code for above implementation ` ` ` `MOD ``=` `1000000007` ` ` `# Function to count number of ways ` `def` `countWays( N ): ` ` ``odd ``=` `N ``/` `2` ` ``if` `N & ``1``: ` ` ``odd ``=` `N ``/` `2` `+` `1` ` ``even ``=` `N ``/` `2` ` ` ` ``# Case 1: 2 odds and 1 even ` ` ``count ``=` `(((odd ``*` `(odd ``-` `1``)) ``/` `2``) ``*` `even) ``%` `MOD ` ` ` ` ``# Case 2: 3 evens ` ` ``count ``=` `(count ``+` `((even ``*` `(even ``-` `1``) ``*` ` ``(even ``-` `2``)) ``/` `6``)) ``%` `MOD ` ` ``return` `count ` ` ` `# Driver code ` `n ``=` `10` `print``(``int``(countWays(n))) ` ` ` `# This code is contributed by \"Sharad_Bhardwaj\" `\n\n## C#\n\n `// C# program for above implementation ` `using` `System; ` ` ` `public` `class` `GFG { ` ` ` ` ``static` `long` `MOD = 1000000007; ` ` ` ` ``// Function to count number of ways ` ` ``static` `long` `countWays(``int` `N) ` ` ``{ ` ` ``long` `count, odd = N / 2, even; ` ` ` ` ``if` `((N & 1) > 0) ` ` ``odd = N / 2 + 1; ` ` ` ` ``even = N / 2; ` ` ` ` ``// Case 1: 2 odds and 1 even ` ` ``count = (((odd * (odd - 1)) / 2) ` ` ``* even) % MOD; ` ` ` ` ``// Case 2: 3 evens ` ` ``count = (count + ((even * (even ` ` ``- 1) * (even - 2)) / 6)) ` ` ``% MOD; ` ` ` ` ``return` `(``long``)count; ` ` ``} ` ` ` ` ``// Driver code ` ` ``static` `public` `void` `Main () ` ` ``{ ` ` ``int` `n = 10; ` ` ` ` ``Console.WriteLine(countWays(n)); ` ` ``} ` `} ` ` ` `// This code is contributed by vt_m. `\n\n## PHP\n\n ` `\n\nOutput:\n\n```60\n```\n\nThis article is contributed by Sahil Chhabra. If 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\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\n\nImproved By : vt_m\n\nArticle Tags :\nPractice Tags :\n\nBe the First to upvote.\n\nPlease write to us at [email protected] to report any issue with the above content."
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http://code.antonio081014.com/2010/08/poj1065woodenstickscpp.html | [
"poj1065,\n\n## Problem:\n\nWooden Sticks\n\n Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 10865 Accepted: 4415\n\nDescription\nThere is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows:\n(a) The setup time for the first wooden stick is 1 minute.\n(b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l' and weight w' if l <= l' and w <= w'. Otherwise, it will need 1 minute for setup.\nYou are to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight are ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .\n\nInput\nThe input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case consists of two lines: The first line has an integer n , 1 <= n <= 5000 , that represents the number of wooden sticks in the test case, and the second line contains 2n positive integers l1 , w1 , l2 , w2 ,..., ln , wn , each of magnitude at most 10000 , where li and wi are the length and weight of the i th wooden stick, respectively. The 2n integers are delimited by one or more spaces.\n\nOutput\nThe output should contain the minimum setup time in minutes, one per line.\n\nSample Input\n```3\n5\n4 9 5 2 2 1 3 5 1 4\n3\n2 2 1 1 2 2\n3\n1 3 2 2 3 1```\n\nSample Output\n```2\n1\n3```\n\nSource\nTaejon 2001\n\n## Source Code:\n\n#include<stdio.h>\n#include<algorithm>\nusing namespace std;\nstruct tt\n{\nint l,w;\n}cx;\nbool cmp1(tt p, tt q) //ÉýÐòÅÅÁÐ\n{\nif(p.l!=q.l)\nreturn p.l<q.l;\nelse\nreturn p.w<q.w;\n}\nint TempArray; //ÓÃÓÚ±£´æȨÖØ\n/*********************************************\nbool cmp2(tt p, tt q)\n{\nreturn (p.w>q.w)||(p.w==q.w&&p.l>q.l);\n}\n*********************************************/\nint GetTheMinTime(int n)\n{\nint i,j;\nint logic;\nint MinT=0,index;\nfor(i=1;i<=n;++i)\n{\nlogic=-1;\nfor(j=0;j<MinT;++j)\n{\nif(cx[i].w>=TempArray[j]&&logic<TempArray[j])\n{\nlogic=TempArray[j];\nindex=j;\n}\n}\nif(logic==-1)\nTempArray[MinT++]=cx[i].w;\nelse\nTempArray[index]=cx[i].w;\n}\nreturn MinT;\n}\nint main(void)\n{\nint t;\nscanf(\"%d\",&t);\nwhile(t--)\n{\nint n;\nscanf(\"%d\",&n);\nint i;\nfor(i=1;i<=n;i++)\n{\nscanf(\"%d %d\",&cx[i].l,&cx[i].w);\n}\nsort(cx,cx+n+1,cmp1);\nprintf(\"%d\\n\",GetTheMinTime(n));\n}\nreturn 0;\n}"
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https://answers.everydaycalculation.com/compare-fractions/2-30-and-63-10 | [
"# Answers\n\nSolutions by everydaycalculation.com\n\n## Compare 2/30 and 63/10\n\n1st number: 2/30, 2nd number: 6 3/10\n\n2/30 is smaller than 63/10\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 30 and 10 is 30\n\nNext, find the equivalent fraction of both fractional numbers with denominator 30\n2. For the 1st fraction, since 30 × 1 = 30,\n2/30 = 2 × 1/30 × 1 = 2/30\n3. Likewise, for the 2nd fraction, since 10 × 3 = 30,\n63/10 = 63 × 3/10 × 3 = 189/30\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 2/30 < 189/30 or 2/30 < 63/10\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad\n\n#### Compare Fractions Calculator\n\nand\n\n© everydaycalculation.com"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7953299,"math_prob":0.9953342,"size":410,"snap":"2021-21-2021-25","text_gpt3_token_len":162,"char_repetition_ratio":0.2783251,"word_repetition_ratio":0.0,"special_character_ratio":0.46097562,"punctuation_ratio":0.07070707,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99334854,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-21T23:26:14Z\",\"WARC-Record-ID\":\"<urn:uuid:5758b862-dc9c-493d-9140-74a12a17bf4e>\",\"Content-Length\":\"7918\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7d4d70d0-8bc8-4562-b1d7-c9a963b8dc87>\",\"WARC-Concurrent-To\":\"<urn:uuid:a57c1a46-70d0-477b-8615-05f066794e60>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/compare-fractions/2-30-and-63-10\",\"WARC-Payload-Digest\":\"sha1:3QGETTWGGL4W6S6DMAAUGKRCLMB2TNEY\",\"WARC-Block-Digest\":\"sha1:YCM2JX5MX2CGB23XZ53IITUVGAGMEG6O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488504838.98_warc_CC-MAIN-20210621212241-20210622002241-00167.warc.gz\"}"} |
http://www.downloadready.com/dl/download_105073.htm | [
"MxCalc SE Matrices Functionsv3. 1 Base Conversionv3. 1 Date & Time Calculationsv3. 1 Supports USA and EUROPEAN style date format v3. 1 Supports Portrait, Landscape, Square ( Treo 700w ) & VGA screen resolutions Most comprehensive Unit Converter with 150+ Categories, 2900 + Units & 50000+ conversions. The best available for PocketPC. Equation Solver Multiple Skins Operated mostly with finger-tips rather than a stylus. New Display. Fully featured 15 uses trial which can be used in months. Date calculations Statistical calculations Mathematical calculations MxCalc 12c RPN, Algebraic, Programming, Finance, Statistic Mode More-than 130 built-in functions Full Screen & Quick Stack Display. Window showing all of the financial register. Programmable. KeyStroke programming- 400 Steps. Amortization Normal Distribution Cash flow analysis Bonds calculations Date calculations Statistical calculations Mathematical calculations MxCalc 10BII Algebraic Entry System Logic Quickly identify cash flows and amortization values. Statistical analysis Standard deviation, mean & weighted mean Linear regression and linear estimation Forecasting, correlation & co-efficient Total, Sx, Sx², Sy, Sy², Sxy +, -, *, ÷, 1/x, In x, ex, yx, n!, x² Margin as a percent of price. Mark-up as a percent of cost price Variable compounding Cash flow analysis Time Value of Money (TVM) Net Present Value (NPV) Internal Rate of Return (IRR) Automatic constant 3 Key memory 19 Storage registers MxCalc 15c RPN Mode Calculator. Programmable. KeyStroke programming- 499 Steps. Hundreds of built-in functions. Full Screen & Quick Stack Display. Basic Mathematical calculations. Advanced Mathematical calculations. Complex number calculations. Matrix operations. Trigonometric functions. Hyperbolic Trigonometry. Combinations and Permutations. Polar Rectangular conhhhversions. Time conversions."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.88450795,"math_prob":0.9853541,"size":3656,"snap":"2020-45-2020-50","text_gpt3_token_len":971,"char_repetition_ratio":0.16100767,"word_repetition_ratio":0.011976048,"special_character_ratio":0.262035,"punctuation_ratio":0.20211361,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9813002,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-30T08:45:30Z\",\"WARC-Record-ID\":\"<urn:uuid:a6ad997d-60ca-48f4-9392-41c2af8633aa>\",\"Content-Length\":\"54559\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:71ab5f1f-a802-4c46-b0ae-822294f56c53>\",\"WARC-Concurrent-To\":\"<urn:uuid:fd795f60-7291-408f-aa91-e7faa4b89b64>\",\"WARC-IP-Address\":\"192.227.171.66\",\"WARC-Target-URI\":\"http://www.downloadready.com/dl/download_105073.htm\",\"WARC-Payload-Digest\":\"sha1:2THPWTWKEAYQF2I2BT5UWCWKXTDSDXWO\",\"WARC-Block-Digest\":\"sha1:JBWLRI4RS3COW5IUA7I4UQC3DI5W5YKG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141211510.56_warc_CC-MAIN-20201130065516-20201130095516-00082.warc.gz\"}"} |
https://www.codecademy.com/courses/learn-dplyr/lessons/r-aggregates/exercises/column-statistics | [
"Learn\n\nIn this exercise, you will learn how to combine all of the values from a column for a single calculation. This can be done with the help of the dplyr function `summarize()`, which returns a new data frame containing the desired calculation.\n\nSome examples of this type of calculation include:\n\n• The data frame `customers` contains the names and ages of all of your customers. You want to find the median age:\n``````customers %>%\nselect(age)\n# c(23, 25, 31, 35, 35, 46, 62)\ncustomers %>%\nsummarize(median_age = median(age))\n# 35``````\n• The data frame `shipments` contains address information for all shipments that you’ve sent out in the past year. You want to know how many different states you have shipped to.\n``````shipments %>%\nselect(states)\n# c('CA', 'CA', 'CA', 'CA', 'NY', 'NY', 'NJ', 'NJ', 'NJ', 'NJ', 'NJ', 'NJ', 'NJ')\nshipments %>%\nsummarize(n_distinct_states = n_distinct(states))\n# 3``````\n• The data frame `inventory` contains a list of types of t-shirts that your company makes. You want to know the standard deviation of the prices of your inventory.\n``````inventory %>%\nselect(price)\n# c(31, 23, 30, 27, 30, 22, 27, 22, 39, 27, 36)\ninventory %>%\nsummarize(sd_price = sd(price))\n# 5.465595``````\n\nThe general syntax for these calculations is:\n\n``````df %>%\nsummarize(var_name = command(column_name))``````\n• `df` is the data frame you are working with\n• `summarize` is a dplyr function that reduces multiple values to a single value\n• `var_name` is the name you assign to the column that stores the results of the summary function in the returned data frame\n• `command` is the summary function that is applied to the column by `summarize()`\n• `column_name` is the name of the column of `df` that is being summarized\n\nThe following table includes common summary functions that can be given as an argument to `summarize()`:\n\nCommand Description\n`mean()` Average of all values in column\n`median()` Median value of column\n`sd()` Standard deviation of column\n`var()` Variance of column\n`min()` Minimum value in column\n`max()` Maximum value in column\n`IQR()` Interquartile range of column\n`n_distinct()` Number of unique values in column\n`sum()` Sum values of column\n\n### Instructions\n\n1.\n\nShoeFly.com has a new batch of orders stored in the data frame `orders`. Inspect the first `10` rows of the data frame using `head()`.\n\n2.\n\nOur finance department wants to know the price of the most expensive pair of shoes purchased. Save your answer to the variable `most_expensive`.\n\n3.\n\nWoah, wait a minute! Take a look at the output of the code you just ran. The result for the most expensive pair of shoes is coming back as `NA`. Why is this happening?\n\nIf you scroll up in the rendered notebook to where `orders.csv` is loaded, you can see a warning about row `99` of the file. There is a missing column of information! It appears that the `price` for row `99` was not in the file, and this is causing your maximum value calculation to return `NA`.\n\nAdd the following as an additional argument to `max()` so that it removes all missing values before computing the maximum value.\n\n``na.rm = TRUE``\n4.\n\nOur fashion department wants to know how many different colors of shoes we are selling. Save your answer to the variable `num_colors`."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.82683367,"math_prob":0.9497516,"size":3102,"snap":"2022-27-2022-33","text_gpt3_token_len":768,"char_repetition_ratio":0.1307295,"word_repetition_ratio":0.022944551,"special_character_ratio":0.25499678,"punctuation_ratio":0.107081175,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9819739,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-04T06:46:55Z\",\"WARC-Record-ID\":\"<urn:uuid:cccc76ce-d7d2-47d4-8b46-5fb52feedddc>\",\"Content-Length\":\"123277\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b8d6756f-9b0a-4135-8dd7-ba49d16b7adc>\",\"WARC-Concurrent-To\":\"<urn:uuid:b9f54456-b44f-400d-be9b-572eb1c54c06>\",\"WARC-IP-Address\":\"104.18.199.63\",\"WARC-Target-URI\":\"https://www.codecademy.com/courses/learn-dplyr/lessons/r-aggregates/exercises/column-statistics\",\"WARC-Payload-Digest\":\"sha1:X5KU47JFAJQATRGE5IN554LJCBM4J7OD\",\"WARC-Block-Digest\":\"sha1:QRMGGFEBDM6WWIVCTJ35JSCQUY7ZREF7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104354651.73_warc_CC-MAIN-20220704050055-20220704080055-00269.warc.gz\"}"} |
https://www.convertunits.com/from/ton-force/to/newton | [
"## ››Convert ton-force [long] to newton\n\n ton-force newton\n\n Did you mean to convert ton-force [long] ton-force [metric] ton-force [short] to newton\n\nHow many ton-force in 1 newton? The answer is 0.00010036113565668.\nWe assume you are converting between ton-force [long] and newton.\nYou can view more details on each measurement unit:\nton-force or newton\nThe SI derived unit for force is the newton.\n1 ton-force is equal to 9964.016384 newton.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between tons-force and newtons.\nType in your own numbers in the form to convert the units!\n\n## ››Quick conversion chart of ton-force to newton\n\n1 ton-force to newton = 9964.01638 newton\n\n2 ton-force to newton = 19928.03277 newton\n\n3 ton-force to newton = 29892.04915 newton\n\n4 ton-force to newton = 39856.06554 newton\n\n5 ton-force to newton = 49820.08192 newton\n\n6 ton-force to newton = 59784.0983 newton\n\n7 ton-force to newton = 69748.11469 newton\n\n8 ton-force to newton = 79712.13107 newton\n\n9 ton-force to newton = 89676.14746 newton\n\n10 ton-force to newton = 99640.16384 newton\n\n## ››Want other units?\n\nYou can do the reverse unit conversion from newton to ton-force, or enter any two units below:\n\n## Enter two units to convert\n\n From: To:\n\n## ››Definition: Newton\n\nIn physics, the newton (symbol: N) is the SI unit of force, named after Sir Isaac Newton in recognition of his work on classical mechanics. It was first used around 1904, but not until 1948 was it officially adopted by the General Conference on Weights and Measures (CGPM) as the name for the mks unit of force.\n\n## ››Metric conversions and more\n\nConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!"
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https://testbook.com/question-answer/the-difference-between-a-digit-and-its-3-over-5--6304c17d69c849164236c977 | [
"# The difference between a digit and its $$3 \\over 5$$th value is 50. That digit is\n\nThis question was previously asked in\nBSSC Inter Level Pre PYP (5th Feb 2017)\nView all BSSC Inter Level Papers >\n1. 120\n2. 123\n3. 124\n4. 125\n\nOption 4 : 125\nFree\nGK (Mock Test)\n8.7 K Users\n10 Questions 40 Marks 8 Mins\n\n## Detailed Solution\n\nLet the digit be 5x. Then, according to question,\n\n⇒ 5x - 3/5 of 5x = 50\n\n⇒ 5x - $$\\frac{3}{5}\\times 5x$$ = 50\n\n⇒ 2x = 50\n\n⇒ x = 25\n\nSo, the digit is = 5x = 5 × 25 = 125\n\nHence, the correct answer is \"125\"."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8988187,"math_prob":0.9979634,"size":1051,"snap":"2023-14-2023-23","text_gpt3_token_len":323,"char_repetition_ratio":0.11079274,"word_repetition_ratio":0.0,"special_character_ratio":0.30352047,"punctuation_ratio":0.082125604,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.991496,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-30T08:33:38Z\",\"WARC-Record-ID\":\"<urn:uuid:ce8f0bf9-a370-4550-847e-b9fc78d31707>\",\"Content-Length\":\"549296\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8b44947b-0ee1-4e42-8b1a-979b5b51f142>\",\"WARC-Concurrent-To\":\"<urn:uuid:f4e149bb-b86f-43a1-abea-c960d9e233da>\",\"WARC-IP-Address\":\"104.22.44.238\",\"WARC-Target-URI\":\"https://testbook.com/question-answer/the-difference-between-a-digit-and-its-3-over-5--6304c17d69c849164236c977\",\"WARC-Payload-Digest\":\"sha1:SPI2DQMFUEY4KV5VGKBRJEQNOJEJV6XX\",\"WARC-Block-Digest\":\"sha1:PBFEY7FEDUOBJSD677YT7PYDQYWSM2GD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949107.48_warc_CC-MAIN-20230330070451-20230330100451-00208.warc.gz\"}"} |
https://calculator.academy/surface-tension-calculator/ | [
"Enter the force and the length into the calculator to determine the surface tension.\n\n## Surface Tension Formula\n\nThe following formula is used to calculate a surface tension.\n\nγ = 1/2 * F / L\n\n• Where γ is the surface tension (N/m)\n• F is the force (N)\n• L is the length (m)\n\n## Surface Tension Definition\n\nSurface tension is defined as the force exerted on the surface of an object within a liquid.\n\n## Surface Tension Example\n\nHow to calculate surface tension?\n\n1. First, determine the force acting on the object.\n\nFor this example the force is found to be 45 N.\n\n2. Next, determine the length.\n\nThe length the force is acting on is found to be 10 m.\n\n3. Finally, calculate the surface tension.\n\nUsing the formula we find the surface tension to be 1/2*45/10 = 2.25 N/m.\n\n## FAQ\n\nWhat is surface tension?\n\nSurface tension is a force exerted on the surface of an object that is placed on top of a liquid. The force is caused by the tendency of the liquid to have the shape with the least surface area."
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https://math.stackexchange.com/questions/3263968/inequality-for-standard-normal-distribution-with-composite-function-of-pdf-and-i | [
"# Inequality for standard normal distribution with composite function of pdf and inverse cdf\n\nI am reading one paper https://arxiv.org/abs/1207.7209 In proposition 4.1 the author mentioned a fact $$p \\sqrt{k_1 \\log (1/p)} \\leq \\phi \\circ \\Phi^{-1}(p)$$ where $$k_1 = 1/2, p \\in (0, 1/2], \\phi, \\Phi^{-1}$$ are pdf and inverse cdf for standard normal distribution respectively.\n\nI'm not sure how this fact comes. Can anyone prove this inequality? Thanks!\n\nTips: the above inequality may follow from $$\\phi(x) - x\\bar{\\Phi}(x) \\geq 0$$ for $$x > 0$$, where $$\\bar{\\Phi}(x) = 1 - \\Phi(x)$$\n\n## 1 Answer\n\nThat is not what is claimed in the paper.\n\nThe paper says\n\nFor $$\\kappa_1=1/2$$, $$p\\in(0,1/2]$$, the fact that $$p\\sqrt{\\kappa_1\\log(1/p)}\\leq\\phi\\circ\\Phi^{\\leftarrow}(p)$$ follows from $$\\phi(x)-x\\overline{\\Phi}(x)\\geq 0$$ for $$x>0$$.\n\nwhere the notation $$\\Phi^{\\leftarrow}$$ is the (generalized) inverse to $$\\Phi$$ (immediately before Definition 2.4) and $$\\overline{\\Phi}$$ means $$1-\\Phi$$ (Definition 2.6) has been introduced previously.\n\n• yes, thanks! fixed the mistakes. Any ideas about how to prove? – Alyssa Jun 16 at 6:53"
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https://iris.unica.it/handle/11584/341532 | [
"Let Bn ⊂ ℝn and Sn ⊂ Rn+1 denote the Euclidean n-dimensional unit ball and sphere, respectively. The extrinsic k-energy functional is defined on the Sobolev space Wk,2 (Bn, Sn) as follows: Ekext(u) = ∫Bn |Δs u|2 dx when k = 2s, and Ekext(u) = ∫Bn|∇ Δs u|2 dx when k = 2s + 1. These energy functionals are a natural higher order version of the classical extrinsic bienergy, also called Hessian energy. The equator map u∗: Bn → Sn, defined by u∗(x) = (x/|x|,0), is a critical point of Ekext(u) provided that n ≥ 2k + 1. The main aim of this paper is to establish necessary and sufficient conditions on k and n under which u∗: Bn → Sn is minimizing or unstable for the extrinsic k-energy.\n\n### On the Stability of the Equator Map for Higher Order Energy Functionals\n\n#### Abstract\n\nLet Bn ⊂ ℝn and Sn ⊂ Rn+1 denote the Euclidean n-dimensional unit ball and sphere, respectively. The extrinsic k-energy functional is defined on the Sobolev space Wk,2 (Bn, Sn) as follows: Ekext(u) = ∫Bn |Δs u|2 dx when k = 2s, and Ekext(u) = ∫Bn|∇ Δs u|2 dx when k = 2s + 1. These energy functionals are a natural higher order version of the classical extrinsic bienergy, also called Hessian energy. The equator map u∗: Bn → Sn, defined by u∗(x) = (x/|x|,0), is a critical point of Ekext(u) provided that n ≥ 2k + 1. The main aim of this paper is to establish necessary and sufficient conditions on k and n under which u∗: Bn → Sn is minimizing or unstable for the extrinsic k-energy.\n##### Scheda breve Scheda completa Scheda completa (DC)",
null,
"2022\nHigher Order Energy; Stability; Equator Map\nFile in questo prodotto:\nFile\n2007.01509.pdf\n\naccesso aperto\n\nDescrizione: Main article\nTipologia: versione pre-print\nDimensione 226.72 kB\nIMRN-FMO.pdf\n\nSolo gestori archivio\n\nTipologia: versione editoriale\nDimensione 235.63 kB\nUtilizza questo identificativo per citare o creare un link a questo documento: `https://hdl.handle.net/11584/341532`\n•",
null,
"ND\n•",
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"1\n•",
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"1"
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https://answers.everydaycalculation.com/multiply-fractions/18-42-times-48-81 | [
"Solutions by everydaycalculation.com\n\n## Multiply 18/42 with 48/81\n\nThis multiplication involving fractions can also be rephrased as \"What is 18/42 of 48/81?\"\n\n18/42 × 48/81 is 16/63.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 18/42 × 48/81 = 18 × 48/42 × 81 = 864/3402\n3. After reducing the fraction, the answer is 16/63\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
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"https://answers.everydaycalculation.com/mathstep-app-icon.png",
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https://classes.engineering.wustl.edu/2009/spring/mase5513/abaqus/docs/v6.6/books/usb/pt03ch07s01aus40.html | [
"### 7.1.1 Solving nonlinear problems",
null,
"Products: ABAQUS/Standard ABAQUS/CAE\n\n### Overview",
null,
"Solving nonlinear problems in ABAQUS/Standard involves:\n\n• a combination of incremental and iterative procedures;\n\n• using the Newton method to solve the nonlinear equations;\n\n• determining convergence;\n\n• defining loads as a function of time; and\n\n• choosing suitable time increments automatically.\n\n### The solution of nonlinear problems",
null,
"The nonlinear load-displacement curve for a structure is shown in Figure 7.1.1–1.",
null,
"The objective of the analysis is to determine this response. In a nonlinear analysis the solution cannot be calculated by solving a single system of linear equations, as would be done in a linear problem. Instead, the solution is found by specifying the loading as a function of time and incrementing time to obtain the nonlinear response. Therefore, ABAQUS/Standard breaks the simulation into a number of time increments and finds the approximate equilibrium configuration at the end of each time increment. Using the Newton method, it often takes ABAQUS/Standard several iterations to determine an acceptable solution to each time increment.\n\n#### Steps, increments, and iterations\n\n• The time history for a simulation consists of one or more steps. You define the steps, which generally consist of an analysis procedure, loading, and output requests. Different loads, boundary conditions, analysis procedures, and output requests can be used in each step. For example:\n\nStep 1: Hold a plate between rigid jaws.\n\nStep 3: Find the natural frequencies of the deformed plate.\n\n• An increment is part of a step. In nonlinear analyses each step is broken into increments so that the nonlinear solution path can be followed. You suggest the size of the first increment, and ABAQUS/Standard automatically chooses the size of the subsequent increments. At the end of each increment the structure is in (approximate) equilibrium and results are available for writing to the restart, data, results, or output database files.\n\n• An iteration is an attempt at finding an equilibrium solution in an increment. If the model is not in equilibrium at the end of the iteration, ABAQUS/Standard tries another iteration. With every iteration the solution that ABAQUS/Standard obtains should be closer to equilibrium; however, sometimes the iteration process may diverge—subsequent iterations may move away from the equilibrium state. In that case ABAQUS/Standard may terminate the iteration process and attempt to find a solution with a smaller increment size.\n\n### Convergence",
null,
"Consider the external forces, P, and the internal (nodal) forces, I, acting on a body (see Figure 7.1.1–2(a) and Figure 7.1.1–2(b), respectively). The internal loads acting on a node are caused by the stresses in the elements that are attached to that node.\n\nFigure 7.1.1–2 Internal and external loads on a body.",
null,
"For the body to be in equilibrium, the net force acting at every node must be zero. Therefore, the basic statement of equilibrium is that the internal forces, I, and the external forces, P, must balance each other:",
null,
"The nonlinear response of a structure to a small load increment,",
null,
", is shown in Figure 7.1.1–3. ABAQUS/Standard uses the structure's tangent stiffness,",
null,
", which is based on its configuration at",
null,
", and",
null,
"to calculate a displacement correction,",
null,
", for the structure. Using",
null,
", the structure's configuration is updated to",
null,
".\n\nFigure 7.1.1–3 First iteration in an increment.",
null,
"ABAQUS/Standard then calculates the structure's internal forces,",
null,
", in this updated configuration. The difference between the total applied load, P, and",
null,
"can now be calculated as",
null,
"where",
null,
"is the force residual for the iteration.\n\nIf",
null,
"is zero at every degree of freedom in the model, point a in Figure 7.1.1–3 would lie on the load-deflection curve and the structure would be in equilibrium. In a nonlinear problem",
null,
"will never be exactly zero, so ABAQUS/Standard compares it to a tolerance value. If",
null,
"is less than this force residual tolerance at all nodes, ABAQUS/Standard accepts the solution as being in equilibrium. By default, this tolerance value is set to 0.5% of an average force in the structure, averaged over time. ABAQUS/Standard automatically calculates this spatially and time-averaged force throughout the simulation. You can change this, and all other such tolerances, by specifying solution controls (see Convergence criteria for nonlinear problems, Section 7.2.3).\n\nIf",
null,
"is less than the current tolerance value, P and",
null,
"are considered to be in equilibrium and",
null,
"is a valid equilibrium configuration for the structure under the applied load. However, before ABAQUS/Standard accepts the solution, it also checks that the last displacement correction,",
null,
", is small relative to the total incremental displacement,",
null,
". If",
null,
"is greater than a fraction (1% by default) of the incremental displacement, ABAQUS/Standard performs another iteration. Both convergence checks must be satisfied before a solution is said to have converged for that time increment.\n\nIf the solution from an iteration is not converged, ABAQUS/Standard performs another iteration to try to bring the internal and external forces into balance. First, ABAQUS/Standard forms the new stiffness,",
null,
", for the structure based on the updated configuration,",
null,
". This stiffness, together with the residual",
null,
", determines another displacement correction,",
null,
", that brings the system closer to equilibrium (point b in Figure 7.1.1–4).\n\nFigure 7.1.1–4 Second iteration.",
null,
"ABAQUS/Standard calculates a new force residual,",
null,
", using the internal forces from the structure's new configuration,",
null,
". Again, the largest force residual at any degree of freedom,",
null,
", is compared against the force residual tolerance, and the displacement correction for the second iteration,",
null,
", is compared to the increment of displacement,",
null,
". If necessary, ABAQUS/Standard performs further iterations. For more details on convergence in ABAQUS/Standard, see Convergence criteria for nonlinear problems, Section 7.2.3.\n\nFor each iteration in a nonlinear analysis ABAQUS/Standard forms the model's stiffness matrix and solves a system of equations. Therefore, the computational cost of each iteration is close to the cost of conducting a complete linear analysis, making the computational expense of a nonlinear analysis potentially many times greater than the cost of a linear analysis. Since it is possible with ABAQUS/Standard to save results at each converged increment, the amount of output data available from a nonlinear simulation can also be much greater than that available from a linear analysis of the same geometry.\n\n### Automatic incrementation control",
null,
"By default, ABAQUS/Standard automatically adjusts the size of the time increments to solve nonlinear problems efficiently. You need to suggest only the size of the first increment in each step of the simulation, after which ABAQUS/Standard automatically adjusts the size of the increments. If you do not provide a suggested initial increment size, ABAQUS/Standard will attempt to apply all of the loads defined in the step in a single increment. For highly nonlinear problems ABAQUS/Standard will have to reduce the increment size repeatedly to obtain a solution, resulting in wasted CPU time. It is advantageous to provide a reasonable initial increment size because only in mildly nonlinear problems can all of the loads in a step be applied in a single increment.\n\nThe number of iterations needed to find a converged solution for a time increment will vary depending on the degree of nonlinearity in the system. With the default incrementation control, the procedure works as follows. If the solution has not converged within 16 iterations or if the solution appears to diverge, ABAQUS/Standard abandons the increment and starts again with the increment size set to 25% of its previous value. It then attempts to find a converged solution with this smaller time increment. If the solution still fails to converge, ABAQUS/Standard reduces the increment size again. This process is continued until a solution is found. If the time increment becomes smaller than the minimum you defined or more than 5 attempts are needed, ABAQUS/Standard stops the analysis.\n\nIf the increment converges in fewer than 5 iterations, this indicates that the solution is being found fairly easily. Therefore, ABAQUS/Standard automatically increases the increment size by 50% if 2 consecutive increments require fewer than 5 iterations to obtain a converged solution.\n\nWhile the default automatic incrementation control is suitable for most analyses, you can change all the defaults when necessary by specifying solution controls; see Commonly used control parameters, Section 7.2.2, and Time integration accuracy in transient problems, Section 7.2.4.\n\n### Automatic stabilization of static problems",
null,
"Nonlinear static problems can be unstable. Such instabilities may be of a geometrical nature, such as buckling, or of a material nature, such as material softening. If the instability manifests itself in a global load-displacement response with a negative stiffness, the problem can be treated as a buckling or collapse problem as described in Unstable collapse and postbuckling analysis, Section 6.2.4. However, if the instability is localized, there will be a local transfer of strain energy from one part of the model to neighboring parts, and global solution methods may not work. This class of problems has to be solved either dynamically or with the aid of (artificial) damping; for example, by using dashpots.\n\nABAQUS/Standard provides an automatic mechanism for stabilizing unstable quasi-static problems through the automatic addition of volume-proportional damping to the model. The mechanism is triggered by including automatic stabilization in any nonlinear quasi-static procedure. Viscous forces of the form",
null,
"are added to the global equilibrium equations",
null,
"where",
null,
"is an artificial mass matrix calculated with unity density, c is a damping factor,",
null,
"is the vector of nodal velocities, and",
null,
"is the increment of time (which may or may not have a physical meaning in the context of the problem being solved).\n\nFor the case of static stabilization the mass matrix for Timoshenko beams is always calculated assuming isotropic rotary inertia, regardless of the type of rotary inertia specified for the beam section definition (Rotary inertia for Timoshenko beams” in “Beam section behavior, Section 23.3.5).\n\nAutomatic stabilization does not carry over automatically to subsequent steps. It needs to be declared for any step in which you want it to be active. ABAQUS/Standard recalculates new values for the damping factor, based on the declared damping intensity and on the solution of the first increment of the step. Therefore, unless you specify the same damping factor directly (see “Using the damping factor” below), an analysis with an unstable step may produce slightly different results from the same analysis with the original step split into two steps. Moreover, if the instabilities in the model have not subsided by the end of a step, viscous forces may be terminated abruptly or modified at the beginning of subsequent steps, potentially causing convergence difficulties if automatic stabilization is not used in the subsequent step. If such a situation arises, it is recommended that the problem be restarted with the damping factor set equal to the value chosen by ABAQUS/Standard (or to the value you specified) in the previous step. This value is printed in the message (.msg) file for the previous step. If it is necessary to have an accurate static equilibrium solution after an instability has occurred (and the model's behavior has returned to a stable regime), the step with automatic stabilization can be followed by a step without such stabilization.\n\n#### Using the dissipated energy fraction\n\nIt is assumed that the problem is stable at the beginning of the step and that instabilities may develop in the course of the step. While the model is stable, viscous forces and, therefore, the viscous energy dissipated are very small. Thus, the additional artificial damping has no effect. If a local region goes unstable, the local velocities increase and, consequently, part of the strain energy then released is dissipated by the applied damping. ABAQUS/Standard can, if necessary, reduce the time increment to permit the process to occur without the unstable response causing very large displacements. ABAQUS/Standard calculates and prints to the message file the damping factor, c, based on the solution of the first increment of a step. In most applications the first increment of the step is stable without the need to apply damping. The damping factor is then determined in such a way that the extrapolated dissipated energy for the step is a small fraction of the extrapolated strain energy. The fraction is called the dissipated energy fraction and has a default value of 2.0 × 10–4.\n\nAlternatively, you can specify the dissipated energy fraction for automatic stabilization directly.\n\n Input File Usage: Use any of the following options to activate automatic stabilization with the default dissipated energy fraction: ```*COUPLED TEMPERATURE-DISPLACEMENT, STABILIZE *SOILS, STABILIZE *STATIC, STABILIZE *STEADY STATE TRANSPORT, STABILIZE *VISCO, STABILIZE```Use any of the following options to specify a nondefault dissipated energy fraction:```*COUPLED TEMPERATURE-DISPLACEMENT, STABILIZE=dissipated energy fraction *SOILS, STABILIZE=dissipated energy fraction *STATIC, STABILIZE=dissipated energy fraction *STEADY STATE TRANSPORT, STABILIZE=dissipated energy fraction *VISCO, STABILIZE=dissipated energy fraction```\n\n ABAQUS/CAE Usage: Step module: Create Step: General: any valid step type: Basic: toggle on Use stabilization with, and select dissipated energy fraction\n\n##### Ensuring that an accurate solution is obtained\n\nThe procedure of computing an appropriate damping factor works well for many applications. However, there are cases where the computed damping factor is either too small, thus not controlling the instability, or too high, thus leading to inaccurate results. This occurs when a poor estimation of the extrapolated strain energy is made during the first increment. For example, consider a sequentially coupled thermal-stress analysis in which a mechanical analysis reads temperatures from a previous transient thermal analysis. Typically the thermal analysis exhibits a diffusive process, where rapid changes in temperature occurs early in the analysis and minor changes in temperature occur once steady state is reached. In such a case ABAQUS will compute the extrapolated strain energy based on the temperatures corresponding to the time of the first increment (in this case there may be a significant change in temperature for the first increment), thus yielding a larger then expected extrapolated strain energy. This in turn leads to a damping factor that is too large, resulting in inaccurate results.\n\nWith automatic stabilization, check the following to ensure that accurate solutions are obtained:\n\n• Check the damping factor printed to the message (.msg) file at the end of the first increment to ensure that a reasonable amount of damping is applied. Unfortunately, the damping factor is problem dependent; therefore, you must rely on experience from previous runs.\n\n• Compare the viscous forces (VF) with the overall forces in the analysis, and ensure that the viscous forces are relatively small compared with the overall forces in the model.\n\n• Compare the viscous damping energy (ALLSD) with the total strain energy (ALLSE), and ensure that the ratio does not exceed the dissipated energy fraction or any reasonable amount. The viscous damping energy may be large if the structure undergoes a large amount of motion.\n\nIf the computed damping factor is inappropriate, specify a damping factor directly as discussed below.\n\n#### Using the damping factor\n\nThere are cases where the first increment is either unstable or singular (due to a rigid body mode). In such cases it is not possible to obtain a solution to the first increment without applying some damping. Therefore, some damping is already applied during the first increment. The damping factor used for the initial increment is chosen such that the average element damping matrix component, divided by the step time, is equal to the average element stiffness matrix component multiplied by the dissipated energy fraction. If the calculated strain energy change in this increment indicates that the solution without damping is stable, the damping factor is recalculated based upon the energy method described in the previous paragraph. However, if the strain energy change indicates that the solution is unstable or singular, the initially calculated damping factor is maintained, and a warning message is issued indicating that the amount of damping applied may not be appropriate. In many cases the amount of damping may actually be rather large, which can affect the solution in ways that are not desirable. Therefore, if the above mentioned warning message is issued, check the viscous forces (VF) and compare them with the expected nodal forces to make sure that the viscous forces do not dominate the solution. If necessary, follow the stabilized step with another step in which stabilization is not used or with a step in which a much smaller damping factor is used.\n\nAlternatively, you can specify the damping factor directly. Unfortunately, it is quite difficult to make a reasonable estimate for the damping factor, unless a value is known from the output of previous runs; the damping factor includes information not only about the amount of damping but also about mesh size and material behavior.\n\n Input File Usage: Use any of the following options to specify the damping factor directly: ```*COUPLED TEMPERATURE-DISPLACEMENT, STABILIZE, FACTOR=damping factor *SOILS, STABILIZE, FACTOR=damping factor *STATIC, STABILIZE, FACTOR=damping factor *STEADY STATE TRANSPORT, STABILIZE, FACTOR=damping factor *VISCO, STABILIZE, FACTOR=damping factor```\n\n ABAQUS/CAE Usage: Step module: Create Step: General: Coupled temp-displacement, Soils, Static, General, or Visco: Basic: toggle on Use stabilization with, and select damping factor"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.876417,"math_prob":0.9386092,"size":16660,"snap":"2022-27-2022-33","text_gpt3_token_len":3392,"char_repetition_ratio":0.15856148,"word_repetition_ratio":0.026771653,"special_character_ratio":0.1852341,"punctuation_ratio":0.11577522,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9807856,"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],"im_url_duplicate_count":[null,null,null,null,null,null,null,2,null,null,null,2,null,2,null,4,null,2,null,8,null,4,null,8,null,8,null,6,null,2,null,6,null,6,null,2,null,null,null,null,null,null,null,null,null,null,null,6,null,6,null,8,null,2,null,8,null,2,null,6,null,null,null,4,null,2,null,4,null,2,null,4,null,4,null,2,null,null,null,null,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-03T18:44:58Z\",\"WARC-Record-ID\":\"<urn:uuid:c88f9793-aa5b-4335-8c7d-4a75e4d18e26>\",\"Content-Length\":\"33966\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:72b133d6-9453-45d8-8add-7ef5253127df>\",\"WARC-Concurrent-To\":\"<urn:uuid:2ed51c72-3057-43a7-9f51-705f9f1754f5>\",\"WARC-IP-Address\":\"128.252.202.244\",\"WARC-Target-URI\":\"https://classes.engineering.wustl.edu/2009/spring/mase5513/abaqus/docs/v6.6/books/usb/pt03ch07s01aus40.html\",\"WARC-Payload-Digest\":\"sha1:ISMQLQW3YMRTYXYTI5ZF52J7AF6FNFAG\",\"WARC-Block-Digest\":\"sha1:NL6XXPGIQDFP3SJWRNPF2Y352RVPYD43\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104248623.69_warc_CC-MAIN-20220703164826-20220703194826-00220.warc.gz\"}"} |
https://leconjugueur.lefigaro.fr/conjugaison/anglais/would+exploit.html | [
"",
null,
"Verbe anglais à conjuguer : Modal : aucun | may | might | can | could | shall | should | will | would | must | ought to\n\n# Conjugaison du verbe anglais WOULD EXPLOIT\n\nVerbe régulier : exploit - exploited - exploited\n\n## Affirmation\n\nForme simple\n\nI would exploit\nyou would exploit\nhe would exploit\nwe would exploit\nyou would exploit\nthey would exploit\n\nForme en V-ing\n\nI would be exploiting\nyou would be exploiting\nhe would be exploiting\nwe would be exploiting\nyou would be exploiting\nthey would be exploiting\n\nPerfect\n\nI would have exploited\nyou would have exploited\nhe would have exploited\nwe would have exploited\nyou would have exploited\nthey would have exploited\n\nPerfect en V-ing\n\nI would have been exploiting\nyou would have been exploiting\nhe would have been exploiting\nwe would have been exploiting\nyou would have been exploiting\nthey would have been exploiting\n\n## Négation\n\nForme simple\n\nI would not exploit\nyou would not exploit\nhe would not exploit\nwe would not exploit\nyou would not exploit\nthey would not exploit\n\nForme en V-ing\n\nI would not be exploiting\nyou would not be exploiting\nhe would not be exploiting\nwe would not be exploiting\nyou would not be exploiting\nthey would not be exploiting\n\nPerfect\n\nI would not have exploited\nyou would not have exploited\nhe would not have exploited\nwe would not have exploited\nyou would not have exploited\nthey would not have exploited\n\nPerfect en V-ing\n\nI would not have been exploiting\nyou would not have been exploiting\nhe would not have been exploiting\nwe would not have been exploiting\nyou would not have been exploiting\nthey would not have been exploiting\n\n## Interrogation\n\nForme simple\n\nwould I exploit?\nwould you exploit?\nwould he exploit?\nwould we exploit?\nwould you exploit?\nwould they exploit?\n\nForme en V-ing\n\nwould I be exploiting?\nwould you be exploiting?\nwould he be exploiting?\nwould we be exploiting?\nwould you be exploiting?\nwould they be exploiting?\n\nPerfect\n\nwould I have exploited?\nwould you have exploited?\nwould he have exploited?\nwould we have exploited?\nwould you have exploited?\nwould they have exploited?\n\nPerfect en V-ing\n\nwould I have been exploiting?\nwould you have been exploiting?\nwould he have been exploiting?\nwould we have been exploiting?\nwould you have been exploiting?\nwould they have been exploiting?\n\n## Interro-négation\n\nForme simple\n\nwould I not exploit?\nwould you not exploit?\nwould he not exploit?\nwould we not exploit?\nwould you not exploit?\nwould they not exploit?\n\nForme en V-ing\n\nwould I not be exploiting?\nwould you not be exploiting?\nwould he not be exploiting?\nwould we not be exploiting?\nwould you not be exploiting?\nwould they not be exploiting?\n\nPerfect\n\nwould I not have exploited?\nwould you not have exploited?\nwould he not have exploited?\nwould we not have exploited?\nwould you not have exploited?\nwould they not have exploited?\n\nPerfect en V-ing\n\nwould I not have been exploiting?\nwould you not have been exploiting?\nwould he not have been exploiting?\nwould we not have been exploiting?\nwould you not have been exploiting?\nwould they not have been exploiting?"
]
| [
null,
"https://w5c2n4b4.stackpathcdn.com/images/logo-EN.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9834279,"math_prob":0.7027098,"size":3190,"snap":"2021-43-2021-49","text_gpt3_token_len":718,"char_repetition_ratio":0.3979912,"word_repetition_ratio":0.30427048,"special_character_ratio":0.20846395,"punctuation_ratio":0.08275862,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95812786,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-02T16:43:14Z\",\"WARC-Record-ID\":\"<urn:uuid:0fdef1f4-fda2-4a4f-8551-97c79228c35c>\",\"Content-Length\":\"65298\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:35d155a3-cc80-41ad-9b18-32fa78c5eda5>\",\"WARC-Concurrent-To\":\"<urn:uuid:25bd7870-35da-4a73-8b5d-c5996edc4be5>\",\"WARC-IP-Address\":\"31.193.142.228\",\"WARC-Target-URI\":\"https://leconjugueur.lefigaro.fr/conjugaison/anglais/would+exploit.html\",\"WARC-Payload-Digest\":\"sha1:SPHOMCXZEC4ZNVXUMMTAHBW3V6IZ2IDR\",\"WARC-Block-Digest\":\"sha1:II3QYBXCODJEH7J7PB5JZM5AMW3M7JGI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362230.18_warc_CC-MAIN-20211202145130-20211202175130-00383.warc.gz\"}"} |
https://an-com.org/37-kp-expression-pictures/ | [
"37+ Kp Expression\nFootage\n. The balanced equation for the response. When kp = (the partial strain of the merchandise raised to exponents equal to their respective coefficients within the equation) / (the partial strain of the the expression of a generic response is"
]
| [
null
]
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https://www.delftstack.com/howto/cpp/check-if-input-is-integer-cpp/ | [
"# Check if Input Is Integer in C++\n\nThis article will demonstrate multiple methods about how to check if the input is an integer in C++.\n\n## Use the `std::find_if` Algorithm to Check if Input Is Integer in C++\n\n`std::find_if` is part of the STL algorithms library defined in the `<alogrithm>` header file, and it can be utilized to search for the specific element in the range. Since the user input is most likely to be a string, we will assume that the input data is stored in a `std::string` object. Note that we implement a function called `isNumber` that takes a reference to `std::string` and returns the`bool` value.\n\nThe prototype of the `std::find_if` function that we used in the following example takes three arguments, the first two of which specify the range elements - `[first, last]`. The third argument is a unary predicate, which is a lambda function that returns the `bool` value from evaluating the inverted value of the `isdigit` function. On the outer layer, the `std::find_if` return value is compared to the `str.end()`, as the `true` value of the expression indicates that no non-digit character was found; thus it’s the number. Additionally, we logically AND the former expression with the `!str.empty` to indicate that string is empty and designate it with `false` return value.\n\n``````#include <iostream>\n#include <string>\n#include <algorithm>\n\nusing std::cout; using std::cin;\nusing std::endl; using std::string;\n\nbool isNumber(const string& str)\n{\nreturn !str.empty() &&\nstd::find_if(str.begin(), str.end(),\n[](unsigned char c) { return !std::isdigit(c); }) == str.end();\n}\n\nint main(){\nstring str1 = \"12345.\";\nstring str2 = \"12312\";\nstring str3 = \"123142.2\";\n\nisNumber(str1) ? cout << \"Number\\n\" : cout << \"Not number\\n\";\nisNumber(str2) ? cout << \"Number\\n\" : cout << \"Not number\\n\";\nisNumber(str3) ? cout << \"Number\\n\" : cout << \"Not number\\n\";\n\nexit(EXIT_SUCCESS);\n}\n``````\n\nOutput:\n\n``````Not number\nNumber\nNot number\n``````\n\n## Use the `std::string::find_first_not_of` Function to Check if Input Is Integer in C++\n\nAlternatively, we can reimplement `isNumber` function using the `find_first_not_of` method that is built-in for the `std::string` object. `find_first_not_of` can take the string value and find the first character that equals none of the characters in the string sequence. If the function does not find such a character, `string::npos` is returned. Thus, we specify all ten decimal digits as `find_first_not_of` argument and check its equality to `npos`, as the expression value is returned from the function.\n\n``````#include <iostream>\n#include <string>\n#include <algorithm>\n\nusing std::cout; using std::cin;\nusing std::endl; using std::string;\n\nbool isNumber(const string& str)\n{\nreturn str.find_first_not_of( \"0123456789\" ) == string::npos;\n}\n\nint main(){\nstring str1 = \"12345.\";\nstring str2 = \"12312\";\nstring str3 = \"123142.2\";\n\nisNumber(str1) ? cout << \"Number\\n\" : cout << \"Not number\\n\";\nisNumber(str2) ? cout << \"Number\\n\" : cout << \"Not number\\n\";\nisNumber(str3) ? cout << \"Number\\n\" : cout << \"Not number\\n\";\n\nexit(EXIT_SUCCESS);\n}\n``````\n\nOutput:\n\n``````Not number\nNumber\nNot number\n``````\n\n## Use the `std::string::find_first_not_of` Function to Check if Input Is Integer in C++\n\nNote though, the previous method does not identify the real numbers and treats them as illegitimate. So, we can add `.` character to the string and allow the function to recognize any sequence of digits with a dot symbol as a valid number. We need to eliminate two cases when the `.` character is the first and the last symbol in the input sequence, which by our convention would not be a valid real number. We can use `string` built-in methods `front` and `back` to verify that input does not begin/end with a dot symbol. Finally, we logically AND all three expressions with each other and return that value.\n\n``````#include <iostream>\n#include <string>\n#include <algorithm>\n\nusing std::cout; using std::cin;\nusing std::endl; using std::string;\n\nbool isNumber3(const string& str)\n{\nreturn str.find_first_not_of( \".0123456789\" ) == string::npos &&\nstr.front() != '.' && str.back() != '.';\n}\n\nint main(){\nstring str1 = \"12345.\";\nstring str2 = \"12312\";\nstring str3 = \"123142.2\";\n\nisNumber(str1) ? cout << \"Number\\n\" : cout << \"Not number\\n\";\nisNumber(str2) ? cout << \"Number\\n\" : cout << \"Not number\\n\";\nisNumber(str3) ? cout << \"Number\\n\" : cout << \"Not number\\n\";\n\nexit(EXIT_SUCCESS);\n}\n``````\n\nOutput:\n\n``````Not number\nNumber\nNumber\n``````\nContribute\nDelftStack is a collective effort contributed by software geeks like you. If you like the article and would like to contribute to DelftStack by writing paid articles, you can check the write for us page.\n\n## Related Article - C++ Int\n\n• Convert Int to Char Array in C++\n• Convert Int to ASCII Char in C++"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6258167,"math_prob":0.962886,"size":4542,"snap":"2021-04-2021-17","text_gpt3_token_len":1134,"char_repetition_ratio":0.17276333,"word_repetition_ratio":0.36953455,"special_character_ratio":0.28577718,"punctuation_ratio":0.20372398,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99267703,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-21T08:20:27Z\",\"WARC-Record-ID\":\"<urn:uuid:87e106d2-c461-4cad-a443-f19e33cfdac3>\",\"Content-Length\":\"40368\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2a6757d4-c808-4615-8e4b-f24dbaa6d7aa>\",\"WARC-Concurrent-To\":\"<urn:uuid:60a5643d-770f-4b00-9039-5f04c9aa9c99>\",\"WARC-IP-Address\":\"3.234.104.255\",\"WARC-Target-URI\":\"https://www.delftstack.com/howto/cpp/check-if-input-is-integer-cpp/\",\"WARC-Payload-Digest\":\"sha1:CIDTBSRFL3EUHBACG4PHJ67B7UBJQ7XV\",\"WARC-Block-Digest\":\"sha1:VRI6XZRTRICDDG5SLB2EAMWKRZJ5SAGE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039526421.82_warc_CC-MAIN-20210421065303-20210421095303-00038.warc.gz\"}"} |
https://www.skillsugar.com/how-to-generate-random-numbers-in-javascript | [
"# How to Generate Random Numbers in JavaScript",
null,
"We can generate random numbers in JavaScript by using the `Math.random()` function. In this tutorial, we will learn how to get random boolean values and then expand on this principle to get random integers between two values.\n\n## Random Boolean Values\n\nThe `Math.random()` method returns random decimals somewhere between 0 and 1. To make the number an integer we can then use the `Math.round()` method which will round the decimal to the nearest integer.\n\n``````var random_bool = Math.round(Math.random());\n\nconsole.log(random_bool);\n``````\n``````1\n``````\n\nIn the example above we are getting the output from `Math.random()` and supplying it directly `Math.round()` and storing the final result in a variable called `random_bool`.\n\n## Random Integer Between 0 and a Maximum Value\n\nBecause `Math.random()` returns random decimals, we can multiply that result by the maximum value and round it to a whole number. Let's generate a random number between 0 and 10.\n\n``````var random_number = Math.round(Math.random() * 10);\n\nconsole.log(random_number);\n``````\n``````7\n``````\n\n## Random Number Between Two Integers\n\nTo get a random number between two values we will create a function that excepts a minimum and a maximum value to get a random integer from.\n\n``````function getRandomNumber(min, max) {\nreturn Math.round(Math.random() * (max - min) + min);\n}\n\nvar min = 10;\nvar max = 20;\n\nrandom_number = getRandomNumber(min, max);\n\nconsole.log(random_number);\n``````\n\nIn the `getRandomNumber()` function, we are multiplying the result of `Math.random()` by `max` minus the `min `then adding `min` to the result before rounding it to a whole number. The minimum and maximum numbers can be anything, even negative values.\n\n## Conclusion\n\nYou now know how to generate random numbers in JavaScript in a variety of different ways depending on your needs.\n\n#### Related Tutorials",
null,
"### How to Generate Random Numbers in Python\n\nSeptember 20, 2020",
null,
"### How to get Random Elements from an Array in JavaScript\n\nSeptember 13, 2020",
null,
"### How to Generate a Random String in Python\n\nJune 17, 2021",
null,
"### How to Generate Random Passwords in Python\n\nFebruary 11, 2022",
null,
"### How to Generate a Random Number of Length N in Python\n\nFebruary 10, 2022",
null,
"June 16, 2021"
]
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null,
"https://www.skillsugar.com/media/image/thumb/poster-default.png",
null,
"https://www.skillsugar.com/media/image/thumb/poster-default.png",
null,
"https://www.skillsugar.com/media/image/thumb/poster-default.png",
null,
"https://www.skillsugar.com/media/image/thumb/poster-default.png",
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"https://www.skillsugar.com/media/image/thumb/poster-default.png",
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"https://www.skillsugar.com/media/image/thumb/poster-default.png",
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https://discuss.interviewbit.com/t/shortest-sweetest-solution/42051 | [
"",
null,
"# Shortest+sweetest solution\n\n#1\n\nUsing binary search. I will approach to the solution gradually. Scroll to the last if you just wanna see the solution quickly.\n\nUnderstand that binary search is just a fast way to do trial and error. If value is too low to be a possible answer, try bigger value. If value is too high, try smaller value.\n\nUse binary search because it saves you from hassling with complicated calculations. Now luckily the time constraints were easy to do binary search. If the time constraints were stricter, you will have to do the calculation manually (as done in the official solutions) instead of doing the trial and error(using binary search).\n\nHere we go:\n\n• In all problems of binary search, first think of target. Our target is to find maximum possible number of fruit basket.\n\n• Binary search is trial and error. So you will want to set the range of numbers you want to try and check. int left, and int right are used to defined the lower limit and upper limit of the range.\n\n• Decide the value of left (lower limit of target(target=no of fruit baskets) = 0)\n\n• Decide the value of right(upper limit of target(target = no of fruit baskets) = explained below how to find this).\n\nExplanation to find upper limit of number of fruit baskets\nYou can use possible upper limits. It can be `right = a` or `right = b` or `right = a + 2b` or `right = 2a + 3b`. But higher values will take more time to execute.\nBut if you think smartly, the number of baskets cannot exceed the value of a and b.\nso `right < a` and `right < b`. So we decide to set the value of right = min(a,b).\n\nSo far I have covered and variables required to setup the binary search. Now i will discuss the process.\n\n• Lets say `mid = (left + right)/2`\nSo we are trying `mid` number of baskets , and see if it possible.\n\n• If there are `mid` baskets, this means there are `a` apples, `b` bananas already inside them. So each basket already has 1 banana, 1 apple. Now the remaining apples bananas and cherries left are `a-mid`, `b-mid`, and `c`.\n\n• These remaining fruits must fill all the baskets such that all baskets have 3 fruits. Since all baskets already have 2 fruits (1 banana and 1 apple), We need 1 extra fruit per basket.\n`if(a-mid+b-mid+c>=mid && a1>=0 && b>=0)`\n`{`\n`low=mid+1;`\n`ans=mid;`\n`}`\n\n• So the condition is fruits left`(a-mid + b-mid + c) >=` fruits needed (`mid`).\n\n• If this condition is satisfied, that means these number of baskets are possible to exist. Since we need max no of baskets, lets try higher number of baskets (`low = mid + 1`). This means set the trial and error range to `mid + 1` to `right`.\n\n• If condition is not satisfied, try lower values. `high = mid-1` does this.\nPosting the entire code for reference.\n\n`if(a==0 || b==0)`\n`return 0;`\n\n• Dont forget these base cases mentioned above.\n\n`int low=1, high=min(a,b),ans=0;`\n`while(low <= high)`\n`{`\n`int mid=low+(high-low)/2;`\n`int a1=a-mid, b1=b-mid;`\n`if(a1+b1+c>=mid && a1>=0 && b>=0)`\n`{`\n`low=mid+1;`\n`ans=mid;`\n`}`\n`else`\n`high=mid-1;`\n\n`}`\n`return ans;`\n\nSorry for the indendation issues, i tried figuring out but to vain.\n\nIf you liked my effort, please give me an endorsement in C++ and algorithms on my LikedIn Profile:"
]
| [
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"https://discuss-files.s3.dualstack.us-west-2.amazonaws.com/original/4X/9/6/6/966130bf03a57db2de73897c626d264b27cc950f.png",
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https://wiki.cns.iu.edu/pages/viewpage.action?pageId=2198690&navigatingVersions=true | [
"##### Child pages\n• Degree Distribution\nGo to start of banner\n\n# Degree Distribution\n\nYou are viewing an old version of this page. View the current version.\n\nVersion 2\n\n###### Description\n\nThe algorithm builds the histogram of the values of the degree of all nodes, which will be delivered in the two output files.\n\n###### Pros & Cons\n\nThe network to analyze must be undirected, otherwise there are no special constraints.\n\n###### Applications\n\nBasic analysis tool, not particular for special disciplines or problems.\n\n###### Implementation Details\n\nThe algorithm requires two inputs, the file where the edges of the network are listed and the number of points one wishes to have in the binned distribution described below. A first read-in of the inputfile will set the values of the number of nodes and edges of the network. In the second read-in the degrees of all nodes will be calculated. Then the distribution is calculated.\nThe program generates two output files, corresponding to two different ways of partitioning the interval spanned by the values of degree. In the first output, the occurrence of any degree value between the minimum and the maximum is estimated and divided by the number of nodes of the network, so to obtain the probability: the output displays all degree values in the interval with their probabilities.\nThe second output gives the binned distribution, i.e. the interval spanned by the values of degree is divided into bins whose size grows while going to higher values of the variable. The size of each bin is obtained by multiplying by a fixed number the size of the previous bin. The program calculates the fraction of nodes whose degree falls within each bin. Because of the different sizes of the bins, these fractions must be divided by the respective bin size, to have meaningful averages.\nThis technique is particularly suitable to study skewed distributions: the fact that the size of the bins grows large for large degree values compensates for the fact that not many nodes have high degree values, so it suppresses the fluctuations that one would observe by using bins of equal size. On a double logarithmic scale, which is very useful to determine the possible power law behavior of the distribution, the points of the latter will appear equally spaced on the x-axis.\nThe program runs in a time O(m), m being the number of edges of the network.\n\n###### Usage Hints\n\nA simple application of this algorithm could be to calculate the degree distribution of networks created by the modeling algorithms of the NWB. For instance, the network file can be created through the Barabasi-Albert model.\n\n###### Acknowledgements\n\nThe algorithm was implemented and documented by S. Fortunato, integrated by S. Fortunato and W. Huang.\n\n###### References\n\nAlbert, R., and Barabasi, A.-L. (2002) Statistical mechanics of complex networks. Review of Modern Physics 74:47-97.\n\nNewman, M.E.J. (2003) The structure and function of complex networks. SIAM Review 45:167-256.\n\nPastor-Satorras, R., Vespignani, A. (2004) Evolution and Structure of the Internet. Cambridge University Press.\n\nBoccaletti, S., Latora, V., Moreno, Y.,Chavez, M., Hwang, D.-U. (2006) Complex networks: Structure and dynamics. Physics Reports 424: 175-308.\n\n• No labels"
]
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https://math.stackexchange.com/questions/80975/sum-of-2-two-digits-numbers | [
"# Sum of $2$ two-digits numbers\n\n$X$ and $Y$ are two-digit numbers. If $Y=2X+2$ and $Y=2X$ in decimal and octal system respectively, and unit digits of $X$ and $Y$ are $5$ and $2$ respectively, then how to find $X+Y$ in decimal number system?\n\nMy attempt:\n\nI tried representing the two numbers in decimal as $(10a+5,10b+2)$ and in octal as $(8a+5,8b+2)$ and then tried to manipulate with according to the conditions $Y=2X+2$ and $Y=2X$, but they only give me one equation $b-2a=1$, how to get $10a+10b+7$(the sum of $X+Y$ in decimals) from these?\n\n• Equations are nice, but why not do a brute force search? You will be able to eliminate possibilities quite quickly. – André Nicolas Nov 11 '11 at 0:18\n• Probably $X=(10a+5)_{10} \\ne (8a+b)_8$ because the a's can be different. – GarouDan Nov 11 '11 at 0:30\n• @AndréNicolas: I find 15, 25, or 35 for X with 32, 52, or 72 for Y (respectively) all work. – Ross Millikan Nov 11 '11 at 0:40\n• @GarouDan: $b$ it the leading digit of $Y$, so the relations are $10b+2=2(10a+5)+2$ and $8b+2=2(8a+5)$, which are redundant as MaX says. – Ross Millikan Nov 11 '11 at 0:42\n\nDo you mean \"unit digits of $X$ and $Y$ are $5$ and $2$ respectively\"? I agree with you that $a$ can be any of $1,2, \\text{or } 3$ (no higher or $Y$ will carry in octal) and there is no single answer.\n\nEdit: (shortened) We are told that the (decimal) units digit of $X$ is $5$. So the only candidates for $X$ are $15$, $25$, and $35$. (Anything bigger, when expressed in octal, then doubled, is not a two-digit octal number.) The specification that $Y$ has units digit $2$ is superfluous.\n\nCheck which ones of $15$, $25$, and $35$ work. They all do.\n\nFor a problem in which the numbers are so nearly pinned down, trying to use \"algebra\" can be a waste of time. Before introducing symbols, it is useful to play with the numbers to get a concrete grip on the problem.\n\n• How$\\space Y<64$? – Quixotic Nov 11 '11 at 0:52\n• @MaX: It isn't! Answer corrected. – André Nicolas Nov 11 '11 at 1:01\n• +1,I like your approach but actually the answer is what I and Ross got there is not unique solution.Thanks :) – Quixotic Nov 12 '11 at 5:11\n• Sorry I didn't clarified myself before, so yes I noticed that, but you are not supporting the algebraic approach which according to the official solution is the best method. However, your answer has it's own value as it invokes lateral thinking than the usual algebra. – Quixotic Nov 12 '11 at 5:51\n• To deal with a big problem, we may need to devise a \"general\" approach. But this is a very small problem. Analogy: to solve the problem roughly how many primes are there less than $10^{30}$, we need to develop general theory. To solve the problem of roughly how many primes are less than $40$, we get our hands dirty and count. – André Nicolas Nov 12 '11 at 6:12"
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https://jeremykun.com/2018/03/05/earthmover-distance/?like_comment=60622&_wpnonce=4fd3b801c5 | [
"# Earthmover Distance\n\nProblem: Compute distance between points with uncertain locations (given by samples, or differing observations, or clusters).\n\nFor example, if I have the following three “points” in the plane, as indicated by their colors, which is closer, blue to green, or blue to red?",
null,
"It’s not obvious, and there are multiple factors at work: the red points have fewer samples, but we can be more certain about the position; the blue points are less certain, but the closest non-blue point to a blue point is green; and the green points are equally plausibly “close to red” and “close to blue.” The centers of masses of the three sample sets are close to an equilateral triangle. In our example the “points” don’t overlap, but of course they could. And in particular, there should probably be a nonzero distance between two points whose sample sets have the same center of mass, as below. The distance quantifies the uncertainty.",
null,
"All this is to say that it’s not obvious how to define a distance measure that is consistent with perceptual ideas of what geometry and distance should be.\n\nSolution (Earthmover distance): Treat each sample set",
null,
"$A$ corresponding to a “point” as a discrete probability distribution, so that each sample",
null,
"$x \\in A$ has probability mass",
null,
"$p_x = 1 / |A|$. The distance between",
null,
"$A$ and",
null,
"$B$ is the optional solution to the following linear program.\n\nEach",
null,
"$x \\in A$ corresponds to a pile of dirt of height",
null,
"$p_x$, and each",
null,
"$y \\in B$ corresponds to a hole of depth",
null,
"$p_y$. The cost of moving a unit of dirt from",
null,
"$x$ to",
null,
"$y$ is the Euclidean distance",
null,
"$d(x, y)$ between the points (or whatever hipster metric you want to use).\n\nLet",
null,
"$z_{x, y}$ be a real variable corresponding to an amount of dirt to move from",
null,
"$x \\in A$ to",
null,
"$y \\in B$, with cost",
null,
"$d(x, y)$. Then the constraints are:\n\n• Each",
null,
"$z_{x, y} \\geq 0$, so dirt only moves from",
null,
"$x$ to",
null,
"$y$.\n• Every pile",
null,
"$x \\in A$ must vanish, i.e. for each fixed",
null,
"$x \\in A$,",
null,
"$\\sum_{y \\in B} z_{x,y} = p_x$.\n• Likewise, every hole",
null,
"$y \\in B$ must be completely filled, i.e.",
null,
"$\\sum_{y \\in B} z_{x,y} = p_y$.\n\nThe objective is to minimize the cost of doing this:",
null,
"$\\sum_{x, y \\in A \\times B} d(x, y) z_{x, y}$.\n\nIn python, using the ortools library (and leaving out a few docstrings and standard import statements, full code on Github):\n\nfrom ortools.linear_solver import pywraplp\n\ndef earthmover_distance(p1, p2):\ndist1 = {x: count / len(p1) for (x, count) in Counter(p1).items()}\ndist2 = {x: count / len(p2) for (x, count) in Counter(p2).items()}\nsolver = pywraplp.Solver('earthmover_distance', pywraplp.Solver.GLOP_LINEAR_PROGRAMMING)\n\nvariables = dict()\n\n# for each pile in dist1, the constraint that says all the dirt must leave this pile\ndirt_leaving_constraints = defaultdict(lambda: 0)\n\n# for each hole in dist2, the constraint that says this hole must be filled\ndirt_filling_constraints = defaultdict(lambda: 0)\n\n# the objective\nobjective = solver.Objective()\nobjective.SetMinimization()\n\nfor (x, dirt_at_x) in dist1.items():\nfor (y, capacity_of_y) in dist2.items():\namount_to_move_x_y = solver.NumVar(0, solver.infinity(), 'z_{%s, %s}' % (x, y))\nvariables[(x, y)] = amount_to_move_x_y\ndirt_leaving_constraints[x] += amount_to_move_x_y\ndirt_filling_constraints[y] += amount_to_move_x_y\nobjective.SetCoefficient(amount_to_move_x_y, euclidean_distance(x, y))\n\nfor x, linear_combination in dirt_leaving_constraints.items():\n\nfor y, linear_combination in dirt_filling_constraints.items():\n\nstatus = solver.Solve()\nif status not in [solver.OPTIMAL, solver.FEASIBLE]:\nraise Exception('Unable to find feasible solution')\n\nreturn objective.Value()\n\n\nDiscussion: I’ve heard about this metric many times as a way to compare probability distributions. For example, it shows up in an influential paper about fairness in machine learning, and a few other CS theory papers related to distribution testing.\n\nOne might ask: why not use other measures of dissimilarity for probability distributions (Chi-squared statistic, Kullback-Leibler divergence, etc.)? One answer is that these other measures only give useful information for pairs of distributions with the same support. An example from a talk of Justin Solomon succinctly clarifies what Earthmover distance achieves",
null,
"Also, why not just model the samples using, say, a normal distribution, and then compute the distance based on the parameters of the distributions? That is possible, and in fact makes for a potentially more efficient technique, but you lose some information by doing this. Ignoring that your data might not be approximately normal (it might have some curvature), with Earthmover distance, you get point-by-point details about how each data point affects the outcome.\n\nThis kind of attention to detail can be very important in certain situations. One that I’ve been paying close attention to recently is the problem of studying gerrymandering from a mathematical perspective. Justin Solomon of MIT is a champion of the Earthmover distance (see his fascinating talk here for more, with slides) which is just one topic in a field called “optimal transport.”\n\nThis has the potential to be useful in redistricting because of the nature of the redistricting problem. As I wrote previously, discussions of redistricting are chock-full of geometry—or at least geometric-sounding language—and people are very concerned with the apparent “compactness” of a districting plan. But the underlying data used to perform redistricting isn’t very accurate. The people who build the maps don’t have precise data on voting habits, or even locations where people live. Census tracts might not be perfectly aligned, and data can just plain have errors and uncertainty in other respects. So the data that district-map-drawers care about is uncertain much like our point clouds. With a theory of geometry that accounts for uncertainty (and the Earthmover distance is the “distance” part of that), one can come up with more robust, better tools for redistricting.\n\nSolomon’s website has a ton of resources about this, under the names of “optimal transport” and “Wasserstein metric,” and his work extends from computing distances to computing important geometric values like the barycenter, computational advantages like parallelism.\n\nOthers in the field have come up with transparency techniques to make it clearer how the Earthmover distance relates to the geometry of the underlying space. This one is particularly fun because the explanations result in a path traveled from the start to the finish, and by setting up the underlying metric in just such a way, you can watch the distribution navigate a maze to get to its target. I like to imagine tiny ants carrying all that dirt.",
null,
"Finally, work of Shirdhonkar and Jacobs provide approximation algorithms that allow linear-time computation, instead of the worst-case-cubic runtime of a linear solver.\n\n## 15 thoughts on “Earthmover Distance”\n\n1.",
null,
"liyuanzhe\n\nThanks for the great post, is there a typo in the third constraint on $z_{xy}$? every hole $y \\in B$ must be completely filled, i.e. \\sum_{y \\in B} z_{x,y} = p_x, should it by p_y at the end?\n\nLike\n\n•",
null,
"j2kun\n\nYes! Thanks for pointing that out, just fixed it.\n\nLike\n\n•",
null,
"clumma\n\n> Likewise, every hole $y \\in B$ must be completely filled\n\nThe $y \\ and B$ are showing up in escaped form on two browsers here. Everything else is getting formatted.\n\nLike\n\n•",
null,
"j2kun\n\nThanks! Fixed both\n\nLike\n\n2.",
null,
"minshallGreg Minshall\n\nhi. non-math comment/question. the rendering of the math is pretty bad on my linux laptop (with both webkit-based and firefox browsers), but fine on my ipad. mathjax normally renders well on my laptop. i’m curious how you produce your html? no mathjax? cheers.\n\nLike\n\n•",
null,
"j2kun\n\nIt’s wordpress’s shitty support for math 😦 they render serverside and display inline images, which some browsers will put in displaystyle\n\nI’ve been considering migrating to a better platform, but I have resolved to finish writing my book before I do that.\n\nLike\n\n3.",
null,
"clumma\n\n> The distance between S and B is the optional solution\nThis should be A and B, right?\n\nLike\n\n4.",
null,
"Daniel\n\nI would expect an optimal solution to this kind of problem to have some property about encoding of positions. If I apply a function f(x,y) to every point then how is the change in the distance? f is some perturbation, translation, rotation, change of scale, symmetry.\n\nWhat are the properties of the earth-mover distance with respect to a family of perturbation of the data?\n\nA good distance should be stable with respect to noise with small variance. If you have prior knoledge of the distribution of noise, then the optimal distance seems to have the property that the variance of the distance between the perturbed points is minimal, that is the variance of X = sum(d((xi+nxi,yi+nyi),(xj+nxj,yj+nyj)) is minimal where nxi = noise asociate with xi, for example nxi = random_normal(0,sigma) if the noise doesn’t depend of the coordinates of pointes.\n\nLike\n\n•",
null,
"j2kun\n\nI agree with your sentiment. For rigid transformations of the data it seems clear to me that the distance is unchanged. Scaling all the underlying point distances by the same constant",
null,
"$c > 0$ should scale the distance by",
null,
"$O(c)$ (with discrepancies depending on whether points overlap?)\n\nI don’t any theorems about noise models off-hand, but I would be surprised if small noise caused a big change in the earthmover distance. I tried to look for a few theorems in the literature about this, but the problem is “stability” and “perturbation” result in wildly different measure-theoretic convergence theorems that I think are unrelated to this question, and probably mask more relevant work on this.\n\nLike\n\n5.",
null,
"Nemean\n\nAre you going to write further articles about the earthmover’s distance or some other thing on-topic? Because this seems too interesting to leave it at one article.\n\nLike\n\n•",
null,
"j2kun\n\nThis post was a break from book writing for me, so I don’t want to promise anything until after I finish my book, hopefully this summer\n\nLike\n\n6.",
null,
"Daniel\n\nNoise related: fast and robust earth-movers distance, http://leibniz.cs.huji.ac.il/tr/1143.pdf, they use saturated distance which provide robust property against noise,\nd_t (a,b) = min(d(a,b),t), they prove is a distance and they give a fast algorithm with applications.\n\nLike\n\n7.",
null,
"odedbd\n\nThank you for an interesting read. The EM distance was all the rage, about a year ago, in the context of training Generative Adverserial Networks – GANs (google for Wasserstein GAN or WGAN).\n\nEM was used in training GANs for the same reason mentioned here, dealing with distributions that (might) have non-overlapping supports. Interestingly, EM was not calculated directly, but rather via it’s dual problem. I am not an expert on the mathematics, but those interested may find a through and well-written discussion here –\nhttps://vincentherrmann.github.io/blog/wasserstein/\n\nLike\n\n8.",
null,
"dyl4n\n\nIsn’t a more succinct description of the Earth Mover’s Distance the absolute distance between two cumulative distribution functions?\n\nLike\n\n•",
null,
"j2kun\n\nA cumulative distribution function implies there’s an ordering of the probability space. This is not true of distributions embedded in, say, the plane.\n\nLike"
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"https://0.gravatar.com/avatar/90b179348780a6e7fe8e502968dc534a",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.90523183,"math_prob":0.97020787,"size":10730,"snap":"2019-35-2019-39","text_gpt3_token_len":2357,"char_repetition_ratio":0.121107586,"word_repetition_ratio":0.0060204696,"special_character_ratio":0.20959926,"punctuation_ratio":0.12185297,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99085915,"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],"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],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-18T21:56:03Z\",\"WARC-Record-ID\":\"<urn:uuid:667cb9c1-2319-4075-baab-c6334a638378>\",\"Content-Length\":\"127537\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:04ba199f-913a-4bf2-bffe-27f1ca34b724>\",\"WARC-Concurrent-To\":\"<urn:uuid:6113ada0-3e29-4e20-a431-b759a630ce92>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://jeremykun.com/2018/03/05/earthmover-distance/?like_comment=60622&_wpnonce=4fd3b801c5\",\"WARC-Payload-Digest\":\"sha1:YUS5BD5TBPCU7GEB4UITGJPIOPOQ6RII\",\"WARC-Block-Digest\":\"sha1:XCD2QLX2OMJ3VKWPX73NWQZS5WVNDBRX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573368.43_warc_CC-MAIN-20190918213931-20190918235931-00277.warc.gz\"}"} |
https://fornoob.com/consider-an-ionic-compound-mx2-composed-of-generic-metal-m/ | [
"# Consider an ionic compound, MX2, composed of generic metal M\n\nConsider an ionic compound, MX2, composed of generic metal M and generic, gaseous halogen X?\n\nGiven:\n\nThe enthalpy of formation of MX2 is ΔHf° = –957 kJ/mol.\n\nThe enthalpy of sublimation of M is ΔHsub = 123 kJ/mol.\n\nThe first and second ionization energies of M are IE1 = 757 kJ/mol and IE2 = 1451 kJ/mol.\n\nThe electron affinity of X is ΔHEA = –313 kJ/mol.\n\nThe bond energy of X2 is BE = 153 kJ/mol.\n\n• Lattice energy U(MX2) to calculate\n\nUse Born-Haber(-Fajans) Cycle\n\nM(s) + X2 → MX2(s) ΔHf°MX2 = -957 kJ mol^-1\n\nM(s) →ΔHsub(M)→ M(g) ΔHsub = 123kJ mol^-1\n\nM(g) →ΔHIE1(M)→ M^+ + e⁻ 757 kJ mol^-1\n\nM^+→ΔHIE2(M)→ M^2+ + e⁻ 1451 kJ mol^-1\n\nX2 →ΔHBE(X2)→ 2X +153 kJ mol^-1\n\nX + e⁻→ΔHEA(X)→ X^- -313 kJ mol^-1\n\nΔHfo = U + ΔHsub(M) + IE1 + IE2 + ΔHdissX2 + 2EA1(X)\n\n-957= U + 123 + 757 + 1451 + 153 – 626 = U + 1858\n\nU = -2815 kJ mol^-1 The –ve sign is often dropped"
]
| [
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.55774665,"math_prob":0.9995043,"size":903,"snap":"2022-40-2023-06","text_gpt3_token_len":400,"char_repetition_ratio":0.13904338,"word_repetition_ratio":0.0,"special_character_ratio":0.43964562,"punctuation_ratio":0.065656565,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9989232,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-30T00:37:07Z\",\"WARC-Record-ID\":\"<urn:uuid:8862937e-8e3b-436c-9a61-de98df40bbbd>\",\"Content-Length\":\"74203\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b5545394-2d58-4913-bee0-a998fb9cb6e5>\",\"WARC-Concurrent-To\":\"<urn:uuid:816878ea-9579-4f25-9d61-c298d6c9d6e2>\",\"WARC-IP-Address\":\"104.21.95.251\",\"WARC-Target-URI\":\"https://fornoob.com/consider-an-ionic-compound-mx2-composed-of-generic-metal-m/\",\"WARC-Payload-Digest\":\"sha1:IN3EIVVUT5S4W77DOPTWNLZZF3FZW6VX\",\"WARC-Block-Digest\":\"sha1:ZUHWHBRRRFJJXQ3WFSMJT2I5H73HFUES\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335396.92_warc_CC-MAIN-20220929225326-20220930015326-00507.warc.gz\"}"} |
https://bmcneurosci.biomedcentral.com/articles/10.1186/1471-2202-12-S1-P325 | [
"Volume 12 Supplement 1\n\n# Model parameter estimation for Channelrhodopsin-2 light gated ion channels\n\nRecently, Gunyadin et al. designed and validated an engineered mutant of channelrhodopsin-2 (ChR2) labeled E123T to accelerate channel kinetics of ChR2 light gated ion channels. They reported a mono-exponential decay for E123T photocurrent after the light is turned off. This is in contrast to the photocurrent characteristics of the wild type ChR2, which not only exhibits slower channel kinetics, but the decay after the light is turned off is bi-exponential. In earlier work, Nikolic et al. had captured the bi-exponential decay of wild type ChR2 in a four state transition rate model.\n\nOur goal in this work is to determine the conditions under which the four state model proposed by Nikolic et al. can exhibit fast mono-exponential decay seen for the E123T mutant of ChR2. We start from the four state transition rate model proposed by Nikolic et al. . This model consists of two closed dark adapted states C1 and C2 and two open light adapted conducting states O1 and O2. The unknown parameters of the model to be estimated from experimental data are: transition rate parameters: {Gd1,Gd2,Gr, e12, e21}, parameters ε1 and ε2 that capture the efficiency of light induced transition of the closed states, and the parameter γ, which captures the relative permeability of the two open states to ion flow. We first identify constraints on the four state model parameters based on the experimental data by Nikolic et al. on wild type ChR2 photo-current generated in hippocampal cells for light stimulation with intensities I={0.9, 11} mW/mm2. We show that the problem of estimating the unknown parameters defined above from the model derived constraints represents an under-determined system. We convert the parameter estimation problem into a two-step procedure and separate the estimation of light independent and dependent parameters. The parameters are estimated using simulated annealing, a global optimization algorithm. In Table 1, we compare our estimated parameters (MPChR2) for the four state-model with those obtained by Nikolic et al. . Our model parameters provide a much better fit to the experimental data on the slow and fast component of the decay current. In particular, Nikolic et al. assumed Gr=0 to derive the constraints on the bi-exponential decay of the ChR2 photocurrent after the light is turned off. We relaxed this assumption and were able to obtain a better fit to the ChR2 decay current kinetics. Specifically, with Nikolic et al. parameters, the ratio Islow/Ifast=0.37 for the amplitudes of the slow and fast component of the bi-exponential decay current while Islow/Ifast=0.194, using our parameters. The experimental value reported for this ratio is 0.2. We have further conducted sensitivity analysis on the estimated parameters to understand the degree of variability in the parameters given uncertainty in the experimental data. Using this framework, we then estimate the four state model parameters for E123T mutant of ChR2. We notice that the mono-exponential decay rate of E123T is similar to the fast component of the bi-exponential decay current of wild-type ChR2. This indicates that for the mutant, the amplitude of the slow component of the decay current is zero, which provides one additional constraint on the model parameters. We again use simulated annealing to determine the four state model parameters for the E123T mutant of ChR2. The results of this analysis are presented in row 4 of Table 1. Our analysis of the resulting model reveals that it fits the experimental data from Nikolic et al. quite well, with our fit error matching that of Nikolic et al. parameters fit error (10E-4).\n\n## Conclusion\n\nWe have systematically identified constraints on the four state transition rate model first proposed by Nikolic et al. to capture the photocurrent kinetics of both wild type ChR2 with bi-exponential decay rate and E123T mutant with mono-exponential decay rate. Using these constraints and global optimization, we have estimated model parameters that fit experimental data well.\n\n## References\n\n1. Gunyadin LA, Yizhar O, Berndt A, Sohal VS, Deisseroth K, Hegemann P: Ultrafast optogenetic control. Nat. Neurosci. 2010\n\n2. Nikolic K, Grossman N, Grubb MS, Burrone J, Toumazou C, Degenaar P: Photocycles of Channelrhodopsin-2. Photochem and Photobiol. 2009\n\nAuthors\n\n## Rights and permissions\n\nReprints and Permissions",
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""
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null,
"https://bmcneurosci.biomedcentral.com/track/article/10.1186/1471-2202-12-S1-P325",
null
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https://research.ipmu.jp/seminar/?seminar_id=2113 | [
"# MS Seminar (Mathematics - String Theory)\n\nSpeaker: Alexander Voronov (U of Minnesota) QuantIzing Deformation Theory Thu, Aug 09, 2018, 15:30 - 17:00 Seminar Room B Classical deformation theory is based on the Classical Master Equation (CME), a.k.a. the Maurer-Cartan Equation: dS + 1/2 [S,S] = 0. Physicists have been using a quantized CME, called the Quantum Master Equation (QME), a.k.a. the Batalin-Vilkovisky (BV) Master Equation: dS + h \\Delta S + 1/2 {S,S} = 0. The CME is defined in a differential graded (dg) Lie algebra, whereas the QME is defined in a space V[[h]] of formal power series or V((h)) of formal Laurent series with values in a dg-BV-algebra V or a bi-dg-Lie algebra. A generalization of classical deformation theory based on the QME may be thought of as quantized deformation theory. Examples include cohomological field theory and string-field theory. Quantum deformation functor and its representability will also be discussed in the talk. Blackboard talk"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.90532726,"math_prob":0.98137057,"size":812,"snap":"2020-34-2020-40","text_gpt3_token_len":213,"char_repetition_ratio":0.13737623,"word_repetition_ratio":0.0,"special_character_ratio":0.23768473,"punctuation_ratio":0.12883435,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97908825,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-10T00:01:40Z\",\"WARC-Record-ID\":\"<urn:uuid:d3349cc6-d185-45ab-aee5-744f0b14f712>\",\"Content-Length\":\"14668\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4880903e-54c7-49ae-a144-696d1ebf403d>\",\"WARC-Concurrent-To\":\"<urn:uuid:2fd27342-fea6-4a11-ac15-1b3b71113515>\",\"WARC-IP-Address\":\"157.82.236.45\",\"WARC-Target-URI\":\"https://research.ipmu.jp/seminar/?seminar_id=2113\",\"WARC-Payload-Digest\":\"sha1:UILIQ5WHYSZRJOQRVDBNTLHCVLISAYDL\",\"WARC-Block-Digest\":\"sha1:CBQEKJA7GFZOSIK56X5QKSXT2DWRYBAV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738595.30_warc_CC-MAIN-20200809222112-20200810012112-00481.warc.gz\"}"} |
https://math.answers.com/basic-math/What_is_the_decimal_of_11_and_15 | [
"",
null,
"",
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"",
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"",
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"0\n\n# What is the decimal of 11 and 15?\n\nUpdated: 4/28/2022",
null,
"Wiki User\n\n7y ago\n\n11/15 = 0.73333 repeating",
null,
"Wiki User\n\n7y ago",
null,
"",
null,
"Study guides\n\n84 cards\n\n## 166\n\n➡️\nSee all cards\n3.83\n224 Reviews",
null,
"Earn +20 pts\nQ: What is the decimal of 11 and 15?\nSubmit\nStill have questions?",
null,
"",
null,
"Related questions\n\n### What is 11 over 15 is a decimal?\n\n11/15 as a decimal is 0.7'3' recurring '3'\n\n### What is 14 15 in decimal form?\n\n0.933314/15:= 11 ÷ 36= 0.9333 in decimal\n\n### What is 11 over 15 in decimal?\n\nIt is: 11/15 = 0.73333 .... recurring 3\n\n### How do you have 15 take away 6 and have 10 left?\n\nWell, that would be true if the first two numbers are in base=11 but the 10 is in base=10. (15)11 = decimal 16 (6)11 = decimal 6 (15)11 - (6)11 = (decimal 16) minus (decimal 6) = decimal 10 (*For the solution to the Poptropica riddle, see the related question below.)\n\n0.733 . . .\n\n0.7333\n\n### What is 11 out of 15 in a percentage?\n\nIt is: 11/15 times 100/1 = 73.33% recurring decimal 3\n\n### What is the decimal equilvalent for 11 out of 15?\n\n0.73333 repeating\n\n### How do you turn 11 over 15 into a decimal?\n\nDivide 11 by 15 untilthe remainder is 0, orthe decimal representation goes into a finite string which keeps repeating.\n\n0.733 . . .\n\n### Is 7 over 10 greater than 11 over 15?\n\nExpressed as a decimal, 7/10 is equal to 0.7.Expressed as a decimal, 11/15 is equal to 0.73 recurring (that is, 0.73333...)Therefore, 7/10 is not bigger than 11/15.\n\n### What is 4 and 11 over 32 into a decimal?\n\n(4+11)/32 = 15/32 = 0.46875"
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https://www.4open-sciences.org/articles/fopen/full_html/2022/01/fopen220023/fopen220023.html | [
"Issue 4open Volume 5, 2022 Statistical Inference in Markov Processes and Copula Models 20 9 Mathematics - Applied Mathematics https://doi.org/10.1051/fopen/2022022 21 December 2022",
null,
"This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.\n\n## Introduction\n\nIn this article, we address the notion of conditional independence and its impact on unconditional dependence. Here, unconditional dependence is being represented by a copula function and the impact of conditional independence on unconditional dependence is studied using the assumption of conditional independence in the copula function. Independence allows us to carry out the usual inferential process, for example, those obtained by calculating the likelihood function, and because of that is a relevant notion in statistical inference. Under the independence assumption is possible to carry out the classic techniques of point estimation, hypothesis testing, and all those procedures based on the likelihood function. As can then be verified, the concept of independence occupies a relevant space in statistics and for this reason, it is widely found in the literature. Several efforts have been done to promote its identification, see for instance García and González-López [1, 2]. In such papers are proposed statistics for independence detection, see also Genest and Rémillard and Hollander et al. . Such statistics work under certain conditions imposed on the nature of the underlying variables. Other authors are concerned with measuring dependence. For example, García et al. proposes an index with such purpose. Despite its enormous use, independence is a very restrictive case within the diversity of types of dependencies that could occur. For example, if we consider the random vector (X, Y) with bivariate Normal distribution, and density function",
null,
"then, the independence only occurs when ρ = 0, which shows that independence is a restrictive assumption. The theory of copulas has been extensively used to investigate such diversity. Nelsen and Joe show the enormous diversity of possible dependencies, addressed from the notion of copulas.\n\nThis article explores conditional independence, seeking to preserve some inferential procedures, and we also seek to characterize the impact of conditional independence on dependence. Conditional independence requires the identification of the entity Θ under which we can declare conditional independence (conditional to Θ), that requirement can be a constraint in practice since there is no clear identification procedure of Θ. The first question we seek to answer is how conditional independence impacts unconditional dependence. For the characterization of unconditional dependence, we will rely on the notion of copula. Secondly, we approach the impact of the notion of infinite exchangeability on conditional independence, based on the fact that such a notion guarantees conditional independence [8, 9]. With such a notion in hand, we return to the investigation of the structure of the copula representing unconditional dependence. That is, in this paper, we investigate the impact of conditional independence on unconditional dependence and extend this question to a stronger notion as is the case of infinite exchangeability.\n\nIndependence is a very restrictive case within the diversity of types of dependencies, on the other hand, conditional independence (to Θ) is a much more flexible concept, and for this reason, it is the condition that we explore in this article. In our approach, we take the entity Θ as being a random variable since in this way we generalize the notion. In addition, we give space so that the reader can access the interpretation that de Finetti offers of such an entity.\n\nHere we describe the content of this paper. In Section 2 we address the notion of conditional independence and we introduce a theorem that shows a representation of the dependence (copula). We also show corollaries addressing situations of interest that result from that theorem. In Section 3 we introduce the notion of infinite exchangeability, which allows us to identify the quantity Θ, guarantying conditional independence. We then show a result offering a representation of the dependence (copula density function), given the property of infinite exchangeability. The final considerations are given in Section 4.\n\n## Conditional independence\n\nWe begin this section by introducing the notion of conditional independence and then exemplifying it. We introduce in the sequence the notion of copula function. From these notions follows our main result, Theorem 2.1 which shows the impact of conditional independence on the dependence, that is, in this case, on the copula structure. We emphasize the consequences that result from Theorem 2.1 in two corollaries and we show also some examples.\n\n## Definition 2.1\n\nGiven two continuous random variables X and Y, X and Y are conditionally independent given Θ, where Θ is a random variable, if",
null,
"with",
null,
"and",
null,
"denoting the conditional density functions, given θ, of (X,Y), X and Y respectively.\n\nThe next example shows a case of conditional independence given Θ.\n\n## Example 2.1\n\nSet the joint density function of X, Y, Θ, as",
null,
"where IA (x) = 1, if x ∈ A, and IA (x) = 0, if x ∉ A. The marginal distributions fX,Y, fX, fY and fΘ are respectively,",
null,
"",
null,
",",
null,
"and",
null,
"Moreover,",
null,
"then X and Y are conditionally independent given Θ = θ, with different conditional distributions, being identically distributed given θ, when k = 1.\n\nFrom the previous example we see that conditional independence does not impose the same conditional distribution. Note that Definition 2.1 allows θ to be considered as a constant value, and in such case θ is a parameter.\n\nThe result that we show in Theorem 2.1 seeks to understand the meaning that Θ plays in the construction of the dependence between X and Y (not conditioned to Θ = θ). For dependence between X and Y we understand the copula between them. In rough terms, the copula is a function that, combined with the marginal cumulative distributions, returns the joint cumulative distribution.\n\nNext we formally introduce the notion of copula explaining how this notion serves to represent the dependence between X and Y.\n\n## Definition 2.2\n\nA copula (in dimension 2) is a function C: [0, 1]2 → [0, 1] with the following properties,\n\n1. u,v ∈ [0, 1], C(u,0) = C(0,v) = 0, and C(u,1) = u,C(1,v) = v;\n\n2. for every ui, vi ∈ [0, 1], i = 1, 2, such that",
null,
"and",
null,
"",
null,
"It is possible to show that every copula given by Definition 2.2 is a cumulative distribution function on [0, 1]2, in the strict sense of the term. We now show how Definition 2.2 is formally related to the dependence between the variables of a random vector.\n\nAccording to Sklar’s theorem , if HX,Y is the joint distribution function of (X, Y) with margins FX and FY, there exists a copula CX,Y (called 2-copula of (X, Y)) such that",
null,
"",
null,
"If FX and FY are continuous, then CX,Y is unique. Conversely, if C is a copula (following Definition 2.2), F and G are distribution functions, then the function H(x,y) = C(F(x),G(y)) is a joint distribution function with margins F and G.\n\nFrom what is given in Sklar’s theorem, the dependence between X and Y is represented by the 2-copula of (X,Y), CX,Y, since the margins FX and FY depend on X and Y, respectively. Then, our purpose is to see the effect of Definition 2.1 in CX,Y.\n\nThe next theorem provides an analytical representation of the copula, knowing that the variables are conditionally independent given the entity Θ.\n\n## Theorem 2.1\n\nGiven X and Y under the assumptions of Definition 2.1, if CX,Y is the 2-copula of (X,Y), Θ the random variable of Definition 2.1 with density function πΘ,",
null,
"(1)\n\nwith CX and CY denoting the 2-copulas of (X,Θ) and (Y,Θ) respectively, and FΘ denoting the cumulative distribution of Θ.\n\nProof. Since X, Y are continuous random variables we can write the cumulative distribution HX,Y between X and Y as follows, applying Definition 2.1,",
null,
"(2)\n\nConsidering the cumulative distribution between X and Θ, by Sklar’s theorem, HX(s,θ) = CX (FX(s), FΘ(θ)) then, the joint density function between X and Θ is given by",
null,
"(3)\n\nwith cX denoting the density function related to the copula CX. Then, from equation (3), we obtain,",
null,
"(4)\n\n□We proceed to integrate the equation (4). Define u = FX(s), then, du = fX (s)ds and",
null,
"(5)\n\nWe have similar expressions for fY(t,θ) and fY|θ(t). Then, applying equation (5) in equation (2), we obtain",
null,
"(6)\n\nConsider u and v in [0, 1], there exist values x and y, such that u = FX(x) and v = FY(y), then, by Sklar’s theorem and equation (6),",
null,
"□\n\nWe use in the following example a family of copulas that corresponds to weak dependence. It is the Farlie–Gumbel–Morgenstern family, and it encompasses the case of independence. See Nelsen .\n\n## Example 2.2\n\nConsider (X,Y) with a 2-copula",
null,
"(7)\n\nwith parameters α ∈ [−1, 1] and β ∈ [−1, 1]. Then, (X,Y) has a Farlie–Gumbel–Morgenstern 2-copula with parameter",
null,
"Consider Θ as a continuous random variable, with Uniform distribution in (0,1).\n\nIf we set the 2-copulas of (X, Θ) and (Y, Θ) as being Farlie–Gumbel–Morgenstern 2-copulas with parameters α ∈ [ 1, 1] and β ∈ [−1, 1], respectively. Since θ ~ U(0,1), we have",
null,
"And, we obtain,",
null,
"Then,",
null,
"As a consequence,",
null,
"One question that arises is whether every expression given by the right side of equation (1) is a copula, where CX and CY are copulas. The Theorem 1, in González-López and Litvinoff Justus , states that the expression is always a copula. That is, postulating two copulas of (X, Θ) and (Y, Θ), we get a copula that could be used to deal with the dependence between X and Y. An example of what is stated in this paragraph appears in the previous example, since taking (X, Θ) and (Y, Θ) with Farlie–Gumbel–Morgenstern copulas, with parameters α and β, respectively, the resulting copula between X and Y (using the right side of equation (1)) is a Farlie–Gumbel–Morgenstern copula, with parameter",
null,
"The following corollary shows the version of Theorem 2.1 in terms of density functions of copulas.\n\n## Corollary 2.1\n\nUnder the assumptions of Theorem 2.1, if cX,Y, cX and cY are the density functions related to the 2-copulas CX,Y, CX and CY, respectively,",
null,
"(8)\n\nProof: Consider that",
null,
"",
null,
"and",
null,
"The results follow differentiating both sides of equation (1) and permuting the integral with",
null,
"□\n\n## Example 2.3\n\nConsidering the Example 2.2. Since",
null,
"",
null,
"Then,",
null,
"(9)\n\nand, according to the equation (7), it is the copula density function of a FGM copula with parameter",
null,
"The following corollary shows the version of Theorem 2.1 under the assumptions of identical conditional margins. The impact of the assumption appears in the integrating term of equation (1). This assumption about margins will become relevant later, in this paper.\n\n## Corollary 2.2\n\nUnder the assumptions of Theorem 2.1, if the variables X and Y are conditionally (to Θ) independent and identically distributed,",
null,
"## Example 2.4\n\nConsidering the Example 2.2, if α = β, the copula of (X,Y) is a Farlie–Gumbel–Morgenstern copula with",
null,
"Then, the restriction (on the margins) imposes on the relationship between X and Y a positive Kendall’s τ coefficient",
null,
".\n\nTheorem 2.1 shows that the conditional independence of X and Y given Θ defines the structure of the copula between X and Y. See equation (1) for the form of the copula and see Corollary 2.1 for the form of the density function of copula between X and Y. We exemplify both results in Examples 2.2 and 2.3, respectively. Corollary 2.2 shows the case covered by Corollary 2.1, under the assumption of identical margins. The latter is exemplified in Example 2.4. The Corollary 2.2 is especially interesting for the situation that we will address in the next section.The results of this section describe the analytical form of the copula between X and Y, no longer depending on Θ. This means we have offered a representation for the predictive copula between X and Y.\n\nThe key to determining conditional independence between X and Y is to know that there exists a variable Θ establishing this independence. This leads us to the following reflections, showing a context where such an entity can be identified. That is the goal of de Finetti’s theorems of representation.\n\n## Exchangeability\n\nWe start the section with the definition of infinite exchangeability. This notion is weaker than independence and makes it possible to identify Θ. Then, we exemplify certain characteristics that result from this imposition. And, we present in Corollary 3.1, the impact of this notion on the dependence structure, that is, on the copula function.\n\n## Definition 3.1\n\nA sequence X1,X2,X3,... of random variables is infinitely exchangeable, if for each collection of n variables",
null,
"the joint distribution is identical for all the permuted set of variables",
null,
"with σ: {1, 2,…,n} → {1, 2,…,n} any permutation function.\n\nThe success in identifying the Θ entity stems from de Finetti’s representation theorem, see de Finetti . de Finetti proves that, a binary sequence X1, X2,... follows Definition 3.1, if and only if, there exists a distribution function FΘ on [0, 1] such that for all n the joint probability mass function",
null,
"where FΘ is the cumulative distribution function of the entity",
null,
"and xi ∈ {0, 1}. As a consequence, the conditional mass probability is decomposed in a product of Bernoulli’s",
null,
"with",
null,
"being θ a value of the variable Θ. Such a result is extended for all types of random variables in Hewitt and Savage . For a sequence X1,X2,... of continuous random variables following Definition 3.1, there exists a variable Θ, such that the conditional density function",
null,
"can be decomposed as",
null,
"The previous relation happens for each value θ of the variable Θ, then, the structure imposed by Definition 3.1 in the sequence X1,X2,⋯ guarantee the existence of the entity Θ.\n\nWe see in the remark to follow (Remark 3.1) that this notion leads us to conditional independence, and implies identical margins. But as we show in Example 3.1, exchangeability coexists with dependence.\n\n## Remark 3.1\n\nInfinitely exchangeable variables (following Definition 3.1) are identically distributed conditionally to θ (value of the variable Θ identified by de Finetti’s theorem). If the infinitely exchangeable sequence is composed by absolutely continuous random variables and Θ has a density function πΘ (·), for each",
null,
"as a consequence,",
null,
"since,",
null,
"The concept of infinite exchangeability could be considered unrealistic, but it can be seen applied in usual inferential strategies. If a sample is available, say X1,,Xn, when constructing the likelihood function its elements are considered independent given a certain parameter. And how does this assumption is related to Definition 3.1? If the distribution of X1,…,Xn is invariant under permutations of the elements of X1,…,Xn, then we say that X1…,Xn is exchangeable. If X1,…,Xn is exchangeable",
null,
", then, we say that X1, X2,… is infinitely exchangeable. If X1,…,Xn can be embedded in an infinitely exchangeable sequence X1, X2,…, then we say that X1,…,Xn is infinitely exchangeably extendable. Bayesian and frequentist approaches, in general, treat observable values as infinitely exchangeably extendable and, as a consequence of de Finetti’s representations, independent and identically distributed conditional on some unknown entity Θ.\n\nAlthough identically distributed independent random variables must be exchangeable, identically distributed exchangeable random variables need not be independent. Consider for instance the bivariate case of Gumbel distribution as shown by the next example.\n\n## Example 3.1\n\nLet (X,Y) be a random vector, H its joint distribution function, given by",
null,
"and H(x,y) = 0, otherwise, with α a parameter in [0, 1]. We see that X and Y are exchangeable. Note also that X and Y have identical marginal distribution (exponential distribution). But, since the 2-copula of (X,Y) is",
null,
"X and Y are not independent.\n\nThe previous example shows us a case where exchangeability occurs, the variables share the same marginal distribution, but X and Y are not independent. Thus we see that exchangeability is a distinctly different notion from the notion of independence. It also follows from Example 3.1 that identically distributed variables can be dependent since every copula is a joint distribution with uniform marginal distributions, in particular, the copula of Example 3.1.\n\nThe next example shows a case of dependent variables, with identical marginal distributions but not exchangeable.\n\n## Example 3.2\n\nLet (X,Y) be a random vector, H its joint distribution function, given by H(x,y= xy − x3y(1 − x)(1 − y), x,y ∈ [0, 1]. Then, H is a copula, since fits the conditions of Definition 2.2. X and Y are identically distributed but, not exchangeable since H is not symmetrical.\n\nBearing in mind the previous example, it is now necessary to mention Theorem 2.7.4 which shows how the margins should be and also the copula to guarantee the exchangeability between a pair of variables. Let X and Y be continuous random variables with joint distribution function H, margins FX and FY, respectively, and copula C. Then X and Y are exchangeable if and only if FX = FY and C is symmetric (C(u,v) = C(v,u), ∀(u,v) ∈ [0, 1]2).\n\nIn the treatment and modeling via copulas, it is very common to operate with the original variables transformed by their margins, so below we summarize the impact of Definition 3.1 on such variables.\n\n## Remark 3.2\n\nA property that is straight from Definition 3.1, is the preservation of the infinite exchangeability from the original sequence to the standardized ones. Namely, if the sequence X1, X2, X3,… of random variables is absolutely continuous, the sequence U1, U2, U3,… is also infinitely exchangeable, where",
null,
"Since, for any collection i1,…in, with arbitrary n,",
null,
"for any permutation σ and arbitrary values u1,…,un ∈ [0, 1].",
null,
"",
null,
"(10)",
null,
"(11)\n\nwhere equation (10) results from Definition 3.1 and equation (11) follows from Remark 3.1 (",
null,
").\n\nWe note that if the variables U1, U2, U3,… of Remark 3.2 are infinitely exchangeable, this does not imply that the variables X1, X2, X3,… are infinitely exchangeable. It is enough that one of the marginal distributions",
null,
"is different from the rest",
null,
"which would violate Remark 3.1.\n\nUnder the framework of Definition 3.1 and assuming the continuity of the variable identified by the de Finetti representation, say Θ, we will have the copula between pairs of variables of the infinitely exchangeable sequence given by the copula between a member of the sequence (the first one is enough) and the variable Θ, let’s see.\n\n## Corollary 3.1\n\nLet X1, X2,… be a sequence of continuous random variables following Definition 3.1. If Θ is the variable identified in the de Finetti representation, producing the conditional independence, and Θ has density function πΘ(·), then, for each pair of variables Xi and Xj, i ≠ j,",
null,
"(12)\n\nwhere",
null,
"and",
null,
"are the density functions related to the 2-copulas",
null,
"and",
null,
"respectively.\n\nProof. Since infinitely exchangeable variables are conditionally independent (to the random variable of de Finetti’s theorem)",
null,
"for each pair Xi and Xj (from Corollary 2.1).\n\nInfinitely exchangeable sequences of absolutely continuous random variables share their unconditional marginal distributions",
null,
"from Remark 3.1. Also, for each θ value of Θ,",
null,
"since,",
null,
"(13)\n\nwhere equation (13) follows from Sklar’s theorem. And, using that",
null,
"we can see that",
null,
"□\n\nHowever, a question that arises is whether there is any restriction on the type of copula that could result from Corollary 3.1, a hint is given in O’ Neill , formulated in the next remark.\n\n## Remark 3.3\n\nLet X1,X2,… be a sequence of continuous random variables following Definition 3.1. Then, from Theorem 2 of O’ Neill , for each pair of variables Xi and Xj, i ≠ j, i, j ≥ 1,",
null,
"with",
null,
"denoting the Pearson’s rho correlation coefficient. Then, from Remark 3.2, since the variables U1,U2,… follow Definition 3.1, for each pair of variables Xi and Xj, i ≠ j, i, j ≥ 1,",
null,
"with",
null,
"denoting the Spearman’s rho correlation coefficient.\n\nThe above remark indicates that the possible copulas identified in Corollary 3.1 cannot correspond to negative correlations, under Definition 3.1.\n\nThe following example shows a case under Definition 3.1, and the resulting predictive copula.\n\n## Example 3.3\n\nLet X1, X2,… be an infinite sequence, under Definition 3.1, of exponential random variables with parameter θ, conditionally independent, given θ, a value of the continuous random variable Θ with exponential distribution and hyperparameter λ. Then, each conditional density function is",
null,
"and the prior density function of Θ is",
null,
"",
null,
"is proportional to",
null,
"for",
null,
"Since,",
null,
"for",
null,
"and",
null,
"Using the previous proportionality we obtain",
null,
"On the other hand,",
null,
"for",
null,
"then, the right side of equation (12) is proportional to",
null,
"Verifying the Corollary 3.1.\n\nThen, since we have",
null,
"the copula density function",
null,
"",
null,
"and the predictive 2-copula is",
null,
"We start the section with some notes and examples so that Corollary 3.1 is a natural consequence of the impact of conditional independence in the framework of infinite exchangeability. Remark 3.1 shows that infinite exchangeability implies conditionally independence and identical margins. While independence implies permutability, the opposite is not valid, see Example 3.1. Also, if the margins are all equal this does not imply permutability, see Example 3.2. Under Definition 3.1, we show in Corollary 3.1 that infinite exchangeability allows to obtain a representation of the copula between pair of elements of the infinite sequence in terms of the copula between each element of the infinite sequence and the entity Θ, quantity identified by de Finetti’s theorem. Also, note that the right side of equation (12) reflects that is only necessary to know the copula between one element of the infinite sequence and Θ to obtain the predictive copula between pairs of elements of the infinite sequence. Example 3.3 exposes how works in practice the corollary.\n\n## Conclusion\n\nIn the framework of absolutely continuous random variables X and Y, we use the notion of conditional independence between X and Y (conditional on Θ), to obtain a representation of the predictive 2-copula between X and Y (Theorem 2.1, see also Corollary 2.1). Such a result tells us that the representation involves the 2-copula between X and Θ, the 2-copula between Y and Θ, and the distribution of Θ. Being the property of identically distributed useful for inferential procedures, in Corollary 2.2 we reveal the impact of such an assumption on the representation of the predictive 2-copula between X and Y. There are indications (see Example 2.4) that the restriction imposed by the condition of identical margins could lead to restrictions on the values of coefficients of dependence, such as Kendall’s τ coefficient, Spearman’s ρ coefficient.\n\nSince the notion of conditional independence requires the identification of the entity Θ making X and Y conditionally independent given Θ, we introduce the concept of infinite and exchangeable sequences, X1, X2, X3.... This condition appears as one of the inferential strategies applicable in the process of manipulating finite samples. It is reasonable to suppose that a finite sequence is part of an infinite sequence. Under Definition 3.1, by representation theorems [8, 9] we can guarantee that Θ exists and that it is a random variable. To give the reader a more precise notion of the meaning of infinite exchangeability, we show that it imposes the same margins on X and Y (conditional or not, see Remark 3.1), but that it does not impose unconditional independence (see Example 3.1) and it does impose conditional independence on Θ. Then, under infinite exchangeability, we revisit the copula density function structure and prove Corollary 3.1, which shows the structure of the predictive 2-copula density function between Xi and Xj, i ≠ j elements of the infinite sequence. We complete our findings with examples and notes for the reader. Corollaries 2.2 and 3.1, although they start from different assumptions, show that the predictive density function of the 2-copula between pairs of variables X and Y can be analytically expressed as a function of the density functions of the copula between only one variable (say X) and Θ. That is to say that the dependence, represented by the predictive 2-copula between X and Y, is only a consequence of the dependence between X and Θ. This fact shows the relevance of the identification of Θ, for the description of the predictive 2-copula.\n\n## Acknowledgments\n\nVinícius Litvinoff Justus gratefully acknowledges the support provided by CNPq with a fellowship of scientific initiation from University of Campinas. The authors wish to thank the referees and editors for their many helpful comments and suggestions on an earlier draft of this paper.\n\n## References\n\n1. García JE, González-López VA (2020), Random permutations, non-decreasing subsequences and statistical independence. Symmetry 12, 9, 1415. https://doi.org/10.3390/sym12091415. [Google Scholar]\n2. García JE, González-López VA (2014), Independence tests for continuous random variables based on the longest increasing subsequence. J Multiv Anal 127, 126–146. https://doi.org/10.1016/j.jmva.2014.02.010. [Google Scholar]\n3. Genest C, Rémillard B (2004), Test of independence and randomness based on the empirical copula process. Test 13, 335–369. https://doi.org/10.1007/BF02595777. [Google Scholar]\n4. Hollander M, Wolfe D, Chicken E (2013), Nonparametric statistical methods, John Wiley & Sons, New York. [Google Scholar]\n5. García JE, González-López VA, Nelsen RB (2013), A new index to measure positive dependence in trivariate distributions. J Multivar Anal 115, 481–495. https://doi.org/10.1016/j.jmva.2012.11.007. [Google Scholar]\n6. Nelsen RB (2006), An introduction to copulas, Springer-Verlag, New York. https://doi.org/10.1007/0-387-28678-0. [Google Scholar]\n7. Joe H (2014), Dependence modeling with copulas, Chapman and Hall/CRC, New York. https://doi.org/10.1201/b17116. [CrossRef] [Google Scholar]\n8. De Finetti B (1929), Funzione caratteristica di un fenomeno aleatorio, in: Atti del Congresso Internazionale dei Matematici: Bologna del 3 al 10 de settembre di 1928 (Comunicazioni, sezione IV (A)–V–VII), vol. 6, pp. 179–190. [Google Scholar]\n9. Hewitt E, Savage LJ (1955), Symmetric measures on Cartesian products. Trans Am Math Soc 80, 2, 470–501. https://doi.org/10.2307/1992999. [Google Scholar]\n10. González-López VA, Litvinoff Justus V, A method for the elicitation of copulas (preprint) [Google Scholar]\n11. O’ Neill B (2009), Exchangeability, Correlation, and Bayes’ Effect. Int Stat Rev 77, 2, 241–250. https://doi.org/10.1111/j.1751-5823.2008.00059.x. [CrossRef] [Google Scholar]\n\nCite this article as: González-López VA & Litvinoff Justus V 2022. Conditional independence and predictive copula. 4open, 5, 20.\n\nCurrent usage metrics show cumulative count of Article Views (full-text article views including HTML views, PDF and ePub downloads, according to the available data) and Abstracts Views on Vision4Press platform.\n\nData correspond to usage on the plateform after 2015. The current usage metrics is available 48-96 hours after online publication and is updated daily on week days."
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https://ramdeoshubham.com/newbieprogrammer/arrays-examples | [
"April 17, 2016\n\n# Arrays Some cool examples\n\nThat’s how we play with arrays, practically !",
null,
"Welcome to the Newbie Programmer Series. In the last two parts, we have discussed the concepts behind the arrays and how to use them. In this part we will discuss some examples to learn more about arrays. So if you are new here, please check out all the previous parts (click here) so that you can easily understand what is happening below.\n\n### Average with array formed during run time\n\nCalculating average of some numbers is one of the easiest examples. Why I put it here in a different way is because here I have used variable array !!! WHAT ? But we have already studied in the last part that variable arrays cannot be made. Then how is it made possible ? Actually it is a trick. It is very true that once the array is created of a certain size, it cannot be change. But before creating the array, we can change the size. So in this example here, user need to set the size of the array. Then the array is created which remains fix. In the following code, program asks for total number of values. Say user enters 5. Then the program asks 5 values to be entered and then prints their average.\n\n``````#include <stdio.h>\nmain()\n{\nint n, k, sum, avg;\nprintf(\"Enter the number of values. \\n\");\nscanf(\"%d\", &n);\nint val[n]; //now array is created.\nprintf(\"Now enter %d values one by one\",n);\n\nfor(k=0; k\n``````\n\nStudy it a little by yourself. The above example is for integers, you can set it for float too. The main point to learn is how we have tricked it with array looking like variable size. If there is any trouble, please comment below.\n\nLet’s Enjoy the output :\n\n``````Enter the Number of values.\n4\nNow enter 4 values one by one.\n10\n20\n30\n40\nThe average of this data is 25\nPress Enter to exit...\n``````\n\n### Rotating Arrays\n\nThere is no need to create any program for this. We will just look at the concept. Suppose we have a table ‘A’ and we want to rotate it to ‘B’ :\n\n`````` (A) (B)\n1 2 3 clockwise 7 4 1\n4 5 6 rotated to 8 5 2\n7 8 9 9 6 3\n``````\n\nHere each element is moved to a different position. In a simple way, we can do it by creating a new array and then reassigning the positions. Let’s say “A[row][column]” is the original array and “B[row][column]” will become the rotated one.\n\n`````` (A) (B)\n\n0 1 2 0 1 2\n----------- -----------\n0 | 1 | 2 | 3 | 0 | 7 | 4 | 1 |\n----------- -----------\n1 | 4 | 5 | 6 | 1 | 8 | 5 | 2 |\n----------- -----------\n2 | 7 | 8 | 9 | 2 | 9 | 9 | 3 |\n----------- -----------\n``````\n\nComparing from the above table, you can equate like :\n\n``````B = A;\nB = A;\nB = A;\nB = A;\nB = A;\nB = A;\nB = A;\nB = A;\nB = A;\n``````\n\nAfter this, the array “B” is the rotated array “A”. This rotation is also called as Transpose. Two invert a table, or rotating a table anticlockwise, instead of doing some thing again, we should make a function for a “single rotation” (as above) and then, for example, for anticlockwise rotation, rotate it clockwise three times ! Well it is a very confusing concept. But later we will learn using pointers which can make everything easy.\n\n### Printing Multiplication Tables\n\nWhile learning loops, (click here) we have already created a multiplication table program. Here let’s learn a little new thing. In that example, we have printed tables like “2 x 1 = 2… 2 x 2 = 4…”. Using loops directly without storing anything in a form of table.\n\nHere, we will use a two-dimensional array and will print the table as it is.\n\nThe point to learn here is that how the table gets printed. So keep your eyes on the table printing code :\n\n``````#include <stdio.h>\nmain()\n{\nint i,k,tab;\nfor(i=0; i<10, i++)\n{\nfor(k=0; k<10; k++)\n{\ntab[i][k] = i * k;\n}\n}\n\nfor(i=0; i<10, i++)\n{\nfor(k=0; k<10; k++)\n{\nprintf(\"%d \",tab[i][k]);\n}\nprintf(\"\\n\");\n}\n\nprintf(\"Press Enter to exit...\\n\");\ngetchar();\nreturn 0;\n}\n``````\n\nNow you would have already understood how the table was formed. Let’s talk about the printing function. The concept is that the INNER loop starts printing with ROW=0, Column=0 to Column=9. When the end of a ROW is reached (column=9), the inner loops ends. And a “new line” character is printed due to the body of the second loop. The Row increases from ROW=0 to ROW=1. Same pattern is followed. All column of a row is printed then comes a new line for the second row. When all the rows are also printed, the loops ends.\n\nI think this is enough. We have discussed some practical ways of using arrays and here we can complete our introduction of arrays. From the next part we will start discussing “structures”. My favorite as we can do programming like treating real physical objects. So please stay connected :)\n\n#### You might also like:\n\n(prev)\n(next)\n\n(rand) How technology affected our Lives? Happy being Busy?\n(rand) Structure some Examples\n(rand) Beginning C programming"
]
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https://www.fullpotentialtutor.com/category/home-school/page/2/ | [
"",
null,
"3 May\n\n## Using The Laws Of Logarithms To Solve The Equation\n\n2021-06-03T01:16:27-04:00May 3rd, 2021|Brain teaser, FPLA Winning Strategies, Home School, Math-e-Magics, SAT, ACT, PERT, HSPT|\n\nIn a previous lesson, you learned how to find a logarithmic base. Therefore, we will assume that you already know all the log laws and their applications. In this tutorial, we will be focusing on how to solve logarithmic equations using those laws that are in your arsenals.\n\n1 May\n\n## How To Find The Base of a Logarithm?\n\n2021-05-25T08:05:10-04:00May 1st, 2021|Brain teaser, FPLA Winning Strategies, Home School, Math-e-Magics, SAT, ACT, PERT, HSPT|\n\nSolving for an unknown base of a logarithm simply means working in a reverse form. In this case, you will have a logarithm of a number without knowing the base of that logarithm. From our knowledge of how to evaluate basic logarithm, this reverse process should not pose any challenge to us.\n\n29 April\n\n## What Are Percentages and How To Calculate Percentages?\n\n2021-06-03T01:28:35-04:00April 29th, 2021|Brain teaser, FPLA Winning Strategies, Home School, Math-e-Magics, SAT, ACT, PERT, HSPT|\n\nA percentage is a different way of representing fractions or decimals. It represents part of a whole thing that is in question. But instead of leaving the result as a fraction or as a decimal, it is expressed as a percentage.\n\n26 April\n\n## Equation of a Line Through Two Points\n\n2021-06-03T01:45:15-04:00April 26th, 2021|Brain teaser, FPLA Winning Strategies, Home School, Math-e-Magics, SAT, ACT, PERT, HSPT|\n\nWe need at least two points to find the equation of a line if we don’t have any information about its gradient. Equation of any line takes the slope-intercept form, \\$\\$y=mx+c\\$\\$ and we can always express any equation in this form if we have two sets of coordinates of any two points on the line.\n\n•",
null,
"Function Composition Examples\nGallery\n\n22 April\n\n## Function Composition Examples\n\n2021-06-11T01:02:45-04:00April 22nd, 2021|Brain teaser, FPLA Winning Strategies, Home School, Math-e-Magics, SAT, ACT, PERT, HSPT|\n\nThe basic definition of a composite function is that it has two functions in one. The inner function is a variable to the outer function. \\$\\$\n\n21 April\n\n## Simplify Algebraic Expressions Involving Complex Fractions\n\n2021-06-11T01:10:33-04:00April 21st, 2021|Brain teaser, FPLA Winning Strategies, Home School, Math-e-Magics, SAT, ACT, PERT, HSPT|\n\nIf a fraction has a fraction or fractions either on the numerator, denominator, or both, then it becomes a complex fraction (aka compound fraction). Simplifying compound fractions is no different from solving other fractions. Such fractions may seem complex, but following the necessary steps makes everything easier.\n\n•",
null,
"How To Solve Rational Equations\nGallery\n\n20 April\n\n## How To Solve Rational Equations\n\n2021-05-26T00:05:11-04:00April 20th, 2021|Brain teaser, FPLA Winning Strategies, Home School, Math-e-Magics, SAT, ACT, PERT, HSPT|\n\nEquations with one or more expressions are rational equations. Such equations may involve additions, subtractions, multiplications, and divisions. Rational expressions are ratios of two numbers, numerator and denominator, where denominator is never zero.\n\n19 April\n\n## Solving Inequalities Involving Absolute Values\n\n2021-06-11T01:17:39-04:00April 19th, 2021|Brain teaser, FPLA Winning Strategies, Home School, Math-e-Magics, SAT, ACT, PERT, HSPT|\n\nHandling inequalities that involve absolute values follow the same steps for solving equations with absolute values. You can refer to our previous tutorial as a prerequisite to this lesson. However, we will again list all the steps here and use them in solving some examples.\n\nGo to Top"
]
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https://www.hindawi.com/journals/cin/2016/2305854/ | [
"#### Abstract\n\nBy combining with sparse kernel methods, least-squares temporal difference (LSTD) algorithms can construct the feature dictionary automatically and obtain a better generalization ability. However, the previous kernel-based LSTD algorithms do not consider regularization and their sparsification processes are batch or offline, which hinder their widespread applications in online learning problems. In this paper, we combine the following five techniques and propose two novel kernel recursive LSTD algorithms: (i) online sparsification, which can cope with unknown state regions and be used for online learning, (ii) and regularization, which can avoid overfitting and eliminate the influence of noise, (iii) recursive least squares, which can eliminate matrix-inversion operations and reduce computational complexity, (iv) a sliding-window approach, which can avoid caching all history samples and reduce the computational cost, and (v) the fixed-point subiteration and online pruning, which can make regularization easy to implement. Finally, simulation results on two 50-state chain problems demonstrate the effectiveness of our algorithms.\n\n#### 1. Introduction\n\nThe least-squares temporal difference (LSTD) learning may be the most popular approach for policy evaluation in reinforcement learning (RL) [1, 2]. Compared with the standard temporal difference (TD) learning, LSTD uses samples more efficiently and eliminates all step-size parameters. However, LSTD also has some drawbacks. First, LSTD requires a matrix-inversion operation at each time step. To reduce computational complexity, Bradtke and Barto proposed a recursive LSTD (RLSTD) algorithm , and Xu et al. proposed a RLSTD algorithm . But these two algorithms still require many features especially for highly nonlinear RL problems, since the RLS approximator assumes a linear model . Second, when the number of features is larger than the number of training samples, LSTD is prone to overfitting. To overcome this problem, Kolter and Ng proposed an -regularized LSTD algorithm called LARS-TD for feature selection , but it is only applicable for batch learning and its implementation is complicated. On this basis, Chen et al. proposed an -regularized RLSTD algorithm . In contrast with LARS-TD, it has an analytical solution, but it cannot obtain a sparse solution. Third, LSTD requires users to design the feature vector manually, and poor design choices can result in estimates that diverge from the optimal value function .\n\nIn the last two decades, kernel methods have been intensively and extensively studied in supervised and unsupervised learning . The basic idea behind kernel methods can be summarized as follows: By using a nonlinear transform, the origin input data can be mapped into a high-dimensional feature space, and an inner product in this space can be interpreted as a Mercer kernel function. Thus, as long as a linear algorithm can be formulated in terms of inner products, there is no need to perform computations in the high-dimensional feature space . Recently, there have also been many research works on kernelizing least-squares algorithms . Here, we only review some works related to our proposed algorithms. One typical work is the sparse kernel recursive least-squares (SKRLS) algorithm with the approximate linear dependency (ALD) criterion . Compared with traditional RLS algorithms, it not only has a good nonlinear approximation ability but also can construct the feature dictionary automatically. Similarly, Chen et al. proposed an -regularized SKRLS algorithm with the online vector quantization . Besides having the good properties of SKRLS-ALD, it can avoid overfitting. In addition, Chen et al. proposed an -regularized SKRLS algorithm with the fixed-point subiteration , which can yield a much sparser dictionary.\n\nIntuitively, we can also bring the benefits of kernel machine learning to LSTD algorithms. In fact, kernel-based RL algorithms have become more and more popular in recent years , and several works have been done for kernelizing LSTD algorithms. In an earlier paper, Xu proposed a sparse kernel-based LSTD (SKLSTD) algorithm with the ALD criterion . Although this algorithm can avoid selecting features manually, it is only applicable for batch learning and its derivation is complicated. After that, Xu et al. proposed an incremental version of the SKLSTD algorithm for policy iteration , but this algorithm still requires a matrix-inversion operation at each time step. Moreover, the feature dictionary is required to be constructed offline, which makes this algorithm only approximate the value function correctly in the area of the state space that is covered by the training samples. Recently, Jakab and Csató proposed a sparse kernel RLSTD (SKRLSTD) algorithm by using a proximity graph sparsification method . Unfortunately, its sparsification process is also offline. In addition, all of these algorithms do not consider regularization, whereas many real problems exhibit noise and the high expressiveness of the kernel matrix can result in overfitting .\n\nIn this paper, we propose two online SKRLSTD algorithms with and regularization, called OSKRLSTD- and OSKRLSTD-, respectively. Compared with the derivation of SKLSTD, our derivation uses Bellman operator along with projection operator and thus is more simple. To cope with unknown state-space regions and avoid overfitting, our algorithms use online sparsification and regularization techniques. Besides, to reduce computational complexity and avoid caching all history samples, our algorithms also use the recursive least-squares and the sliding-window technique. Moreover, different from LARS-TD, OSKRLSTD- uses the subiteration and online pruning to find the fixed point. These techniques make our algorithms more suitable for online RL problems with a large or continuous state space. The rest of this paper is organized as follows. In Section 2, we present preliminaries and review the LSTD algorithm. Section 3 contains the main contribution of this paper: we derive OSKRLSTD- and OSKRLSTD- algorithms in detail. In Section 4, we demonstrate the effectiveness of our algorithms for two 50-state chain problems. Finally, we conclude the paper in Section 5.\n\n#### 2. Background\n\nIn this section, we introduce the basic definitions and notations, which will be used throughout the paper without any further mention. We also review the LSTD algorithm, which is needed to establish our algorithms described in Section 3.\n\n##### 2.1. Preliminaries\n\nIn RL and dynamic programming (DP), an underlying sequential decision-making problem is often modeled as a Markov decision process (MDP). An MDP can be defined as a tuple , where is a set of states, is a set of actions, is a state transition probability function where denotes the probability of transitioning to state when taking action in state , is a reward function, is the discount factor, and is an initial state distribution. For simplicity of presentation, we assume that and are finite. Given an MDP and a policy , the sequence is a Markov reward process , where and .\n\nRL and DP often use the state-value function to evaluate how good the policy is for the agent to be in state . For an MDP, can be defined as , which must obey the Bellman equation ,or be expressed in vector form,If and are known, can be solved analytically; that is,where is the identity matrix.\n\nHowever, different from the case in DP, and are unknown in RL. The agent has to estimate by exploring the environment. Furthermore, many real problems have a large or continuous state space, which makes hard to be expressed explicitly. To overcome this problem, we often resort to linear function approximation; that is,where is a parameter vector, is the feature vector of state , and is a feature matrix. Unfortunately, when approximating in this manner, there is usually no way to satisfy the Bellman equation exactly, because may lie outside the span of .\n\n##### 2.2. LSTD Algorithm\n\nThe LSTD algorithm presents an efficient way to find such that “approximately” satisfies the Bellman equation . By solving the least-squares problem , we can find a closest approximation in the span of to replace . Then, from (2) and (4), we can use for approximating . That means we can find by solving the fixed-point equation:where is a nonnegative diagonal matrix indicating a distribution over states. Nevertheless, since and are unknown and since is too large to form anyway in a large or continuous state space, we cannot solve (5) exactly. Instead, given a trajectory following policy , LSTD uses , , and to replace , , and , respectively. Then, (5) can be approximately rewritten asLet ; we haveThus, the fixed point can be found by\n\n#### 3. Regularized OSKRLSTD Algorithms\n\nTo overcome the weaknesses of the previous kernel-based LSTD algorithms, we propose two regularized OSKRLSTD algorithms in this section.\n\n##### 3.1. OSKRLSTD- Algorithm\n\nNow, we use regularization and online sparsification to derive the first OSKRLSTD algorithm, which is called OSKRLSTD-.\n\nFirst, we use the kernel trick to kernelize (6). Suppose the feature dictionary , and let denote the corresponding feature matrix. By the Representer Theorem , and can be expressed as follows:where and are the coefficient vector of and , respectively. Then, from (6), we haveBy the Mercer Theorem , the inner product of two feature vectors can be calculated by . Thus, we can define , , and . On this basis, (10) can be rewritten as\n\nSecond, we try to derive the -regularized solution of (11). Add an -norm penalty into (11); that is,where is a regularization parameter. Let ; we haveSince , we easily have from (9). Then, the above equation can be rewritten aswhere is the identity matrix. Thus, can be analytically solved aswhere and denotewhere and .\n\nThird, we derive the recursive formulas of and . Under online sparsification, there are two cases: () , , , , and ; () , , , , where , and is expanded aswhere is the dimensional zero vector.\n\nFor the first case, (16) can be rewritten as follows:Applying the matrix-inversion lemma for , we getThus, plugging (19) and (20) into (15), we obtain\n\nFor the second case, (16) can be rewritten as follows:where and are the same as the updated and when the feature dictionary keeps unchanged, , , , and . However, computing , , , and requires caching all history samples, and the computational cost will become more and more expensive as increases. Inspired by the work of Van Vaerenbergh et al. , we introduce a sliding window to deal with these problems. Let , where is the window size. We only use the samples in to evaluate , , , and ; that is,Then, similar to those in the first case, and can be derived as follows:where and is the same as the updated when the dictionary keeps unchanged.\n\nFinally, we summarize the whole algorithm in Algorithm 1.\n\n ) Input: to be evaluated, , , , () for do () if then () Initialize , () Take given by , and observe , () Initialize () Initialize () Initialize () else () Take given by , and observe , () Update , () Update , by (21) and (20) () if satisfies the sparsification condition then () , , () Compute , , and by (23) () Update , by (25) and (24) () end if () end if () () end for\n\nRemark 1. Here, we do not restrict the OSKRLSTD- algorithm to a specific online sparsification method. That means it can be combined with many popular sparsification methods such as the novelty criterion (NC) and the ALD criterion.\n\nRemark 2. Although the OSKRLSTD- algorithm is designed for infinite horizon tasks, it can be modified for episodic tasks. When is an absorbing state, it only requires setting temporarily and setting as the start state of next episode.\n\nRemark 3. Our simulation results show that a big sliding window cannot help improve the convergence performance of the OSKRLSTD- algorithm. Thus, to save memory and reduce the computational cost, should be set to a small integer.\n\n##### 3.2. OSKRLSTD- Algorithm\n\nIn this subsection, we use regularization and online sparsification to derive the second OSKRLSTD algorithm, which is called OSKRLSTD-.\n\nFirst, we try to derive the -regularized solution of (11). Add an -norm penalty into (11); that is,where is a regularization parameter. However, is not differentiable. Similar to Painter-Wakefield and Parr in , we resort to the subdifferential of ; that is,where is the set-valued function defined component-wise asLet , so thatSince , we also have from (9). Then, the above equation can be rewritten aswhere has the same meaning as . To avoid the singularity of and further reduce the complexity of the subsequent derivation, we introduce into both sides; that is,where is a regularization parameter. Obviously, the left hand side of (31) is the same as that of (14). Thus, from (16), the above equation can be rewritten asThen, we have the following fixed-point equation:where denotesUnfortunately, here, cannot be solved analytically.\n\nSecond, we investigate how to find the fixed point of (33). In -regularized LSTD algorithms [5, 29], researchers often used the LASSO method to tackle this problem. However, the LASSO method is inherently a batch method and is unsuitable for online learning. Instead, we resort to the fixed-point subiteration method introduced in . We first use the sign function to replace in (33). Then, we can construct the following subiteration:where denotes the th subiteration and is initialized to since the fixed point will be close to if and are small. If the subiteration number reaches a preset value or is less than or equal to a preset threshold , the subiteration will stop. From (32) and (28), if , should be 0. Obviously, the replacement of makes lose the ability to select features. To remedy this situation, after the whole subiteration, we remove the weakly dependent elements from according to the magnitude of ; that is,where denotes the operation to remove the elements indexed by the set , which is determined bywhere is a preset threshold. Note that we do not remove the last element of , since is probably very small, especially when is just added to . Similarly, we perform and to remove the weakly dependent coefficients. From (16), also requires removing some rows and columns. Unfortunately, we cannot use the method in to do this like Chen et al. in , since is not a symmetry matrix. Considering that will remove the corresponding elements if is pruned, we directly perform to remove the rows and columns indexed by . Although this method may bring some bias into , our simulation results show that it is feasible and effective. The whole fixed-point subiteration and online pruning algorithm are summarized in Algorithm 2.\n\n ) Input: , , , , , , , () Initialize: () for to do () Update by (35) () if then () Break out of the loop () end if () end for () () Determine the index set by (37) () Perform , , and\n\nRemark 4. Our simulation results show that Algorithm 2 will converge in few iterations. Thus, Algorithm 2 does not become the computational bottleneck of the OSKRLSTD- algorithm, and the maximum subiteration number can be set to a small positive integer.\n\nThird, we derive the recursive formulas of and . Although the dictionary can be pruned by using Algorithm 2, it still has the risk of rapidly growing if new samples are allowed to be added continually. Thus, the conventional sparsification method is also required to be considered here. Similar to Section 3.1, there are two cases under online sparsification. Since and have the same definitions as and in the OSKRLSTD- algorithm, we can directly use (20) and (24) for updating and rewrite (21) and (25) for updating . If dissatisfies the sparsification condition, will be updated byOtherwise, will be updated bywhere , , , and are also calculated by (23). Since , , and will be pruned by Algorithm 2 after the update, it is important to note that and in (39) denote and updated by but unpruned by . Likewise, when (24) is used here, has the same meaning.\n\nFinally, we summarize the whole algorithm in Algorithm 3. For episodic tasks, the modification is the same as Remark 2. In addition, similar to Remark 3, the sliding-window size should also be set to a small integer.\n\n ) Input: to be evaluated, , , , , , , , () for do () if then () Initialize , () Take given by , and observe , () Initialize () Initialize () Initialize () Perform Algorithm 2 () else () Take given by , and observe , () Update , () Update , by (38) and (20) () , () Perform Algorithm 2 () if satisfies the sparsification condition then () , , () Compute , , and by (23) () Update , by (39) and (24) () Perform Algorithm 2 () end if () end if () () end for\n\nRemark 5. By pruning the weakly dependent features, the OSKRLSTD- algorithm can yield a much sparser solution than the OSKRLSTD- algorithm.\n\n#### 4. Simulations\n\nIn this section, we use a nonnoise chain and a noise chain [2, 20, 31] to demonstrate the effectiveness of our proposed algorithms. For comparison purposes, RLSTD and SKRLSTD algorithms are also tested in the simulations. To analyze the effect of regularization and online pruning on the performance of our algorithms, the OSKRLSTD- algorithm with and the OSKRLSTD- algorithm with (called OSKRLSTD-0 and OSKRLSTD-, resp.) are tested here, too. In addition, the effects of the sliding-window size on the performance of our algorithms and OSKRLSTD- are evaluated as well.\n\n##### 4.1. Simulation Settings\n\nAs shown in Figure 1, in both chain problems, each chain consists of 50 states, which are numbered from 1 to 50. For each state, there are two actions available, that is, “left” (L) and “right” (R). Each action succeeds with probability 0.9, changing the state in the intended direction, and fails with probability 0.1, changing the state in the opposite direction. The two boundaries of each chain are dead-ends, and the discount factor of each chain is set to 0.9. For the nonnoise chain, the reward is 1 only in states 10 and 41, whereas, for the noise chain, the reward is corrupted by an additive Gaussian noise . Due to the symmetry, the optimal policy for both chains is to go right in states 1–9 and 26–41 and left in states 10–25 and 42–50. Here, we use it as the policy to be evaluated. Note that the state transition probabilities are available only for solving the true state-value functions , and they are assumed to be unknown for all algorithms compared here.\n\nIn the implementations of all tested algorithms for both chain problems, the settings are summarized as follows: (i) For all OSKRLSTD algorithms, the Mercer kernel is defined as , the sparsification condition is defined as , and the sliding-window size is set to 5. Besides, for the OSKRLSTD- algorithm, the regularization parameters and are set to 0.8 and 0.3, the maximum subiteration number is set to , the precision threshold is set to 0.1, and the pruning threshold is set to 0.4; for the OSKRLSTD- algorithm, , , and are the same as those in the OSKRLSTD- algorithm; for the OSKRLSTD- algorithm, is set to 1. (ii) For the SKRLSTD algorithm, the Mercer kernel and the sparsification condition are the same as those in each OSKRLSTD algorithm. (iii) For the RLSTD algorithm, the feature vector consists of 19 Gaussian radius basis functions (GRBFs) plus a constant term 1, resulting in a total of 20 basis functions. The GRBF has the same definition as the Mercer kernel used in each OSKRLSTD algorithm, and the centers of GRBFs are uniformly distributed over . In addition, the variance matrix of RLSTD is initialized to 0.4, where is the 20 × 20 identity matrix. (iv) In the simulations, each algorithm performs 50 runs, each run includes 100 episodes, and each episode is truncated after 100 time steps. In particular, the SKRLSTD algorithm requires an extra run for offline sparsification before each regular run.\n\n##### 4.2. Simulation Results\n\nWe first report the comparison results of all tested algorithms with the simulation settings described in Section 4.1. Their learning curves are shown in Figure 2. At each episode, the root mean square error (RMSE) of each algorithm is calculated by , where is solved by (1) and is the approximate value of the th run. From Figure 2, we can observe that (i) OSKRLSTD- and OSKRLSTD- can obtain the similar performance as RLSTD and converge much faster than SKRLSTD. (ii) Without regularization, the performance of OSKRLSTD-0 becomes very poor, especially in the noise chain. In contrast, OSKRLSTD- and OSKRLSTD- still perform well. (iii) The performance of OSKRLSTD- is only slightly better than that of OSKRLSTD-, which indicates that online pruning has little effect on the performance. Figure 3 illustrates approximated by all tested algorithms at the final episode. Clearly, OSKRLSTD-0 has lost the ability to approximate of the noise chain. Figure 4 shows the dictionary growth curves of all tested algorithms. Compared with RLSTD and SKRLSTD, all OSKRLSTD algorithms can construct the dictionary automatically, and OSKRLSTD- yields a much sparser dictionary. Figure 5 shows the average subiterations per time step in OSKRLSTD- and OSKRLSTD-. As episodes increase, the subiterations decline gradually. In addition, online pruning can reduce the subiterations significantly. Even in the noise chain, the subiterations are small. Finally, the main simulation results of all tested algorithms at the final episode are summarized in Table 1.\n\nNext, we evaluate the effect of the sliding-window size on our proposed algorithms and OSKRLSTD- with . The logarithmic RMSEs of each algorithm at the final episode are illustrated in Figure 6. Note that the parameter settings of these algorithms are the same as those described in Section 4.1 except for . From Figure 6, OSKRLSTD- and OSKRLSTD- obviously become worse rather than better as the window size increases, and OSKRLSTD- has a strong adaptability to different window sizes. The reason for this result is analyzed as follows: From the derivation of our algorithms, the influence of the window size is mainly manifest in . Since here is calculated by recursive update instead of matrix inversion and samples are used one by one, using too many history samples together may increase the calculation error. In OSKRLSTD-, a moderate regularization parameter can relieve the influence of this error. In contrast, in OSKRLSTD- and OSKRLSTD-, the subiteration may expand the influence. Especially for OSKRLSTD-, online pruning can introduce the new error, which further worsens the convergence performance. To verify the above analysis, we reset , , and for OSKRLSTD- and OSKRLSTD- and reevaluate the effect of the window size. The new results are illustrated in Figure 7. As expected, OSKRLSTD- and OSKRLSTD- can also adapt to . Nevertheless, there is still no proof that a big window size can help improve the convergence performance of OSKRLSTD- and OSKRLSTD-. Thus, as stated in Remark 3, is suggested to be set to a small integer in practice.\n\n#### 5. Conclusion\n\nAs an important approach for policy evaluation, LSTD algorithms can use samples more efficiently and eliminate all step-size parameters. But they require users to design the feature vector manually and often require many features to approximate state-value functions. Recently, there are some works on these issues by combining with sparse kernel methods. However, these works do not consider regularization and their sparsification processes are batch or offline. In this paper, we propose two online sparse kernel recursive least-squares TD algorithms with and regularization, that is, OSKRLSTD- and OSKRLSTD-. By using Bellman operator along with projection operator, our derivation is more simple. By combining online sparsification, and regularization, recursive least squares, a sliding window, and the fixed-point subiteration, our algorithms not only can construct the feature dictionary online but also can avoid overfitting and eliminate the influence of noise. These advantages make them more suitable for online RL problems with a large or continuous state space. In particular, compared with the OSKRLSTD- algorithm, the OSKRLSTD- algorithm can yield a much sparser dictionary. Finally, we illustrate the performance of our algorithms and compare them with RLSTD and SKRLSTD algorithms by several simulations.\n\nThere are also some interesting topics to be studied in future work: (i) How to select proper regularization parameter should be investigated. (ii) A more thorough simulation analysis is needed, including an extension of our algorithms to learning control problems. (iii) Eligibility traces would be combined for further improving the performance of our algorithms. (iv) The convergence and prediction error bounds of our algorithms will be analyzed theoretically.\n\n#### Competing Interests\n\nThe authors declare that there are no competing interests regarding the publication of this paper.\n\n#### Acknowledgments\n\nThis work is supported in part by the National Natural Science Foundation of China under Grant nos. 61300192 and 11261015, the Fundamental Research Funds for the Central Universities under Grant no. ZYGX2014J052, and the Natural Science Foundation of Hainan Province, China, under Grant no. 613153."
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https://www.hackmath.net/en/examples/equation?tag_id=13 | [
"# Equation + right triangle - examples\n\n1. Find parameters",
null,
"Find parameters of the circle in the plane - coordinates of center and radius: ?\n2. Trapezoid MO",
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"The rectangular trapezoid ABCD with right angle at point B, |AC| = 12, |CD| = 8, diagonals are perpendicular to each other. Calculate the perimeter and area of the trapezoid.\n3. Garden",
null,
"Area of square garden is 6/4 of triangle garden with sides 56 m, 35 m and 35 m. How many meters of fencing need to fence a square garden?\n4. Right Δ",
null,
"Right triangle has length of one leg 28 cm and length of the hypotenuse 53 cm. Calculate the height of the triangle.\n5. Short cut",
null,
"Imagine that you are going to the friend. That path has a length 330 meters. Then turn left and go another 2000 meters and you are at a friend's. The question is how much the journey will be shorter if you go direct across the field?\n6. River",
null,
"From the observatory 14 m high and 32 m from the river bank, river width appears in the visual angle φ = 20°. Calculate width of the river.\n7. Right triangle Alef",
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"The obvod of a right triangle is 84 cm, the hypotenuse is 37 cm long. Determine the lengths of the legs.\n8. Proof PT",
null,
"Can you easy prove Pythagoras theorem using Euclidean theorems? If so, do it.\n9. Rectangle",
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"In rectangle ABCD with sides |AB|=19, |AD|=16 is from point A guided perpendicular to the diagonal BD, which intersects at point P. Determine the ratio ?.\n10. R triangle",
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"Calculate the area of a right triangle whose longer leg is 6 dm shorter than the hypotenuse and 3 dm longer than the shorter leg.\n11. Medians",
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"Calculate the sides of a right triangle if the length of the medians to the legs are ta = 21 cm and tb=12 cm.\n12. ISO triangle",
null,
"Calculate the area of an isosceles triangle KLM if the length of its sides are in the ratio k:l:m = 4:4:3 and has perimeter 377 mm.\n13. Leg and height",
null,
"Solve right triangle with height v = 9.6 m and shorter cathetus b = 17.3 m.\n14. Hypotenuse and height",
null,
"In a right triangle is length of the hypotenuse c = 56 cm and height hc = 4 cm. Determine the length of both trangle legs.\n15. RT - hypotenuse and altitude",
null,
"Right triangle BTG has hypotenuse g=117 m and altitude to g is 54 m. How long are hypotenuse segments?\n16. Rhombus",
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"Internal angles of rhombus is in ratio 2:3. How many times is the shorter diagonal longer than side of rhombus?\n17. Nice prism",
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"Calculate the surface of the cuboid if the sum of its edges is a + b + c = 19 cm and the body diagonal size u = 13 cm.\n18. Goat",
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"Meadow is a circle with radius r = 19 m. How long must a rope to tie a goat to the pin on the perimeter of the meadow to allow goat eat half of meadow?\n19. Column",
null,
"Perpendicular pole high 8 m tall broke and its toe fell 2.7 m from the bottom of the pole. At what height above the ground pole broke?\n20. Triangle",
null,
"Calculate the sides of the triangle if its area S = 630 and the second cathethus is shorter by 17.\n\nDo you have an interesting mathematical example that you can't solve it? Enter it, and we can try to solve it.\n\nTo this e-mail address, we will reply solution; solved examples are also published here. Please enter e-mail correctly and check whether you don't have a full mailbox.\n\nDo you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? See also our right triangle calculator."
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https://forum.image.sc/t/2d-array-workaround/35139 | [
"",
null,
"# 2D array workaround?\n\nHi,\nI have 6 different arrays (CIELAB coordinates) and an equation (Euclidean Distance) linking them all together:\nLref[j], aref[j], bref[j] (GROUP 1) - Fixed length (reference values).\nL[i], a[i], b[i] (GROUP 2) - Variable length (experimental values).\nEucDist = sqrt(pow(L[i]-Lref[j], 2) + pow(a[i]-aref[j], 2) + pow(b[i]-bref[j], 2));\n\nI need to generate a table/matrix (XY = X213) with X columns and the result of “EucDist” in each cell of the 213 rows of each column. Is there anyway for doing this? I’ve tried loops with two variables (i, j), loops within loops, etc. but I can’t get this… I guess I need 2D arrays and I’m aware ImageJ doesn’t allow this. Is there any workaround for this case?\n\nI attached an Example in Excel.\nExample.zip (13.3 KB)\n\nThanks.\n\nLee\n\nHey @insertsk8,\n\ndo you want to share the code you tried? It might then be easier for us to point you in the right direction",
null,
"Cheers,\nRobert\n\n2 Likes\n\nHello Lee -\n\nThe ImageJ Macro language (IJM) doesn’t support 2D arrays, but\nImageJ itself does.\n\nThe best “workaround” would be to switch to jython (python) or one\nof the other scripting languages supported by ImageJ (or use java,\nthe language in which ImageJ is written).\n\nIf you wish to stick with IJM, there are three common workarounds:\n\n Zero: Use “index arithmetic” to pack a 2D array into a 1D array.\nThus:\n\n``````for (i = 0; i < nRows; i++) {\nfor (j = 0; j < nColumns; j++) {\nmatrix_entry_ij = matrix[i * nColumns + j];\n// process matrix_entry_ij\n}\n}\n``````\n\nThis is an old-school multi-dimensional matrix idiom.\n\nFirst: Use a bunch of row variables, each of which is an IJM array\ncontaining the fields in that row, and access them with a big nested\nif statement. Thus: `row1`, `row2`, …, `row213`, and\n\n``````for (i = 1; i <= 213; i++) {\nif (i == 1) // process row1\nelse if (i == 2) // process row2\n...\nelse if (i == 213) // process row213\n}\n``````\n\nOf course, this will be rather unwieldy, given the size of your matrix.\n\nSecond: Pack one of your dimensions into strings that you then\nparse. Thus, `matrix` is an IJM (1D) array, and\n\n``````matrix = \"1.1, 1.2, ..., 9.9\";\n...\nfor (i = 0; i < lengthOf (matrix); i++) {\nvals = split (matrix[i], \",\");\nfor (j = 0; j < lengthOf (vals); j++) {\nmatrix_entry_ij = vals[j]\n// process matrix_entry_ij\n}\n}\n``````\n\nThis is also somewhat annoying, but works.\n\n( You can also use the Results Table as a 2D array, but it has\nnamed columns, rather than indexed columns, so you’re not really\nany better off than using a bunch of named row (or named column)\nvariables.)\n\nMy recommendation would be jython, rather than one of these\nworkarounds.\n\nThanks, mm\n\n2 Likes\n\nThis helped me a lot!\nI also tried slicing the 1D array depending on the number of columns I have.\nPrinting the array and saving it in .csv, and then opening the file again and splitting it in rows and the delimiter “,” allowed me to get a pseudo 2D result table.\n\n``````Array.print(val1);\nfor(i = 1; i<codeL; i++){\ninitial = i*ExpL;\nfinal = (i+1)*ExpL;\nval2 = Array.slice(num0,initial,final);\nArray.print(val2);\n}\n``````\n\nThanks!\nLee"
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"https://aws1.discourse-cdn.com/business4/uploads/imagej/original/2X/a/a8bb0afa608549b70f9141516502fe7fb4693171.png",
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"https://sjc3.discourse-cdn.com/business4/images/emoji/twitter/wink.png",
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http://dict.antkh.com/khmer/dictionaries/sum.aspx | [
"# Sum\n\nnoun\n1.\n~ (of something) (a total, a particular amount of money, a problem in arithmetic, gist summary, the whole amount totality) ចំនួនរួម, ទាំងស្រុង, សរុប, សង្ខេប, arithmetical sum ផលបូក, sum of \\$100 ចំនួនប្រាក់ in sum ដោយសង្ខេបទៅ\nExample: He was fined the sum of £ 200.\n2.\n(បច្ចេកទេស) ចំនួនសរុប ចំនួនទឹកប្រាក់\nverbpast tense: summed ; past participle: summed ; present participle: summing ;\n1.\nបូកចំនួន, សរុបចំនួន\n2.\n(TRANSITIVE) (Get or come to a total) sum a column of figures បូកចូលគ្នា, បូករួម\nExample: The sum of 4 an 2 is 6, Sophy sometimes as trouble doing his sums, Makara has as malisum for pocket money each week, to sum up what we had learned during the day.sum up (a column) បូករួម, sum up (an argument) ធ្វើសង្ខេប, សេចក្ដី, sum up (the situation) ប្រមើល\nExample: Now sum up (your views) in a few words.\n• capital sum\n- ចំនួនប្រាក់ដែលក្រុមហ៊ុនធានារ៉ាប់រងត្រូវបង់សងសរុប (ដុល)\n• in sum\n- យ៉ាងខ្លី, យ៉ាងសង្ខេប\n• inclusive sum\n- ថ្លៃឈ្នួលសរុប\n• Sum certain\n- ចំនួនទឹកប្រាក់ជាក់ច្បាស់\n• sum up\n- សង្ខេប បង្ហាញមតិឬចំណាប់អារម្មណ៍\nENGLISH MEANING\nnoun\n1.\nThe aggregate of two or more numbers, magnitudes, quantities, or particulars; the amount or whole of any number of individuals or particulars added together; as, the sum of 5 and 7 is 12.\n2.\nA quantity of money or currency; any amount, indefinitely; as, a sum of money; a small sum, or a large sum.\n3.\nThe principal points or thoughts when viewed together; the amount; the substance; compendium; as, this is the sum of all the evidence in the case; this is the sum and substance of his objections.\n4.\nHeight; completion; utmost degree.\n5.\nA problem to be solved, or an example to be wrought out.\nverbpast tense: summed ; past participle: summed ; present participle: summing ;\n1.\n(TRANSITIVE) To bring together into one whole; to collect into one amount; to cast up, as a column of figures; to ascertain the totality of; -- usually with up.\n2.\n(TRANSITIVE) To bring or collect into a small compass; to comprise in a few words; to condense; -- usually with up.\n3.\n(TRANSITIVE) To have (the feathers) full grown; to furnish with complete, or full-grown, plumage.\n1.\nadd, affix, append, attach, augment, increase, total\n2.\naggregate, entirety, total, whole\n1.\ndetach, reduce, remove, subtract\n2.\nfraction, ingredient, part, sample"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8751422,"math_prob":0.7906172,"size":2144,"snap":"2020-24-2020-29","text_gpt3_token_len":833,"char_repetition_ratio":0.11168224,"word_repetition_ratio":0.07253886,"special_character_ratio":0.2761194,"punctuation_ratio":0.21073559,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9803701,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-27T15:55:39Z\",\"WARC-Record-ID\":\"<urn:uuid:5cb998ef-2913-4b7c-b14d-339aed1d184c>\",\"Content-Length\":\"13039\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:79edb668-145f-4df5-a2cf-8a972791afa2>\",\"WARC-Concurrent-To\":\"<urn:uuid:833c5d6b-aacd-4c1c-a97a-d4966e6d0d8a>\",\"WARC-IP-Address\":\"64.209.142.231\",\"WARC-Target-URI\":\"http://dict.antkh.com/khmer/dictionaries/sum.aspx\",\"WARC-Payload-Digest\":\"sha1:35YSKVBUBD2YPBAZQOD7XDPPFKZRGHH3\",\"WARC-Block-Digest\":\"sha1:I4BNUI4I5A6CF3SEPBKB5HYKIF6322A6\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347394756.31_warc_CC-MAIN-20200527141855-20200527171855-00031.warc.gz\"}"} |
https://levlafayette.com/node/629 | [
"# Chapter 2: The IS–LM model\n\nAims of the chapter\n\nThe focus of the chapter is on the IS–LM model [\"investment-savings\" (IS) and \"liquidity preference-money supply\" (LM)], set out by Hicks (1937) in response to Keynes’ The general theory of employment, interest and money (1936) for analysing the aggregate demand and macroeconomic stabilisation policy in a closed economy and with fixed prices. The IS–LM model integrates the goods and the money markets and considers under what conditions output and the interest rate are simultaneously in equilibrium; what factors may determine short-run fluctuations; and how macroeconomic policy can be employed to stabilise the economy in the short run.\n\n• list and discuss the key equations of the IS–LM model\n• describe the determinants of the intercept and the slope of both the IS and the LM curves\n• calculate the equilibrium level of output and the interest rate in a closed economy with fixed wages and prices\n• evaluate how any change in the variables and the parameters of the IS–LM model alters the equilibrium levels of output and the interest rate\n• show how fiscal and monetary policies contribute to the determination of output and the interest rate in the short run, and their use as tools for macroeconomic stabilisation\n• illustrate and exlain the Keynesian view of short-run fluctuations in economic activity.\n\nIntroduction\n\nThe IS–LM model combines the goods market, which determines the equilibrium level of output, with the money market, which determines the equilibrium level of real money, in order to determine under what conditions they are simultaneously in equilibrium, and to assess the role that fiscal and monetary policies may have in explaining real output fluctuations.\n\nThe model focuses on the assessment of aggregate demand in the short run, and in a closed economy. The short run is defined as the period of time during which the price level is fixed (there is no inflation), in the sense that it does not change in response to variations in other macroeconomic variables. Recall that the price level can be thought of as sluggish or sticky variable, since it does not respond immediately to a macroeconomic disturbance. Instead the adjustment builds up gradually, and it is fully completed after several years. In this respect, the long run is a period of time long enough for prices to be considered fully flexible, whereas the medium run is the period of time over which the price level completes its adjustment to macroeconomic disturbances. The closed\neconomy assumption means that the model is studied without taking into account the effect of foreign trade on domestic aggregate demand.\n\nTherefore, the IS–LM model is useful in analysing the effects of stabilisation policy on real output and the real interest rate, but it cannot be employed to assess issues related to inflation, long-run unemployment, and the exchange rate. Because great attention is given to fiscal policy, it is important that you have a clear understanding of some basic assumptions about the role of the government in the economy, the links existing among different fiscal variables, and the standard fiscal policy jargon. Here is a quick summary. The government provides public goods and services (e.g. health, education, roads, defence) that the private sector would either under-produce or not produce at all. The government also redistributes resources among households and firms. These activities can be financed either by levying taxes on households and firms, or by issuing debt, or by issuing money. The relationship between public spending, taxes, debt and money is summarised by the government budget constraint (GBC), an accounting identity that for every generic period t can be written as:\n\nG_t + i_t B_t–1 – T_t = ∆B_t + ∆M_t\n\nwhere G_t is government spending in goods and services during year t; B_t–1 is government debt at the end of year t–1 or, equivalently, at the beginning of year t; i_t B_t–1 measures the interest payments on government debt made by the end of period t at the interest rate i_t ; 1 T_t computes tax revenue net of transfers to households and firms during year t; the term ∆B t = B_t – B_t–1 is the change in debt during year t; and the term ∆M_t = M_t – M_t–1 is the change in money supply during year t. In the rest of this chapter we will assume that the government cannot create money to finance\npublic spending, hence we will set ∆M_t = 0.\n\nThe difference between government spending and tax revenue, G_t – T_t , is referred to as the primary budget deficit. The term G_t + i_t B_t–1 – T_t is instead called the total budget deficit, as it includes interest spending. The standard interpretation of the GBC is that any difference between total spending and tax revenue must be balanced with a corresponding change in the stock of debt. The government runs a balanced primary budget when public spending is entirely financed through taxes, that is G_t = T_t .\n\nWe say that a balanced budget expansion or contraction of the fiscal sector occurs when taxes and government spending either increase or decrease by the same amount.\n\nThe IS–LM model: key assumptions\n\nThe IS–LM model determines under what conditions output and the interest rate are simultaneously in equilibrium in the short run.\n\nOutput is determined in the goods market, which refers to the trade market of all goods and services produced in an economy. The goods market equilibrium is described under the following assumptions:\n\n• All firms produce the same goods, which are then used by consumers for consumption and residential investment, by firms for fixed assets investment, or by the government.\n• Firms are willing to supply any amount of goods at the existing price level.\n• The economy is closed.\n\nThe second assumption is important because it implies that in the IS–LM model aggregate supply has no effect on the equilibrium level of income, which is determined by aggregate demand.\n\nThe interest rate is determined in the money market. The Keynesian Liquidity Preference theory, which you have already encountered in describes money market equilibrium under the following assumptions:\n\n• The financial market includes only two assets: money and bonds.\n• Money is a liquid asset, in the sense that it can be used to purchase goods and services, and pays no interest.\n• Bonds are issued by the government, cannot be used for transactions, but pay a positive nominal interest rate i.\n• Real wealth is fixed in the short run.\n\nBonds are defined as assets that pay a fixed amount of money (face value) at a specific period of time (maturity). Individuals buy bonds at a price lower than the face value. The difference between the price and the face value of a bond is the gain for the bond holder. Since the face value is fixed, the lower the price of a bond the higher is the gain for the holder. The rate of return from investment in bonds, the nominal interest rate, is calculated by dividing the gain from bonds by the current price. Analytically, the relationship between the interest rate and the price of a\nbond is written as:\n\ni = FP_B − P_B/P_B = (FP_B/P_B) -1\n\nwhere FP_B indicates the face value of the bond and P B is the current price of bonds. Thus, the interest rate and the price of bonds are, by construction, inversely related: an increase in the price of bond, P_B , is equivalent to a reduction in the interest rate.\n\nIt is important that you clearly understand the difference between income and wealth. Individuals earn income essentially from three sources: labour, in the form of wages, real assets, in the form of rent, and financial assets, in the form of interest and dividends. Income can either be spent to purchase goods and services or saved to buy new assets. Wealth is defined as the net value of all assets held by an individual, including their human capital. If the level of consumption of an individual exceeds their income, then this individual is defined as a net borrower.\n\nFinally, it is important that you note the equilibrium conditions in the financial market. These state that the demand for money, M^d , must be equal to the supply of money, M^s :\nM^d = M^s\n\nand that the demand for bonds, B^d, must be equal to the supply of bonds, B^s:\nB^d = B^s\n\nIn general, we know from Walras’ law that if there are n markets and the first n–1 of these are in equilibrium, then the nth market must be in equilibrium too. This means that we can choose to ignore one market in the model, knowing that equilibrium there will be assured whenever it holds generally elsewhere. For this reason, we ignore the bond market in what follows, and make use only of the money market equilibrium condition.\n\nThe IS–LM model: behavioural equations and identities\n\nThe IS–LM model is based upon six behavioural equations, each describing the determinants of one of the macroeconomic variables considered by the model: consumption, investment, tax revenue, government spending, money demand and money supply.\n\nConsumption, C, is defined as the value of goods and services purchased by households. The linear consumption function, which depends positively on income, Y, and negatively on taxes, T, can be written as:\n\nC = c0 + c1(Y–T) c0 ≥ 0, 0\n\nwhere the term Y–T indicates disposable income. The parameter c0 measures the level of consumption affected by factors other than disposable income, such as borrowing, sale of real and financial assets, etc. The value of c0 depends also upon consumers’ spending habits and consumer confidence about current and future spending opportunities.\n\nThe parameter c 1 is the marginal propensity to consume. In turn, the value 1– c 1 measures the marginal propensity to save. Since 0\n\nInvestment spending includes three broad categories: fixed business investment carried out by firms, residential investment carried out by households, and inventory investment (i.e. unsold goods and services and/or unused materials).\n\nDemand for investment depends positively on income (sales) and negatively on the real interest rate, r, and it can be written as:\n\nI = I+b1Y – b2r, b1 ≥ 0, b2 ≥ 0\n\nWhere Ī is a constant that computes the effect on investment of any variable other than income and the interest rate, whereas b1 and b2 measure the sensitivity of investment to income and the interest rate, respectively.\n\n[Note that in economics, the symbol i normally indicates the nominal interest rate, whereas the symbol r is used to indicate the real interest rate.]\n\nNominal and real interest rates are linked by the Fisher equation: i = r + p which shows that the nominal interest rate is equal to the sum of the real interest rate and the rate of inflation, p.\n\nSince the Keynesian theory assumes prices to be constant in the short run, the rate of inflation is equal to zero, which implies that the real and the nominal interest rates are equal in the short run, i = r. For this reason we can replace r with i in the demand for investment equation when solving the model.\n\nTax revenue depends upon the structure of the tax system (lum sum, Tfix, proportional income, proportional consumption) . The linear tax function can be written as:\n\nT = Tfix + τ1Y + τ2C, τ1 ≥ 0, τ2 ≥ 0\n\nThis is an approximate version of the equation only.\n\nGovernment spending refers to the demand for goods and services of the public sector and it can be generally described by the equation:\n\nG = Gfix + g1Y + g2C ,\n\nwhere Gfix represents the fixed component of government spending, while g1 and g2 measure changes in public spending proportional to variations in income and consumption, respectively.\n\nThe demand for money, Md , is positively related to income and negatively related to the nominal interest rate. The higher the level of transactions, the more money is demanded for consumption and investment by individuals, firms and the government. The interest rate can be regarded as the opportunity cost of holding money. The higher the interest rate, the higher is the cost of holding money rather than bonds, and thus the lower is money demand. Analytically, real money demand is described as:\n\nM^d/Pfix = = h0 + h1Y − h2i, h1 ≥ 0, h2 ≥ 0\n\nwhere P indicates the price level, which is constant in the IS–LM model, h0 measures the level of demand for money independent of income and the interest rate, and the parameters h1 and h2 measure the sensitivity of money demand to income and the interest rate respectively.\n\nNote that equation shows that demand for money is ultimately a demand for real balances, M^d/Pfix , rather than nominal balances, M^d . The underlying assumption behind this specification is that individuals are free from money illusion, namely the tendency of individuals to think of currency in nominal terms rather than taking into account its purchasing power (**COMMENT** c.f., above on factors of production **COMMENT**).\n\nThe supply of money, Ms , is assumed to be independent from the interest rate and directly controlled by the central bank. The central bank can change money supply through open market operations. An expansionary open market operation occurs when the central bank buys bonds to increase the money supply. When the central bank buys bonds from the private sector, the excess demand for bonds raises the price of bonds, in turn reducing the interest rate. This reduction in the opportunity cost of holding money simutaneously increases money demand. In contrast, a contractionary open market operation occurs when the central bank sells bonds. As a result of excess supply the bond price falls, the interest rate increases, and money demand decreases.\n\nAnalytically, real money supply, MsSqrtP is written as:\n\nMs/P = M/P'\n\nWhere M is the level of nominal money supply chosen by the central bank.\n\nIncome accounting in a closed economy\n\nThe equilibrium output in the goods market can be alternatively retrieved from the equality between saving and investment in the loanable funds market. Consider again the income identity in equation (2.9):\n\nY = C + I + G.\n\nNext, move C and G to the left-hand side of the equation and add and\nsubtract T to obtain:\n\n(Y – T – C) + (T – G) = I.\n\nThe term on the left hand side is the total saving of the economy. In particular, S = Y – T – C indicates private saving, measured by the excess of income over taxes and consumption, whereas T – G is public saving, measured by the primary budget surplus. 8\n\nTherefore, the equilibrium output in the goods market can be alternatively determined by the condition ‘saving equals investment’: (2.10)\n\nS + T – G = I\n\nAt this stage, make sure you are familiar with the concept, already encountered in EC1002 Introduction to economics, of the ‘paradox of thrift’ (or ‘paradox of saving’) which states that an exogenous increase in saving, equivalent to lower consumption at any given income level, leads to a lower equilibrium level of output and unchanged aggregate saving so long as the interest rate stays constant.\n\nEquation (2.10) gives an alternative way of writing the income identity in a closed economy, by relating private saving to the budget deficit and investment. The identity states that, in a closed economy, investment can be financed by a combination of private and public saving. In addition, changes in any of these three variables must necessarily affect at least one of the other two. However, the identity is silent about the determinants of changes in saving and investment, as well as the extent of their effect on other macroeconomic variables.\n\nThe IS curve\n\nThe IS curve represents combinations of income and the interest rate, such that the goods market is in equilibrium.\n\nAnalytically the IS curve is computed by replacing in the income identity (2.9) the behavioural equations for consumption, investment, tax revenue and government spending. The resulting equation has to be solved for the interest rate as a function of income, as the IS curve is always plotted in the income–interest rate space.\n\nBefore deriving the IS curve, I will simplify equations (2.3) and (2.4) by assuming that government spending is exogenous (g 1 = g 2 = 0) whereas taxes do not depend on consumption (τ 2 = 0).\n\ni^* (1/b^2)(c_0 - c_1 barT + barI + barG) - (((1-c_1)(1-τ_1)-b)/(b^2))Y^*\n\nIn the above expression the symbols i* and Y* denote equilibrium values. Note we have used the assumption of fixed prices to replace the real interest rate r with the nominal rate i. The slope of the IS curve in absolute terms is negatively related to the marginal propensity to consume, c1 , the responsiveness of investment to output, b1 , and the interest rate, b2 ; it is positively related to the tax rate, τ1.\n\nThe larger the marginal propensity to consume, the flatter the IS curve, since a given increase in investment caused by a reduction in the interest rate, will then deliver a larger increase in income via the multiplier. Similarly, the more interest-sensitive investment spending is, the flatter is the IS curve, since a given reduction in the interest rate causes a relatively large increase in investment spending, which results in a larger increase in equilibrium income. An increase in b1 makes the IS curve flatter, since investment spending responds more to income for any given interest rate, and this amplifies the multiplier effect. Conversely, tax rate increases make the IS curve steeper as they dampen the multiplier effect of higher aggregate income on consumer expenditure.\n\nThe intercept of the IS curve depends upon the level of autonomous spending, c0 − c1 T + I + G . An increase (reduction) of autonomous spending shifts the IS curve upward (downward). Note also that the position of the IS curve is affected by the responsiveness of investment to the interest rate: the larger b 2 the smaller the intercept of the IS curve.\n\nIf the central bank controls the interest rate, the equilibrium level of income moves along the IS curve. An increase (decrease) in government spending raises aggregate demand, thus shifting upwards (downwards) the IS curve. The effect of changes in taxation depends upon the structure of the tax system. If taxes are levied as lump-sums, then tax changes can only change the position of the IS curve, without altering its slope.\n\nHowever, if taxes are levied as a proportion of consumers’ income, any tax policy change has an effect on the slope of the IS curve.\n\nThe LM curve\n\nThe LM curve comprises combinations of the interest rate and income, for which the money market is in equilibrium.\n\nAnalytically, the LM curve is computed combining the equations for money demand (2.5) and supply (2.6). The resulting equation has to be solved for the interest rate as a function of income, since the LM curve is always plotted in the income-interest rate space. This yields the following\nequation:\n\ni^* = (1/h_2)(h_0 - (M/P) + (h_1/h_2) Y^*\n\nThe slope of the LM curve depends on the sensitivity of money demand to income and the interest rate, as measured by the coefficient h1 / h2 . The more money demand is sensitive to income, relative to the interest rate, the steeper the LM curve. If money demand does not respond to the interest rate, h2 = 0, the LM curve is vertical. A vertical LM curve is often referred to as the classical case. In contrast, if money demand is very sensitive to the interest rate, h2 = ∞ , the LM curve is horizontal. In this case it is not possible for the central bank to change the equilibrium interest rate through variations in the money supply. The case of a horizontal LM curve is often referred to as a liquidity trap.\n\nMonetary policy affects the position of the LM curve. An expansionary monetary policy shifts the LM curve downwards, since it increases the liquidity in the money market and reduces the interest rate for any given level of income. In contrast, a contractionary monetary policy shifts the LM curve upwards, as it reduces the liquidity in the money market and increases the interest rate at any given level of income.\n\nIt may be useful to recall that, at this stage of the subject, the conduct of monetary policy is based upon the following simplifying assumptions:\n1. The central bank has direct control over money supply through open market operations.\n2. The government issues bonds on behalf of the central bank; the central bank does not directly issue bonds, but can only create money to buy bonds issued by the government.\n3. Individuals are always willing to trade bonds at some price (i.e. bond demand from the private sector is unlimited).\n\nTo understand the link between monetary policy and money supply, you may find it convenient to consider the simplified central bank balance sheet in Table 2.1. The central bank’s assets are represented by bonds, while the central bank’s liabilities are represented by the currency (money) held by the public. To increase the money supply, the central bank has to purchase new bonds. This increases both assets (through the additional bonds) and liabilities (through the new currency created and exchanged for bonds). To reduce the money supply, the central bank sells bonds for existing currency. This operation reduces assets (through the sale of bonds) and liabilities (through the reduction of currency held by the general public).\n\n Central Bank Assets Liabilities Bonds Money\n\nTable 2.1: Central bank balance sheet in a closed economy.\n\nA more accurate description of the central bank’s balance sheet and the mechanism of money creation will be provided in Chapter 11 of the subject guide.\n\nEquilibrium in the IS–LM model\n\nThe IS–LM model determines combinations of the interest rate and income that simultaneously satisfy the equilibrium condition in the goods market and in the money market. Analytically, the equilibrium level of output and the interest rate are computed by combining the IS equation in (2.11) with the LM equation in (2.12). For instance, if we replace the interest rate in the IS equation with the right-hand side of the LM equation, the equilibrium level of income is calculated as:",
null,
"Note that the policy variables G and M both increase the equilibrium level of income. Tax policy affects Y* negatively through the term c1 T and the term c1 τ1 . Therefore, expansionary fiscal and monetary policies increase the equilibrium level of income. Conversely, fiscal and monetary contractions reduce the equilibrium level of income.\n\nAlternatively, you can compute the equilibrium level of the interest rate by replacing the income level in the IS equation with the level of income from the LM equation. This yields the following expression for the equilibrium interest rate:",
null,
"Fiscal expansions and monetary contractions increase the interest rate, whereas fiscal contractions and monetary expansions reduce the interest rate.\n\n Fiscal policy Expansionary Fiscal policy Contractionary Monetary policy Expansionary Monetary policy Contractionary Output + - + - Interest rate + - - +\n\nThe table shows that the effect of fiscal policy, on output and the interest rate, is symmetric in the sense that an expansionary fiscal policy increases both output and the interest rate, while a contractionary fiscal policy reduces both variables. In contrast, monetary policy has an asymmetric effect on output and the interest rate: an expansionary monetary policy increases output while reducing the interest rate, whereas a contractionary monetary policy reduces output while increasing the interest rate.\n\nFiscal and monetary policy in the IS–LM model\n\nFigure 2.1 provides a graphical illustration of the effect of fiscal and monetary policy in the IS–LM model. In each panel, the initial equilibrium is the point E 1 , which corresponds to the level of income Y 1 and interest rate i 1 . Panel A shows that a fiscal expansion affects both the goods and the money market. In the goods market, the increase in aggregate demand resulting from the fiscal expansion raises the equilibrium level of income. Since the money supply is fixed, the increase in income increases money demand, which can only be accommodated by an increase in the interest rate. In turn, the increase in the interest rate causes a fall in investment spending (crowding out effect), which partially offsets the initial increase in income. The final equilibrium position is indicated in the graph by point E 2 , but, in general, depends upon the slope of the LM curve relative to the IS curve.\n\nConversely, Panel B shows that a fiscal contraction reduces the equilibrium levels of income and interest rate. Note that the fiscal contraction reduces aggregate demand and income in the goods market. The fall in income causes a contraction in money demand, and money market equilibrium is restored only by a fall in the interest rate. In turn, the fall in the interest rate stimulates investment spending and contributes to partially offsetting the initial reduction in income.\n\nPanel C shows that a monetary expansion shifts the LM curve downwards. At the initial level of income, the interest rate drops from i 1 to i 3 , since the expansionary monetary policy operation raises the price of bonds and reduces the interest rate. The fall in the interest rate stimulates investment spending and increases output. Ultimately, the equilibrium converges to point E 2 . In general, the flatter the IS curve, the greater the monetary policy stimulus on output. Panel D shows how the IS–LM equilibrium adjusts as a result of a monetary contraction.",
null,
"The effectiveness of monetary and fiscal policy – on the interest rate, income, and unemployment – crucially depends on the relative slopes of the IS and the LM curves. Broadly speaking, the flatter (steeper) the IS curve is, the more (less) effective is monetary policy on output and unemployment, relative to the interest rate. On the other hand, the flatter (steeper) the LM curve is, the more (less) effective is fiscal policy on output and unemployment relative to the interest rate.\n\nFigure 2.2 illustrates four extreme cases. A liquidity trap occurs when the public is prepared to hold any amount of money at the current interest rate. This implies that money demand is horizontal in the real money–interest rate space, and monetary policy is ineffective because changes in the money supply do not alter income and the interest rate. As a consequence, the LM curve is horizontal. If the economy is in a liquidity trap, fiscal policy is very effective, as there is no crowding out effect on investment following a fiscal expansion (Panel A). The classical case occurs when money demand is entirely unresponsive to the interest rate, so that the LM curve is vertical (Panel B). This is consistent with the classical quantity theory of money, which states that nominal income, PY, is entirely determined by the money supply. In this case, if we assume that the price level is fixed then a monetary expansion has a one-to-one effect on real income, whereas fiscal policy has no effect on income because the fiscal expansion causes the interest rate to rise, reducing investment spending one-for-one with the rise in government spending. 9\n\nThe effectiveness of monetary policy depends upon the slope of the IS curve. If investment is fully sensitive to the interest rate, then the IS curve is horizontal. This implies that an expansionary monetary policy will exert all its effect (Panel C). Vice versa, if investment is completely insensitive to the interest rate the IS curve is vertical, which implies monetary policy ineffectiveness (Panel D).",
null,
"Policy mixes\nThe IS–LM model shows that fiscal and monetary policies can be used either in isolation, or simultaneously, to influence key macroeconomic variables, such as the real interest rate and output. Since output and employment are positively correlated, that is employment increases (decreases) when output increases (decreases), fiscal and monetary policy also influence the unemployment rate.\n\nIt is important that you know how to employ the IS–LM model to prescribe the appropriate policy mix to achieve a specific macroeconomic outcome. For instance, if the government and the central bank plan to increase output, without changing the interest rate, this goal can be achieved through a combination of expansionary fiscal policy and expansionary monetary policy. The expansionary fiscal policy increases both output and the interest rate, while the expansionary monetary policy increases output, but reduces the interest rate (Figure 2.3, Panel A). Another example occurs when policy-makers want to reduce the interest rate, while keeping the level of output unaffected. This goal can be achieved through a combination of expansionary monetary policy, which reduces the interest rate and increases output, and a contractionary fiscal policy, which will further contribute to the reduction in the interest rate while reducing output (Figure 2.3, Panel B).",
null,
"## Blanchard, O., Macroeconomics (Fifth Edition), Pearson International Edition, 2009.\n\nChapter 3\n\nProduction = Sales + Inventory; Inventory = Production - Sales; Sales = Production - Inventory\n\nTotal demand for goods Z := C + I + G + X - IM\n\nC_0 is what you must spend (even if income is 0). C_1 is marginal propensity to consume. Consumption function, y axis = Consumption, x axis = disposable income, C_0 is starting point on y at x = 0, C_1 is upward slope, less than 45 deg (usually?) C = c_0 + c_1 Y_D (disposable income). Y_D = Y-T\n\nEquibrium in the goods market sugggests that production equals demand. Demand depends on income which is equal to production.\n\nY = C + I + G; Y - T - C = I + G - T; Y - T - C = S = I + G - T; I = S + (T - G)\n\nChapter 4\n\nSimplify money that pays not interest; bonds as all interest-bearing accounts.\n\nA low interest rate increases the demand for money. An increase in nominal income increases interest rate (M^S is vertical M^D downward sloping, y-axis interest, x-axis money demand)\n\nChapter 5\n\nFiscal Policy Shift of IS Shift in LM Output Interest Rate\nIncrease Taxes Left None Down Down\nDecrease Taxes Right None Up Up\nIncrease Spending Right None Up Up\nDecrease Spending Left None Down Down\nIncrease Money None Down Up Down\nDecrease Money None Up Down Up\n\n## Dornbusch R., Fischer, S., Startz, R., Macroeconomics (9th edition, international), McGraw-Hill, 2004\n\nChapter 9\n\nAggregate demand is the total amount of good demanded in the economy: AD = C + I + G + NX\n\nOutput is at an equilibrium when the quantity produced is equal to the amount demanded: Y = AD = C + I + G + NX\n\nThe equilibrium level of output is higher the larger the marginal propensity to consume, c, and the higher the level of autonomous spending barA (or c_1, c_0 in Mankiw)\n\nChapter 10\n\nThe IS-LM model finds the values of GDP and the interedst rate which simultaneously clear the goods and money markets.\n\nThe IS curve (or schedule) show combinations of interest rates and levels of output such that planned spending equals income.\n\nThe smaller the sensitivity of investment spending to the interest rate the smalleer the multiplier, the steeper the IS curve.\n\nThe LM curve (or schedule) show combinations of interest rates and levels of output such that money demand equals money supply.\n\nThe greater the responsiveness of the demand for money to income the lower the responsiveness or the demand for money to the interest rate, the steeper the LM curve will be.\n\nChapter 11\n\nWhen the LM curve is vertical monetary policy has a maximal effect on the level of income, and fiscal policy has no effect on income.\n\nCrowding out occurs when expansionary fiscal policy causes interest rates to rise, thereby reducing private spending, particularly investment.\n\n Interest Rate Consumption Investment GDP Income Tax Cut + + - + Government Spending + + -+ Investment Subsidy + + ++\n\nReal interest rate is nominal rate minus inflation\n\n## Mankiw N.G., Taylor M.P., Macroeconomics (European Edition), Worth, 2008\n\nChapter 9\n\n\"*In the long run, prices are flexible and can respond to changes in supply and demand. In the short run, many prices are 'sticky' at some predetermined level*\". p276\n\nIn classic macroeconomic theory output depends on the ability to supply, which depends on capital and labour. When prices are sticky output also depends on demand. Hence a model of aggregate supply and aggregate demand. p278.\n\n\"**Aggregate demand** (AD) is the relationship between the quantity of output demanded and tha aggregate price level\" p278\n\nAD slopes downward, Y-axis P (price level), X-axis income, output Y. High price, low demand/output, low price, high demand/output. Reduction in money supply moves aggregate demand to the left, increase moves to the right.\n\nMV = PY (money supply, M; V velocity of money; P price level, Y is output). Also;\nM/P = (M/P)^d = kY, where KY = 1/v, a parameter determining how much money people want to hold for every euro of income, supply of real money balances M/P equats the deamnd for real money balances (M/P)^d\n\n\"**Aggregate Supply** (AS) is the relationshiup between the quantity of goods and services supplied and the price level\" p281.\n\nAS slopes upward, Y- axis P (price level), X-axis output Y. Differentiate between Long-Run Aggregate Supply (LRAS) and Short-Run Aggregate Supply (SRAS).\n\nIn LRAS firms can change prices to adapt, but in SRAS prices are sticky. LRAS curve is vertical, barY; output is measured by Kapital and Labour (K&L). A fall in AD will lower the price in LRAS, but the output will remain the same.\n\nIn SRAS firms cannot change prices immediately; SRAS curve is horizontal. Regardless of changes in demand, price remains the same. A fall in AD still results in the same price P.\n\nThe long-run equilibrium is where LRAS, SRAS, and AD intersect.\n\nA shift to the AD is called a demand shock, a shift to AS is called a supply shock. Stablisation policy aims to minimuse the severity of short-run fluctuations.\n\ne.g., readily available credit cards reduces money demand with an equivalent increase in money velocity (1/k). Spending rises and a there's an economic boom. Over time, pulls up wages and prices. A stablisation policy to offset the increase in velocity would be to reduce money supply. An adverse supply shock (e.g., destruction of crops) or a favourable price shock (e.g., break up of an oil cartel). Adverse cause SRAS to shift upwards, increasing prices and demand falls (stagflation, increasing prices, falling output). Stablisation will be by raising aggregate demand, which stablises output but at a higher price level.\n\nWhen GDP declines so does consumption (by a smaller amount) and investment (by a larger amount), and unemployment rises.\n\nChapter 10\n\nTwo parts of IS-LM are investment-saving and liquidity-money. Interest rate influences both and links the two variables.\n\nKeynesian Cross: Actual expenditure versus desired Planned expenditure.\n\nIn a closed economy E = C + I + G (planned expenditure = consumption plus investment, government expenditure). Assume fixed I and G and fixed taxes, becomes E = C(Y-barT) + barI + barG.\n\nPlanned expenditure depends on income, because higher income leads to higher consumption.\n\nThe economy is in equilibrium when planned expenditure = actual expenditure, E = Y, 45 deg on the Keynesian cross, E = planned expenditure, y axis, Y equals income, output X axis. Planned starts higher but has lower gradiant. Too much Y (actual) over E (planned) means inventory accumulation, incomes fall, too much E (planned) over Y (actual) means inventory declines, incomes rise.\n\nA change in G causes a greater change in Y; deltaY/deltaG > 1. Higher income means high econsumption, which raises income, which raises consumption etc. How much? Depends on Marginal Propensity to Consume (MPC).\n\ndeltaY/deltaG = 1/(1-MPC) (limit on geometric series)\n\nLikewise a decrease in taxes will have the same effect (a \"tax multiplier\"). A decrease in taxes raises effective income and output which leads to an increase in interest rates.\n\nFiscal policy, however, is also sticky! Government expenditure takes time, tax cuts take time etc. Interest rates to the rescue! An increase in interest rates reduces investment which shifts planned expenditure downward, and a fall in income and output.\n\nThe IS Curve is derived from the downward sloping investment Curve and the Keynesian Cross.\n\nA loanable funs interpretation of the IS Curve: Y - C -G = I; S = I. Y - C(Y-T) - G = I(r)\n\nAn increase in income raises saving causing interest rates to drop.\n\nLiquidity Preference: In the short-run the interest rate adjusts to balance the supply and demand for the economy's most liquid asset - money. Keynesian cross builds the IS curve, liquidity preference the LM curve.\n\nAssume fixed supply of real money balances (M/P)^s = barM/baP; Money supply is exogenuous variable chosen by the central bank. Supply (barM/barP) is fixed. Demand, downward slopin L(r). Y axis interest rate, r, x axis real money balances M/P. High interest rate, low M/P, low interest rate, high M/P. Shift right change in barM/barP (expansion) will increase real money balances, reduce interest rates, shift left (contract) will reduce, increase interest rates.\n\nWhen income is high, expenditure is high, more transactions, greater money demand; (M/P)^d = L(r,Y). The quanity of real money balances is negatively related to the interest rate and positively to inocme.\n\nAn increase in income raises money demand an increases interest rate which is derived to the upward sloping LM curve. A reduction in money supply shifts the LM curve upward, increasing interest rates.\n\nThe whole equation:\n\nY = C(Y-T) + I(r) + G is IS downward sloping, interest r (Y axis), income, output Y (x axis)\nM/P = L(r, Y) is LM is upward sloping."
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"http://levlafayette.com/files/EC2065_equilibriumincome.png",
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"http://levlafayette.com/files/EC2065_equilibriuminterest.png",
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https://byjus.com/vector-projection-calculator/ | [
"",
null,
"# Vector Projection Calculator\n\nEnter the First Vector A (a1i+b1j) = i + j\n\nEnter the Second Vector B (a2i+b2j) = i + j\n\nScalar Projection =\n\nVector Projection = (, )\n\nVector Projection Calculator is a free online tool that displays the vector projection for the given two vectors. BYJU’S online vector projection calculator tool makes the calculation faster, and it displays the vector projection in a fraction of seconds.\n\n## How to Use the Vector Projection Calculator?\n\nThe procedure to use the vector projection calculator is as follows:\n\nStep 1: Enter the coefficients of the vector components in the input field\n\nStep 2: Now click the button “Find Vector Projection” to get the result\n\nStep 3: Finally, the vector projection will be displayed in the output field\n\n### What is Meant by the Vector Projection?\n\nThe vector projection is used to find the component of the vectors along with the direction. We know that vectors have both magnitude and direction.\n\nAlso, check:\n\nIn other words, the vector projection is defined as a vector in which one vector is resolved into two component vectors. Among the two-component vectors, one is parallel to the second vector, and the other one is perpendicular to the second vector. In this parallel vector is called the vector projection."
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"https://www.facebook.com/tr",
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https://groups.google.com/a/groups.riscv.org/g/isa-dev/c/_CA4k785q9s | [
"# RISC-V - Function like representation of Instructions\n\n48 views\n\n### Jim Kenua\n\nApr 19, 2023, 9:05:40 AMApr 19\nto RISC-V ISA Dev\nHello,\n\nI am working on a function-like representation of the instructions, each time I had to do changes on a core I had to look into the ISA manuals and read quite a number of pages just to remember the actual functions done by each instruction.\n\nIf someone knows a source from where I can extract a similar form it would great if it could be shared.\n\nPlease let me know if the below functions are miss wrote or if someone is interested in helping to write these.\n\nThank you all!\n\n#######\nLUI -> IRF[rd] = {SignExt(imm),12'd0};\nAUIPC -> IRF[rd] = PC + {SignExt(imm),12'd0};\n#######\n\n#######\nJAL -> IRF[rd] = PC + 4; PC = (PC + {SignExt(imm),1'b0});\nJALR -> IRF[rd] = PC + 4; PC = (IRF[rs1] + {SignExt(imm),1'b0});\n#######\n\n#######\nBEQ -> if (Signed / Unsigned(IRF[rs1]) == Signed / Unsigned(IRF[rs2])) PC = (PC + {SignExt(imm),1'b0});\nBNE -> if (Signed / Unsigned(IRF[rs1]) != Signed / Unsigned(IRF[rs2])) PC = (PC + {SignExt(imm),1'b0});\nBLT -> if (Signed (IRF[rs1]) < Signed (IRF[rs2])) PC = (PC + {SignExt(imm),1'b0});\nBGE -> if (Signed (IFR[rs1]) >= Signed (IRF[rs2])) PC = (PC + {SignExt(imm),1'b0});\nBLTU -> if ( Unsigned(IRF[rs1]) < Unsigned(IRF[rs2])) PC = (PC + {SignExt(imm),1'b0});\nBGEU -> if ( Unsigned(IRF[rs1]) >= Unsigned(IRF[rs2])) PC = (PC + {SignExt(imm),1'b0});\n#######\n\n#######\nLB -> IRF[rd] = SignExt(MEM[IRF[rs1] + SignExt(imm)][ 7:0]) // 1 Byte\nLBU -> IRF[rd] = ZeroExt(MEM[IRF[rs1] + SignExt(imm)][ 7:0]) // 1 Byte\nLH -> IRF[rd] = SignExt(MEM[IRF[rs1] + SignExt(imm)][15:0]) // 2 Bytes\nLHU -> IRF[rd] = ZeroExt(MEM[IRF[rs1] + SignExt(imm)][15:0]) // 2 Bytes\nLW -> IRF[rd] = SignExt(MEM[IRF[rs1] + SignExt(imm)][31:0]) // 4 Bytes\nLWU -> IRF[rd] = ZeroExt(MEM[IRF[rs1] + SignExt(imm)][31:0]) // 4 Bytes\nLD -> IRF[rd] = (MEM[IRF[rs1] + SignExt(imm)][63:0]) // 8 Bytes\n#######\n\n#######\nSB -> MEM[IRF[rs1] + SignExt(imm)] = IRF[rs2][ 7:0]; // 1 Byte\nSH -> MEM[IRF[rs1] + SignExt(imm)] = IRF[rs2][15:0]; // 2 Bytes\nSW -> MEM[IRF[rs1] + SignExt(imm)] = IRF[rs2][31:0]; // 4 Bytes\nSD -> MEM[IRF[rs1] + SignExt(imm)] = IRF[rs2][63:0]; // 8 Bytes\n#######\n\n#######\nSLTI -> IRF[rd] = ZeroExt( Signed(IRF[rs1]) < Signed(SignExt(imm)));\nSLTIU -> IRF[rd] = ZeroExt(Unsigned(IRF[rs1]) < Unsigned(SignExt(imm)));\n\nSLT -> IRF[rd] = ZeroExt( Signed(IRF[rs1]) < Signed(IRF[rs2]));\nSLTU -> IRF[rd] = ZeroExt(Unsigned(IRF[rs1]) < Unsigned(IRF[rs2]));\n#######\n\n#######\nSLLIW -> IRF[rd] = SignExt((IRF[rs1] << imm[4:0])[31:0]);\nSRLIW -> IRF[rd] = SignExt((IRF[rs1] >> imm[4:0])[31:0]);\nSRAIW -> IRF[rd] = SignExt((IRF[rs1] >>> imm[4:0])[31:0]);\nSLLI -> IRF[rd] = IRF[rs1] << imm[5:0];\nSRLI -> IRF[rd] = IRF[rs1] >> imm[5:0];\nSRAI -> IRF[rd] = IRF[rs1] >>> imm[5:0];\n\nSLLW -> IRF[rd] = SignExt((IRF[rs1] << IRF[rs2][5:0])[31:0]);\nSRLW -> IRF[rd] = SignExt((IRF[rs1] >> IRF[rs2][5:0])[31:0]);\nSRAW -> IRF[rd] = SignExt((IRF[rs1] >>> IRF[rs2][5:0])[31:0]);\nSLL -> IRF[rd] = IRF[rs1] << IRF[rs2][5:0];\nSRL -> IRF[rd] = IRF[rs1] >> IRF[rs2][5:0];\nSRA -> IRF[rd] = IRF[rs1] >>> IRF[rs2][5:0];\n#######\n\n#######\nADDI -> IRF[rd] = IRF[rs1] + SignExt(imm);\nXORI -> IRF[rd] = IRF[rs1] XOR SignExt(imm);\nORI -> IRF[rd] = IRF[rs1] OR SignExt(imm);\nANDI -> IRF[rd] = IRF[rs1] AND SignExt(imm);\n#######\n\n#######\nADDW -> IRF[rd] = SignExt((IRF[rs1] + IRF[rs2])[31:0]);\nSUBW -> IRF[rd] = SignExt((IRF[rs1] - IRF[rs2])[31:0]);\n\nADD -> IRF[rd] = IRF[rs1] + IRF[rs2];\nSUB -> IRF[rd] = IRF[rs1] - IRF[rs2];\nXOR -> IRF[rd] = IRF[rs1] XOR IRF[rs2];\nOR -> IRF[rd] = IRF[rs1] OR IRF[rs2];\nAND -> IRF[rd] = IRF[rs1] AND IRF[rs2];\n#######\n\n#######\nMULW -> IRF[rd] = SignExt((Signed(IRF[rs1][31:0]) * Signed(IRF[rs2][31:0]))[31:0]);\n\nMUL -> IRF[rd] = ( Signed(IRF[rs1]) * Signed(IRF[rs2]))[ XLEN -1:0 ];\n\nMULH -> IRF[rd] = ( Signed(IRF[rs1]) * Signed(IRF[rs2]))[(XLEN*2)-1:XLEN];\nMULHU -> IRF[rd] = (Unsigned(IRF[rs1]) * Unsigned(IRF[rs2]))[(XLEN*2)-1:XLEN];\nMULHSU -> IRF[rd] = ( Signed(IRF[rs1]) * Unsigned(IRF[rs2]))[(XLEN*2)-1:XLEN];\n#######\n\n#######\nDIV -> if (IRF[rs1] == -2**63 & IRF[rs2] == -1) IRF[rd] = -2**63 ; else\nif (IRF[rs2] == 0 ) IRF[rd] = -1 ; else\nIRF[rd] = Signed(IRF[rs1]) / Signed(IRF[rs2]); // Round towards Zero\n\nDIVW -> if (IRF[rs1] == -2**31 & IRF[rs2] == -1) IRF[rd] = -2**31 ; else\nif (IRF[rs2] == 0 ) IRF[rd] = -1 ; else\nIRF[rd] = SigneExt(Signed(IRF[rs1][31:0]) / Signed(IRF[rs2][31:0])); // Round towards Zero\n\nDIVU -> if (IRF[rs2] == 0) IRF[rd] = (2**64)-1 ; else\nIRF[rd] = Unsigned(IRF[rs1]) / Unsigned(IRF[rs2]); // Round towards Zero\n\nDIVUW -> if (IRF[rs2] == 0) IRF[rd] = (2**32)-1 ; else\nIRF[rd] = SigneExt(Unsigned(IRF[rs1][31:0]) / Unsigned(IRF[rs2][31:0])); // Round towards Zero\n#######\n\n#######\nREM -> if (IRF[rs1] == -2**31 & IRF[rs2] == -1) IRF[rd] = 0 ; else\nif (IRF[rs2] == 0 ) IRF[rd] = IRF[rs1] ; else\nIRF[rd] = Signed(IRF[rs1] % IRF[rs2]); // Sign of the result is equal to the dividend (rs1)\n\nREMW -> if (IRF[rs1] == -2**31 & IRF[rs2] == -1) IRF[rd] = 0 ; else\nif (IRF[rs2] == 0 ) IRF[rd] = IRF[rs1] ; else\nIRF[rd] = SigneExt(Signed(IRF[rs1][31:0]) % Signed(IRF[rs2][31:0]));\n\nREMU -> if (IRF[rs2] == 0) IRF[rd] = IRF[rs1] ; else\nIRF[rd] = Unsigned(IRF[rs1]) % Unsigned(IRF[rs2]);\n\nREMUW -> if (IRF[rs2] == 0) IRF[rd] = IRF[rs1] ; else\nIRF[rd] = SigneExt(Unsigned(IRF[rs1][31:0]) % Unsigned(IRF[rs2][31:0]));\n#######\n\n####### CSR read accesses are classified as device input (I), and CSR write accesses are classified as device output (O).\nCSRRW -> CSR_FILE[csr] = IRF[rs1] ; if (rd != 0) IRF[rd] = ZeroExt(CSR_FILE[csr]);\nCSRRWI -> CSR_FILE[csr] = ZeroExt(uimm); if (rd != 0) IRF[rd] = ZeroExt(CSR_FILE[csr]);\n\nCSRRS -> if (rs1 != 0) CSR_FILE[csr] = IRF[rs1] | CSR_FILE[csr]; IRF[rd] = ZeroExt(CSR_FILE[csr]);\nCSRRSI -> if (uimm != 0) CSR_FILE[csr] = ZeroExt( uimm) | CSR_FILE[csr]; IRF[rd] = ZeroExt(CSR_FILE[csr]);\n\nCSRRC -> if (rs1 != 0) CSR_FILE[csr] = ~IRF[rs1] & CSR_FILE[csr]; IRF[rd] = ZeroExt(CSR_FILE[csr]);\nCSRRCI -> if (uimm != 0) CSR_FILE[csr] = ZeroExt(~uimm) & CSR_FILE[csr]; IRF[rd] = ZeroExt(CSR_FILE[csr]);\n#######\n\n### Tommy Murphy\n\nApr 19, 2023, 9:18:47 AMApr 19\nto Jim Kenua, RISC-V ISA Dev\nWhat about GCC's machine description files?\n\nFor example:\n\nFrom: Jim Kenua <[email protected]>\nSent: Wednesday, April 19, 2023 2:05:40 PM\nTo: RISC-V ISA Dev <[email protected]>\nSubject: [isa-dev] RISC-V - Function like representation of Instructions\n\n--\nYou received this message because you are subscribed to the Google Groups \"RISC-V ISA Dev\" group.\nTo unsubscribe from this group and stop receiving emails from it, send an email to [email protected].\n\n### Kito Cheng\n\nApr 19, 2023, 9:27:48 AMApr 19\nto Tommy Murphy, Jim Kenua, RISC-V ISA Dev\n\n### Rishiyur Nikhil\n\nApr 19, 2023, 10:09:51 AMApr 19\nto Kito Cheng, Tommy Murphy, Jim Kenua, RISC-V ISA Dev\nI second the recommendation and link in the previous message by Kito Cheng,\ni.e., the Sail RISC-V ISA formal spec.\nNikhil"
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https://www.homeworkmarket.com/content/computegrade-python | [
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null,
"lkatie\n\nRewrite the grade program from the previous chapter using a function called computegrade that takes a score as its parameter and returns a grade as a string.\n\n Score Grade > 0.9 A > 0.8 B > 0.7 C > 0.6 D <= 0.6 F\n\nHINT:\n\n1. Write a custom function with the definition def computegrade(<your score variable>):\n\n2. The function should compute the grade using the score variable based on the table above. You can use if-elif statements to accomplish this\n\n3. Ensure your function has a “return” statement to return the computed grade back to the calling program\n\n4. Once the function is defined, proceed to writing the part of the program that takes in user input for score.\n\n5. Validate that the score is numeric within a try-except block.\n\n6. Invoke the above-defined computegrade function with the user-supplied score. Ensure you capture the return value within your program into a variable placeholder, or else use the print statement to print the return value directly\n\n8. Remember your Python code will be graded according to our class Rubric.\n\n9. Submit your Python code file. Name it “XXXX-score-function.py” where XXXX is your name.\n\n10. Submit also a Word document showing screen shots of the various testing conditions. Good programmers test all cases, so you should make sure you show tests for each possible grade level (A through F) plus error messages.\n\nhere is the code I did. run it, it works.\n\njust need to reflect the computegrade and make that work too\n\n# Given a given a numerical grade between 0.0 and 1.0\n\n#This program converts the numerical score in letter grade.\n\n#Prompt user for score between 0.0 and 1.0\n\nprompt1 = \"Please enter a score between 0.0 and 1.0: \"\n\n#convert input to float, because the input statement accepts user input as a string by default\n\n#Wrap within a try-except code block to exit gracefully as the conversion will fall if the user supplied a non-numeric input!\n\ntry:\n\nscore = input (prompt1)\n\nscore = float(score)\n\n#Print the result\n\nif score <=1.0:\n\nif 0.9 <= score <= 1.0:\n\nelif score >= 0.8:\n\nelif score >= 0.7:\n\nelif score >= 0.6:\n\nelif score < 0.6:\n\nelse:\n\nprint (\"Error, score cannot be greater than 1.0\")\n\nexcept:\n\nprint (\"Error, please enter a number\")\n\n#End of Script\n\n• 3 years ago\n• 7\n•",
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https://digitalcommons.kettering.edu/mech_eng_facultypubs/11/ | [
"## Mechanical Engineering Publications\n\n#### Title\n\nQuantifying R2 bias in the presence of measurement error\n\nArticle\n\n4-1-2010\n\n#### Publication Title\n\nJournal of Applied Statistics\n\n#### Abstract\n\nMeasurement error (ME) is the difference between the true unknown value of a variable and the data assigned to that variable during the measuring process. The multiple correlation coefficient quantifies the strength of the relationship between the dependent and independent variable(s) in regression modeling. In this paper, we show that ME in the dependent variable results in a negative bias in the multiple correlation coefficient, making the relationship appear weaker than it should. The adjusted R 2 provides regression modelers an unbiased estimate of the multiple correlation coefficient. However, due to the ME induced bias in the multiple correlation coefficient, the otherwise unbiased adjusted R 2 under-estimates the variance explained by a regression model. This paper proposes two statistics for estimating the multiple correlation coefficient, both of which take into account the ME in the dependent variable. The first statistic uses all unbiased estimators, but may produce values outside the [0,1] interval. The second statistic requires modeling a single data set, created by including descriptive variables on the subjects used in a gage study. Based on sums of squares, the statistic has the properties of an R 2: it measures the proportion of variance explained; has values restricted to the [0,1] interval; and the endpoints indicate no variance explained and all variance explained respectively. We demonstrate the methodology using data from a study of cervical spine range of motion in children.\n\n37\n\n4\n\n667\n\n677\n\n#### DOI\n\ndoi: 10.1080/02664760902814542\n\n#### ISSN\n\nPrint ISSN: 0266-4763 Online ISSN: 1360-0532"
]
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https://access.openupresources.org/curricula/our6-8math/en/grade-7/unit-5/lesson-9/index.html | [
"# Lesson 9Multiplying Rational Numbers\n\nLet's multiply signed numbers.\n\n### Learning Targets:\n\n• I can explain what it means when time is represented with a negative number in a situation about speed and direction.\n• I can multiply two negative numbers.\n\n## 9.1Before and After\n\nWhere was the girl\n\n1. 5 seconds after this picture was taken? Mark her approximate location on the picture.\n2. 5 seconds before this picture was taken? Mark her approximate location on the picture.\n\n## 9.2Backwards in Time\n\nA traffic safety engineer was studying travel patterns along a highway. She set up a camera and recorded the speed and direction of cars and trucks that passed by the camera. Positions to the east of the camera are positive, and to the west are negative.\n\n1. Here are some positions and times for one car:\n\n position (feet) time (seconds) -180 -120 -60 0 60 120 -3 -2 -1 0 1 2\n1. In what direction is this car traveling?\n2. What is its velocity?\n1. What does it mean when the time is zero?\n\n2. What could it mean to have a negative time?\n\n2. Here are the positions and times for a different car whose velocity is -50 feet per second:\n\n position (feet) time (seconds) 0 -50 -100 -3 -2 -1 0 1 2\n1. Complete the table with the rest of the positions.\n2. In what direction is this car traveling? Explain how you know.\n3. Complete the table for several different cars passing the camera.\n\nvelocity\n(meters per\nsecond)\ntime after\npassing\ncamera\n(seconds)\nending\nposition\n(meters)\nequation\ncar C +25 +10 +250\ncar D -20 +30\ncar E +32 -40\ncar F -35 -20\ncar G -15 -8\n1. If a car is traveling east when it passes the camera, will its position be positive or negative 60 seconds before it passes the camera?\n2. If we multiply a postive number and a negative number, is the result positive or negative?\n1. If a car is traveling west when it passes the camera, will its position be positive or negative 60 seconds before it passes the camera?\n2. If we multiply two negative numbers, is the result positive or negative?\n\n## 9.3Cruising\n\nAround noon, a car was traveling -32 meters per second down a highway. At exactly noon (when time was 0), the position of the car was 0 meters.\n\n1. Complete the table.\n\n time (s) position (m) -10 -7 -4 -1 2 5 8 11\n2. Graph the relationship between the time and the car's position.\n1. What was the position of the car at -3 seconds?\n2. What was the position of the car at 6.5 seconds?\n\n### Are you ready for more?\n\nFind the value of these expressions without using a calculator.\n\n## 9.4Rational Numbers Multiplication Grid\n\nLook at the patterns along the rows and columns and continue those patterns to complete the table. When you have filled in all the boxes you can see, click on the \"More Boxes\" button.\n\nWhat does this tell you about multiplication by a negative?\n\n## Lesson 9 Summary\n\nWe can use signed numbers to represent time relative to a chosen point in time. We can think of this as starting a stopwatch. The positive times are after the watch starts, and negative times are times before the watch starts.\n\nIf a car is at position 0 and is moving in a positive direction, then for times after that (positive times), it will have a positive position. A positive times a positive is positive.\n\nIf a car is at position 0 and is moving in a negative direction, then for times after that (positive times), it will have a negative position. A negative times a positive is negative.\n\nIf a car is at position 0 and is moving in a positive direction, then for times before that (negative times), it must have had a negative position. A positive times a negative is negative.\n\nIf a car is at position 0 and is moving in a negative direction, then for times before that (negative times), it must have had a positive position. A negative times a negative is positive.\n\nHere is another way of seeing this:\n\nWe can think of as , which has a value of .\n\nWe can think of as , which has a value of .\n\nWe can multiply positive numbers in any order:\n\nIf we can multiply signed numbers in any order, then .\n\nWe can find two ways:\n\n• (this is the distributive property)\n\nThat means that\n\nWhich is the same as\n\nSo\n\nThere was nothing special about these particular numbers. This always works!\n\n• A positive times a positive is always positive\n• A negative times a positive or a positive times a negative is always negative\n• A negative times a negative is always positive\n\n## Lesson 9 Practice Problems\n\n1. Fill in the missing numbers in these equations\n\n2. A weather station on the top of a mountain reports that the temperature is currently and has been falling at a constant rate of per hour. Find each temperature. Explain or show your reasoning.\n\n1. If it continues fall at this rate, what will the temperature be:\n\n1. in 2 hours?\n2. in 5 hours?\n3. in half an hour?\n1. What was the temperature:\n\n1. 1 hour ago?\n2. 3 hours ago?\n3. 4.5 hours ago?\n3. Find the value of each expression.\n\n4. To make a specific hair dye, a hair stylist uses a ratio of oz of red tone, oz of gray tone, and oz of brown tone.\n\n1. If the stylist needs to make 20 oz of dye, how much of each dye color is needed?\n\n2. If the stylist needs to make 100 oz of dye, how much of each dye color is needed?\n\n1. Here are the vertices of rectangle : . Find the perimeter of this rectangle. If you get stuck, try plotting the points on a coordinate plane.\n2. Find the area of the rectangle .\n3. Here are the coordinates of rectangle : . Find the perimeter and area of this rectangle. See if you can figure out its side lengths without plotting the points."
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