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[ "# A note on the approximability of the tenacity of graphs\n\nDocument Type : Research Paper\n\nAuthors\n\n1 University of Tehran, Department of Algorithms and Computation.\n\n2 University of Tehran, College of Engineering, Faculty of Engineering Science\n\nAbstract\n\nIn this paper we show that, if $NP\\neq ZPP$, for any $\\epsilon > 0$, the tenacity of graph\nwith $n$ vertices is not approximable in polynomial time within a factor of\n$\\frac{1}{2} \\left( \\frac{n-1}{2} \\right) ^{1-\\epsilon}$.\n\nKeywords" ]

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[ "# Congruent triangles", null, "", null, "Main Article Discussion Related Articles  [?] Bibliography  [?] External Links  [?] Citable Version  [?] This editable Main Article is under development and subject to a disclaimer. [edit intro]\n\nIn Euclidean geometry, two triangles are congruent if there is a rigid motion which brings one triangle exactly onto the other (\"superposition\"). Since properties of Euclidean geometry are determined by the Euclidean distance, which in turn determines angles, two triangles are congruent when their configuration is described by the same set of distances and angles. It is a matter of convention whether triangles are regarded as congruent if the motion that transforms one into the other is orientation-reversing, such as a reflection, but this is usually the case.\n\n## Criteria for congruence\n\nThere are a number of traditional criteria for congruence in terms of measures of corresponding sides or angles. We may regard these as stating that the data are sufficient to construct the triangle unambiguously.\n\n• SSS (side-side-side): the lengths of corresponding sides are equal in the two triangles;\n• ASA (angle-side-angle): two angles and the side they have in common are equal;\n• SAS (side-angle-side): two sides and the enclosed angle are equal;\n• RHS (right angle-hypotenuse-side): the triangles are right-angled and the hypotenuse and another side are equal;\n\nThere is a well-known fallacious criterion, ASS, when two sides and an angle not enclosed are equal. In general there are two non-congruent triangles corresponding to this data, unless the angle is a right angle, when we have the valid RHS criterion.\n\nWe may also mention that AAA is a criterion for similarity of triangles, but not congruence, as it fails to prescribe any scale factor." ]

[ null, "https://s9.addthis.com/button1-share.gif", null, "https://en.citizendium.org/images/4/4f/Statusbar2.png", null ]

[ "Discover a lot of information on the number 42502: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!\n\n## Mathematical properties of 42502\n\nIs 42502 a prime number? No\nIs 42502 a perfect number? No\nNumber of divisors 8\nList of dividers 1, 2, 79, 158, 269, 538, 21251, 42502\nSum of divisors 64800\nPrime factorization 2 x 79 x 269\nPrime factors 2, 79, 269\n\n## How to write / spell 42502 in letters?\n\nIn letters, the number 42502 is written as: Forty-two thousand five hundred and two. And in other languages? how does it spell?\n\n42502 in other languages\nWrite 42502 in english Forty-two thousand five hundred and two\nWrite 42502 in french Quarante-deux mille cinq cent deux\nWrite 42502 in spanish Cuarenta y dos mil quinientos dos\nWrite 42502 in portuguese Quarenta e dois mil quinhentos dois\n\n## Decomposition of the number 42502\n\nThe number 42502 is composed of:\n\n1 iteration of the number 4 : The number 4 (four) is the symbol of the square. It represents structuring, organization, work and construction.... Find out more about the number 4\n\n2 iterations of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2\n\n1 iteration of the number 5 : The number 5 (five) is the symbol of freedom. It represents change, evolution, mobility.... Find out more about the number 5\n\n1 iteration of the number 0 : ... Find out more about the number 0\n\nOther ways to write 42502\nIn letter Forty-two thousand five hundred and two\nIn roman numeral\nIn binary 1010011000000110\nIn octal 123006\nIn US dollars USD 42,502.00 (\\$)\nIn euros 42 502,00 EUR (€)\nSome related numbers\nPrevious number 42501\nNext number 42503\nNext prime number 42509\n\n## Mathematical operations\n\nOperations and solutions\n42502*2 = 85004 The double of 42502 is 85004\n42502*3 = 127506 The triple of 42502 is 127506\n42502/2 = 21251 The half of 42502 is 21251.000000\n42502/3 = 14167.333333333 The third of 42502 is 14167.333333\n425022 = 1806420004 The square of 42502 is 1806420004.000000\n425023 = 76776463010008 The cube of 42502 is 76776463010008.000000\n√42502 = 206.16013193632 The square root of 42502 is 206.160132\nlog(42502) = 10.657306412629 The natural (Neperian) logarithm of 42502 is 10.657306\nlog10(42502) = 4.6284093669568 The decimal logarithm (base 10) of 42502 is 4.628409\nsin(42502) = 0.57041449737591 The sine of 42502 is 0.570414\ncos(42502) = -0.82135698766334 The cosine of 42502 is -0.821357\ntan(42502) = -0.69447816959428 The tangent of 42502 is -0.694478" ]

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[ "# Why does this derived table improve performance?\n\nI have a query which takes a json string as a parameter. The json is an array of latitude,longitude pairs. An example input might be the following.\n\n``````declare @json nvarchar(max)= N'[[40.7592024,-73.9771259],[40.7126492,-74.0120867]\n,[41.8662374,-87.6908788],[37.784873,-122.4056546]]';\n``````\n\nIt calls a TVF that calculates the number of POIs around a geographical point, at 1,3,5,10 mile distances.\n\n``````create or alter function [dbo].[fn_poi_in_dist](@geo geography)\nreturns table\nwith schemabinding as\nreturn\nselect count_1 = sum(iif(LatLong.STDistance(@geo) <= 1609.344e * 1,1,0e))\n,count_3 = sum(iif(LatLong.STDistance(@geo) <= 1609.344e * 3,1,0e))\n,count_5 = sum(iif(LatLong.STDistance(@geo) <= 1609.344e * 5,1,0e))\n,count_10 = count(*)\nfrom dbo.point_of_interest\nwhere LatLong.STDistance(@geo) <= 1609.344e * 10\n``````\n\nThe intent of the json query is to bulk call this function. If I call it like this the performance is very poor taking nearly 10 seconds for just 4 points:\n\n``````select row=[key]\n,count_1\n,count_3\n,count_5\n,count_10\nfrom openjson(@json)\ncross apply dbo.fn_poi_in_dist(\ngeography::Point(\nconvert(float,json_value(value,'\\$'))\n,convert(float,json_value(value,'\\$'))\n,4326))\n``````\n\nHowever, moving the construction of the geography inside a derived table causes the performance to improve dramatically, completing the query in about 1 second.\n\n``````select row=[key]\n,count_1\n,count_3\n,count_5\n,count_10\nfrom (\nselect [key]\n,geo = geography::Point(\nconvert(float,json_value(value,'\\$'))\n,convert(float,json_value(value,'\\$'))\n,4326)\nfrom openjson(@json)\n) a\ncross apply dbo.fn_poi_in_dist(geo)\n``````\n\nThe plans look virtually identical. Neither uses parallelism and both use the spatial index. There is an additional lazy spool on the slow plan that I can eliminate with the hint `option(no_performance_spool)`. But the query performance does not change. It still remains much slower.\n\nRunning both with the added hint in a batch will weigh both queries equally.\n\nSql server version = Microsoft SQL Server 2016 (SP1-CU7-GDR) (KB4057119) - 13.0.4466.4 (X64)\n\nSo my question is why does this matter? How can I know when I should calculate values inside a derived table or not?\n\n• Looking at IO stats they are identical too. Both do 358306 logical reads on the `point_of_interest` table, both scan the index 4602 times, and both generate a worktable and workfile. The estimator believes these plans are identical yet performance says otherwise. – Michael B May 2 '19 at 14:59\n• P.S. You may get even more performance improvement by doing a simple arithmetic box before doing the straight-line distance calculation. That is, filter first for those where the value `|LatLong.Lat - @geo.Lat| + |LatLong.Long - @geo.Long| < n` before you do the more complicated `sqrt((LatLong.Lat - @geo.Lat)^2 + (LatLong.Long - @geo.Long)^2)`. And even better, calculate the upper and lower bounds first, then `LatLong.Lat > @geoLatLowerBound && LatLong.Lat < @geoLatUpperBound && LatLong.Long > @geoLongLowerBound && LatLong.Long < @geoLongUpperBound`. (This is pseudocode, adapt appropriately.) – ErikE May 2 '19 at 18:07\n• P.S. Noting that lat & long are not the same dimensions, pick boundary boxes that are bigger than needed to get a correct/safe pre-filter in all conditions. Still, if there are many points being selected from, this can be a huge speed improvement. Last, if the built-in geo methods support returning points within a bounding box natively, use that instead of my suggested manual bounding box, and definitely instead of calculating the distance to all possible points and then filtering on the calculated distance—because you don't actually care what numeric distance is for unwanted points. – ErikE May 2 '19 at 18:11\n• I tried various bounding boxs (precomputed and with an NC on them) and tried using a buffered geometry with an intersection as the input. It seems with SQL server 2016 this is the fastest approach. I think the only thing better could be precomputing bounding box geography for 10 miles to benefit from the potentially faster STIntersect. – Michael B May 2 '19 at 18:42\n\nI can give you a partial answer that explains why you are seeing the performance difference - though that still leaves some open questions (such as can SQL Server produce the more optimal plan without introducing an intermediate table expression that projects the expression as a column?)\n\nThe difference is that in the fast plan the work needed to parse the JSON array elements and create the Geography is done 4 times (once for each row emitted from the `openjson` function) - whereas it is done more than 100,000 times that in the slow plan.\n\nIn the fast plan...\n\n``````geography::Point(\nconvert(float,json_value(value,'\\$'))\n,convert(float,json_value(value,'\\$'))\n,4326)\n``````\n\nIs assigned to `Expr1000` in the compute scalar to the left of the `openjson` function. This corresponds to `geo` in your derived table definition.", null, "In the fast plan the filter and stream aggregate reference `Expr1000`. In the slow plan they reference the full underlying expression.\n\nStream aggregate properties", null, "The filter is executed 116,995 times with each execution requiring an expression evaluation. The stream aggregate has 110,520 rows flowing into it for aggregation and creates three separate aggregates using this expression. `110,520 * 3 + 116,995 = 448,555`. Even if each individual evaluation takes 18 microseconds this adds up to 8 seconds additional time for the query as a whole.\n\nYou can see the effect of this in the actual time statistics in the plan XML (annotated in red below from the slow plan and blue for the fast plan - times are in ms)", null, "The stream aggregate has an elapsed time 6.209 seconds greater than its immediate child. And the bulk of the child time was taken up by the filter. This corresponds to the extra expression evaluations.\n\nBy the way.... In general it is not a sure thing that underlying expressions with labels like `Expr1000` are only calculated once and not re-evaluated but clearly in this case from the execution timing discrepancy this happens here.\n\n• As an aside, if I switch the query to use a cross apply to generate the geography, I also get the fast plan. `cross apply(select geo=geography::Point( convert(float,json_value(value,'\\$')) ,convert(float,json_value(value,'\\$')) ,4326))f` – Michael B May 2 '19 at 16:39" ]

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[ "# (7,2)-torus knot\n\n71 knot", null, "Arf invariant0\nBraid length7\nBraid no.2\nBridge no.2\nCrosscap no.1\nCrossing no.7\nGenus3\nHyperbolic volume0\nStick no.9\nUnknotting no.3\nConway notation\nA–B notation71\nDowker notation8, 10, 12, 14, 2, 4, 6\nLast /Next6372\nOther\nalternating, torus, fibered, prime, reversible\n\nIn knot theory, the 71 knot, also known as the septoil knot, the septafoil knot, or the (7, 2)-torus knot, is one of seven prime knots with crossing number seven. It is the simplest torus knot after the trefoil and cinquefoil.\n\n## Properties[]\n\nThe 71 knot is invertible but not amphichiral. Its Alexander polynomial is\n\n$\\Delta (t)=t^{3}-t^{2}+t-1+t^{-1}-t^{-2}+t^{-3},\\,$", null, "its Conway polynomial is\n\n$\\nabla (z)=z^{6}+5z^{4}+6z^{2}+1,\\,$", null, "and its Jones polynomial is\n\n$V(q)=q^{-3}+q^{-5}-q^{-6}+q^{-7}-q^{-8}+q^{-9}-q^{-10}.\\,$", null, "## Example[]", null, "Play media\nAssembling of 71 knot." ]

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[ "# Noncontext free for quotient\n\nHave such exercise: Let $L_1=\\{a^{2n}b^n|n\\geq1\\}^*$\n\nand $L_2=\\{b^na^n|n\\geq 1\\}^*b$\n\nProve, that $L_1, L_2$ are context free, while quotient $L_1/L_2$ is not context-free.\n\n(This is home exercise, which I am not sure what to make of it. 3rd one)\n\nAnyway, proving $L_1, L_2$ are CFL is trivial.\n\nFor $L_1$:\n\nS->aaSb|$\\epsilon$\n\nFor $L_2$\n\nS->Ab\n\nA->bAa|$\\epsilon$\n\nSo far so good, grammar can be constructed, i.e CFL. Fine.\n\nBut quotient.. Lets list few words for $L_1$: $\\epsilon$ then: aab, aabaab, aabaabaab, $\\dots$ then aaaabb, aaaabbaaaabb, $\\dots$ and so on - string under Kleene, which contains twice as much $a$ than $b$\n\nNow about $L_2$ words: b then bab,babab,bababab,$\\dots$ then bbaab,bbaabbaab,$\\dots$, ie - $(b^na^n)^*b$\n\nBut.. Only words in $L_1$ are of format which ends in suffix ...ab are where $n=1$: $aab,aabaab,\\dots$, moreover - if we take string from $L_1$, for example $aabaab$, we kinda remove trailing $b$ ($aabaa$), after which we can't really remove anything else, in this particular case we would have to remove suffix $bbaa$, but there is no such suffix in word from $L_1$\n\nSo, all in all - quotient, to my understanding is $L_1$, with only difference that there is very last b always removed. And it feels to me like CF, what am I doing wrong?\n\nEdit: According to comment, I have mistake in my thoughts, which is reasonable, we dont have to Kleenize $n$ in each substring.\n\nIe $aaaabbaab$ for $L_1$ is possible. Ok, now lets look exactly at $aaaabbaab$ - first remove trailing $b$, get $aaaabbaa$. Now $bbaa$ - resulting in $aaaa$.. So all in all, quotient is still $L_1$, which has some arbitrary suffix removed. Still feels CF to me\n\n• $L_1$ can also contain strings of the form $aabaaaabbaab$. Similarly $L_2$ can contain strings of the form $babbaabbbaaab$. Nov 20, 2016 at 17:31\n• Ah.. yea. Except that now I am even more confused, but - this is valid point Nov 20, 2016 at 17:35\n• What exactly is your question? It's hard for me to tell what question you are looking to have answered and what kind of answer you are looking for.\n– D.W.\nNov 20, 2016 at 18:18\n• Yes, I might be talking too much. Exercise 3, users.utu.fi/jkari/automata/hw10.pdf Nov 20, 2016 at 18:26\n• Not dumb, but it wasn't really a question. Both grammars are disregarding the Kleene's stars. Nov 20, 2016 at 23:53\n\nWork backwards, from right to left. Try to find which strings of the form $a^*$ belong to $L_1/L_2$. So find matches of strings $w_1\\in L_1$ and $w_2\\in L_2$ such that at the start only a string of the form $a^*$ sticks out: $w_1= a^k w_2$.\n• Why only strings of the form $a^*$? Nov 21, 2016 at 6:11\n• @skankhunt42 Good question. To me it seems more easier to handle. When restricting to $a$'s only we get a simple solution, and a very clear non-context-free language. Technically this means we intersected the quotient that we obtain by the regular language $a^*$: this does not change the (non-)context-freeness. Nov 21, 2016 at 23:12\n• That doesn't seem right to me. Suppose $L = \\{b^p \\mid p \\text{ is a prime number}\\} \\cup L(a^*)$. Then if we take intersection with $L(b^*)$, we get a non-context-free language. However, intersection with $L(a^*)$ gives us a context-free language. Nov 22, 2016 at 8:38\n• @skankhunt42 Formulated that way you are right. What I meant here is that context-free languages are closed under intersection with regular languages. If we intersect with $a^*$ in this construction we get a non-context-free language, so indeed the original language must be non-context-free either. Nov 23, 2016 at 2:02" ]

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[ "Excel 多條件變數 countif 函數 sumproduct and or - 通達人驛站\n\n# Excel 多條件變數 countif 函數 sumproduct and or\n\n#### 在 SUMPRODUCT 內使用 and", null, "=SUMPRODUCT((B2:B11=\"Sales\")*(C2:C11=\"A\"))", null, "SUMPRODUCT 內的「*」就是 and 的意思。\n\n#### 在 SUMPRODUCT 內使用 or\n\n=SUMPRODUCT((B2:B11=\"R&D\")*((C2:C11=\"A\")+(C2:C11=\"B\"))\n\nSUMPRODUCT 內的「+」就是 or 的意思。\n\n#### 將 SUMPRODUCT True/False 陣列轉換為 1/0 陣列\n\n=SUMPRODUCT(--(B2:B11=\"Sales\"),--(C2:C11=\"A\"))\n\n=SUMPRODUCT(--(B2:B11=\"R&D\"),--((C2:C11=\"A\")+(C2:C11=\"B\"))\n\n=ARRAYFORMULA(SUMPRODUCT((B2:B11=\"Sales\")*(C2:C11=\"A\")))\n\n#### 注意事項\n\n1. mrexcel -《COUNTIF with two Variables?\n", null, "" ]

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[ "# Parametric and nonparametric test pdf\n\nNonparametric methods still use traditional statistical. Parametric and nonparametric statistics phdstudent. The friedman test is a non parametric test w hich was developed and implemented by milton friedman. The parametric test process mainly depends on assumptions related to the shape of the normal distribution in the underlying population and about the parameter forms of the assumed distribution. Gibbons1993 observed that ordinal scale data, notably likerttype scales, are very common in social sciences and argued these should be analyzed with nonparametric tests. Knowing the difference between parametric and nonparametric test will help you chose the best test for your research. Wilcoxonmannwhitney u test and wilcoxon rank sum test 2 equivalent tests wilcoxon rank sum.\n\nNon parametric tests do not make as many assumptions about the distribution of the data as the parametric such as t test do not require data to be normal good for data with outliers nonparametric tests based on ranks of the data work well for ordinal data data that have a defined order, but for which averages may not make sense. Nonparametric tests are distributionfree and, as such, can be used for nonnormal variables. However, if one or more of the assumptions have been violated, then some but not all statisticians advocate transforming the data into a format that is compatible with the appropriate nonparametric test. A statistical test used in the case of nonmetric independent variables, is called nonparametric test. Difference between parametric and nonparametric test with. We have covered a number of testing scenarios and a parametric and nonparametric test for each of those scenarios. One sample single set of observations the sign test is used to test the null hypothesis that the median of a distribution is equal to some value. They are perhaps more easily grasped by illustration than by definition. When data are collected from more than two populations, the multiple sample analysis procedure can test for significant differences between the population medians using either a kruskalwallis test. Parametric tests and analogous nonparametric procedures as i mentioned, it is sometimes easier to list examples of each type of procedure than to define the terms.\n\nPdf a comparison of parametric and nonparametric statistical tests. Spss nonparametric tests are mostly used when assumptions arent met for other tests such as anova or t tests. A nonparametric statistical test is a test whose model does not specify conditions about the parameters of the population from which the sample was drawn. Because the distribution from which the sample is taken is speci. Comparative analysis of parametric and nonparametric tests. In parametric tests, data change from scores to signs or ranks.\n\nParametric tests are said to depend on distributional assumptions. For this reason, categorical data are often converted to. Alternative nonparametric tests of dispersion viii. The parametric test uses a mean value, while the nonparametric one uses a median value. Parametric data is data that clusters around a particular point, with fewer outliers as the distance from that point increases. Nonparametric tests nonparametric methods i many nonparametric methods convert raw values to ranks and then analyze ranks i in case of ties, midranks are used, e. Nonparametric statistics is the branch of statistics that is not based solely on parametrized families of probability distributions common examples of parameters are the mean and variance. A statistical test, in which specific assumptions are made about the population parameter is known as parametric test. Denote this number by, called the number of plus signs. Nonparametric methods apply in all other instances. Parametric tests are those that make assumptions about the parameters of the population distribution from which the sample is drawn. Howard is a clinical psychologist and a professional writer and he has been partnering with patients to create positive.\n\nA parametric test is used on parametric data, while nonparametric data is examined with a nonparametric test. Parametric and nonparametric tests blackwell publishing. Nonparametric statistics portland state university. A comparison of parametric and nonparametric methods. If a nonparametric test is required, more data will be needed to make the same conclusion. Additional examples illustrating the use of the siegeltukey test for equal variability test 11. The nonparametric tests option of the analyze menu offers a wide range of nonparametric tests, as illustrated in figure 5. Some of the most common statistical tests and their nonparametric analogs. A parametric test is a hypothesis testing procedure based on the assumption that observed data are distributed according to some distributions of wellknown form e.\n\nStrictly, most nonparametric tests in spss are distribution free tests. The two methods of statistics are presented simultaneously, with indication of their use in data analysis. Oddly, these two concepts are entirely different but often used interchangeably. Nonparametric methods nonparametric statistical tests. Parametric tests and analogous nonparametric procedures. There are two types of test data and consequently different types of analysis. The pdf for a test statistic is called the sampling distribution of the statistic. It is worth repeating that if data are approximately normally distributed then parametric tests as in the modules on hypothesis testing are more appropriate. Non parametric tests non parametric methods i many non parametric methods convert raw values to ranks and then analyze ranks i in case of ties, midranks are used, e. The second drawback associated with nonparametric tests is that their results are often less easy to interpret than the results of. This book comprehensively covers all the methods of parametric and nonparametric statistics such as correlation and regression, analysis of variance, test construction, onesample test to ksample tests, etc. Choosing between parametric and nonparametric tests.\n\nAs implied by the name, nonparametric statistics are not based on the parameters of the normal curve. The term parametric is intended to refer to statistical tests that make assumptions about particular population parameters e. This is often the assumption that the population data are normally distributed. A comparison of parametric and nonparametric statistical. Most nonparametric tests apply to data in an ordinal scale, and some apply to data in nominal scale. Chapter nonparametric statistics mit opencourseware. In statistical inference, or hypothesis testing, the traditional tests are called parametric tests because they depend on the speci. Assume the following test scores have been obtained. As i mentioned, it is sometimes easier to list examples of each type of procedure than to define the. The intervention was treatment with betamethasone, 12 mg intramuscularly daily for two consecutive days at 3436 weeks of pregnancy. The parametric tests will be applied when normality and homogeneity of variance assumptions are satisfied otherwise the equivalent nonparametric test will.\n\nNonparametric methods transportation research board. The parametric approach requires previous knowledge about the population, contrary to the nonparametric approach. In the following, a sample 7 observations will be used to illustrate how, when, and with what consequences nonparametric procedures can be used. Researchers investigated the effectiveness of corticosteroids in reducing respiratory disorders in infants born at 3436 weeks gestation. Statistical test these are intended to decide whether a hypothesis about distribution of one or more populations should be rejected or accepted. The model structure of nonparametric models is not specified a priori. Do not require measurement so strong as that required for the parametric tests. Non parametric tests do not make as many assumptions about the distribution of the data as the parametric such as t test do not require data to be normal good for data with outliers non parametric tests based on ranks of the data work well for ordinal data data that have a defined order, but for which averages may not make sense. Parametric and nonparametric tests in spine research. Friedmans test the friedmans test is the nonparametric test equivalent to the repeated measures anova, and an extension of the wilcoxon test it allows the comparison of more than two dependent groups two or more conditions. Although this difference in efficiency is typically not that much of an issue, there are instances where we do need to consider which method is more efficient. Nonparametric statistical tests if you have a continuous outcome such as bmi, blood pressure, survey score, or gene expression and you want to perform some sort of statistical test, an important consideration is whether you should use the standard parametric tests like ttests or anova vs. The onesample t test applies when the population is normally distributed with unknown mean and variance.\n\nHere the variances must be the same for the populations. Remember that when we conduct a research project, our goal is to discover some truth about a population and the effect of an intervention on that population. For example, a psychologist might be interested in the depressant effects of certain recreational drugs. Nonparametric statistics includes nonparametric descriptive statistics, statistical models, inference, and statistical tests. Explanations social research analysis parametric vs. Participants were 320 women at 3436 weeks of pregnancy who. However, if normality assumptions meet then the parametric tests are more efficient. The mannwhitney u test can be used to determine if two samples of unpaired data have different median values. Parametric and non parametric test linkedin slideshare. Therefore, if your data violate the assumptions of a usual parametric and nonparametric statistics might better define the data, try running. Table 1 contains the names of several statistical procedures you might be familiar with and categorizes each one as parametric or nonparametric. Given the small numbers of bins involved n 4 ranks, tests of normality of distribution such as the. Mannwhitney test the mannwhitney test is used in experiments in which there are two conditions and different subjects have been used in each condition, but the assumptions of parametric tests are not tenable. This type of test is used for the comparison of three or more dependent.\n\nParametric and nonparametric tests for comparing two or. As the table below shows, parametric data has an underlying normal distribution which allows for more conclusions to be drawn as the shape can be mathematically described. Pdf this paper explains, through examples, the application of nonparametric methods in hypothesis testing. As discussed in chapter 5, the ttest and the varianceratio test make certain assumptions about the. Parametric and nonparametric tests are broad classifications of statistical testing procedures. Discussion of some of the more common nonparametric tests follows. Many times parametric methods are more efficient than the corresponding nonparametric methods. Nonparametric statistics is based on either being distributionfree or having a specified distribution but with the distributions parameters unspecified. However, there are situations in which assumptions for a parametric test are violated and a nonparametric test is more appropriate. Important probability density functions for test statistics are the t pdf for the t test statistic, the f pdf for the f test statistic, and the. Textbook of parametric and nonparametric statistics sage. Non parametric tests are distributionfree and, as such, can be used for nonnormal variables. A parametric equivalent is the twosample unpaired data students ttest the. A randomised placebo controlled trial was performed.\n\n58 873 158 847 1347 693 392 991 90 671 1533 204 165 1494 741 293 1227 540 1457 1482 1215 1453 486 1396 577 1187 462 529 95 359 528 1176 1340 1459 1093 121 537 1491 542 1073 49" ]

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[ "``````#include<iostream>\nusing namespace std;\n\nclass area\n{\ndouble dim1, dim2;\npublic:\nvoid setarea(double d1, double d2)\n{\ndim1= d1;\ndim2 = d2;\n}\nvoid getdim(double &d1, double &d2)\n{\nd1 = dim1;\nd2 = dim2;\n}\nvirtual double getarea()\n{\ncout<<\"You must override this functionn\";\nreturn 0.0;\n}\n};\n\nclass rectangle : public area\n{\npublic:\ndouble getarea()\n{\ndouble d1, d2;\ngetdim(d1, d2);\n\nreturn d1*d2;;\n}\n};\n\nclass triangle : public area\n{\npublic:\ndouble getarea()\n{\ndouble d1, d2;\ngetdim(d1, d2);\nreturn 0.5*d1*d2;\n}\n};\n\nint main()\n{\narea *p;\nrectangle r;\ntriangle t;\n\nr.setarea(3.3, 4.5);\nt.setarea(4.0, 5.0);\n\np=&r;\ncout<<\"Rectangle has area: \"<<p->getarea()<<'n';\n\np=&t;\ncout<<\"Triangle has area: \"<<p->getarea()<<'n';\n}\n``````\n\nI only have problem with the `getdim()` function within \"class area\". I don't understand what is the necessity of this function.\nFuthermore, in the derived class when the virtual function overridden, `getdim` function is used outside the \"class area\". But the `getdim` function is not identified in the derived class.\nIs it always possible to use a function to another class where the function is not identified??\n\nThanks in advance\n\n## Recommended Answers\n\nHonestly, this code looks like it may be an overly-explanatory example from a text. The code contains many things that you generally won't see in well-designed/written code. I suggest you read the chapter it's part of a little closer. Here is a highly generalized summary (I want to avoid doing …\n\n`getdim()` is declared as public in the base class, and the inheritance is public, so the derived classes will have access to this method (as public).\n\nThis method retrieves the values of the `dim1` and `dim2` variables by reference. In `triangle` and `rectangle` , the values of `d1`\n\n## All 7 Replies\n\nHonestly, this code looks like it may be an overly-explanatory example from a text. The code contains many things that you generally won't see in well-designed/written code. I suggest you read the chapter it's part of a little closer. Here is a highly generalized summary (I want to avoid doing your homework for you).\n\nRe-posting code with [code] code tags [/code] for readability and discussion:\n\n``````#include<iostream>\nusing namespace std;\n\nclass area\n{\ndouble dim1, dim2;\npublic:\nvoid setarea(double d1, double d2)\n{\ndim1= d1;\ndim2 = d2;\n}\nvoid getdim(double &d1, double &d2)\n{\nd1 = dim1;\nd2 = dim2;\n}\nvirtual double getarea()\n{\ncout<<\"You must override this function\\n\";\nreturn 0.0;\n}\n};\n\nclass rectangle : public area\n{\npublic:\ndouble getarea()\n{\ndouble d1, d2;\ngetdim(d1, d2);\n\nreturn d1*d2;;\n}\n};\n\nclass triangle : public area\n{\npublic:\ndouble getarea()\n{\ndouble d1, d2;\ngetdim(d1, d2);\nreturn 0.5*d1*d2;\n}\n};\n\nint main()\n{\narea *p;\nrectangle r;\ntriangle t;\n\nr.setarea(3.3, 4.5);\nt.setarea(4.0, 5.0);\n\np=&r;\ncout<<\"Rectangle has area: \"<<p->getarea()<<'\\n';\n\np=&t;\ncout<<\"Triangle has area: \"<<p->getarea()<<'\\n';\n}``````\n\nThis code includes examples of inheritance and polymorphism.\n\nObserve Lines 25 & 37:\n\n...\n25. class rectangle : public area\n...\n37. class triangle : public area\n...\n\nThese lines say that the rectangle and triangle classes inherit properties from the area class. Because they inherit from area, they have access to the members of the area class.\n\nIn actuality, because the functions/methods are declared as part of the \"public\" section of area, any piece of code can access them as long as there is an object of that type available to call them through.\n\n`getdim()` is declared as public in the base class, and the inheritance is public, so the derived classes will have access to this method (as public).\n\nThis method retrieves the values of the `dim1` and `dim2` variables by reference. In `triangle` and `rectangle` , the values of `d1` and `d2` are passed into the method, and are assigned the values of `dim1` and `dim2` respectively.\n\n`getdim()` is declared as public in the base class, and the inheritance is public, so the derived classes will have access to this method (as public).\n\nThis method retrieves the values of the `dim1` and `dim2` variables by reference. In `triangle` and `rectangle` , the values of `d1` and `d2` are passed into the method, and are assigned the values of `dim1` and `dim2` respectively.\n\nThanks a lot Jonsca.\n\nBut can't I output the same things without getdim() function?? This is not my homework but I just need the concept to understand some fundamental things of c++.\n\n>>But can't I output the same things without getdim() function??\n\nUnfortunately no. The variables area::dim1 and area::dim2 are private. This means that only members of the actual area class can access them directly. In order to allow rectangle and triangle to access them directly, they need to be re-classified as protected.\n\nThis is just one of the atypical things that I alluded to in my previous post.\n\n>>But can't I output the same things without getdim() function??\n\nUnfortunately no. The variables area::dim1 and area::dim2 are private. This means that only members of the actual area class can access them directly. In order to allow rectangle and triangle to access them directly, they need to be re-classified as protected.\n\nThis is just one of the atypical things that I alluded to in my previous post.\n\nWhat would if I declare the dim1 and dim2 as public inside the class area?? Then can I delete the getdim() function??\n\nIs it always necessary to declare the variable as private?? Everywhere I see that the variable always declare as private....what is the advantage of this and what is the disadvantage of declaring the variable inside the public??\n\nRead Wikipedia's explanation of encapsulation to see some of the philosophy behind it.\n\nYou could certainly make it public, but what if you want people using your classes not to have to muck around with those things. As an example, what if you wanted to change the way you stored the member variables? If they are private, all you'd have to do is modify `getdim()` and you'd be good to go. Everyone could access the values like they always did. If the member variables were public, you'd need to modify each time they are used in the code.\n\nThanks a lot from my heart. All the things of the code now I can understand.\n\nBe a part of the DaniWeb community\n\nWe're a friendly, industry-focused community of 1.21 million developers, IT pros, digital marketers, and technology enthusiasts learning and sharing knowledge." ]

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[ "# Set-Builder Notation\n\nA Set is a collection of things (often numbers).\nExample: {2, 3, 5} is a set.\n\nHere is a simple example of set-builder notation:\n\nGeneral Form: {formula for elements: restrictions} or\n\n{formula for elements| restrictions}\n\nExamples\n\n{x | x < 4} or {x : x < 4}\n\nDetails\n\n Symbol Details Note { } the set of x all x just a place-holder, it could be anything : or | Such that x < 4 restrictions\n\nSo {x | x < 4} or {x : x < 4} says “the set of all x less than 4”\n\n## Show the Type of Number\n\nYou can show what type of number x is.\n\nExamples\n\n{ x ∈ R | x ≠ 7}\n\n Symbol Details ∈ Element of R Real Number\n\nSo it says “the set of all real numbers except 7”\n\n## Number Types\n\n Symbols Details N Natural Numbers Z Integers Q Rational Numbers R Real Numbers I Imaginary Numbers C Complex Numbers\n\n## Defining a Domain\n\nSet Builder Notation is really useful for defining a domain of a function. The domain is the set of all the values that go into a function.\n\nExamples\n\n f(x) Domain Notation 1/x all the Real Numbers, except 0(because 1/x is undefined at x = 0) {x ∈ R | x ≠ 0} √x all the Real Numbers from 0 onwards(because there’s no square root of a negative number) {x ∈ R | x > 0} 1/(x2 − 1) all the Real Numbers, except -1 and 1(because 1/(x2 − 1) is undefined at x = -1 or x = 1) {x ∈ R | x ≠ -1, x ≠1}" ]

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[ "Wallis wrote in a letter in 1669: \"In a dark night in bed, without pen, ink or paper or anything equivalent, I did by memory extract the square root of 3,00000, 00000, 00000, 00000, 00000, 00000, 00000, 00000 which I found to be 1,73205, 08075, 68077, 29353, feré, and did the next day commit to writing.\"\n\n\"February 18th 1670, Joannes Georgius Pelshower giving me a visit, and desiring an example of the like, I did that night propose to myself in the dark without help to my memory a number in 53 places: 24681, 35791, 01112, 14111, 31516, 18201, 71921, 22242, 62830, 23252, 72931 of which I extracted the square root in 27 places: 15,71030, 16871, 48280, 58171, 52171 proxima.\"\n\nJohn Wallis (1616-1703) was a brilliant mathematician and a contemporary of Isaac Newton. In addition to his works on pure mathematics he also wrote on astronomy, the tides, the laws of motion, botany, physiology, music, geology etc.\n\nChapter 5\n\nAll From 9 and the Last From 10\n\nThis chapter shows a surprisingly easy way of multiplying numbers near a base, near different bases, or near multiples of a base, and has a considerable range. The use of negative numbers which can enormously simplify calculations is introduced, and applications in addition and subtraction are also included.\n\nNumbers Just Below a Base\n\nIn the conventional system of mathematics a sum like 88 × 98 is considered especially difficult because of the large figures, 8 and 9.\n\nBut since the numbers 88 and 98 are close to the base of 100 we may think that there ought to be an easy way to find such a product.\n\nIn the Vedic system this kind of sum is extremely easy however. We simply note that both numbers are close to 100, and that 88 is 12 below 100, and 98 is 2 below 100, and we just give the answer:", null, "Example 1", null, "The deficiencies (12 and 2) have been written above the numbers (on the flag), the minus signs indicating that the numbers are below 100.\n\nThe answer 8624 is in two parts: 86 and 24.\nThe 86 is found by taking one of the deficiencies from the other number.\nThat is:\n\n88 - 2 = 86 or 98 - 12 = 86 (whichever you like),\n\nand the 24 is simply the product of the deficiencies: 12 × 2 = 24.\nIt could hardly be easier.", null, "Example 2\n\nSimilarly", null, "The differences from 100 are 7 and 4,\n93 - 4 = 89 or 96 - 7 = 89,\nand 7 × 4 = 28.", null, "Example 3\n\nAlso", null, "Note the zero inserted here: the numbers being multiplied are near to 100, so two digits are required on the right, as in the other examples.\n\nQuestion Comment !\n Tutorial Title Question Title No Questions 0 Questions Asked 0 Skipped Questions 0 Correctly Answered 0 Wrongly Answered 0 Total Question Attempts 0 Current Question Attempts 0\n\nSkip Question\n\nReset Test\n\nHalt Test\n\n No Questions 0 Questions Asked 0 Skipped Questions 0 Correctly Answered 0 Wrongly Answered 0 Total Question Attempts 0\n\nTime Taken\n\nClose", null, "Exercise A", null, "Example 4", null, "Here the numbers are each 11 below 100, and 11 × 11 = 121, a 3-figure number. The hundreds digit of this is therefore carried over to the left.\n\nSo the most efficient mental procedure is to take one number and subtract the other deficiency from it. Then multiply the deficiencies together, mentally adjusting the first part of the answer if there is a carry figure.", null, "Exercise B\n\nFind 92 × 196 by\n\na) thinking of 196 as1/96,\nb) using proportionally and All from 9 and the Last from 10,\nc) using proportionally and By One More than the One Before", null, "Example 5\n\n568 × 998 = 566864\n\nWe see here that the numbers 568 and 998 are conveniently close to 1000, so we allow 3 figures on the right. The differences of the numbers from 1000 are 432 and 2.\n\nThe method is just the same:\n568 - 2 = 566,\n432 × 2 = 864.\n\nHowever in the case of 568 here the deficiency is not so obvious as in the previous examples, and this is exactly where the Sutra of the present chapter comes in.\n\nIf All From 9 and the Last From 10 is applied to the digits of 568 we get 432:\n\ntake the 5 from 9 and get 4,\ntake the 6 from 9 and get 3,\nwe take the 8 from 10 and get 2.\n\nThis formula gives the deficiency of any number from the next highest base. It could have been applied in the previous examples too: for 88, in the first example, we take the first 8 from 9 to get 1, and the last from 10 to get 2. This gives the deficiency of 12 below the base of 100. For 98 we get 02, or just 2.\n\nSubtraction\n\nThus All From 9 and the Last From 10 provides us also with a very effective method of subtraction from a base number:\n\n1000 - 587 = 413,\n\n10000 - 785 = 10000 - 0785 = 9215, and so on.\n\nAlso 7000 - 111 = 6889, the 7 is reduced to a 6 because 111 is to be taken from one of the 7 thousands, so only 6 thousands are left, and the Sutra is applied to 111 to get 889.\n\nThus frequent subtraction problems involving money etc. are quickly solved by this method:\n\nfor example £70.00 - £1.11 = £68.89.\n\nWe may also write 7000 - 111 = 7111 in which we put a bar (called a \"vinculum\") over the 111 to show that it is negative.\n\nThis subtraction method is completely general, covering all types of subtraction:\n\n7654 - 1928 = 6334 (since 7-1= 6, 6-9 = 3, 5-2 = 3, 4-8 = 4)\n\nand 6334 = 63/34 (since 60-3 = 57, 30-4 = 26).\n\nWe will see subtractions like this coming up in some of the later multiplication devices.", null, "Exercise C", null, "Example 6\n\n58776 × 99998 = 58774/82448\n\n58776 - 2 = 58774, and the formula gives the deficiencies 41224, 2 which are multiplied together to get 82448.", null, "Example 7\n\nSimilarly 857 × 994 = 851858\n\nwhere 851 = 857 - 6, and 858 = 143 × 6.", null, "Example 8\n\n7 × 8 = 56\n\nMultiplication tables above 5 × 5 are not essential in the Vedic system.\n\nHere the deficiencies are 3 and 2 as we take 7 and 8 from 10 (...Last from 10):\n7 - 2 = 5 and 3 × 2 = 6, therefore 7 × 8 = 56.", null, "Exercise D\n\nDivision by numbers near a base (above or below the base) includes the use of the Sutra Vertically and Crosswise as well as the Sutra of the present chapter. It is therefore shown at the end of the next chapter.\n\nWe now extend this simple multiplication technique in several different directions.\n\nNumbers Above A Base\n\nFirst let us suppose that the numbers being multiplied are both above a base, rather than below it.", null, "Example 9\n\n103 × 104 = 10712\n\nThis is even easier than the previous examples, but the method is just the same. The deficiencies are +3 and +4: positive now because the numbers are above the base.\n\n103 + 4 = 107 or 104 + 3 = 107, and 4 × 3 = 12.", null, "Example 10\n\n12 × 13 = 156 (12+3=15, 2×3=6)", null, "Example 11\n\n1234 × 1003 = 1237702 (1234+3=1237, 234×3=702)", null, "Example 12\n\n10021 × 10002 = 100230042 (10021+2=10023, 0021×2=0042)", null, "Exercise E\n\nOne Number Above and One Number Below the Base", null, "Example 13\n\n124 × 98 = 12248 = 12152\n\nHere the base is 100 and the deficiencies from 100 are +24 and -2.\n\nApplying the usual procedure we find 124 - 2 = 122 or 98 + 24 = 122.\nSo 122 is the left-hand part of the answer.\nThen multiplying the deficiencies we get -48, written 48 (since a plus times a minus gives a minus). This gives the answer as 12248.\nTo remove the negative portion of the answer we just take 48 from one of the hundreds in the hundreds column. This simply means reducing the hundreds column by 1 and applying All From 9 and the Last From 10 to 48. Thus 122 becomes 121 and 48 becomes 52.\nSo 12248 = 12152.", null, "Example 14\n\n1003 × 987 = 990/039 = 989/961\n\nSimilarly, we first get 1003-13 = 990 or 987+3 = 990,\nand +3 × -13 = 039 (3 figures required here as the base is 1000).\nThen 990 is reduced by 1 to 989, and applying the formula to 039 gives 961.", null, "Example 15", null, "Here we have a minus one to carry over to the left so that the 112 is reduced by 2 altogether.", null, "Exercise F\n\nProportionally\n\nHere we bring in the formula from Chapter 2.", null, "Example 16\n\n213 × 203 = 2 × 216/39 = 43239\n\nWe observe here that the numbers are not near any of bases used before: 10, 100, 1000 etc. But they are close to 200, with deficiencies of 13 and 3.\nThe usual procedure gives us 216/39 (213+3=216, 13×3=39).\nNow since our base is 200 which is 100×2 we multiply only the left-hand part of the answer by 2 to get 43239.", null, "Example 17\n\n29 × 28 = 3×27/2 = 812\n\nThe base is 30 (3×10), and the deficiencies are -1 and -2.\n29-2 = 27, 1×2 = 2 and 3×27 = 81.", null, "Example 18\n\n311 × 298 = 3×309/22 = 926/78\n\nHere the numbers are above and below 300: we multiply the left-hand side by 3 before deducting 1 to deal with the negative right-hand portion.\n\nThus the Proportionately formula extends considerably the range of the method.\nThe only additional step being the multiplication of the left-hand part of the answer. One further application of this formula may also be noted:", null, "Example 19\n\n88 × 49 = half of (88×98) = half of (8624) = 4312", null, "Exercise G\n\nNumbers Near Different Bases", null, "Example 20\n\n9998 × 94 = 9398/12\n\nThe bases here are 10,000 and 100 and the deficiencies are -2 and -6.\nThe answer is in two parts: 9398 and 12.\nIn subtracting the deficiency 6 from the first number, imagine the numbers are under each other:\n\n9998\n94\n\nand subtract the 6 in the column indicated by the last figure of the smaller number, that is, the second column from the left here.\nSo 9998 becomes 9398.\nThen multiply the deficiencies together: 2×6 = 12.\n\nNote that the number of figures in the right-hand part of the answer corresponds to the base of the lower number (94 is near 100, therefore there are 2 figures on the right).", null, "Example 21\n\n10007 × 1003 = 10037021\n\nLining the numbers up:", null, "we see that the deficiency 3 is added in the 4th column, giving 10037.\n\nThe product of the deficiencies is 7×3=21, but as the base of the smaller number is 1000 we need 3 figures on the right, so we put 021.", null, "Example 22\n\n1032 × 98 = 1012/64 = 101136\n\nNote here that because 98 = 102 the deficiency 2 is deducted from the 3 to give 1012 on the left.", null, "Exercise H\n\nMultiplying Three Numbers Simultaneously", null, "Example 23\n\n98 × 97 × 96 = 91/26/24 = 912576\n\nThese numbers are all close to 100, their deficiencies being 2, 3, 4.\nThe answer is in 3 parts as indicated by the oblique lines.\nFirst take one number and take both of the other deficiencies from it.\nFor example 98-3-4 = 91. This is the first part of the answer.\n\nThen multiply the deficiencies together in pairs and add the results up:\n2×3 + 2×4 + 3×4 = 6+8+12 = 26.\nThis is the middle part of the answer.\n\nFinally just multiply all 3 deficiencies together: 2 × 3 × 4 = 24.\nWe write 24 because the deficiencies are actually negative.\nThe 24 is then removed as in Examples 13 and 14 earlier.", null, "Example 24\n\n1022 × 1002 × 1003 = 1027/116/132\n\n1022+2+3 = 1027, 22×2 + 22×3 + 2×3 = 116, 22×2×3 = 132.", null, "Exercise I\n\nSquaring Numbers Near A Base\n\nThis is an especially easy case under the present formula, which is described by the sub-formula Reduce (or increase) by the Deficiency and also set up the square.", null, "Example 25\n\n962 = 92/16\n\n96 is 4 below 100, so we reduce 96 by 4, which gives us the first part of the answer. The last part is just 42 = 16 as the formula says.", null, "Example 26\n\n10062 = 1012/036\n\nHere 1006 is increased by 6 to 1012, and 62 = 36: but with a base of 1000 we need 3 figures on the right, so we put 036.", null, "Example 27\n\n3042 = 3×308/16 = 92416\n\nThis is the same but because our base is 300 the left-hand part of the answer is multiplied by 3.\n\nNote the following alternative method: if we look at the number split so that\n3042 = 9/24/16, then we may see that 9 = 32,\n24 = twice 3 × 4, and 16 = 42.", null, "Exercise J\n\nSquaring Numbers Near 50\n\nIt is worth noting this case, which also comes under the above formula.", null, "Example 28\n\n542 = 29/16\n\nSince 502 = 2500 and we have 542, which is 4 more,\nwe have 25 + 4 = 29,\nand 42 = 16.", null, "Example 29\n\n482 = 2304\n\n25 - 2 = 23 (25, as before, minus the deficiency, 2)\n22 = 4 (the deficiency squared)", null, "Exercise K\n\nMultiplication By Nines\n\nThe Vedic formula By One Less Than the One Before, which is the converse of the formula of the previous chapter, comes in here in combination with All From 9 and the Last From 10.", null, "Example 30\n\n763 × 999 = 762/237\n\nThe number being multiplied by 9's is first reduced by 1:763-1 = 762, then All From 9 and the Last From 10 is applied to 763 to get 237.", null, "Example 31\n\n1867 × 99999 = 1866/98133\n\nHere, as 1867 has 4 figures, and 99999 has 5 figures, we suppose 1867 to be 01867. This is reduced by 1 to give 1866,\nand applying All From 9.... to 01867 gives 98133.", null, "Exercise L", null, "Example 32\n\n77 + 19 = 96\n\nIf asked to add 19 to a number we would probably add 20 and take one away, because 19 = 20-1.", null, "Example 33\n\n365 + 177 = 542\n\nNote that we get 23 by applying All From 9... to 77.", null, "Exercise M\n\nSubtraction", null, "Example 34\n\n77 - 19 = 58\n\nSimilarly here, the natural thing to do is to subtract 20 and add 1 back on.", null, "Example 35\n\n5040 - 1688 = 3352\n\nWe subtract 2000 and add 312: the All From 9... value of 688.", null, "Example 36\n\n7222 - 333 = 7000 - 111 = 6889\n\nHere the natural thing to do is probably to observe that we can easily subtract 222 of the 333, leaving 111 to be subtracted from 7000.\nThis last step also involves All From 9... as discussed at the end of Example 5.", null, "Exercise N", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "" ]

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[ "", null, "Financial Terms Markowitz model\n\nInformation about financial, finance, business, accounting, payroll, inventory, investment, money, inventory control, stock trading, financial advisor, tax advisor, credit.\n\n# Definition of Markowitz model", null, "## Markowitz model\n\nA model for selecting an optimum investment portfolio,\ndevised by H. M. markowitz. It uses a discrete-time, continuous-outcome\napproach for modeling investment problems, often called the mean-variance\nparadigm. See Efficient frontier.\n\n# Related Terms:\n\n## Efficient frontier\n\nA graph representing a set of portfolios that maximizes\nexpected return at each level of portfolio risk. See markowitz model.\n\n## Arbitrage-free option-pricing models\n\nYield curve option-pricing models.\n\n## Asset pricing model\n\nA model for determining the required rate of return on an asset.\n\n## Asset pricing model\n\nA model, such as the Capital Asset Pricing model (CAPM), that determines the required\nrate of return on a particular asset.\n\n## Binomial model\n\nA method of pricing options or other equity derivatives in\nwhich the probability over time of each possible price follows a binomial\ndistribution. The basic assumption is that prices can move to only two values\n(one higher and one lower) over any short time period.\n\n## Binomial option pricing model\n\nAn option pricing model in which the underlying asset can take on only two\npossible, discrete values in the next time period for each value that it can take on in the preceding time period.\n\n## Black-Scholes model\n\nThe first complete mathematical model for pricing\noptions, developed by Fischer Black and Myron Scholes. It examines market\nprice, strike price, volatility, time to expiration, and interest rates. It is limited\nto only certain kinds of options.", null, "## Black-Scholes option-pricing model\n\nA model for pricing call options based on arbitrage arguments that uses\nthe stock price, the exercise price, the risk-free interest rate, the time to expiration, and the standard deviation\nof the stock return.\n\n## Capital asset pricing model (CAPM)\n\nAn economic theory that describes the relationship between risk and\nexpected return, and serves as a model for the pricing of risky securities. The CAPM asserts that the only risk\nthat is priced by rational investors is systematic risk, because that risk cannot be eliminated by diversification.\nThe CAPM says that the expected return of a security or a portfolio is equal to the rate on a risk-free security\nplus a risk premium.\n\n## Capital Asset Pricing Model (CAPM)\n\nA model for estimating equilibrium rates of return and values of\nassets in financial markets; uses beta as a measure of asset risk\nrelative to market risk\n\n## capital asset pricing model (CAPM)\n\nTheory of the relationship between risk and return which states that the expected risk\npremium on any security equals its beta times the market risk premium.\n\n## constant-growth dividend discount model\n\nVersion of the dividend discount model in which dividends grow at a constant rate.\n\n## Constant-growth model\n\nAlso called the Gordon-Shapiro model, an application of the dividend discount\nmodel which assumes (1) a fixed growth rate for future dividends and (2) a single discount rate.\n\n## Deterministic models\n\nLiability-matching models that assume that the liability payments and the asset cash\nflows are known with certainty. Related: Compare stochastic models\n\n## Discounted dividend model (DDM)\n\nA formula to estimate the intrinsic value of a firm by figuring the\npresent value of all expected future dividends.\n\n## dividend discount model\n\nComputation of today’s stock price which states that share value equals the present value of all expected future dividends.\n\n## Dividend discount model (DDM)\n\nA model for valuing the common stock of a company, based on the\npresent value of the expected cash flows.\n\n## Dividend growth model\n\nA model wherein dividends are assumed to be at a constant rate in perpetuity.\n\n## economic components model\n\nAbrams’ model for calculating DLOM based on the interaction of discounts from four economic components.\nThis model consists of four components: the measure of the economic impact of the delay-to-sale, monopsony power to buyers, and incremental transactions costs to both buyers and sellers.\n\n## Extrapolative statistical models\n\nmodels that apply a formula to historical data and project results for a\nfuture period. Such models include the simple linear trend model, the simple exponential model, and the\nsimple autoregressive model.\n\n## Factor model\n\nA way of decomposing the factors that influence a security's rate of return into common and\nfirm-specific influences.\n\n## Garmen-Kohlhagen option pricing model\n\nA widely used model for pricing foreign currency options.\n\n## Gordon model\n\npresent value of a perpetuity with growth.\nThe end-ofyear Gordon model formula is: 1/(r - g)\nand the midyear formula is: SQRT(1 + r)/(r - g).\n\n## Index model\n\nA model of stock returns using a market index such as the S&P 500 to represent common or\nsystematic risk factors.\n\n## Internet business model\n\na model that involves\n(1) few physical assets,\n(2) little management hierarchy, and\n(3) a direct pipeline to customers\n\n## log size model\n\nAbrams’ model to calculate discount rates as a function of the logarithm of the value of the firm.\n\n## Market model\n\nThis relationship is sometimes called the single-index model. The market model says that the\nreturn on a security depends on the return on the market portfolio and the extent of the security's\nresponsiveness as measured, by beta. In addition, the return will also depend on conditions that are unique to\nthe firm. Graphically, the market model can be depicted as a line fitted to a plot of asset returns against\nreturns on the market portfolio.\n\n## Markowitz diversification\n\nA strategy that seeks to combine assets a portfolio with returns that are less than\nperfectly positively correlated, in an effort to lower portfolio risk (variance) without sacrificing return.\nRelated: naive diversification\n\n## Markowitz efficient frontier\n\nThe graphical depiction of the markowitz efficient set of portfolios\nrepresenting the boundary of the set of feasible portfolios that have the maximum return for a given level of\nrisk. Any portfolios above the frontier cannot be achieved. Any below the frontier are dominated by\nmarkowitz efficient portfolios.\n\n## Markowitz efficient portfolio\n\nAlso called a mean-variance efficient portfolio, a portfolio that has the highest\nexpected return at a given level of risk.\n\n## Markowitz efficient set of portfolios\n\nThe collection of all efficient portfolios, graphically referred to as the\nmarkowitz efficient frontier.\n\n## Modeling\n\nThe process of creating a depiction of reality, such as a graph, picture, or mathematical\nrepresentation.\n\n## percentage of sales models\n\nPlanning model in which sales forecasts are the driving variables and most other variables are\nproportional to sales.\n\n## Pie model of capital structure\n\nA model of the debt/equity ratio of the firms, graphically depicted in slices of\na pie that represent the value of the firm in the capital markets.\n\n## QMDM (quantitative marketability discount model)\n\nmodel for calculating DLOM for minority interests r the discount rate\n\n## Simple linear trend model\n\nAn extrapolative statistical model that asserts that earnings have a base level and\ngrow at a constant amount each period.\n\n## Single factor model\n\nA model of security returns that acknowledges only one common factor.\nSee: factor model.\n\n## Single index model\n\nA model of stock returns that decomposes influences on returns into a systematic factor,\nas measured by the return on the broad market index, and firm specific factors.\n\n## Single-index model\n\nRelated: market model\n\n## Stochastic models\n\nLiability-matching models that assume that the liability payments and the asset cash flows\nare uncertain. Related: Deterministic models.\n\n## Two-factor model\n\nBlack's zero-beta version of the capital asset pricing model.\n\n## Two-state option pricing model\n\nAn option pricing model in which the underlying asset can take on only two\npossible (discrete) values in the next time period for each value it can take on in the preceding time period.\nAlso called the binomial option pricing model.\n\n## Value-at-Risk model (VAR)\n\nProcedure for estimating the probability of portfolio losses exceeding some\nspecified proportion based on a statistical analysis of historical market price trends, correlations, and volatilities.\n\n## Yield curve option-pricing models\n\nmodels that can incorporate different volatility assumptions along the\nyield curve, such as the Black-Derman-Toy model. Also called arbitrage-free option-pricing models.\n\nRelated to : financial, finance, business, accounting, payroll, inventory, investment, money, inventory control, stock trading, financial advisor, tax advisor, credit." ]

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[ "## Name\n\nremquo, remquof, remquol — remainder and part of quotient\n\n## Synopsis\n\n```#include <math.h>\n```\n ```double remquo(``` double x, double y, int *quo`)`;\n\n ```float remquof(``` float x, float y, int *quo`)`;\n\n ```long double remquol(``` long double x, long double y, int *quo`)`;", null, "Note\nFeature Test Macro Requirements for glibc (see feature_test_macros(7)):\n`remquo`(), `remquof`(), `remquol`():\n`_XOPEN_SOURCE` >= 600 || `_ISOC99_SOURCE` || `_POSIX_C_SOURCE` >= 200112L;\nor cc `-std=c99`\nNote", null, "Link with `−lm`.\n\n## DESCRIPTION\n\nThese functions compute the remainder and part of the quotient upon division of `x` by `y`. A few bits of the quotient are stored via the `quo` pointer. The remainder is returned as the function result.\n\nThe value of the remainder is the same as that computed by the remainder(3) function.\n\nThe value stored via the `quo` pointer has the sign of x / y and agrees with the quotient in at least the low order 3 bits.\n\nFor example, remquo(29.0, 3.0) returns −1.0 and might store 2. Note that the actual quotient might not fit in an integer.\n\n## RETURN VALUE\n\nOn success, these functions return the same value as the analogous functions described in remainder(3).\n\nIf `x` or `y` is a NaN, a NaN is returned.\n\nIf `x` is an infinity, and `y` is not a NaN, a domain error occurs, and a NaN is returned.\n\nIf `y` is zero, and `x` is not a NaN, a domain error occurs, and a NaN is returned.\n\n## ERRORS\n\nSee math_error(7) for information on how to determine whether an error has occurred when calling these functions.\n\nThe following errors can occur:\n\nDomain error: `x` is an infinity or `y` is 0, and the other argument is not a NaN\n\nAn invalid floating-point exception (`FE_INVALID`) is raised.\n\nThese functions do not set `errno`.\n\n## VERSIONS\n\nThese functions first appeared in glibc in version 2.1.\n\n## CONFORMING TO\n\nC99, POSIX.1-2001.\n\nThis page is part of release 3.52 of the Linux `man-pages` project. A description of the project, and information about reporting bugs, can be found at http://www.kernel.org/doc/man−pages/." ]

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[ "# Sorting Algorithms in Ruby\n\nHaving enjoyed excellent posts on sorting algorithms I wanted to share my implementation of common sorting algorithms in Ruby. There are some great visualizations of sorting algorithms especially on Mike Bostock's blog such as the Fisher-Yates Shuffle, merge sort, and quick sort. Taking a leaf out of Jesse La Russo's blog below are my implementations of common sorting algorithms in Ruby.\n\n### Insertion Sort\n\n1. Begin at the second element. Compare the element to the element prior. Swap if less than the element prior.\n2. Increment by one element. Compare this element to prior two elements. Insert accordingly.\n3. Continue incrementing by one element and comparing that element to all prior sorted elements until the list is fully sorted.\n``````def insertion_sort(list)\n(1...list.length).each do |i|\nk = i\nwhile k > 0 && list[k] < list[k-1]\nlist[k], list[k-1] = list[k-1], list[k]\nk -= 1\nend\nend\nlist\nend\n``````\n\n### Selection Sort\n\n1. Start at first element of unsorted list. Look for the smallest element in the list. Swap with left most unsorted element.\n2. Move to the second element. Swap with smallest element of this unsorted list.\n``````def selection_sort(list)\n(0...list.length).each do |i|\nk = i\n\n(i+1...list.length).each do |j|\nk = j if list[j] < list[k]\nend\n\nif k != i\nlist[i],list[k] = list[k],list[i]\nend\nend\nlist\nend\n``````\n\n### Bubble Sort\n\n1. Start at first element. Compare adjacent pairs. Swap if out of order.\n2. Iterate through the list repeatedly. Each iteration requires one less comparison.\n``````def bubble_sort(list)\nbegin\n\nswapped = false\n(1...list.length).each do |i|\n\nif list[i] < list[i-1]\nlist[i], list[i-1] = list[i-1],list[i]\nswapped = true\nend\n\nend\n\nend until !swapped\nlist\nend\n``````\n\n### Shell Sort\n\n(Insertion sort on sublists or sublists of elements allowing swap of elements far apart.)\n\n1. Determine sublists by splitting the list in half repeatedly.\n2. Insertion sort of the elements in each sublist.\n3. When the sublist is one or zero elements long then the list is sorted.\n``````def shell_sort(list)\ngap = list.length/2\nwhile gap > 0\n\n# Insertion Sort\n(gap...list.length).each do |i|\nk = i\nwhile k >= gap && list[k] < list[k-gap]\nlist[k], list[k-gap] = list[k-gap], list[k]\nk -= gap\nend\nend\n\ngap /= 2\nend\nlist\nend\n``````\n\n### Merge Sort\n\n1. Divide the unsorted list into sublists around a pivot.\n2. Recursively call merge sort on the sublists.\n3. Merge the results of the recursive calls.\n``````def merge_sort(list)\nreturn list if list.length <= 1\nleft = []\nright = []\npivot = list.length / 2\n\nlist.each_with_index do |e,i|\nleft << e if i < pivot\nend\n\nlist.each_with_index do |e,i|\nright << e if i >= pivot\nend\n\nleft = merge_sort(left)\nright = merge_sort(right)\n\nmerge(left, right)\nend\n\ndef merge(left, right)\nresult = []\nwhile left.length > 0 || right.length > 0\nif left.length > 0 && right.length > 0\nresult << (left <= right ? left.shift : right.shift)\nelsif left.length > 0\nresult << left.shift\nelse\nresult << right.shift\nend\nend\nresult\nend\n``````\n\n### Heap Sort\n\n1. Build a heap out of the data. The heap is a tree data structure where the parent node is always greater than or equal to its child nodes across the entire heap.\n2. Remove the root of the heap and insert in sorted array. Reconstruct heap and repeat.\n``````def heap_sort(list)\ncount = list.length\n\nheapify(list, count)\n\nlast = count - 1\nwhile last > 0\np \"#{list[last]}, #{list}\"\nlist[last], list = list, list[last]\nlast -= 1\nsift_down(list,0,last)\nend\n\nlist\nend\n\ndef heapify(list, count)\nstart = (count - 2) / 2\n\nwhile start >= 0\nsift_down(list,start,count-1)\nstart -= 1\nend\nend\n\ndef sift_down(list, start, last)\nroot = start\n\nwhile root * 2 + 1 <= last\nchild = root * 2 + 1\nswap = root\n\nswap = child if list[swap] < list[child]\nif child + 1 <= last && list[swap] < list[child+1]\nswap = child + 1\nend\nif swap != root\nlist[root], list[swap] = list[swap], list[root]\nroot = swap\nelse\nreturn\nend\nend\nend\n``````\n\n### Quick Sort\n\n1. Divide list into two smaller lists.\n2. Pick a pivot.\n3. Reorder list so that all elements less than the pivot come before it and all elements greater than the pivot come after it.\n4. Recursively apply steps 2 and 3 to the sublists until base case of size 0 or 1 lists is reached.\n``````def quick_sort(list)\n\nreturn list if list.length <= 1\n\npivot = list[list.length/2]\n\nless, greater = list.partition {|e| e < pivot}\n\nquick_sort(less) + [pivot] + quick_sort(greater)\n\nend\n``````" ]

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[ "Periodic Temperature Measurements using the DS1820\n\n' DS1820_2.Bas (BX-24)\n'\n' Performs a temperature measurement using a DS1820 every 15 seconds and\n' displays. Performs an 8-bit cyclic redundancy check on the data\n'\n' Illustrates the use of 1-Wire commands and use of the on-board real time\n' clock to perfrom periodic measurements.\n'\n' Note that Timer is used to fetch the elapsed time since midnight and thus\n' periodicity is achieved using;\n'\n' do\n' TimeCurrent = Timer()\n' until (TimeCurrent >= TimeOut)\n'\n' However, a problem occurs when TimeOut >= 86400.0. That is, TimeOut\n' is in a new day. In this case, TimeOut = TimeOut - 86400.0 and the\n' timing is performed in two steps;\n'\n' do\n' TimeCurrent = Timer()\n' until (TimeCurrent <= TimeOut)\t' wait for clock to roll over\n'\n' do\n' TimeCurrent = Timer()\n' until (TimeCurrent >= TimeOut)\n'\n' For a more complete discussion of the implemenation of the calculation of\n' the 8-bit CRC, see http://www.phanderson.com/PIC\n'\n' BX24\t\t\t\t\tDS1820\n'\t\t\t+5\n'\t\t\t|\n'\t\t\t* 4.7K\n' \t\t\t|\n' RA.7 (Term 13) ---------------------- DQ (term 2)\n'\t\t\t\t\tTerms 1 and 3 to GRD\n'\n'\n' copyright, Peter H. Anderson, Baltimore, MD, Dec, '99\n\nConst TEST as Byte = 1\n\nSub Main()\n\nDim Str as String *15\nDim T_C as Single\nDim TimeCurrent as Single, TimeOut as Single, Tperiod as Single\n\nTperiod = 15.0\n\nCall OpenSerialPort(1, 19200)\nCall PutTime(0, 0, 0.0)\nTimeOut = Timer()\n\nDo\t\t' continually perform measurements\nT_C = MakeTempMeas(13)\nIf (T_C < -80.00) Then\nStr=\"CRC Error\"\nCall PutStr(str)\nElse\nCall PutS(T_C)\nEnd If\nCall NewLine()\n\nTimeOut = TimeOut + Tperiod\n\nIf (TimeOut >= 86400.0) Then\t' rollover\nTimeOut = TimeOut - 86400.0\nDo\t\t\t\t' wait for RTC to roll over\nTimeCurrent = Timer()\t' 86398, 86399, 0\nLoop Until (TimeCurrent <= TimeOut)\nEnd If\nDo\nTimeCurrent = Timer()\nLoop Until (TimeCurrent >= TimeOut)\nLoop\nEnd Sub\n\nFunction MakeTempMeas(ByVal Pin as Byte) as Single\n' Returns temperature. Returns -89.99 if CRC error\n\nDim N as Integer\nDim Dat(1 To 9) as Byte, CRC as Byte\nDim T_C as Single\n\nCall Init_1W(Pin)\nCall OutByte_1W(Pin, &Hcc)\t' skip ROM\n\nCall OutByte_1W(Pin, &H44)\t' perform temperature conversion\nCall StrongPullup_1W(Pin)\t' strong pullup\n\nCall Init_1W(Pin)\nCall OutByte_1W(Pin, &Hcc)\t' skip ROM\n\nCall OutByte_1W(Pin, &Hbe)\t' get temperature data\n\nFor N = 1 to 9\nDat(N) = InByte_1W(Pin)\t' fetch the nine bytes\nNext\n\nIf (TEST <> 0) Then\nFor N = 1 to 9\nCall PutHexB(Dat(N))\nCall PutByte(Asc(\" \"))\nNext\nCall NewLine()\nEnd If\n\nCRC = CalcCRC(Dat, 9)\t' calculate CRC of the 9 bytes\n\nIf (CRC <>0) Then\t\t' CRC failure\nT_C = -89.99\nElse\nIf (Dat(2) = 0) Then\t' its postive\nT_C = CSng(Dat(1)) / 2.0\nElse\nDat(1) = (Dat(1) XOR &Hff) + 1\t' takes the 2's comp\nT_C = CSng(Dat(1)) / 2.0\nEnd If\nEnd If\n\nMakeTempMeas = T_C\n\nEnd Function\n\nSub Init_1W(ByVal Pin as Byte) ' bring Pin low for 500 usecs and then back\n' high\n\nDim N as Integer\nCall PutPin(Pin, 2)\t' be sure DQ is an input\n\nCall PutPin(Pin, 0)\n\nFor N = 1 to 3\t' adjust for 500 usec delay\nNext\n\nCall PutPin(Pin, 2)\n\nFor N = 1 to 3\nNext\n\nEnd Sub\n\nFunction InByte_1W(ByVal Pin as Byte) as Byte\n\nDim N as Integer, IByte as Byte, B as Byte\nFor N =1 to 8\nB = Get1Wire(Pin)\nIf (B=1) then\nIByte = (IByte\\2) OR bx10000000\nElse\nIByte = IByte\\2\nEnd If\nNext\n\nInByte_1W = IByte\n\nEnd Function\n\nSub OutByte_1W(ByVal Pin as Byte, ByVal OByte as Byte)\n\nDim N as Integer, B as Byte\nFor N = 1 to 8\nB = OByte AND bx00000001\nIf (B=1) Then\nCall Put1Wire(Pin, 1)\nElse\nCall Put1Wire(Pin, 0)\nEnd If\nOByte = OByte \\ 2\nNext\nEnd Sub\n\nSub StrongPullUp_1W(ByVal Pin as Byte)\n' Provide a hard logic one for 0.5 secs\nCall PutPin(Pin, 1)\nCall Sleep(0.5)\nCall PutPin(Pin, 2)\nEnd Sub\n\nSub PutHexB(ByVal X as Byte)\t' display a byte in hex format\nDim Y as Byte\n\nY= X \\ 16 \t\t\t' convert high nibble to character\nIf (Y < 10) then\nY = Y + Asc(\"0\")\nElse\nY = Y - 10 + Asc (\"A\")\nEnd If\nCall PutByte(Y)\n\nY= X And bx00001111 \t' same for low nibble\nIf (Y < 10) then\nY = Y + Asc(\"0\")\nElse\nY = Y - 10 + Asc (\"A\")\nEnd If\nCall PutByte(Y)\nEnd Sub\n\nFunction CalcCRC(ByRef Buff() as Byte, _\nByVal NumVals as Integer) as Byte\n\nDim ShiftReg as Byte, SRlsb as Byte, DataBit as Byte, V as Byte\nDim FBbit as Byte, I as Integer, J as Integer\n\nShiftReg = 0\t\t\t' initialize the shift regsiter\n\nFor I = 1 to NumVals\t\t' for each byte in the array\nV = Buff(I)\nFor J = 1 to 8\t\t' for each bit\nDataBit = V AND &H01\t' isolate least sign bit\nSRlsb = ShiftReg AND &H01\nFBbit = (DataBit XOR SRlsb) AND &H01\n' calculate the feed back bit\nShiftReg = ShiftReg \\ 2\t' shift right\nIf (FBbit = 1) then\nShiftReg = ShiftReg XOR &H8c\nEnd if\n'Call PutB(Shift_Reg)\t' for debugging\n'Call NewLine()\nV = V \\ 2\t\t' next bit now in least sig bit position\nNext\n' Call PutB(Shift_Reg)\n' Call NewLine()\nNext\nCalcCRC = ShiftReg\t\t' return the result\n\nEnd Function" ]

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[ "M550: Introduction to Numerical Methods for Partial Differential Equations\nSpring 2012  Reference Number: CRN 14828\nInstructor: Dr. I. Oprea\nhttp://www.math.colostate.edu/~juliana\nMWF 8-8:50AM, Room: Weber 201(see comments below)\n\nOverview\nAnalytic solutions exist only for the most elementary partial differential equations (PDEs); the rest must be tackled with numerical methods.  This course will cover numerical solution of PDEs: the method of lines, finite differences, finite element and spectral methods, to an extent necessary for successful numerical modeling of physical phenomena. We shall construct and characterize the behavior of computational methods for PDEs based on the physical meaning and the mathematical analysis of the underlying continuous equations\n\nThis course is an entry-level graduate course. The intended audience is graduate students and advanced undergraduates. The design philosophy of the course is to cover the mathematical theory applied to the simplest examples so as to minimize technical details, yet provide a strong foundation that students can carry back to their particular fields. Applications and examples will be correlated to students interests.\n\nSchedule - MWF 8-8:50AM, but can be adjusted to fit everyone's' schedule\n\nCoursework\nThe course work consists of a mixture of mathematical problem sets and a few computational projects. The computational projects emphasize experimentation with mathematical issues such as convergence, stability and accuracy. Grading is based on HW/Projects - 70% and a Take-Home Final Exam-30%.\n\nPrerequisites\nThe course assumes familiarity with basic linear algebra and ordinary differential equations. Exposure to solutions of the classic models in partial differential equations is a plus.  The use of MATLAB is strongly encouraged.\n\nRecommended Textbook\nJichun Li; Yi-Tung Chen, University of Nevada, Las Vegas,  - Computational Partial Differential Equations Using MATLAB , Series: Chapman & Hall/CRC Applied Mathematics & Nonlinear Science; Chapman and Hall/CRC -\nSupplementary Lecture:\n1. Computational Differential Equations, K. Eriksson, D. Estep, P. Hansbo, and C. Johnson, Cambridge University Press, 1996\n\n2. Numerical Solution of Partial Differential Equations: An Introduction, K. Morton and D. Mayers, Cambridge University Press, 2-nd edition\n\nTopics to be Covered\n1. Brief Introduction to Partial Differential Equations and Basic Numerical Analysis\n- Interpolation theory Numerical quadrature, The need for numerical solutions of differential equations\n\n2. Spectral Methods\n\n- an overview\n\n2.Elliptic Problems and the Finite Element Method\n- Models involving conservation of heat, behavior of solutions\n- Two-point boundary value problems and the Laplace and Poisson equations\n- The variational formulation and the Galerkin finite element method\n- Convergence in the energy norm, a priori convergence, order of convergence\n- Quadrature in the finite element method and the finite difference method\n- Brief overview of complications that can occur in realistic models\n3. Parabolic Problems and the Method of Lines\n- Explicit and implicit method of lines using finite elements in space and finite differences in time\n- Numerical stability, stiffness and dissipativity, convergence\n4. Hyperbolic Problems and the Finite Difference Method\n- The transport equation and wave equations, characteristics and the transport of information, behavior of solutions\n- Finite difference schemes, consistency\n- Stability, dissipativity, dispersion, the CFL condition, convergence\n5. Miscellaneous Topics as Time Permits : Error estimation, computational error estimation and adaptive schemes, conservation laws" ]

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[ "# Critical frontier of the Potts and percolation models on triangular-type and kagome-type lattices. II. Numerical analysis.\n\n```@article{Ding2010CriticalFO,\ntitle={Critical frontier of the Potts and percolation models on triangular-type and kagome-type lattices. II. Numerical analysis.},\nauthor={Chengxiang Ding and Zhe Fu and Wenan Guo and F. Y. Wu},\njournal={Physical review. E, Statistical, nonlinear, and soft matter physics},\nyear={2010},\nvolume={81 6 Pt 1},\npages={\n061111\n}\n}```\n• Published 10 January 2010\n• Mathematics\n• Physical review. E, Statistical, nonlinear, and soft matter physics\nIn the preceding paper, one of us (F. Y. Wu) considered the Potts model and bond and site percolation on two general classes of two-dimensional lattices, the triangular-type and kagome-type lattices, and obtained closed-form expressions for the critical frontier with applications to various lattice models. For the triangular-type lattices Wu's result is exact, and for the kagome-type lattices Wu's expression is under a homogeneity assumption. The purpose of the present paper is twofold: First…\n23 Citations\nThe critical curves of the q-state Potts model can be determined exactly for regular two-dimensional lattices G that are of the three-terminal type. This comprises the square, triangular, hexagonal\n• Physics\nPhysical review. E, Statistical, nonlinear, and soft matter physics\n• 2012\nThe exact critical frontier of the Potts model on bowtie lattices is given, and the critical frontier yields the thresholds of bond percolation on these lattices, which are exactly consistent with the results given by Ziff et al.\n• Physics\n• 2015\nWe compute critical polynomials for the q-state Potts model on the Archimedean lattices, using a parallel implementation of the algorithm of Jacobsen (2014 J. Phys. A: Math. Theor 47 135001) that\n• Physics\nPhysical review. E, Statistical, nonlinear, and soft matter physics\n• 2011\nIt is verified that the Ising model on the kagome lattice belongs to a certain class of spin systems on two-dimensional infinite strips and an upper bound for the critical coupling strength K(c)(q) for the q-state Potts model is suggested from exact calculations of the internal energy for the two smallest strip widths.\n• Mathematics\n• 2012\nWe give a conditional derivation of the inhomogeneous critical percolation manifold of the bow-tie lattice with five different probabilities, a problem that does not appear at first to fall into any\n• Mathematics\n• 2012\nPercolation thresholds have recently been studied by means of a graph polynomial PB(p), henceforth referred to as the critical polynomial, that may be defined on any periodic lattice. The polynomial\n• Mathematics\n• 2012\nAny two-dimensional infinite regular lattice G can be produced by tiling the plane with a finite subgraph B⊆G; we call B a basis of G. We introduce a two-parameter graph polynomial PB(q, v) that\n• Mathematics\n• 2013\nIn our previous work we have shown that critical manifolds of the q-state Potts model can be studied by means of a graph polynomial PB(q, v), henceforth referred to as the critical polynomial.\nAlthough every exactly known bond percolation critical threshold is the root in [0,1] of a lattice-dependent polynomial, it has recently been shown that the notion of a critical polynomial can be" ]

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[ "Solutions by everydaycalculation.com\n\n## What is the GCF of 900 and 1008?\n\nThe GCF of 900 and 1008 is 36.\n\n#### Steps to find GCF\n\n1. Find the prime factorization of 900\n900 = 2 × 2 × 3 × 3 × 5 × 5\n2. Find the prime factorization of 1008\n1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7\n3. To find the GCF, multiply all the prime factors common to both numbers:\n\nTherefore, GCF = 2 × 2 × 3 × 3\n4. GCF = 36\n\nMathStep (Works offline)", null, "Download our mobile app and learn how to find GCF of upto four numbers in your own time:\nAndroid and iPhone/ iPad\n\n#### GCF Calculator\n\nEnter two numbers separate by comma. To find GCF of more than two numbers, click here.\n\nThe greatest common factor (GCF) is also known as greatest common divisor (GCD) or highest common factor (HCF)." ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "# Is 92 Divisible By Anything?\n\nOkay, so when we ask if 92 is divisible by anything, we are looking to see if there are any WHOLE numbers that can be divided into 92 that will result in a whole number as the answer. In this short guide, we'll walk you through how to figure out whether 92 is divisible by anything. Let's go!\n\nFun fact! All whole numbers will have at least two numbers that they are divisible by. Those would be the actual number in question (in this case 92), and the number 1.\n\nSo, the answer is yes. The number 92 is divisible by 6 number(s).\n\nLet's list out all of the divisors of 92:\n\n• 1\n• 2\n• 4\n• 23\n• 46\n• 92\n\nWhen we list them out like this it's easy to see that the numbers which 92 is divisible by are 1, 2, 4, 23, 46, and 92.\n\nYou might be interested to know that all of the divisor numbers listed above are also known as the Factors of 92.\n\nNot only that, but the numbers can also be called the divisors of 92. Basically, all of those numbers can go evenly into 92 with no remainder.\n\nAs you can see, this is a pretty simple one to calculate. All you need to do is list out all of the factors for the number 92. If there are any factors, then you know that 92 is divisible by something.\n\nGive this a go for yourself and try to calculate a couple of these without using our calculator. Grab a pencil and a piece of paper and pick a couple of numbers to try it with.\n\nIf you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!\n\n• \"Is 92 Divisible By Anything?\". VisualFractions.com. Accessed on June 4, 2023. http://visualfractions.com/calculator/divisible-by-anything/is-92-divisible-by-anything/.\n\n• \"Is 92 Divisible By Anything?\". VisualFractions.com, http://visualfractions.com/calculator/divisible-by-anything/is-92-divisible-by-anything/. Accessed 4 June, 2023." ]

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[ "×\n×\n\n# In a road-paving process, asphalt mix is delivered to the", null, "ISBN: 9780321629111 32\n\n## Solution for problem 35E Chapter 4\n\nProbability and Statistics for Engineers and the Scientists | 9th Edition\n\n• Textbook Solutions\n• 2901 Step-by-step solutions solved by professors and subject experts\n• Get 24/7 help from StudySoup virtual teaching assistants", null, "Probability and Statistics for Engineers and the Scientists | 9th Edition\n\n4 5 1 240 Reviews\n10\n1\nProblem 35E\n\nIn a road-paving process, asphalt mix is delivered to the hopper of the paver by trucks that haul the material from the batching plant. The article ?\"Modeling of Simultaneously Continuous and Stochastic Construction Activities for Simulation\" U. of Construction Engr. and Mgmnt., 2013: 1037-1045) proposed a normal distribution with mean value 8.46 min and standard deviation .913 min for the rv X = truck haul time. a. What is the probability that haul time will be at least 10 min? Will exceed 10 min? b. What is the probability that haul time will exceed 15 min? c. What is the probability that haul time will be between 8 and 10 min? d. What value c is such that 98% of all haul times are in the interval from 8.46 — c to 8.4.6 + c? e. If four haul times are independently selected, what is the probability that at least one of them exceeds 10 min?\n\nStep-by-Step Solution:\nStep 2 of 3\n\nAnswer: Step1 of 6: Given a normal distribution with a mean value 8.46 min and standard deviation 0.913 min. The random variable X = truck haul time. Step2 of 6: a). We need to find the probability that haul time will be i).at least 10 min and ii). Will exceed 10 min. That is, x 108.46 i). P(X 10) = P( 0.913 ) = P(Z1.69) = P(Z -1.69) = 0.0455 ( this value from standard normal table) Thus,the probability that haul time will be at least 10 min is 0.0455. ii). To determine the probability that haul time will exceed 10 min. That is, P(X > 10) = P(X 10) = 0.0455. Thus,the probability that haul time will exceed 10 min. Step2 of 6: b). To determine the probability that haul time will exceed 15 min. x 158.46 That is, P(X > 15) = P( > 0.913) = P(Z >7.1632) (from standard normal table) = 0. Thus,the probability that haul time will be at least 15 min is zero.\n\nStep 3 of 3\n\n##### ISBN: 9780321629111\n\nUnlock Textbook Solution" ]

[ null, "https://studysoup.com/cdn/8cover_2450434", null, "https://studysoup.com/cdn/8cover_2450434", null ]

[ "User Real IP - 3.236.126.69\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => Array\n(\n => 95.142.117.29\n)\n\n => Array\n(\n => 150.129.60.107\n)\n\n => Array\n(\n => 94.205.243.22\n)\n\n => Array\n(\n => 115.42.71.143\n)\n\n => Array\n(\n => 117.217.195.59\n)\n\n => Array\n(\n => 182.77.112.56\n)\n\n => Array\n(\n => 182.77.112.108\n)\n\n => Array\n(\n => 41.80.69.10\n)\n\n => Array\n(\n => 117.5.222.121\n)\n\n => Array\n(\n => 103.11.0.38\n)\n\n => Array\n(\n => 202.173.127.140\n)\n\n => Array\n(\n => 49.249.249.50\n)\n\n => Array\n(\n => 116.72.198.211\n)\n\n => Array\n(\n => 223.230.54.53\n)\n\n => Array\n(\n => 102.69.228.74\n)\n\n => Array\n(\n => 39.37.251.89\n)\n\n => Array\n(\n => 39.53.246.141\n)\n\n => Array\n(\n => 39.57.182.72\n)\n\n => Array\n(\n => 209.58.130.210\n)\n\n => Array\n(\n => 104.131.75.86\n)\n\n => Array\n(\n => 106.212.131.255\n)\n\n => Array\n(\n => 106.212.132.127\n)\n\n => Array\n(\n => 223.190.4.60\n)\n\n => Array\n(\n => 103.252.116.252\n)\n\n => Array\n(\n => 103.76.55.182\n)\n\n => Array\n(\n => 45.118.166.70\n)\n\n => Array\n(\n => 103.93.174.215\n)\n\n => Array\n(\n => 5.62.62.142\n)\n\n => Array\n(\n => 182.179.158.156\n)\n\n => Array\n(\n => 39.57.255.12\n)\n\n => Array\n(\n => 39.37.178.37\n)\n\n => Array\n(\n => 182.180.165.211\n)\n\n => Array\n(\n => 119.153.135.17\n)\n\n => Array\n(\n => 72.255.15.244\n)\n\n => Array\n(\n => 139.180.166.181\n)\n\n => Array\n(\n => 70.119.147.111\n)\n\n => Array\n(\n => 106.210.40.83\n)\n\n => Array\n(\n => 14.190.70.91\n)\n\n => Array\n(\n => 202.125.156.82\n)\n\n => Array\n(\n => 115.42.68.38\n)\n\n => Array\n(\n => 102.167.13.108\n)\n\n => Array\n(\n => 117.217.192.130\n)\n\n => Array\n(\n => 205.185.223.156\n)\n\n => Array\n(\n => 171.224.180.29\n)\n\n => Array\n(\n => 45.127.45.68\n)\n\n => Array\n(\n => 195.206.183.232\n)\n\n => Array\n(\n => 49.32.52.115\n)\n\n => Array\n(\n => 49.207.49.223\n)\n\n => Array\n(\n => 45.63.29.61\n)\n\n => Array\n(\n => 103.245.193.214\n)\n\n => Array\n(\n => 39.40.236.69\n)\n\n => Array\n(\n => 62.80.162.111\n)\n\n => Array\n(\n => 45.116.232.56\n)\n\n => Array\n(\n => 45.118.166.91\n)\n\n => Array\n(\n => 180.92.230.234\n)\n\n => Array\n(\n => 157.40.57.160\n)\n\n => Array\n(\n => 110.38.38.130\n)\n\n => Array\n(\n => 72.255.57.183\n)\n\n => Array\n(\n => 182.68.81.85\n)\n\n => Array\n(\n => 39.57.202.122\n)\n\n => Array\n(\n => 119.152.154.36\n)\n\n => Array\n(\n => 5.62.62.141\n)\n\n => Array\n(\n => 119.155.54.232\n)\n\n => Array\n(\n => 39.37.141.22\n)\n\n => Array\n(\n => 183.87.12.225\n)\n\n => Array\n(\n => 107.170.127.117\n)\n\n => Array\n(\n => 125.63.124.49\n)\n\n => Array\n(\n => 39.42.191.3\n)\n\n => Array\n(\n => 116.74.24.72\n)\n\n => Array\n(\n => 46.101.89.227\n)\n\n => Array\n(\n => 202.173.125.247\n)\n\n => Array\n(\n => 39.42.184.254\n)\n\n => Array\n(\n => 115.186.165.132\n)\n\n => Array\n(\n => 39.57.206.126\n)\n\n => Array\n(\n => 103.245.13.145\n)\n\n => Array\n(\n => 202.175.246.43\n)\n\n => Array\n(\n => 192.140.152.150\n)\n\n => Array\n(\n => 202.88.250.103\n)\n\n => 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199.58.164.135\n)\n\n => Array\n(\n => 101.53.228.151\n)\n\n => Array\n(\n => 117.230.50.57\n)\n\n => Array\n(\n => 223.225.138.84\n)\n\n => Array\n(\n => 110.225.67.65\n)\n\n => Array\n(\n => 47.15.200.39\n)\n\n => Array\n(\n => 39.42.20.127\n)\n\n => Array\n(\n => 117.97.241.81\n)\n\n => Array\n(\n => 111.119.185.11\n)\n\n => Array\n(\n => 103.100.5.94\n)\n\n => Array\n(\n => 103.25.137.69\n)\n\n => Array\n(\n => 47.15.197.159\n)\n\n => Array\n(\n => 223.188.176.122\n)\n\n => Array\n(\n => 27.4.175.80\n)\n\n => Array\n(\n => 181.215.43.82\n)\n\n => Array\n(\n => 27.56.228.157\n)\n\n => Array\n(\n => 117.230.19.19\n)\n\n => Array\n(\n => 47.15.208.71\n)\n\n => Array\n(\n => 119.155.21.176\n)\n\n => Array\n(\n => 47.15.234.202\n)\n\n => Array\n(\n => 117.230.144.135\n)\n\n => Array\n(\n => 112.79.139.199\n)\n\n => Array\n(\n => 116.75.246.41\n)\n\n => Array\n(\n => 117.230.177.126\n)\n\n => Array\n(\n => 212.103.48.134\n)\n\n => Array\n(\n => 102.69.228.78\n)\n\n => Array\n(\n => 117.230.37.118\n)\n\n => Array\n(\n => 175.143.61.75\n)\n\n => Array\n(\n => 139.167.56.138\n)\n\n => Array\n(\n => 58.145.189.250\n)\n\n => Array\n(\n => 103.255.5.65\n)\n\n => Array\n(\n => 39.37.153.182\n)\n\n => Array\n(\n => 157.43.85.106\n)\n\n => Array\n(\n => 185.209.178.77\n)\n\n => Array\n(\n => 1.39.212.45\n)\n\n => Array\n(\n => 103.72.7.16\n)\n\n => Array\n(\n => 117.97.185.244\n)\n\n => Array\n(\n => 117.230.59.106\n)\n\n => Array\n(\n => 137.97.121.103\n)\n\n => Array\n(\n => 103.82.123.215\n)\n\n => Array\n(\n => 103.68.217.248\n)\n\n => Array\n(\n => 157.39.27.175\n)\n\n => Array\n(\n => 47.31.100.249\n)\n\n => Array\n(\n => 14.171.232.139\n)\n\n => Array\n(\n => 103.31.93.208\n)\n\n => Array\n(\n => 117.230.56.77\n)\n\n => Array\n(\n => 124.182.25.124\n)\n\n => Array\n(\n => 106.66.191.242\n)\n\n => Array\n(\n => 175.107.237.25\n)\n\n => Array\n(\n => 119.155.1.27\n)\n\n => Array\n(\n => 72.255.6.24\n)\n\n => Array\n(\n => 192.140.152.223\n)\n\n => Array\n(\n => 212.103.48.136\n)\n\n => Array\n(\n => 39.45.134.56\n)\n\n => Array\n(\n => 139.167.173.30\n)\n\n => Array\n(\n => 117.230.63.87\n)\n\n => Array\n(\n => 182.189.95.203\n)\n\n => Array\n(\n => 49.204.183.248\n)\n\n => Array\n(\n => 47.31.125.188\n)\n\n => Array\n(\n => 103.252.171.13\n)\n\n => Array\n(\n => 112.198.74.36\n)\n\n => Array\n(\n => 27.109.113.152\n)\n\n => Array\n(\n => 42.112.233.44\n)\n\n => Array\n(\n => 47.31.68.193\n)\n\n => Array\n(\n => 103.252.171.134\n)\n\n => Array\n(\n => 77.123.32.114\n)\n\n => Array\n(\n => 1.38.189.66\n)\n\n => Array\n(\n => 39.37.181.108\n)\n\n => Array\n(\n => 42.106.44.61\n)\n\n => Array\n(\n => 157.36.8.39\n)\n\n => Array\n(\n => 223.238.41.53\n)\n\n => Array\n(\n => 202.89.77.10\n)\n\n => Array\n(\n => 117.230.150.68\n)\n\n => Array\n(\n => 175.176.87.60\n)\n\n => Array\n(\n => 137.97.117.87\n)\n\n => Array\n(\n => 132.154.123.11\n)\n\n => Array\n(\n => 45.113.124.141\n)\n\n => Array\n(\n => 103.87.56.203\n)\n\n => Array\n(\n => 159.89.171.156\n)\n\n => Array\n(\n => 119.155.53.88\n)\n\n => Array\n(\n => 222.252.107.215\n)\n\n => Array\n(\n => 132.154.75.238\n)\n\n => Array\n(\n => 122.183.41.168\n)\n\n => Array\n(\n => 42.106.254.158\n)\n\n => Array\n(\n => 103.252.171.37\n)\n\n => Array\n(\n => 202.59.13.180\n)\n\n => Array\n(\n => 37.111.139.137\n)\n\n => Array\n(\n => 39.42.93.25\n)\n\n => Array\n(\n => 118.70.177.156\n)\n\n => Array\n(\n => 117.230.148.64\n)\n\n => Array\n(\n => 39.42.15.194\n)\n\n => Array\n(\n => 137.97.176.86\n)\n\n => Array\n(\n => 106.210.102.113\n)\n\n => Array\n(\n => 39.59.84.236\n)\n\n => Array\n(\n => 49.206.187.177\n)\n\n => Array\n(\n => 117.230.133.11\n)\n\n => Array\n(\n => 42.106.253.173\n)\n\n => Array\n(\n => 178.62.102.23\n)\n\n => Array\n(\n => 111.92.76.175\n)\n\n => Array\n(\n => 132.154.86.45\n)\n\n => Array\n(\n => 117.230.128.39\n)\n\n => Array\n(\n => 117.230.53.165\n)\n\n => Array\n(\n => 49.37.200.171\n)\n\n => Array\n(\n => 104.236.213.230\n)\n\n => Array\n(\n => 103.140.30.81\n)\n\n => Array\n(\n => 59.103.104.117\n)\n\n => Array\n(\n => 65.49.126.79\n)\n\n => Array\n(\n => 202.59.12.251\n)\n\n => Array\n(\n => 37.111.136.17\n)\n\n => Array\n(\n => 163.53.85.67\n)\n\n => Array\n(\n => 123.16.240.73\n)\n\n => Array\n(\n => 103.211.14.183\n)\n\n => Array\n(\n => 103.248.93.211\n)\n\n => Array\n(\n => 116.74.59.127\n)\n\n => Array\n(\n => 137.97.169.254\n)\n\n => Array\n(\n => 113.177.79.100\n)\n\n => Array\n(\n => 74.82.60.187\n)\n\n => Array\n(\n => 117.230.157.66\n)\n\n => Array\n(\n => 169.149.194.241\n)\n\n => Array\n(\n => 117.230.156.11\n)\n\n => Array\n(\n => 202.59.12.157\n)\n\n => Array\n(\n => 42.106.181.25\n)\n\n => Array\n(\n => 202.59.13.78\n)\n\n => Array\n(\n => 39.37.153.32\n)\n\n => Array\n(\n => 177.188.216.175\n)\n\n => Array\n(\n => 222.252.53.165\n)\n\n => Array\n(\n => 37.139.23.89\n)\n\n => Array\n(\n => 117.230.139.150\n)\n\n => Array\n(\n => 104.131.176.234\n)\n\n => Array\n(\n => 42.106.181.117\n)\n\n => Array\n(\n => 117.230.180.94\n)\n\n => Array\n(\n => 180.190.171.5\n)\n\n => Array\n(\n => 150.129.165.185\n)\n\n => Array\n(\n => 51.15.0.150\n)\n\n => Array\n(\n => 42.111.4.84\n)\n\n => Array\n(\n => 74.82.60.116\n)\n\n => Array\n(\n => 137.97.121.165\n)\n\n => Array\n(\n => 64.62.187.194\n)\n\n => Array\n(\n => 137.97.106.162\n)\n\n => Array\n(\n => 137.97.92.46\n)\n\n => Array\n(\n => 137.97.170.25\n)\n\n => Array\n(\n => 103.104.192.100\n)\n\n => Array\n(\n => 185.246.211.34\n)\n\n => Array\n(\n => 119.160.96.78\n)\n\n => Array\n(\n => 212.103.48.152\n)\n\n => Array\n(\n => 183.83.153.90\n)\n\n => Array\n(\n => 117.248.150.41\n)\n\n => Array\n(\n => 185.240.246.180\n)\n\n => Array\n(\n => 162.253.131.125\n)\n\n => Array\n(\n => 117.230.153.217\n)\n\n => Array\n(\n => 117.230.169.1\n)\n\n => Array\n(\n => 49.15.138.247\n)\n\n => Array\n(\n => 117.230.37.110\n)\n\n => Array\n(\n => 14.167.188.75\n)\n\n => Array\n(\n => 169.149.239.93\n)\n\n => Array\n(\n => 103.216.176.91\n)\n\n => Array\n(\n => 117.230.12.126\n)\n\n => Array\n(\n => 184.75.209.110\n)\n\n => Array\n(\n => 117.230.6.60\n)\n\n => Array\n(\n => 117.230.135.132\n)\n\n => Array\n(\n => 31.179.29.109\n)\n\n => Array\n(\n => 74.121.188.186\n)\n\n => Array\n(\n => 117.230.35.5\n)\n\n => Array\n(\n => 111.92.74.239\n)\n\n => Array\n(\n => 104.245.144.236\n)\n\n => Array\n(\n => 39.50.22.100\n)\n\n => Array\n(\n => 47.31.190.23\n)\n\n => Array\n(\n => 157.44.73.187\n)\n\n => Array\n(\n => 117.230.8.91\n)\n\n => Array\n(\n => 157.32.18.2\n)\n\n => Array\n(\n => 111.119.187.43\n)\n\n => Array\n(\n => 203.101.185.246\n)\n\n => Array\n(\n => 5.62.34.22\n)\n\n => Array\n(\n => 122.8.143.76\n)\n\n => Array\n(\n => 115.186.2.187\n)\n\n => Array\n(\n => 202.142.110.89\n)\n\n => Array\n(\n => 157.50.61.254\n)\n\n => Array\n(\n => 223.182.211.185\n)\n\n => Array\n(\n => 103.85.125.210\n)\n\n => Array\n(\n => 103.217.133.147\n)\n\n => Array\n(\n => 103.60.196.217\n)\n\n => Array\n(\n => 157.44.238.6\n)\n\n => Array\n(\n => 117.196.225.68\n)\n\n => Array\n(\n => 104.254.92.52\n)\n\n => Array\n(\n => 39.42.46.72\n)\n\n => Array\n(\n => 221.132.119.36\n)\n\n => Array\n(\n => 111.92.77.47\n)\n\n => Array\n(\n => 223.225.19.152\n)\n\n => Array\n(\n => 159.89.121.217\n)\n\n => Array\n(\n => 39.53.221.205\n)\n\n => Array\n(\n => 193.34.217.28\n)\n\n => Array\n(\n => 139.167.206.36\n)\n\n => Array\n(\n => 96.40.10.7\n)\n\n => Array\n(\n => 124.29.198.123\n)\n\n => Array\n(\n => 117.196.226.1\n)\n\n => Array\n(\n => 106.200.85.135\n)\n\n => Array\n(\n => 106.223.180.28\n)\n\n => Array\n(\n => 103.49.232.110\n)\n\n => Array\n(\n => 139.167.208.50\n)\n\n => Array\n(\n => 139.167.201.102\n)\n\n => Array\n(\n => 14.244.224.237\n)\n\n => Array\n(\n => 103.140.31.187\n)\n\n => Array\n(\n => 49.36.134.136\n)\n\n => Array\n(\n => 160.16.61.75\n)\n\n => Array\n(\n => 103.18.22.228\n)\n\n => Array\n(\n => 47.9.74.121\n)\n\n => Array\n(\n => 47.30.216.159\n)\n\n => Array\n(\n => 117.248.150.78\n)\n\n => Array\n(\n => 5.62.34.17\n)\n\n => Array\n(\n => 139.167.247.181\n)\n\n => Array\n(\n => 193.176.84.29\n)\n\n => Array\n(\n => 103.195.201.121\n)\n\n => Array\n(\n => 89.187.175.115\n)\n\n => Array\n(\n => 137.97.81.251\n)\n\n => Array\n(\n => 157.51.147.62\n)\n\n => Array\n(\n => 103.104.192.42\n)\n\n => Array\n(\n => 14.171.235.26\n)\n\n => Array\n(\n => 178.62.89.121\n)\n\n => Array\n(\n => 119.155.4.164\n)\n\n => Array\n(\n => 43.250.241.89\n)\n\n => Array\n(\n => 103.31.100.80\n)\n\n => Array\n(\n => 119.155.7.44\n)\n\n => Array\n(\n => 106.200.73.114\n)\n\n => Array\n(\n => 77.111.246.18\n)\n\n => Array\n(\n => 157.39.99.247\n)\n\n => Array\n(\n => 103.77.42.132\n)\n\n => Array\n(\n => 74.115.214.133\n)\n\n => Array\n(\n => 117.230.49.224\n)\n\n => Array\n(\n => 39.50.108.238\n)\n\n => Array\n(\n => 47.30.221.45\n)\n\n => Array\n(\n => 95.133.164.235\n)\n\n => Array\n(\n => 212.103.48.141\n)\n\n => Array\n(\n => 104.194.218.147\n)\n\n => Array\n(\n => 106.200.88.241\n)\n\n => Array\n(\n => 182.189.212.211\n)\n\n => Array\n(\n => 39.50.142.129\n)\n\n => Array\n(\n => 77.234.43.133\n)\n\n => Array\n(\n => 49.15.192.58\n)\n\n => Array\n(\n => 119.153.37.55\n)\n\n => Array\n(\n => 27.56.156.128\n)\n\n => Array\n(\n => 168.211.4.33\n)\n\n => Array\n(\n => 203.81.236.239\n)\n\n => Array\n(\n => 157.51.149.61\n)\n\n => Array\n(\n => 117.230.45.255\n)\n\n => Array\n(\n => 39.42.106.169\n)\n\n => Array\n(\n => 27.71.89.76\n)\n\n => Array\n(\n => 123.27.109.167\n)\n\n => Array\n(\n => 106.202.21.91\n)\n\n => Array\n(\n => 103.85.125.206\n)\n\n => Array\n(\n => 122.173.250.229\n)\n\n => Array\n(\n => 106.210.102.77\n)\n\n => Array\n(\n => 134.209.47.156\n)\n\n => Array\n(\n => 45.127.232.12\n)\n\n => Array\n(\n => 45.134.224.11\n)\n\n => Array\n(\n => 27.71.89.122\n)\n\n => Array\n(\n => 157.38.105.117\n)\n\n => Array\n(\n => 191.96.73.215\n)\n\n => Array\n(\n => 171.241.92.31\n)\n\n => Array\n(\n => 49.149.104.235\n)\n\n => Array\n(\n => 104.229.247.252\n)\n\n => Array\n(\n => 111.92.78.42\n)\n\n => Array\n(\n => 47.31.88.183\n)\n\n => Array\n(\n => 171.61.203.234\n)\n\n => Array\n(\n => 183.83.226.192\n)\n\n => Array\n(\n => 119.157.107.45\n)\n\n => Array\n(\n => 91.202.163.205\n)\n\n => Array\n(\n => 157.43.62.108\n)\n\n => Array\n(\n => 182.68.248.92\n)\n\n => Array\n(\n => 157.32.251.234\n)\n\n => Array\n(\n => 110.225.196.188\n)\n\n => Array\n(\n => 27.71.89.98\n)\n\n => Array\n(\n => 175.176.87.3\n)\n\n => Array\n(\n => 103.55.90.208\n)\n\n => Array\n(\n => 47.31.41.163\n)\n\n => Array\n(\n => 223.182.195.5\n)\n\n => Array\n(\n => 122.52.101.166\n)\n\n => Array\n(\n => 103.207.82.154\n)\n\n => Array\n(\n => 171.224.178.84\n)\n\n => Array\n(\n => 110.225.235.187\n)\n\n => Array\n(\n => 119.160.97.248\n)\n\n => Array\n(\n => 116.90.101.121\n)\n\n => Array\n(\n => 182.255.48.154\n)\n\n => Array\n(\n => 180.149.221.140\n)\n\n => Array\n(\n => 194.44.79.13\n)\n\n => Array\n(\n => 47.247.18.3\n)\n\n => Array\n(\n => 27.56.242.95\n)\n\n => Array\n(\n => 41.60.236.83\n)\n\n => Array\n(\n => 122.164.162.7\n)\n\n => Array\n(\n => 71.136.154.5\n)\n\n => Array\n(\n => 132.154.119.122\n)\n\n => Array\n(\n => 110.225.80.135\n)\n\n => Array\n(\n => 84.17.61.143\n)\n\n => Array\n(\n => 119.160.102.244\n)\n\n => Array\n(\n => 47.31.27.44\n)\n\n => Array\n(\n => 27.71.89.160\n)\n\n => Array\n(\n => 107.175.38.101\n)\n\n => Array\n(\n => 195.211.150.152\n)\n\n => Array\n(\n => 157.35.250.255\n)\n\n => Array\n(\n => 111.119.187.53\n)\n\n => Array\n(\n => 119.152.97.213\n)\n\n => Array\n(\n => 180.92.143.145\n)\n\n => Array\n(\n => 72.255.61.46\n)\n\n => Array\n(\n => 47.8.183.6\n)\n\n => Array\n(\n => 92.38.148.53\n)\n\n => Array\n(\n => 122.173.194.72\n)\n\n => Array\n(\n => 183.83.226.97\n)\n\n => Array\n(\n => 122.173.73.231\n)\n\n => Array\n(\n => 119.160.101.101\n)\n\n => Array\n(\n => 93.177.75.174\n)\n\n => Array\n(\n => 115.97.196.70\n)\n\n => Array\n(\n => 111.119.187.35\n)\n\n => Array\n(\n => 103.226.226.154\n)\n\n => Array\n(\n => 103.244.172.73\n)\n\n => Array\n(\n => 119.155.61.222\n)\n\n => Array\n(\n => 157.37.184.92\n)\n\n => Array\n(\n => 119.160.103.204\n)\n\n => Array\n(\n => 175.176.87.21\n)\n\n => Array\n(\n => 185.51.228.246\n)\n\n => Array\n(\n => 103.250.164.255\n)\n\n => Array\n(\n => 122.181.194.16\n)\n\n => Array\n(\n => 157.37.230.232\n)\n\n => Array\n(\n => 103.105.236.6\n)\n\n => Array\n(\n => 111.88.128.174\n)\n\n => Array\n(\n => 37.111.139.82\n)\n\n => Array\n(\n => 39.34.133.52\n)\n\n => Array\n(\n => 113.177.79.80\n)\n\n => Array\n(\n => 180.183.71.184\n)\n\n => Array\n(\n => 116.72.218.255\n)\n\n => Array\n(\n => 119.160.117.26\n)\n\n => Array\n(\n => 158.222.0.252\n)\n\n => Array\n(\n => 23.227.142.146\n)\n\n => Array\n(\n => 122.162.152.152\n)\n\n => Array\n(\n => 103.255.149.106\n)\n\n => Array\n(\n => 104.236.53.155\n)\n\n => Array\n(\n => 119.160.119.155\n)\n\n => Array\n(\n => 175.107.214.244\n)\n\n => Array\n(\n => 102.7.116.7\n)\n\n => Array\n(\n => 111.88.91.132\n)\n\n => Array\n(\n => 119.157.248.108\n)\n\n => Array\n(\n => 222.252.36.107\n)\n\n => Array\n(\n => 157.46.209.227\n)\n\n => Array\n(\n => 39.40.54.1\n)\n\n => Array\n(\n => 223.225.19.254\n)\n\n => Array\n(\n => 154.72.150.8\n)\n\n => Array\n(\n => 107.181.177.130\n)\n\n => Array\n(\n => 101.50.75.31\n)\n\n => Array\n(\n => 84.17.58.69\n)\n\n => Array\n(\n => 178.62.5.157\n)\n\n => Array\n(\n => 112.206.175.147\n)\n\n => Array\n(\n => 137.97.113.137\n)\n\n => Array\n(\n => 103.53.44.154\n)\n\n => Array\n(\n => 180.92.143.129\n)\n\n => Array\n(\n => 14.231.223.7\n)\n\n => Array\n(\n => 167.88.63.201\n)\n\n => Array\n(\n => 103.140.204.8\n)\n\n => Array\n(\n => 221.121.135.108\n)\n\n => Array\n(\n => 119.160.97.129\n)\n\n => Array\n(\n => 27.5.168.249\n)\n\n => Array\n(\n => 119.160.102.191\n)\n\n => Array\n(\n => 122.162.219.12\n)\n\n => Array\n(\n => 157.50.141.122\n)\n\n => Array\n(\n => 43.245.8.17\n)\n\n => Array\n(\n => 113.181.198.179\n)\n\n)\n```\n\\$20K Update : Banker Gurl's More Bang For Your Buck!\n << Back to all Blogs Login or Create your own free blog Layout: Blue and Brown (Default) Author's Creation\nHome > \\$20K Update", null, "", null, "", null, "# \\$20K Update\n\nSeptember 25th, 2017 at 02:29 pm\n\nI received an unexpected rebate on one of my medications. It was only \\$10.45, but it was nice to get something!\n\nCurrent Balance \\$18,300.00\n+\\$10.45 Rx Rebate\n+\\$44.00 FB Sales\n+\\$.55 Even Out Amount\nBalance \\$18,355.00\n\nLocal Savings \\$7040.00\nGC Savings \\$2525.00\nCapital 360 Savings (US) \\$3610.00\nCapital 360 Savings (Mom Loan Payback Account)\\$2415.00\nCapital 360 Savings (Rental Deposit Account #1)\\$1265.00\nCapital 360 Savings (Rental Deposit Account #2)\\$1500.00\nTotal Savings = \\$18,355.00\n\n52 Week Challenge\n2017 Total = \\$892.00(We have paid weeks 1-28 and 44-52)\n\n### 1 Responses to “\\$20K Update ”\n\n1. rob62521 Says:\n\nNice rebate. Even if it isn't huge, far better than paying that amount out.\n\n(Note: If you were logged in, we could automatically fill in these fields for you.)\n Name: * Email: Will not be published. Subscribe: Notify me of additional comments to this entry. URL: Verification: * Please spell out the number 4.  [ Why? ]\n\nvB Code: You can use these tags: [b] [i] [u] [url] [email]" ]

[ null, "https://www.savingadvice.com/blogs/images/search/top_left.php", null, "https://www.savingadvice.com/blogs/images/search/top_right.php", null, "https://www.savingadvice.com/blogs/images/search/bottom_left.php", null ]

[ "# GATE | GATE-CS-2002 | Question 40\n\nThe Newton-Raphson iteration Xn + 1 = (Xn/2) + 3/(2Xn) can be used to solve the equation\n(A) X2 = 3\n(B) X3 = 3\n(C) X2 = 2\n(D) X3 = 2\n\nExplanation: In Newton-Raphson’s method, We use the following formula to get the next value of f(x). f'(x) is derivative of f(x).", null, "Option (A)\n\n```X2 = 3\nf(x) = X2 - 3\n\nXn + 1 = Xn - (Xn2 - 3) / (2*Xn)\n= (Xn/2) + 3/(2xn) ```\nMy Personal Notes arrow_drop_up\nArticle Tags :\n\nBe the First to upvote.\n\nPlease write to us at [email protected] to report any issue with the above content." ]

[ null, "https://media.geeksforgeeks.org/wp-content/cdn-uploads/gq/2013/12/af2d6f780d8673d64e8cc328ae52631d.png", null ]

[ "Adjusting a traverse (also known as balancing a traverse) is used to distributed the closure error back into the angle and distance measurements.\n\nSumming the latitudes and departures for the raw field traverse:", null, "Equations D-5 and D-6", null, "Figure E-1\nLoop Traverse Misclosure", null, "Equations D-3 and D-4", null, "Figure E-2\n\nThe condition for an adjusted traverse is that the adjusted Lats and Deps sum to 0.00. As with other survey adjustments, the method used to balance a traverse should reflect the expected error behavior and be repeatable. Table E-1 lists primary adjustment methods with their respective advantages and disadvantages.\n\n Table E-1 Method Premise Advantage Disadvantage Ignore Don't adjust anything. Simple; repeatable Ignores error Arbitrary Place error in one or more measurements Simple Not repeatable; ignores error behavior Compass Rule Assumes angles and distances are measured with equal accuracy so error is applied to each. Simple; repeatable; compatible with contemporary measurement methods. Treats random errors systematically Transit Rule Assumes angles are measured more accurately than distances; distances receive greater adjustment. Simple; repeatable; compatible with older transit-tape surveys. Treats random errors systematically; not compatible with contemporary measurement methods. Crandall Method Quasi-statistical approach. Angles are held and errors are statistically distributed into the distances. Allows some random error modeling; repeatable. Models only distance errors, not angle errors. Least squares Full statistical approach. Allows full random error modeling; repeatable; can mix different accuracy and precision measurements; provides measurement uncertainties. Most complicated method\n\nThe Compass Rule works sufficiently well for simple surveying projects and is the one we will apply.\n\n2. Compass Rule\n\nThe Compass Rule (also known as the Bowditch Rule) applies a proportion of the closure error to each line.\n\nFor any line IJ, Figure E-3,", null, "Figure E-3 Adjusted Latitude and Departure", null, "Equations E-1 and E-2", null, "Equations E-3 and E-4\n\nThe Compass Rule distributes closure error based on the proportion of a line's length to the entire distance surveyed.\n\nRegardless of the adjustment method applied, changing a line's Lat and Dep will in turn change the length and direction of the line.", null, "Figure E-4 Adjusted Length and Direction\n\nThe adjusted length can be computed from the Pythagorean theorem:", null, "Equation E-5\n\nComputing direction is a two-step process: (1) Determine β, the angle from the meridian to the line (2) Convert β into a direction based on the line's quadrant.\n\nTo determine β:", null, "Equation E-6\n\nβ falls in the range of -90° to +90°.\n\nThe sign on β indicates the direction of turning from the meridian: (+) is clockwise, (-) is counter-clockwise. The combined signs on the adjusted Lat and Dep will identify the line's quadrant.\n\nFigure E-5 shows the quadrant and direction computation for the various mathematic combinations of the adjusted Lat and Dep:", null, "Figure E-5 Converting ß to a Direction\n\n4. Examples\n\nThese examples are a continuation of those from the Latitudes and Departures chapter.\n\na. Traverse with Bearings", null, "Figure E-6 Bearing Traverse Example\n\n Line Bearing Length (ft) Lat (ft) Dep (ft) AB S 68°05'35\"W 472.68 -176.357 -438.548 BC N 19°46'00\"W 216.13 +203.395 -73.093 CD N 45°55'20\"E 276.52 +192.357 +198.651 DA S 54°59'15\"E 382.24 -219.312 +313.065 sums: 1347.57 +0.083 +0.075 Distance Lat err too far N Dep err too far E\n\n(1) Adjust the Lats and Deps\n\nSetup Equations E-1 and E-2:", null, "Now solve Equations E-3 and E-4 for each line:\n\nLine AB", null, "Line BC", null, "Line CD", null, "Line DA", null, "Check the closure condition\n\n Adjusted Line Lat (ft) Dep (ft) AB -176.386 -438.574 BC +203.382 -73.105 CD +192.340 +198.635 DE -219.336 +313.044 sums: 0.000 0.000 check check\n\nA common mistake is to forget to negate Lat err and Dep err in the correction equations. If that happens, the closure condition will be twice what it originally was as the corrections were applied in the wrong direction.\n\n(2) Compute adjusted lengths and directions\n\nUse Equations E-5 and E-6 along with Figure E-5 to compute the new length and direction for each line.\n\nLine AB\n\nAdj Lat = -176.386 <- South\nAdj Dep = -438.574 <- West", null, "", null, "Because it's the SW quadrant, Brng =S 68°05'27.4\" W.\n\nLine BC\n\nAdj Lat = +203.382 <- North\nAdj Dep = -73.105 <- West", null, "", null, "Because it's the NW quadrant, Brng = N 19°46'14.9\" W\n\nLine CD\n\nAdj Lat = +192.340 <- North\nAdj Dep = +198.635 <- East", null, "", null, "Because it's the NE quadrant, Brng = N 45°55'20.7\" E\n\nLine DA\n\nAdj Lat = -219.336 <- South\nAdj Dep = +313.044 <- East", null, "", null, "Because it's the SE quadrant, Brng = S 54°58'58.0\" E\n\n Adjusted Adjusted Line Lat (ft) Dep (ft) Length Bearing AB -176.386 -438.574 472.715 S 68°05'27.4\" W BC +203.382 -73.105 216.122 N 19°46'14.9\" W CD +192.340 +198.635 276.479 N 45°55'20.7\" E DE -219.336 +313.044 382.237 S 54°58'58.0\" E\n\nb. Traverse with Azimuths", null, "Figure E-7 Azimuth Traverse Example\n\n Line Azimuth Length (ft) Lat (ft) Dep (ft) ST 309°05'38\" 347.00 +218.816 -269.311 TU 258°34'22\" 364.55 -72.226 -357.324 UV 128°04'44\" 472.74 -291.560 +372.123 VS 60°21'26\" 292.94 +144.885 +254.602 sums: 1477.23 -0.085 +0.090 Distance Lat err too far S Dep err too far E\n\n(1) Adjust the Lats and Deps\n\nSetup Equations E-1 and E-2:", null, "Solve Equations E-3 and E-4 for each line:\n\nLine ST", null, "Line TU", null, "Line UV", null, "Line VS", null, "Check the closure condition\n\n Adjusted Line Lat (ft) Dep (ft) ST +218.836 -269.332 TU -72.205 -357.346 UV -291.533 +372.094 VS +144.902 +254.584 sums: 0.000 0.000 check check\n\n(2) Compute adjusted lengths and directions\n\nUse Equations E-5 and E-6 along with Figure E-5 to compute the new length and direction for each line.\n\nLine ST\n\nAdj Lat = +218.836 <- North\nAdj Dep = -269.332 <- West", null, "", null, "Because it's in the NW quadrant: Az = 360°00'00\"+(-50°54'20.4\") =309°05'39.6\"\n\nLine TU\n\nAdj Lat = -72.205 <- South\nAdj Dep = -357.346 <- West", null, "", null, "Because it's in the SW quadrant: Az = 180°00'00\"+(78°34'36.0\") = 258°34'36.0\"\n\nLine UV\n\nAdj Lat = -291.533 <- South\nAdj Dep = +372.094 <- East", null, "", null, "Because it's in the SE quadrant: Az = 180°00'00\"+(-51°55'17.6\") = 128°04'42.4\"\n\nLine VS\n\nAdj Lat = +144.902 <- North\nAdj Dep = +254.584 <- East", null, "", null, "Because it's in the NE quadrant: Az = 60°21'09.7\"\n\n Adjusted Adjusted Line Lat (ft) Dep (ft) Length (ft) Azimuth ST +218.836 -269.332 347.029 309°05'39.6\" TU -72.205 -357.346 364.568 258°34'36.0\" UV -291.533 +372.094 472.700 128°04'42.4\" VS +144.902 +254.584 292.933 60°21'09.7\"\n\nc. Crossing Loop Traverse\n\nAs long as a traverse closes back on its beginning point, it can be adjusted the same as any other loop traverse.", null, "Figure E-8 Crossing Loop Traverse Example\n\n Line Azimuth Length (ft) Lat (ft) Dep (ft) EF 133°02'45\" 455.30 -310.780 +332.737 FG 24°33'35\" 228.35 +207.691 +94.912 GH 241°05'15\" 422.78 -204.403 -370.084 HE 349°25'20\" 312.85 +307.534 -57.430 sums: 1419.28 +0.042 +0.135 Dist Lat err too far N Dep err too far E\n\n(1) Adjust and recompute each line.\n\nSetup Equations E-1 and E-2:", null, "Solve Equations E-3 and E-4 for each line:\n\nLine EF", null, "", null, "", null, "Because it's in the SE quadrant: Az = 180°00'00\"+(-46°56'57.1\") = 133°03'02.9\"\n\nLine FG", null, "", null, "", null, "Because it's in the NE quadrant: Az = 24°33'19.7\"\n\nLine GH", null, "", null, "", null, "Because it's in the SW quadrant: Az = 180°00'00\"+(61°05'18.8\") =241°05'18.8\"\n\nLine HE", null, "", null, "", null, "Because it's in the NW quadrant: Az = 360°00'00\"+(-10°35'00.5\") = 349°24'59.5\"" ]

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[ "# VBA - Arrays\n\nWe know very well that a variable is a container to store a value. Sometimes, developers are in a position to hold more than one value in a single variable at a time. When a series of values are stored in a single variable, then it is known as an array variable.\n\n## Array Declaration\n\nArrays are declared the same way a variable has been declared except that the declaration of an array variable uses parenthesis. In the following example, the size of the array is mentioned in the brackets.\n\n```'Method 1 : Using Dim\nDim arr1()\t'Without Size\n\n'Method 2 : Mentioning the Size\nDim arr2(5) 'Declared with size of 5\n\n'Method 3 : using 'Array' Parameter\nDim arr3\narr3 = Array(\"apple\",\"Orange\",\"Grapes\")\n```\n• Although, the array size is indicated as 5, it can hold 6 values as array index starts from ZERO.\n\n• Array Index cannot be negative.\n\n• VBScript Arrays can store any type of variable in an array. Hence, an array can store an integer, string, or characters in a single array variable.\n\n## Assigning Values to an Array\n\nThe values are assigned to the array by specifying an array index value against each one of the values to be assigned. It can be a string.\n\n### Example\n\n```Private Sub Constant_demo_Click()\nDim arr(5)\narr(0) = \"1\" 'Number as String\narr(1) = \"VBScript\" 'String\narr(2) = 100 \t\t 'Number\narr(3) = 2.45 \t\t 'Decimal Number\narr(4) = #10/07/2013# 'Date\narr(5) = #12.45 PM# 'Time\n\nmsgbox(\"Value stored in Array index 0 : \" & arr(0))\nmsgbox(\"Value stored in Array index 1 : \" & arr(1))\nmsgbox(\"Value stored in Array index 2 : \" & arr(2))\nmsgbox(\"Value stored in Array index 3 : \" & arr(3))\nmsgbox(\"Value stored in Array index 4 : \" & arr(4))\nmsgbox(\"Value stored in Array index 5 : \" & arr(5))\nEnd Sub\n```\n\nWhen you execute the above function, it produces the following output.\n\n```Value stored in Array index 0 : 1\nValue stored in Array index 1 : VBScript\nValue stored in Array index 2 : 100\nValue stored in Array index 3 : 2.45\nValue stored in Array index 4 : 7/10/2013\nValue stored in Array index 5 : 12:45:00 PM\n```\n\n## Multi-Dimensional Arrays\n\nArrays are not just limited to a single dimension, however, they can have a maximum of 60 dimensions. Two-dimensional arrays are the most commonly used ones.\n\n### Example\n\nIn the following example, a multi-dimensional array is declared with 3 rows and 4 columns.\n\n```Private Sub Constant_demo_Click()\nDim arr(2,3) as Variant\t' Which has 3 rows and 4 columns\narr(0,0) = \"Apple\"\narr(0,1) = \"Orange\"\narr(0,2) = \"Grapes\"\narr(0,3) = \"pineapple\"\narr(1,0) = \"cucumber\"\narr(1,1) = \"beans\"\narr(1,2) = \"carrot\"\narr(1,3) = \"tomato\"\narr(2,0) = \"potato\"\narr(2,1) = \"sandwitch\"\narr(2,2) = \"coffee\"\narr(2,3) = \"nuts\"\n\nmsgbox(\"Value in Array index 0,1 : \" & arr(0,1))\nmsgbox(\"Value in Array index 2,2 : \" & arr(2,2))\nEnd Sub\n```\n\nWhen you execute the above function, it produces the following output.\n\n```Value stored in Array index : 0 , 1 : Orange\nValue stored in Array index : 2 , 2 : coffee\n```\n\n## ReDim Statement\n\nReDim statement is used to declare dynamic-array variables and allocate or reallocate storage space.\n\n### Syntax\n\n```ReDim [Preserve] varname(subscripts) [, varname(subscripts)]\n```\n\n## Parameter Description\n\n• Preserve − An optional parameter used to preserve the data in an existing array when you change the size of the last dimension.\n\n• Varname − A required parameter, which denotes the name of the variable, which should follow the standard variable naming conventions.\n\n• Subscripts − A required parameter, which indicates the size of the array.\n\n### Example\n\nIn the following example, an array has been redefined and then the values preserved when the existing size of the array is changed.\n\nNote − Upon resizing an array smaller than it was originally, the data in the eliminated elements will be lost.\n\n```Private Sub Constant_demo_Click()\nDim a() as variant\ni = 0\nredim a(5)\na(0) = \"XYZ\"\na(1) = 41.25\na(2) = 22\n\nREDIM PRESERVE a(7)\nFor i = 3 to 7\na(i) = i\nNext\n\n'to Fetch the output\nFor i = 0 to ubound(a)\nMsgbox a(i)\nNext\nEnd Sub\n```\n\nWhen you execute the above function, it produces the following output.\n\n```XYZ\n41.25\n22\n3\n4\n5\n6\n7\n```\n\n## Array Methods\n\nThere are various inbuilt functions within VBScript which help the developers to handle arrays effectively. All the methods that are used in conjunction with arrays are listed below. Please click on the method name to know about it in detail.\n\nSr.No. Function & Description\n1 LBound\n\nA Function, which returns an integer that corresponds to the smallest subscript of the given arrays.\n\n2 UBound\n\nA Function, which returns an integer that corresponds to the largest subscript of the given arrays.\n\n3 Split\n\nA Function, which returns an array that contains a specified number of values. Split based on a delimiter.\n\n4 Join\n\nA Function, which returns a string that contains a specified number of substrings in an array. This is an exact opposite function of Split Method.\n\n5 Filter\n\nA Function, which returns a zero based array that contains a subset of a string array based on a specific filter criteria.\n\n6 IsArray\n\nA Function, which returns a boolean value that indicates whether or not the input variable is an array.\n\n7 Erase\n\nA Function, which recovers the allocated memory for the array variables." ]

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[ "You are on page 1of 12\n\n# 2\n\nELECTROSTATICS\n\nCharge\nCharging a body implies transfer of charge (electrons) from one body to another. Positively charged\nbody means loss of electrons, i.e., deficiency of electrons. Negatively charged body means\nexcess of electrons.\nConservation of charge: In an isolated system, total charge (sum of positive and negative) remains\nconstant.\nQuantization of charge: Charge on any body always exists in integral multiples of a fundamental\nunit of electric charge. This unit is equal to the magnitude of charge on electron (i.e. = 1.6 x 10\n-\n19\ncoulomb). So Q = ne, where n is an integer and e is the charge of the electron.\n1. Repulsion is a sure test of electrification. A charged body may attract a neutral body or an\noppositely charged body but it always repels a similarly charged body.\n2. A body can be charged by means of (a) friction (b) conduction and (c) induction.\n3. In charging by induction, if q be the inducing charge, then charge induced on body having\ndielectric constant K is given by: |\n.\n|\n\n\\\n|\n= '\nK\nq q\n1\n1 . For a conductor induced charge is: q' = q (as K\n= for a conductor). Charging a body by means of induction is preferable since the same\ncharged body can be used to charge any number of bodies without loss of charge.\n4. A charge at rest produces only electric field around itself; a charge having un-accelerated\nmotion produces electric as well as magnetic field around itself while a charge having\naccelerated motion emits electromagnetic radiation also in addition to producing electric and\nmagnetic fields.\n5. Surface charge density: Surface charge density on a charged conductor o 1/r (r is radius of\ncurvature)\nCoulomb's law\nThe force of attraction or repulsion between two point charges is given by:\n(a) For air or vacuum\n2\n2 1\n0\nair\nr\nq q\n4\n1\nF\ntc\n=\n(b) For any other medium |\n.\n|\n\n\\\n|\nc tc\n= |\n.\n|\n\n\\\n|\ntc\n=\n2\n2 1\nr 0\n2\n2 1\nmedium\nr\nq q\n4\n1\nr\nq q\n4\n1\nF , So\nmedium\nair\nr\nF\nF\n= c where, cr = (c/c0) =\nrelative permittivity of the medium and it is also known as dielectric constant (K) of the\nmedium. It has got no units and no dimensions. The dielectric constant of a metal is infinity.\n2. If a charge Q is to be divided into two parts such that the force between them is to be maximum,\nthen each part is equal to Q/2.\n3. When two identical conductors having charges q1 and q2 are put in contact and then separated,\nthen each has a charge equal to (q1 + q2)/2. On the other hand, if charges are q1 and q2, then\neach has a charge (q1 q2)/2.\n\n3\n4. Two charges Q each are at a distance r from each other. If a third charge q equal to ( Q/4) is\nplaced at a distance (r/2) from each charge, then all the charges will be in equilibrium.\n5. The force of interaction (attraction or repulsion) between two charges increases with decrease\nof dielectric constant and is maximum when the medium between them is air.\nElectric field and electric field intensity:\nThe electric field intensity E at any point is equal in magnitude to the force experienced per unit\n(test) charge placed at that point and is directed along the direction of the force experienced,\ni.e.,\n0\nq\nF\nE=\n1. Electric field intensity E is a vector quantity. Electric field intensity due to a positive charge is\nalways away from the charge and that due to a negative charge is always towards the charge.\n2. Unit of E is newton/coulomb or volt/metre and its dimensional formula is [MLT\n-3\nA\n-1\n].\n3. Electric field due to a point charge q at a distance r from it is: r\nr\nq\n4\n1\nE\n2\n0\ntc\n=\n4. Intensity of the electric field inside a charged spherical (hollow or solid) conductor is zero, since\ncharge resides on the outer surface of the conductor.\n5. The intensity of the electric field outside the charged spherical (hollow or solid) conductor is:\n\n2\n0\nr\nQ\n4\n1\nE\ntc\n= where r > R (R = radius of spherical conductor)\n6. Electric field due to a non-conducting charged sphere:\n(a) ,\nr\nQ\n4\n1\nE\n2\n0\nout\ntc\n= (b)\n2\n0\nsurface\nR\nQ\n4\n1\nE\ntc\n= , (c)\n3\n0\nin\nR\nQr\nK 4\n1\nE\nc t\n=\n7. Intensity of the electric field at some point (distant x from centre) on the axis of uniformly\ncharged ring (Radius R) is given by:\n2 / 3 2 2\n0\n) R x (\nqx\n4\n1\nE\n+ tc\n= i.e.E x, for a small distance, but it is\nzero at the centre of ring.\n8. Electric field due to a plane sheet (infinite dimensions) of charge having surface charge\ndensity o is: E = (o/2c0). For a positively charged sheet, the field is directed towards outward\nnormal and does not depend on the distance of point from the sheet.\n9. For a charged metal plate, Einside = 0 and Eout = (o/c0).\n10. Intensity of the electric field in between the plates of a charged parallel plate capacitor is E =\n(o/c0).\n11. Electric field at a distance r from a line charge (density ) is: |\n.\n|\n\n\\\n|\n=\ntc\n\n=\nr\n2\nK\nr 2\nE\n0\n. If the line charge\nis a cylinder of radius R,\n(a) then the electric field outside: E = /2tc0r\n(b) electric field on the surface; E = /2tc0R\n(c) and electric field inside at a distance r from the axis:\n2\n0\nR 2\nr\nE\ntc\n\n=\n\n4\nElectric lines of force\nThe imaginary line or a curve, tangent to which at any point gives the direction of field at that point,\nis known as electric line of force.\n1. Electric field lines emerge from a positive charge and terminate into a negative charge.\n2. Electric lines of force never cross each other.\n3. Electric field lines do not exist inside a conductor.\n4. Electric field lines are always perpendicular to charged conducting surface.\n5. Electric field lines never form closed loops.\n6. Electric field lines are always perpendicular to equipotential surface.\n7. In a uniform electric field, the electric lines of force are equidistant, parallel straight lines.\n8. If the lines of force are closer, the intensity of electric field is more and if the lines of force are\nfar apart, the intensity of electric field is less.\n9. If a metallic solid sphere is placed in a uniform electric field, then the lines of force are normal\nto the surface at every point but they cannot pass through the conductor.\nElectric dipole\nTwo equal and opposite charges separated by a small distance constitute an electric dipole.\nElectric dipole moment p is a vector quantity whose magnitude is equal to the product of\nmagnitude of one charge and the distance between the two charges. It is directed from\nnegative charge to positive charge.\nElectric field at some axial point: at a distance r from midpoint of dipole of length 2a, the\nmagnitude of field is:\n2 2 2\n0\n] a r [\npr 2\n4\n1\nE\n\ntc\n= The direction of E is along axis and parallel to p . For\nshort dipole r >> 2a |\n.\n|\n\n\\\n|\ntc\n=\n3\n0\nr\np 2\n4\n1\nE\nElectric field at some equatorial point: at a distance r from midpoint of dipole of length 2a, the\nmagnitude of electric field is:\n2 / 3 2 2\n0\n] a r [\np\n4\n1\nE\n+ tc\n= The direction of E is opposite to that of p .\nFor short dipole r >> 2a |\n.\n|\n\n\\\n|\ntc\n=\n3\n0\nr\np\n4\n1\nE\n1. If u be the angle between the direction of uniform electric field E and the axis of dipole, then the\ntorque acting on the dipole is: t = pE sin u or E p = t and potential energy U = - pE cos u\n2. The work done in deflecting the dipole through an angle u1 to u2 is given by: W = pE ( cos u1 -\ncos u2)\nElectric flux\nElectric flux is defined as dS E d = o or\n\n## = o dS E and for a closed surface,\n\n= o dS E . In a\nuniform electric field E , flux through a plane surface of area S is: u = = o cos ES S E . The flux is\npositive, if field lines leave the area, negative if field lines enter the area.\n\n5\nGauss law: The total or net outward electric flux through a closed surface is equal to the total\ncharge enclosed by the surface divided by c0, i.e.,.\n\n## = o dS E = (q/c0). Even if total net flux\n\nthrough a closed surface is zero, the electric field E at the Gaussian surface may be nonzero.\n(e.g. A conductor placed in uniform electric field)\nForce per unit area on a charged conductor (or electric pressure):\nIf o is the surface charge density, then electric pressure dF/dA = (o\n2\n/2c0). The force is always\ndirected outward as (o)\n2\nis positive, i.e., whether body is charged positively or negatively,\nthis force will always try to expand the charged body. A soap bubble or rubber balloon\nexpands on giving a positive or negative charge to it.\n\nElectric potential\nThe absolute electric potential at any point in the electric field is defined as the work done per unit\npositive charge required to move the test charge from infinity to that point. Potential\ndifference between two points A and B in an electric field is defined as the work required to\nmove a unit positive charge from the point A to the point B against the intensity of the electric\nfield.\ni.e.\n0\nq\nW\nV V\nAB\nA B\n= (where q0 is the test charge)\n(a) If WAB is +ve, VB > VA (b) if WAB is ve, VB < VA and (c) if WAB is zero, VB = VA.\n1. The electric potential at a distance r from a point charge q is given by:\nr\nq\n4\n1\nV\n0\ntc\n=\n2. The electric potential at a point due to a group of point charges q1, q2,.., qn, which are at\ndistances r1, r2, ...,rn from the point, is given by;\nn 3 2 1\nV . .......... V V V V + + + =\n3. The potential at any point due to a dipole is given by:\n2\n0\nr\ncos p\n4\n1\nV\nu\ntc\n= (where p = 2lq = electric\ndipole moment)\n4. The potential on the axial line of the dipole: |\n.\n|\n\n\\\n|\ntc\n=\n2\n0\nr\np\n4\n1\nV and the potential on the equator of\nthe dipole is zero\n5. Potential due to a charged hollow spherical conductor: Electric potential is constant from\nthe centre to the surface and is equal to |\n.\n|\n\n\\\n|\ntc\n=\nR\nq\n4\n1\nV\n0\n. Outside the conductor, the potential is\ninversely proportion to the distance from the centre.\n6. If the charge distribution is continuous rather than being a collection of point charges, then\n\ntc\n= =\nr\ndq\n4\n1\ndV V\n0\n\nRelation between electric field intensity and electric potential difference\n\n6\nIf A and B are two points in an electric field separated by an infinitesimal distance dx so that the\npotential difference between them is dV and the electric field intensity is E, then E = (dV/dx).\nThe negative of the potential gradient is known as intensity of the electric field. The negative\nsign shows that E points in the direction of decreasing V.\nElectric potential energy\n1. The electric potential energy of a system of fixed point charges is equal to the work done by an\nexternal agent to assemble the system, bringing each charge in from an infinite distance. The\npotential energy may be positive or negative depending on whether the work is done against\nthe electric force or by the electric force during transport respectively.\n2. A charged particle placed in an electric field has potential energy because of its interaction with\nthe electric field.\n3. If two charges q1 and q2 are separated by a distance r, then the potential energy of system\nr\nq q\n4\n1\n2 1\n0\ntc\n=\n4. If three charges q1, q2 and q3 are arranged at the three vertices of an equilateral triangle, then\nthe PE of the system is given by:\n\n+ +\ntc\n=\nr\nq q\nr\nq q\nr\nq q\n4\n1\nU\n3 1 3 2 2 1\n0\n\nEquipotential surface\nThe locus of all points which are at the same potential is known as equipotential surface.\n1. No work is required to move a charge from one point to another on equipotential surface.\n2. Near an isolated point charge the equipotential surface is a sphere.\n3. The work required to move a unit positive charge around a charge Q along a circle of radius r is\nequal to zero.\n4. The electric lines of force are always normal to the equipotential surface, since E should not\nhave a component along the equipotential surface.\n5. The surface of a charged conductor is always an equipotential surface whatever may be its\nshape.\n\nMotion of charge particle in an electric field\nWhen a positive charge is placed in an electric field, it experiences a force which drives it from\npoints of higher potential to points of lower potential. On the other hand, a negative charge\nexperiences a force driving it from lower to higher potential.\n1. The work done in moving a charge between two points in an electric field is independent of the\npath followed between these two points, since the electric field is a conservative field.\n2. If a charged particle is accelerated through a potential difference of V volts, then the kinetic\nenergy acquired by the particle KE = eV.\n3. The ratio of the velocities acquired by two charged particles accelerated from rest through the\nsame potential difference are in the ratio:\n|\n|\n.\n|\n\n\\\n|\n=\n2 2\n1 1\n2\n1\nm / q\nm / q\nv\nv\n\nTrajectory of a charged particle in a uniform electric field\n\n7\n1. A charged particle thrown horizontally into a uniform electric field\nacting vertically downwards is similar to a body thrown\nhorizontally from the top of a tower. Therefore, it follows a\nparabolic path.\n2. If t is the time taken by the charged particle to cross the length l of\nthe electric field with constant velocity vx then t = (l/vx) and\nvelocity gained by the charged particle along y-direction at the point of exit is:\nx\ny\nv\nl\nm\neE\nat v = =\n3. Deviation suffered by the charged particle along y-axis, as it just comes out of the electric field,\nis given by:\n2\n2\n2\n2\n1\n2\n1\nx\nv\nl\nm\neE\nat y = =\nElectric capacitance\n1. The capacitance of a conductor is defined as the ratio of the charge given to the conductor to the\npotential raised due to it, i.e., C = Q/V (Q is measured in coulomb and V in volt). The SI unit of\ncapacitance is farad.\n2. A capacitor or condenser consists of two conductors separated by an insulator or dielectric. The\nconductors carry equal and opposite charges Q. It is an electrical device that stores electric\ncharge.\nElectric potential energy stored in a charged capacitor\n1. The capacitor stores charge as well as electric energy, which is given by:\n2\n2\n2\n1\n2\n1\n2\nCV QV\nC\nQ\nU = = =\n2. The energy is stored in the electric field between the plates of a charged capacitor.\n3. The energy stored per unit volume in the electric field between the plates is called energy\ndensity (u). It is given by:\n0\n2\n2\n0\n2 2\n1\nc\no\n= c = E u where o represents surface charge density on the\nplates of capacitor.\n4. Force of attraction between oppositely charged plates of capacitor is\nA\nQ\nF\n0\n2\n2c\n= or A E F\n2\n0\n2\n1\nc =\nCapacitance formulae for different types of capacitors\n1. The electrical capacitance of a conducting sphere is 4tc0R, where R is the radius of conductor.\n2. Capacity of a parallel plate condenser with air in between the plates is given by: ,\n0\n0\nd\nA\nC\nc\n= where\nA is the effective area of the plates, d is the distance between the plates and c0 is the permittivity\nof free space.\n3. Capacitance of a parallel plate capacitor with a dielectric slab of dielectric constant K,\ncompletely filled between the plates of the capacitor, is given by:\nd\nA\nd\nA K\nC\nr\nmed\nc c\n=\nc\n=\n0 0\n.\n\n4. If a dielectric slab of thickness t and dielectric constant K is introduced between the plates, then\nthe capacity of the condenser is given by:\n|\n.\n|\n\n\\\n|\n\nc\n= '\nK\nt d\nA\nC\ndielectric\n1\n1\n0\n. If the dielectric slab is replaced\n\n8\nby a metallic slab of same thickness, then\nt d\nA\nC\nmetal\n\nc\n= '\n0\n(because K is infinite for metal). It is\ninteresting to note here that C' metal > C' dielectric\n5. If a number of slabs of thickness t1, t2,..., tn and dielectric constants K1, K2,.., Kn are completely\nfill the space between the plates, then the capacity is given by:\n\n+ + + +\nc\n= '\nn\nn\nK\nt\nK\nt\nK\nt\nK\nt\nA\nC\n. .......... ..........\n3\n3\n2\n2\n1\n1\n0\n\n6. If the upper plate of the parallel plate condenser is connected to the lower plate, then its\nelectrical capacitance becomes infinity.\n7. For an air filled spherical capacitor:\n) (\n4\n0\n0\na b\nab\nC\n\ntc\n= .\n8. For an air filled cylindrical capacitor:\n) / ( log\n2\n0\n0\na b\nL\nC\ne\ntc\n=\nGrouping of capacitors\n1. In series grouping,\n3 2 1\n1 1 1 1\nC C C C\nS\n+ + = and i.e., Q1 = Q2 = Q3 . Here, effective capacitance CS, is\nalways less than the least of the individual capacitances.\n2. In parallel grouping, Cp = C1 + C2 + C3 and i.e., V1 = V2 = V3\nRegrouping of condensers\nTwo condensers of capacity C1 and C2 are charged separately to potentials V1 and V2. If the\ncondensers are now connected with plates of same polarity together, then\ncommon potential (V) =\n2 1\n2 2 1 1\nC C\nV C V C\n+\n+\nand final charges on two condensers is q1' = C1V, q2' =\nC2V.\nTwo condensers of capacities C1 and C2 are charged separately to potentials V1 and V2. If they are\nreconnected with plates of opposite polarity together, then the common potential is given by:\n2 1\n2 2 1 1\nC C\nV C V C\nV\n+\n\n=\nEffect of dielectric\n1. Effect of inserting dielectric slab on capacitance, potential difference, charge, intensity of the\nelectric field and energy stored with battery still in connection:\n(a) C = KC0 (b) V = V0 (c) Q = KQ0 (d) E = E0 (e) U = KU0\n2. Battery is disconnected and dielectric slab is inserted:\n(b) C = KC0 (b) Q = Q0 (c) V = V0/K (d) E = E0/ K (e) U = U0/K\n3. Capacitor is charged and then battery is disconnected: If the distance between the two plates is\nincreased by insulating handles, then\n(a) capacity decreases (b) potential difference increases (c) intensity of electric field remains same\n(d) charge on the plates remains same and (e) energy stored in the capacitor increases\n\n9\n\nASSIGNMENT\n\n1. When a solid body is negatively charged by friction, it\nmeans that the body has\n(a) acquired some of electrons\n(b) lost some protons\n(c) acquired some electrons and lost a lesser number\nof protons\n(d) lost some positive ions.\n2. Two positive charges of magnitudes 2 and 3 coulombs\nare placed 10 cm apart. The electric potential at a\ndistance of 10 cm from the middle point on the right\nbisector of the line joining the two charges is :\n(a) 5 x 10\n11\nV (b) 4 x 10\n9\nV\n(c) 4 x 10\n11\nV (d) 5 x 10\n9\nV\n3. A and B are two points in an electric field. If the work\ndone in carrying 4.0 coulomb of electric charge from\nA to B is 16.0 joule, the potential difference between\nA and B is\n(a) zero (b) 2.0 V (c) 4.0 V (d) 16.0 V\n4. The work done in moving a positive charge on an\nequipotential surface is\n(a) finite and positive (b) infinite\n(c) finite and negative (d) zero\n5. Two charges are placed at a distance apart. If a glass\nslab is placed between them, force between them will\n(a) remain the same (b) increase\n(c) decrease (d) be zero\n6. Electrons are caused to fall through a p.d. of 1500\nvolts. If they were initially at rest. their final speed is\n(a) 4.6 x 10\n7\nm/s (b) 2.3 x 10\n7\nm/s\n(c) 0.23 x 10\n7\nm/s (d) 5.1 x 10\n7\nm/s.\n7. The potential inside a hollow spherical conductor\n(a) is constant\n(b) varies directly as the distance from the centre\n(c) varies inversely as the distance from the centre\n(d) varies inversely as the square of the distance from\nthe centre.\n8. Two plates are 1 cm apart and the potential difference\nbetween them is 10 volt. The electric field between the\nplates is\n(a) 10 N/C (b) 250 N/C\n(c) 500 N/C (d) 1000 N/C\n9. A and B are two spherical conductors of the same\nextent and size. A is solid and B is hollow. Both are\ncharged to the same potential. If the charges on A and\nB are Q\nA\nand Q\nB\nrespectively, then\n(a) Q\nA\nis less than Q\nB\n\n(b) Q\nA\nis greater than Q\nB\nbut not double\n(c) Q\nA\n= Q\nB\n\n10\n(d) Q\nA\n= 2Q\nB\n\n10. Two capacitors of capacities C\n1\nand C\n2\nare\nconnected in parallel. If a charge Q is given to the\nassembly, it gets shared. The ratio of the charge on\ncapacitor C\n1\nto the charge on capacitor C\n2\nis given\nby\n(a) C\n1\n/C\n2\n(b) C\n2\n/C\n1\n(c) C\n1\n2\n/C\n2\n2\n(d) C\n2\n2\n/C\n1\n2\n\n11. A capacitor connected to a 10 V battery collects a\ncharge of 40 micro coulomb with air as dielectric\nand 100 micro coulomb with oil as dielectric. The\ndielectric constant of the oil is :\n(a) 4 (b) 2.5 (c) 0.4 (d) 1.0\n12. A given charge situated at a certain distance from a\nshort electric dipole in the end-on position\nexperiences a force F. If distance of the charge is\ndoubled, the force acting on the charge will be\n(a) 2F (b) F/2 (c) F/4 (d) F/8\n13. A charge Q is placed at each of the two opposite\ncorners of a square. A charge q is placed at each of\nthe other two corners. If the resultant force on Q is\nzero, then\n(a) Q = \\2q (b) Q = - \\2q\n(c) Q =2\\2q (d) Q = -2\\2q\n14. The electric field in a region of space is given by\nj\n\n2 i\n\n## 5 E + = N/C. The electric flux due to this field\n\nthrough an area 2m\n2\nlying in the YZ plane, in S.I.\nunits, is\n(a) 10 (b) 20 (c) 10\\2 (d) 2\\29\n15. An infinite number of charges, each equal to q\ncoulomb, are placed along the X axis at x (in\nmetres) = 1, 2, 4, 8, ... and so on. The potential and\nfield in SI units at x = 0 due to this set of charges are\nrespectively l/4tc\n0\ntimes\n(a) 2q, 4q (b) 2q/5, 4q\n(c) 2q/3, 4q/3 (d) 2q, 4q/3\n16. Three identical charges are placed at the corners of\nan equilateral triangle. If the force between any two\ncharges is F, then the net force on each will be\n(a) \\2F (b) 2F (c) \\3F (d) 3F\n17. An electric dipole placed in a uniform electric field\nwill have minimum potential energy when the dipole\nmoment is inclined to the field at an angle\n(a) t (b) t/2 (c) zero (d) 3t/2\n18. Two charged conducting spheres of radii R\n1\nand R\n2\n\nseparated by a large distance, are connected by a\nlong wire. The ratio of the charges on them is\n(a) R\n1\n/R\n2\n(b) R\n2\n/R\n1\n(c) R\n1\n2\n/R\n2\n2\n(d) R\n2\n2\n/R\n1\n2\n\n19. Two concentric metallic spherical shells are given\npositive charges. Then\n(a) the outer sphere is always at a higher potential\n(b) the inner sphere is always at a higher potential\n(c) both the spheres are at the same potential\n(d) no prediction can be made about their potentials\nunless the actual values of charges arid radii are\nknown.\n20. In the network shown below, all the capacitors are of\n1 microfarad. The\nequivalent capacitance\nbetween P and Q in\nmicrofarads is\n(a) 4 (b)\n(c) (d) 4/3\n21. A capacitor having a capacity of 2.0 microfarad is\ncharged up to 200 V and its plates are joined to a wire.\nThe heat produced in joule will be\n(a) 4 x 10\n4\n(b) 4 x 10\n-10\n\n(c) 4 x 10\n-2\n(d) 2 x 10\n-2\n\n22. In the circuit given,\nthe charge in micro-\ncoulomb, on the\ncapacitor having\ncapacity 5\nmicrofarad is\n(a) 4.5 (b) 9\n(c) 7 (d) 15\n23. Small drops of the same size are charged to V volt\neach, If n such drops coalesce to form a single large\ndrop, then its potential will be\n(a) nV (b) V/n (c) Vn\n1/3\n(d) Vn\n2/3\n\n24. Each of the four capacitors in the given circuit is 50\nuF. The charge on\neach capacitor is\n(a) 5 x 10\n3\nC\n(b) 5 x 10\n-3\nC\n(c) 2.5 x 10\n3\nC\n(d) 2.5 x 10\n-3\nC\n25. A capacitor of 20 uF, charged to 500 V, is connected\nin parallel with another capacitor of 10uF charged to\n200 V. The common potential difference across the\ncombination is\n(a) 300 V (b) 350 V (c) 400 V (d) 700 V\n26. Five identical capacitors connected in series have an\nequivalent capacitance of 4uF. If all of them are\nconnected in parallel across a 400 V source, the total\nenergy stored in them is\n(a) 2J (b) 4 J (c) 8 J (d) 16 J\n27. Four metallic plates, each having area A, are placed as\nshown. The distance between the consecutive plates is\nd. Alternate plates are connected to points A and B.\nThe equivalent capacitance\nof the system is\n(a) e\n0\nA/d (b) 2e\n0\nA/d\n(c) 3e\n0\nA/d (d) 4e\n0\nA/d\n28. A parallel plate capacitor with air between the plates\nis charged to a p.d. of 500 V and then insulated. A\nplastic plate is inserted between the plates filling the\nwhole gap. The p.d. between the plates now becomes\n75 V. The dielectric constant of plastic is\n(a) 10/3 (b) 5 (c) 20/3 (d) 10\n29. Two slabs of the same dimensions, having dielectric\nconstants K\n1\nand K\n2\n,\n\n11\ncompletely fill the space between the plates of a\nparallel plate capacitor as shown in the figure. If C is\nthe original capacitance of the capacitor, the new\ncapacitance is\n(a) C\n2\nK K\n2 1\n|\n.\n|\n\n\\\n| +\n(b) C\nK K\nK K 2\n2 1\n2 1\n|\n|\n.\n|\n\n\\\n|\n+\n\n(c) ( )C K K\n2 1\n+ (d) C\nK K\nK K\n2 1\n2 1\n|\n|\n.\n|\n\n\\\n|\n+\n\n30. The figure shows two identical parallel plate\ncapacitors connected to a battery with switch S\nclosed. The switch is now opened and the free space\nbetween the plates of the capacitors is filled with a\ndielectric of constant 3.\nThe ratio of the total\nenergy stored in both\ncapacitors before and after\nthe introduction of the\ndielectric is\n(a) 5/3 (b) 3/5 (c) 5/2 (d) 2/5\n31. ,A parallel plate capacitor is charged and the\ncharging battery is then disconnected. If the plates of\nthe capacitor are moved farther apart by means of\ninsulating handles\n(a) the charge on the capacitor increases\n(b) the voltage across the plates increases\n(c) the capacitance increases\n(d) None of these.\n32. An infinite number of capacitors, having capacitances\n1uF, 2uF, 4uF, 8uF,. are connected in series. The\nequivalent capacitance of the system is\n(a) infinite (b) 0.25 uF (c) 0.5 uF (d) 2 uF\n33. The radius of a hollow metallic sphere is R. If the\npotential difference between its surface and a point at\na distance of 3R from its centre is V, then the electric\nfield intensity at a distance of 3R from its centre is\n(a) V/2R (b) V/3R (c) V/4R (d) V/6R\n34. A point charge q is placed at the midpoint of a cube of\nside L. The electric flux emerging from the cube is\n(a) q/e\n0\n(b) q/6L\n2\ne\n0\n(c) 6qL\n2\n/e\n0\n\n(d) zero\n35. Electric charges q, q and - 2q are placed at the three\ncorners of an equilateral triangle of side l. The\nmagnitude of the electric dipole moment of the system\nis\n(a) ql (b) 2 ql (c) \\3ql (d) 4 ql\n36. A metallic sphere is placed in a uniform electric field.\nThe lines of force follow the path(s) shown in the\nfigure as\n(a) 1 (b) 2 (c) 3 (d) 4\n\nANSWERS\n1a ,2c ,3c ,4d ,5c ,6b ,7a ,8d ,9c ,10a ,11b ,12d\n13d ,14a ,15d ,16c ,17c ,18a ,19b ,20d ,21c ,22b\n23c ,24b ,25c ,26c ,27c ,28c ,29a ,30b ,31b ,32c\n33d ,34a ,35c ,36d" ]

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[ "# 2D Heat equation: inconsistent boundary and initial conditions\n\nI'm attempting to use NDSolve on a 2D boundary value problem with initial conditions. Upon running my code, I get the following message:\n\n\"NDSolve::ibcinc: Warning: Boundary and initial conditions are inconsistent.\"\n\nAfter much head-scratching, I can't seem to find my mistake. It seems to me that my initial condition is consistent with my boundary conditions:\n\nk = 1 / (5*(Pi^2));\nsoln = NDSolve[\n{\n(* PDE *)\nD[u[x, y, t], t] == k*(D[u[x, y, t], x, x] + D[u[x, y, t], y, y]),\n\n(* initial condition *)\nu[x, y, 0] == y + Cos[Pi*x] Sin[2*Pi*y],\n\n(* boundary conditions *)\nu[x, 0, t] == 0,\nu[x, 1, t] == 1,\n(D[u[x, y, t], x] /. x -> 0) == 0,\n(D[u[x, y, t], x] /. x -> 1) == 0\n},\nu,\n{x, 0, 1},\n{y, 0, 1},\n{t, 0, 1}\n]\n\n\nRick\n\n• Please see lots of related questions using this search. Mar 6, 2013 at 3:36\n• The solution does appear fine ... and afaict the boundary conditions are indeed consistent with the initial cond Mar 6, 2013 at 3:46\n• @ Szabolcs: thanks for having a look.\n– Rick\nMar 6, 2013 at 6:25\n• Somewhat irrelevant, have you tried separation of variables? I think you can express the solution using fourrier series ... Mar 6, 2013 at 18:55\n• @Spawn1701D: I'm writing my own implementation of FTCS to solve the problem. Separation of variables gives an exact answer, but I wanted to solve it first with Mathematica so that I can compare my output to the result as a cross-check.\n– Rick\nMar 7, 2013 at 5:04\n\nThe documentation has a full section dedicated to inconsistent boundary conditions in PDEs.\n\nQuoting it,\n\nOccasionally, NDSolve will issue the NDSolve::ibcinc message warning about inconsistent boundary conditions when they are actually consistent. This happens due to discretization error in approximating Neumann boundary conditions or any boundary condition that involves a spatial derivative. The reason this happens is that spatial error estimates (see \"Spatial Error Estimates\") used to determine how many points to discretize with are based on the PDE and the initial condition, but not the boundary conditions. The one-sided finite difference formulas that are used to approximate the boundary conditions also have larger error than a centered formula of the same order, leading to additional discretization error at the boundary. Typically this is not a problem, but it is possible to construct examples where it does occur.\n\nThen an example follows, and a possible solution using the Method option's \"TensorProductGrid\" suboption, which we can also apply to your problem.\n\nWhen the boundary conditions are consistent, a way to correct this error is to specify that NDSolve use a finer spatial discretization.\n\nk = 1/(5*(Pi^2));\nsoln = NDSolve[{\nD[u[x, y, t], t] == k*(D[u[x, y, t], x, x] + D[u[x, y, t], y, y]),\nu[x, y, 0] == y + Cos[Pi*x] Sin[2*Pi*y],\nu[x, 0, t] == 0,\nu[x, 1, t] == 1,\n(D[u[x, y, t], x] /. x -> 0) == 0,\n(D[u[x, y, t], x] /. x -> 1) == 0},\n\nu, {x, 0, 1}, {y, 0, 1}, {t, 0, 1},\n\nMethod -> {\"MethodOfLines\",\n\"SpatialDiscretization\" -> {\"TensorProductGrid\", \"MinPoints\" -> 20}}]\n\n\nIn this instance \"MinPoints\" -> 20 was sufficient to make the problem go away.\n\nThe same problem was discussed here. I vaguely remembered it, but it took me a while to find it again ...\n\n• Yep. This kind of quirks may cause madness Mar 6, 2013 at 20:28\n• Very nice find. I don't want to admit to how much time I spent last night looking for something in the Advanced Documentation.. regarding that issue. And I even knew, more or less, what to look for (error from discretization and numerical differencing). Oh well. Mar 6, 2013 at 20:58\n• @Nasser I don't suppose you have a pdf version of that note of yours available? Mar 6, 2013 at 20:59\n• @Daniel It would be really good to have a link to that advanced tutorial on the NDSolve::ibcinc doc page (what you get when you click >>). I sent a suggestion to support@wolfram half an hour ago, but I don't know if such user suggestions make a difference or not ... Mar 6, 2013 at 20:59\n• @Szabolcs I echoed your request to some people involved in NDSolve development. Of course that does not mean it will carry any more weight than your note to Support... Mar 6, 2013 at 21:03" ]

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[ "# [offtopic] iTunes Library Scripts\n\nSo I was trying to get my pc laptop setup for fudcon which meant a triple boot setup between windows, ubuntu and fedora. I had some troubles importing various important information about my music within iTunes.\n\nHere is a script to remove duplicate songs in an iTunes folder:\n\n```import hashlib\nimport os\nimport re\nimport sys\n\ndef hash(fnam):\nfobj = open(fnam, \"r\")\nhobj = hashlib.sha256()\n\nwhile (1):\n\nif (not data):\nbreak\n\nhobj.update(data)\n\nreturn hobj.hexdigest()\n\nfdic = {}\n\nwhile (1):\n\nif (not file):\nbreak\n\nfile = file.strip()\nuniq = hash(file)\n\nif (uniq in fdic.keys()):\nprint(\"removing:\",file)\n\nelse:\nprint(\"found:\",file)\nfdic[uniq] = file\n\n```\n\nThis script attempts to merge an old iTunes library xml file with a newly imported one. The result is a merged iTunes library xml file that can then be re-imported into iTunes thus restoring some various meta-data about your music. Note: Before you re-import the merged xml playlist file, make sure you delete the current one and disable the check-box stating to copy songs into the media folder.\n\n```import os\nimport re\nimport sys\n\nif (len(sys.argv) < 3):\nprint(\"Usage: %s <source> <merge>\" % (sys.argv))\nsys.exit(0)\n\ndef outp(ordr, dict, line):\nfor item in ordr:\nsys.stdout.write(dict[item])\n\nsys.stdout.write(line)\nsys.stdout.flush()\n\ndef xmlr(fnam, sdic={}):\nalen = len(sys.argv)\nslen = len(sdic.keys())\nxord = []; xdic = {}\nxobj = open(fnam, \"r\")\n\nxdat = {}\n\nwhile (1):\n\nif (not xlin):\nbreak\n\nslin = xlin.strip(\"\\0\\t\\r\\n \")\nxreg = re.match(\"^<key>([^<]+)</key>(<[^>]+>[^<]+<[^>]+>)\\$\", slin)\n\ntry:\nxkey = (xdic[\"Artist\"] + xdic[\"Album\"] + xdic[\"Name\"])\n\nexcept:\nxkey = \"\"\n\nif (slin == \"<dict>\"):\nif (slen > 0):\noutp(xord, xdic, xlin)\n\nxord = []; xdic = {}\n\nelif (xreg):\nif (xreg.group(1) not in xord):\nxord.append(xreg.group(1))\n\nxdic[xreg.group(1)] = [xreg.group(2), xlin]\n\nelif (slin == \"</dict>\"):\nif (slen < 1):\nxdat[xkey] = xdic\n\nelse:\nfor x in range(3, alen):\ntry:\nxdic[sys.argv[x]] = sdic[xkey][sys.argv[x]]\n\nif (sys.argv[x] not in xord):\nxord.append(sys.argv[x])\n\nexcept:\n#print(\"error:\",sys.exc_info())\npass\n\noutp(xord, xdic, xlin)\n\nxord = []; xdic = {}\n\nelse:\nif (slen > 0):\noutp(xord, xdic, xlin)\n\nxord = []; xdic = {}\n\nreturn xdat\n\nsdat = xmlr(sys.argv)\nddat = xmlr(sys.argv, sdic=sdat)\n\n```\n\nHere’s an example of how to use the command:\n\n```python.exe /cygdrive/c/Users/jon/Desktop/itml.py /cygdrive/g/tmp/itunes/iTunes\\ Music\\ Library0.xml /cygdrive/g/tmp/itunes/iTunes\\ Music\\ Library.xml 'Date Added' 'Play Count' | unix2dos.exe | tee /cygdrive/g/tmp/itunes/iTunes\\ Music\\ Library2.xml && cp /cygdrive/g/tmp/itunes/iTunes\\ Music\\ Library2.xml /cygdrive/g/tmp/itunes/iTunes\\ Music\\ Library3.xml\n```" ]

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[ "src/HOL/Tools/ATP/res_clasimpset.ML\n author haftmann Tue May 09 10:09:37 2006 +0200 (2006-05-09) changeset 19599 a5c7eb37d14f parent 19480 868cf5051ff5 child 19675 a4894fb2a5f2 permissions -rw-r--r--\n``` 1 (* ID: \\$Id\\$\n```\n``` 2 Author: Claire Quigley\n```\n``` 3 Copyright 2004 University of Cambridge\n```\n``` 4 *)\n```\n``` 5\n```\n``` 6 signature RES_CLASIMP =\n```\n``` 7 sig\n```\n``` 8 val blacklist : string list ref (*Theorems forbidden in the output*)\n```\n``` 9 val whitelist : thm list ref (*Theorems required in the output*)\n```\n``` 10 val use_simpset: bool ref\n```\n``` 11 val get_clasimp_atp_lemmas :\n```\n``` 12 Proof.context ->\n```\n``` 13 Term.term list ->\n```\n``` 14 (string * Thm.thm) list ->\n```\n``` 15 (bool * bool * bool) -> bool -> string Array.array * (thm * (string * int)) list\n```\n``` 16 end;\n```\n``` 17\n```\n``` 18 structure ResClasimp : RES_CLASIMP =\n```\n``` 19 struct\n```\n``` 20 val use_simpset = ref false; (*Performance is much better without simprules*)\n```\n``` 21\n```\n``` 22 (*The rule subsetI is frequently omitted by the relevance filter.*)\n```\n``` 23 val whitelist = ref [subsetI];\n```\n``` 24\n```\n``` 25 (*In general, these produce clauses that are prolific (match too many equality or\n```\n``` 26 membership literals) and relate to seldom-used facts. Some duplicate other rules.\n```\n``` 27 FIXME: this blacklist needs to be maintained using theory data and added to using\n```\n``` 28 an attribute.*)\n```\n``` 29 val blacklist = ref\n```\n``` 30 [\"Datatype.prod.size\",\n```\n``` 31 \"Divides.dvd_0_left_iff\",\n```\n``` 32 \"Finite_Set.card_0_eq\",\n```\n``` 33 \"Finite_Set.card_infinite\",\n```\n``` 34 \"Finite_Set.Max_ge\",\n```\n``` 35 \"Finite_Set.Max_in\",\n```\n``` 36 \"Finite_Set.Max_le_iff\",\n```\n``` 37 \"Finite_Set.Max_less_iff\",\n```\n``` 38 \"Finite_Set.max.f_below_strict_below.below_f_conv\", (*duplicates in Orderings.*)\n```\n``` 39 \"Finite_Set.max.f_below_strict_below.strict_below_f_conv\", (*duplicates in Orderings.*)\n```\n``` 40 \"Finite_Set.Min_ge_iff\",\n```\n``` 41 \"Finite_Set.Min_gr_iff\",\n```\n``` 42 \"Finite_Set.Min_in\",\n```\n``` 43 \"Finite_Set.Min_le\",\n```\n``` 44 \"Finite_Set.min_max.below_inf_sup_Inf_Sup.inf_Sup_absorb\",\n```\n``` 45 \"Finite_Set.min_max.below_inf_sup_Inf_Sup.sup_Inf_absorb\",\n```\n``` 46 \"Finite_Set.min.f_below_strict_below.below_f_conv\", (*duplicates in Orderings.*)\n```\n``` 47 \"Finite_Set.min.f_below_strict_below.strict_below_f_conv\", (*duplicates in Orderings.*)\n```\n``` 48 \"IntDef.Integ.Abs_Integ_inject\",\n```\n``` 49 \"IntDef.Integ.Abs_Integ_inverse\",\n```\n``` 50 \"IntDiv.zdvd_0_left\",\n```\n``` 51 \"List.append_eq_append_conv\",\n```\n``` 52 \"List.hd_Cons_tl\", (*Says everything is [] or Cons. Probably prolific.*)\n```\n``` 53 \"List.in_listsD\",\n```\n``` 54 \"List.in_listsI\",\n```\n``` 55 \"List.lists.Cons\",\n```\n``` 56 \"List.listsE\",\n```\n``` 57 \"Nat.less_one\", (*not directional? obscure*)\n```\n``` 58 \"Nat.not_gr0\",\n```\n``` 59 \"Nat.one_eq_mult_iff\", (*duplicate by symmetry*)\n```\n``` 60 \"NatArith.of_nat_0_eq_iff\",\n```\n``` 61 \"NatArith.of_nat_eq_0_iff\",\n```\n``` 62 \"NatArith.of_nat_le_0_iff\",\n```\n``` 63 \"NatSimprocs.divide_le_0_iff_number_of\", (*too many clauses*)\n```\n``` 64 \"NatSimprocs.divide_less_0_iff_number_of\",\n```\n``` 65 \"NatSimprocs.equation_minus_iff_1\", (*not directional*)\n```\n``` 66 \"NatSimprocs.equation_minus_iff_number_of\", (*not directional*)\n```\n``` 67 \"NatSimprocs.le_minus_iff_1\", (*not directional*)\n```\n``` 68 \"NatSimprocs.le_minus_iff_number_of\", (*not directional*)\n```\n``` 69 \"NatSimprocs.less_minus_iff_1\", (*not directional*)\n```\n``` 70 \"NatSimprocs.less_minus_iff_number_of\", (*not directional*)\n```\n``` 71 \"NatSimprocs.minus_equation_iff_number_of\", (*not directional*)\n```\n``` 72 \"NatSimprocs.minus_le_iff_1\", (*not directional*)\n```\n``` 73 \"NatSimprocs.minus_le_iff_number_of\", (*not directional*)\n```\n``` 74 \"NatSimprocs.minus_less_iff_1\", (*not directional*)\n```\n``` 75 \"NatSimprocs.mult_le_cancel_left_number_of\", (*excessive case analysis*)\n```\n``` 76 \"NatSimprocs.mult_le_cancel_right_number_of\", (*excessive case analysis*)\n```\n``` 77 \"NatSimprocs.mult_less_cancel_left_number_of\", (*excessive case analysis*)\n```\n``` 78 \"NatSimprocs.mult_less_cancel_right_number_of\", (*excessive case analysis*)\n```\n``` 79 \"NatSimprocs.zero_le_divide_iff_number_of\", (*excessive case analysis*)\n```\n``` 80 \"NatSimprocs.zero_less_divide_iff_number_of\",\n```\n``` 81 \"OrderedGroup.abs_0_eq\", (*duplicate by symmetry*)\n```\n``` 82 \"OrderedGroup.diff_eq_0_iff_eq\", (*prolific?*)\n```\n``` 83 \"OrderedGroup.join_0_eq_0\",\n```\n``` 84 \"OrderedGroup.meet_0_eq_0\",\n```\n``` 85 \"OrderedGroup.pprt_eq_0\", (*obscure*)\n```\n``` 86 \"OrderedGroup.pprt_eq_id\", (*obscure*)\n```\n``` 87 \"OrderedGroup.pprt_mono\", (*obscure*)\n```\n``` 88 \"Parity.even_nat_power\", (*obscure, somewhat prolilfic*)\n```\n``` 89 \"Parity.power_eq_0_iff_number_of\",\n```\n``` 90 \"Parity.power_le_zero_eq_number_of\", (*obscure and prolific*)\n```\n``` 91 \"Parity.power_less_zero_eq_number_of\",\n```\n``` 92 \"Parity.zero_le_power_eq_number_of\", (*obscure and prolific*)\n```\n``` 93 \"Parity.zero_less_power_eq_number_of\", (*obscure and prolific*)\n```\n``` 94 \"Power.zero_less_power_abs_iff\",\n```\n``` 95 \"Relation.diagI\",\n```\n``` 96 \"Relation.ImageI\",\n```\n``` 97 \"Ring_and_Field.divide_cancel_left\", (*fields are seldom used & often prolific*)\n```\n``` 98 \"Ring_and_Field.divide_cancel_right\",\n```\n``` 99 \"Ring_and_Field.divide_divide_eq_left\",\n```\n``` 100 \"Ring_and_Field.divide_divide_eq_right\",\n```\n``` 101 \"Ring_and_Field.divide_eq_0_iff\",\n```\n``` 102 \"Ring_and_Field.divide_eq_1_iff\",\n```\n``` 103 \"Ring_and_Field.divide_eq_eq_1\",\n```\n``` 104 \"Ring_and_Field.divide_le_0_1_iff\",\n```\n``` 105 \"Ring_and_Field.divide_le_eq_1_neg\", (*obscure and prolific*)\n```\n``` 106 \"Ring_and_Field.divide_le_eq_1_pos\", (*obscure and prolific*)\n```\n``` 107 \"Ring_and_Field.divide_less_0_1_iff\",\n```\n``` 108 \"Ring_and_Field.divide_less_eq_1_neg\", (*obscure and prolific*)\n```\n``` 109 \"Ring_and_Field.divide_less_eq_1_pos\", (*obscure and prolific*)\n```\n``` 110 \"Ring_and_Field.eq_divide_eq_1\", (*duplicate by symmetry*)\n```\n``` 111 \"Ring_and_Field.field_mult_cancel_left\",\n```\n``` 112 \"Ring_and_Field.field_mult_cancel_right\",\n```\n``` 113 \"Ring_and_Field.inverse_le_iff_le_neg\",\n```\n``` 114 \"Ring_and_Field.inverse_le_iff_le\",\n```\n``` 115 \"Ring_and_Field.inverse_less_iff_less_neg\",\n```\n``` 116 \"Ring_and_Field.inverse_less_iff_less\",\n```\n``` 117 \"Ring_and_Field.le_divide_eq_1_neg\", (*obscure and prolific*)\n```\n``` 118 \"Ring_and_Field.le_divide_eq_1_pos\", (*obscure and prolific*)\n```\n``` 119 \"Ring_and_Field.less_divide_eq_1_neg\", (*obscure and prolific*)\n```\n``` 120 \"Ring_and_Field.less_divide_eq_1_pos\", (*obscure and prolific*)\n```\n``` 121 \"Ring_and_Field.one_eq_divide_iff\", (*duplicate by symmetry*)\n```\n``` 122 \"Set.Diff_eq_empty_iff\", (*redundant with paramodulation*)\n```\n``` 123 \"Set.Diff_insert0\",\n```\n``` 124 \"Set.disjoint_insert_1\",\n```\n``` 125 \"Set.disjoint_insert_2\",\n```\n``` 126 \"Set.empty_Union_conv\", (*redundant with paramodulation*)\n```\n``` 127 \"Set.insert_disjoint_1\",\n```\n``` 128 \"Set.insert_disjoint_2\",\n```\n``` 129 \"Set.Int_UNIV\", (*redundant with paramodulation*)\n```\n``` 130 \"Set.Inter_iff\", (*We already have InterI, InterE*)\n```\n``` 131 \"Set.Inter_UNIV_conv_1\",\n```\n``` 132 \"Set.Inter_UNIV_conv_2\",\n```\n``` 133 \"Set.psubsetE\", (*too prolific and obscure*)\n```\n``` 134 \"Set.psubsetI\",\n```\n``` 135 \"Set.singleton_insert_inj_eq'\",\n```\n``` 136 \"Set.singleton_insert_inj_eq\",\n```\n``` 137 \"Set.singletonD\", (*these two duplicate some \"insert\" lemmas*)\n```\n``` 138 \"Set.singletonI\",\n```\n``` 139 \"Set.Un_empty\", (*redundant with paramodulation*)\n```\n``` 140 \"Set.Union_empty_conv\", (*redundant with paramodulation*)\n```\n``` 141 \"Set.Union_iff\", (*We already have UnionI, UnionE*)\n```\n``` 142 \"SetInterval.atLeastAtMost_iff\", (*obscure and prolific*)\n```\n``` 143 \"SetInterval.atLeastLessThan_iff\", (*obscure and prolific*)\n```\n``` 144 \"SetInterval.greaterThanAtMost_iff\", (*obscure and prolific*)\n```\n``` 145 \"SetInterval.greaterThanLessThan_iff\", (*obscure and prolific*)\n```\n``` 146 \"SetInterval.ivl_subset\"]; (*excessive case analysis*)\n```\n``` 147\n```\n``` 148 (*These might be prolific but are probably OK, and min and max are basic.\n```\n``` 149 \"Orderings.max_less_iff_conj\",\n```\n``` 150 \"Orderings.min_less_iff_conj\",\n```\n``` 151 \"Orderings.min_max.below_inf.below_inf_conv\",\n```\n``` 152 \"Orderings.min_max.below_sup.above_sup_conv\",\n```\n``` 153 Very prolific and somewhat obscure:\n```\n``` 154 \"Set.InterD\",\n```\n``` 155 \"Set.UnionI\",\n```\n``` 156 *)\n```\n``` 157\n```\n``` 158 (*The \"name\" of a theorem is its statement, if nothing else is available.*)\n```\n``` 159 val plain_string_of_thm =\n```\n``` 160 setmp show_question_marks false\n```\n``` 161 (setmp print_mode []\n```\n``` 162 \t(Pretty.setmp_margin 999 string_of_thm));\n```\n``` 163\n```\n``` 164 (*Returns the first substring enclosed in quotation marks, typically omitting\n```\n``` 165 the [.] of meta-level assumptions.*)\n```\n``` 166 val firstquoted = hd o (String.tokens (fn c => c = #\"\\\"\"))\n```\n``` 167\n```\n``` 168 fun fake_thm_name th =\n```\n``` 169 Context.theory_name (theory_of_thm th) ^ \".\" ^ firstquoted (plain_string_of_thm th);\n```\n``` 170\n```\n``` 171 fun put_name_pair (\"\",th) = (fake_thm_name th, th)\n```\n``` 172 | put_name_pair (a,th) = (a,th);\n```\n``` 173\n```\n``` 174 (*Hashing to detect duplicate and variant clauses, e.g. from the [iff] attribute*)\n```\n``` 175\n```\n``` 176 exception HASH_CLAUSE and HASH_STRING;\n```\n``` 177\n```\n``` 178 (*Catches (for deletion) theorems automatically generated from other theorems*)\n```\n``` 179 fun insert_suffixed_names ht x =\n```\n``` 180 (Polyhash.insert ht (x^\"_iff1\", ());\n```\n``` 181 Polyhash.insert ht (x^\"_iff2\", ());\n```\n``` 182 Polyhash.insert ht (x^\"_dest\", ()));\n```\n``` 183\n```\n``` 184 fun make_banned_test xs =\n```\n``` 185 let val ht = Polyhash.mkTable (Polyhash.hash_string, op =)\n```\n``` 186 (6000, HASH_STRING)\n```\n``` 187 fun banned s = isSome (Polyhash.peek ht s)\n```\n``` 188 in app (fn x => Polyhash.insert ht (x,())) (!blacklist);\n```\n``` 189 app (insert_suffixed_names ht) (!blacklist @ xs);\n```\n``` 190 banned\n```\n``` 191 end;\n```\n``` 192\n```\n``` 193\n```\n``` 194 (*** a hash function from Term.term to int, and also a hash table ***)\n```\n``` 195 val xor_words = List.foldl Word.xorb 0w0;\n```\n``` 196\n```\n``` 197 fun hashw_term ((Const(c,_)), w) = Polyhash.hashw_string (c,w)\n```\n``` 198 | hashw_term ((Free(_,_)), w) = w\n```\n``` 199 | hashw_term ((Var(_,_)), w) = w\n```\n``` 200 | hashw_term ((Bound _), w) = w\n```\n``` 201 | hashw_term ((Abs(_,_,t)), w) = hashw_term (t, w)\n```\n``` 202 | hashw_term ((P\\$Q), w) = hashw_term (Q, (hashw_term (P, w)));\n```\n``` 203\n```\n``` 204 fun hashw_pred (P,w) =\n```\n``` 205 let val (p,args) = strip_comb P\n```\n``` 206 in\n```\n``` 207 \tList.foldl hashw_term w (p::args)\n```\n``` 208 end;\n```\n``` 209\n```\n``` 210 fun hash_literal (Const(\"Not\",_)\\$P) = Word.notb(hashw_pred(P,0w0))\n```\n``` 211 | hash_literal P = hashw_pred(P,0w0);\n```\n``` 212\n```\n``` 213\n```\n``` 214 fun get_literals (Const(\"Trueprop\",_)\\$P) lits = get_literals P lits\n```\n``` 215 | get_literals (Const(\"op |\",_)\\$P\\$Q) lits = get_literals Q (get_literals P lits)\n```\n``` 216 | get_literals lit lits = (lit::lits);\n```\n``` 217\n```\n``` 218\n```\n``` 219 fun hash_term term = Word.toIntX (xor_words (map hash_literal (get_literals term [])));\n```\n``` 220\n```\n``` 221 fun hash_thm thm = hash_term (prop_of thm);\n```\n``` 222\n```\n``` 223 fun eq_thm (thm1,thm2) = Term.aconv(prop_of thm1, prop_of thm2);\n```\n``` 224 (*Create a hash table for clauses, of the given size*)\n```\n``` 225 fun mk_clause_table n =\n```\n``` 226 Polyhash.mkTable (hash_thm, eq_thm)\n```\n``` 227 (n, HASH_CLAUSE);\n```\n``` 228\n```\n``` 229 (*Use a hash table to eliminate duplicates from xs*)\n```\n``` 230 fun make_unique ht xs =\n```\n``` 231 (app (ignore o Polyhash.peekInsert ht) xs; Polyhash.listItems ht);\n```\n``` 232\n```\n``` 233 fun mem_thm thm [] = false\n```\n``` 234 | mem_thm thm ((thm',name)::thms_names) = eq_thm (thm,thm') orelse mem_thm thm thms_names;\n```\n``` 235\n```\n``` 236 fun insert_thms [] thms_names = thms_names\n```\n``` 237 | insert_thms ((thm,name)::thms_names) thms_names' =\n```\n``` 238 if mem_thm thm thms_names' then insert_thms thms_names thms_names'\n```\n``` 239 else insert_thms thms_names ((thm,name)::thms_names');\n```\n``` 240\n```\n``` 241 fun display_thms [] = ()\n```\n``` 242 | display_thms ((name,thm)::nthms) =\n```\n``` 243 let val nthm = name ^ \": \" ^ (string_of_thm thm)\n```\n``` 244 in Output.debug nthm; display_thms nthms end;\n```\n``` 245\n```\n``` 246 (*Write out the claset, simpset and atpset rules of the supplied theory.*)\n```\n``` 247 (* also write supplied user rules, they are not relevance filtered *)\n```\n``` 248 fun get_clasimp_atp_lemmas ctxt goals user_thms (use_claset, use_simpset', use_atpset) run_filter =\n```\n``` 249 let val claset_thms =\n```\n``` 250 \t if use_claset then\n```\n``` 251 \t\tmap put_name_pair (ResAxioms.claset_rules_of_ctxt ctxt)\n```\n``` 252 \t else []\n```\n``` 253 val simpset_thms =\n```\n``` 254 \t if (!use_simpset andalso use_simpset') then (* temporary, may merge two use_simpset later *)\n```\n``` 255 \t\tmap put_name_pair (ResAxioms.simpset_rules_of_ctxt ctxt)\n```\n``` 256 \t else []\n```\n``` 257 val atpset_thms =\n```\n``` 258 \t if use_atpset then\n```\n``` 259 \t map put_name_pair (ResAxioms.atpset_rules_of_ctxt ctxt)\n```\n``` 260 \t else []\n```\n``` 261 val _ = if !Output.show_debug_msgs then (Output.debug \"ATP theorems: \"; display_thms atpset_thms) else ()\n```\n``` 262 val user_rules =\n```\n``` 263 \t case user_thms of (*use whitelist if there are no user-supplied rules*)\n```\n``` 264 \t [] => map (put_name_pair o ResAxioms.pairname) (!whitelist)\n```\n``` 265 \t | _ => map put_name_pair user_thms\n```\n``` 266 val banned = make_banned_test (map #1 (user_rules@atpset_thms@claset_thms@simpset_thms))\n```\n``` 267 fun ok (a,_) = not (banned a)\n```\n``` 268 val claset_cls_thms =\n```\n``` 269 if run_filter then ResAxioms.cnf_rules_pairs (filter ok claset_thms)\n```\n``` 270 else ResAxioms.cnf_rules_pairs claset_thms\n```\n``` 271 val simpset_cls_thms =\n```\n``` 272 \t if run_filter then ResAxioms.cnf_rules_pairs (filter ok simpset_thms)\n```\n``` 273 \t else ResAxioms.cnf_rules_pairs simpset_thms\n```\n``` 274 val atpset_cls_thms =\n```\n``` 275 \t if run_filter then ResAxioms.cnf_rules_pairs (filter ok atpset_thms)\n```\n``` 276 \t else ResAxioms.cnf_rules_pairs atpset_thms\n```\n``` 277 val user_cls_thms = ResAxioms.cnf_rules_pairs user_rules (* no filter here, because user supplied rules *)\n```\n``` 278 val cls_thms_list = make_unique (mk_clause_table 2200)\n```\n``` 279 (List.concat (user_cls_thms@atpset_cls_thms@simpset_cls_thms@claset_cls_thms))\n```\n``` 280 val relevant_cls_thms_list =\n```\n``` 281 \t if run_filter\n```\n``` 282 \t then ReduceAxiomsN.relevance_filter (ProofContext.theory_of ctxt) cls_thms_list goals\n```\n``` 283 \t else cls_thms_list\n```\n``` 284 val all_relevant_cls_thms_list = insert_thms (List.concat user_cls_thms) relevant_cls_thms_list (*ensure all user supplied rules are output*)\n```\n``` 285 in\n```\n``` 286 \t(Array.fromList (map fst (map snd all_relevant_cls_thms_list)), all_relevant_cls_thms_list)\n```\n``` 287 end;\n```\n``` 288\n```\n``` 289\n```\n``` 290\n```\n` 291 end;`" ]

[ null ]

[ "Documentation\n\n# waveguideSlotted\n\nCreate slotted waveguide antenna\n\n## Description\n\nThe `waveguideSlotted` object creates a slotted waveguide antenna. There are different types of slotted waveguides, including longitudinal slots, transversal slots, center inclined slots, inclined slots, and inclined slots cut into a narrow wall. Slotted waveguide antennas are used in navigation radar as an array fed by a waveguide.", null, "## Creation\n\n### Syntax\n\n``ant = waveguideSlotted``\n``ant = waveguideSlotted(Name,Value)``\n\n### Description\n\nexample\n\n````ant = waveguideSlotted` creates a slotted waveguide antenna on the X-Y plane. The circumference of the antenna is chosen for an operating frequency of 2.45 GHz.```\n\nexample\n\n````ant = waveguideSlotted(Name,Value)` sets properties using one or more name-value pairs. For example, `ant = waveguideSlotted('Height',1)` creates a slotted waveguide with a height of 1 meter.```\n\n### Output Arguments\n\nexpand all\n\nSlotted waveguide antenna, returned as a `waveguideSlotted` object.\n\n## Properties\n\nexpand all\n\nLength of the waveguide (n times lambda), specified as a real-valued scalar in meters. n is the number of slots in the waveguide.", null, "Example: `'Length',0.760`\n\nExample: `ant.Length = 0.760`\n\nData Types: `double`\n\nWidth of the waveguide (a), specified as a real-valued scalar in meters.", null, "Example: `'Width',0.0840`\n\nExample: `ant.Width = 0.0840`\n\nData Types: `double`\n\nHeight of the waveguide (b), specified as a real-valued scalar in meters. Please see image in `Width` property.\n\nExample: `'Height',0.0340`\n\nExample: `ant.Height = 0.0340`\n\nData Types: `double`\n\nNumber of slots (n), specified as a scalar integer.\n\nExample: `'Numslots',7`\n\nExample: `ant.Numslots = 7`\n\nData Types: `double`\n\nShape of waveguide slot, specified as one of the following objects: `antenna.Circle`, `antenna.Polygon`, `antenna.Rectangle`.\n\nExample: `'Slot',antenna.rectangle['Length',0.035]`\n\nExample: ```ant.Slot = antenna.rectangle['Length',0.035]```\n\nData Types: `double`\n\nDistance from the closed face edge to the top slot center, specified as a real-valued scalar in meters.\n\nExample: `'SlotToTop',0.0503`\n\nExample: `ant.SlotToTop = 0.0503`\n\nData Types: `double`\n\nSpace between the centers of two adjacent slots, specified as a real-valued scalar in meters.\n\nExample: `'SlotSpacing',0.0906`\n\nExample: `ant.SlotSpacing = 0.0906`\n\nData Types: `double`\n\nSlot displacement from the centreline of the width of the waveguide to the center of the slot, specified as a real-valued scalar in meters.\n\nExample: `'SlotOffset',0.0560`\n\nExample: `ant.SlotOffset = 0.0560`\n\nData Types: `double`\n\nSlot angle, specified as a real-valued scalar in degrees or a two-element vector with each element unit in degrees. In slotted waveguide the slots are in pairs. You use a two-element vector when you want one slot in the pair to be tilted at a different angle form the other.\n\nExample: `'SlotAngle',[20 10]`\n\nExample: `ant.SlotOffset = [20 10]`\n\nData Types: `double`\n\nPlate to close the open-ended side, specified as `0` for open waveguide and `1` for closed waveguide.\n\nExample: `'ClosedWaveguide',1`\n\nExample: `ant.ClosedWaveguide = 1`\n\nData Types: `double`\n\nHeight of the feed, specified as a real-valued scalar in meters.\n\nExample: `'FeedHeight',0.0210`\n\nExample: `ant.FeedHeight = 0.0210`\n\nData Types: `double`\n\nWidth of the feed, specified as a real-valued scalar in meters.\n\nExample: `'FeedWidth',0.0300`\n\nExample: `ant.FeedWidth = 0.0300`\n\nData Types: `double`\n\nSigned distances from the origin measured along the length and width of the waveguide, specified as a two-element vector with each element in meters.\n\nExample: `'FeedOffset',[-0.3627 0]`\n\nExample: `ant.FeedOffset = [-0.3627 0]`\n\nData Types: `double`\n\nLumped elements added to the antenna feed, specified as a lumped element object handle. You can add a load anywhere on the surface of the antenna. By default, the load is at the feed. For more information, see `lumpedElement`.\n\nExample: `'Load',lumpedelement`. `lumpedelement` is the object handle for the load created using `lumpedElement`.\n\nExample: ```ant.Load = lumpedElement('Impedance',75)```\n\nTilt angle of the antenna, specified as a scalar or vector with each element unit in degrees. For more information, see Rotate Antennas and Arrays.\n\nExample: `'Tilt',90`\n\nExample: `'Tilt',[90 90]`,`'TiltAxis',[0 1 0;0 1 1]` tilts the antenna at 90 degree about two axes, defined by vectors.\n\nData Types: `double`\n\nTilt axis of the antenna, specified as:\n\n• Three-element vectors of Cartesian coordinates in meters. In this case, each vector starts at the origin and lies along the specified points on the X-, Y-, and Z-axes.\n\n• Two points in space, each specified as three-element vectors of Cartesian coordinates. In this case, the antenna rotates around the line joining the two points in space.\n\n• A string input describing simple rotations around one of the principal axes, 'X', 'Y', or 'Z'.\n\nExample: `'TiltAxis',[0 1 0]`\n\nExample: `'TiltAxis',[0 0 0;0 1 0]`\n\nExample: `ant.TiltAxis = 'Z'`\n\n## Object Functions\n\n `show` Display antenna or array structure; Display shape as filled patch `axialRatio` Axial ratio of antenna `beamwidth` Beamwidth of antenna `charge` Charge distribution on metal or dielectric antenna or array surface `current` Current distribution on metal or dielectric antenna or array surface `design` Design prototype antenna or arrays for resonance at specified frequency `EHfields` Electric and magnetic fields of antennas; Embedded electric and magnetic fields of antenna element in arrays `impedance` Input impedance of antenna; scan impedance of array `mesh` Mesh properties of metal or dielectric antenna or array structure `meshconfig` Change mesh mode of antenna structure `pattern` Radiation pattern and phase of antenna or array; Embedded pattern of antenna element in array `patternAzimuth` Azimuth pattern of antenna or array `patternElevation` Elevation pattern of antenna or array `returnLoss` Return loss of antenna; scan return loss of array `sparameters` S-parameter object `vswr` Voltage standing wave ratio of antenna\n\n## Examples\n\ncollapse all\n\nCreate and view a slotted waveguide antenna with default property values.\n\n`ant = waveguideSlotted`\n```ant = waveguideSlotted with properties: Length: 0.8060 Width: 0.0857 Height: 0.0428 NumSlots: 8 Slot: [1x1 antenna.Rectangle] SlotToTop: 0.0403 SlotSpacing: 0.0806 SlotOffset: 0.0123 SlotAngle: 0 FeedWidth: 0.0020 FeedHeight: 0.0310 FeedOffset: [-0.3627 0] ClosedWaveguide: 0 Tilt: 0 TiltAxis: [1 0 0] Load: [1x1 lumpedElement] ```\n`show(ant)`", null, "Plot the radiation pattern of the antenna at 2.45 GHz.\n\n`pattern(ant, 2.45e9)`", null, "Create a slotted waveguide antenna with the following dimensions.\n\n``` ant = waveguideSlotted('Length',806e-3,'Width',94e-3, 'NumSlots',8,... 'Height',44e-3,'Slot',antenna.Rectangle('Length',53e-3,'Width',6.5e-3),'SlotToTop',40.3e-3,... 'SlotSpacing',80.6e-3,'SlotOffset',10e-3,'FeedHeight',31e-3, ... 'FeedOffset',[-362.7e-3 0],'FeedWidth',2e-3); show (ant) ```", null, "Plot impedance and S-parameters from 2.2 GHz to 2.8 GHz.\n\n```freq = 2.2e9:0.025e9:2.8e9; figure; impedance(ant,freq);```", null, "```s = sparameters(ant,freq); figure; rfplot(s);```", null, "Perovic, Una. \" Investigation of Rectangular, Unidirectional, Horizontally Polarized Waveguide Antenna with Longitudinal Slotted Arrays Operating at 2.45 GHz\"." ]

[ null, "https://ch.mathworks.com/help/antenna/ref/waveguideslotted.png", null, "https://ch.mathworks.com/help/antenna/ref/waveguideslotted_dimensions_2.png", null, "https://ch.mathworks.com/help/antenna/ref/waveguideslotted_dimensions_1.png", null, "https://ch.mathworks.com/help/examples/antenna/win64/DefaultSlottedWaveguideAntennaAndRadiationPatternExample_01.png", null, "https://ch.mathworks.com/help/examples/antenna/win64/DefaultSlottedWaveguideAntennaAndRadiationPatternExample_02.png", null, "https://ch.mathworks.com/help/examples/antenna/win64/ImpedanceAndSParametersOfCustomSlottedWaveguideAntennaExample_01.png", null, "https://ch.mathworks.com/help/examples/antenna/win64/ImpedanceAndSParametersOfCustomSlottedWaveguideAntennaExample_02.png", null, "https://ch.mathworks.com/help/examples/antenna/win64/ImpedanceAndSParametersOfCustomSlottedWaveguideAntennaExample_03.png", null ]

[ "Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript.\n\n# Evidence for an atomic chiral superfluid with topological excitations\n\n## Abstract\n\nTopological superfluidity is an important concept in electronic materials as well as ultracold atomic gases1. However, although progress has been made by hybridizing superconductors with topological substrates, the search for a material—natural or artificial—that intrinsically exhibits topological superfluidity has been ongoing since the discovery of the superfluid 3He-A phase2. Here we report evidence for a globally chiral atomic superfluid, induced by interaction-driven time-reversal symmetry breaking in the second Bloch band of an optical lattice with hexagonal boron nitride geometry. This realizes a long-lived Bose–Einstein condensate of 87Rb atoms beyond present limits to orbitally featureless scenarios in the lowest Bloch band. Time-of-flight and band mapping measurements reveal that the local phases and orbital rotations of atoms are spontaneously ordered into a vortex array, showing evidence of the emergence of global angular momentum across the entire lattice. A phenomenological effective model is used to capture the dynamics of Bogoliubov quasi-particle excitations above the ground state, which are shown to exhibit a topological band structure. The observed bosonic phase is expected to exhibit phenomena that are conceptually distinct from, but related to, the quantum anomalous Hall effect3,4,5,6,7 in electronic condensed matter.\n\n## Main\n\nQuantum simulation involves the use of a comparably simple and precisely controlled synthetic quantum system to mimic poorly understood, isolated phenomena of a far more complex but less-controlled quantum system, while excluding the superimposed secondary structure that would impede a clear understanding8. Optical lattices—that is, ultracold neutral atoms trapped in laser-induced periodic potentials9,10,11—are well established systems that simulate an elementary class of many-body lattice model, the conventional s-band Hubbard model12,13. However, a large part of the physics that is relevant in electronic many-body systems is related to the coupling of electrons to magnetic fields via the Lorentz force, and hence remains inaccessible to quantum simulations in conventional optical lattices. Extensive research has been undertaken to establish a similar mechanism for neutral atoms in optical lattices, giving rise to so-called artificial magnetic fields or gauge fields14,15 using dynamical techniques—similar to what is sometimes summarized as Floquet engineering for condensed-matter physics16. This has enabled the single-particle band structures of optical lattices to be endowed with built-in local17 or even global18,19 magnetic flux and the resultant single-particle topological properties; however, this comes at the cost of notable decoherence and markedly reduced lifetimes. An alternative approach towards optical lattice simulators beyond conventional s-band Hubbard physics is the implementation of orbital degrees of freedom by making use of higher Bloch bands20,21,22,23,24. This recent direction has led to interesting new multi-orbital quantum phases, including examples that have local angular momentum. An intriguing, but as-yet unobtained, goal of this approach is to achieve a multi-orbital optical lattice quantum simulator that can, for example, capture bosonic versions of the topological superfluidity of paired electrons25, or of quantum-Hall-related physics26.\n\nHere we take a step in this direction by demonstrating interaction-induced, spontaneous time-reversal symmetry (TRS) breaking and formation of a chiral atomic superfluid in an orbital optical lattice with global angular momentum and topological excitations and edge states. This metastable state exhibits remarkable robustness with a lifetime of hundreds of milliseconds. As an unequivocal signature of the prevalence of global angular momentum, characteristic momentum spectra are observed using time-of-flight spectroscopy. Our experimental observations directly show the spontaneous breaking of TRS in accordance with a mean-field tight-binding model and exact band calculations. The system is expected to exhibit the bosonic counterpart of the quantum anomalous Hall effect3,4,5,6,7; however, not as a result of an engineered band structure, as in the famous Haldane model27. Rather, as shown by theoretical considerations (Supplementary Information), it is the contact interaction between degenerate p orbitals that leads to a spontaneous breaking of TRS, the formation of global angular momentum, a topologically non-trivial excitation band structure and the existence of topological edge states.\n\nOur experiments use a boron nitride optical lattice that is composed of alternating shallow and deep potential wells—denoted A and B, respectively—arranged on a hexagonal lattice. As indicated in Fig. 1a, the lattice potential is formed by three laser beams operating at a wavelength λ of 1,064 nm. The beams propagate within the x–y plane and intersect at an angle of 120°. Each beam comprises two spectral components at frequencies ω1 and ω2, with linear polarizations in the z-direction. Each set of spectral components with the same frequency forms a triangular lattice with adjustable amplitude. By tuning the frequency difference ω1 − ω2, the intensity patterns of both triangular lattices are shifted with respect to each other such that their sum produces a boron nitride lattice geometry with A and B wells. The relative potential offset of the two classes of well, denoted ΔV, can be rapidly tuned. An optical dipole trap provides weak harmonic confinement with regard to the z-direction. For details, see Methods and Supplementary information.\n\nInitially, a Bose–Einstein condensate of 87Rb atoms was loaded into the lowest Bloch band in the centre of the first Brillouin zone, denoted Γ in Fig. 1b. The overall trap depth along the vertical z-axis—including the lattice potential, the dipole trap and gravitation—is 221 nK. As a key feature of our preparation protocol, we applied additional cooling by means of evaporation after the atoms were loaded into the lattice. The dipole trap depth decreases within 15 ms, such that the overall trap depth along the z-direction is reduced to 41 nK and the energetic atoms can escape. At this stage the atomic wavefunction is composed of s orbitals residing in the deep B wells of the lattice. In a subsequent rapid quench the atoms are excited to the second band (compare with Fig. 1b) while remaining at the Γ point, which constitutes a dynamically unstable energy maximum28. This is accomplished by tuning ΔV in approximately 100 μs according to the diagram in Fig. 2a, until the local s orbitals in the B wells come to lie between the s and p orbitals in the A wells and hence belong to the second band. This also leads to a further reduction, to 24 nK, of the overall trap depth with respect to the z-direction, and thus an additional boost of evaporative cooling. Typical momentum spectra (Fig. 2c) recorded shortly (0.5 ms) after the quench show that a large number of Bloch states of the second band are populated with no apparent coherence28. Related excitation protocols without additional evaporation have been used previously with bipartite square lattices22,23 and with hexagonal lattices29,30. For details, see Methods and Supplementary Information.\n\nThe second band gives rise to a symmetric double-well scenario in quasi-momentum space. It possesses two inequivalent global energy minima located at the corners of the first Brillouin zone—denoted KΔ and K in Fig. 1b—which exhibit perfect degeneracy, protected by TRS. A central observation of this work is that, in a re-condensation process, TRS is spontaneously broken and a condensate forms in either of the two K points with equal probability. This is shown in Fig. 2c, which displays momentum spectra recorded after variable holding times of between 0.5 ms and 755 ms. As time proceeds, sharp Bragg resonances grow at the KΔ and K points. These resonances indicate the build-up of long-range coherence, and hence the formation of condensate fractions at KΔ and K. Notably, these condensate fractions—according to the observations—are generally not equally sized, as is shown in Fig. 2c, in which the three rows show examples with either a dominant KΔ component (top), a dominant K component (bottom) or both components of similar size (middle).\n\nWe next point out that the observation of a dominant condensate fraction in either of the K points constitutes clear evidence of broken TRS. Exact band theory shows that the momentum spectra observed in the top and bottom rows of Fig. 2c are very well reproduced by the calculated momentum spectra of the Bloch functions associated with KΔ and K, respectively. For example, the calculated momentum spectrum of the Bloch function at KΔ in Fig. 2b is plotted for comparison in the top row of Fig. 2c. Hence, the observation of population in a single K point unequivocally indicates the presence of a wavefunction that is well approximated by the Bloch function associated with that K point. Note that these Bloch functions, $${\\Phi }_{{{\\rm{K}}}_{\\Delta }}$$ and $${\\Phi }_{{{\\rm{K}}}_{\\nabla }}$$, inherently break TRS, as stated in Fig. 1c. The first panel schematically illustrates the composition of $${\\Phi }_{{{\\rm{K}}}_{\\Delta }}$$ in terms of local s orbitals in the shallow wells and px and py orbitals in the deep wells, which form p+ = px + ipy hybrid orbitals (p in case of K). The three distinct colours indicate local phases that differ by 2π/3, which results from an inherent vortical phase texture of $${\\Phi }_{{{\\rm{K}}}_{\\Delta }}$$. In the second and third panels of Fig. 1c, $$|{\\Phi }_{{K}_{\\Delta }}|$$, which is derived from exact band theory, is plotted for two different settings of ΔV, showing that different relative populations at A and B wells can be adjusted. The fourth panel shows the vortical mass current associated with $${\\Phi }_{{{\\rm{K}}}_{\\Delta }}$$ for the same choice of ΔV as in the third panel. Note that the vortices residing at each deep well across the entire lattice all share the same sense of rotation—an interesting property that is associated with the threefold rotation symmetry of the hexagonal lattice in connection with the twofold degeneracy of the p orbital manifold in two dimensions.\n\nWe define an experimental measure of chirality χ = (nΔ − n)/(nΔ + n) as the imbalance between $${\\Phi }_{{{\\rm{K}}}_{\\Delta }}$$ and $${\\Phi }_{{{\\rm{K}}}_{\\nabla }}$$ condensate populations, nΔ and n, respectively. This quantity is observed to be random in each experimental implementation with nearly zero average over many shots. However, although for short holding times—that is, in the initial phase of the condensation process—a distribution of χ with a maximum at zero is found, for long times beyond 100 ms this distribution develops pronounced maxima around ±0.5. This is shown in Fig. 3: in Fig. 3a, the average of the modulus of the chirality $$\\langle |\\chi |\\rangle$$ is plotted against the holding time, whereas in Fig. 3b histograms of the distribution of χ are shown for holding times ranging from 35 ms to 505 ms. As is seen in Fig. 3a, $$\\langle |\\chi |\\rangle$$ is small for short holding times of a few ms, and continuously increases until it reaches a maximum at around 0.4 for a holding time of about 265 ms, where the distribution of χ (compare with Fig. 3b) shows a pronounced bimodal structure. According to these observations, the condensation process develops in two stages: the first stage spontaneously breaks U(1) symmetry, forming a condensate that is approximately described by an equal superposition of Bloch states $${\\Phi }_{{{\\rm{K}}}_{\\Delta }}$$ and $${\\Phi }_{{{\\rm{K}}}_{\\nabla }}$$, thus globally preserving TRS and zero net angular momentum. Alternatively, as in the one-dimensional double-well scenario of ref. 31, domains could be formed, such that locally TRS is broken but the net angular momentum remains at zero. However, the narrow width of the observed Bragg peaks—seen in the momentum spectra with equally occupied K points in Fig. 2c—indicate well established coherence over large parts of the lattice, which is not compatible with a fine-grain domain structure. The second condensation stage breaks the global TRS by spontaneously bifurcating to form either of the two states $${\\Phi }_{{{\\rm{K}}}_{\\Delta }}$$ or $${\\Phi }_{{{\\rm{K}}}_{\\nabla }}$$, with equal probability.\n\nTheoretical considerations (detailed in the Supplementary Information) suggest that the second stage of the condensation process is triggered by the ferromagnetic collisional interactions within the manifold of degenerate p orbitals in the deep lattice wells. In the presence of such collisions, the ground state according to mean-field theory is given by a condensate in either of the two K points rather than a superposition of both (see Supplementary Information). The atoms are self-organized into an order of synchronized vortices centred at each of the deep wells across the entire boron nitride lattice, as illustrated in Fig. 1c. The fraction of atoms populating the p+ orbitals give rise to a non-zero global angular momentum. To access the ground state, which requires build-up of global orbital angular momentum, angular momentum must be exchanged with the environment. Indeed, a signature of this process is observed in the experiment. As shown in Fig. 3c, during the second condensation phase, past about 10 ms holding time, we observe a stream of atoms leaving the lattice perpendicular to the lattice plane (x–y plane) along the direction of gravity (−z-direction). This supports the interpretation that kinetic energy and angular momentum are transferred out of the lattice plane into the weakly confined z-direction, and finally escape from the lattice. It is of interest to note that, for stronger collisional interactions, under the assumption of quasi-momentum conservation, a condensate at the M points (Fig. 1b) is predicted to have a lower energy (Supplementary Fig. 5). However, a phase-separated mixed state with equal contributions from both K points might nevertheless be energetically more favourable.\n\nThe plots in Fig. 2c are obtained with ΔV adjusted during step 1 in Fig. 2a, such that most of the atoms in the second band reside in local s orbitals in the shallow wells and thus do not contribute notable orbital angular momentum. This corresponds to the wavefunction illustrated in the second panel of Fig. 1c. We now discuss how, by means of a slightly modified preparation protocol, a state with large global orbital angular momentum can be experimentally formed, which possesses a wavefunction according to the third panel of Fig. 1c. By adding an additional step (step 2 in Fig. 4a) 205 ms after the preparation protocol of step 1 in Fig. 2a, one may relocate a substantial portion of the atomic population towards the vortical p orbitals in the deep wells. This step consists of adiabatically tuning ΔV (in 1 ms) until the local s orbitals in the shallow wells lie energetically above the p orbitals of the deep wells (see Methods and Supplementary Information). The absolute value of the corresponding wavefunction is shown in the third panel of Fig. 1c. In Fig. 4c we observe the resulting faster band relaxation that notably depletes the second band after only 10 ms. This observation indicates that the collision dynamics is now dominated by atoms in the p orbitals of the deep wells, which provide faster decay due to increased overlap with the s orbitals of the lowest band23. As a direct signature of dominant population of the p orbitals, their additional nodes lead to destructive interference, thus preventing the occurrence of Bragg resonances within the areas enclosed by red dashed circles in Fig. 4c. This observation follows the exact band calculations in Fig. 4b. Note the contrast with the analogous spectra in Fig. 2b and with the corresponding experimental spectra in the top row of Fig. 2c. A large global orbital angular momentum arises, to which each atom in the p+ orbitals contributes a portion ħ. A quantitative analysis of the dependence of angular momentum on ΔV is shown in Supplementary Fig. 4.\n\nSymmetric double-well scenarios in quasi-momentum space that lead to an interaction-induced spontaneous breaking of TRS and the formation of phases with staggered chiral order have been previously reported22,31,32. The long-lived atomic orbital superfluid devised in this work exhibits global angular momentum of the ground state wavefunction, although the Hamiltonian preserves TRS. As a consequence, basic Bogoliubov-de Gennes analysis verifies the emergence of bosonic topological excitations and edge modes (Supplementary Information). This opens up the prospect of studying these signatures experimentally. A direct comparison between experiment and theory would be enabled with regards to fundamental concepts that are related to, but have no previous analogue in, condensed-matter electronic and spin materials. Previous calculations have shown that, in a checkerboard square lattice with alternating shallow and deep wells and local non-zero angular momentum order, interaction-induced gaps should open in Bogoliubov quasiparticle spectra, and edge states protected by topological invariants should occur33,34. The proposed Zeeman-like bias is, however, an experimental challenge to implement for orbital degrees of freedom. Here, the boron nitride lattice resolves this challenge by spontaneously breaking TRS globally (driven by intrinsic interaction) with an emerging global angular momentum order. This paves the way to study dynamically controlled quasiparticles as exotic as a possible counterpart of Majorana fermions in a bosonic superfluid. The latter is widely sought-after in electronic topological superconductors, and the Supplementary Information presents calculations that support this expectation.\n\n## Methods\n\n### Realization of optical lattice\n\nA two-dimensional hexagonal boron nitride (BN) optical lattice potential is created by three laser beams, propagating in the xy plane and intersecting at 120° angles. Each laser beam comprises two frequency components ω1 and ω2—which are both linearly polarized along the z direction—with wavelength λ ≈ 1,064 nm and therefore with negative detuning with respect to the relevant atomic transitions of rubidium atoms at 780 nm and 795 nm. The two frequency components are derived from two independent lasers. Experimentally, we first combine the two laser beams at frequencies ω1 and ω2, respectively, before splitting them into three beams. The total electric field for all beams is then written as\n\n$$\\begin{array}{l}{\\bf{E}}({\\bf{r}},t)={E}_{1}{{\\bf{e}}}_{z}\\sum _{j}\\cos ({{\\bf{k}}}_{j}\\cdot {\\bf{r}}-{\\omega }_{1}t+{\\theta }_{j})\\\\ +{E}_{2}{{\\bf{e}}}_{z}\\sum _{j}\\cos ({{\\bf{k}}}_{j}^{{\\prime} }\\cdot {\\bf{r}}-{\\omega }_{2}t+{\\theta }_{j}^{{\\prime} }).\\end{array}$$\n\nHere, the wave vectors are given by $${{\\bf{k}}}_{1}={k}_{{\\rm{L}}}(\\,-\\sqrt{3}/2,1/2)$$, $${{\\bf{k}}}_{2}={k}_{{\\rm{L}}}(\\,\\sqrt{3}/2,1/2)$$, $${{\\bf{k}}}_{3}={k}_{{\\rm{L}}}(0,-1)$$, and $${{\\bf{k}}}_{j}^{{\\prime} }=({\\omega }_{2}/{\\omega }_{1}){{\\bf{k}}}_{j}$$, where $${k}_{{\\rm{L}}}={\\omega }_{1}/c$$. $${\\theta }_{j}={\\omega }_{1}{L}_{j}/c+{\\theta }_{0}$$ and $${\\theta }_{j}^{{\\prime} }={\\omega }_{2}{L}_{j}/c+{\\theta }_{0}^{{\\prime} }$$, where Lj denotes the optical path length of the jth beam from the splitting point to the centre of the lattice. The corresponding laser intensity I(r) is proportional to the time averaging of the square of the electric field according to\n\n$$\\begin{array}{ll}I({\\bf{r}})\\,\\propto \\, & \\,\\frac{3}{2}{E}_{1}^{2}+{E}_{1}^{2}\\sum _{\\langle i,j\\rangle }\\cos [({{\\bf{k}}}_{i}-{{\\bf{k}}}_{j})\\cdot {\\bf{r}}+{\\theta }_{i}-{\\theta }_{j}]\\\\ & +\\frac{3}{2}{E}_{2}^{2}+{E}_{2}^{2}\\sum _{\\langle i,j\\rangle }\\cos [({{\\bf{k}}}_{i}^{{\\prime} }-{{\\bf{k}}}_{j}^{{\\prime} })\\cdot {\\bf{r}}+{\\theta }_{i}^{{\\prime} }-{\\theta }_{j}^{{\\prime} }],\\end{array}$$\n\nwhere the summation $$\\langle i,j\\rangle$$ is limited to $$\\langle 1,2\\rangle$$, $$\\langle 2,3\\rangle$$, $$\\langle 3,1\\rangle$$. The generated optical lattice potential is proportional to the laser intensity and takes the form\n\n$$\\begin{array}{ll}{V}_{{\\rm{BN}}}({\\bf{r}})\\,=\\, & -{V}_{1}\\{3+2\\sum _{\\langle i,j\\rangle }\\cos [({{\\bf{k}}}_{i}-{{\\bf{k}}}_{j})\\cdot {\\bf{r}}+({\\theta }_{i}-{\\theta }_{j})]\\}\\\\ & -{V}_{2}\\{3+2\\sum _{\\langle i,j\\rangle }\\cos [({{\\bf{k}}}_{i}^{{\\prime} }-{{\\bf{k}}}_{j}^{{\\prime} })\\cdot {\\bf{r}}+({\\theta }_{i}^{{\\prime} }-{\\theta }_{j}^{{\\prime} })]\\},\\end{array}$$\n\nwhere V1,2 ≥ 0 for the relevant case of red detuning. Each of the two spectral components creates a triangular lattice potential, which sum up to form the total potential VBN. No interference terms arise because the frequency difference Δωω1 – ω2 is chosen in the range of a few GHz, which exceeds by far all relevant timescales of the atom dynamics. The relative position of the two triangular lattices is determined by the phase differences\n\n$$\\Delta {\\theta }_{ij}=({\\theta }_{i}-{\\theta }_{j})-({\\theta }_{i}^{{\\prime} }-{\\theta }_{j}^{{\\prime} })=\\frac{{\\omega }_{1}-{\\omega }_{2}}{c}({L}_{i}-{L}_{j}).$$\n\nNote that with Δω ≈ 2π × 3 GHz, a change of the lengths Li of the order of 10 μm corresponds to irrelevant changes of Δθij of the order of 10−4 × 2π. Hence, Δθij can be readily adjusted without the need for interferometric control of the lengths Li. Experimentally, we choose convenient values for the lengths L1, L2 and L3 and lock the frequency difference between the two lasers accordingly to fine-tune the relative position of the two triangular lattices appropriately to generate the desired boron nitride optical lattice. We set $$({L}_{1}-{L}_{2},{L}_{2}-{L}_{3},{L}_{3}-{L}_{1})=(\\,-6.04,3.02,3.02)$$cm and $$\\Delta \\omega ={\\omega }_{1}-{\\omega }_{2}=2{\\rm{\\pi }}\\times 3.308$$ GHz, which leads to (Δθ12, Δθ23, Δθ31) = $$(\\,-4{\\rm{\\pi }}/3,2{\\rm{\\pi }}/3,2{\\rm{\\pi }}/3)$$ and hence the boron nitride lattice potential\n\n$$\\begin{array}{l}{V}_{{\\rm{BN}}}({\\bf{r}})\\,=\\,-\\,{V}_{1}\\left\\{3+2\\sum _{\\langle i,j\\rangle }\\cos \\left[({{\\bf{k}}}_{i}-{{\\bf{k}}}_{j})\\cdot {\\bf{r}}-\\frac{2{\\rm{\\pi }}}{3}\\right]\\right\\}\\\\ \\,-\\,{V}_{2}\\left\\{3+2\\sum _{\\langle i,j\\rangle }\\cos \\left[({{\\bf{k}}}_{i}-{{\\bf{k}}}_{j})\\cdot {\\bf{r}}+\\frac{2{\\rm{\\pi }}}{3}\\right]\\right\\}.\\end{array}$$\n\nThe first (second) term describes the triangular lattice potential that gives rise to the A sites (B sites) in the combined boron nitride lattice shown in Fig. 1a. The potential difference between A wells and B wells can be readily adjusted on the microsecond timescale by tuning the ratio V1/V2. To avoid deformations of the lattice potential, the relative frequency difference between the two sets of triangular lattices and the laser intensities are carefully stabilized.\n\nA Bose–Einstein condensate of typical 40,000 87Rb atoms in the state $$|F=1,{m}_{F}=-1\\rangle$$ (where F is the quantum number of total angular momentum and mF = magnetic quantum number) is prepared in an optical dipole trap formed by two crossed laser beams with trapping frequencies of $$\\{{\\omega }_{x},{\\omega }_{y},{\\omega }_{z}\\}=2{\\rm{\\pi }}\\times \\{26,27,71\\}$$ Hz. Including the gravitational force (pointing into the −z direction) the trap depth along the −z direction is 34 nK. A bias magnetic field of 1 G is applied along the z axis. Within 120 ms, the lattice beam intensity is ramped up to V1 = 7.04ER and V2 = 8.03ER, where $${E}_{{\\rm{R}}}={h}^{2}/2m{\\lambda }^{2}$$. At this stage the overall trap depth along the −z direction is 221 nK. To provide additional evaporative cooling, after 5 ms, the depth of the optical dipole trap is ramped down in 15 ms, such that the overall trap depth along the −z direction is reduced to 41 nK. Excitation into the second band is obtained by swapping the depths of the A and B sites via linearly changing (V1,V2) to (7.81,7.23)ER rapidly in 0.1 ms. The trap depth along the −z direction is thereby further reduced to 24 nK, which gives rise to further evaporation.\n\nWe recorded momentum spectra via time-of-flight spectroscopy or performed band mapping that enabled us to observe the quasi-momentum distribution. These techniques were used to derive the data in Fig. 2c. To obtain the data in Fig. 4c, the experimental protocol was slightly extended. After excitation to the second band and a subsequent holding time of 205 ms, the p orbitals in the A wells are lowered by continuously increasing V1 to 8.35ER in 1 ms. Therefore, atoms are transferred from the s orbitals in the shallow B wells to the p orbitals in adjacent deep A wells. Momentum spectra of the atoms in the xy plane were obtained by switching off all potentials in less than 1 μs and, after a 20-ms-long ballistic expansion, performing absorption imaging. For band mapping, we decreased the intensity of the lattice exponentially with a time constant of 260 μs followed by 20 ms of ballistic expansion before performing absorption imaging. For the atom loss data shown in Fig. 3c, we performed in-situ absorption imaging of a plane perpendicular to the x–y plane, after the atoms were excited to the second band and held there for a variable time.\n\n## Data availability\n\nAll data presented in the figures are available upon request from the corresponding authors.\n\n## Code availability\n\nThe numerical simulations were performed with MATLAB R2019b. Codes are available upon request from the corresponding authors.\n\n## References\n\n1. 1.\n\nSato, M. & Ando, Y. Topological superconductors: a review. Rep. Prog. Phys. 80, 076501 (2017).\n\n2. 2.\n\nWheatley, J. C. Experimental properties of superfluid 3He. Rev. Mod. Phys. 47, 415–470 (1975).\n\n3. 3.\n\nYu, R. et al. Quantized anomalous Hall effect in magnetic topological insulators. Science 329, 61–64 (2010).\n\n4. 4.\n\nChang, C.-Z. et al. Experimental observation of the quantum anomalous Hall effect in a magnetic topological insulator. Science 340, 167–170 (2013).\n\n5. 5.\n\nLiu, C.-X., Zhang, S.-C. & Qi, X.-L. The quantum anomalous hall effect: theory and experiment. Annu. Rev. Condens. Matter Phys. 7, 301–321 (2016).\n\n6. 6.\n\nDeng, Y. et al. Quantum anomalous Hall effect in intrinsic magnetic topological insulator MnBi2Te4. Science 367, 895–900 (2020).\n\n7. 7.\n\nSerlin, M. et al. Intrinsic quantized anomalous Hall effect in a moiré heterostructure. Science 367, 900–903 (2020).\n\n8. 8.\n\nFeynman, R. P. Simulating physics with computers. Int. J. Theor. Phys. 21, 467–488 (1982).\n\n9. 9.\n\nVerkerk, P. et al. Dynamics and spatial order of cold cesium atoms in a periodic optical potential. Phys. Rev. Lett. 68, 3861–3864 (1992).\n\n10. 10.\n\nHemmerich, A. & Hänsch, T. W. Two-dimesional atomic crystal bound by light. Phys. Rev. Lett. 70, 410–413 (1993).\n\n11. 11.\n\nGrynberg, G. & Robilliard, C. Cold atoms in dissipative optical lattices. Phys. Rep. 355, 335–451 (2001).\n\n12. 12.\n\nBloch, I. Ultracold quantum gases in optical lattices. Nat. Phys. 1, 23–30 (2005).\n\n13. 13.\n\nLewenstein, M. et al. Ultracold atomic gases in optical lattices: mimicking condensed matter physics and beyond. Adv. Phys. 56, 243–379 (2007).\n\n14. 14.\n\nDalibard, J., Gerbier, F., Juzeliūnas, G. & Öhberg, P. Colloquium: artificial gauge potentials for neutral atoms. Rev. Mod. Phys. 83, 1523–1543 (2011).\n\n15. 15.\n\nGalitski, V., Juzeliūnas, G. & Spielman, I. B. Artificial gauge fields with ultracold atoms. Phys. Today 72, 38–44 (2019).\n\n16. 16.\n\nOka, T. & Kitamura, S. Floquet engineering of quantum materials. Annu. Rev. Condens. Matter Phys. 10, 387–408 (2019).\n\n17. 17.\n\nJotzu, G. et al. Experimental realization of the topological Haldane model with ultracold fermions. Nature 515, 237–240 (2014).\n\n18. 18.\n\nAidelsburger, M. et al. Realization of the Hofstadter Hamiltonian with ultracold atoms in optical lattices. Phys. Rev. Lett. 111, 185301 (2013).\n\n19. 19.\n\nMiyake, H., Siviloglou, G. A., Kennedy, C. J., Burton, W. C. & Ketterle, W. Realizing the Harper Hamiltonian with laser-assisted tunneling in optical lattices. Phys. Rev. Lett. 111, 185302 (2013).\n\n20. 20.\n\nLiu, W. V. & Wu, C. Atomic matter of nonzero-momentum Bose-Einstein condensation and orbital current order. Phys. Rev. A 74, 013607 (2006).\n\n21. 21.\n\nLewenstein, M. & Liu, W. V. Orbital dance. Nat. Phys. 7, 101–103 (2011).\n\n22. 22.\n\nWirth, G., Ölschläger, M. & Hemmerich, A. Evidence for orbital superfluidity in the P-band of a bipartite optical square lattice. Nat. Phys. 7, 147–153 (2011).\n\n23. 23.\n\nKock, T., Hippler, C., Ewerbeck, A. & Hemmerich, A. Orbital optical lattices with bosons. J. Phys. At. Mol. Opt. Phys. 49, 042001 (2016).\n\n24. 24.\n\nLi, X. & Liu, W. V. Physics of higher orbital bands in optical lattices: a review. Rep. Prog. Phys. 79, 116401 (2016).\n\n25. 25.\n\nQi, X. L. & Zhang, S. C. Topological insulators and superconductors. Rev. Mod. Phys. 83, 1057 (2011).\n\n26. 26.\n\nvon Klitzing, K. et al. 40 years of the quantum Hall effect. Nat. Rev. Phys. 2, 397–401 (2020).\n\n27. 27.\n\nHaldane, F. D. M. Model for a quantum Hall effect without Landau levels: condensed-matter realization of the “parity anomaly”. Phys. Rev. Lett. 61, 2015–2018 (1988).\n\n28. 28.\n\nSharma, V., Choudhury, S. & Mueller, E. J. Dynamics of Bose-Einstein recondensation in higher bands. Phys. Rev. A 101, 033609 (2020).\n\n29. 29.\n\nWeinberg, M., Staarmann, C., Ölschläger, C., Simonet, J. & Sengstock, K. Breaking inversion symmetry in a state-dependent honeycomb lattice: artificial graphene with tunable band gap. 2D Mater. 3, 024005 (2016).\n\n30. 30.\n\nJin, S. et al. Evidence of Potts-Nematic superfluidity in a hexagonal sp2 optical lattice. Phys. Rev. Lett. 126, 035301 (2021).\n\n31. 31.\n\nParker, C. V., Ha, L.-C. & Chin, C. Direct observation of effective ferromagnetic domains of cold atoms in a shaken optical lattice. Nat. Phys. 9, 769–774 (2013).\n\n32. 32.\n\nStruck, J. et al. Quantum simulation of frustrated classical magnetism in triangular optical lattices. Science 333, 996–999 (2011).\n\n33. 33.\n\nXu, Z.-F., You, L., Hemmerich, A. & Liu, W. V. π-Flux dirac bosons and topological edge excitations in a bosonic chiral p-wave superfluid. Phys. Rev. Lett. 117, 085301 (2016).\n\n34. 34.\n\nDi Liberto, M., Hemmerich, A. & Morais Smith, C. Topological varma superfluid in optical lattices. Phys. Rev. Lett. 117, 163001 (2016).\n\n## Acknowledgements\n\nThis work is supported by the National Key R&D Program of China (grant no. 2018YFA0307200), by the Key-Area Research and Development Program of Guangdong Province (grant no. 2019B030330001), by the NSFC (grant no. U1801661), by the high-level special funds from SUSTech (grant no. G02206401), by a fund from Guangdong province (grant no. 2019ZT08X324 (X.-Q.W., G.-Q.L., J.-Y.L. and Z.-F.X.)), by AFOSR grant no. FA9550-16-1-0006, by MURI-ARO grant no. W911NF-17-1-0323 through UC Santa Barbara, Shanghai Municipal Science and Technology Major Project (grant no. 2019SHZDZX01) (W.V.L.) and by DFG-He2334/17-1 (A.H.).\n\n## Funding\n\nOpen access funding provided by Universität Hamburg.\n\n## Author information\n\nAuthors\n\n### Contributions\n\nX.-Q.W. and J.-Y.L. carried out the experiments. G.-Q.L., Z.-F.X. and W.V.L. performed the calculations. Z.-F.X. devised the experiment with input by A.H. The lattice design was contributed by A.H. The manuscript was prepared by Z.-F.X, W.V.L. and A.H. All authors discussed the research.\n\n### Corresponding authors\n\nCorrespondence to W. Vincent Liu, Andreas Hemmerich or Zhi-Fang Xu.\n\n## Ethics declarations\n\n### Competing interests\n\nThe authors declare no competing interests.\n\nPeer review information Nature thanks the anonymous reviewers for their contribution to the peer review of this work. Peer reviewer reports are available.\n\nPublisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.\n\n## Supplementary information\n\n### Supplementary Information\n\nThis file contains supplementary text, supplementary equations s1 – s31, supplementary figures s1 – s7 and supplementary references.\n\n## Rights and permissions", null, "" ]

[ null, "http://www-nature-com-s.caas.cn/platform/track/article/s41586-021-03702-0", null ]

[ "Effective probability when there is a guarantee after X attempts\n\nOne slightly brute-force-y way is to write a full Markov chain with 120 states. Each state has a 99% chance of going to the next, and an 1% chance of going to the start (including the start itself), except for the final one, which has 100% chance of resetting to start. Then calculate the expected number of visits to the start state. The transition matrix is ergodic, so we have no problems calculating its highest-eigenvalue (1) eigenvector which gives the stationary distribution (fraction of time spent in each state), and you can invert the first value to find the visitaton frequency.\n\nInstead of creating a big matrix with 120 x 120 entries, most of them zero, you can do this same calculation with a bit of thinking. The stationary distribution must be such that each state has 99% times the previous state's fraction of time spent in it, except the first one which we'll call x. So you have 120 values, x + 0.99x + 0.99^(2)x + ... + 0.99^(119)x, adding up to 1. Factor to x(1 + 0.99 + 0.99^(2) + ... + 0.99^(119)) = 1, then use the formula for sums of geometric series to get x(1-0.99^(120))/(1-0.99) = 1, or 1/x = (1-0.99^(120))/(1-0.99) = approximately 70.06.\n\nFor time between visits, we're looking for 1/x, so we're done. The effective probability of getting a prize each step is just the probability of being in the first state each step x, or 1.43ish%. The time between visits of the last state (and therefore pity prizes/resets), then, is 1/(fraction of time spent there) = 1/(0.99^(119)x) = approx 231.68.\nI have spent a bit of time thinking about this and I think it is a rather tricky problem, actually. Suppose we win with probability p, unless we lost all of the previous k rounds, in which case we win with probability 1. (So in your hypothetical problem, p = 0.01 and k = 119.) We can derive a rather ugly sort of recurrence relation for the probability a\\_n of winning on round n for n>k+1:\n\na\\_n = pa\\_(n-1) + (1-a\\_(n-1))(pa\\_(n-2) + (1-a\\_(n-2))(pa\\_(n-3) + (1-a\\_(n-3))( ... (1-a\\_(n-k+1))(pa\\_(n-k) + 1*(1-a\\_(n-k))...)\n\nFor n < k+1, it's pretty easy since everything is still independent: a\\_n = p. And for n = k+1, it's not bad: we get\n\na\\_k = 1 \\* (1-p)^(k) + p(1 - (1-p)^(k)).\n\nBut I have no idea how to get a closed form for these, much less how to compute the long-running \"expected probability of winning.\" To define this meaningfully, I think we'd need to formalize the n-th attempt in the lottery as a Bernoulli random variable X\\_n that is equal to 1 with probability a\\_n and 0 with probability 1-a\\_n, but with dependence on previous X\\_i: X\\_n is 1 with probability p provided that X\\_(n-1) through X\\_(n-k) were not all 0; if X\\_(n-1) through X\\_(n-k) were all 0 then X\\_n is 1 with probability 1. Then we'd be interested in the limit\n\nlimit_(n->infinity) ( (sum from i = 1 to n of X\\_i) / n)\n\nBut this is hard to compute; ordinarily we'd use a law of large numbers, but these X\\_i are neither independent nor identically distributed, so while there may be some LLN that applies, I certainly do not know about it.\n\nPerhaps this question is sophisticated enough to get interesting answers (or references to answers) on math stack exchange—you might want to try posting it there?" ]

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[ "## ››Convert foot water [4 °C] to dyne/square centimetre\n\n foot water dyne/square centimeter\n\n## ››More information from the unit converter\n\nHow many foot water in 1 dyne/square centimeter? The answer is 3.3455256555148E-5.\nWe assume you are converting between foot water [4 °C] and dyne/square centimetre.\nYou can view more details on each measurement unit:\nfoot water or dyne/square centimeter\nThe SI derived unit for pressure is the pascal.\n1 pascal is equal to 0.00033455256555148 foot water, or 10 dyne/square centimeter.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between feet water and dynes/square centimeter.\nType in your own numbers in the form to convert the units!\n\n## ››Quick conversion chart of foot water to dyne/square centimeter\n\n1 foot water to dyne/square centimeter = 29890.669 dyne/square centimeter\n\n2 foot water to dyne/square centimeter = 59781.338 dyne/square centimeter\n\n3 foot water to dyne/square centimeter = 89672.007 dyne/square centimeter\n\n4 foot water to dyne/square centimeter = 119562.676 dyne/square centimeter\n\n5 foot water to dyne/square centimeter = 149453.345 dyne/square centimeter\n\n6 foot water to dyne/square centimeter = 179344.014 dyne/square centimeter\n\n7 foot water to dyne/square centimeter = 209234.683 dyne/square centimeter\n\n8 foot water to dyne/square centimeter = 239125.352 dyne/square centimeter\n\n9 foot water to dyne/square centimeter = 269016.021 dyne/square centimeter\n\n10 foot water to dyne/square centimeter = 298906.69 dyne/square centimeter\n\n## ››Want other units?\n\nYou can do the reverse unit conversion from dyne/square centimeter to foot water, or enter any two units below:\n\n## Enter two units to convert\n\n From: To:\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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[ "Average Table Height Typical Dining Table Height Average Table Size Average Dining Room Table Size Average Dining Table Size Typical Dining Table Height Normal End Table Height", null, "average table height typical dining table height average table size average dining room table size average dining table size typical dining table height normal end table height.\n\naverage table height uk top,how tall is the average table height,average table height mms,average table height in meters is measured,average table height in meters to feet,average table height mmr,average table height cm conversion,average height of a table lamp,average table height ukm,average table height restaurants,average table height cm feet." ]

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[ "# Adding complex numbers as vectors\n\nRecall that adding complex numbers simply involves adding the real parts and imaginary parts separately. When represented as vectors on the complex plane, this is the same as adding vectors “tip to tail” and forming a parallelogram.\n\nExample. To add the complex numbers $$2-i$$ and $$1+3i$$ we draw the vectors tip to tail. It doesn’t matter which order you add the vectors, the result is still $$3+2i$$. Geometrically, the addition of two vectors forms a parallelogram.\n\nIt is often easier to write out vector additions in arrow notation, where a vector from point $$A$$ to point $$B$$ is written as $$\\overrightarrow{AB}$$.\n\nExample. Let $$z=2+i$$ and $$w=1+2i$$. Let $$A$$, $$B$$ and $$C$$ be the points representing $$z$$, $$w$$ and $$z+w$$ respectively. Let $$O$$ represent the origin. Then the point representing $$z+w$$ is represented by the vector $$\\overrightarrow{OC}$$, and \\begin{align*} \\overrightarrow{OC} &= \\overrightarrow{OA}+\\overrightarrow{AC} \\\\ &=\\overrightarrow{OA}+\\overrightarrow{OB} \\\\ &= (2+i)+(1+2i) = 3+3i \\end{align*}\n\n## Problems\n\n1. Add the following complex numbers as vectors on the complex plane.\n1. $$1+i$$ and $$2+i$$\n2. $$-1-2i$$ and $$5-i$$\n2. Let $$z_1$$, $$z_2$$ and $$z_3$$ be complex numbers, represented by the points $$A$$, $$B$$ and $$C$$ on the complex plane. Let $$O$$ be the origin. What are the conditions required on $$z_1$$, $$z_2$$ and $$z_3$$ for $$OABC$$ to be a parallelogram? What about a rhombus?" ]

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[ "I wanted to create a graphic consisting of a circle divided precisely into three segments of equal size. I wanted to learn some of the basic drawing commands native to SVG (Scalable Vector Graphics) and HTML5 Canvas, to compare the procedure.\n\n## SVG\n\nDrawing graphics with SVG is quite straightforward. Rather like the HTML5 canvas element, the SVG tag requires height and width properties with numerical values to set the drawing area.\n\nIn the examples that follow (see below), I set the SVG area to 400 x 400.\n\nThe circle requires an (x,y) origin, or centre point, specified as cx and cy coordinates, with a numerical value for each. It also requires a radius (r) with a numerical value.\n\nAs I want my circle to use the maximum size of the SVG area, I set the x,y origin of the circle to 200,200. And I set the radius to 200.\n\nTo make my circle easy to see, I set the fill colour, using a hexadecimal value: `fill=\"#6a29ea\"`\n\nSo far, the result is this:\n\nNow to divide my circle into three equal segments, using three lines to mark the divisions.\n\nDrawing a line is easy enough: we specify the start point coordinates as x1,y1 and the end point coordinates as x2,y2 , and supply the values we need.\n\nI know that each line has to begin at the centre of the circle, which is 200,200. So my three lines each begin with:\n\n``````x1=\"200\" y1=\"200\"\n\n``````\n\nNext, I need to determine the end point of each line as a pair of coordinates on the edge of the circle. To do this, I need to make some cosine and sine calculations. The forumlas I need for the end-point coordinates of a line:\n\nExplanation:\n\nWhatever angle I choose for my line, I use cosine and sine to plot the final point x,y coordinates at the edge of the circle. However, the area of a computer screen or web page does not use standard Cartesian coordinates, so I have to compensate by adding the radius to the result of each sine and cosine computation.\n\nTo divide the circle (360 degrees) into three, I will divide the circle into 3 x 120 degrees. Whatever first angle I choose for my first line, I then add 120 to each subsequent calculation for the other two lines.\n\nI want my first line to be vertical, so I use 90 degrees as my starting angle.\n\nTo get the line’s end position x coordinate: ((cos90 * 200) + 200). The result is 200.\n\nTo get the line’s end position y coordinate: ((sin90 * 200) + 200). This equals 400.\n\nSo I know my first line x2,y2 coordinates are 200,400.\n\nThus, my first line entry is:\n\n``````\n<line x1=\"200\" y1=\"200\" x2=\"200\" y2=\"400\" style=\"stroke:rgb(255,255,255);stroke-width:6\" />\n\n``````\n\nI then repeat this process twice more, using the angles of (90+120) = 210; and (210+120) = 330.\n\nThe result looks like this:\n\n### More to do\n\nJavascript can be implemented with SVG to calculate the coordinates for each line, rather than inputting the pre-calculated values.\n\nHere is the code to do this for the above example:\n\n``````<svg height=\"400\" width=\"400\">\n<circle cx=\"200\" cy=\"200\" r=\"200\" stroke=\"black\" stroke-width=\"0\" fill=\"#6a29ea\" />\n<line id=\"line0\" x1=\"200\" y1=\"200\" x2=\"0\" y2=\"0\" style=\"stroke:rgb(255,255,255);stroke-width:6\" />\n<line id=\"line1\" x1=\"200\" y1=\"200\" x2=\"0\" y2=\"0\" style=\"stroke:rgb(255,255,255);stroke-width:6\" />\n<line id=\"line2\" x1=\"200\" y1=\"200\" x2=\"0\" y2=\"0\" style=\"stroke:rgb(255,255,255);stroke-width:6\" />\nSorry, your browser does not support inline SVG.\n</svg>\n\n<script>\nvar angles = [90, 210, 330];\nvar i;\nvar x;\nvar y;\nfunction drawCircle() {\nfor(i = 0; i < angles.length; i+=1){\nradians = angles[i] / 180 * Math.PI;\nx = (x * 200) + 200;\ny = (y * 200) + 200;\ndocument.getElementById(\"line\"+i).setAttribute(\"x2\", x );\ndocument.getElementById(\"line\"+i).setAttribute(\"y2\", y );\n}\n}\ndrawCircle();\n</script>\n\n``````\n\nAs can be noted in the above code extract, I have set the initial x2,y2 coordinates for each line to 0. The Javascript function then computes the x,y values, using Math.cos() and Math.sin() methods, then assigns the values to the x2,y2 attributes of each line, using a loop control structure.\n\nTo conclude, SVG works nicely with JavaScript.\n\n## HTML5 Canvas\n\nHTML5 canvas creates bitmap (raster) images, rather than scalable vectors. The pros and cons of each are complex and not within the scope of this simple introduction. However, to illustrate, here is a similar result using the HTML5 canvas object.\n\nYour browser does not support the canvas element.\n\nThe code for the HTML5 Canvas image above is:\n\n``````\n<canvas id=\"myCanvas\" width=\"400\" height=\"400\"\nstyle=\"border:1px solid #d3d3d3;\">\nYour browser does not support the canvas element.\n</canvas>\n\n<script>\nvar canvas = document.getElementById(\"myCanvas\");\nvar ctx = canvas.getContext(\"2d\");\nctx.beginPath();\nctx.arc(200,200,200,0,2*Math.PI);\nctx.stroke();\nctx.fillStyle = \"blue\";\nctx.fill();\n\nfunction drawCircle() {\nvar angles = [1*(360 / 3), 2*(360 / 3), 3*(360 / 3)];\n\nfor (i=0; i < angles.length; i++){\nvar radians = angles[i] / 180 * Math.PI;\n\nvar x = (x * 200) + 200;\nvar y = (y * 200) + 200;\n\nvar canvas = document.getElementById(\"myCanvas\");\nvar ctx = canvas.getContext(\"2d\");\nctx.moveTo(200, 200);\nctx.lineTo(x, y);\nctx.lineWidth = 5;\nctx.strokeStyle = 'white';\nctx.stroke();\n}\n}\ndrawCircle();\n\n</script>\n\n``````" ]

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[ "# What is the divisibility rules for 2 5 and 10?\n\n## What is the divisibility rules for 2 5 and 10?\n\nDivsibility by 2, 5 and 10\n\n• Look at the last digit of the number.\n• If the digit is a 2, 4, 6, 8 or 0, the number is divisible by 2.\n• If the digit is a 5 or a 0, the number is divisible by 5.\n• If the digit is a 0, the number is divisible by 10.\n\n## What is the rules in divisibility rule by 3 6 and 9?\n\nA number is divisible by 3 if the sum of its digits is divisible by 3. A number is divisible by 9 if the sum of its digits is divisible by 9. And a number is divisible by 6 if it is divisible by 2 (even number) and by 3. Back to the game.\n\nWhich of the following is not divisible by 2 5 10?\n\n125370\nUnit digit of this number does not end with either 0 or 5, hence not divisible by 5. Unit digit of this number is not 0, hence not divisible by 10. Hence, 125370 is not divisible by 2, 5 and 10.\n\n### Can you do long division with no remainder?\n\nBy using this long division calculator, users can perform division with remainder or without remainder which comprises large numbers. What is the Quotient and Remainder for 9452 divided by 11 using long division method?\n\n### What are the divisibility rules for 2, 4, 5, and 10?\n\nDivisibility Rules: 2, 3, 4, 5, 6, 9, and 10 A number a is divisible by the number b if a div b has a remainder of zero ( 0 ). For example, 15 divided by 3 is exactly 5 which implies that its remainder is zero.\n\nIs the number 4, 608 divisible by 2 or 3?\n\nLet’s do that by adding all the digits of 4,608 which is 4 + 6+ 0 + 8 = 18. Obviously, the sum of the digits is divisible by 3 because 18 ÷ 3 = 6. Since the number 4,608 is both divisible by 2 and 3 then it must also be divisible by 6.\n\n#### How to do long division with 2 digits?\n\nLong Division with 2 Digits How to do Long Division with Decimal Just supply the values of dividend, divisor and hit on ENTER button to find the Quotient & Remainder in decimal. The step by step work reveals how to do long division between different combination of dividend and divisor." ]

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[ "Home | Menu | Get Involved | Contact webmaster", null, "", null, "", null, "", null, "", null, "0 / 12\n\n# Number 13968697\n\nthirteen million nine hundred sixty eight thousand six hundred ninety seven\n\n### Properties of the number 13968697\n\n Factorization 13968697 Divisors 1, 13968697 Count of divisors 2 Sum of divisors 13968698 Previous integer 13968696 Next integer 13968698 Is prime? YES (908211th prime) Previous prime 13968683 Next prime 13968709 13968697th prime 255598271 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 110101010010010100111001 Octal 65222471 Duodecimal 48178a1 Hexadecimal d52539 Square 195124495877809 Square root 3737.4720065841 Natural logarithm 16.452329455587 Decimal logarithm 7.145155897016 Sine 0.89413121677267 Cosine 0.44780505489848 Tangent 1.9966974624156\nNumber 13968697 is pronounced thirteen million nine hundred sixty eight thousand six hundred ninety seven. Number 13968697 is a prime number. The prime number before 13968697 is 13968683. The prime number after 13968697 is 13968709. Number 13968697 has 2 divisors: 1, 13968697. Sum of the divisors is 13968698. Number 13968697 is not a Fibonacci number. It is not a Bell number. Number 13968697 is not a Catalan number. Number 13968697 is not a regular number (Hamming number). It is a not factorial of any number. Number 13968697 is a deficient number and therefore is not a perfect number. Binary numeral for number 13968697 is 110101010010010100111001. Octal numeral is 65222471. Duodecimal value is 48178a1. Hexadecimal representation is d52539. Square of the number 13968697 is 195124495877809. Square root of the number 13968697 is 3737.4720065841. Natural logarithm of 13968697 is 16.452329455587 Decimal logarithm of the number 13968697 is 7.145155897016 Sine of 13968697 is 0.89413121677267. Cosine of the number 13968697 is 0.44780505489848. Tangent of the number 13968697 is 1.9966974624156\n\n### Number properties\n\n0 / 12\nExamples: 3628800, 9876543211, 12586269025" ]

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[ "1852930000000 (number)\n\n1,852,930,000,000 (one trillion eight hundred fifty-two billion nine hundred thirty million) is an even thirteen-digits composite number following 1852929999999 and preceding 1852930000001. In scientific notation, it is written as 1.85293 × 1012. The sum of its digits is 28. It has a total of 16 prime factors and 256 positive divisors. There are 734,832,000,000 positive integers (up to 1852930000000) that are relatively prime to 1852930000000.\n\nBasic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 13\n• Sum of Digits 28\n• Digital Root 1\n\nName\n\nShort name 1 trillion 852 billion 930 million one trillion eight hundred fifty-two billion nine hundred thirty million\n\nNotation\n\nScientific notation 1.85293 × 1012 1.85293 × 1012\n\nPrime Factorization of 1852930000000\n\nPrime Factorization 27 × 57 × 127 × 1459\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 16 Total number of prime factors rad(n) 1852930 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 1,852,930,000,000 is 27 × 57 × 127 × 1459. Since it has a total of 16 prime factors, 1,852,930,000,000 is a composite number.\n\nDivisors of 1852930000000\n\n256 divisors\n\n Even divisors 224 32 16 16\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 256 Total number of the positive divisors of n σ(n) 4.65374e+12 Sum of all the positive divisors of n s(n) 2.80081e+12 Sum of the proper positive divisors of n A(n) 1.81787e+10 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.36122e+06 Returns the nth root of the product of n divisors H(n) 101.929 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 1,852,930,000,000 can be divided by 256 positive divisors (out of which 224 are even, and 32 are odd). The sum of these divisors (counting 1,852,930,000,000) is 4,653,738,086,400, the average is 18,178,664,400.\n\nOther Arithmetic Functions (n = 1852930000000)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 734832000000 Total number of positive integers not greater than n that are coprime to n λ(n) 5103000000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 68074455961 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 734,832,000,000 positive integers (less than 1,852,930,000,000) that are coprime with 1,852,930,000,000. And there are approximately 68,074,455,961 prime numbers less than or equal to 1,852,930,000,000.\n\nDivisibility of 1852930000000\n\n m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 2 0 1\n\nThe number 1,852,930,000,000 is divisible by 2, 4, 5 and 8.\n\n• Arithmetic\n• Abundant\n\n• Polite\n• Practical\n\n• Frugal\n\nBase conversion (1852930000000)\n\nBase System Value\n2 Binary 11010111101101011001111000000010010000000\n3 Ternary 20120010201212200202111001\n4 Quaternary 122331223033000102000\n5 Quinary 220324300040000000\n6 Senary 3535120240301344\n8 Octal 32755317002200\n10 Decimal 1852930000000\n12 Duodecimal 25b13a611854\n20 Vigesimal 3c7c0ca000\n36 Base36 nn82oi9s\n\nBasic calculations (n = 1852930000000)\n\nMultiplication\n\nn×y\n n×2 3705860000000 5558790000000 7411720000000 9264650000000\n\nDivision\n\nn÷y\n n÷2 9.26465e+11 6.17643e+11 4.63232e+11 3.70586e+11\n\nExponentiation\n\nny\n n2 3433349584900000000000000 6361756446348757000000000000000000000 11787889372133002308010000000000000000000000000000 21842133854306403966580969300000000000000000000000000000000000\n\nNth Root\n\ny√n\n 2√n 1.36122e+06 12282.5 1166.71 284.166\n\n1852930000000 as geometric shapes\n\nCircle\n\n Diameter 3.70586e+12 1.16423e+13 1.07862e+25\n\nSphere\n\n Volume 2.66481e+37 4.31447e+25 1.16423e+13\n\nSquare\n\nLength = n\n Perimeter 7.41172e+12 3.43335e+24 2.62044e+12\n\nCube\n\nLength = n\n Surface area 2.06001e+25 6.36176e+36 3.20937e+12\n\nEquilateral Triangle\n\nLength = n\n Perimeter 5.55879e+12 1.48668e+24 1.60468e+12\n\nTriangular Pyramid\n\nLength = n\n Surface area 5.94674e+24 7.4974e+35 1.51291e+12" ]

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[ "# Intel® Extension for PyTorch* optimizations for quantization\n\nThe quantization functionality in Intel® Extension for PyTorch* currently only supports post-training quantization. This tutorial introduces how the quantization works in the Intel® Extension for PyTorch* side.\n\nWe fully utilize Pytorch quantization components as much as possible, such as PyTorch Observer method. To make a PyTorch user be able to easily use the quantization API, API for quantization in Intel® Extension for PyTorch* is very similar to those in PyTorch. Intel® Extension for PyTorch* quantization supports a default recipe to automatically decide which operators should be quantized or not. This brings a satisfying performance and accuracy tradeoff.\n\n## Static Quantization\n\n```import intel_extension_for_pytorch as ipex\nfrom intel_extension_for_pytorch.quantization import prepare, convert\n```\n\n### Define qconfig\n\nUsing the default qconfig(recommended):\n\n```qconfig = ipex.quantization.default_static_qconfig\n# equal to\n# QConfig(activation=HistogramObserver.with_args(reduce_range=False),\n# weight=PerChannelMinMaxObserver.with_args(dtype=torch.qint8, qscheme=torch.per_channel_symmetric))\n```\n\nor define your own qconfig as:\n\n```from torch.ao.quantization import MinMaxObserver, PerChannelMinMaxObserver, QConfig\nqconfig = QConfig(activation=MinMaxObserver.with_args(qscheme=torch.per_tensor_affine, dtype=torch.quint8),\nweight=PerChannelMinMaxObserver.with_args(dtype=torch.qint8, qscheme=torch.per_channel_symmetric))\n```\n\nNote: we fully use PyTorch observer methonds, so you can use a different PyTorch obsever methond to define the QConfig. For weight observer, we only support torch.qint8 dtype now.\n\nSuggestion:\n\n1. For activation observer, if using qscheme as torch.per_tensor_affine, torch.quint8 is preferred. If using qscheme as torch.per_tensor_symmetric, torch.qint8 is preferred. For weight observer, setting qscheme to torch.per_channel_symmetric can get a better accuracy.\n\n2. If your CPU device doesn’t support VNNI, seting the observer’s reduce_range to True can get a better accuracy, such as skylake.\n\n### Prepare Model and Do Calibration\n\n```# prepare model, do conv+bn folding, and init model quant_state.\nuser_model = ...\nuser_model.eval()\nexample_inputs = ..\nprepared_model = prepare(user_model, qconfig, example_inputs=example_inputs, inplace=False)\n\nfor x in calibration_data_set:\nprepared_model(x)\n\n# Optional, if you want to tuning(performance or accuracy), you can save the qparams as json file which\n# including the quantization state, such as scales, zero points and inference dtype.\n# And then you can achange the json file's settings, loading the changed json file\n# to model which will override the model's original quantization's settings.\n#\n# prepared_model.save_qconf_summary(qconf_summary = \"configure.json\")\n```\n\n### Convert to Static Quantized Model and Deploy\n\n```# make sure the example_inputs's size is same as the real input's size\nconvert_model = convert(prepared_model)\ntraced_model = torch.jit.trace(convert_model, example_input)\ntraced_model = torch.jit.freeze(traced_model)\n# for inference\ny = traced_model(x)\n\n# or save the model to deploy\n\n# traced_model.save(\"quantized_model.pt\")\n# quantized_model = torch.jit.freeze(quantized_model.eval())\n# ...\n```\n\n## Dynamic Quantization\n\n```import intel_extension_for_pytorch as ipex\nfrom intel_extension_for_pytorch.quantization import prepare, convert\n```\n\n### Define QConfig\n\nUsing the default qconfig(recommended):\n\n```dynamic_qconfig = ipex.quantization.default_dynamic_qconfig\n# equal to\n# QConfig(activation=PlaceholderObserver.with_args(dtype=torch.float, compute_dtype=torch.quint8),\n# weight=PerChannelMinMaxObserver.with_args(dtype=torch.qint8, qscheme=torch.per_channel_symmetric))\n```\n\nor define your own qconfig as:\n\n```from torch.ao.quantization import MinMaxObserver, PlaceholderObserver, QConfig\ndynamic_qconfig = QConfig(activation = PlaceholderObserver.with_args(dtype=torch.float, compute_dtype=torch.quint8),\nweight = MinMaxObserver.with_args(dtype=torch.qint8, qscheme=torch.per_tensor_symmetric))\n```\n\nNote: For weight observer, it only supports dtype torch.qint8, and the qscheme can only be torch.per_tensor_symmetric or torch.per_channel_symmetric. For activation observer, it only supports dtype torch.float, and the compute_dtype can be torch.quint8 or torch.qint8.\n\nSuggestion:\n\n1. For weight observer, setting qscheme to torch.per_channel_symmetric can get a better accuracy.\n\n2. If your CPU device doesn’t support VNNI, seeting the observer’s reduce_range to True can get a better accuracy, such as skylake.\n\n### Prepare Model\n\n```prepared_model = prepare(user_model, dynamic_qconfig, example_inputs=example_inputs)\n```\n\n## Convert to Dynamic Quantized Model and Deploy\n\n```# make sure the example_inputs's size is same as the real input's size\nconvert_model = convert(prepared_model)\n# Optional: convert the model to traced model\n# traced_model = torch.jit.trace(convert_model, example_input)\n# traced_model = torch.jit.freeze(traced_model)\n\n# or save the model to deploy\n# traced_model.save(\"quantized_model.pt\")\n# quantized_model = torch.jit.freeze(quantized_model.eval())\n# ...\n# for inference\ny = convert_model(x)\n```\n\nNote: we only support the following ops to do dynamic quantization:\n\n• torch.nn.Linear\n\n• torch.nn.LSTM\n\n• torch.nn.GRU\n\n• torch.nn.LSTMCell\n\n• torch.nn.RNNCell\n\n• torch.nn.GRUCell" ]

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[ "# Statistical Pulse Measurements Using USB Power Sensors\n\nParent Category: 2016 HFE\n\nBy Orwill Hawkins\n\nToday’s modern USB Power Sensors are capable of many advanced power measurements. These Power Sensors are capable of demodulating the signal and processing the video information into usable measurements.\n\nStatistical pulse measurement capability is just one of the many features these new power sensors can deliver to the engineer at a modest cost. This article will discuss the benefits and explain the methodology used to make statistical pulse measurements in two state pulse waveforms. Its purpose is to explain to the power sensor user the methodology in making the measurements so that the user can make better measurements with confidence.\n\nPulse Measurements: Three Methods\n\nPulse measurements are usually made by one of three methods. The first and simplest method begins by measuring average power. Once the average power is known, the user calculates pulse power from an assumed duty cycle. This simple time-tested method is relatively inexpensive; however it relies on an assumed duty cycle which can add error into the measurement.\n\nGraphical analysis is another method that is commonly used. Here, the user or software process acquires, parses and analyzes a time domain representation of the signal. This requires more expensive equipment and additional training. Data analysis can be slower and relies on triggering which can make acquisition difficult. Its greatest value is being able to directly viewing the signal.\n\nThe last method and the subject of this article is statistical analysis in which samples are taken over a period of time and then analyzed. This method collects more data, and can provide better accuracy. This method can produce repeatable accurate results, and requires less training. It is also cost-effective. This approach is usually less reliant on skill or assumptions than the first two methods.\n\nWhile this method has been employed heavily on radar pulses, it can also provide significant value on today’s communication signals. Size and weight reduction in PA (Power Amp) circuits for portable devices is a challenge. Statistical measurements can be utilized to directly measure important signal parameters. These tests can help the engineer confirm device specifications and optimize data density on signals with complex modulation without exceeding component design specifications. Statistical pulse measurements excel in this area and can be used to verify an engineer’s design and test for component failures. A gated measurement can be made on an actual data borne signal with non-repetitive pulse shapes and the statistical measurement will accurately report crest factor and other parameters.\n\nSpecifying Parameters\n\nTo explain how measurements are made, pulse parameters must first be specified. In this article, pulse definitions from IEEE Standard 181-2011 will be utilized. This specification defines all necessary pulse parameters in detail. Only a portion of the defined parameters will be used in this paper, further, an added parameter, Peak Power, will be specified. Refer to Figure 1, Pulse Amplitude Specifications along with the following explanation.", null, "Figure 1 • Pulse Amplitude Specifications.\n\nPulse Specifications\n\nThis article used IEEE Standards 194-1977, 181-1977, 181-2003 and 181-2011. With revisions of these standards, some traditional terminology has been deprecated and or replaced. The statistical measurement principles discussed here remain essentially the same. Table 1 lists the important terminology changes utilized in this article. Descriptions listed for these terms may vary—for example 10% and 90% are not always used as the reference points. Some of these deprecated terms are in use in the industry and may be referenced here along with the new terminology. See Table 1 for a list of some of the new terms along with the deprecated terms.", null, "Table 1 • Terminology.\n\nReferring to Figure 1, Peak Power is added and is not generally referenced in the IEEE Standards. This is usually the first overshoot. The LB479A power sensor used in this article reports this measurement.\n\nState 2 or Top Magnitude Power is often referred to as Pulse Power or Pulse Top Power, and will be referenced as Pulse Power in this document. Occasionally, this measurement is confused with Peak Power; however they are very different. The Power Sensor user should make sure they fully understand their sensor’s measurement. This is an important measurement, often needed by engineers designing PA circuits which operate at or very near specific limits.\n\nThe bisecting points of the 10%, 50% and 90% References are used to locate time related information such as Transition Durations (Rise and Fall Times), Pulse Duration (Pulse Width) etc. Only the 50% Reference will be utilized here. This point is defined as the 50% of the State 2 Power, and may also be referred to as the 3dB down point by some users.\n\nThe Base State represents no power, and is often equated to the noise floor on a typical Power Sensor such as the LB479A used in this paper. With this sensor this level is well below -60 dBm. Base State is the minimum power the system can measure. In some instances, such as the case where a switch with leakage is implemented for pulse generation, the State 1 will represent the presence of power as is indicated in Figure 1; this power is referred to as Offset Power. A typical Power Sensor will represent State 2 Power and Peak Power as absolute power (measured from the Base State), since that is the measurement of interest in most cases. Any Base State Power will simply be included in the average power measurement.\n\nSpecific Measurements\n\nUsing the above Pulse definitions, let’s look at how the specific measurements can be made. The statistical measurements discussed here are done without consideration to pulse timing information, and will not provide Pulse Duration or Pulse Period; however Duty Factor measurements will be made. The only two time considerations are 1) The digitized samples should be taken on a consistent repetitive time basis so that there is the same amount of time between samples. Of course sufficient bandwidth and sampling rate for the given waveform must be available, and 2) The waveform epoch (entire period of time that samples are taken within) is sufficient to collect all required signal information. More on this later.\n\nTo make Statistical Pulse measurements, samples are taken repetitively in accordance with their power level. Referring to Figure 2, this can be visualized by laying the pulse over a sheet of graph paper with the X axis representing point a sample is taken, and the Y axis representing its power level. Note that the Y axis is limited to the measurement power level capability while the X axis is only limited by the number of samples desired, a very significant number of samples may be taken if necessary. In the graph shown in Figure 2, each vertical line represents the occurrence of a sample, while each horizontal line represents a power level. It should be noted that so long as consistency and associated math rules are utilized, the units can be liner or log.\n\nAt this point, it is easy to see the Peak Power. It is simply the highest measured power level.\n\nState 2 or Pulse Power, shown in Figure 2, must be calculated and represent an average of the top of the pulse. Here, the IEEE Standards are utilized. One method is to use Mean of Density Distribution. This is a good method for a basic pulse such as the one in the Figures and is suggested in the original standard. Newer standards reduce the graphical explanation of the method. Visualization is utilized here to help explain the process.", null, "Figure 2 • Pulse with Grid.\n\nHistogram Data\n\nReferring to the graph in Figure 3, note the added histogram data at right. This graph represents the sum of binned samples. As the individual samples are taken across the waveform, each data sample is evaluated, and its associated bin incremented. This “occurrence density distribution” algorithm, develops the shown histogram. Note that the bins do not contain power levels. The contents are the sum of the presence of values within a power level range. For example, if a sample value is 5dBm, the 5dBm bin would be incremented. On pulsed waveforms such as the example in the Figures, this method will develop two distinct modes, State 2 (f2) and State 1 (f1). These are noted on the horizontal peaks shown in the histogram at right in Figure 3. Since the pulse has not been repeated in this diagram, and only one pulse is shown with no clarity of the number of samples taken between pulses, f1 is unknown at this point. It (f1), the count of matching power levels could be much larger than shown in cases such as a radar pulse with significant time between pulses.", null, "Figure 3 • Pulse with Grid and Histogram.\n\nNOTE: The algorithm may be automatically modified or a different algorithm altogether may be used for distorted pulses or other shapes. The LB479A sensor, for example, employs this and several additional algorithms to evaluate statistical power. For further discussion, refer to the IEEE standards listed earlier.\n\nAverage Power\n\nAverage Power can be calculated by simply averaging the sampled power levels (converted to linear units if in log). As previously mentioned, the image shown in Figure 3 does not necessarily include a complete wave form, rendering an average of only the data shown inaccurate. This is no different than any typical average power measurement, many samples should be averaged for accuracy and stability. In this case we are examining specific pulses; however, the averaging calculations are the same. For statistical measurement work, averaging for extended periods of time (lots of samples) increase the accuracy on all fronts. The pulse stream in Figure 4 shows many pulses, the result is an increase in all of the values in the histogram, resulting in increased accuracy and a good average power measurement. The measurement epoch must include many cycles to minimize error.", null, "Figure 4 • Multiple Cycles with Expanded Histogram.\n\nCrest Factor\n\nOnce Average Power is known, Crest Factor (CF) can be calculated since Peak Power has already been determined. Crest Factor is normally reported in db. The following formula can be used to calculate Crest Factor if units are linear:\n\nCF(db) = 10log10 (Peak Power / Average Power)\n\nPulse Power\n\nPulse Power can be found by identifying the State 2 Mode (f2). Recall that while each bin contains the number of samples at a specific power level, it does represent a specific power level. Pulse Power is the power represented by the State 2 Mode (f2).\n\nDuty Factor\n\nDuty Factor (Duty Cycle) can be expressed as the percentage of one period in which the signal is active. Here is the basic formula for Duty factor:\n\nDF(%)= A/P * 100\n\nWhere DF is the duty Factor, A is the time the signal is active, and P is the total period the result is multiplied by 100 so that it is expressed as a percentage.\n\nCalculating Duty Factor requires locating the beginning and end of the pulse. This will be done using the 50% References. Refer to Figures 3 and 5. IEEE 181 specifies that the 50% Reference is Base State Power plus 50% of the difference between State 1 and State 2. This calculation removes the out-of-wave Offset Power from the timing calculation making it more accurate; then adds the Offset back in so that Power is expressed from the absolute reference. Refer to Figures 3 and 5. Note that in Figure 5 Left, the data above and below the 50% Reference appears balanced, this is because the pulse has a duty factor that is fairly close to 50%. Figure 5 Right depicts a pulse stream with similar power levels and a somewhat lower Duty Factor. This can be seen by the additional bin volume below the 50% Reference and less above, indicating that the signal was at State 1 (f1) for a longer period of time and State 2(f2) for less time.", null, "Figure 5 • Histograms with Mesial Line.\n\nAssuming the power levels have been converted to linear units, the following formula can be utilized to calculate the 50% Reference Power Level.\n\nP50%= Base State + (0.5*(State 2 - State 1))\n\nThis Power Level is recognized as the power at the leading and trailing pulse edges. Once this is known, Duty Factor can be calculated. To calculate Duty Factor, we won’t actually use the power level at the 50% Reference because the calculation will be done with bin counts. Since the 50% reference represents the leading and trailing edges of the pulse, all bin counts above the Reference represent time that the Pulse is Active.\n\nReferring to Figures 4 and 5, and the Duty Factor formula above, the total sample count inside the measurement epoch represents the time that that pulses were taken within, and is represented by P. The number of samples taken that lie above the 50% Reference represent the Active time and is representative of A. Note that while units of time are referenced with A and P, any such time units cancel out, leaving only the bin count, and the time information is irrelevant to the calculation. The sensor’s sampling rate could be utilized to associate actual time if desired, however it is unnecessary for the calculation of Duty Factor.\n\nFor example, if during the measurement window, 1,276,412 samples were taken, and it is determined that 437,567 lie above the 50% Reference, the Duty Factor is:\n\nDF = A/P*100 = 437,567/1,276,412*100 = 34.28%\n\nActual Measurement\n\nNow let’s see what a real power sensor can give us as a measurement. Using the pulse information and algorithms found in the IEEE specifications, the LB479A sensor makes economically accurate measurements such as those shown below.\n\nFigure 6 shows the pulse measurement using an LB479A power sensor. Note the Pulse Power in the main display and the Peak Power are nearly identical, indicating a pulse with a flat top and little overshoot or tilt (droop). The Duty Cycle is 11.11% and average power is -3.657 dBm. This measurement took very little time to setup and is highly accurate.", null, "Figure 6 • Statistical Pulse Measurement.\n\nFor comparison purposes, the image in Figure 7 was made with an LB480A power sensor with Option 004 on the same pulse stream. The measurement is edge triggered and trace based, plus it includes time related information such as pulse width. Note that the measurements from the two different sensors, while calculated in very different ways are essentially the same verifying that fast, economical, high accuracy pulse measurements can be achieved using statistical methods.", null, "Figure 7 • Pulse Profiling Display.\n\nConclusion\n\nCompact USB Power Sensors offer a variety of measurements including advanced analytical information. Using USB Power sensors also offers several distinct advantages over traditional Power Meter plus sensing Head technology. The small sensors are easily built into ATE equipment, use little power and they offer equal or greater accuracy. ATE system builders may benefit from Statistical Pulse measurements, which can provide easy to setup numeric measurements and can often eliminate the need for graphical analysis in manufacturing." ]

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[ "# Using csv module to read the data in Pandas\n\nThe so-called CSV (Comma Separated Values) format is the most common import and export format for spreadsheets and databases. There were various formats of CSV until its standardization. The lack of a well-defined standard means that subtle differences often exist in the data produced and consumed by different applications. These differences can make it annoying to process CSV files from multiple sources. For that purpose, we will use Python’s `csv` library to read and write tabular data in CSV format.\n\nCode #1: We will use `csv.DictReader()` function to import the data file into the Python’s enviornment.\n\n `# importing the csv module ` `import` `csv ` ` `  `# Now let's read the file named 'auto-mpg.csv' ` `# After reading as a dictionary convert ` `# it to Python's list ` `with ``open``(``'auto-mpg.csv'``) as csvfile: ` `    ``mpg_data ``=` `list``(csv.DictReader(csvfile)) ` ` `  `# Let's visualize the data ` `# We are printing only first three elements ` `print``(mpg_data[:``3``]) `\n\nOutput :", null, "As we can see the data is stored as a list of ordered dictionary. Let’s perform some operations on the data for better understanding.\n\nCode #2:\n\n `# Let's find all the keys in the dictionary ` `print``(mpg_data[``0``].keys) ` ` `  `# Now we would like to find out the number of  ` `# unique values of cylinders in the car in our dataset ` `# We create a set containing the cylinders value ` `unique_cyl ``=` `set``(data[``'cylinders'``] ``for` `data ``in` `mpg_data) ` ` `  `# Let's print the values ` `print``(unique_cyl) `\n\nOutput :", null, "", null, "As we can see in the output, we have 5 unique values of cylinders in our dataset.\n\nCode #3: Now let’s find out the value of average mpg for each value of cylinders.\n\n `# Let's create an empty list to store the values ` `# of average mpg for each cylinder ` `avg_mpg ``=` `[] ` ` `  `# c is the current cylinder size ` `for` `c ``in` `unique_cyl: ` `    ``# for storing the sum of mpg ` `    ``mpgbycyl ``=` `0` `    ``# for storing the sum of cylinder ` `    ``# in each category ` `    ``cylcount ``=` `0` ` `  `    ``# iterate over all the data in mpg ` `    ``for` `x ``in` `mpg_data: ` `        ``# Check if current value matches c ` `        ``if` `x[``'cylinders'``]``=``=` `c: ` `            ``# Add the mpg values for c ` `            ``mpgbycyl ``+``=` `float``(x[``'mpg'``]) ` `            ``# increment the count of cylinder ` `            ``cylcount ``+``=` `1` ` `  `    ``# Find the average mpg for size c ` `    ``avg ``=` `mpgbycyl``/``cylcount ` `    ``# Append the average mpg to list ` `    ``avg_mpg.append((c, avg)) ` ` `  `# Sort the list ` `avg_mpg.sort(key ``=` `lambda` `x : x[``0``]) ` ` `  `# Print the list ` `print``(avg_mpg) `\n\nOutput :", null, "As we can see in the output, the program has successfully returned a list of tuples containing the average mpg for each unique cylinder type in our dataset.\n\nMy Personal Notes arrow_drop_up", null, "Check out this Author's contributed articles.\n\nIf you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.\n\nPlease Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below.\n\nArticle Tags :\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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[ "# DSP - Understanding Formulas\n\nI am a beginner in DSP but get confused when some formulas like below are referred:\n\ncos(φt) · { A·cos(φt+Δ) } = ½·A·{ cos(Δ) + cos(2φt+Δ) }\nx(n) = Acos(ω0n+φ)\nT= 2kπ/ω\n...........\n\n\nThe problem is that I do not know what terms like 'φ' or How T is equal to 2kπ/ω.\n\nI know that I am lacking some elementary knowledge - but please let me know what are the topics / subjects I should hence first cover up such that when I read DSP related formulas, I understand them properly.\n\n• Maybe it's a good idea to start from some elementary book or doing online course in DSP instead of posting lot's of low-quality questions? – jojek Jul 3 '14 at 6:51\n• The formula that you have shown here has little to do with DSP but is a basic trigonometric identity. DSP is a fairly math intensive topic and it maybe a good idea brush up on Algebra and Calculus before diving in. You should at least be comfortable with complex numbers and phasor math, which are essential to DSP – Hilmar Jul 3 '14 at 12:02\n• Personally I feel no question is of low-quality as long as it clarifies one's doubt. Thanks for your valuable comment. I would start with Trig, Algebra, Calculus, Complex Numbers and Phasor Mathematics – Programmer Jul 3 '14 at 13:05\n\nFirst, the formulas by itself are from sine/cosine waves, so you should read a little about them first. Here.\n\nFor the rest, you only have to keep in mind the unities $\\phi$ in $\\frac{rad}{s}$, $\\Delta$ in $rad$ and finally in x(n) is a discrete sine-wave (or sampled) that is why the $n$ in the formula. It is basically the same thing as with the $t$ but in discrete, or sampled, time.\n\nThe $T$ stands for the Period of the wave. You can perform simply math there to check the following:\n\n$$T = \\frac{2k\\pi}{\\omega}$$ $$T = \\frac{1}{f}$$ $$\\omega = 2\\pi f= \\frac{2\\pi}{T}$$\n\nThen you seen why we can obtain:\n\n$$T = \\frac{2\\pi}{\\omega}$$\n\nFinally, the $k$ stands, normally, for each of the cycle the wave is repeated. Then you can have $k$ = 0, 1, 2, ..., N\n\nTo understand more how to work with this on DPS you only should read about Sine-waves and its representation. Basic knowledge in shifting and characteristics as Period, Frequency, Angular velocity and phase is a must.\n\nBests!\n\n• Thanks for the explanation, it gives me a good insight of how to approach to self educate DSP. Thanks a lot – Programmer Jul 3 '14 at 13:04" ]

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[ "### 7.1.2 CFG recognition using difference lists\n\nA more efficient implementation can be obtained by making use of difference lists. This is a sophisticated (and, once you've understood it, beautiful) Prolog technique that can be used for a variety of purposes. We won't discuss the idea of difference lists in any depth: we'll simply show how they can be used to rewrite our recognizer more efficiently.\n\nThe key idea underlying difference lists is to represent the information about grammatical categories not as a single list, but as the difference between two lists. For example, instead of representing a woman shoots a man as `[a,woman,shoots,a,man]` we might represent it as the pair of lists\n\n`[a,woman,shoots,a,man] [].  `\n\nThink of the first list as what needs to be consumed (or if you prefer: the input list), and the second list as what we should leave behind (or: the output list). Viewed from this (rather procedural) perspective the difference list\n\n`[a,woman,shoots,a,man] []. `\n\nrepresents the sentence a woman shoots a man because it says: If I consume all the symbols on the left, and leave behind the symbols on the right, I have the sentence I am interested in.\n\nThat is: the sentence we are interested in is the difference between the contents of these two lists.\n\nDifference representations are not unique. In fact, we could represent a woman shoots a man in infinitely many ways. For example, we could also represent it as\n\n`[a,woman,shoots,a,man,ploggle,woggle]  [ploggle,woggle].`\n\nAgain the point is: if we consume all the symbols on the left, and leave behind the symbols on the right, we have the sentence we are interested in.\n\nThat's all we need to know about difference lists to rewrite our recognizer. If we bear the idea of `consuming something, and leaving something behind' in mind', we obtain the following recognizer:\n\n`s(X,Z) :- np(X,Y), vp(Y,Z). np(X,Z) :- det(X,Y), n(Y,Z). vp(X,Z) :-  v(X,Y), np(Y,Z). vp(X,Z) :-  v(X,Z). det([the|W],W).det([a|W],W). n([woman|W],W).n([man|W],W). v([shoots|W],W).`\n\nThe `s` rule says: I know that the pair of lists `X` and `Z` represents a sentence if (1) I can consume `X` and leave behind a `Y`, and the pair `  X` and `Y` represents a noun phrase, and (2) I can then go on to consume `Y` leaving Z behind, and the pair `Y` `Z` represents a verb phrase.\n\nThe idea underlying the way we handle the words is similar. The code\n\n`n([man|W],W).`\n\nmeans we are handling man as the difference between `[man|W]` and `W`. Intuitively, the difference between what I consume and what I leave behind is precisely the word `man`.\n\nNow, at first this is probably harder to grasp than our previous recognizer. But we have gained something important: we haven't used `append`. In the difference list based recognizer, they simply aren't needed, and as we shall see, this makes a big difference.\n\nHow do we use such grammars? Here's how to recognize sentences:\n\n`s([a,woman,shoots,a,man],[]).yes`\n\nThis asks whether we can get an `s` by consuming the symbols in `[a,woman,shoots,a,man]`, leaving nothing behind.\n\nSimilarly, to generate all the sentences in the grammar, we ask\n\n`s(X,[]).`\n\nThis asks: what values can you give to `X`, such that we get an `s` by consuming the symbols in `X`, leaving nothing behind?\n\nThe queries for other grammatical categories also work the same way. For example, to find out if a woman is a noun phrase we ask:\n\n`np([a,woman],[]).`\n\nAnd we generate all the noun phrases in the grammar as follows:\n\n`np(X,[]).`\n\nYou should trace what happens when this program analyses a sentence such as a woman shoots a man. As you will see, it is a lot more efficient than our `append` based program. Moreover, as no use is made of `append`, the trace is a lot easier to grasp. So we have made a big step forward.\n\nOn the other hand, it has to be admitted that the second recognizer is not as easy to understand, at least at first, and it's a pain having to keep track of all those difference list variables. If only it were possible to have a recognizer as simple as the first and as efficient as the second. And in fact, it is possible: this is where DCGs come in.\n\nPatrick Blackburn, Johan Bos and Kristina Striegnitz\nVersion 1.2.5 (20030212)" ]

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[ "Categories\n\n# Converting Types in Go\n\nWhile programming, we often need type conversions. If you come from a language that usually does implicit conversions, converting types in Go can be a bit of a hassle.\n\nFirst of all, let’s start with some basics. Go is a statically and strongly typed programming language. This means that, variable types must be known at compile time. Additionally, type conversions must be made explicitly.\n\n### Type Conversions\n\nIn Go, a simple type conversion can be performed in a form like:\n\n``````\nt = TypeToConvert(Value)\n``````\n\nAs you can see in the examples below, all conversions are done explicitly and exactly to the type we want. We can easily convert numeric types as long as they represent each other. However, keep in mind that precision loss may occur in these transformations.\n\n``````\na := 5;\nfmt.Printf(\"type: %T, value: %v \\n\", a,a)\n\nb := float32(a);\nfmt.Printf(\"type: %T, value: %v \\n\", b,b)\n\nc := 5.5\nfmt.Printf(\"type: %T, value: %v \\n\", c,c)\n\nd := int(c)\nfmt.Printf(\"type: %T, value: %v \\n\", d,d)\n``````\n``````\ntype: int, value: 5\ntype: float32, value: 5\ntype: float64, value: 5.5\ntype: int, value: 5\n``````\n\n### How about conversions between int and string?\n\nIn Go, int and string types do not directly represent each other. For this reason, we cannot apply the method we have just done.\n\n``````\nx:= int(\"5\")\n``````\n``````\ncannot convert \"5\" (untyped string constant) to type int\n``````\n\nFortunately, strconv package provides built-in methods for such conversions.\n\n``````\nx := \"5\"\nfmt.Printf(\"type: %T, value: %v \\n\", x,x)\n\ny, _ := strconv.Atoi(x) // (ascii string to int)\nfmt.Printf(\"type: %T, value: %v \\n\", y,y)\n\nz := strconv.Itoa(y) // (int to ascii string)\nfmt.Printf(\"type: %T, value: %v \\n\", z,z)\n``````\n``````\ntype: string, value: 5\ntype: int, value: 5\ntype: string, value: 5\n``````\n\nGo is fairly strict about types, for example you can’t mix numeric types in an expression. You can’t even mix 32-bit type int and int32. This strictness prevents many bugs from occurring, but requires you to explicitly make the necessary conversions at all times.\n\n### Untyped constants?\n\nWhen it comes to constants, Go introduced a concept called untyped constants. This means that if you create a constant that does not have an explicit type, it will not have a fixed type unless a defined type is given.\n\n``````\nconst a = \"Hi\" //no explicit type is defined\nfmt.Printf(\"type: %T, value: %v \\n\", a,a)\n\nvar b string = a // b has a defined type string\nfmt.Printf(\"type: %T, value: %v \\n\", b,b)\n\ntype NewString string\nvar c NewString = a // c has a defined type NewString\nfmt.Printf(\"type: %T, value: %v \\n\", c,c)\n``````\n``````\ntype: string, value: Hi\ntype: string, value: Hi\ntype: main.NewString, value: Hi\n``````\n\nIn the example above, although const a has a default type of string(determined by its syntax), it does not have a fixed type. This allows you to assign it to another variable with a defined type. Of course, if it had a predefined type, you couldn’t assign it to any other type without doing the necessary conversion.\n\n``````\ntype NewString string\nconst a NewString = \"Hi\"\nfmt.Printf(\"type: %T, value: %v \\n\", a,a)\n//var b string = a // does not compile\nvar b string = string(a) // conversion is needed\nfmt.Printf(\"type: %T, value: %v \\n\", b,b)\n``````\n``````\ntype: main.NewString, value: Hi\ntype: string, value: Hi\n``````\n\nOnce a constant has a defined type, the rule of not mixing different types continues to apply.\n\nIn the example below, const a is untyped, but is inferred as NewString when creating c . On the other hand, d has a defined typed of string that cannot be mixed with other types such as NewString.\n\n``````\nconst a = \"Ha\"\nfmt.Printf(\"type: %T, value: %v \\n\", a,a)\n\ntype NewString string\nconst b NewString = \"Ha\"\n\nvar c = b + a // a is untyped and inferred as NewString\nfmt.Printf(\"type: %T, value: %v \\n\", c,c)\n\nconst d string = \"Ha\" // d has a defined type string\n// var e = b + d // does not compile, type Mismatch\n``````\n``````\ntype: string, value: Ha\ntype: main.NewString, value: HaHa\n``````\n\n### Type Assertions\n\nWhen converting interface value’s to the underlying concrete types, Go provides us with a mechanism called type assertions. The following form is used for such assertions.\n\n``````\nt,conversionResult := interface.(TypeToConvert)\n``````\n\nIf you have used java before, you can think of it like the “instanceOf” method.\n\nIn the following example, since interface i has a concrete type of A, we can convert it to A by assertion.\n\n``````\ntype A struct{\nName string\n}\n\nvar i interface{} = A{Name:\"myName\"}\nfmt.Printf(\"type: %T, value: %v \\n\", i,i)\n\nif a,ok := i.(A); ok{ // can be converted to A\nfmt.Printf(\"type: %T, value: %v \\n\", a,a.Name)\n}\n\nif aStr,ok := i.(string); ok{ //cannot convert to string\nfmt.Printf(\"type: %T, value: %v \\n\", aStr,aStr)\n}else{\nfmt.Println(\"error\")\n}\n``````\n``````\ntype: main.A, value: {myName}\ntype: main.A, value: myName\nerror\n``````\n\nAs a result, in Go, we use type conversion for basic types and type assertion for interfaces. Additionally, some built-in methods allow us to convert types those do not directly represent each other. In conversions of constants, it may be necessary to pay attention to whether the type is explicitly defined or not." ]

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[ "# Getting Started\n\nHere you will find both basic and advanced tutorials to help you get started using Piccolo2D.  All tutorials provide examples in both Java and C#.  This section assumes you have read Piccolo2D Patterns and have a basic understanding of the concepts presented there.\n\nShow examples in", null, "Java", null, "C#\n\n# Graph Editor\n\nThis tutorial will illustrate how you might build a graph editor interface using Piccolo2D.  Moving the mouse over a node will highlight that node.  And, dragging a node will move that node around, keeping the edges connected.", null, "Play with the interface.\n\n## Overall Architecture\n\nIn this example, we will use the `PPath` node provided by Piccolo2D for both our nodes and edges.  Rather than creating our own data structure to keep track of the connections, we will use `PNode`'s `Tag` property in .NET and its `getAttribute` and `addAttribute` methods in Java.  Both of these mechanisms provide the ability to attach extra information to a node without subclassing it.  For each node in the graph, we will attach a list of its connected edges.  And for each edge in the graph, we will attach a list of its connected nodes.\n\nWe will then create a reusable GraphEditor component that extends PCanvas. This component will add the nodes and edges to the scene-graph.  We will also create an event handler class to drag the nodes and update the edges.  Finally, we will create a wrapper window called GraphEditorTester and add our new component to the window.\n\n## 1. Create a Graph Canvas\n\nWe will make a reusable component that extends PCanvas and initializes the graph.\n\n1. We extend PCanvas and add some random nodes and edges.  Add the following class to your project.\n\n```public class GraphEditor extends PCanvas {\npublic GraphEditor(int width, int height) {\nsetPreferredSize(new Dimension(width, height));\nint numNodes = 50;\nint numEdges = 50;\n\n// Initialize, and create a layer for the edges\n// (always underneath the nodes)\nPLayer nodeLayer = getLayer();\nPLayer edgeLayer = new PLayer();\nRandom random = new Random();\n\n// Create some random nodes\n// Each node's attribute set has an\n// ArrayList to store associated edges\nfor (int i = 0; i < numNodes; i++) {\nfloat x = random.nextInt(width);\nfloat y = random.nextInt(height);\nPPath node = PPath.createEllipse(x, y, 20, 20);\n}\n\n// Create some random edges\n// Each edge's attribute set has an\n// ArrayList to store associated nodes\nfor (int i = 0; i < numEdges; i++) {\nint n1 = random.nextInt(numNodes);\nint n2 = random.nextInt(numNodes);\n\n// Make sure we have two distinct nodes.\nwhile (n1 == n2) {\nn2 = random.nextInt(numNodes);\n}\n\nPNode node1 = nodeLayer.getChild(n1);\nPNode node2 = nodeLayer.getChild(n2);\nPPath edge = new PPath();\nupdateEdge(edge);\n}\n\n// Create event handler to move nodes and update edges\n}\n\npublic void updateEdge(PPath edge) {\n// Note that the node's \"FullBounds\" must be used\n// (instead of just the \"Bounds\") because the nodes\n// have non-identity transforms which must be included\n// when determining their position.\n\nPNode node1 = (PNode) ((ArrayList)edge.getAttribute(\"nodes\")).get(0);\nPNode node2 = (PNode) ((ArrayList)edge.getAttribute(\"nodes\")).get(1);\nPoint2D start = node1.getFullBoundsReference().getCenter2D();\nPoint2D end = node2.getFullBoundsReference().getCenter2D();\nedge.reset();\nedge.moveTo((float)start.getX(), (float)start.getY());\nedge.lineTo((float)end.getX(), (float)end.getY());\n}\n}\n```\n```public class GraphEditor : PCanvas {\npublic GraphEditor(int width, int height) {\nthis.Size = new Size(width, height);\nint numNodes = 50;\nint numEdges = 50;\n\n// Initialize, and create a layer for the edges\n// (always underneath the nodes)\nPLayer nodeLayer = this.Layer;\nPLayer edgeLayer = new PLayer();\nRandom rnd = new Random();\n\n// Create some random nodes\n// Each node's Tag has an ArrayList\n// used to store associated edges\nfor (int i=0; i<numNodes; i++) {\nfloat x = (float)(this.ClientSize.Width * rnd.NextDouble());\nfloat y = (float)(this.ClientSize.Height * rnd.NextDouble());\nPPath path = PPath.CreateEllipse(x, y, 20, 20);\npath.Tag = new ArrayList();\n}\n\n// Create some random edges\n// Each edge's Tag has an ArrayList\n// used to store associated nodes\nfor (int i=0; i<numEdges; i++) {\nint n1 = rnd.Next(numNodes);\nint n2 = n1;\n\n// Make sure we have two distinct nodes.\nwhile (n2 == n1) {\nn2 = rnd.Next(numNodes);\n}\n\nPNode node1 = nodeLayer[n1];\nPNode node2 = nodeLayer[n2];\nPPath edge = new PPath();\nedge.Tag = new ArrayList();\nUpdateEdge(edge);\n}\n\n// Create event handler to move nodes and update edges\n}\n\npublic static void UpdateEdge(PPath edge) {\n// Note that the node's \"FullBounds\" must be used\n// (instead of just the \"Bounds\") because the nodes\n// have non-identity transforms which must be included\n// when determining their position.\n\nArrayList nodes = (ArrayList)edge.Tag;\nPNode node1 = (PNode)nodes;\nPNode node2 = (PNode)nodes;\nPointF start = PUtil.CenterOfRectangle(node1.FullBounds);\nPointF end = PUtil.CenterOfRectangle(node2.FullBounds);\nedge.Reset();\n}\n}\n\n```\n\nFirst we create some randomly positioned ellipses, for our nodes.  Next, we need to attach a list to each node, to store the connected edges.  Piccolo2D.NET provides the ability to add an object reference as a tag whereas Piccolo2D.Java uses a set of named attributes instead.  In the Java version, we use `addAttribute` to attach an ArrayList.  In the .NET version, we set each node's `Tag` property to an ArrayList.\n\nNext, we create some lines for our edges. Note, we use a separate layer added underneath the main layer, to insure that the nodes are always on top.  For each edge, we randomly choose two nodes to connect.  We add the edge to each node's list of edges and we add the nodes to the edge's list of nodes.  The `UpdateEdge` method is called to position the line's endpoints to the center points of each node.\n\nFinally, we add an event listener, defined below, to make our graph interactive.\n\n## 2. Create a Drag Event Handler\n\nWe will create an event listener class to handle all of the interaction of our application, including highlighting and dragging nodes.\n\n1. We extend `PDragSequenceEventHandler` to create an event listener that performs some operation during a drag sequence.  Add the following code to your project.  For the Java version, you should add the anonymous event listener class to the constructor.\n\n```nodeLayer.addInputEventListener(new PDragEventHandler() {\n{\nPInputEventFilter filter = new PInputEventFilter();\nsetEventFilter(filter);\n}\n\npublic void mouseEntered(PInputEvent e) {\nsuper.mouseEntered(e);\nif (e.getButton() == MouseEvent.NOBUTTON) {\ne.getPickedNode().setPaint(Color.RED);\n}\n}\n\npublic void mouseExited(PInputEvent e) {\nsuper.mouseExited(e);\nif (e.getButton() == MouseEvent.NOBUTTON) {\ne.getPickedNode().setPaint(Color.WHITE);\n}\n}\n\nprotected void startDrag(PInputEvent e) {\nsuper.startDrag(e);\ne.setHandled(true);\ne.getPickedNode().moveToFront();\n}\n\nprotected void drag(PInputEvent e) {\nsuper.drag(e);\n\nArrayList edges = (ArrayList) e.getPickedNode().getAttribute(\"edges\");\nfor (int i = 0; i < edges.size(); i++) {\nGraphEditor.this.updateEdge((PPath) edges.get(i));\n}\n}\n});\n```\n```class NodeDragHandler : PDragEventHandler {\npublic override bool DoesAcceptEvent(PInputEventArgs e) {\nreturn e.IsMouseEvent &&\n(e.Button != MouseButtons.None || e.IsMouseEnterOrMouseLeave);\n}\n\npublic override void OnMouseEnter(object sender, PInputEventArgs e) {\nbase.OnMouseEnter (sender, e);\nif (e.Button == MouseButtons.None) {\ne.PickedNode.Brush = Brushes.Red;\n}\n}\n\npublic override void OnMouseLeave(object sender, PInputEventArgs e) {\nbase.OnMouseLeave (sender, e);\nif (e.Button == MouseButtons.None) {\ne.PickedNode.Brush = Brushes.White;\n}\n}\n\nprotected override void OnStartDrag(object sender, PInputEventArgs e) {\nbase.OnStartDrag(sender, e);\ne.Handled = true;\ne.PickedNode.MoveToFront();\n}\n\nprotected override void OnDrag(object sender, PInputEventArgs e) {\nbase.OnDrag (sender, e);\n\nArrayList edges = (ArrayList)e.PickedNode.Tag;\nforeach (PPath edge in edges) {\nGraphEditor.UpdateEdge(edge);\n}\n}\n}\n\n```\n\nWhen the mouse enters a node, we set the node's fill color to red to highlight the node.  When the mouse leaves a node, we set it's fill color back to white.  We only do this when no mouse buttons are pressed because we don't want to highlight a node when the mouse is dragged overtop of it.  And, we consume the drag event, so that we will not pan when a node is dragged.\n\nWe also need to update the edges.  During a drag sequence, we iterate over the dragged node's edges and call `UpdateEdge` on each one.  Recall this method gets the two nodes each edge connects, and then repositions the edge to the center points of each connected node.\n\n## 3. Add the Canvas to a Window\n\nNow we are ready to add our new component to the window.\n\n1. We create a `JFrame` in Java or a `Form` in .NET as a wrapper for our component. Add the following class to your project.\n\n```public class GraphEditorTester extends JFrame {\npublic GraphEditorTester() {\nsetTitle(\"Piccolo2D Graph Editor\");\nsetDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);\nGraphEditor graphEditor = new GraphEditor(500, 500);\npack();\nsetVisible(true);\n}\n\npublic static void main(String args[]) {\nnew GraphEditorTester();\n}\n}\n```\n```public class GraphEditorTester : System.Windows.Forms.Form {\npublic GraphEditorTester() {\nGraphEditor graphEditor = new GraphEditor(this.ClientSize.Width,\nthis.ClientSize.Height);" ]

[ null, "http://piccolo2d.org/learn/images/check.png", null, "http://piccolo2d.org/learn/images/check.png", null, "http://piccolo2d.org/learn/images/grapheditor.jpg", null ]

[ "# Help on contour integral on another answer on this site\n\nDo you mind expanding on the part along the diagonal in the first answer by Robjohn to this question proof? Particularly how to achieve (3).\n\nI am trying use a parameterization for $z=xe^{i \\pi/4}$ for x from $0$ to R and I am not getting the correct result.\n\n$$\\int_{diagonal}e^{-z^2}=\\int_0^R e^{-(xe^{i \\pi/4})^2}e^{i\\pi/4}dx=\\int_0^R e^{-x^2e^{i \\pi/2}}e^{i\\pi/4}dx$$\n\n• $e^{i\\pi/2}=i$ and the integral is from $Re^{i\\pi/4}$ to 0 so you have to change the sign? – Bob Jul 20 '18 at 20:14\n• Oh you might be right. Let me take a look at that in a few. Thanks. :) – MathIsHard Jul 20 '18 at 20:39\n• Thank you. I see it now. I appreciate the help. – MathIsHard Jul 20 '18 at 20:50\n• One of you should post that as an answer so that the question doesn't remain unanswered. – joriki Jul 20 '18 at 23:06\n• Yes, please post that Bob or let me know if you want me to do it. Thanks! – MathIsHard Jul 20 '18 at 23:34\n\n## 1 Answer\n\nParametrize the segment from $0$ to $R e^{i\\pi / 4}$ by: $$\\gamma :[0,R]\\rightarrow \\mathbb{C}, t\\mapsto te^{i(\\pi/4)}.$$ Since you want to integrate from $Re^{i\\pi/4}$ to $0$ and not from $0$ to $Re^{i\\pi/4}$, you have to switch the orientation of this curve, with the result that the integral changes sign, so the value of the integral you're looking for is $$-\\int_\\gamma e^{-z^2}dz= -\\int_0 ^ R e^{-\\gamma(t)^2}\\gamma'(t)dt = -\\int_0 ^{R} e^{-(te^{i\\pi/4})^2}e^{i\\pi/4}dt=\\\\-e^{i\\pi/4}\\int_0 ^{R} e^{-t^2e^{i\\pi/2}}dt=-e^{i\\pi/4}\\int_0 ^{R} e^{-it^2}dt,$$ where we used the well known fact that $e^{i\\pi/2}=i.$ Now, letting $R\\rightarrow\\infty$ you get the result claimed in the linked answer.\n\n• Thanks Bob :) appreciate it. – MathIsHard Jul 21 '18 at 5:03" ]

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[ "# What is Electric Current: the basics\n\n### Electric current results when electric charges move - these may be negatively charged electrons or positive charge carriers - positive ions.\n\nElectric Current Tutorial Includes:\nWhat is electric current     Current unit - Ampere     AC & DC\n\nElectric current is one of the most basic concepts that exists within electrical and electronic science - electric current is at the core of the science of electricity.\n\nWhether it is an electrical heater, a large electrical grid system, a mobile phone, computer, remote sensor node or whatever, the concept of electrical current is central to its operation.\n\nHowever current as such cannot normally be seen, although its effects can be seen, heard and felt all the time, and as a result it is sometimes difficult to gain a view of what it really is.", null, "Lightning strike is an impressive show of electrical current flow Photo taken from top of Petronas Towers in Kuala Lumpur Malaysia\n\n## Electrical current definition\n\n### Electric current definition:\n\nAn electric current is a flow of electric charge in a circuit. More specifically, the electric current is the rate of charge flow past a given point in an electric circuit. The charge can be negatively charged electrons or positive charge carriers including protons, positive ions or holes.\n\nThe magnitude of the electric current is measured in coulombs per second, the common unit for this being the Ampere or amp which is designated by the letter ‘A’.\n\nThe Ampere or amp is widely used within electrical and electronic technology along with the multipliers like milliamp (0.001A), microamp (0.000001A), and so forth.\n\nRead more about . . . . the Ampere, the unit of current.\n\nCurrent flow in a circuit is normally designated by the letter ‘I’, and this letter is used in equations like Ohms law where V=I⋅R.\n\n## What is electric current: the basics\n\nThe basic concept of current is that it is the movement of electrons within a substance. Electrons are minute particles that exist as part of the molecular structure of materials. Sometimes these electrons are held tightly within the molecules and other times they are held loosely and they are able to move around the structure relatively freely.\n\nOne very important point to note about the electrons is that they are charged particles - they carry a negative charge. If they move then an amount of charge moves and this is called current.\n\nIt is also worth noting that the number of electrons that able to move governs the ability of a particular substance to conduct electricity. Some materials allow current to move better than others.\n\nThe motion of the free electrons is normally very haphazard - it is random - as many electrons move in one direction as in another and as a result there is no overall movement of charge.\n\nIf a force acts on the electrons to move them in a particular direction, then they will all drift in the same direction, although still in a somewhat haphazard fashion, but there is an overall movement in one direction.\n\nThe force that acts on the electrons is called and electromotive force, or EMF, and its quantity is voltage measured in volts.\n\nTo gain a little more understanding about what current is and how it acts in a conductor, it can be compared to water flow in a pipe. There are limitations to this comparison, but it serves as a very basic illustration of current and current flow.\n\nThe current can be considered to be like water flowing through a pipe. When pressure is placed on one end it forces the water to move in one direction and flow through the pipe. The amount of water flow is proportional to the pressure placed on the end. The pressure or force placed on the end can be likened to the electro-motive force.\n\nWhen the pressure is applied to the pipe, or the water is allowed to flow as a result of a tap being opened, then the water flows virtually instantaneously. The same is true for the electrical current.\n\nTo gain an idea of the flow of electrons, it takes 6.24 billion, billion electrons per second to flow for a current of one ampere.\n\n## Conventional current and electron flow\n\nThere is often a lot of misunderstanding about conventional current flow and electron flow. This can be a little confusing at first but it is really quite straightforward.\n\nThe particles that carry charge along conductors are free electrons. The electric field direction within a circuit is by definition the direction that positive test charges are pushed. Thus, these negatively charged electrons move in the direction opposite the electric field.\n\nThis came about because the initial investigations in static and dynamic electric currents was based upon what we would now call positive charge carriers. This meant that then early convention for the direction of an electric current was established as the direction that positive charges would move. This convention has remained and it is still used today.\n\nIn summary:\n\n• Conventional current flow:   The conventional current flow is from positive to the negative terminal and indicates the direction that positive charges would flow.\n• Electron flow:   The electron flow is from negative to positive terminal. Electrons are negatively charged and are therefore attracted to the positive terminal as unlike charges attract.\n\nThis is the convention that is used globally to this day, even though it may seem a little odd and out-dated.\n\n## Speed of electron or charge movement\n\nThe speed of the transmission of electrical current is very different to that of the speed of the actual electron movement. The electron itself bounces around in the conductor, and possibly only makes progress along the conductor at the rate of a few millimetres a second. This means that in the case of alternating current, where the current changes direction 50 or 60 times per second, most of the electrons never make it out of the wire.\n\nTo take a different example, in the near-vacuum inside a cathode ray tube, the electrons travel in near-straight lines at about a tenth of the speed of light.\n\n## Effects of current\n\nWhen an electric current flows through a conductor there are a number of signs which tell that a current is flowing.\n\n• Heat is dissipated:   Possibly the most obvious is that heat is generated. If the current is small then the amount of heat generated is likely to be very small and may not be noticed. However if the current is larger then it is possible that a noticeable amount of heat is generated. An electric fire is a prime example showing how a current causes heat to be generated. The actual amount of heat is governed not only be the current, but also be the voltage and the resistance of the conductor.\n• Magnetic effect:   Another effect which can be noticed is that a magnetic field is built up around the conductor. If a current is flowing in conductor then it is possible to detect this. By placing a compass close to a wire carrying a reasonably large direct current, the compass needle can be seen to be deflect. Note this will not work with mains because the field is alternating too fast for the needle to respond and the two wires (live and neutral) close together in the same cable will cancel out the field.\n\nThe magnetic field generated by a current is put to good use in a number of areas. By winding a wire into a coil, the effect can be increased, and an electro-magnet can be made. Relays and a host of other items use the effect. Loudspeakers also use a varying current in a coil to cause vibrations to occur in a diaphragm which enable the electronic currents to be converted into sounds.\n\n## How to measure current\n\nOne important aspect of current is knowing the amount of current that may be flowing in a conductor. As electric current is such a key factor in electrical and electronic circuits, knowing what current is flowing is very important.\n\nThere are many different ways is measuring current. One of the easiest is to use a multimeter.\n\n#### How to measure current with a DMM:\n\nUsing a DMM, digital multimeter it is easy to measure current by placing the DMM actually in the circuit carrying the current. The DMM will then give an accurate reading of the current flowing in the circuit\n\nAlthough there are other methods of measuring current, this is the most common.\n\nCurrent is one of the most important and fundamental elements within electrical and electronic technology. The current flowing in a circuit can be used in a variety of ways from generating heat to causing circuits to switch, or information to be stored in an integrated circuit.\n\nMore Basic Electronics Concepts & Tutorials:\nVoltage     Current     Power     Resistance     Capacitance     Inductance     Transformers     Decibel, dB     Kirchoff's Laws     Q, quality factor     RF noise" ]

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[ "# Elastic solution of axisymmetric thick truncated conical shells based on first-order shear deformation theory\n\n• G. H. Rahimi\n\n## Abstract\n\nIn this paper, based on first-order shear deforma-tion theory (FSDT), and the virtual work principle, the differential equations governing axisymmetric thick trun-cated conical shells have been derived. The governing equations are a system of ordinary differential equations with variable coefficients. Using the matched asymptotic method (MAM) of the perturbation theory, these equations could be converted into a system of algebraic equations and two systems of differential equations with constant coefficients. The derived systems of equations have closed form solutions. For different conical angles, displacements and stresses along the radius and length have been plotted." ]

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[ "# Tag Info\n\n48\n\nAs I see it, the key intuition is passing from the equator orthogonal to a single vector to looking at a whole orthonormal basis. Suppose we pick a random unit vector $(x_1,\\dots,x_n)$. What we want to know is why $x_1$ is probably near zero, since this is equivalent to being near the equator relative to the first basis vector. But this feels intuitively ...\n\n42\n\nThis question was explored here: Lenhart, William J., and Sue H. Whitesides. \"Reconfiguring closed polygonal chains in Euclidean $d$-space.\" Discrete & Computational Geometry 13, no. 1 (1995): 123-140; DOI: 10.1007/BF02574031, eudml. From the Abstract: \"It is shown that in three or more dimensions, reconfiguration is always possible, but that in ...\n\n30\n\nJames Wenk and I just finished a paper proving Zalgaller's sphere inspection conjecture for closed curves: Shortest closed curve to inspect a sphere. We show that in $R^3$ any closed curve $\\gamma$ which inspects the unit sphere $S^2$, i.e., lies outside $S^2$ and contains $S^2$ within its convex hull, has length $L(\\gamma)\\geq 4\\pi$. Equality holds only ...\n\n27\n\nThat isn't a conjecture but a routine exercise assigned after the students learn about Bang's solution of the Tarski plank problem. The proof goes in 2 steps: 1) Consider all sums $\\sum_j \\varepsilon_i u_i$ with $\\varepsilon_i=\\pm 1$ and choose the longest one. Replacing some $u_j$ with $-u_j$ if necessary, we can assume WLOG that it is $y=\\sum_i u_i$. ...\n\n25\n\nIn dimensions $3$ or greater, it is false. Take some non-coplanar points. You can connect them with a nonconvex polyhedron with arbitrarily small surface area by thickening a spanning tree, for example. The convex hull has surface area at least as great as the surface area of the convex hull of the points. In dimension $2$, it is true. In two dimensions, ...\n\n25\n\nThere has been a bunch of work along these lines, and I think the idea has been rediscovered several times. I suggest looking at the papers of Anna Romanowska, who refers to them as \"barycentric algebras\", to get an idea of what's known. Her book with Smith, \"Modes\", covers this as well as generalizations where $t$ is not required to be in $[0,1]$. Here ...\n\n24\n\nLet me address the specific complaint of that review. The situation is the following. Our (bounded, open) convex set is denoted $K\\subseteq\\mathbb R^n$ with closure $\\overline K$, and we consider the \"distance to $p$\" function $d_p:\\partial K\\to\\mathbb R$ for $p\\in\\overline K$. Let $V\\subseteq\\overline K$ be the set of $p\\in\\overline K$ for which $d_p$ ...\n\n23\n\nThe numbers that are not of the form $a^2-b^2-c^2$ with $b$ and $c$ positive and $a>b+c$ are precisely the idoneal numbers apart from $7$, $28$, $112$, $15$, $60$, and $240$. As noted in the problem, the paper by Hertel and Richter shows that the numbers not of this form are necessarily idoneal numbers, and a quick calculation shows that $7$, $28$, $112$,...\n\n23\n\nVisualize first, for comparison, the 2-dimensional unit sphere in 3-dimensional Euclidean space (something that I can visualize!), and imagine it cut, by circles of latitude (perpendicular to the $z$-axis), into narrow zones. Of course, the zones closer to the poles have smaller radii, and therefore smaller circumferences, than the zones near the equator. ...\n\n23\n\nConsider the sphere with equator 4. Divide it into spherical cubes, the central projections from an inscribed cube. Note that the exponential map from tangent plane to the sphere is short. Note also that if one maps a unit cube centered at the origin by the exponential map it will cover the spherical cube. It is easy to modify the map to get a short ...\n\n21\n\nThe following proposition answers OP's question regarding the upper bound of $$\\tau(C) \\Doteq f(C)/\\lambda^2(C).$$ Let $B_n$ be the closed Euclidean unit ball of $\\mathbb{R^n}$ centred at $0$, that is $$B_n = \\{ (x_1,\\dots, x_n) \\in \\mathbb{R^n} \\, \\vert \\,\\, x_1^2 + \\cdots + x_n^2 \\le 1\\},$$ and let $\\tau_n = \\tau(B_n)$. Proposition. The Euclidean ...\n\n20\n\nI have recently finished a paper called The length, width, and inradius of space curves where it is shown that the length $L$ of any closed curve $\\gamma\\colon[a,b]\\to \\mathbf{R}^3$ inspecting the unit sphere $\\mathbf{S}^2$ must be at least $$6\\sqrt{3}\\approx 10.3923,$$ which is almost $83$% of the conjectured lower bound $4\\pi\\approx 12.5664$ by ...\n\n19\n\nThis fails already for $d=3$. Consider a tetrahedron, e.g. the convex hull of the points $v_1,v_2,v_3,v_4$. Let $K$ be the closed subset consisting of $\\sum_{i=1}^4 a_i v_i$ with $\\sum_{i=1}^4 a_i=1$ and $a_3 +a_4 \\leq 1/2-\\epsilon$. Let $L$ be defined similarly, but $a_1+a_2 \\leq 1/2-\\epsilon$. Clearly the convex hull of $K$ and $L$ is this tetrahedron. ...\n\n17\n\nYes, if the convex body is \"sufficiently round\". If it is not, the resulting \"closeness\" to the boundary of a convex set is in absolute terms rather than relative. I don't know whether it can be improved, but Bill Johnson's remarks suggest that it can't. Let $X=\\{x_i\\}$, $i=1,\\dots,k$, be the set in question and $K$ the convex body whose boundary contains $... 17 For$n=3$this question was asked in 1996 by James Propp, conjecturing that the answer is Yes. (I got the reference from this page, which concerns the very special case of a rectangular box in${\\bf R}^3$; this is already a nontrivial problem, as Jim Propp noted, and the cuboid page reports calculations claimed to prove it in that case and to locate the ... 17 This won't be a complete answer to Q2, but something of a starting point, at least for all two-dimensional convex$P$. Assuming for simplicity that the area of$P$is 1,$P$contains a parallelogram of area at least$1/2$(triangular$P$-s are extreme in this respect). This is quite easy to prove, and for polygonal$P$, an algorithm can be produced to find ... 16 No, the Loewner ellipsoid is not monotone w.r.t. inclusion. Let$K$be a square, whose Loewner ellipsoid is its circumcircle. Let$L$be any other ellipse through the four vertices of$K$. The Loewner ellipsoid of$L$is$L$itself but it does not contain the circle. (I assume that the Loewner ellipsoid is the minimal circumscibed one. If you meant the ... 16 I don't understand your reference to the model (since the geometry of the hyperbolic plane does not depend on any model), but, in fact, the Beltrami-Klein model demonstrates that any qualitative statement about convex sets in the Euclidean plane holds in the Hyperbolic plane and vice versa, since the model maps convex sets to convex sets. EDIT This has (... 15 A convex body$K\\subset\\mathbb{R}^n$all of whose$(n-1)$-dimensional projections have the same$n-1$-content is known as a body of constant brightness, by analogy with bodies of constant width. The theory is very similar to that of bodies of constant width. The surface area measure$dS_K(\\mathbf{x})$takes the place of the support function$h_K$. The ... 15 The answer seems to be$\\frac{1}{2\\pi}$, using a semi circle. See Moran, P. A. P. \"On a problem of S. Ulam.\" Journal of the London Mathematical Society 1.3 (1946): 175-179. 15 Take iid Gaussian random variables$X_1,\\ldots,X_d$with mean$0$and variance$1/d$. Normalizing the vector$X=(X_1,\\ldots,X_d)$will produce a random point on the unit sphere, but it's already close to having unit norm, so we will avoid this for the sake of intuition. For each unit vector$v$, there is an equator given by $$\\{x\\in S^{d-1}:\\langle x,v\\... 15 [This is an attempt to explain the details in Anton Petrunin's answer to this question, since the comments suggest that a number of people have found it hard to understand as it was written.] Let C be the surface of the unit cube in \\mathbb{R}^d (for d\\geq 2); when considered as a metric space, it is endowed with the distance-on-the-surface: so the ... 14 I think the identity you want is$$2\\inf_x f(x)=\\inf_x(f^\\ast(x)+f^\\ast(-x))\\mbox{.}$$(I'm skipping a bunch of conditions required of f to make this hold. We'll need convexity at least.) Let's use \\oplus for infimal convolution and let g(x)=f(-x). By definition (f\\oplus g)(x)=\\inf_y(f(x-y)+g(y)). Infimal convolution gives us (f\\oplus g)^\\ast=f^... 14 The statement for C^1 regularity is true, but with \"dimension-2 projections\" instead of \"codimension-1 projections''. This is even stronger, if d\\ge 3. On the other hand, for d=2 the statement with \"hyperplane projections\" fails, since 1 dimensional projections are just closed intervals, whose boundary is certainly smooth. Also, an analogous ... 14 Take a path that joins the antipodes and concatenate it with its symmetric image. Get a centrally symmetric closed path on the boundary of the cube. If this path avoids one of the facets of the cube (and hence the antipodal facet as well), then we can project it to the boundary of the cube one dimension less and get a centrally symmetric path of at most the ... 14 Denote the diameter by d and distance by |x-y|. Then there are y,z such that d=|y-z| and we have by triangle inequality for every x:$$d=|y-z|\\leq |y-x|+|x-z|\\leq 2f(x),$$so we obtain your inequality. Notice that I did not use convexity, or any other of your assumptions, only the triangle inequality. 13 The problem seems to be still open even for$n=3$: Weisstein, Eric W. \"Tetrahedron Circumscribing.\" 13 As it was noticed by Abhinav Kumar, you only can hope for equality up to$\\mathrm{GL}_n(\\mathbb Z)$transformations. The following picture shows two$\\mathrm{GL}_n(\\mathbb Z)$-distinct plane figures which have the same number of integer points after any rescaling. It still might be true that such bodies are$\\mathrm{GL}_n(\\mathbb Z)$-equidecomposable. P.S.... 12 The baseball stitches curve suggested by Gjergji Zaimi appears in another paper of Zalgaller: V. A. Zalgaller. Extremal problems on the convex hull of a space curve. Algebra i Analiz, 8(3):1–13, 1996. Here Zalgaller also conjectures that this curve should be the minimizer. 12 I will try to answer your question 2. The answer is negative. There must be at least two antipodal points with parallel tangent planes. Here is a sketch of a proof. The idea of that proof is that points you are looking for have variational nature. Proof works in any dimension. Denote the boundary of your body by$\\Gamma\\$. Now we fix a Euclidean structure on ...\n\nOnly top voted, non community-wiki answers of a minimum length are eligible" ]

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[ "# Use Fraction Equivalence to Add Tenths and Hundredths\n\nExamples, solutions, and videos to help Grade 4 students learn how to apply understanding of fraction equivalence to add tenths and hundredths.\n\nTopic D: Addition with Tenths and Hundredths\nCommon Core Standards: 4.NF.5, 4.NF.6, 4.NF.3c, 4.MD.1\n\n### NYS Common Core Grade 4 Module 6, Lesson 12\n\nLesson 12 Concept Development\n\nProblem 1: Add tenths and hundredths written in unit form using pictorial models.\n\nProblem 2: Add tenths and hundredths by converting using multiplication. Express the sum as a decimal.\n\nProblem 3: Add tenths and hundredths with sums greater than 1. Express the sum as a decimal.\n\nLesson 12 Problem Set\n\n1. Complete the number sentence by expressing each part using hundredths. Model using the place value chart, as shown in Part (a).\na. 1 tenth + 5 hundredths = ______ hundredths\nb. 2 tenths + 1 hundredth = ______ hundredths\nc. 1 tenth + 12 hundredths = ______ hundredths\n2. Solve by converting all addends to hundredths before solving.\na. 1 tenth + 3 hundredths = ______ hundredths + 3 hundredths = ______ hundredths\nb. 5 tenths + 12 hundredths = ______ hundredths + ______ hundredths = ______ hundredths\nc. 7 tenths + 27 hundredths = ______ hundredths + ______ hundredths = ______ hundredths\nd. 37 hundredths + 7 tenths = ______ hundredths + ______ hundredths = ______ hundredths\n3. Find the sum. Convert tenths to hundredths as needed. Write your answer as a decimal.\n5. Beaker A has 63/100 liter of iodine. It is filled the rest of the way with water up to 1 liter. Beaker B has 4/10 liter of iodine. It is filled the rest of the way with water up to 1 liter. If both beakers are emptied into a large beaker, how much iodine will be in the large beaker?\n\nLesson 12 Homework\n\n1. Complete the number sentence by expressing each part using hundredths. Model using the place value chart, as shown in Part (a).\na. 1 tenth + 5 hundredths = _____ hundredths\nb. 2 tenths + 3 hundredths = _____ hundredths\nc. 1 tenth + 14 hundredths = _____ hundredths\n2. Solve by converting all addends to hundredths before solving.\na. 1 tenth + 2 hundredths = ______ hundredths + 2 hundredths = ______ hundredths\nb. 4 tenths + 11 hundredths = ______ hundredths + ______ hundredths = ______ hundredths\nc. 8 tenths + 25 hundredths = ______ hundredths + ______ hundredths = ______ hundredths\nd. 43 hundredths + 6 tenths = ______ hundredths + ______ hundredths = ______ hundredths\n3. Find the sum. Convert tenths to hundredths as needed. Write your answer as a decimal.", null, "" ]

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[ "## Precalculus (6th Edition) Blitzer", null, "Step 1. Rewrite the equation as $\\frac{x^2}{5}+\\frac{y^2}{6}=1$; we can identify $a^2=6, b^2=5$. Thus $c=\\sqrt {a^2-b^2}=1$. The ellipse is centered at $(0,0)$ with a vertical major axis. Step 2. We can graph the equation as shown in the figure, and the foci can be located at $(0,\\pm1)$" ]

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[ "Search a number\nBaseRepresentation\nbin100011000000\n310001222\n4203000\n532430\n614212\n76350\noct4300\n93058\n102240\n111757\n121368\n131034\n14b60\n159e5\nhex8c0\n\n2240 has 28 divisors (see below), whose sum is σ = 6096. Its totient is φ = 768.\n\nThe previous prime is 2239. The next prime is 2243. The reversal of 2240 is 422.\n\nIt is a tau number, because it is divible by the number of its divisors (28).\n\nIt is a Harshad number since it is a multiple of its sum of digits (8).\n\nIt is a d-powerful number, because it can be written as 27 + 43 + 211 + 0 .\n\nIt is a plaindrome in base 12.\n\nIt is a nialpdrome in base 8 and base 14.\n\nIt is not an unprimeable number, because it can be changed into a prime (2243) by changing a digit.\n\nIt is a pernicious number, because its binary representation contains a prime number (3) of ones.\n\nIt is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 317 + ... + 323.\n\n22240 is an apocalyptic number.\n\n2240 is a gapful number since it is divisible by the number (20) formed by its first and last digit.\n\n2240 is a droll number since its even prime factors and its odd prime factors have the same sum.\n\nIt is an amenable number.\n\nIt is a practical number, because each smaller number is the sum of distinct divisors of 2240, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (3048).\n\n2240 is an abundant number, since it is smaller than the sum of its proper divisors (3856).\n\nIt is a pseudoperfect number, because it is the sum of a subset of its proper divisors.\n\n2240 is an equidigital number, since it uses as much as digits as its factorization.\n\n2240 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 24 (or 14 counting only the distinct ones).\n\nThe product of its (nonzero) digits is 16, while the sum is 8.\n\nThe square root of 2240 is about 47.3286382648. The cubic root of 2240 is about 13.0842652408.\n\nAdding to 2240 its reverse (422), we get a palindrome (2662).\n\nIt can be divided in two parts, 2 and 240, that added together give a palindrome (242).\n\nThe spelling of 2240 in words is \"two thousand, two hundred forty\", and thus it is an iban number." ]

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[ "# Banach spaces with no reflexive complemented subspaces\n\nIf $X$ is a Banach space with the Dunford Pettis Property (DPP), then no infinite reflexive subspace can be complemented. Suppose now that the Banach space has the property, that no infinite reflexive subspace is complemented. Is it true that $X$ has DPP?\n\n• Thank you Bill, do you know additional conditions on the space $X$, such that the question is true? Mar 15, 2014 at 21:34\n• Oh, you saw before I deleted--my answer was wrong. Tomek gave in his answer what is probably the simplest example. Mar 15, 2014 at 21:51\n• I do not know a reasonable extra condition on the space that would make the answer affirmative. Mar 15, 2014 at 21:54\n• The example with the Schreier space actually shows that a space failing DPP can have no reflexive subspaces at all. Mar 15, 2014 at 21:56\n\nNo. The Schreier space $S$ is $c_0$-saturated (every closed infinite-dimensional subspace of $S$ contains a copy of $c_0$) as it embeds into $C[0,\\omega^\\omega]$ which has this property, yet if fails the Dunford-Pettis Property (you will find an easy proof here (p. 168))." ]

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[ "시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율\n1 초 512 MB 8 3 2 66.667%\n\n## 문제\n\nJamshid, a great king of ancient Persia, is looking for the cup of divination, a miraculous cup through which one could observe all over the universe. He has asked Shahrasb, a great wizard who lives in Alborz mountains, for his help.\n\nShahrasb told Jamshid that the cup is hidden somewhere in the Great Salt Desert, a large desert in the middle of ancient Persia, but he doesn't know its exact location. Furthermore, Jamshid can ask him several questions. In each question, Jamshid selects a point anywhere in Persia (inside or outside of the desert), and Shahrasb can use his magical powers to find the Katouzian distance between the cup and the selected point.\n\nEach point in Persia has integer $x$ and $y$ coordinates in the range $[-10^9, 10^9]$. The desert is a square region in the center, with $x$ and $y$ coordinates in the range $[-5 \\times 10^8, 5 \\times 10^8]$. The Katouzian distance between two points $(x, y)$ and $(p, q)$ is calculated as $|x - p| \\oplus |y - q|$, where $|x - p|$ is the absolute value of $(x-p)$, and $\\oplus$ indicates bitwise XOR (exclusive OR).\n\n## 구현\n\nThere are $T$ different scenarios, numbered $0$ through $T-1$. The coordinates of the cup in scenario $i$ (for $0 \\leq i \\leq T-1$) is $(a[i], b[i])$. You should implement the following procedure:\n\nint[] find_cup()\n\n• This procedure will be called $T$ times, once for each scenario.\n• In scenario $i$ (for $0 \\leq i \\leq T-1$), the procedure must return an array $c$ of length $2$, such that $c=a[i]$ and $c=b[i]$.\n\nTo implement the above procedure, you can call the following procedure:\n\nint ask_shahrasb(int x, int y)\n\n• $x$ and $y$: the coordinates of the selected point. Both coordinates should be integer values in the range $[-10^9,10^9]$.\n• This procedure returns the Katouzian distance between the cup and the point $(x, y)$.\n\n## 예제\n\nConsider find_cup() is called, and the cup is hidden at the point $(1,3)$. The implementation of find_cup() makes the following procedure calls:\n\n• ask_shahrasb(4, 1) returns $1$.\n• ask_shahrasb(0, 2) returns $0$.\n• ask_shahrasb(-1, 0) returns $1$.\n• Now the location of the cup is uniquely determined, and find_cup() returns $[1,3]$.\n\n## 제한\n\n• $1 \\leq T \\leq 1000$,\n• $-5 \\times 10^8 \\leq a[i], b[i] \\leq 5 \\times 10^8$.\n\n## 점수\n\nYour score will be $0$ if the return value of find_cup() is incorrect for any of the scenarios. Otherwise, your score will be computed as below ($Q$ is the maximum number of questions asked among all scenarios).\n\nCondition Score\n$1000 < Q$ $0$\n$104 < Q \\leq 1000$ $20$\n$70 < Q \\leq 104$ $30$\n$39 < Q \\leq 70$ $61$\n$32 < Q \\leq 39$ $132-Q$\n$Q \\leq 32$ $100$\n\n## 샘플 그레이더\n\n• line $1$: $\\;\\;T$\n• line $2 + i$ (for $0 \\le i \\le T-1$): $\\;\\;a[i]\\;\\;b[i]$\n\nFor each scenario, the sample grader prints a single integer in a separate line: the number of calls to ask_shahrasb() in the scenario, or $-1$ if the return value of find_cup() was incorrect.\n\n## 제출할 수 있는 언어\n\nC++17, C++14, C++20, C++14 (Clang), C++17 (Clang), C++20 (Clang)\n\n## 채점 및 기타 정보\n\n• 100점 이상을 획득해야 를 받는다.\n• 예제는 채점하지 않는다." ]

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[ "# Simplifying rational expressions and restrictions\n\n### Simplifying rational expressions and restrictions\n\nA rational expression is a fraction that its numerator and/or denominator are polynomials. In this lesson, we will first learn how to find the non-permissible values of the variable in a rational expression. Then, we will how to simplify rational expressions.\n\n#### Lessons\n\n$\\cdot$ multiplication rule: $x^a \\cdot x^b=x^{a+b}$\n\n$\\cdot$ division rule: $\\frac{x^a}{x^b}=x^{a-b}$\n• Introduction\nWhy is it important to determine the non-permissible values prior to simplifying a rational expression?\n\n• 1.\nFor each rational expression:\ni) determine the non-permissible values of the variable, then\nii) simplify the rational expression\na)\n$\\frac{{6{x^3}}}{{4x}}$\n\nb)\n$\\frac{{5 - {x}}}{{{x^2} - 8x + 15}}$\n\nc)\n$\\frac{{{x^2} + 13x + 40}}{{{x^2} - 25}}$\n\n• 2.\nFor each rational expression:\ni) determine the non-permissible values of the variable, then\nii) simplify the rational expression\na)\n$\\frac{{9{t^3} - 16t}}{{3{t^2} + 4t}}$\n\nb)\n$\\frac{{{x^2} + 2x - 3}}{{{x^4} - 10{x^2} + 9}}$\n\n• 3.\nFor each rational expression:\ni) determine the non-permissible values of the variable, then\nii) simplify the rational expression\na)\n$\\frac{{x - 3}}{{3 - x}}$\n\nb)\n$\\frac{{5{y^3} - 10{y^2}}}{{30 - 15y}}$\n\nc)\n$\\frac{{1 - 9{x^2}}}{{6{x^2} - 7x - 3}}$\n\n• 4.\nThe area of a rectangular window can be expressed as $4{x^2} + 13x + 3$, while its length can be expressed as $4x + 1$.\na)\nFind the width of the window.\n\nb)\nIf the perimeter of the window is 68 $m$, what is the value of $x$?\n\nc)\nIf a cleaning company charges \\$3/$m^2$ for cleaning the window, how much does it cost to clean the window?\n\n• 5.\nFor each rational expression:\ni) determine the non-permissible values for $y$ in terms of $x$ , then\nii) simplify, where possible.\na)\n$\\frac{{2x + y}}{{2x - y}}$\n\nb)\n$\\frac{{x - 3y}}{{{x^2} - 9{y^2}}}$" ]

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[ "Contents\n\n# Relax Loop-Carried Dependency\n\n## Example 1: Multiply Chain\n\nBased on the feedback from the optimization report, you can relax a loop-carried dependency by allowing the compiler to infer a shift register and increase the dependence distance. Increase the dependence distance by increasing the number of loop iterations that occur between the generation of a loop-carried value and its use.\nConsider the following code example:\n``````constexpr int N = 128;\nqueue.submit([&](handler &cgh) {\naccessor result(result_buf, cgh, write_only);\nfloat mul = 1.0f;\nfor (int i = 0; i < N; i++)\nmul *= A[i];\n\nresult = mul;\n});\n});``````\nThe optimization report shows that the\nIntel® oneAPI\nDPC++/C++\nCompiler\ninfers pipelined execution for the loop successfully. However, the loop-carried dependency on the variable\nmul\ncauses loop iterations to launch every six cycles. In this case, the floating-point multiplication operation on line 8 (that is,\nmul *= A[i]\n) contributes the largest delay to the computation of the variable\nmul\n.\nTo relax the loop-carried data dependency, instead of using a single variable to store the multiplication results, operate on\nM\ncopies of the variable, and use one copy every\nM\niterations:\n1. Declare multiple copies of the variable\nmul\n(for example, in an array called\nmul_copies\n).\n2. Initialize all copies of\nmul_copies\n.\n3. Use the last copy in the array in the multiplication operation.\n4. Perform a shift operation to pass the last value of the array back to the beginning of the shift register.\n5. Reduce all copies to\nmul\nand write the final value to the result.\nThe following code illustrates the restructured kernel:\n``````constexpr int N = 128;\nconstexpr int M = 8;\nqueue.submit([&](handler &cgh) {\naccessor result(result_buf, cgh, write_only);\nfloat mul = 1.0f;\n\n// Step 1: Declare multiple copies of variable mul\nfloat mul_copies[M];\n\n// Step 2: Initialize all copies\nfor (int i = 0; i < M; i++)\nmul_copies[i] = 1.0f;\n\nfor (int i = 0; i < N; i++) {\n// Step 3: Perform multiplication on the last copy\nfloat cur = mul_copies[M-1] * A[i];\n\n// Step 4a: Shift copies\n#pragma unroll\nfor (int j = M-1; j > 0; j--)\nmul_copies[j] = mul_copies[j-1];\n\n// Step 4b: Insert updated copy at the beginning\nmul_copies = cur;\n}\n\n// Step 5: Perform reduction on copies\n#pragma unroll\nfor (int i = 0; i < M; i++)\nmul *= mul_copies[i];\n\nresult = mul;\n});\n});``````\n\n## Example 2: Accumulator\n\nSimilar to Example 1, this example also applies the same technique and demonstrates how to write single work-item kernels that carry out double precision floating-point operations efficiently by inferring a shift register.\nConsider the following example:\n``````queue.submit([&](handler &cgh) {\naccessor result(result_buf, cgh, write_only);\ndouble temp_sum = 0;\nfor (int i = 0; i < *N; ++i)\ntemp_sum += arr[i];\nresult = temp_sum;\n});\n});``````\nThe kernel\nunoptimized\nis an accumulator that sums the elements of a double precision floating-point array\narr[i]\n. For each loop iteration, the\nIntel® oneAPI\nDPC++/C++\nCompiler\ntakes 10 cycles to compute the result of the addition and then stores it in the variable\ntemp_sum\n. Each loop iteration requires the value of\ntemp_sum\nfrom the previous loop iteration, which creates a data dependency on\ntemp_sum\n.\nTo relax the data dependency, infer the array\narr[i]\nas a shift register.\nThe following is the restructured kernel\noptimized\n:\n``````//Shift register size must be statically determinable\nconstexpr int II_CYCLES = 12;\nqueue.submit([&](handler &cgh) {\naccessor result(result_buf, cgh, write_only);\n//Create shift register with II_CYCLE+1 elements\ndouble shift_reg[II_CYCLES+1];\n\n//Initialize all elements of the register to 0\nfor (int i = 0; i < II_CYCLES + 1; i++) {\nshift_reg[i] = 0;\n}\n\n//Iterate through every element of input array\nfor(int i = 0; i < N; ++i){\n//Load ith element into end of shift register\n//if N > II_CYCLE, add to shift_reg to preserve values\nshift_reg[II_CYCLES] = shift_reg + arr[i];\n\n#pragma unroll\n//Shift every element of shift register\nfor(int j = 0; j < II_CYCLES; ++j)\n{\nshift_reg[j] = shift_reg[j + 1];\n}\n}\n\n//Sum every element of shift register\ndouble temp_sum = 0;\n\n#pragma unroll\nfor(int i = 0; i < II_CYCLES; ++i)\n{\ntemp_sum += shift_reg[i];\n}\n\nresult = temp_sum;\n});\n}); ``````\n\n#### Product and Performance Information\n\n1\n\nPerformance varies by use, configuration and other factors. Learn more at www.Intel.com/PerformanceIndex." ]

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[ "", null, "StatLect\n\n# Model selection criteria\n\nModel selection criteria are rules used to select a statistical model among a set of candidate models, based on observed data. Typically, the criteria try to minimize the expected dissimilarity, measured by the Kullback-Leibler divergence, between the chosen model and the true model (i.e., the probability distribution that generated the data).\n\nIn this lecture we focus on the selection of models that have been estimated by the maximum likelihood method.", null, "## Competing models\n\nFirst of all, we need to define precisely what we mean by statistical model.\n\nA statistical model is a set of probability distributions that could have generated the data we are analyzing.\n\nExample Suppose we observe", null, "data points", null, "which have all been independently drawn from the same probability distribution (in technical terms, they are IID draws). If we assume that the draws come from a normal distribution, then we are formulating a statistical model: we are restricting our attention to the set of all normal distributions and we are ruling out all the probability distributions that are not normal. Note that the normal distribution has two parameters, the mean", null, "and the variance", null, ", so that the set of distributions we are considering (the statistical model) includes many normal distributions: one for each possible couple", null, ". If instead we assume that the data has been drawn from an exponential distribution, then we are formulating an alternative model. The exponential distribution has one parameter", null, ", called rate parameter. Our statistical model is a set including many possible distributions: one for each possible value of the parameter", null, ".\n\nThe previous example, although admittedly unrealistic, introduces in a simple manner the problem that we are going to deal with: how do we select one model (normal vs exponential distribution in the example) if we deem that two or more alternative models are plausible?\n\n## Notation and main assumptions\n\nLet us denote the vector of observed data by", null, ". We assume that the data is continuous, and that a model for", null, "is a family of joint probability density functions", null, "parametrized by a parameter vector", null, "for each model", null, ".\n\nWe focus on continuous distributions in order to simplify the discussion, but everything we say is valid also for discrete distributions, with straightforward modifications (replace probability densities with probability mass functions).\n\nExample In the example above the vector", null, "contains the", null, "data points:", null, "The number of models is", null, ". The two parameter vectors are", null, "for the normal distribution and", null, "for the exponential distribution. The joint probability density function for the first model is", null, "because the joint density of a vector of independent random variables is equal to the product of their marginal densities. The joint probability density function for the second model is", null, "where", null, "is an indicator function (equal to 1 if", null, "and to 0 otherwise).\n\nWe assume that model parameters are estimated by maximum likelihood (ML). We denote by", null, "the ML estimates of the parameters of the", null, "models.\n\nIf you want to see some examples of how ML estimates are derived, you can have a look at these two lectures:\n\nFinally, we will denote by", null, "the unknown probability distribution that generated the data, and by", null, "the index of the model selected by a model selection criterion. Clearly,", null, "can range between", null, "and", null, ".\n\n## The general criterion\n\nAkaike (1973) was the first to propose a general criterion for selecting models estimated by maximum likelihood. He proposed to minimize the expected dissimilarity between the chosen model", null, "at the ML estimate and the the true distribution", null, ".\n\nThe dissimilarity between an estimated model and the true distribution is measured by the Kullback-Leibler divergence", null, "where the expected value is with respect to the true density", null, "The expected dissimilarity is computed as", null, "where the expectation is over the sampling distribution of", null, ", which, being a function of the sample", null, ", is regarded as stochastic.\n\nIdeally, we would like to select the model that minimizes the expected dissimilarity:", null, "However, the expected dissimilarity cannot be computed exactly because the true distribution", null, "and the sampling distribution of", null, "are unknown.\n\nAkaike (1973) proposed an approximation to the expected dissimilarity that can be easily computed, giving rise to the so-called Akaike Information Criterion (AIC).\n\nAs proved, for example, by Burnham and Anderson (2004), other popular selection criteria such as the AIC corrected for small-sample bias (AICc; Sugiura 1978, Hurvich and Tsai 1989) and the Bayesian Information Criterion (BIC; Schwarz 1978) are based on different approximations of the same measure of expected dissimilarity.\n\n## Popular criteria\n\nWe briefly present here the most popular selection criteria.\n\n### Akaike Information Criterion (AIC)\n\nAccording to the Akaike Information Criterion, the selected model", null, "solves the minimization problem", null, "where the value of the", null, "-th model is", null, "where", null, "is the number of parameters to be estimated in the", null, "-th model.\n\nNote that any linear transformation applied to all model values does not change the selected model. As a matter of fact, many references define the value of the", null, "-th model as", null, "### Corrected Akaike Information Criterion (AIC)\n\nAn approximation that is more precise in small samples is the so-called corrected Akaike Information Criterion (AICc), according to which the value to be minimized is", null, "where", null, "is the size of the sample being used for estimation.\n\n### Bayesian Information Criterion (BIC)\n\nAnother popular criterion is the Bayesian Information Criterion, according to which the selected model is the one that achieves the minimum value of", null, "## The penalty for complexity\n\nAs you might have noticed, all of these criteria penalize the dimension of the model: the higher the number of parameters", null, "is, the more model", null, "is penalized.\n\nThis penalty for complexity is typical of model selection criteria: a model with many parameters is more likely to over-fit, that is, to have a spuriously high value of the log-likelihood", null, ". For a discussion of over-fitting see the lecture on the R squared of a linear regression.\n\nThe complexity penalty is also related to the so-called bias-variance trade-off: by increasing model complexity, we usually decrease the bias and increase the variance; beyond a certain degree of complexity, increases in variance are larger than reductions in bias, and, as a consequence, the quality of our inferences becomes worse.\n\n## References\n\nAkaike, H., 1973. Information theory as an extension of the maximum likelihood principle. In: Petrov, BN and Csaki, F. In Second International Symposium on Information Theory. Akademiai Kiado, Budapest, pp. 276-281.\n\nBurnham, K.P. and Anderson, D.R., 2004. Multimodel inference: understanding AIC and BIC in model selection. Sociological methods & research, 33(2), pp. 261-304.\n\nHurvich, C.M. and Tsai, C.L., 1989. Regression and time series model selection in small samples. Biometrika, 76(2), pp. 297-307.\n\nSchwarz, G., 1978. Estimating the dimension of a model. The annals of statistics, 6(2), pp. 461-464.\n\nSugiura, N., 1978. Further analysis of the data by Akaike's information criterion and the finite corrections, in Statistics Theory and Methods, 7(1), pp. 13-26.\n\nThe book\n\nMost of the learning materials found on this website are now available in a traditional textbook format.\n\nGlossary entries\nShare" ]

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[ "Introduction >\n\n# A Comprehensive Guide to Mean Absolute Percentage Error (MAPE)", null, "Noa Azaria\n5 min read Aug 12, 2023\n\nToday we’re going to delve into a vital metric called Mean Absolute Percentage Error, or MAPE for short. Understanding MAPE is crucial if you are dealing with forecasting models. So let’s get started!\n\n## What is MAPE?\n\nBefore jumping into calculations, let’s understand what Mean Absolute Percentage Error (MAPE) really is. In simple terms, it is a statistical measure that helps you determine how accurate your predictions or forecasts are in relation to the actual values. In forecasting models, such as time series analysis, it’s crucial to gauge the performance of your model, and MAPE offers a handy means to do just that. It expresses the error as a percentage, making it easier for you to interpret and communicate the model’s accuracy.\n\n## How to calculate MAPE\n\nKnowing the importance of MAPE, the next step is to learn how to calculate it. It’s surprisingly simple!\n\n### MAPE Formula\n\nThe formula for MAPE is:\n\nWhere:\n\n• MAPE is the Mean Absolute Percentage Error\n• n’ is the number of data points\n• Ai is the actual value for the ith data point\n• Fi is the forecasted value for the ith data point\n\nBy iterating through each data point, calculating the absolute percentage error, and then averaging them, you obtain the MAPE. The result is a percentage indicating the average deviation of the predicted values from the actual values.\n\n## Measuring forecasting accuracy with Mean Absolute Percentage Error\n\nNow that you understand how to calculate MAPE, let’s explore how you can use it to measure the accuracy of your forecasting models.\n\n1. Evaluate Model Performance: By comparing the MAPE of different models, you can evaluate which model performs better in terms of forecasting accuracy.\n2. Interpreting MAPE: Lower values of MAPE indicate higher accuracy, while higher values indicate lower accuracy. A MAPE of 5% implies that, on average, the predictions are 5% off from the actual values.\n3. Communicate Performance: Since MAPE is expressed as a percentage, it’s easy to communicate the performance of your model to stakeholders who might not be as technically inclined.\n\nHowever, keep in mind that MAPE has its limitations. It cannot be used when the actual values have instances of zero, as this would lead to division by zero in the formula. Additionally, MAPE can sometimes favor models that under-forecast.\n\nIn order to illustrate how MAPE can favor models that under-forecast, especially when forecast values approach zero, consider two scenarios:\n\nScenario 1 (Over-forecast with small actual value):\n\n• Actual value: 10\n• Predicted value (over-forecast): 20\n\nThe MAPE in this case would be:\n\nMAPE = (|10 - 20| / 10) * 100% = 100%\n\nScenario 2 (Under-forecast with small actual value):\n\n• Actual value: 10\n• Predicted value (under-forecast): 0\n\nThe MAPE in this case would be:\n\nMAPE = (|10 - 0| / 10) * 100% = 100%\n\nIn these two scenarios, the over-forecast has the same percentage error as the under-forecast.\n\nHowever, if the under-forecast is very close to zero but not exactly zero, the MAPE is still high.\n\nLet’s consider an extra scenario to illustrate this:\n\nScenario 3 (Under-forecast close to zero):\n\n• Actual value: 10\n• Predicted value (under-forecast): 0.0001\n\nThe MAPE in this case would be:\n\nMAPE = (|10 - 0.0001| / 10) * 100% = 99.999%\n\nBut if the under-forecast is exactly zero, the MAPE is zero:\n\nScenario 4 (Under-forecast exactly zero):\n\n• Actual value: 10\n• Predicted value (under-forecast): 0\n\nThe MAPE in this case would be:\n\nMAPE = (|10 - 0| / 10) * 100% = 100%\n\nDespite the actual value being 10, an under-forecast of exactly zero results in a MAPE of 100%. This is the reason why MAPE can sometimes favor models that under-forecast, especially when the forecast values are very close or equal to zero.\n\n## Limitations of Mean Absolute Percentage Error\n\nAs with any metric, MAPE comes with its own set of limitations. Being aware of these will help you make better-informed decisions regarding when to use MAPE.\n\n1. Inaccuracy with Low Actual Values: As the actual values approach zero, the percentage errors can become extremely large even if the forecasted values are close to the actual values. This can distort the MAPE.\n2. Cannot Handle Zero Actual Values: MAPE is undefined when actual values are zero, as this leads to division by zero in the formula.\n3. Scale Dependence: Being a percentage error, MAPE can sometimes be tricky to interpret when comparing across different scales or units.\n4. Not Ideal for Comparing Across Different Datasets: Due to its scale dependence, it is not the best metric for comparing the forecasting accuracy across datasets that have different scales or units.\n\nIn light of these limitations, it is advisable to use MAPE in conjunction with other evaluation metrics such as Mean Absolute Error (MAE) or Root Mean Squared Error (RMSE) to get a more comprehensive understanding of your model’s performance. Additionally, it is important to consider the nature of your data and the specific requirements of your project when deciding on the metrics to use.\n\n## Using Mean Absolute Percentage Error in model monitoring\n\nMAPE is not only useful for evaluating model performance but also for continuous monitoring of a model after deployment.\n\n1. Setting Thresholds: You can set a threshold for MAPE, which, if crossed, triggers an alert. This helps in keeping an eye on the model’s performance and ensuring that it doesn’t degrade over time.\n2. Detecting Data Shifts: Significant changes in MAPE could be indicative of changes in the underlying data distribution. Keeping track of MAPE could help in detecting these shifts early on.\n3. Model Retraining: Regularly monitoring the MAPE can inform you when it might be time to retrain your model with new data to improve its accuracy.\n\n## Wrapping up Mean Absolute Percentage Error\n\nUnderstanding Mean Absolute Percentage Error (MAPE) is crucial for anyone in predictive modeling, particularly for evaluating and monitoring forecasting models. However, it’s imperative to recognize its limitations, such as inaccuracy near zero values and asymmetry. Thus, incorporating complementary metrics like Mean Absolute Error (MAE) or Root Mean Squared Error (RMSE) is advisable for a more comprehensive evaluation.\n\nBeing vigilant regarding your data’s nature and the specific requirements of your forecasting tasks will empower you to make informed decisions about which metrics to employ. I hope this article has shed light on MAPE, and I encourage you to stay curious and continue learning. Happy monitoring!", null, "" ]

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[ "# Shared Parameter Models\n\nIt is possible to summarise the steps involved in drawing inference from incomplete data as (Daniels and Hogan (2008)):\n\n• Specification of a full data model for the response and missingness indicators $$f(y,r)$$\n\n• Specification of the prior distribution (within a Bayesian approach)\n\n• Sampling from the posterior distribution of full data parameters, given the observed data $$Y_{obs}$$ and the missingness indicators $$R$$\n\nIdentification of a full data model, particularly the part involving the missing data $$Y_{mis}$$, requires making unverifiable assumptions about the full data model $$f(y,r)$$. Under the assumption of the ignorability of the missingness mechanism, the model can be identified using only the information from the observed data. When ignorability is not believed to be a suitable assumption, one can use a more general class of models that allows missing data indicators to depend on missing responses themselves. These models allow to parameterise the conditional dependence between $$R$$ and $$Y_{mis}$$, given $$Y_{obs}$$. Without the benefit of untestable assumptions, this association structure cannot be identified from the observed data and therefore inference depends on some combination of two elements:\n\n1. Unverifiable parametric assumptions\n\n2. Informative prior distributions (under a Bayesian approach)\n\nWe show some simple examples about how these nonignorable models can be constructed, identified and applied. In this section, we specifically focus on the class of nonignorable models known as Shared Parameter Models(SPM).\n\n## Shared Parameter Models\n\nThe shared parameter model approach consists in an explicit multilevel specification, where random effects $$b$$ are modelled jointly with $$Y$$ and $$R$$ (Wu and Carroll (1988)). The general form of the full data modelling using a SPM approach is\n\n$f(y,r \\mid \\omega) = \\int f(y, r, b \\mid \\omega)db.$\n\nNext, specific SPMs are formulated by making assumptions about the joint distribution under the integral sign. Main advantages of this models is that they are quite easy to specify and that, through the use of random effects, high-dimensional or multilevel data modelling is relatively easy to accomplish. The main drawback is that the underlying missingness mechanism is often difficult to understand and may not have even a closed form.\n\n### Example random coefficients selection model\n\nWu and Carroll (1988) specified a SPM assuming the response follow a linear random effects model\n\n$Y_i \\mid x_i,b_i \\sim N(x_i\\beta + w_ib_i, \\Sigma_i(\\phi)),$\n\nwhere $$w_i$$ are the random effects covariates with rows $$w_i=(1,t_{ij})$$, therefore implying that each individual has a random slope and intercept. The random effects $$b_i=(b_{i1},b_{i2})$$ are assumed to follow a bivariate normal distribution\n\n$b_i \\sim N(0,\\Omega),$\n\nwhile the hazard of dropout is Bernoulli with\n\n$R_{ij} \\mid R_{ij-1}=1,b_i \\sim Bern(\\pi_{ij}),$\n\nwhich depends on the random effects via\n\n$g(\\pi_{ij}) = \\psi_0 + \\psi_1b_{i1} + \\psi_2b_{i2}.$\n\nThe model can be seen as a special case of the general SPM formulation by noticing that the joint distribution under the integral sign can be factored as\n\n$f(y,r,b \\mid x, \\omega) = f(r \\mid b,x,\\psi)f(y \\mid b,x,\\beta,\\phi)f(b \\mid \\Omega)$\n\nunder the assumption that $$R$$ is independent of both $$Y_{obs}$$ and $$Y_{mis}$$, conditionally on $$b$$. However, integrating over the random effects, dependence between $$R$$ and $$Y_{mis}$$, given $$Y_{obs}$$, is induced and therefore the model characterises a Missing Not At Random(MNAR) mechanism.\n\nThe conditional linear model (Wu and Bailey (1989)) can also be seen as a version of the SPM, which is formulated as\n\n$f(y,r,b \\mid x) = f(y \\mid r,b,x)f(b \\mid r,x)f(r \\mid x).$\n\n## Conlcusions\n\nTo summarise, shared parameter models are very useful for characterizing joint distributions of repeated measures and event times, and can be particularly useful as a method of data reduction when the dimension of $$Y$$ is high. Nonetheless, their application to the problem of making full data inference from incomplete longitudinal data should be made with caution and with an eye toward justifying the required assumptions. Sensitivity analysis is an open area of research for these models." ]

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[ "### Merge branch 'more_samples_for_sandwiches' into 'NIFTy_4'\n\n```Allow inverse samples from sandwiches which are diagonal\n\nSee merge request ift/NIFTy!258```\nparents 23b0ed81 7b4b15b0\nPipeline #29554 passed with stages\nin 4 minutes and 39 seconds\n ... ... @@ -31,7 +31,7 @@ d = R(s_x) + n R_p = R * FFT * A j = R_p.adjoint(N.inverse(d)) D_inv = ift.SandwichOperator(R_p, N.inverse) + S.inverse D_inv = ift.SandwichOperator.make(R_p, N.inverse) + S.inverse N_samps = 200 ... ...\n ... ... @@ -80,12 +80,12 @@ if __name__ == \"__main__\": IC = ift.GradientNormController(name=\"inverter\", iteration_limit=500, tol_abs_gradnorm=1e-3) inverter = ift.ConjugateGradient(controller=IC) D = (ift.SandwichOperator(R, N.inverse) + Phi_h.inverse).inverse D = (ift.SandwichOperator.make(R, N.inverse) + Phi_h.inverse).inverse D = ift.InversionEnabler(D, inverter, approximation=Phi_h) m = HT(D(j)) # Uncertainty D = ift.SandwichOperator(aHT, D) # real space propagator D = ift.SandwichOperator.make(aHT, D) # real space propagator Dhat = ift.probe_with_posterior_samples(D.inverse, None, nprobes=nprobes) sig = ift.sqrt(Dhat) ... ...\n ... ... @@ -51,7 +51,7 @@ if __name__ == \"__main__\": IC = ift.GradientNormController(name=\"inverter\", iteration_limit=500, tol_abs_gradnorm=0.1) inverter = ift.ConjugateGradient(controller=IC) D = (ift.SandwichOperator(R, N.inverse) + Sh.inverse).inverse D = (ift.SandwichOperator.make(R, N.inverse) + Sh.inverse).inverse D = ift.InversionEnabler(D, inverter, approximation=Sh) m = D(j) ... ...\n ... ... @@ -46,7 +46,7 @@ class PoissonEnergy(Energy): R1 = Instrument*Rho*ht self._grad = (phipos + R1.adjoint_times((lam-d)/(lam+eps))).lock() self._curv = Phi_h.inverse + SandwichOperator(R1, W) self._curv = Phi_h.inverse + SandwichOperator.make(R1, W) def at(self, position): return self.__class__(position, self._d, self._Instrument, ... ...\n ... ... @@ -39,5 +39,5 @@ def WienerFilterCurvature(R, N, S, inverter): inverter : Minimizer The minimizer to use during numerical inversion \"\"\" op = SandwichOperator(R, N.inverse) + S.inverse op = SandwichOperator.make(R, N.inverse) + S.inverse return InversionEnabler(op, inverter, S.inverse)\n ... ... @@ -16,31 +16,46 @@ # NIFTy is being developed at the Max-Planck-Institut fuer Astrophysik # and financially supported by the Studienstiftung des deutschen Volkes. import numpy as np from .diagonal_operator import DiagonalOperator from .endomorphic_operator import EndomorphicOperator from .scaling_operator import ScalingOperator import numpy as np class SandwichOperator(EndomorphicOperator): \"\"\"Operator which is equivalent to the expression `bun.adjoint*cheese*bun`. Parameters ---------- bun: LinearOperator the bun part cheese: EndomorphicOperator the cheese part \"\"\" def __init__(self, bun, cheese=None): def __init__(self, bun, cheese, op, _callingfrommake=False): if not _callingfrommake: raise NotImplementedError super(SandwichOperator, self).__init__() self._bun = bun self._cheese = cheese self._op = op @staticmethod def make(bun, cheese=None): \"\"\"Build a SandwichOperator (or something simpler if possible) Parameters ---------- bun: LinearOperator the bun part cheese: EndomorphicOperator the cheese part \"\"\" if cheese is None: self._cheese = ScalingOperator(1., bun.target) self._op = bun.adjoint*bun cheese = ScalingOperator(1., bun.target) op = bun.adjoint*bun else: self._cheese = cheese self._op = bun.adjoint*cheese*bun op = bun.adjoint*cheese*bun # if our sandwich is diagonal, we can return immediately if isinstance(op, (ScalingOperator, DiagonalOperator)): return op return SandwichOperator(bun, cheese, op, _callingfrommake=True) @property def domain(self): ... ... @@ -54,8 +69,11 @@ class SandwichOperator(EndomorphicOperator): return self._op.apply(x, mode) def draw_sample(self, from_inverse=False, dtype=np.float64): # Inverse samples from general sandwiches is not possible if from_inverse: raise NotImplementedError( \"cannot draw from inverse of this operator\") # Samples from general sandwiches return self._bun.adjoint_times( self._cheese.draw_sample(from_inverse, dtype))\nMarkdown is supported\n0% or .\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!" ]

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[ "# Definition:Parenthesis\n\n## Definition\n\nParenthesis is a syntactical technique to disambiguate the meaning of a logical formula.\n\nIt allows one to specify that a logical formula should (temporarily) be regarded as being a single entity, being on the same level as a statement variable.\n\nSuch a formula is referred to as being in parenthesis.\n\nTypically, a formal language, in defining its formal grammar, ensures by means of parenthesis that all of its well-formed words are uniquely readable.\n\nGenerally, brackets are used to indicate that certain formulas are in parenthesis.\n\nThe brackets that are mostly used are round ones, the left (round) bracket $($ and the right (round) bracket $)$.\n\n## Parenthesis in Natural Language\n\nWhen parenthesis is needed in natural language, it is usual to employ a number of different techniques.\n\nIt is often the case that ambiguity is avoided by taking care with the word order.\n\n## Also denoted as\n\nThere is no universal convention as to exactly what shaped brackets are used for parentheses, but (usually) round brackets \"$\\paren \\;$\" are used.\n\nA notable counterexample is the elegantly-presented 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability, which uses square ones: \"$\\sqbrk \\;$\".\n\nSome authors, when writing complicated statements with nested parentheses, use differently shaped brackets, either square brackets $\\sqbrk \\;$ or braces $\\set \\;$ for each different parenthesis, in an attempt to make it clearer which brackets go with which substatements.\n\nHowever, some have the opinion that this does not actually aid comprehension and can add unnecessary confusion -- especially when particular bracket styles are being used for particular mathematical tasks, as they frequently are.\n\nIt also happens, unfortunately, that square brackets do not render well in all browsers when they have been automatically scaled by our rendering software.\n\nTherefore it is recommended that on $\\mathsf{Pr} \\infty \\mathsf{fWiki}$ round brackets are used throughout for parenthesis.\n\n## Examples\n\n### Example 1\n\nConsider the following this formula of propositional logic:\n\n$p \\land q \\lor r$\n\nThis can mean either:\n\nthe conjunction of $p$ with $q \\lor r$\n\nor:\n\nthe disjunction of $p \\land q$ with $r$.\n\nUsing parenthesis, the ambiguity is removed by presenting what is required either as:\n\n$p \\land \\paren {q \\lor r}$\n\nor:\n\n$\\paren {p \\land q} \\lor r$\n\n### Example 2\n\nConsider the following this formula of propositional logic:\n\n$p \\lor q \\implies \\neg \\, r \\implies p \\land q$\n\nThis can be interpreted in several different ways:\n\nIf either $p$ or $q$ is true, then it is not the case that the truth of $r$ implies the truth of both $p$ and $q$.\nEither $p$ is true, or if $q$ is true, then it is not the case that the truth of $r$ implies the truth of both $p$ and $q$.\nand so on.\n\nSo we need a way, for such a formula, to determine which of these interpretations is the one intended.\n\nIn the example above, the two different interpretations will be written in the style we have chosen as:\n\n$\\paren {p \\lor q} \\implies \\paren {\\neg \\paren {r \\implies \\paren {p \\land q} } }$\n$p \\lor \\paren {q \\implies \\paren {\\neg \\paren {r \\implies \\paren {p \\land q} } } }$\n\nIn these expressions, $\\paren {p \\lor q}$ and $\\paren {\\neg \\paren {r \\implies \\paren {p \\land q} } }$ are examples of formulas in parenthesis.\n\nNote that while the latter expressions may in fact be WFFs of propositional logic, the ambiguous expression they were derived from is not.\n\n## Historical Note\n\nRound brackets $\\paren \\;$ first appeared in $1544$.\n\nSquare brackets $\\sqbrk \\;$ and braces $\\set \\;$ were used by François Viète in around $1593$.\n\n1910: Alfred North Whitehead and Bertrand Russell: Principia Mathematica idiosyncratically use dots \"$.$\" for no immediate discernible benefit. The rules governing their use are complex and clumsy.\n\n## Linguistic Note\n\nThe plural of parenthesis is parentheses.\n\nThe correct pronunciation of parenthesis is par-en-te-sis (or par-en-the-sis), while parentheses is pronounced par-en-te-sees (or par-en-the-sees).\n\nIt also needs to be pointed out that US English uses the term parentheses to mean the brackets $\\paren \\ldots$ themselves (specifically the round ones), rather than their content. The word brackets is generally reserved for square $\\sqbrk \\ldots$ and curly $\\set \\ldots$ versions (although the technical term for the latter is braces).\n\nWhile this is common in natural language, such usage is discouraged in $\\mathsf{Pr} \\infty \\mathsf{fWiki}$, as it is more useful to have a word which can be specifically used to unambiguously refer to the content. It is also worth pointing out that use of parentheses to mean the brackets is considered by much of the rest of the world as ignorant." ]

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[ "Search a number\nBaseRepresentation\nbin110111001\n3121100\n412321\n53231\n62013\n71200\noct671\n9540\n10441\n11371\n12309\n1327c\n14237\n151e6\nhex1b9\n\n441 has 9 divisors (see below), whose sum is σ = 741. Its totient is φ = 252.\n\nThe previous prime is 439. The next prime is 443. The reversal of 441 is 144.\n\nMultipling 441 by its product of digits (16), we get a square (7056 = 842).\n\nAdding to 441 its reverse (144), we get a palindrome (585).\n\nMultipling 441 by its reverse (144), we get a square (63504 = 2522).\n\nIt can be divided in two parts, 44 and 1, that added together give a triangular number (45 = T9).\n\n441 = T20 + T21.\n\n441 = 13 + 23 + ... + 63.\n\nThe square root of 441 is 21.\n\nIt is a perfect power (a square), and thus also a powerful number.\n\n441 is nontrivially palindromic in base 4.\n\n441 is an esthetic number in base 4, because in such base its adjacent digits differ by 1.\n\nIt is an interprime number because it is at equal distance from previous prime (439) and next prime (443).\n\nIt is a tau number, because it is divible by the number of its divisors (9).\n\nIt is not a de Polignac number, because 441 - 21 = 439 is a prime.\n\nIt is a Harshad number since it is a multiple of its sum of digits (9).\n\nIt is an Ulam number.\n\nIt is a Curzon number.\n\nIt is a plaindrome in base 13 and base 14.\n\nIt is a nialpdrome in base 9 and base 10.\n\nIt is an inconsummate number, since it does not exist a number n which divided by its sum of digits gives 441.\n\nIt is not an unprimeable number, because it can be changed into a prime (443) by changing a digit.\n\nIt is a polite number, since it can be written in 8 ways as a sum of consecutive naturals, for example, 60 + ... + 66.\n\n441 is the 21-st square number.\n\n441 is the 11-th centered octagonal number.\n\nIt is an amenable number.\n\n441 is a deficient number, since it is larger than the sum of its proper divisors (300).\n\n441 is a wasteful number, since it uses less digits than its factorization.\n\n441 is an evil number, because the sum of its binary digits is even.\n\nThe sum of its prime factors is 20 (or 10 counting only the distinct ones).\n\nThe product of its digits is 16, while the sum is 9.\n\nThe cubic root of 441 is about 7.6116626110.\n\nThe spelling of 441 in words is \"four hundred forty-one\", and thus it is an aban number and an iban number.\n\nDivisors: 1 3 7 9 21 49 63 147 441" ]

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[ "English | Español\n\n# Try our Free Online Math Solver!", null, "Online Math Solver\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. 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[ "", null, "DifferentialGeometry/Tensor/SectionalCurvature - Maple Help\n\nTensor[SectionalCurvature] - calculate the sectional curvature for a metric \n\nCalling Sequences\n\nSectionalCurvature(g, R, X, Y)\n\nParameters\n\ng    - a metric tensor on the tangent bundle of a manifold\n\nR    - the curvature tensor of the metric $g,$calculated from the Christoffel symbol of $g$\n\nX, Y - a pair of vectors", null, "Description\n\n • Let $M$ be an $n$-dimensional manifold with metric $g$. The sectional curvature of the metric $g$ at a point $p\\in M$ is the Gaussian curvature $K$ (at $p$) of the geodesic surface whose tangent space at is spanned by vectors ${X}_{}$ and ${Y}_{}$. If $R$ is the covariant form of the curvature tensor (that is, $R$ is a tensor of type $\\left(\\genfrac{}{}{0}{}{0}{4}\\right)$), then\n\n.\n\n • If $K$ is independent of the choice of the vectors $X$ and $Y$ then $K=\\frac{S}{n\\left(n-1\\right)}$, where $S$ is the Ricci scalar of $g$.\n • This command is part of the DifferentialGeometry:-Tensor package, and so can be used in the form SectionalCurvature(...) only after executing the command with(DifferentialGeometry) and with(Tensor) in that order.  It can always be used in the long form DifferentialGeometry:-Tensor:- SectionalCurvature(...).", null, "Examples\n\n > $\\mathrm{with}\\left(\\mathrm{DifferentialGeometry}\\right):$$\\mathrm{with}\\left(\\mathrm{Tensor}\\right):$\n\nExample 1.\n\nFirst create a 2 dimensional manifold $\\mathrm{M1}$ and define a metric $\\mathrm{g1}$ on $\\mathrm{M1}$.\n\n M4 > $\\mathrm{DGsetup}\\left(\\left[x,y\\right],\\mathrm{M1}\\right)$\n ${\\mathrm{frame name: M1}}$ (2.1)\n M1 > $\\mathrm{g1}≔\\mathrm{evalDG}\\left(\\frac{1\\left(\\mathrm{dx}&t\\mathrm{dx}+\\mathrm{dy}&t\\mathrm{dy}\\right)}{{y}^{2}}\\right)$\n ${\\mathrm{g1}}{:=}\\frac{{\\mathrm{dx}}{}{\\mathrm{dx}}}{{{y}}^{{2}}}{+}\\frac{{\\mathrm{dy}}{}{\\mathrm{dy}}}{{{y}}^{{2}}}$ (2.2)\n\nCompute the sectional curvature determined by the coordinate basis vectors ${\\partial }_{x}$ and ${\\partial }_{y}$ .\n\n M1 > $\\mathrm{R1}≔\\mathrm{CurvatureTensor}\\left(\\mathrm{g1}\\right)$\n ${\\mathrm{R1}}{:=}{-}\\frac{{\\mathrm{D_x}}{}{\\mathrm{dy}}{}{\\mathrm{dx}}{}{\\mathrm{dy}}}{{{y}}^{{2}}}{+}\\frac{{\\mathrm{D_x}}{}{\\mathrm{dy}}{}{\\mathrm{dy}}{}{\\mathrm{dx}}}{{{y}}^{{2}}}{+}\\frac{{\\mathrm{D_y}}{}{\\mathrm{dx}}{}{\\mathrm{dx}}{}{\\mathrm{dy}}}{{{y}}^{{2}}}{-}\\frac{{\\mathrm{D_y}}{}{\\mathrm{dx}}{}{\\mathrm{dy}}{}{\\mathrm{dx}}}{{{y}}^{{2}}}$ (2.3)\n M1 > $K≔\\mathrm{SectionalCurvature}\\left(\\mathrm{g1},\\mathrm{R1},\\mathrm{D_x},\\mathrm{D_y}\\right)$\n ${K}{:=}{-}{1}$ (2.4)\n\nFor 2-dimensional manifolds the sectional curvature coincides with the Gaussian curvature ${R}_{1212}}{\\mathrm{det}\\left(g\\right)}$. Let us check this formula.\n\n M1 > $R≔\\mathrm{RaiseLowerIndices}\\left(\\mathrm{g1},\\mathrm{R1},\\left[1\\right]\\right)$\n ${R}{:=}{-}\\frac{{\\mathrm{dx}}{}{\\mathrm{dy}}{}{\\mathrm{dx}}{}{\\mathrm{dy}}}{{{y}}^{{4}}}{+}\\frac{{\\mathrm{dx}}{}{\\mathrm{dy}}{}{\\mathrm{dy}}{}{\\mathrm{dx}}}{{{y}}^{{4}}}{+}\\frac{{\\mathrm{dy}}{}{\\mathrm{dx}}{}{\\mathrm{dx}}{}{\\mathrm{dy}}}{{{y}}^{{4}}}{-}\\frac{{\\mathrm{dy}}{}{\\mathrm{dx}}{}{\\mathrm{dy}}{}{\\mathrm{dx}}}{{{y}}^{{4}}}$ (2.5)\n M1 > $\\mathrm{R1212}≔{\\mathrm{Tools}:-\\mathrm{DGinfo}\\left(R,\"CoefficientList\",\\left[\\left[1,2,1,2\\right]\\right]\\right)}_{1}$\n ${\\mathrm{R1212}}{:=}{-}\\frac{{1}}{{{y}}^{{4}}}$ (2.6)\n M1 > $\\frac{\\mathrm{R1212}}{\\mathrm{MetricDensity}\\left(\\mathrm{g1},2\\right)}$\n ${-}{1}$ (2.7)\n\nExample 2.\n\nFirst create a 3 dimensional manifold $\\mathrm{M2}$ and define a metric $\\mathrm{g2}$ on $\\mathrm{M2}$.\n\n M1 > $\\mathrm{DGsetup}\\left(\\left[x,y,z\\right],\\mathrm{M2}\\right)$\n ${\\mathrm{frame name: M2}}$ (2.8)\n M2 > $\\mathrm{g2}≔\\mathrm{evalDG}\\left(\\frac{{a}^{2}\\left(\\mathrm{dx}&t\\mathrm{dx}+\\mathrm{dy}&t\\mathrm{dy}+\\mathrm{dz}&t\\mathrm{dz}\\right)}{{\\left({k}^{2}+{x}^{2}+{y}^{2}+{z}^{2}\\right)}^{2}}\\right)$\n ${\\mathrm{g2}}{:=}\\frac{{{a}}^{{2}}{}{\\mathrm{dx}}{}{\\mathrm{dx}}}{{\\left({{k}}^{{2}}{+}{{x}}^{{2}}{+}{{y}}^{{2}}{+}{{z}}^{{2}}\\right)}^{{2}}}{+}\\frac{{{a}}^{{2}}{}{\\mathrm{dy}}{}{\\mathrm{dy}}}{{\\left({{k}}^{{2}}{+}{{x}}^{{2}}{+}{{y}}^{{2}}{+}{{z}}^{{2}}\\right)}^{{2}}}{+}\\frac{{{a}}^{{2}}{}{\\mathrm{dz}}{}{\\mathrm{dz}}}{{\\left({{k}}^{{2}}{+}{{x}}^{{2}}{+}{{y}}^{{2}}{+}{{z}}^{{2}}\\right)}^{{2}}}$ (2.9)\n\nDefine a pair of vectors which span a generic tangent plane.\n\n M2 > $X≔\\mathrm{evalDG}\\left(\\mathrm{D_x}+r\\mathrm{D_y}+s\\mathrm{D_y}\\right)$\n ${X}{:=}{\\mathrm{D_x}}{+}\\left({s}{+}{r}\\right){}{\\mathrm{D_y}}$ (2.10)\n M2 > $Y≔\\mathrm{evalDG}\\left(\\mathrm{D_y}+t\\mathrm{D_z}\\right)$\n ${Y}{:=}{\\mathrm{D_y}}{+}{t}{}{\\mathrm{D_z}}$ (2.11)\n\nCalculate the curvature and sectional curvature. Note that the sectional curvature is independent of the parameters $r,s,t$ appearing in the vector fields $X$ and $Y$.\n\n M2 > $\\mathrm{R2}≔\\mathrm{CurvatureTensor}\\left(\\mathrm{g2}\\right):$\n M2 > $\\mathrm{K2}≔\\mathrm{SectionalCurvature}\\left(\\mathrm{g2},\\mathrm{R2},X,Y\\right)$\n ${\\mathrm{K2}}{:=}\\frac{{4}{}{{k}}^{{2}}}{{{a}}^{{2}}}$ (2.12)\n\nSince the metric $\\mathrm{g2}$ has constant sectional curvature and the dimension of $\\mathrm{M2}$ is $3$, the sectional curvature is 1/6 the Ricci scalar.\n\n M2 > $\\mathrm{S2}≔\\mathrm{RicciScalar}\\left(\\mathrm{g2},\\mathrm{R2}\\right)$\n ${\\mathrm{S2}}{:=}\\frac{{24}{}{{k}}^{{2}}}{{{a}}^{{2}}}$ (2.13)\n\nExample 3.\n\nWe re-work the previous example in an orthonormal frame.\n\n M2 > $f≔\\frac{a}{{k}^{2}+{x}^{2}+{y}^{2}+{z}^{2}}$\n ${f}{:=}\\frac{{a}}{{{k}}^{{2}}{+}{{x}}^{{2}}{+}{{y}}^{{2}}{+}{{z}}^{{2}}}$ (2.14)\n M2 > $\\mathrm{FR}≔\\mathrm{FrameData}\\left(\\left[f\\mathrm{dx},f\\mathrm{dy},f\\mathrm{dz}\\right],\\mathrm{M3}\\right):$\n M2 > $\\mathrm{DGsetup}\\left(\\mathrm{FR}\\right)$\n ${\\mathrm{frame name: M3}}$ (2.15)\n M3 > $\\mathrm{g3}≔\\mathrm{evalDG}\\left(\\mathrm{Θ1}&t\\mathrm{Θ1}+\\mathrm{Θ2}&t\\mathrm{Θ2}+\\mathrm{Θ3}&t\\mathrm{Θ3}\\right)$\n ${\\mathrm{g3}}{:=}{\\mathrm{Θ1}}{}{\\mathrm{Θ1}}{+}{\\mathrm{Θ2}}{}{\\mathrm{Θ2}}{+}{\\mathrm{Θ3}}{}{\\mathrm{Θ3}}$ (2.16)\n\nCalculate the sectional curvature.\n\n M3 > $\\mathrm{R3}≔\\mathrm{CurvatureTensor}\\left(\\mathrm{g3}\\right):$\n M3 > $\\mathrm{K3}≔\\mathrm{SectionalCurvature}\\left(\\mathrm{g3},\\mathrm{R3},\\mathrm{E1}+r\\mathrm{E2}+t\\mathrm{E3},\\mathrm{E2}+t\\mathrm{E3}\\right)$\n ${\\mathrm{K3}}{:=}\\frac{{4}{}{{k}}^{{2}}}{{{a}}^{{2}}}$ (2.17)\n\nExample 4.\n\nFirst create a 3 dimensional manifold $\\mathrm{M4}$ and define a metric $\\mathrm{g4}$ on $\\mathrm{M4}$.\n\n M3 > $\\mathrm{DGsetup}\\left(\\left[x,y,z\\right],\\mathrm{M4}\\right)$\n ${\\mathrm{frame name: M4}}$ (2.18)\n M4 > $\\mathrm{g4}≔\\mathrm{evalDG}\\left(y\\mathrm{dx}&t\\mathrm{dx}+\\mathrm{dy}&t\\mathrm{dy}+\\mathrm{dz}&t\\mathrm{dz}\\right)$\n ${\\mathrm{g4}}{:=}{y}{}{\\mathrm{dx}}{}{\\mathrm{dx}}{+}{\\mathrm{dy}}{}{\\mathrm{dy}}{+}{\\mathrm{dz}}{}{\\mathrm{dz}}$ (2.19)\n\nDefine a pair of vectors which span a generic tangent plane.\n\n M4 > $X≔\\mathrm{evalDG}\\left(\\mathrm{D_x}+r\\mathrm{D_y}+s\\mathrm{D_z}\\right)$\n ${X}{:=}{\\mathrm{D_x}}{+}{r}{}{\\mathrm{D_y}}{+}{s}{}{\\mathrm{D_z}}$ (2.20)\n M4 > $Y≔\\mathrm{evalDG}\\left(\\mathrm{D_y}+t\\mathrm{D_z}\\right)$\n ${Y}{:=}{\\mathrm{D_y}}{+}{t}{}{\\mathrm{D_z}}$ (2.21)\n\nCalculate the curvature and sectional curvature. In this example, the sectional curvature is dependent on the parameters $r,s,t$ appearing in the vector fields $X$ and $Y$.\n\n M4 > $\\mathrm{R4}≔\\mathrm{CurvatureTensor}\\left(\\mathrm{g4}\\right):$\n M4 > $\\mathrm{K4}≔\\mathrm{SectionalCurvature}\\left(\\mathrm{g4},\\mathrm{R4},X,Y\\right)$\n ${\\mathrm{K4}}{:=}\\frac{{1}}{{4}{}{y}{}\\left({{s}}^{{2}}{+}{{t}}^{{2}}{}{{r}}^{{2}}{+}{y}{}{{t}}^{{2}}{+}{y}{-}{2}{}{t}{}{s}{}{r}\\right)}$ (2.22)", null, "See Also" ]

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[ "# AQA A Level Maths: Statistics复习笔记3.1.1 Calculating Probabilities & Events\n\n### Probability Basics\n\n#### What do I need to know about probability for AS and A level Mathematics?\n\n• The language used in probability can be confusing so here are some definitions of commonly misunderstood terms\n• An experiment is a repeatable activity that has a result that can be observed or recorded; it is what is happening in a question\n• An outcome is the result of an experiment\n• All possible outcomes can be shown in a sample space – this may be a list or a table and is particularly useful when it is difficult to envisage all possible outcomes in your head\n\ne.g.  The sample space below is for two fairfour-sided spinners whose outcomes are the product of the sides showing when spun.", null, "• An event is an outcome or a collection of outcomes; it is what we are interested in happening\n• Do note how this could be more than one outcome\ne.g. For the spinners above,\nthe event “the product is -2” has one outcome but\nthe event “the product is negative” has 6 outcomes\n\n####", null, "How do I solve A level probability questions?\n\n•  The big difference with probability at A level is the language and the notation used\n•", null, "Be aware of whether you are using theoreticalprobabilities or probabilities based on the results of several experiments (relativefrequency). You may have to compare the two and make a judgement as to whether there is bias in the experiment.\n\ne.g.        The outcomes from rolling a fair dice have theoretical probabilities but the outcomes from a football match would be based on previous results between the two teams\n\n• Ensure you can interpret common ways of displaying data – from frequency tables, histograms, box plots and other ways to illustrate data\n• See Revision Notes       2.1.2 Frequency Tables\n2.2.1 Data Presentation\n2.2.2 Box Plots & Cumulative Frequency\n2.2.3 Histograms\n• Be particularly careful when using histograms\n• These use frequency density, not frequency\n• Using parts of bars may be required due to where class boundaries fall so values will be estimates (using the proportion of the bar needed, sometimes called interpolation)\n\n#### Worked Example\n\n100 skydivers took part in an all-day charity event, with the altitude of the aeroplane at which they jumped from summarised in the histogram below.", null, "(a)Use the histogram to find the probability that a randomly chosen skydiver jumped from the aeroplane at an altitude\n(i)between 14 000 and 16 000 feet,\n(ii)between 16 000 and 20 000 feet.\n(b)Estimate the probability that a randomly chosen skydiver jumped from the aeroplane at an altitude between 13 000 and 15 000 feet.", null, "", null, "#### Exam Tip\n\n• Most probability questions are in context so can be long and wordy; go back and re-read the question, several times, whenever you need to\n• Try to get immersed in the context of the question to help understand a problem\n\n### Independent & Mutually Exclusive Events\n\n#### What are independent events?\n\n• Independent eventsdo not affect each other\n• For two independent events, the probability of one event happening is unaffected by the outcome of the other event\n\ne.g.    The events “rolling a 6 on a dice” and “flipping heads on a coin” are    independent - the outcome “rolling a 6” does not affect the probability of the outcome “heads” (and vice versa)\n\n• For two independent events, A and B", null, "• Independent events could refer to events from different experiments\n\n####", null, "How do I solve problems involving independent and mutually exclusive events?\n\n• Make sure you know the statistical terms – independent and mutually exclusive\n• Remember\n•", null, "Solving problems will require interpreting the information given and the application of the appropriate formula\n• Information may be explained in words or by diagram(s)\n\n(including Venn diagrams – see Revision Note 3.1.2 Venn Diagrams)\n\n• Showing or determining whether two events are independent or mutually exclusive are also common\n• To do this you would show the relevant formula is true\n\nJust for fun …\n\n• A well-known sports TV broadcaster used to advertise their football matches as either “Live and exclusive” or “Exclusively live” – can you tell the difference?\n• “Live and exclusive” meant that the broadcaster was airing the football match live and was the only broadcaster allowed to air any of the match at any time.\n“Exclusively live” meant that the broadcaster was the only one airing the football match live, but other broadcasters would be able to air any of the match afterwards.\n\n#### Worked Example", null, "", null, "#### Exam Tip\n\n• Try to rephrase questions in your head in terms of AND and/or OR !\ne.g.      A fair six-sided die is rolled and a fair coin is flipped.\n“Find the probability of obtaining a prime number with heads.”\n\nwould be\n\n“Find the probability of rolling a 2 OR a 3 OR a 5 AND heads.”", null, "" ]

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[ "5\n\n# Is/pearl/Downloads/maths\"o2OdiffpdfQuestion 3. (Total matks: 16 Find tle: solution_ ineluding domain of Ihe initial value problemwith V = \"bahMake suTe&#x...\n\n## Question\n\n###### Is/pearl/Downloads/maths\"o2OdiffpdfQuestion 3. (Total matks: 16 Find tle: solution_ ineluding domain of Ihe initial value problemwith V = \"bahMake suTe' explain the se]k of JUIu sulutiola clueck Hol reqquirexi but_ Iot full Matky WIsI juclul cuough words Aldl symlals make VT solutiou cle;tActnvate . 06pafcn\n\nIs/pearl/Downloads/maths\"o2Odiffpdf Question 3. (Total matks: 16 Find tle: solution_ ineluding domain of Ihe initial value problem with V = \"bah Make suTe' explain the se]k of JUIu sulutiola clueck Hol reqquirexi but_ Iot full Matky WIsI juclul cuough words Aldl symlals make VT solutiou cle;t Actnvate . 0 6pafcn", null, "", null, "#### Similar Solved Questions\n\n##### 5.13 Suppose that we want to fit the no-F weighted least squares ~intercept Assume model y Bx + using but have unequal variances. that the observations are uncorrelated Find general formula for the of B. weighted least-squares estimator What is the variance of the weighted least Suppose ~squares estimator? that Var(y) tional-to the 7Cri, that is, the variance of ~corresponding xi ~Using Yt is propor- find the weighted least-= the results of parts & and b, this estimator: ~squares estimator\n5.13 Suppose that we want to fit the no-F weighted least squares ~intercept Assume model y Bx + using but have unequal variances. that the observations are uncorrelated Find general formula for the of B. weighted least-squares estimator What is the variance of the weighted least Suppose ~squares e...\n##### MATH 249Worksheet 20Questions for today's in-class discussion _ Turn in today; 12/3 For Questions # and 2, use the matrixA =Find the eigenvalues and eigenvectors of the matrix A_ 2. Find the solution of the system of first order linear differential equations y (t) = Ay(t) , y(0) = (91(0) , 92(0))T = (1,-1)T\nMATH 249 Worksheet 20 Questions for today's in-class discussion _ Turn in today; 12/3 For Questions # and 2, use the matrix A = Find the eigenvalues and eigenvectors of the matrix A_ 2. Find the solution of the system of first order linear differential equations y (t) = Ay(t) , y(0) = (91(0) , ...\n##### Point) Let Pz be the vector space of all polynomials of degree 2 or less, and let H be the subspace spanned by &x 4x2 7, Sx - 6+2 1 and 3x x2The dimension of the subspace H is{8x ~ 4x2 _ 7,Sx - 6x2 _ 1,3x-x2 _ 3} basis for Pz choose can explain and justify your answer:Be sure youA basis for the subspace H is comma separated Iist of polynomials_Enter polynomial\npoint) Let Pz be the vector space of all polynomials of degree 2 or less, and let H be the subspace spanned by &x 4x2 7, Sx - 6+2 1 and 3x x2 The dimension of the subspace H is {8x ~ 4x2 _ 7,Sx - 6x2 _ 1,3x-x2 _ 3} basis for Pz choose can explain and justify your answer: Be sure you A basis for ...\n##### Y(s) = 98 Solve e! 0 2 for Y(s), of view 1 11 Homework: the 9y = cos 2t the Laplace pts table properties sin 2t, y(O) 1 transform of the =3, Section of Laplace ~oion y(t) to 7.5 transforms. the initial value problem below:\nY(s) = 98 Solve e! 0 2 for Y(s), of view 1 11 Homework: the 9y = cos 2t the Laplace pts table properties sin 2t, y(O) 1 transform of the =3, Section of Laplace ~oion y(t) to 7.5 transforms. the initial value problem below:...\n##### Cld Question (1 point) A10* 10-2 Maqueous solution - CalOHl at 25,0 \"Chas & pH of (First calculate hydroxide Ion concentration; concentration using fror hete calculate hydroniurn ion H,0*] x [OH ] = Kw =L.0x 10-14PH = -log[H3ot])C A) 5.0 x 10-13B) 12.30C) 12.00DI 1.70MacBook Air30888% 5532.0 * 10-2\nCld Question (1 point) A10* 10-2 Maqueous solution - CalOHl at 25,0 \"Chas & pH of (First calculate hydroxide Ion concentration; concentration using fror hete calculate hydroniurn ion H,0*] x [OH ] = Kw =L.0x 10-14 PH = -log[H3ot]) C A) 5.0 x 10-13 B) 12.30 C) 12.00 DI 1.70 MacBook Air 30 88...\n##### QUESTION4Imaginc thatthe diugram below free energy diagram for the chemical reaction would vou expect tne dharam for the closely-related reaction ADP dlffer from the first one?CFIn what WaY61couid decreasecauld decreatcgn '0l 2 could(evensedtneabovn coud mpocn\nQUESTION4 Imaginc thatthe diugram below free energy diagram for the chemical reaction would vou expect tne dharam for the closely-related reaction ADP dlffer from the first one? CF In what WaY 6 1 couid decrease cauld decreatc gn '0l 2 could (evensed tneabovn coud mpocn...\n##### Suppose you want to test the hypothesis that arandom variable has a mean for one group that is higher than the meanit has for another group. Based on the SPSS output you should never reject the hypothesis if:The p-value is less than 0.05_Half the p-value is less than 0.05_The mean for the first group is higher than for the second group.The mean for the first group is lower than for the second group.\nSuppose you want to test the hypothesis that arandom variable has a mean for one group that is higher than the meanit has for another group. Based on the SPSS output you should never reject the hypothesis if: The p-value is less than 0.05_ Half the p-value is less than 0.05_ The mean for the first g...\n##### Suppose for a particular commodity, € 1.3 and that currently 30,000 units are selling for 814 each: How does a 1 % increase in the price affect the revenue?A 1 % increase in price decreasesthe total revenue by\nSuppose for a particular commodity, € 1.3 and that currently 30,000 units are selling for 814 each: How does a 1 % increase in the price affect the revenue? A 1 % increase in price decreases the total revenue by...\n##### A pharmaceutical development corporation tests a new pain reliever on 78 people with head colds. Of the 78 people; 58 are cured. If we assume that 20% of the head colds would have subsided naturally without the drug, what is the probability that 58 or more would be cured without the pain reliever?\nA pharmaceutical development corporation tests a new pain reliever on 78 people with head colds. Of the 78 people; 58 are cured. If we assume that 20% of the head colds would have subsided naturally without the drug, what is the probability that 58 or more would be cured without the pain reliever?...\n##### Ped 8Dolemine Ine elecinca proe Inai two piotons tne nucleus 0f helium alom exert on ach other when separated by L.8 * 101 LI Exprose Youf wlth tho apercprlate unlla:Fia 27ValueUnitsSubmliBqauaL AnNd\nPed 8 Dolemine Ine elecinca proe Inai two piotons tne nucleus 0f helium alom exert on ach other when separated by L.8 * 101 LI Exprose Youf wlth tho apercprlate unlla: Fia 27 Value Units Submli BqauaL AnNd...\n##### Let u = (1, 2, 3),= (4,4, -2), and w = (3, 0, -3). Find 2uSTEP 1: Multiply each vector by scalar:XSTEP 2: Add the results from step\nLet u = (1, 2, 3), = (4,4, -2), and w = (3, 0, -3). Find 2u STEP 1: Multiply each vector by scalar: X STEP 2: Add the results from step...\n##### Here we consider the function f: R + R that is given by f(c) = [sin()1: a) Show, that f (x) is ZT-periodic: Use next Eulers equations to show that the complex f(c) = Cn Cneinz Fourier series is given by 1 Cn for n € Z T 1 4n2 (hint: you can with advantage justify, that Cn can be calculated by integration over the interval [0, 2n] 4, & € L?([-T,w]) b) In the space L? ( [T, ~n]) is the inner product given for arbitrary by (p,+) = p(c)v(c) dx. Determine whether the Fourier series for f (x)\nHere we consider the function f: R + R that is given by f(c) = [sin()1: a) Show, that f (x) is ZT-periodic: Use next Eulers equations to show that the complex f(c) = Cn Cneinz Fourier series is given by 1 Cn for n € Z T 1 4n2 (hint: you can with advantage justify, that Cn can be calculated by ...\n##### I +13. f(z) = (2' + 3)(24 72)f(c) = 2+2-2\nI +1 3. f(z) = (2' + 3)(24 72) f(c) = 2+2-2...\n##### A) Study the convergence ofIn (2) dx, 25/2and if the integral is convergent, give its value_ A.1) The integral is Click for ListA.2) If the integral converges, its exact value is Input 333 if the integral is divergent. B) Study the convergence of1n (& dx 25/2and if the integral is convergent, give its value_ B.1) The integral is Click for ListB.2) If the integral converges, its exact value is Input 333 if the integral is divergent.\nA) Study the convergence of In (2) dx, 25/2 and if the integral is convergent, give its value_ A.1) The integral is Click for List A.2) If the integral converges, its exact value is Input 333 if the integral is divergent. B) Study the convergence of 1n (& dx 25/2 and if the integral is convergen...\n##### A mass of 2 kg is left free on an inclined plane of o = 30 ° andin contact with a spring k = 98 N / M and negligible mass, which isundeformed in the position x = 0. Determine if the mass is placedwithout velocity and the friction between the mass and the plane iszero: to.a. an expression for potential energyb. the equilibrium position\na mass of 2 kg is left free on an inclined plane of o = 30 ° and in contact with a spring k = 98 N / M and negligible mass, which is undeformed in the position x = 0. Determine if the mass is placed without velocity and the friction between the mass and the plane is zero: to. a. an expression for...\n##### Let A =Determine which of the following is FALSE.liuea combination of the colums of A211 + T2 AI =cAx=bis consistent [or b of the fOrmThe columns of A does not span R3.none 0f these.\nLet A = Determine which of the following is FALSE. liuea combination of the colums of A 211 + T2 AI = cAx=bis consistent [or b of the fOrm The columns of A does not span R3. none 0f these...." ]

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[ "# Gaussian Antenna as Approximation for Spiral Antenna\n\nThis example shows how a Gaussian antenna can approximate the radiation pattern of a spiral antenna. Spiral antennas are known for their wideband behavior and are often used in wideband communications. The Gaussian antenna approximation is widely used in the literature.\n\nCreate a spiral antenna using Antenna Toolbox™. The antenna is in the y-z plane and has inner and outer radii of 0.65 mm and 40 mm, respectively.\n\n```se = spiralArchimedean( ... InnerRadius=0.65e-3,OuterRadius=40e-3, ... Tilt=90,TiltAxis=\"y\"); show(se)```", null, "Compute the radiation pattern for the spiral antenna at an operating frequency of 4 GHz. Specify a range of azimuth angles from –90° to 90° and zero elevation. Normalize the pattern so that its maximum value is 0 dB.\n\n```fc = 4e9; az = -90:0.5:90; sePattern = pattern(se,fc,az,0, ... CoordinateSystem=\"rectangular\",Type=\"powerdb\"); sePatternNorm = sePattern - max(sePattern);```\n\nUse Phased Array System Toolbox™ to create a Gaussian element with the same beamwidth as the spiral antenna. Compute its radiation pattern, which is normalized by construction.\n\n```seBw = beamwidth(se,fc,az,0); ge = phased.GaussianAntennaElement(Beamwidth=seBw); gePattern = pattern(ge,fc,az,0, ... CoordinateSystem=\"rectangular\",Type=\"powerdb\");```\n\nFind the smallest positive azimuth angle at which the patterns differ by about 3 dB.\n\n```idx = find((abs(sePatternNorm.' - gePattern) >= 3) & (az' >= 0),1); az3dB = az(idx);```\n\nPlot the patterns for the spiral and Gaussian antennas. Overlay the 3-dB points. The Gaussian antenna pattern matches the spiral antenna pattern well out to about 75 degrees and thus can be used as an excellent approximation of a spiral antenna.\n\n```plot(az,sePatternNorm,az,gePattern) xline([-az3dB az3dB],'--') xlabel(\"Azimuth Angle, az (degrees)\") ylabel(\"Power (dB)\") legend(\"Spiral\",\"Gaussian\",\"3-dB Point\")```", null, "" ]

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[ "# Isoparametric hypersurfaces with four principal curvatures\n\n### Abstract\n\nLet $M$ be an isoparametric hypersurface in the sphere $S^n$ with four distinct principal curvatures. Münzner showed that the four principal curvatures can have at most two distinct multiplicities $m_1, m_2$, and Stolz showed that the pair $(m_1,m_2)$ must either be $(2,2)$, $(4,5)$, or be equal to the multiplicities of an isoparametric hypersurface of FKM-type, constructed by Ferus, Karcher and Münzner from orthogonal representations of Clifford algebras. In this paper, we prove that if the multiplicities satisfy $m_2 \\geq 2m_1 – 1$, then the isoparametric hypersurface $M$ must be of FKM-type. Together with known results of Takagi for the case $m_1 = 1$, and Ozeki and Takeuchi for $m_1 = 2$, this handles all possible pairs of multiplicities except for four cases, for which the classification problem remains open." ]

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[ "# Parallel integer sorting using small operations\n\nRamachandran Vaidyanathan, Carlos R.P. Hartmann, Pramod K. Varshney\n\nResearch output: Contribution to journalArticle\n\n4 Scopus citations\n\n### Abstract\n\nWe consider the problem of sorting n integers in the range [0, nc-1], where c is a constant. It has been shown by Rajasekaran and Sen that this problem can be solved \"optimally\" in O(log n) steps on an EREW PRAM with O(n) n-bit operations, for any constant ∈>O. Though the number of operations is optimal, each operation is very large. In this paper, we show that n integers in the range [0, nc-1] can be sorted in O(log n) time with O(nlog n)O(1)-bit operations and O(n) O(log n)-bit operations. The model used is a non-standard variant of an EREW PRAMtthat permits processors to have word-sizes of O(1)-bits and Θ(log n)-bits. Clearly, the speed of the proposed algorithm is optimal. Considering that the input to the problem consists of O (n log n) bits, the proposed algorithm performs an optimal amount of work, measured at the bit level.\n\nOriginal language English (US) 79-92 14 Acta Informatica 32 1 https://doi.org/10.1007/BF01185406 Published - Jan 1 1995\n\n### ASJC Scopus subject areas\n\n• Software\n• Information Systems\n• Computer Networks and Communications" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "RE: Books on learning mathematics with Mathematica\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg69278] RE: [mg69240] Books on learning mathematics with Mathematica\n• From: \"David Park\" <djmp at earthlink.net>\n• Date: Tue, 5 Sep 2006 05:30:43 -0400 (EDT)\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```Alex,\n\nDo you actually have Mathematica?\n\nIf you are serious enough about learning mathematics so that it is worth the\ncost, then Mathematica will certainly do the job for you.\n\nThere are good books to get started with Mathematica such as \"The Beginner's\nGuide to Mathematica\" by Jerry Glynn and Theodore Gray.\n\nThe best way to learn Mathematica is to first read Stephen Wolfram's\n'Suggestions about Learning Mathematica' in the front of The Mathematica\nBook and then work through most of Part I, at least, in the book to become\nfamiliar with the syntax and more common commands.\n\nThen pick a non-Mathematica book, say a high school book on algebra and\ntrigonometry and try to work through it with Mathematica.\n\nMathematica works best when you take a 'mathematical' as opposed to a\n'programming' approach. Much of this is encapsulated in the functional\nprogramming and rule-based programming that is also a part of Mathematica as\nopposed to procedural programming. You will soon get the hang of it and\nthink much more in mathematical terms.\n\nDavid Park\n\nFrom: Alex Polite [mailto:notmyprivateemail at gmail.com]\nTo: mathgroup at smc.vnet.net\n\nI'm (amongst other things) an autodidact programmer. Quite often I'll\nrun into computing problems that I realize are well understood, I'll\ndo a google and find some papers outlining algorithms that tackles the\nproblem. Sprinkled in the text there's infallibly some mathematical\nformulas. That's were my troubles begin ;)\n\nOne part of the problem is simply not being familiar mathematical\nnotation. Once I grasp what the formula express I usually have no\nproblem understanding the concept. If the formula had been expressed\nin pseudo code, I'd have an easier time following along.\n\nI've recently found out about Mathematica and it seems like it could\nbe a great learning tool for someone like me. If I understand things\ncorrectly, Mathematica will let me enter formulas in the syntax of a\nprogramming language and then render them in standard mathematical\nnotation, right?\n\nNow, to optimize this self study program, all I need is a great text\nbook that uses Mathematica to teach math. Preferably the book should" ]

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[ "# Is the stability matrix of a linearised RG flow always diagonalisable?\n\nThis is a follow up on \"Why are the eigenvalues of a linearized RG transformation real?\".\n\nMy question is simple: Is there some physical (or mathematical) reason for the stability matrix of Renormalisation Group flows close to fixed points to be diagnoalisable? What is it? If there isn't: Are there known counter examples? How do we deal with them?\n\n## 1 Answer\n\nI don't believe there is a mathematical reason, especially if there is latitude in reverse-engineering the field theory or stat mech system to evince such a behavior. Indeed, if Lorentz-nonivariant systems are examined, things like limit cycles , e.g. this one are not hard to concoct. As for physical reasons, they might well be easy to bypass/moot if one argued for them. I don't know of any systems, however, with this property, which might not say much.\n\nAs a mathematical wisecrack, I could manufacture a simple toy system with two couplings, x and y and logarithmic scale variable t : $$\\dot{x}=-x + ay, \\qquad \\dot{y}= -y ,$$ with evident solutions stable around the fixed point (0,0), $$y= e^{-t}, \\qquad x= (c +at) e^{-t} .$$ The stability matrix of the ODE system is $$\\left( \\begin{array}{cc} -1 & a \\\\ 0 & -1 \\\\ \\end{array} \\right)$$ which is not diagonalizable, with only one eigenvector, $$\\left( \\begin{array}{c} 1 \\\\ 0 \\\\ \\end{array} \\right)$$ of eigenvalue -1. This is not to say the system is not stable, however, if one could solve the ODE, somehow, as here." ]

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[ "4.0\n\n# Increasing Performance when Assigning Subsets of Arrays\n\nBarrett Sather\n\nWhen working with multi-dimensional data, as is the case with images, it is sometimes necessary to replace a chunk of data with another chuck of the same size. With images, this could be a spatial area that needs to be replaced because of clouds, or a spectral band that needs to be replaced.\n\nIn the case of a spatial replacement, say we want to replace a section of data with an array of the same size. For this example, the subset is 50 rows and 50 columns, and the larger data is 250 rows, and 250 columns.\n\nIDL> array = bytarr(250, 250) + 127B\n\nIDL> sub = indgen(50, 50)\n\nTo replace a section of data in IDL, one might go about itthis way:\n\nIDL> array[50:99, 100:149] = sub\n\nThis is an acceptable method, but it is not the faster way of doing this. With the method of replacement done above, IDL will replace each individual element one at a time. To replace the entire subset at once, IDL only requires the first index where the subset needs to be inserted:\n\nIDL> array[50, 100] = sub\n\nIDL will fill in data until the entirety of the subset is in the larger data. Do be cautious though, if you run out of room IDL will either give unexpected results, or throw and out-of-bounds error. Here is the image of the array we just manipluated:\n\nIDL> i = image(array)", null, "An Array Within an Array\n\nThe advantage though, is that this is much faster than explicitly calling out the array elements to replace, or using a wildcard (*) character. Let's take a look at using wildcard characters for the next example - replacing a band.\n\nIf we wanted to take an image, and decrease the amount of red in that image, we could do that by multiplying the red band by a number smaller than one. First, let's get some image data:\n\nIDL> file = file_which('rose.jpg')\n\nIDL> help, data\n\nDATA           BYTE      = Array[3, 227, 149]\n\nIn order to get the red band we will have to use wildcards. However, this will not be a performance issue, as we are assigning a chuck of data to a variable. Nothing is being replaced. As a good rule of thumb, it is best to avoid using wildcards on the left-hand side of an assignment when possible, but wildcards on the right-hand side of an assignment are okay.\n\nIDL> red = data[0,*,*]\n\nNow let's assign the old red band to the original multiplied by 0.6. We could use wildcards again to do this:\n\nIDL> data[0,*,*]= byte(red * 0.6)\n\nBut we now know that this is not the most efficient way to assign this band in IDL. It only needs the first index of the data that we want to replace:\n\nIDL> data[0,0,0] = byte(red* 0.6)\n\nBecause the red band is still the same size, the entire band will be replaced all at once instead of one element at a time. For these small arrays, the speed increase is not huge, but with larger datasets, it becomes important to use this practice to achieve high levels of performance.\n\nIDL> i = image(data)", null, "" ]

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[ "", null, "Home  - Math_Discover - Boolean Algebra\ne99.com Bookstore\n Images Newsgroups\n Page 1     1-20 of 93    1  | 2  | 3  | 4  | 5  | Next 20\n\nBoolean Algebra:     more books (100)\n1. Schaum's Outline of Modern Abstract Algebra (Schaum's) by Frank Ayres, 1965-06-01\n2. Boolean Algebra and Its Applications (Dover Books on Mathematics) by J. Eldon Whitesitt, 1995-03-27\n3. Schaum's Outline of Boolean Algebra and Switching Circuits by Elliott Mendelson, 1970-06-01\n4. Ones and Zeros: Understanding Boolean Algebra, Digital Circuits, and the Logic of Sets (IEEE Press Understanding Science & Technology Series) by John R. Gregg, 1998-03-16\n5. Boolean Algebra by R. L. Goodstein, 2007-01-15\n6. Sets and Boolean algebra, (Mathematical studies: a series for teachers and students, no. 4) by Marcel Rueff, 1970\n7. Handbook of Boolean Algebras, Volume Volume 2 by MONK, 1989-03-01\n8. Boolean Algebras in Analysis (Mathematics and Its Applications) by D.A. Vladimirov, 2002-03-31\n9. Digital Circuits: Numbering Systems, Binary Codes, Logic Gates, Boolean Algebra (Engineer's Tutor Series) by Amalou Abdelilah, 1989-08\n10. Boolean Algebra for Computer Logic by Harold E. Ennes, 1978-08\n11. Applied Boolean Algebra an Elementar 2ND Edition by Franz Hohn, 1966\n12. Lectures on Boolean Algebras by Paul R. Halmos, 1974-01\n13. BOOLEAN ALGEBRA AND ITS APPLICATION INCLUDING BOOLEAN MATRIX ALGEBRA by H. Graham Flegg, 1964\n14. Cardinal Functions on Boolean Algebras (Lectures in Mathematics) by J.Donald Monk, 1990-12", null, "", null, "", null, "", null, "", null, "", null, "", null, "1. The Mathematics Of Boolean Algebra\nJUL 5 2002. The Mathematics of boolean algebra. On the other hand, the theory of a boolean algebra with a distinguished subalgebra is undecidable.\nhttp://plato.stanford.edu/entries/boolalg-math/\nversion history\nHOW TO CITE\n\nTHIS ENTRY\n##### Stanford Encyclopedia of Philosophy\nA B C D ... Z\nThis document uses XHTML-1/Unicode to format the display. Older browsers and/or operating systems may not display the formatting correctly. last substantive content change\nJUL\n##### 1. Definition and simple properties\nA Boolean algebra (BA) is a set A together with binary operations + and and a unary operation , and elements 0, 1 of A such that the following laws hold: commutative and associative laws for addition and multiplication, distributive laws both for multiplication over addition and for addition over multiplication, and the following special laws: x + (x y) = x\nx (x + y) = x\nx + (-x) = 1\nx (-x) = These laws are better understood in terms of the basic example of a BA, consisting of a collection A of subsets of a set X closed under the operations of union, intersection, complementation with respect to\n\n2. Boolean Algebra\nboolean algebra. The following table contains just a few rules that hold in a boolean algebra, written in both set and logic notation.\nhttp://www.math.csusb.edu/notes/sets/boole/boole.html\n##### Boolean Algebra\nA Boolean algebra is a set with two binary operations, and , that are commutative, associative and each distributes over the other, plus a unary operation . Also required are identity elements and U for the binary operations that satisfy and for all elements A in the set. One interpretation of Boolean algebra is the collection of subsets of a fixed set X . We take and U to be set union, set intersection, complementation, the empty set and the set X respectively. Equality here means the usual equality of sets. Another interpretation is the calculus of propositions in symbolic logic. Here we take and U to be disjunction, conjunction, negation, a fixed contradiction and a fixed tautology respectively. In this setting equality means logical equivalence. It is not surprising then that we find analogous properties and rules appearing in these two areas. For example, the axiom of the distributive properties says that for sets we have while is a familiar equivalence in logic. From the axioms above one can prove DeMorgan's Laws (in some axiom sets this is included as an axiom). The following table contains just a few rules that hold in a Boolean algebra, written in both set and logic notation. Rows 3 and 4 are DeMorgan's Laws. Note that the two versions of these rules are identical in structure, differing only in the choice of symbols.\n\n3. What's So Logical About Boolean Algebra?\nboolean algebra, a simple explanation\nhttp://www.kerryr.net/pioneers/boolean.htm\n People ASCII-HTML Binary Boolean ... Home What's So Logical About Boolean Algebra? George Boole believed in what he called the ‘process of analysis’, that is, the process by which combinations of interpretable symbols are obtained. It is the use of these symbols according to well-determined methods of combination that he believed presented ‘true calculus’. Today, all our computers employ Boole's logic system - using microchips that contain thousands of tiny electronic switches arranged into logical ‘gates’ that produce predictable and reliable conclusions. The basic logic gates comprise of AND OR and NOT . It is these gates, used in differing combinations, that allow the computer to execute its operations using binary language Each gate assesses various information (consisting of high or low voltages) in accordance with predetermined rules, and produces a single high or low voltage logical conclusion. The voltage itself represents the binary yes-no, true-false, one-zero concept. AND gates will only yield a TRUE result (that is, a binary 1) if all input is TRUE. Therefore, the top two gates will produce a FALSE (binary 0) result. OR gates are less fussy. An\n\n4. Boolean Algebra\nboolean algebra. boolean algebra is defined as the study of the manipulation of symbols representing operations A Brief Overview of Logic. boolean algebra. Boole, George (18151864)\nhttp://www.programcpp.com/chapter04/4_1_4.html\nTOPIC 4.1.4\nBoolean Algebra Boolean algebra is defined as the study of the manipulation of symbols representing operations according to the rules of logic. For more information on logic or Boolean Algebra, consult the following links:\n• A Brief Overview of Logic Boolean Algebra\n##### Boole, George (1815-1864)\n• A excelent summary by Jones Telecomunications and Multimedia Encyclopedia, with many related sources, and sources listed.\n##### Boole\n• History of a Mathematician\nAlso see Fuzzy Logic (On the Net Topic 7.1.1) for related information.\n\n5. Boolean Algebra\nAddition Negative Numbers and Binary Subtraction Multiplexer Decoder/Demultiplexer boolean algebra and known today as boolean algebra. The rules of boolean algebra are\nhttp://www.play-hookey.com/digital/boolean_algebra.html\n Home www.play-hookey.com Tue, 06-01-2004 Digital Logic Families Digital Experiments Analog ... Test HTML Direct Links to Other Digital Pages: Combinational Logic: Basic Gates Derived Gates The XOR Function Binary Addition ... Boolean Algebra Sequential Logic: RS NAND Latch RS NOR Latch Clocked RS Latch RS Flip-Flop ... Converting Flip-Flop Inputs Alternate Flip-Flop Circuits: D Flip-Flop Using NOR Latches CMOS Flip-Flop Construction Counters: Basic 4-Bit Counter Synchronous Binary Counter Synchronous Decimal Counter Frequency Dividers ... The Johnson Counter Registers: Shift Register (S to P) Shift Register (P to S) The 555 Timer: 555 Internals and Basic Operation 555 Application: Pulse Sequencer Boolean Algebra One of the primary requirements when dealing with digital circuits is to find ways to make them as simple as possible. This constantly requires that complex logical expressions be reduced to simpler expressions that nevertheless produce the same results under all possible conditions. The simpler expression can then be implemented with a smaller, simpler circuit, which in turn saves the price of the unnecessary gates, reduces the number of gates needed, and reduces the power and the amount of space required by those gates. One tool to reduce logical expressions is the mathematics of logical expressions, introduced by George Boole in 1854 and known today as\n\n6. Boolean Algebra\nboolean algebra. boolean algebra provides the ability to work with many interesting functions will informally prove that boolean algebra can describe all boolean functions this\nhttp://cs-people.bu.edu/jconsidi/teaching/notes/cs210/node6.html\nNext: Desirable Circuit Properties Up: Logic Design Previous: Truth Tables\n##### Boolean Algebra\nBoolean algebra provides the ability to work with many interesting functions while often using significantly less space than a truth table. We will informally prove that boolean algebra can describe all boolean functions - this justifies our use of gates since they correspond to the building blocks of boolean algebra. Boolean expressions have the following grammar.\nThat is, a boolean expression can be a variable (representing an input wire), the negation of a boolean expression (NOT and ), the conjunction of two boolean expressions (AND and ) or the disjunction of two boolean expressions (OR and ). Boolean expressions and circuits are interchangeable though circuits are generally smaller since they may share common subexpressions. Given a truth table for a function (one always exists), we can generate a boolean expression for the same function. We can do so in the following fashion. Each line of the truth table corresponds to a particular assignment of variables. For each assignment with output true (or one), generate an expression that is true for that assignment and no others. For example, F F F T T This assignment corresponds to the expression . After generating all of these expressions, form the disjunction (OR) of all these expressions. This disjunction is the same as the function of the truth table.\n\n7. Logic Gates And Boolean Algebra\nLogic Gates and boolean algebra. Created by Mark Mamo and Shane Bauman. The following is a set of resources for a unit on Logic Gates and boolean algebra.\nhttp://educ.queensu.ca/~compsci/units/BoolLogic/titlepage.html\n Logic Gates and Boolean Algebra Created by Mark Mamo and Shane Bauman The following is a set of resources for a unit on Logic Gates and Boolean Algebra. Introduction to Boolean Logic an outline of an activity to get students thinking about situations using Boolean logic. This activity also serves as an introduction to the AND and OR logic gates. Black Box Circuits an interesting hands-on activity that investigates different gate combinations as well as introduces NAND, NOR. XOR and XNOR Summary of Logic Gates a convenient hand-out summarizing the basic logic gates, their Boolean algebra notation and their truth tables Sample Questions on Logic Gates, Circuits and Truth Tables a handout for students to complete to reinforce the ideas of logic gates, circuits, truth tables and the relationships between them Discovering the Rules of Boolean Algebra a series of worksheets to help students discover the rules of Boolean algebra for themselves Simplifying Boolean Expressions a worksheet which helps students to discover the value of simplifying Boolean expressions and the role it plays in designing circuits\n\n 8. Boolean Algebra From MathWorld boolean algebra from MathWorld A mathematical structure which is similar to a Boolean ring, but which is defined using the meet and join operators instead of the usual addition andhttp://rdre1.inktomi.com/click?u=http://mathworld.wolfram.com/BooleanAlgebra.htm\n\n9. Discovering The Basic Rules Of Boolean Algebra\nDiscovering the Basic Rules of boolean algebra. These two rules that you have discovered are known as the associative laws of boolean algebra. Example 1.\nhttp://educ.queensu.ca/~compsci/units/BoolLogic/assocdistrib.html\n##### Discovering the Basic Rules of Boolean Algebra\n• Consider the Boolean expression A + B + C. Does it matter which OR you evaluate first? Verify your answer using truth tables and then express your discovery using Boolean algebra notation. Now consider the Boolean expression A B C. Does it matter which AND is evaluated first? Once again verify your answer using truth tables and then express your discovery using Boolean algebra notation.\n• These two rules that you have discovered are known as the associative laws of Boolean algebra. Example 1\n• To get into a physics program in university, Samantha needs to have OAC physics and either OAC algebra or OAC calculus. Assign Boolean variables to the conditions and write a Boolean expression for the program requirements.\n• We will get you started: Let P represent whether or not Samantha has OAC physics.\n• Another way of stating the conditions for the physics program is that Samantha needs OAC physics and OAC algebra, or OAC physics and OAC calculus. Using the same Boolean variables as above, write a Boolean expression for the program requirements. Since both of these expressions refer to the same situation the Boolean expressions must be equal. Verify this statement by comparing the truth tables for the expressions.\n• 10. A Brief History Of Algebra And Computing: An Eclectic Oxonian View\nBy Oxford professor, Jonathan Bowen. Discusses origins in ancient Greece, Arabia and England, analytical machines, boolean algebra, and recent developments in the field.\nhttp://vmoc.museophile.org/algebra/\n\nNext: The Origins of Algebra\nUp: The Virtual Museum of Computing\n##### Jonathan P. Bowen\nEx Oxford University Computing Laboratory\nWolfson Building, Parks Road, Oxford UK Dedicated to Prof. C.A.R. Hoare , F RS James Martin Professor of Computing\nat the Oxford University Computing Laboratory\nText originally completed on his 60th birthday, 11th January 1994\nIf you are faced by a difficulty or a controversy in science, an ounce of algebra is worth a ton of verbal argument. J.B.S. Haldane That excellent woman knew no more about Homer than she did about Algebra, but she was quite contented with Pen's arrangements ... and felt perfectly confident that her dear boy would get the place which he merited. Pendennis (1848-50), by William Makepeace Thackeray\n(The story of the progress of an Oxford student.) O h h eck- a nother h our o f a lgebra!\nPrinted version published in IMA Bulletin , pages 6-9, January/February 1995. See also longer version in\n\n 11. Redirect... To New Location boolean algebra assistant program. The program allows minimizing Boolean function by a graphic method of Karnaugh maps. Export Map to HTML table to easier creation of synthesis reports.http://www.elistel.net/~shurik/karnaugh/\n\n12. Volume IV - Digital :: Chapter 7: BOOLEAN ALGEBRA\nBoolean arithmetic. boolean algebraic identities. boolean algebraic properties. Ask questions and help answer others. Check it out! Chapter 7 boolean algebra.\n Volume I - DC Volume II - AC Volume III - Semiconductors Volume IV - Digital ... Converting truth tables into Boolean expressions Search All Volumes Volume I - DC Volume II - AC Volume III - Semiconductors Volume IV - Digital Volume V - Reference Volume VI - Experiments Check out our new Electronics Forums Ask questions and help answer others. Check it out! Chapter 7: BOOLEAN ALGEBRA All About Circuits Volume IV - Digital Chapter 7: BOOLEAN ALGEBRA All About Electronic Circuits ... Contact\n\n13. Design, Analysis & Circuit Theory\nBasic DC analysis, analogue coupling circuits, antenna theory, boolean algebra, BJT configurations, basic DC theory BJT bias, basic AC theory, BJT bias analysis. An extensive array of information.\n Welcome to my new unified circuit analysis, design and theory page. As all three subjects are all closely integrated, then it makes sense to provide a common reference page for navigation. Remember,only underlined articles are complete. Basic DC Analysis Analogue Coupling Circuits Antenna Theory and Transmitting Boolean Algebra Examples BJT Configurations Basic DC Theory Frequency Response BJT Bias Circuits Basic AC Theory BJT Bias Analysis Negative Feedback in amplifiers Boolean Algebra Colpitts Oscillator Transistor as a Switch Karnaugh Maps Transmitter Distance Multistage Bias Circuits Ohms Law for AC Circuits Small-Signal Analysis FET Configurations Logic Truth Tables Measuring Input Output Impedance Low Noise Design Techniques Transistor Modelling Over Voltage Protection Thevevin and Norton Networks Removing \"DC thump\" from Audio Tuned Circuits ... Switch De-Bouncing Mesh Analysis Unregulated Power Supply Design\n\n14. Boolean Algebraic Identities - Chapter 7: BOOLEAN ALGEBRA - Volume IV - Digital\nboolean algebraic identities. Like ordinary algebra, boolean algebra has its own unique identities based on the bivalent states of Boolean variables.\nVolume I - DC Volume II - AC Volume III - Semiconductors Volume IV - Digital ... Back to Chapter Index Search All Volumes Volume I - DC Volume II - AC Volume III - Semiconductors Volume IV - Digital Volume V - Reference Volume VI - Experiments\nCheck out our new Electronics Forums\n\nBoolean algebraic identities\nVolume IV - Digital Chapter 7: BOOLEAN ALGEBRA Boolean algebraic identities\n##### Boolean algebraic identities\nIn mathematics, an identity is a statement true for all possible values of its variable or variables. The algebraic identity of x + = x tells us that anything (x) added to zero equals the original \"anything,\" no matter what value that \"anything\" (x) may be. Like ordinary algebra, Boolean algebra has its own unique identities based on the bivalent states of Boolean variables. The first Boolean identity is that the sum of anything and zero is the same as the original \"anything.\" This identity is no different from its real-number algebraic equivalent: No matter what the value of A, the output will always be the same: when A=1, the output will also be 1; when A=0, the output will also be 0.\n\n15. Mai Gehrke's Curriculum Vita\nNew Mexico State University Nonstandard mathematics, operators on boolean algebras, fuzzy mathematics, universal algebra, general topology, posets and lattices.\nhttp://www.math.nmsu.edu/mgehrke/mgehrke.html\n##### Mai Gehrke\nDepartment of Mathematical Sciences Phone: (505] 646-4218 New Mexico State University Fax: (505) 646-1064 Las Cruces, NM 88003 [email protected] Office Location: Science Hall Room 232 House for rent while on sabbatical 2004-2005 Return to Faculty Page Return to Main Index\n##### PROFESSIONAL EXPERIENCE\nProfessor, New Mexico State University, Las Cruces, New Mexico.\n1/99Present Part-time consulting, Physical Science Laboratory, Las Cruces, New Mexico.\n1/975/97 Visiting Professor, Vanderbilt University, Nashville, Tennessee.\n8/9612/96 Visiting Lektor, University of Copenhagen, Copenhagen, Denmark.\n8/93 5/00 Associate Professor, New Mexico State University, Las Cruces, New Mexico.\n8/905/93 Assistant Professor, New Mexico State University, Las Cruces, New Mexico.\n8/885/90 2-year position as Assistant Professor, Vanderbilt University, Nashville, Tennessee.\n1/837/87 Teaching Assistant, University of Houston, Houston, Texas.\n##### RESEARCH INTERESTS\nLogic and its Applications\nUniversal Algebra and Lattice Theory\n##### Mathematical Research Publications\n• A new proof of completeness of S4 with respect to the real line , with G. Bezhanishvili, submitted to\n• 16. Elements Of Boolean Algebra\nboolean algebra. Laws of boolean algebra. Table 2 shows the basic Boolean laws. Note that every law has two expressions, (a) and (b). This is known as duality.\nhttp://www.ee.surrey.ac.uk/Projects/Labview/boolalgebra/\n##### Boolean Algebra\nIntroduction Laws of Boolean Algebra\n• Commutative Law\n• Associative Law ... On-line Quiz\n##### Introduction\nThe most obvious way to simplify Boolean expressions is to manipulate them in the same way as normal algebraic expressions are manipulated. With regards to logic relations in digital forms, a set of rules for symbolic manipulation is needed in order to solve for the unknowns.\nA set of rules formulated by the English mathematician George Boole describe certain propositions whose outcome would be either true or false . With regard to digital logic, these rules are used to describe circuits whose state can be either, 1 (true) or (false) . In order to fully understand this, the relation between the AND gate OR gate and NOT gate operations should be appreciated. A number of rules can be derived from these relations as Table 1 demonstrates.\n• P1: X = or X = 1\n\nTable 1: Boolean Postulates\n##### Laws of Boolean Algebra\nTable 2 shows the basic Boolean laws. Note that every law has two expressions, (a) and (b). This is known as duality . These are obtained by changing every AND(.) to OR(+), every OR(+) to AND(.) and all 1's to 0's and vice-versa.\n\n17. Web Page Experiment\nResearch group in set theoretic topology and boolean algebra. Members, research interests.\nhttp://www.math.ukans.edu/faculty/roitman/seminar.html\n Research group in set theoretic topology and Boolean algebra Permanent faculty members Bill Fleissner, Jack Porter, Judy Roitman Post-dctoral faculty member : Tetsuya Ishiu Advanced graduate students: Nate Carlson, Jila Niknejad Seminar : nearly every Monday and Wednesday 306 Snow Hall For pictures from the Spring 2001 AMS Special Session in Set Theory, Topology, and Boolean Algebra, held at KU, click here Faculty research interests William Fleissner started his research in set theory, in particular, consistency results in general topology. After focusing on the normal Moore space conjecture and related topics, he became interested in many areas of set theoretic topology. Recently, he has studied \"projective properties\" for example, the questions, \"If all continuous Tychonoff images of space are realcompact, must the space be Lindelof?\" and \"What can be said about spaces all of whose regular images are normal?\" Two topics of current research are D-spaces and subspaces of the product of finitely many ordinals. Jack Porter 's research is focused on spaces in which a given space is dense (extensions), and on the related notion of\n\n18. BOOLEAN ALGEBRA QUIZ\nboolean algebra Quiz. Give the relationship that represents the dual of the Boolean property A + 1 = 1? (Note * = AND, + = OR and = NOT)\nhttp://www.ee.surrey.ac.uk/Projects/Labview/boolalgebra/quiz/\n##### Boolean Algebra Quiz\n• Give the relationship that represents the dual of the Boolean property A + 1 = 1?\n(Note: * = AND, + = OR and ' = NOT)\n• A * 1 = 1\n• A * = 0\n• A + = 0\n• A * A = A ...\n• A * 1 = 1\n• Give the best definition of a literal?\n• A Boolean variable\n• The complement of a Boolean variable\n• 1 or 2\n• A Boolean variable interpreted literally ...\n• The actual understanding of a Boolean variable\n• Simplify the Boolean expression (A+B+C)(D+E)' + (A+B+C)(D+E) and choose the best answer.\n• A + B + C\n• D + E\n• A'B'C'\n• D'E' ...\n• None of the above\n• Which of the following relationships represents the dual of the Boolean property x + x'y = x + y?\n• x'(x + y') = x'y'\n• x(x'y) = xy\n• x*x' + y = xy\n• x'(xy') = x'y' ...\n• x(x' + y) = xy\n• Given the function F(X,Y,Z) = XZ + Z(X'+ XY), the equivalent most simplified Boolean representation for F is:\n• Z + YZ\n• Z + XYZ\n• XZ\n• X + YZ ...\n• None of the above\n• Which of the following Boolean functions is algebraically complete?\n• F = xy\n• F = x + y\n• F = x'\n• F = xy + yz ...\n• F = x + y'\n• Simplification of the Boolean expression (A + B)'(C + D + E)' + (A + B)' yields which of the following results?\n• A + B\n• A'B'\n• C + D + E\n• C'D'E' ...\n• A'B'C'D'E'\n• Given that F = A'B'+ C'+ D'+ E', which of the following represent the only correct expression for F'?\n• 19. Redirect... To New Location boolean algebra assistant program is an interactive program extremely easy to use. A musthave tool for the freshmen electrical engineering student. Shows output in either SOP(DNF) or POS(CNF) format. Win 98/ME/NT/2000/XPhttp://www.etel.dn.ua/~shurik/karnaugh/\n\n20. Boolean Algebra - Wikipedia, The Free Encyclopedia\nboolean algebra. Specifically, boolean algebra was an attempt to use algebraic techniques to deal with expressions in the propositional calculus.\nhttp://en.wikipedia.org/wiki/Boolean_algebra" ]

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[ "Guru\n\n# Find the sum of all 3 digit natural numbers which are divisible by 7.\n\n• 0\n\nOne of the most important question from ML Aggarwal from Arithmetic Progression is to find the sum of all 3 digit natural numbers which are divisible by 7.\n\nArithmetic Progression chapter 9 question no 21 ii\n\nShare\n\n1. ## All three-digit natural numbers which are divisible by 7 are\n\n105,112,119,126,...,994 which are in AP with a=105,d=7,l=994\nLet the number of these terms be n. Then,\nan=994\na+(n1)d=994\n105+(n1)×7=994\nn=128\nNow, Sn=2n(a+l)\n\nSn=2128(105+994)=70336.\n\n• 0" ]

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[ "# Opciós vételár képlet, Navigációs menü", null, "", null, "The definition of volatility Calculating volatility Volatility is the variability of the return. This variability is measured by the standard deviation of the return with continuous interest rate payments.", null, "Volatility is usually calculated from the daily closing prices, but weekly, monthly, hourly, or even minutely data could be used. It is important to note at this point that the n number of observations does not equal to how many parts one wants to divide the analysed period. The higher the volatility the higher the fluctuation in the market prices.", null, "Statistical volatility is the percentage value of the price fluctuations determined by statistical methods. Traders and analysts examine the normal distribution of the closing market prices in a given period to determine the value of the statistical volatility.\n\n• Bejó Ágnes\n• Modern vállalati pénzügyek | Digitális Tankönyvtár\n• Fajtái[ szerkesztés ] Call vételi jog A vételi opció vételi jogot biztosít jogosultjának vevőjénekmíg az opció kiírója eladója kötelezettséget vállal az eladásra.\n• Bevonatos opciók\n\nWhen assuming the price cannot be zero or opciós vételár képlet, lognormal distribution may be a better choice, but it requires more data processing. The The sequence can be formed from the closing prices on 20 days when this period can realistically reflect the market price fluctuations. The table below summarises the statistical volatility for different periods.", null, "However, the The decreasing volatility is accompanied by an increasing share price. Historical volatility.", null, "" ]

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[ "# List of Physical Quantities (with Symbols, SI Unit, and Dimension)", null, "List of Physical Quantities\nA physical quantity is a physical property of a phenomenon, object, or substance that can be measured. This list contains two tables. First table lists the fundamental or base quantities defined in SI system, whereas second table lists the physical quantities which can be expressed in terms of base quantities.\n\nBase quantitySymbolSI unitDimension\nLengthlmetre (m)L\nMassmkilogram (kg)M\nTimetsecond (s)T\nElectric currentIampere (A)I\nTemperatureTkelvin (K)Θ\nAmount of substancenmole (mol)N\nLuminous intensityLcandela (cd)J\n\nDerived quantity Symbol SI unit Dimension Acceleration a m s-2 L T−2 Angular acceleration α rad s-2 T-2 Angular momentum L kg m2 s-1 M L2 T-1 Angular speed (or angular velocity) ω rad s-1 T-1 Area A m2 L2 Area density ρA kg m-2 M L-2 Capacitance C farad (F = A2 s4 kg-1 m−2) M-1 L-2 T4 I2 Catalytic activity katal (kat = mol s-1) T-1 N Chemical potential μ J mol-1 M L2 T-2 N-1 Current density J → A m−2 L-2 I Dynamic viscosity η Pa s M L-1 T-1 Electric charge Q coulomb (C = A s) T I Electric displacement D C m−2 L-2 T I Electric field strength E→ V m-1 M L T-3 I-1 Electrical conductance G siemens (S = A2 s3 kg-1 m−2) M-1 L−2 T3 I2 Electrical conductivity σ S m-1 M-1 L-3 T3 I2 Electric potential V volt (V = kg M2 A-1 s-3) M L2 T-3 I-1 Electrical resistance R ohm (Ω = kg M2 A-2 s-3) M L2 T-3 I-2 Electrical resistivity ρ ohm metre (Ω⋅m = kg M3 A-2 s-3) M L3 T-3 I-2 Energy E joule (J = kg m2 s-2) M L2 T-2 Energy density ρE J m-3 M L-1 T-2 Entropy S J K-1 M L2 T-2 Θ-1 Force F→ newton (N = kg m s−2) M L T-2 Frequency f hertz (Hz = s-1) T-1 Fuel efficiency L−2 Half-life t1/2 s T Heat Q joule (J) M L2 T-2 Heat capacity Cp J K-1 M L2 T-2 Θ-1 Heat flux density ϕQ W m−2 M T-3 Illuminance Ev lux (lx = cd sr m-2) L−2 J Impedance Z ohm (Ω = kg M2 A-2 s-3) M L2 T-3 I-2 Impulse J newton second (N⋅s = kg m s-1) M L T-1 Inductance L henry (H = kg M2 A-2 s−2) M L2 T-2 I-2 Irradiance E W m-2 M T-3 Intensity I W m-2 M T-3 Jerk j→ m s-3 L T-3 Luminous flux (or luminous power) F lumen (lm = cd sr) J Mach number (or mach) M unitless 1 Magnetic field strength H A m-1 L-1 I Magnetic flux Φ weber (Wb = kg M2 A-1 s−2) M L2 T-2 I-1 Magnetic flux density B tesla (T = kg A-1 s−2) M T-2 I-1 Magnetization M A m-1 L-1 I Mass fraction x kg/kg 1 (Mass) Density (or volume density) ρ kg m-3 M L-3 Mean lifetime τ s T Molar concentration C mol m-3 L-3 N Molar energy J mol-1 M L2 T-2 N-1 Molar entropy J K-1 mol-1 M L2 T-2 Θ-1 N-1 Molar heat capacity c J K-1 mol-1 M L2 T-2 Θ-1 N-1 Moment of inertia I kg m2 M L2 Momentum p→ N s M L T-1 Permeability μ H m-1 M L T-2 I-2 Permittivity ε F m-1 M-1 L-3 T4 I2 Plane angle θ radian (rad) 1 Power P watt (W) M L2 T-3 Pressure p pascal (Pa = kg m-1 s−2) M L-1 T-2 (Radioactive) Activity A becquerel (Bq = s-1) T-1 Radiance ys W m−2 sr-1 M T-3 Refractive index n unitless 1 Solid angle Ω steradian (sr) 1 Speed v m s-1 L T-1 Specific energy J kg-1 L2 T-2 Specific heat capacity c J kg-1 K-1 L2 T-2 Θ-1 Specific volume v M3 kg-1 M-1 L3 Strain ε unitless 1 Stress σ Pa M L-1 T-2 Surface tension γ N m-1 or J m−2 M T-2 Thermal conductivity k W m-1 K-1 M L T-3 Θ-1 Torque τ newton metre (N m) M L2 T-2 Velocity v→ m s-1 L T-1 Volume V m3 L3 Wavelength λ m L Wavenumber k m-1 L-1 Weight w newton (N = kg m s−2) M L T-2 Work W joule (J = kg M2 s−2) M L-22 T-2 Young's modulus E pascal (Pa = kg m-1 s−2) M L-1 T-2 Thermal conductivity k W m-1 K-1 M L T-3 Θ-1 Torque τ newton metre (N m) M L2 T-2 Velocity v→ m s-1 L T-1 Volume V m3 L3 Wavelength λ m L Wavenumber k m-1 L-1 Wavevector k→ m-1 with direction L-1 Weight w newton (N = kg m s−2) M L T-2 Work W joule (J = kg M2 s-2) M L2 T-2 Young's modulus E pascal (Pa = kg m-1 s−2) M L-1 T-2" ]

[ null, "https://2.bp.blogspot.com/-T945MqVXwac/W2_qRRblDXI/AAAAAAAAJBw/TSkqjSkpaP4Zo_NU9GMsiSaF0-Pq_vhnQCEwYBhgL/s320/measurement%2Binstrument.jpg", null ]

[ "", null, "Line joining any two points of the circle  is the chord of the circle.\n\nHere,\n\n(AB) , (CD) ,  (EF) ,  (GH)  are chords of the circle.", null, "What if a chord passes through center of circle?\n\nA chord through center of circle is the diameter of the circle.\n\nAnd diameter is the longest chord.\n\nHere, (AB) is the diameter of the circle\n\n1. Chapter 4 Class 6 Basic Geometrical Ideas\n2. Concept wise\n3. Circles\n\nCircles", null, "" ]

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[ "", null, "### 7.1 Use gravity sensor to control ball movement\n\nRPiSparkModule has already a way for us to access the gravity sensor. With the RPiSparkModule.RPiSpark.Attitude property we can easily use RPi-Spark's gravity sensor, not only that, We can also use gyro sensors and temperature sensors at the same time. ( see the SDK manual get more information )\n\nOpen the gravity sensor with RPiSparkModule.RPiSpark.Attitude.openWith(temp=True, accel=True, gyro=False) and get the data using RPiSparkModule.RPiSpark.Attitude.getAccelData(). We will get the data measured by the gravity sensor, which is x, y, z for the three-axis, for example: {\"x\":1.2342342, \"y\":3.434535, \"z\": 7.2342425} . Here we use the x,y data to control the direction of the ball movement, x to control the left and right movement, and y to control the up and down movement. Just enter the data directly to change the position of the ball, for example: self.ball.move( x, y )\n\nWe open BallBox_6_4.py and to perform the above steps to complete this gravity sensing control ball:\n\n``````from JMRPiFoundations.Skeleton.RPiSparkModule import RPiSparkModule\nfrom Ball import Ball\n\nclass BallBox(RPiSparkModule):\nball = None\nvelocity = None\n\ndef _drawVel(self):\nself.RPiSpark.Screen.Canvas.rectangle( (2, 2, 38, 16), 0, 0 )\nself.RPiSpark.Screen.write(\"VEL:{:1d}\".format( self.velocity ), xy=(0,4))\n\ndef _changeVel(self, offsetVel):\n\"\"\"\nChange velocity of ball move\noffsetVel can be > 0 to increase velocity or < 0 to reduce velocity\nvelocity is limited between 1 and 9, default: 3\n\"\"\"\nself.velocity += offsetVel\nif self.velocity < 1:\nself.velocity = 1\n\nif self.velocity > 9:\nself.velocity = 9\n\ndef onKeyButtonUp(self, channel):\n# Press SW_A to reduce velocity\nif channel == self.RPiSparkConfig.BUTTON_ACT_A:\nself._changeVel(-1)\nreturn\n\n# Press SW_B to increase velocity\nif channel == self.RPiSparkConfig.BUTTON_ACT_B:\nself._changeVel(1)\nreturn\n\ndef setup(self):\nself.velocity = 3\nself.ball = Ball(64, 32, 5)\n# setup all key buttons to INT mode, same time query work fine\nself.initKeyButtons(\"INT\")\n# Open accel\nself.RPiSpark.Attitude.openWith(temp=True, accel=True, gyro=False)\n\ndef run(self):\n\nwhile True:\n# Check exit key status ( JOY_UP + SW_A )\nif self.readExitButtonStatus():\nbreak;\n\n# Get Accel data from attitude sensor\naccelData = self.RPiSpark.Attitude.getAccelData()\nself.ball.move( -accelData[\"x\"], accelData[\"y\"] )\n\n# Move the ball up\nif self.readKeyButton(self.RPiSparkConfig.BUTTON_JOY_UP):\nself.ball.move(0, -self.velocity)\n\n# Move the ball down\nif self.readKeyButton(self.RPiSparkConfig.BUTTON_JOY_DOWN):\nself.ball.move(0, self.velocity)\n\n# Move the ball left\nif self.readKeyButton(self.RPiSparkConfig.BUTTON_JOY_LEFT):\nself.ball.move(-self.velocity, 0)\n\n# Move the ball right\nif self.readKeyButton(self.RPiSparkConfig.BUTTON_JOY_RIGHT):\nself.ball.move(self.velocity, 0)\n\n# Move the ball to center of screen\nif self.readKeyButton(self.RPiSparkConfig.BUTTON_JOY_OK):\nself.ball.reset()\n\n# Drawing the ball on the screen\nself.RPiSpark.Screen.clear()\nself._drawVel()\nself.RPiSpark.Screen.Canvas.ellipse( self.ball.getXY(), 1, 1 )\nself.RPiSpark.Screen.refresh()\n\nprint(\"BallBox is done.\")``````\n\nSave as BallBox_7_1.py and execute the following command in the terminal:\n\n``\\$> rspk BallBox_7_1 -f``\n\nAt this time you can control the movement of the ball by changing the attitude of RPi-Spark.\n\nWas this article helpful?\n\nYES     |     NO\n\nEnjoying the project? Spotted a mistake? Any opinions on the website? Let us know!" ]

[ null, "https://mobinrg.com/images/logo/logo_183x64.png", null ]

[ "# Conversion of Temperature\n\nIn conversion of temperature from one scale into another the given temperature in °C we can convert it into °F and also the temperature in °F we can convert it into °C.\n\nThe rules which are used in this conversion are given below:\n\n1. When the temperature is given in degree Celsius:\n\nStep I: Multiply the given temperature in degree by 9\n\nStep II: The product we obtained from step I divide it by 5.\n\nStep III: Atlast add 32 with the quotient we obtained from step II to get the temperature in degree Fahrenheit.\n\n2. When the temperature is given in degree Fahrenheit:\n\nStep I: Subtract 32 from the given temperature in degree\n\nStep II: The difference we obtained from step I multiply it by 5.\n\nStep III: Atlast the product we obtained from step II divide it by 9 to get the temperature in degree Celsius.\n\nSolved examples to convert the temperature from one scale into another:\n\n1. Convert into degree Fahrenheit:\n\n(i) 45° C\n\nThe temperature is given in 45° C.\n\nTo convert into degree Fahrenheit first multiply 45 by 9 and we get 405.\n\nThen, divide 405 by 5 we get 81.\n\nAtlast we need to add 32 and 81 we get 113.\n\nTherefore, 45° C = 113° F\n\n(ii) 100° C\n\nThe temperature is given in 100° C. Now to convert into degree Fahrenheit first multiply 100 by 9 we get 900 then, divide 900 by 5 we get 180. Atlast we need to add 32 and 180 then we get 212.\n\nTherefore, 100° C = 212° F\n\n(iii) 50° C\n\nTherefore, 50° C = 122° F\n\n2. Convert into degree Celsius:\n\n(i) 98.4° F\n\nThe temperature is given in 98.4° F.\n\nTo convert into degree Celsius first we need to subtract 32 from 98.4 and we get 66.4.\n\nThen, multiply 66.4 by 5 we get 332.\n\nAtlast divide 332 by 9 and we get 36.9.\n\nTherefore, 98.40° F = 36.9° C\n\n(ii) 112° F\n\nThe temperature is given in 112° F. Now to convert into degree Celsius first we need to subtract 32 from 112 and we get 80 then, multiply 80 by 5 we get 400. Atlast divide 400 by 9 and we get 44.4.\n\nTherefore, 112° F = 44.4° C\n\n(iii) 212° F\n\nTherefore, 212° F = 100° C\n\nNow let us understand:\n\nOn the Celsius Scale            On the Fahrenheit Scale\n\nWater freezes at                            0° C                                    32° F\n\nWater boils at                             100° C                                  212° F\n\nNormal body temperature                37° C                                 98.6° F\n\nTemperature.\n\nConverting the Temperature from Celsius to Fahrenheit.\n\nConverting the Temperature from Fahrenheit to Celsius.\n\nClinical Thermometer.\n\nWorksheet on Temperature." ]

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[ "", null, "Search for dijet mass resonances at CDF (1.1 fb-1 analysis)", null, "Primary contacts: Kenichi Hatakeyama1, Anwar Bhatti1, Robert Harris2\n\nRockefeller University1\nFermilab2\n\nfor the CDF Collaboration\n\nSubmitted to PRD: arXiv:0812.4036\n\nAbstract\nWe present a search for a dijet mass resonance in p-pbar collisions at √s=1.96 TeV using data collected by the CDF Collaboration in Fermilab Tevatron Run II corresponding to an integrated luminosity of 1.13 fb-1. Jets are reconstructed using the Midpoint algorithm with R=0.7, and the dijet mass spectrum is formed for dijet events with the dijet mass above 180 GeV/c2 and |yjet1,2|<1.0. No significant evidence of a dijet mass resonance is observed, so we interpret the results to exclude new massive particles decaying into dijets.\n\nPlots in the paper.", null, "FIG. 1: (a) The measured dijet mass spectrum for both jets to have |y| < 1 compared to the NLO pQCD prediction obtained using the CTEQ6.1 PDFs. (b) The ratio of the data to the NLO pQCD prediction. The experimental systematic uncertainties, theoretical uncertainties from PDF, the ratio of MRST2004/CTEQ6.1, and the dependence on the choice of renormalization and factorization scales are also shown. An additional 6% uncertainty in the determination of the luminosity is not shown. [eps| gif]", null, "FIG. 2: (a) The measured dijet mass spectrum (points) fitted to Eq. (2) (dashed curve). The bin-by-bin unfolding corrections is not applied. Also shown are the predictions from the excited quark, q*, simulations for masses of 300, 500, 700, 900, and 1100 GeV/c2, respectively (solid curves). (b) The fractional difference between the measured dijet mass distri- bution and the fit (points) compared to the predictions for q* signals divided by the fit to the measured dijet mass spec- trum (curves). The inset shows the expanded view in which the vertical scale is restricted to +/-0.04. [eps| gif]", null, "FIG. 3: Dijet mass distributions for simulated signals of the q*, RS graviton, W', and Z' with the mass of 800 GeV/c2. [eps| gif]", null, "FIG. 4: Observed 95% C.L. upper limits on new particle production cross sections times the branching fraction to dijets obtained with the signal shapes from (a) W', (b) Z', (c) RS graviton, and (d) q* production. Also shown are the cross section predictions for the production of W', Z', RS graviton, #rhoT8, q*, axigluon, flavor-universal coloron, and E6 diquark for the set of parameters described in the text. The limits and theoretical predictions are for events in which both of the leading two jets have |y| < 1. [eps| gif]\n\nDijet mass distributions from QCD and expected signals", null, "Excited quark:\nDijet mass spectrum expected from QCD and excited quark production with excited quark mass=300, 500, 700, 900, and 1100 GeV/c2. The excited quark decaying to a quark-gluon pait is simulated with Pythia with f=f'=fs=1.\n[eps| gif]\nW':\nThe W' production is simulated with Pythia with SM couplings. The W' production cross section is multiplied by a factor of 1.3 to account for the k-factor. [eps| gif]\nZ':\nThe Z' production is simulated with Pythia with SM couplings. The Z' production cross section is multiplied by a factor of 1.3 to account for the k-factor. [eps| gif]\nRandall-Sundrum graviton:\nThe R-S graviton production is simulated with Pythia with k/MPL=0.1. The R-S graviton production cross section is multiplied by a factor of 1.3 to account for the k-factor. [eps| gif]\nDijet mass distribution from four models that produce dijet mass resonances", null, "Dijet mass distribution from four models that produce dijet mass resonances.\n[eps| gif]\nTest of a parametrization form used for modeling QCD production with Pythia and Herwig dijet MC samples", null, "", null, "The dijet mass spectra from Pythia and Herwig Monte Carlo and fits to the parametrization form:", null, "Pythia [eps| gif]\nHerwig [eps| gif]\n(Note: these spetra are the \"detector-level\" dijet mass distributions, not corrected to the particle-level.)\nDijet mass distribution from data and fit to a parametrization form", null, "The measured dijet mass spectrum and results of a fit to the parametrization form:", null, "[eps| gif]\nNo significant indication of resonant structure is observed.\n\n(Note: this measured spetrum is the \"detector-level\" dijet mass distribution, not corrected to the particle-level.)\nEffect of systematic uncertainties on limits for new particle production decaying into dijets.", null, "", null, "The four parameters in the parametrization form (shown above) are not a priori determined. When forming likelihood distributions as function of signal cross sections at each new particle mass, the likelihood is maximized keeping these four parameters free (profiling).\n\nThe other systematic uncertainties from (1) jet energy scale (JES), (2) jet energy resolution, (3) luminosity, (4) trigger prescale, and (5) trigger efficiency are folded into the likelihood by the Bayesian approach.\n\nThe effects of these systematic uncertainties on the limits are shown in:\ntop [eps| gif]\nbottom [eps| gif]\nLimits on production of new particles decaying into dijets.", null, "", null, "[top] 95% C.L. limits on the Randall-Sundrum graviton and color-octet technirho producion (red), and limits on the excited quark, axigluon, flavor-universal coloron, and E6 diquark (black), compared with the theoretical predictions for these particle production.\n[eps| gif]\n[bottom] 95% C.L. limits on the W' and Z' production compared with their theoretical predictions.\n[eps| gif]\nThe shown cross section predictions and limits are for the total particle production cross section times branching fraction to dijets times the kinematic acceptance that the leading two particle-level jets are within |y(jet1,2)|<1.\nMass exclusion:\n Observed mass exclusion range Model description 260-870 GeV/c2 Excited quark (f=f'=fs=1) 260-1100 GeV/c2 Color-octet technirho [top-color-assisted-technicolor (TC2) couplings, M'8=0, M(pi228)=5M(rho)/6, M(pi221)=M(pi228)/2, M8=5M(rho)/6] 260-1250 GeV/c2 Axigluon and flavor-universal coloron (mixing of two SU(3)'s, cot(theta)=1) 290-630 GeV/c2 E6 diquark 280-840 GeV/c2 W' (SM couplings) 320-740 GeV/c2 Z' (SM couplings)" ]

[ null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/cdf_ii_logo2.gif", null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/logo_exotic.gif", null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/plots_paper/mjj_data_vs_NLO_prd.gif", null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/plots_paper/mjj_data_fit_resonance.gif", null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/plots_paper/plot_resonance_800.gif", null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/plots_paper/plot_limit_theory_vs_mass_prd.gif", null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/plots/QCD_plus_exq.gif", null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/plots/plot_resonance_Cal_800.gif", null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/plots/mjj_py_fit.gif", null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/plots/mjj_hw_fit.gif", null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/plots/mjj_param_form.gif", null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/plots/mjj_data_fit.gif", null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/plots/mjj_param_form.gif", null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/plots/plot_limits_vs_mass_sys_diff.gif", null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/plots/plot_limits_vs_mass_each_sys_diff.gif", null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/plots/plot_limit_theory_vs_mass.gif", null, "https://www-cdf.fnal.gov/physics/exotic/r2a/20080214.mjj_resonance_1b/plots/plot_limit_theory_vs_mass_wpmzpm.gif", null ]

[ "Cody\n\n# Problem 2311. Vector Magnitude Calculator\n\nSolution 868026\n\nSubmitted on 7 Apr 2016 by Pritom Kumar Mondal\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\nx = 5; y = 12; mm = 13; assert(isequal(vector_magnitude(x, y),mm))\n\n2   Pass\nx = 3; y = 4; mm = 5; assert(isequal(vector_magnitude(x, y),mm))\n\n3   Pass\nx = 12; y = 35; mm = 37; assert(isequal(vector_magnitude(x, y),mm))" ]

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[ "››Convert centihg to ton/square inch [long]\n\n centihg ton/square inch\n\n Did you mean to convert centihg to ton/square inch [long] ton/square inch [short]\n\nHow many centihg in 1 ton/square inch? The answer is 11584.142998937.\nWe assume you are converting between centihg and ton/square inch [long].\nYou can view more details on each measurement unit:\ncentihg or ton/square inch\nThe SI derived unit for pressure is the pascal.\n1 pascal is equal to 0.00075006156130264 centihg, or 6.4748990181794E-8 ton/square inch.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between centihg and tons/square inch.\nType in your own numbers in the form to convert the units!\n\n››Want other units?\n\nYou can do the reverse unit conversion from ton/square inch to centihg, or enter any two units below:\n\nEnter two units to convert\n\n From: To:\n\n››Definition: Centihg\n\nThe SI prefix \"centi\" represents a factor of 10-2, or in exponential notation, 1E-2.\n\nSo 1 centihg = 10-2 hg.\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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[ "Documentation\n\n## HDL Optimized QPSK Receiver with Captured Data\n\nThis example shows how to optimize the QPSK receiver modeled in QPSK Transmitter and Receiver example for HDL code generation and hardware implementation. The HDL-optimized model shows a QPSK receiver that addresses real-world communications issues like carrier recovery and timing recovery in a hardware-friendly manner.\n\n### Overview\n\nThe HDL Optimized QPSK Receiver with Captured Data example provides a hardware-friendly solution that performs baseband processing to handle a time-varying frequency offset and a time-varying symbol delay. Specifically, this example provides an HDL-optimized reference design of a practical digital receiver to mitigate the above-mentioned impairments, and includes coarse frequency compensation, PLL-based fine frequency compensation, timing recovery with fixed-rate resampling, bit stuffing/skipping, frame synchronization, and phase ambiguity resolution.\n\nCompared with the implementation of the receiver in the QPSK Transmitter and Receiver example, three major modifications have been made for efficient HDL code generation:\n\n• Streaming Input and Output: The HDL optimized QPSK receiver processes data one sample at a time. The captured real-world signal is streamed into the receiver front-end. The streaming output of the HDL optimized receiver is buffered and passed to the text message decoder.\n\n• Fixed-point: The QPSK receiver logic operates on fixed-point data.\n\n• HDL optimized architecture: Several blocks have been redesigned to use hardware efficient algorithms and architectures.\n\n### Structure of the Example\n\nThe top-level structure of the QPSK receiver model is shown in the following figure. The HDLRx subsystem has been optimized for HDL code generation.", null, "The input data is captured using two USRP® devices and the Communications Toolbox Support Package for USRP® Radio running the transmitter model and the receiver model respectively. The captured data represents the baseband received signal with a sampling rate of 200 kHz. The data is sample-based and has a length of 200001, which corresponds to a period of 1 s.\n\nThe following diagram shows the detailed structure of the HDLRx subsystem.", null, "The subsystems within are further described in the following sections.\n\n1. Automatic Gain Control (AGC) - Adjusts the received signal amplitude to a desired level\n\n2. Root Raised Cosine Receive Filter - Uses a rolloff factor of 0.5, and decimates the input signal by two\n\n3. Coarse Frequency Compensation (CFC) - Estimates an approximate frequency offset of the received signal and corrects it\n\n4. Fine Frequency Compensation (FFC) - Compensates for the residual frequency and phase offset\n\n5. Timing Recovery - Resamples the input signal according to a recovered timing strobe so that symbol decisions are made at the optimum sampling instants\n\n6. Data Decoding - Aligns the frame boundaries, resolves the carrier phase ambiguity caused by the Fine Frequency Compensation subsystem, and demodulates the signal\n\nThe structure of the Text Message Decoding subsystem is shown below.", null, "This subsystem is expected to be run in software, therefore, it processes frame-based signals to speed up the computation. The HDLRx subsystem outputs three sample-based Boolean signals: bit1, bit2, and dValid. Given that the downstream processing requires a frame signal, the task of converting sample-based signals to frame-based counterparts is accomplished by the dataframer block. The demodulated bit pair, bit1 and bit2, is valid only when dValid is set high. The dataframer block uses the dValid signal to properly fill up a delay line with bit1 and bit2. The Descramble and Print subsystem processes the received data only when its enable signal goes high. This occurs when both the delay line accumulates exactly 200 valid demodulated bits and the RxGo signal is set high. While the simulation is running, the Descramble and Print subsystem outputs the string \"Hello world ###\" to the MATLAB® command window, where '###' is a repeating sequence of '000', '001, '002', ..., '099'.\n\nThe Reference Frequency Offset Estimation subsystem provides an accurate estimation of the frequency offset for diagnostic purposes.\n\n1. AGC\n\nThe AGC ensures a stable input to the frequency and timing recovery subsystems. It sets the amplitude of the Coarse Frequency Compensation subsystem input as 1/Upsampling Factor , so that the equivalent gains of the phase and timing error detectors stay constant over time. The AGC is placed before the Root Raised Cosine Receive Filter so that the signal amplitude can be measured with an oversampling factor of four, thus improving the accuracy of the estimate.\n\nThe AGC structure is shown in the following diagram, and pipeline registers are shown in green throughout the model.", null, "2. Root Raised Cosine Receive Filter\n\nThe Root Raised Cosine Receive Filter decimates the input signal by a factor of two, with a rolloff factor of 0.5. It provides matched filtering for the transmitted waveform to boost the signal to noise ratio and facilitate the downstream signal processing.\n\nThe Root Raised Cosine Receive Filter is implemented using a fully parallel architecture.\n\n3. Coarse Frequency Compensation\n\nThe Coarse Frequency Compensation subsystem corrects the input signal with a rough estimate of the frequency offset. The following diagram shows the Coarse Frequency Compensation subsystem.", null, "This subsystem estimates the frequency and phase offsets of the baseband QPSK signal. First, the subsystem raises the input signal to the power of four. This is implemented by cascading two product blocks. Then, from the modulation-independent signal, it estimates the tone at four times the frequency offset. After dividing the estimate by four, the so-obtained frequency offset is corrected in the original signal. There is usually a residual frequency offset even after the CFC, which would cause a slow rotation of the constellation. The Fine Frequency Compensation subsystem compensates for this residual frequency.\n\nComparing the implementation of the Coarse Frequency Compensation subsystem here with those in QPSK Transmitter and Receiver and QPSK Receiver with USRP® Hardware, we can see several modifications:\n\n• The model implements a correlation based algorithm, also known as the Luise algorithm [ 1 ], for frequency estimation. This algorithm saves hardware resources compared with an FFT algorithm. Pipeline registers are used in the data path of the Luise algorithm to ensure the circuit speed. To learn more about the CFC algorithm, refer to the Communications Toolbox documentation.", null, "• The", null, "function, which constitutes a key component in the Luise algorithm, is computed using the Complex to Magnitude-Angle HDL Optimized block. This block computes the phase using the hardware friendly CORDIC algorithm. To learn more about the Complex to Magnitude-Angle HDL Optimized block, refer to the DSP System Toolbox™ documentation.\n\n• The detected phase offset is sent to an NCO to generate a complex exponential signal that is used to correct the phase offset in the original signal. The NCO HDL Optimized block maps the lookup table into a ROM, and provides a lookup table compression option to significantly reduce the lookup table size. To learn more about the NCO HDL Optimized block, refer to the DSP System Toolbox documentation.\n\n4. Fine Frequency Compensation\n\nThe Fine Frequency Compensation subsystem, shown in the following figure, implements a phase-locked loop (PLL), described in Chapter 7 of [ 2 ], to track the residual frequency offset and the phase offset in the input signal.", null, "A maximum likelihood Phase Error Detector (PED), described in Chapter 7.2.2 of [ 2 ], generates the phase error. A tunable proportional-plus-integral Loop Filter, described in Appendix C.2 of [ 2 ], filters the error signal and then feeds it into the Phase Calculation block. The Phase Calculation block generates a complex exponential signal that is used to correct the residual frequency and phase offsets in the output of the CFC. The Loop Filter allows tuning of Loop Bandwidth (normalized by the sample rate) and Loop Damping Factor . The default normalized loop bandwidth is set to 0.13 and the default damping factor is set to 2.5 (over damping), so that the PLL quickly locks to the intended phase while introducing little phase noise. To learn more about the FFC algorithm, refer to the Communications Toolbox documentation.\n\n5. Timing Recovery\n\nThe Timing Recovery subsystem is shown in the following diagram.", null, "The Timing Recovery subsystem implements a PLL, described in Chapter 8 of [ 2 ], to correct the timing error in the received signal. On average, the Timing Recovery subsystem generates one output sample for every two input samples.\n\nThe Interpolation Control subsystem implements a decrementing modulo-1 counter, described in Chapter 8.4.3 of [ 2 ], to generate the control signal to facilitate the Data Decoding subsystem to properly select the interpolants of the Interpolation Filter. This control signal also enables the Timing Error Detector (TED), so that it calculates the timing errors at the correct timing instants. The Interpolation Control subsystem updates the timing difference for the Interpolation Filter, generating interpolants at optimum sampling instants.\n\nThe Interpolation Filter is a Farrow parabolic filter with", null, "as described in Chapter 8.4.2 of [ 2 ]. The filter uses an", null, "of 0.5 so that all the filter coefficients become 1, -1/2 and 3/2, which significantly simplifies the interpolator structure.\n\nBased on the interpolants, timing errors are generated by a zero-crossing Timing Error Detector as described in Chapter 8.4.1 of [ 2 ], filtered by a tunable proportional-plus-integral Loop Filter as described in Appendix C.2 of [ 2 ], and fed into the Interpolation Control for a timing difference update. The Loop Filter allows tuning of Loop Bandwidth (normalized by the sample rate) and Loop Damping Factor. The default normalized loop bandwidth is set to 0.01 and the default damping factor is set to unity so that the PLL quickly locks to the correct timing while introducing little phase noise.\n\nWhen the timing error (delay) reaches symbol boundaries, there is one extra or missing interpolant in the output. The TED implements bit stuffing or skipping to handle the extra or missing interpolants. You can refer to Chapter 8.4.4 of [ 2 ] for details of bit stuffing/skipping.\n\nThe timing recovery loop normally generates one output symbol for every two input samples. It also outputs a timing strobe (dValid signal) that runs at the input sample rate. Under normal circumstances, the strobe value is simply a sequence of alternating ones and zeros. However, this occurs only when the relative delay between transmitter and receiver contains some fractional part of one symbol period and the integer part of the delay (in symbols) remains constant. If the integer part of the relative delay changes, the strobe value can have two consecutive zeros or two consecutive ones.\n\n6. Data Decoding\n\nThe Data Decoding subsystem performs frame synchronization, carrier phase ambiguity resolution, and QPSK demodulation. Its structure is shown in the diagram below:", null, "• Frame synchronization: The Matched Filter subsystem uses a QPSK-modulated Barker code as a reference to correlate against the received symbols. The modulus of the matched filter output is calculated in the Modulus subsystem and then compared with a threshold. Frame synchronization is declared if the modulus output exceeds the threshold. The threshold for frame synchronization is tunable: a large value increases the miss probability whereas a small value increases the probability of false alarm. In this example, the threshold value is set to 16.\n\n• Phase ambiguity resolution: The carrier phase PLL of the Fine Frequency Compensation subsystem may lock to the unmodulated carrier with a phase shift of 0, 90, 180, or 270 degrees, which can cause a phase ambiguity. For details of phase ambiguity and its resolution, refer to Chapter 7.2.2 and 7.7 in [ 2 ]. The angle of the matched filter output determines the extra phase shift. The Matched Filter output is fed into the conjugate block to negate the extra phase shift. Once frame synchronization is achieved, the conjugated version of the matched filter output is frozen and multiplied with all the symbols in a frame to effectively resolve the phase ambiguity issue.\n\n• QPSK demodulation: Each corrected symbol is demodulated and mapped to a pair of bits based on the symbol mapping of QPSK constellation.\n\n### Performance Analysis\n\nThe following figure shows the Bit Error Rate (BER) for this example. The captured data in this example has a small frequency offset ranging from -120 to -90Hz. Extra offset is added using the Frequency Offset block in the model.", null, "The BER plot shows that using CFC followed by FFC ensures a low BER for a wide frequency offset range, while using the FFC (without CFC) can only correct small offsets (smaller than 1200Hz with the parameter settings in this example). Using both CFC and FFC is recommended, especially to avoid poor BER performance when the frequency offset drifts out of the range the FFC can track.\n\nThe CFC may introduce small offset in the system, which could lead to a slight performance degradation when the actual offset is close to 0. In this case, using just the FFC may be a better choice. The Reference Frequency Offset Estimation subsystem uses the FFT-based Coarse Frequency Compensation block from the Communications Toolbox to provide an accurate estimation of the frequency offset. This information can be used to help user making design decisions.", null, "When frequency compensation subsystems cannot estimate and correct the frequency and phase offset, it is difficult for the Timing Recovery to correct timing errors. BER equal to 1 means the enable signal of the Data Decoding subsystem is always low and there is no data decoded.\n\n### Results and Displays\n\nWhen running the simulation, the model displays two scatter plots to show the constellation of the FFC output and the Timing Recovery output respectively.\n\nThe following diagram shows the constellation plot of the FFC output. The cluster is scattered around, mainly due to two reasons:\n\n• The timing error between the clocks at the transmitter and receiver\n\n• The signals are oversampled by a factor of two. Therefore, half of the symbols are in the transition state between QPSK symbols.", null, "The following diagram shows the constellation of the Timing Recovery output. One observes four concentrated clusters around the true 4-point constellation for QPSK modulation. This verifies the effectiveness of the Timing Recovery subsystem. However, as mentioned before, the Fine Frequency Compensation subsystem may lock the signal with a phase shift of 0, 90, 180, or 270 degree. The phase ambiguity issue is fixed in the Data Decoding subsystem.", null, "### HDL Code Generation\n\nPipeline registers (shown in green) have been added throughout the model to make sure the HDLRx subsystem does not have a long critical path. The HDL code generated from the HDLRx subsystem was synthesized using Xilinx® ISE on a Virtex6 (XC6VLX240T-1FFG1156) FPGA, and the circuit ran at about 97 MHz.\n\nTo check and generate the HDL code referenced in this example, you must have an HDL Coder™ license.\n\nYou can use the commands makehdl and makehdltb to generate HDL code and testbench for subsystems in HDLRx. To generate the HDL code, use the following command:\n\n` makehdl('commqpskrxhdl/HDLRx')`\n\nTo generate testbench, use the following command:\n\n` makehdltb('commqpskrxhdl/HDLRx')`\n\n### References\n\n1. M. Luise and R. Reggiannini, \"Carrier frequency recovery in all-digital modems for burst-mode transmissions,\" IEEE Trans. Communications, pp. 1169-1178, 1995.\n\n2. Michael Rice, \"Digital Communications - A Discrete-Time Approach\", Prentice Hall, April 2008." ]

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[ "# #013cbc Color Information\n\nIn a RGB color space, hex #013cbc is composed of 0.4% red, 23.5% green and 73.7% blue. Whereas in a CMYK color space, it is composed of 99.5% cyan, 68.1% magenta, 0% yellow and 26.3% black. It has a hue angle of 221.1 degrees, a saturation of 98.9% and a lightness of 37.1%. #013cbc color hex could be obtained by blending #0278ff with #000079. Closest websafe color is: #0033cc.\n\n• R 0\n• G 24\n• B 74\nRGB color chart\n• C 99\n• M 68\n• Y 0\n• K 26\nCMYK color chart\n\n#013cbc color description : Strong blue.\n\n# #013cbc Color Conversion\n\nThe hexadecimal color #013cbc has RGB values of R:1, G:60, B:188 and CMYK values of C:0.99, M:0.68, Y:0, K:0.26. Its decimal value is 81084.\n\nHex triplet RGB Decimal 013cbc `#013cbc` 1, 60, 188 `rgb(1,60,188)` 0.4, 23.5, 73.7 `rgb(0.4%,23.5%,73.7%)` 99, 68, 0, 26 221.1°, 98.9, 37.1 `hsl(221.1,98.9%,37.1%)` 221.1°, 99.5, 73.7 0033cc `#0033cc`\nCIE-LAB 31.505, 36.692, -70.664 10.704, 6.868, 48.336 0.162, 0.104, 6.868 31.505, 79.622, 297.44 31.505, -13.254, -93.965 26.207, 27.041, -91.008 00000001, 00111100, 10111100\n\n# Color Schemes with #013cbc\n\n• #013cbc\n``#013cbc` `rgb(1,60,188)``\n• #bc8101\n``#bc8101` `rgb(188,129,1)``\nComplementary Color\n• #019abc\n``#019abc` `rgb(1,154,188)``\n• #013cbc\n``#013cbc` `rgb(1,60,188)``\n• #2301bc\n``#2301bc` `rgb(35,1,188)``\nAnalogous Color\n• #9abc01\n``#9abc01` `rgb(154,188,1)``\n• #013cbc\n``#013cbc` `rgb(1,60,188)``\n• #bc2301\n``#bc2301` `rgb(188,35,1)``\nSplit Complementary Color\n• #3cbc01\n``#3cbc01` `rgb(60,188,1)``\n• #013cbc\n``#013cbc` `rgb(1,60,188)``\n• #bc013c\n``#bc013c` `rgb(188,1,60)``\n• #01bc81\n``#01bc81` `rgb(1,188,129)``\n• #013cbc\n``#013cbc` `rgb(1,60,188)``\n• #bc013c\n``#bc013c` `rgb(188,1,60)``\n• #bc8101\n``#bc8101` `rgb(188,129,1)``\n• #012470\n``#012470` `rgb(1,36,112)``\n• #012c89\n``#012c89` `rgb(1,44,137)``\n• #0134a3\n``#0134a3` `rgb(1,52,163)``\n• #013cbc\n``#013cbc` `rgb(1,60,188)``\n• #0144d5\n``#0144d5` `rgb(1,68,213)``\n• #014cef\n``#014cef` `rgb(1,76,239)``\n• #0c58fe\n``#0c58fe` `rgb(12,88,254)``\nMonochromatic Color\n\n# Alternatives to #013cbc\n\nBelow, you can see some colors close to #013cbc. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #016bbc\n``#016bbc` `rgb(1,107,188)``\n• #015bbc\n``#015bbc` `rgb(1,91,188)``\n• #014cbc\n``#014cbc` `rgb(1,76,188)``\n• #013cbc\n``#013cbc` `rgb(1,60,188)``\n• #012cbc\n``#012cbc` `rgb(1,44,188)``\n• #011dbc\n``#011dbc` `rgb(1,29,188)``\n• #010dbc\n``#010dbc` `rgb(1,13,188)``\nSimilar Colors\n\n# #013cbc Preview\n\nThis text has a font color of #013cbc.\n\n``<span style=\"color:#013cbc;\">Text here</span>``\n#013cbc background color\n\nThis paragraph has a background color of #013cbc.\n\n``<p style=\"background-color:#013cbc;\">Content here</p>``\n#013cbc border color\n\nThis element has a border color of #013cbc.\n\n``<div style=\"border:1px solid #013cbc;\">Content here</div>``\nCSS codes\n``.text {color:#013cbc;}``\n``.background {background-color:#013cbc;}``\n``.border {border:1px solid #013cbc;}``\n\n# Shades and Tints of #013cbc\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #00040c is the darkest color, while #f8faff is the lightest one.\n\n• #00040c\n``#00040c` `rgb(0,4,12)``\n• #000a20\n``#000a20` `rgb(0,10,32)``\n• #001033\n``#001033` `rgb(0,16,51)``\n• #001747\n``#001747` `rgb(0,23,71)``\n• #001d5a\n``#001d5a` `rgb(0,29,90)``\n• #01236e\n``#01236e` `rgb(1,35,110)``\n• #012981\n``#012981` `rgb(1,41,129)``\n• #013095\n``#013095` `rgb(1,48,149)``\n• #0136a8\n``#0136a8` `rgb(1,54,168)``\n• #013cbc\n``#013cbc` `rgb(1,60,188)``\n• #0142d0\n``#0142d0` `rgb(1,66,208)``\n• #0148e3\n``#0148e3` `rgb(1,72,227)``\n• #014ff7\n``#014ff7` `rgb(1,79,247)``\n• #0e59fe\n``#0e59fe` `rgb(14,89,254)``\n• #2167fe\n``#2167fe` `rgb(33,103,254)``\n• #3574fe\n``#3574fe` `rgb(53,116,254)``\n• #4882fe\n``#4882fe` `rgb(72,130,254)``\n• #5c8ffe\n``#5c8ffe` `rgb(92,143,254)``\n• #6f9cfe\n``#6f9cfe` `rgb(111,156,254)``\n• #83aafe\n``#83aafe` `rgb(131,170,254)``\n• #96b7fe\n``#96b7fe` `rgb(150,183,254)``\n• #aac5ff\n``#aac5ff` `rgb(170,197,255)``\n• #bdd2ff\n``#bdd2ff` `rgb(189,210,255)``\n• #d1dfff\n``#d1dfff` `rgb(209,223,255)``\n• #e4edff\n``#e4edff` `rgb(228,237,255)``\n• #f8faff\n``#f8faff` `rgb(248,250,255)``\nTint Color Variation\n\n# Tones of #013cbc\n\nA tone is produced by adding gray to any pure hue. In this case, #585c65 is the less saturated color, while #013cbc is the most saturated one.\n\n• #585c65\n``#585c65` `rgb(88,92,101)``\n• #515a6c\n``#515a6c` `rgb(81,90,108)``\n• #4a5773\n``#4a5773` `rgb(74,87,115)``\n• #42547b\n``#42547b` `rgb(66,84,123)``\n• #3b5182\n``#3b5182` `rgb(59,81,130)``\n• #344f89\n``#344f89` `rgb(52,79,137)``\n• #2d4c90\n``#2d4c90` `rgb(45,76,144)``\n• #254998\n``#254998` `rgb(37,73,152)``\n• #1e479f\n``#1e479f` `rgb(30,71,159)``\n• #1744a6\n``#1744a6` `rgb(23,68,166)``\n``#1041ad` `rgb(16,65,173)``\n• #083fb5\n``#083fb5` `rgb(8,63,181)``\n• #013cbc\n``#013cbc` `rgb(1,60,188)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #013cbc is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "### Extended GiNaC interface to inum\n\nparent 14e345d6\n ... ... @@ -413,7 +413,7 @@ CONTAINS if (what .eq. 0) then evalt = G(arr(1:l)) elseif (what.eq.1) then evalt = geval(arr(1:l),l) evalt = geval(toinum(arr(1:l)),l) endif end function ... ...\n ... ... @@ -5,18 +5,25 @@ using namespace GiNaC; #include typedef struct {double r,i;} complex_t; typedef struct {complex_t c; signed char i0;} inum_t; extern \"C\"{ complex_t geval_(complex_t * z, int* n); complex_t geval_(inum_t * z, int* n); }; complex_t geval_(complex_t * z, int* n) { complex_t geval_(inum_t * z, int* n) { cln::cl_inhibit_floating_point_underflow = true; lst w; lst w,s; for(long i=0;i<(*n)-1;i++) { w.append((z->r)+(z->i)*I); ex zz; if (abs(z->c.i) < 1e-15) w.append((z->c.r)); else w.append((z->c.r)+(z->c.i)*I); s.append(z->i0); z++; } ex ans = G(w,z->r).evalf(); ex ans = G(w,s,z->c.r).evalf(); return { .r = ex_to(evalf(real_part(ans))).to_double(), .i = ex_to(evalf(imag_part(ans))).to_double() ... ...\n• Owner\n\nThere is a subtlety in GiNaC. The explicitly given ieps prescription for `value`is ignored if\n\n`` cln::instanceof(value, cln::cl_R_ring)``\n\nOf course it would make sense to ignore the explicit value if `value` is indeed complex (cf. 63b4cc59). However, even if it is real, my old C++ interface\n\n`````` for(long i=0;i<(*n)-1;i++)\n{\nw.append((z->r)+(z->i)*I);\nz++;\n}``````\n\nwould create an `cln::cl_R_ring`. We now check whether `z` is real and act accordingly.\n\nMarkdown is supported\n0% or\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!" ]

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[ "# Invariant measure\n\nIn mathematics, an invariant measure is a measure that is preserved by some function. Ergodic theory is the study of invariant measures in dynamical systems. The Krylov–Bogolyubov theorem proves the existence of invariant measures under certain conditions on the function and space under consideration.\n\n## Definition\n\nLet (X, Σ) be a measurable space and let f be a measurable function from X to itself. A measure μ on (X, Σ) is said to be invariant under f if, for every measurable set A in Σ,\n\n$\\mu \\left(f^{-1}(A)\\right)=\\mu (A).$", null, "In terms of the push forward, this states that f(μ) = μ.\n\nThe collection of measures (usually probability measures) on X that are invariant under f is sometimes denoted Mf(X). The collection of ergodic measures, Ef(X), is a subset of Mf(X). Moreover, any convex combination of two invariant measures is also invariant, so Mf(X) is a convex set; Ef(X) consists precisely of the extreme points of Mf(X).\n\nIn the case of a dynamical system (X, T, φ), where (X, Σ) is a measurable space as before, T is a monoid and φ : T × X  X is the flow map, a measure μ on (X, Σ) is said to be an invariant measure if it is an invariant measure for each map φt : X  X. Explicitly, μ is invariant if and only if\n\n$\\mu \\left(\\varphi _{t}^{-1}(A)\\right)=\\mu (A)\\qquad \\forall t\\in T,A\\in \\Sigma .$", null, "Put another way, μ is an invariant measure for a sequence of random variables (Zt)t≥0 (perhaps a Markov chain or the solution to a stochastic differential equation) if, whenever the initial condition Z0 is distributed according to μ, so is Zt for any later time t.\n\nWhen the dynamical system can be described by a transfer operator, then the invariant measure is an eigenvector of the operator, corresponding to an eigenvalue of 1, this being the largest eigenvalue as given by the Frobenius-Perron theorem.\n\n## Examples\n\n• Consider the real line R with its usual Borel σ-algebra; fix aR and consider the translation map Ta : RR given by:\n$T_{a}(x)=x+a.$", null, "Then one-dimensional Lebesgue measure λ is an invariant measure for Ta.\n• More generally, on n-dimensional Euclidean space Rn with its usual Borel σ-algebra, n-dimensional Lebesgue measure λn is an invariant measure for any isometry of Euclidean space, i.e. a map T : RnRn that can be written as\n$T(x)=Ax+b$", null, "for some n × n orthogonal matrix A O(n) and a vector b Rn.\n• The invariant measure in the first example is unique up to trivial renormalization with a constant factor. This does not have to be necessarily the case: Consider a set consisting of just two points ${\\boldsymbol {\\rm {S}}}=\\{A,B\\}$", null, "and the identity map $T={\\rm {Id}}$", null, "which leaves each point fixed. Then any probability measure $\\mu :{\\boldsymbol {\\rm {S}}}\\rightarrow {\\boldsymbol {\\rm {R}}}$", null, "is invariant. Note that S trivially has a decomposition into T-invariant components {A} and {B}.\n• The measure of circular angles in degrees or radians is invariant under rotation. Similarly, the measure of hyperbolic angle is invariant under squeeze mapping.\n• Area measure in the Euclidean plane is invariant under 2 × 2 real matrices with determinant 1, also known as the special linear group SL(2,R).\n• Every locally compact group has a Haar measure that is invariant under the group action." ]

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[ "If you randomly select a card from a well-shuffled standard deck of 52 cards, what is the probability that the card you select is a spade or 6? (Your answer must be in the form of a reduced fraction.)\n\n1. 👍 0\n2. 👎 0\n3. 👁 126\n1. There are 52 cards in a deck.\nThere are 4 Sixes\n13+4-1=16\n16/52=\n8/26=\n4/13\n\n1. 👍 0\n2. 👎 0\n\n## Similar Questions\n\n1. ### Probability\n\nIf you randomly select a card from a well-shuffled standard deck of 52 cards, what is the probability that the card you select is a heart or King?\n\nasked by vinicio on March 25, 2016\n2. ### math\n\nIf you randomly select a card from a well-shuffled standard deck of 52 cards, what is the probability that the card you select is a Queen or Jack?\n\nasked by Anonymous on March 28, 2016\n3. ### Cards\n\nIf you randomly select a card from a well-shuffled standard deck of 52 cards, what is the probability that the card you select is a club or 5? (Your answer must be in the form of a reduced fraction.)\n\nasked by Anonymous on October 1, 2016\n4. ### math\n\nIf you randomly select a card from a well-shuffled standard deck of 52 cards, what is the probability that the card you select is a Jack or 2? (Your answer must be in the form of a reduced fraction.)\n\nasked by Anonymous on March 28, 2016\n5. ### Probability\n\nIf you randomly select a card from a well-shuffled standard deck of 52 cards, what is the probability that the card you select is a 3 or King? (Your answer must be in the form of a reduced fraction.)\n\nasked by vinicio on March 25, 2016\n6. ### Math\n\nIf you randomly select a card from a well-shuffled standard deck of 52 cards, what is the probability that the card you select is a heart or 10?\n\nasked by Anonymous on March 31, 2016\n7. ### math\n\nIf you randomly select a card from a well-shuffled standard deck of 52 cards, what is the probability that the card you select is a diamond or 6?\n\nasked by Anonymous on March 28, 2016\n8. ### math\n\nIf you randomly select a card from a well-shuffled standard deck of 52 cards, what is the probability that the card you select is a spade or 6?\n\nasked by Anonymous on March 28, 2016\n9. ### Math\n\nIf you randomly select a card from a well-shuffled standard deck of 52 cards, what is the probability that the card you select is a heart or 10?\n\nasked by Anonymous on March 31, 2016\n10. ### Math\n\nIf you randomly select a card from a well-shuffled standard deck of 52 cards, what is the probability that the card you select is a heart or 10?\n\nasked by Anonymous on March 31, 2016\n\nMore Similar Questions" ]

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[ "# How much force is required to start a block moving on level ground?\n\n## Homework Statement\n\nThe Egyptian's sled system had a rather high coefficient of static friction (0.7) but a much lower coefficient of kinetic friction (0.3) How much force is required to start a block moving on level ground?\nWe know the block is 1980 kg\n\nand after that:\nAssume that one person is able to apply 100N the block. Using this value, how many people does it take to pull a block at constance velocity along flat ground?\n\n## The Attempt at a Solution\n\nDo i do Fs - Fk to find the answer to the first question? or is it just Fs?\n\nJust Fs .\n\nand is the second one just fk?\n\nWell why do you think static and kinetic friction are different forces? They are just names of different phases of the same force. So use Fs when block does not move and Fk when it has started moving.\n\nthanks!\nAssume that a comfortable walking speed for an average person is 1.33m/s Assume that the workers can only walk at half this speed while they were pulling a block, but that they would walk at this speed on the trip from the pyramid back to the quarry. At these speeds, how many blocks can one crew of 59 people, move from the quarry (5km away) to the pyramid in one ten hour work day?\n\nLast edited:\nI am guessing that this one is related to the last. You can use simple kinematics for this- time for one round trip, how many stones for the crew in one round trip and then for the day.You have everything just go for it.\n\nwhat forumula do I use? I'm confused what to do with the velocity\n\nWell v=distance/t, what else?\n\n:S and then?\n\nLast edited:\nAre you overcomplicating? Start with time for one round trip for one block. Since you have already calculated the people required for one block, you can calculate the no. of blocks a group of 59 would move in one round trip.The time for one trip divided by total working time and multiplied by no. of blocks in one trip is your answer.\n\nwow, yes i was over complicating it, thank you so much!\n\nsorry maybe i just dont understand what this question is asking but:\nHow long does it take a crew to accelerate a block to the walking speed?" ]

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[ "# Ionic Bonding\n\n## Introduction\n\nCations and anions are formed in such a way that the \"OctetRule\" is satisfied.\nExample:\nNa(s) + ½Cl2 → NaCl(s)\nΔH°r = − 410.9 Kj (highly exothermic)\nBoth Na and Cl in NaCl have inert gas configuration.", null, "### Energetics of Ionic Bond Formation\n\nDefinition: Lattice energy is the energy required to completely separate an ionic solid into its gaseous ions. E1 = κ(Q1Q2)/d\nWhere: κ is a constant = 8.99 x 109 J·m/C2, Q1 and Q2 are the charges on the ions, d is the distance between ions.\n\nLattice energy depends on:\n• Charges on the ions (Q1 and Q2)\n• Sizes of the ions (d)\nLattice energy increases as:\n• Charges on the ions increase\n• The distance between the ions decreases\n\n### Exercise: Energetics of Ionic Bond Formation\n\nExercise on Energetics of Ionic Bond Formation\n\nCheck your answers here: Solution to the Exercise on Energetics of Ionic Bond Formation\n\nFor more details, please contact me here.\nDate of last modification: Summer , 2019" ]

[ null, "https://www.initiatewebdevelopment.com/Chemistry/Screenshots/ionic-bonding.jpg", null ]

[ "Home\n\nDividing Polynomials by Monomials worksheet\n\nDivision of Polynomials by Monomials Worksheet\n\n• Division of Polynomials by Monomials Worksheets How to Divide a Polynomial by a Monomial - The division in polynomials and monomials is conducted in the same manner as they are done in numerals. If we need to divide a polynomial by a monomial, the method is simple, consider the monomial as the divisor and the polynomial as the dividend\n• Student Name: _____ Score: Free Math Worksheets @ http://www.mathworksheets4kids.co\n• g and makes the work more fun. The coloring key includes 3 extra answers that won't be used\n• Homework 4 Dividing Polynomials By Monomials - Displaying top 8 worksheets found for this concept.. Some of the worksheets for this concept are Division of polynomials by monomials, Dividing polynomials by monomials, Dividing polynomials, Dividing monomials activities, Algebra 1, Dividing monomials answer key, Power of monomials homework practice answers, Study guide and intervention dividing.\n• Division of Polynomials Worksheets Incorporate this extensive range of dividing polynomials worksheet pdfs featuring exercises to divide monomials by monomials, polynomials by monomials and polynomials by polynomials employing methods like factorization, synthetic division, long division and box method\n• Answer Key. From Factors to Equations . SInce foc1ored founs . of . equaflons con be used to delermlne ,8 solutions, solutio . be . used to d91ermlne the equation\n• Displaying top 8 worksheets found for - Kutadividing By Monomial. Some of the worksheets for this concept are Dividing polynomials date period, Multiplying a polynomial and a monomial, Exponents and division, Division of polynomials by monomials, Multiplying dividing monomials, Multiplying polynomials date period, Dividing polynomials date period, Dividing polynomials by monomials\n• -3-Answers to Dividing Polynomials 1) 2 r4 + 4r3 + 3r2 2) x3 + x2 + 5x 3) n 5 + 2 + 1 10n 4) 1 3 + 1 9v + 2 9v2 5) 5v + 2 + 4 9v 6) n + 1 9 + 1 3n 7) 3r + 1 5 + 3 r 8) 3k2m 2 + kn 2 + 9m2 9) 2p2\n• This monomial and polynomial worksheet will produce ten problems per page. This monomial and polynomial worksheet will produce ten problems per page. This monomials worksheet is a good resource for students in the 5th Grade, 6th Grade, 7th Grade, and 8th Grade. Multiplying Binomials Worksheets This monomial and polynomial worksheet will produce.\n\nFirst a quick look at dividing monomials: Example 1: (a) Law of exponents; Cancellation. Identity property property. of multiplication (b) Example 1 demonstrates a general rule: If m and n are nonnegative integers, and m > n, then . Dividing two polynomials. To divide two polynomials, we first must write each polynomial in standard form Worksheet by Kuta Software LLC Grade 8 Math Multiplying & Dividing Monomials Name_____ ID: 1 ©K f2P0X1N7O cKQurtpaY QSro\\fItWwWaIrLe` xLiLACH.\\ V EA\\lklp irEidgjhtttsw ArveiskeNrLvmeAdt.-1-Simplify. Your answer should contain only positive exponents. 1) 2n2 × 8n5 2) 5m6 × 8m2 3) 10x4 × 4x4 × 5x6 4) 6k5 × 11k abraham lincoln worksheets. visual division worksheets. reading strategies worksheets. french possessive adjectives worksheet. parts of a book worksheet. horizontal and vertical lines worksheet. make up worksheets. multiple transformations worksheets. Dividing polynomials worksheet answers resource plans monomials. Polynomial division simple steps dividing polynomials monomials worksheet Displaying top 8 worksheets found for - Dividing Polynomials By Monomial. Some of the worksheets for this concept are Division of polynomials by monomials, Dividing polynomials by monomials, 6 dividing a polynomial by a monomial, Multiplying and dividing monomials, Algebra 1, Dividing polynomials, Dividing polynomials remainder and factor theorems, Dividing polynomials 1\n\nDividing Polynomials By A Monomial Worksheets & Teaching\n\n50 Multiplying Monomials Worksheet Answers In 2020 Worksheet Maker Worksheets Hone your skills in dividing polynomials by monomials by splitting the polynomial expression term by term and dividing each term with the monomial Multiplication and division of polynomials exercises with answers. Type keywords and hit enter. Pre-Algebra Worksheets | Monomials and Polynomials Worksheets #70165. Dividing polynomials by binomials #70166. Galois' Theorem and Polynomial Arithmetic #70167. Long Division of Polynomials #70168 About Division of monomials by monomials worksheet Division of monomials by monomials worksheet : Here we are going to see some practice questions on division of monomial by another monomials. Division of monomials by monomials worksheet - practice questions (1) Simplify 5a 2 b 2 c 2 ÷ 15abc (2) Simplify 16x 4 ÷ 32\n\nHomework 4 Dividing Polynomials By Monomials Worksheets\n\n1. Dividing Polynomials Notes and Worksheet is designed to help guide students in their understanding of dividing polynomials by a monomial (long division is not used), binomial, or trinomial.Joke worksheets are a fun way for students to practice a concept they have learned. Through the completion of\n2. dividing polynomials worksheets incorporate this extensive range of dividing polynomials worksheet pdfs featuring exercises to divide monomials by monomials polynomials by monomials and polynomials by polynomials employing methods like factorization synthetic division long division and box method exercises in the word format are included for.\n3. Dividing Monomials If your middle school student is struggling with math, help him learn about monomials with this series of worksheets. A monomial is a product of a power of variables\n4. ©n p2C031 B2f tK au GtDaF bS Ao5f ptlw Gaur meI 4LbLSCt. s O LARljl g DrPi zg 5hvt Ss1 mrNeusfe mrEvDexdt. q f zM ba Kdje o RwJiAtNhG eIBn4fbi hn DiFt 4eh zA El9g BeIb jr TaH U1h.D Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Dividing Polynomials Date_____ Period____ Divide\n5. After changing the signs, +3x 2 and -3 x 2 will get canceled. By simplifying, we get 2x + 10. Step 2 : In the second step again we are going to divide the first term that is 2x by the first term of divisor that is x\n6. Factoring & Dividing Polynomials 8th 10th Grade from dividing polynomials by monomials worksheet , image source: www.lessonplanet.com. Gallery of 50 Dividing Polynomials by Monomials Worksheet\n7. Dividing polynomials Shows how to divide a polynomial by a monomial and a polynomial by a binomial, includes example with missing descending terms\n\nFree worksheet(pdf) and answer key on multiplying monomials . Over 25 scaffolded questions that start relatively easy and end with some real challenges. Plus model problems explained step by ste The lesson called Dividing Polynomials with Long and Synthetic Division: Practice Problems is a great resource you can use to learn more about this mathematical concept. In this lesson you will.\n\nDividing Binomials - Displaying top 8 worksheets found for this concept.. Some of the worksheets for this concept are Dividing polynomials date period, Division of binomials by monomials, Dividing polynomials by binomials, Dividing polynomials, Dividing polynomials by monomials, Dividing monomials 1, Dividing monomials, Multiplying polynomials date period Dividing Polynomials By Monomials To divide a polynomial by a monomial, divide each term of the polynomial by the monomial. Dividing Polynomials Words To divide a polynomial by a monomial, divide each term of the polynomial by the monomial. Symbols a + b _ c = a _ c + b _ c Key Concept EXAMPLE 1 Divide a Polynomial by a Monomial Divide. a. (9 b. Dividing Polynomials by Monomials Jefferson Davis Learning Center, Sandra Peterson Simplify. Answers 1. 2 4n+8 1. 2n+4 2. 3 9 12 − −g + 2. 3g −4 3. 4 16x2 −12x+8 3. 4x2 −3x+2 4 Multiplying and Dividing Polynomials by Monomials Step 4 Staple the three booklets you made into the Foldable from Step 1 as shown. 7.1 Divide s Using a Model 7.1 Divide s Using Symbols 7.1 Multiply s 7.1 Multiply s Using Symbols Key Words: monomial polynomial binomial distributive propert Dividing monomials by polynomials worksheet This tutorial will help you accomplish the whole problem: Divide the following using the long division algorithm: (y5 + 5y4 + 9y3 + 11y2 + 12y + 13) ÷ (y + 2) This expression is then placed under its similar terms and subtracted from it, the same process is repeated with the rest of it until the expression can be furthe", null, "Free worksheet(pdf) and answer key on multiplying monomials . Over 25 scaffolded questions that start relatively easy and end with some real challenges. Plus model problems explained step by ste HW 4 Polynomial Operations _____ I will be able to add, subtract, multiply, and divide polynomials. Name Pe Here are the steps required for Dividing by a Polynomial by a Monomial: Step 1: If you are dividing by a monomial, you can split the problem into pieces by putting each term in the numerator over the denominator. Step 2: Simplify each term. Use the rules for exponents to simplify the variables in each term and reduce the fractions In case you call for assistance with math and in particular with dividing polynomials by monomials solver or equivalent fractions come pay a visit to us at Algebra-equation.com. We offer a tremendous amount of quality reference tutorials on subjects starting from value to adding and subtracting rationa\n\nPolynomial Worksheets Algebra I. These worksheets focus on the topics typically covered in Algebra I. Multiplying Monomials Worksheet Multiplying and Dividing Monomials Sheet Adding and Subtracting Polynomials worksheet Dividing Polynomials with Long and Synthetic Division: Practice Problems 10:11 Operations with Polynomials in Several Variables 6:09 Go to 6th-8th Grade Algebra: Monomials & Polynomials This Divide the Polynomials by Monomials Worksheet is suitable for 8th - 10th Grade. In this algebra practice learning exercise, students use their problem solving skills to solve 7 problems that require them to divide polynomials by monomials dividing polynomials solver ; buy holt middle school math course 2 florida edition ; adding integers worksheet ; bridge to algebra test of genius ; find the answers in math of permutation ; algebra 1 formulas ; integer worksheets order of operations ; 8th polynomials worksheet ; converting lineal metres to square metres ; trigonometry math test. In this multiplying and dividing monomials worksheet, students solve 105 short answer problems. Kids multiply and divide monomials. Students simplify their answers using exponent rules such as product to a power or quotient to a power\n\nSome of the worksheets displayed are dividing polynomials date period dividing polynomials dividing polynomials by binomials dividing polynomials multiplying and dividing polynomials work division of binomials by monomials addition and subtraction when adding multiplying By Easy Worksheet. 12) 2x3 - 5x2+ 3x. x. 13) -2x5 - 4x4 _ 4x3. 2x2. y Easy Worksheet. By Easy Worksheet. 14) -10x8 + 20x4 - x4-5x3. 15) 5x6 + 2x5 -4x4-x2. y. Easy Worksheetsy Workshee 16) Dividing a polynomial by a monomial. How to divide a Polynomial by a Monomial, examples and step by step solutions, Grade 9, 10, 11 and 12. Divide a Polynomial by a Monomial . Related Topics: More Lessons for Grade 9 Math Math Worksheets Examples, solutions, videos, worksheets, games, and activities to help Algebra students learn about dividing a polynomial by a monomial Enter the expression you want to divide into the editor. The polynomial division calculator allows you to take a simple or complex expression and find the quotient and remainder instantly. Step 2: Click the blue arrow to submit and see the result\n\nmultiplying monomials, binomials and what would make math worksheets for algebra monomials before when tab out of these worksheets. Figure out the individual worksheet you very much for? Polynomials using the addition and about monomials, in these division problems. Given polynomial and different step by multiplying the example of monomials Divide the polynomials worksheets. Algebra 1 worksheets. Polynomial worksheets. Answers to Dividing Polynomials Worksheets. Divide Polynomials Worksheet-1. Divide Polynomials Worksheet-2. Divide Polynomials Worksheet-3. Divide Polynomials Worksheet-4. All worksheets are created by experienced and qualified teachers Multiplying and dividing monomials . 3. Multiplying polynomials by monomials. 4. Dividing polynomials by monomials. 5. Multiplying monomial by monomial. 6. Multiplying monomial by binomial. 7. Multiplying binomial by binomial. 8. Multiplying polynomial by polynomial. 9. Applications of polynomials. 10. Solving polynomial equations. 11. Word. Multiplying polynomials with monomials Use the distributive property to simplify or solve these expressions ID: 1474021 More Polynomials interactive worksheets. Adding polynomials by jamin: Multiply -divide monomials by jamin: Subtracting polynomials by jamin: Polynomials by jcross: Special Products by Jenno Polynomial Long Division Worksheet Koogra from Polynomial Long Division Worksheet, source:koogra.com. Multiplying Monomials And Polynomials With Two Factors Mixed from Polynomial Long Division Worksheet, source:koogra.com. Polynomial Long Division Worksheet from Polynomial Long Division Worksheet, source:guillermotull.co\n\nMar 9, 2021 - Divide Polynomials Worksheet 2 Polynomials Math Word Problems Math Word Multiplying Monomials Challenge: Challenge: NOT a basic skills worksheet. Students must extend thinking to solve problems involving variable exponents. Some problems require converting to exponential notation prior to solving (i.e. rewrite 9 as 3 squared) etc\n\nDividing Polynomials Worksheet\n\nRight from dividing monomials calculator online to denominator, we have got every aspect discussed. Come to Algebra-net.com and study adding and subtracting rational, division and several additional algebra subject Dividing Polynomials By Monomials Worksheet from Dividing Polynomials Worksheet, source:guillermotull.com. Dividing Polynomials Worksheet 6 10 from Dividing Polynomials Worksheet, source:youtube.com. Make Polynomial Division Simple with these Steps from GradeA from Dividing Polynomials Worksheet, source:gradeamathhelp.com\n\nKutadividing By Monomial Worksheets - Learny Kid\n\n1. Welcome to The Multiplying Monomials and Polynomials with Two Factors Mixed Questions (A) Math Worksheet from the Algebra Worksheets Page at Math-Drills.com. This math worksheet was created on 2015-03-19 and has been viewed 90 times this week and 387 times this month. It may be printed, downloaded or saved and used in your classroom, home school, or other educational environment to help.\n2. ator. Then, cross out each pair of numbers or variables that appears both in the numerator and the deno\n3. 1: To divide monomials use the laws of exponents in division. 2: To divide a polynomial by a monomial, we use (a + b) / c = a/c + b/c. 3: The last rule is to divide a polynomial by another polynomial with at least two terms. This type of division is applied only when the degree of the polynomial in the numerator is greater than or equal to the.\n\nfraction add subtract multiply divide worksheet ; model+solving problem+Quadratic equation ; examples of factoring algebraic expressions ; how to put in n roots in a ti-83 ; the purpose of completing the square in the real world ; mathematical trivia ; ti-89 heaviside function ; adding and subtracting polynomial worksheet ; The definition of a. Dividing polynomials by binomials: To divide polynomials by binomials, we must use long division. This process looks confusing at first, but once you get the hang of it, it's actually pretty easy. The steps match the steps you take to do a long division problem with numbers\n\nPre-Algebra Worksheets Monomials and Polynomials Worksheet\n\nWorksheet by Kuta Software LLC Algebra 2 7.1: Polynomial Long Division Name_____ Period____ ©E b2g0u1v5y EKSuStVaO oSdoofAtWwjavraeu CLHLpCs.w X rA`lYlq ZrOiVgGhVt_sj FrzefsPearSvleqdr.-1-Divide the following polynomials. Check to see if the result can be factored further. 1) (m3 - 6m2 - 16m + 63) ¸ (m. Q Worksheet by Kuta Software LLC Kuta Software - Infinite Pre-Algebra Name_____ Multiplying a Polynomial and a Monomial Date_____ Period____ Find each product. 1) 8 x(6x + 6) 48 x2 + 48 x 2) 7n(6n + 3) 42 n2 + 21 n 3) 3r(7r − 8) 21 r2 − 24 r 4) 8(8k − 8) 64 k − 64 5) 10 a(a − 10 b) 10 a2 − 100 ab 6) 2(9x − 2y Polynomial Long Division Calculator. The calculator will perform the long division of polynomials, with steps shown. Show Instructions. In general, you can skip the multiplication sign, so `5x` is equivalent to `5*x`. In general, you can skip parentheses, but be very careful: e^3x is `e^3x`, and e^(3x) is `e^(3x)` Polynomial long division worksheet answers. N p2c031 b2f tk au gtdaf bs ao5f ptlw gaur mei 4lblsct. Divide 2 2 3 8 2 9 2 x x x x using long division. Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol next multiply or distribute the answer obtained in the previous. Polynomials - Negative Exponents Objective: Simplify expressions with negative exponents using the properties of exponents. There are a few special exponent properties that deal with exponents that are not positive. The first is considered in the following example, which is worded out 2 different ways: Example 1. a3 a3.\n\nDividing Polynomials By Monomials Worksheet - snowtanye\n\n1. By the way, related with Dividing Polynomials by Monomials Worksheet, scroll the page to see several similar images to give you more ideas. 6th grade long division worksheets, adding polynomials worksheet and factoring polynomials worksheet are some main things we want to present to you based on the gallery title. Continue with more related.\n2. If perhaps you actually demand help with math and in particular with practice worksheets on dividing monomials or radicals come pay a visit to us at Factoring-polynomials.com. We provide a whole lot of really good reference tutorials on subjects ranging from percents to adding and subtracting rational expression\n3. e if d(x) is a factor of f(x). 1) f (x) x x Create your own worksheets like this one with Infinite Precalculus. Free trial available at KutaSoftware.com. Title: document1 Author: Mik\n\nDividing Polynomials By Monomial Worksheets - Learny Kid\n\n1. Worksheets are multiplying a polynomial and a monomial, multiplying polynomials date period, model practice challenge problems vi, multiplying monomials and powers of monomials, multiplying monomials single variable s1, multiplying polynomials, multiplying dividing monomials. 100%100% found this document useful, mark this document as useful.\n2. Dividing polynomials using synthetic division use synthetic division to divide the polynomial by the linear factor. Dividing polynomials long synthetic division worksheet. When dividing polynomials we can use either long division or synthetic division to arrive at an answer. 7 8r3 55 r2 44 r 12\n3. monomial worksheets with answers - pachislot.info #31815 Monomial X Polynomial Worksheet Gina Wilson | Wiring Library #31816 Multi Step Multiplication Word Problems Grade Worksheets.\n\nMultiplying Monomials and Polynomials Worksheet Along with Polynomial Functions Worksheets Algebra 2 Worksheets Download by size: Handphone Tablet Desktop (Original Size) There are many different types of methods for dividing these into several parts Dividing polynomials by monomials worksheet with answers doc 10m2n - 15mn - 25mn2 by (-2/5)mn (vi) 21a3b3 + 35a4b2 - 56a2b4 by -7a2b2 3. Divide the following polynomial with monomial and write the answer in the simplest form: (i) 7x3 - 35x2 + 49x with 7x (ii) 18u2v2 - 3uv2 + 9uv3 s 3uv (iii) 8m2n - 5mn2 - 20mn by (20mn by.\n\nShowing top 8 worksheets in the category - Dividing Polynomials By Monomials. Some of the worksheets displayed are Division of polynomials by monomials, Factoring polynomials work with answers, Dividing polynomials 1, Dividing polynomials, Dividing monomials, Dividing polynomials date period, Algebra 1, Dividing polynomials remainder and factor theorems Sep 5, 2020 - Dividing Polynomials by Monomials Worksheet. 20 Dividing Polynomials by Monomials Worksheet. Divide Polynomials Worksheet 2 with Image\n\nDividing Monomials Worksheet - workshee\n\nTake your practice beyond books with our printable dividing polynomials worksheets. Instruct high school students to divide each term of the polynomial individually by the monomial divisor. Reduce the fractions in each term, and apply the quotient rule that says the exponents should be subtracted when dividing two powers with the same base Dividing by Monomials 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too Dividing Monomial Worksheets And Answer Sheets? Here is a number of keywords that visitors entered recently in order to get to our site. How is this helpful ? identify the keyword you are looking (i.e. Dividing Monomial Worksheets And Answer Sheets) in the leftmost column belo Example 2: Dividing Polynomials by Monomials. 1. Have students re-write the problem as a fraction with the second term in the denominator; 2. Have students re-write the fraction as 3 separate mini-fractions or mini-problems, with the operations in the middle. This helps students break the problem into individual monomial divided by. Continue with more related things such dividing polynomials by monomials worksheet, algebra factoring polynomials worksheet and multiplying and dividing monomials worksheet. Our goal is that these Algebra Monomial Worksheet photos gallery can be useful for you, give you more references and most important: make you have an amazing day\n\nMultiplication and division of polynomials - Worksheet\n\n• Practice Worksheet (key) Adding and Subtracting Polynomials Guided Notes (blank) Guided Notes (completed) A/S Polynomials Practice Worksheet Practice Worksheet (key) Lessons 1-4 Review Review Key Multiplying a Polynomial by a Monomial Guided Notes (blank) Guided Notes (completed) Multiplying a Polynomial by a Monomial Practice Worksheet\n• Here is a set of practice problems to accompany the Dividing Polynomials section of the Polynomial Functions chapter of the notes for Paul Dawkins Algebra course at Lamar University\n• Answers for the worksheet on dividing monomials are given below to check the exact answers of the above questions on division of monomials. Some of the worksheets displayed are multiplying dividing monomials model practice challenge problems vi multiplying a polynomial and a monomial multiplying polynomials date period multiplication and.\n• Improve your math knowledge with free questions in Divide a polynomial by a monomial and thousands of other math skills\n• Dividing Polynomials - The first problems that we are going to work on are binomials that are being divided by monomials. Remember that a binomial is two terms connected with either addition or subtraction, and that a monomial is one term. Here is an example: - On Board: (6x2 - 7) / 2\n• Play this game to review Algebra I. Divide the polynomial by the monomial. Preview this quiz on Quizizz. Divide the polynomial by the monomial. Dividing Polynomials by Monomials DRAFT. 9th grade. 0 times. Mathematics. 0% average accuracy. 3 days ago. ejackson_87153. 0. Save. Edit. Edit. Dividing Polynomials by Monomials DRAFT\n\nLook at this example of division using factors. When you review the strategy you use in Arithmetic, algebra will make more sense. Simply show the factors, cancel out the factors (which is division) and you will be left with your solution. Follow the steps through to fully understand the sequence involved to divide monomials Polynomial Worksheets On this page you will find: a complete list of all of our math worksheets relating to Polynomials.Choose a specific addition topic below to view all of our worksheets in that content area. You will find addition lessons, worksheets, homework, and quizzes in each section divide a polynomial by a monomial, divide each term in the polynomial by the monomial, and then write each quotient in lowest terms. Example 1: Divide 9x4 + 3x2 - 5x + 6 by 3x. Solution: Step 1: Divide each term in the polynomial 9x4 + 3x2 - 5x + 6 by the . monomial 3x. 93 569 3 5 642 4 2 3333 xxx x x x\n\nDividing Polynomials Coloring Page from Dividing Polynomials Worksheet, source: pinterest.com. Polynomial Word Problem Educational Cool Tools from Dividing Polynomials Worksheet, source: pinterest.com. Product Monomials Worksheet With Answers Multiplying Two from Dividing Polynomials Worksheet, source: gigidiaries.co multiplying and dividing monomials worksheet with answers To find the product of powers that have the same base, add their.Multiplying and Dividing Polynomials Worksheet Answer Key. multiplying and dividing polynomials worksheet kuta 3x - 42x 3x - 42. 6x2 - 8x 6x - 8. 6x2 - 2x - 8.Multiplying and Dividing Polynomials Worksheet Unit 5 Worksheet. Polynomial Division. Divide each of the polynomials using long division. 1. (4x2 - 9) ÷ (2x + 3) 2. (x2 - 4) ÷ (x + 4 Monomial Worksheets. Resources Academic Maths Algebra Polynomials Monomial Worksheets. Learn Maths from the best. First Lesson Free! Emma. February 2, 2021. Chapters. Exercise 1; Dividing Polynomials. Algebraic Fractions. Adding or Subtracting Algebraic Fractions. Binomial Theorem. Adding Polynomials\n\nHow to divide polynomials by monomials? To divide a polynomial by a monomial, separately divide each term of the polynomial by the monomial and add each operation's quotient to get the answer. Let's try a few examples here. Example 5. Divide 24x 3 - 12xy + 9x by 3x. Solution (24x 3 -12xy + 9x)/3x (24x 3 /3x) - (12xy/3x) + (9x/3x) = 8x. Resources Academic Maths Algebra Polynomials In this section, you will find exercises and worksheets to review the theory of polynomials. A polynomial is an algebraic expression consisting of an algebraic sum of monomials connected by the signs + (plus) and - (minus) Dividing Monomials Calculator . A monomial is a polynomial which has only one term. Dividing a monomial by a monomial is similar to multiplying a monomial with a monomial, except a small difference", null, "Division of monomials by monomials workshee\n\nDivide polynomials by monomials (with remainders) Practice: Divide polynomials by monomials (with remainders) Dividing polynomials with remainders. Practice: Divide polynomials with remainders. This is the currently selected item. Next lesson. Solving equations by graphing Videos by Julie Harland http://yourmathgal.comExplains and shows examples of how to Divide a Polynomial by a Monomial\n\nDividing. Polynomials can sometimes be divided using the simple methods shown on Dividing Polynomials. But sometimes it is better to use Long Division (a method similar to Long Division for Numbers) Numerator and Denominator. We can give each polynomial a name: the top polynomial is the numerator; the bottom polynomial is the denominato 4. Polynomial times polynomial: To multiply two polynomials where at least one has more than two terms, distribute each term in the first polynomial to each term in the second. Examples: a. ˆ b. DIVISION: 1. Division by Monomial: Each term of the polynomial is divided by the monomial and it is simplified as individual fractions. Examples: a. ˙ �", null, "", null, "Dividing A Polynomial By A Monomial Guided Notes\n\nShould you actually demand advice with algebra and in particular with simplify monomials calculator or geometry come pay a visit to us at Algebra-net.com. We maintain a large amount of quality reference information on matters ranging from solving exponential to numbe adding, subtracting, multiplying, dividing integers adding integers subtracting dividing polynomials polynomial multiplying monomials addition division trinomial binomial explanation numbers nominal tutorials2 softmat", null, "", null, "", null, "", null, "", null, "Adding Polynomials Subtracting Polynomials Multiplication (Monomial by Polynomial) Multiplication using FOIL (Binomial by Binomial) Multiplication (Polynomial by Polynomial) Dividing a Monomial by a Monomial Dividing a Polynomial by a Monomial Mixing Addition, Multiplication, and Division in one problem. A mix of all the different operation. multiplying and dividing polynomials worksheet doc Dividing Polynomials with Long Division Worksheets. 4x2 −3x+2 4. Answers to Dividing Polynomials 1) 2 r4 + 4r3 + 3r2 2) x3 + x2 + 5x 3) n 5 + 2 + 1 10n 4) 1 3 + 1 9v + 2 9v2 5) 5v + 2 + 4 9v 6) n + 1 9 + 1 3n 7) 3r + 1 5 + 3 r 8) 3k2m 2 + Polynomials class 9 worksheet pdf. Aug 7, 2013 - These Algebra 1 Worksheets allow you to produce. Adding and Subtracting Polynomials: Worksheets and Answers Adding and Subtracting Polynomials Worksheets This polynomial worksheet will produce problems for adding and subtracting polynomials. You may select which type of polynomials problem to use and the range of numbers to use as the constants View Homework Help - multiplying-monomials-worksheet.pdf from MATH 101 at Chapel Hill High School. I. Model Problems. II. Practice III. Challenge Problems VI. Answer Key We Some of the worksheets displayed are multiplying a polynomial and a monomial multiplying monomials and powers of monomials multiplying polynomials date period multiplying polynomials multiply the binomials and monomials addition and subtraction when adding multiply the monomials multiplying dividing monomials Divide polynomials by x (with remainders) CCSS.Math: HSA.APR.D.6 they're all logical good things to do to simplify this thing now once you have it here now we just have a bunch of monomials that we're just dividing by 6x and here we can just use the exponent properties this first one over here we can take the coefficients and divide them 18.\n\n• Xbox 360 disc tray stuck partially open.\n• Thermostat wiring diagram Honeywell.\n• Can I use lava rocks in a charcoal grill.\n• How to help a stuttering child at home.\n• How to use address book in Thunderbird.\n• 12px to PT.\n• Psychodynamic approach to phobias.\n• Bead Crochet pattern maker online.\n• Gmail going to Archive instead of Inbox Mac.\n• Artichoke nih.\n• Cat afraid of tin foil.\n• Computer power supply for audio amplifier.\n• Xopenex generic name.\n• Profit percentage formula Class 8.\n• Html conditional statements if else.\n• PNC record deleted.\n• Descending order of fractions with same numerator.\n• Chances of Down syndrome with high hCG.\n• Songs written by Bruno Mars for other artists.\n• What is an IRQ.\n• CarPlay Apple Music vs Spotify.\n• How to clean toenails with hydrogen peroxide.\n• Ribbit ribbit frog song.\n• Jangho me fungal infection ka ilaj.\n• Maddy from Beauty and the Geek illness.\n• Mark to market swap.\n• Brooklyn Brew Shop replacement parts.\n• OST vs PST.\n• What makes me happy essay for class 4.\n• Segway tours Victoria.\n• ATM withdrawal charges from other Bank in Pakistan 2020.\n• Hyundai i10 Price in Kerala.\n• Comedy Store San Diego.\n• Wheelchair Attendant (airport).\n• Event planner certification online free.\n• Jonas Bevacqua family.\n• Baked lemon pepper chicken recipe.\n• MAC address on iPhone 6.\n• How do I find My SMTP server on Windows 10.\n• Trip Expense Manager online.\n• What is Clear seeded glass." ]

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[ "[This article was first published on R on I Should Be Writing: Est. 1641, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)\nWant to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nThis spooky post was written in collaboration with Yoav Kessler (@yoav_kessler) and Naama Yavor (@namivor)..\n\nExperimental psychology is moving away from repeated-measures-ANOVAs, and towards linear mixed models (LMM1). LMMs have many advantages over rmANOVA, including (but not limited to):\n\n• Analysis of single trial data (as opposed to aggregated means per condition).\n• Specifying more than one random factor (typically crossed random intercepts of subject and item).\n• The use of continuous variables as predictors.\n• Making you look like you know what you’re doing.\n• Defeating the un-dead / reviewer 2.\n• The ability to specify custom models.2\n\nThis post will focus on this last point. Specifically, why you should always include main-effects when modeling interactions, and what happens if you don’t (spooky).\n\n## Fitting the Right (yet oh so wrong) Model\n\nSay you’ve finally won that grant you submitted to study candy consumption during ghostly themed holidays. As part of your first study, you decide to measure the effects of costume type (scary / cute) and level of neighborhood decor (high / low levels of house decorations) on the total weight of collected candy (in Kg). A simple, yet informative 2-by-2 design.\n\nBeing the serious scientist you are, you have several hypotheses:\n\n1. A main effect for decor level – neighborhoods with more decorations will overall give out more candy.\n2. No main effect for costume – overall, children with cute and scary costumes will receive the same amount of candy (in Kg).\n3. A decor level $$\\times$$ costume interaction – high decor neighborhoods will favor scary costumes, while low decor neighborhoods will favor cute costumes.\n\nIt would only make sense to specify your statistical model accordingly – after all, why shouldn’t your model represent your hypotheses?\n\nIn R, such a model is described as candy_kg ~ decor + decor:costume, instructing R to model candy_kg as a function of the effect for decor + the interaction decor:costume.\n\nAnd so, you fit the model:\n\noptions(contrasts = c('contr.sum', 'contr.poly')) # set effects coding (just once)\n\nfit <- aov(candy_kg ~ decor + decor:costume, data = spooky_data)\nTerm df SS MS F p-value $$\\eta^2$$\ndecor 1 30.00 30.00 23.64 <0.001 0.10\ndecor:costume 2 120.00 60.00 47.28 <0.001 0.40\nResiduals 116 147.20 1.27\n\nAs predicted, you find both a significant main effect for decor and the interaction decor $$\\times$$ costume, with the interaction explaining 40% of the variance in collected candy weight. So far so good - your results reflect your hypotheses!\n\nBut then you plot your data, and to your horror you find…", null, "It looks like there is no interaction at all! Your interaction was nothing more than a ghost! An apparition! How is this possible?? Where has all of variance explained by it gone???", null, "What IS This??\n\nIn fact, had you fit the full model, you would have found:\n\nfit <- aov(candy_kg ~ decor * costume, data = spooky_data)\nTerm df SS MS F p-value $$\\eta^2$$\ndecor 1 30.00 30.00 23.64 <0.001 0.10\ncostume 1 120.00 120.00 94.56 <0.001 0.40\ndecor:costume 1 0.00 0.00 0.00 1.000 0.00\nResiduals 116 147.20 1.27\n\nThe interaction actually explains 0% of the variance! And the effect of costume is the one that explains 40% of the variance!3 How could this be?? Have we angered Fisher’s spirit somehow?\n\nWhat happened was that because we did not account for costume in our model, the variance explained by costume was swallowed by the interaction decor $$\\times$$ costume!\n\n## The Math\n\nIf you find math too scary, feel free to skip to conclusion.\n\nTravel back to Intro to Stats, and recall that the interaction’s sum-of-squares - $$SS_{A\\times B}$$ - is calculated as:\n\n$$SS_{A\\times B} = (\\bar{x}_{ij} - \\bar{x}_{i.} - \\bar{x}_{.j} + \\bar{\\bar{x}}_{..})^2$$\n\nThis is a simplification of the following equation:\n\n$$SS_{A\\times B} = \\sum \\sum (\\bar{x}_{ij} - (\\bar{x}_{i.} - \\bar{\\bar{x}}_{..}) - (\\bar{x}_{.j} - \\bar{\\bar{x}}_{..}) + \\bar{\\bar{x}}_{..})^2$$\n\nWhere $$(\\bar{x}_{i.} - \\bar{\\bar{x}}_{..})$$ represents the main effect for $$A$$ and $$(\\bar{x}_{.j} - \\bar{\\bar{x}}_{..})$$ represents the main effect for $$B$$. We can see that $$SS_{A\\times B}$$ represents the deviation from the additive model - i.e., it is the degree by which the observed cells’ means deviate from what would be expected if there were only the two main effects.\n\nWhen we exclude the main effect of $$B$$ from out model, we are telling our model that there is no need to estimate the main effect. That is, we set $$(\\bar{x}_{.j} - \\bar{\\bar{x}}_{..})=0$$. The resulting $$SS_{A\\times B}$$ is computed not as above, but as:\n\n$$SS_{A\\times B} = \\sum \\sum (\\bar{x}_{ij} - (\\bar{x}_{i.} - \\bar{\\bar{x}}_{..}) + \\bar{\\bar{x}}_{..})^2$$\n\nThis formula represents the degree by which the observed cells’ means deviate from what would be expected if there was only the main effect of $$A$$. But now if the cells’ means deviate in a way that would otherwise have been part of a main effect for $$B$$, the cells’ deviations from the main effect for $$A$$ will now include the deviations that would otherwise have been accounted for by a main effect of $$B$$! This results in the main effect for $$B$$ essentially getting “pooled” into $$SS_{A\\times B}$$. Furthermore, had you also excluded a main effect for $$A$$, this effect too would have been “pooled” into the so-called $$A\\times B$$ interaction.\n\nIn other words:\n\nWhen we don’t estimate (model) main effects, we change the meaning of interactions - they no longer represents a deviation from the additive model.\n\n## Conclusion\n\nSure, you can specify a model with no main effect and only interactions, but in such a case the interactions no longer mean what we expect them to mean. If we want interactions to represent deviation from the additive model, our model must also include the additive model!\n\nFor simplicity’s sake, this example has focused on a simple 2-by-2 between subject design, but the conclusions drawn here are relevant for any design in which a factor interacts with or moderates the effect of another factor or continuous variable.\n\n1. Or hierarchical linear models (HLM)… or mixed linear models (MLM)…\n\n2. Whereas in an AVONA analysis with 4 factors you always have: Four main effects + Six 2-way interaction + Four 3-way interaction + One 4-way interaction.\n\n3. Note also that the $$df_{residual}$$ is the same for both models, indicating the same number of parameters overall have been estimated in both. E.g., while in the full model we would have 3 parameters - one for each main effect + one for the interaction, in the misspecified model we have one for the main effect, and two for the interaction. That is, no matter how you tell the model to split the $$SS$$s, the number of parameters needed to model 4 cells will always be 3." ]

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[ "# What does 25th percentile mean?\n\n## What does 25th percentile mean?\n\n25th Percentile – Also known as the first, or lower, quartile. The 25th percentile is the value at which 25% of the answers lie below that value, and 75% of the answers lie above that value. 50th Percentile – Also known as the Median. Half of the answers lie below the median and half lie above the median.\n\n## What is the 25th percentile of the scores?\n\nThe 25th percentile is also called the first quartile. The 50th percentile is generally the median (if you’re using the third definition—see below). The 75th percentile is also called the third quartile.\n\nIs 25th percentile normal?\n\nIf your child is in the 25th percentile for weight, this means he’s heavier than 25 percent of boys his age, and less heavy than 75 percent of boys his age. Kids, like adults, come in all shapes and sizes, and some fall on the heavier side, some on the lighter side.\n\nIs scoring in the 25th percentile good?\n\nThe 25th and 75th percentiles mark the boundaries for the middle 50% of admitted students. Half of students scored either above or below these numbers. Having a score above the 75th percentile does not guarantee admission. Having a score below the 25th percentile does not mean you should not apply.\n\n### What is 25th percentile salary?\n\nUS Percentile Wages The 25th percentile salaries in the U.S. were \\$22,150 per year. At the 75th percentile, salaries for Americans were \\$54,250 per year. The annual 90th percentile salary rate was \\$83,140. The 50th percentile or median salary for all jobs in the U.S. was \\$33,840 per year.\n\n### What is the 25th 50th and 75th percentile?\n\nThe 25th percentile is also known as the first quartile (Q1), the 50th percentile as the median or second quartile (Q2), and the 75th percentile as the third quartile (Q3).\n\nWhat does 25th percentile mean in salary?\n\nTwenty-fifth percentile\n25th%(Twenty-fifth percentile) The lowest quarter of salaries for this job fall below the twenty-fifth percentile. The “middle half” of people in this job have salaries that fall between the 25th and 75th percentile. 50th%(Fiftieth percentile or median)\n\nIs a 25 on the ACT bad?\n\nIs 25 ACT Score Good? With a score of 25, you are in the 79th percentile of all test takers. More likely than not, with a score of 25, you’ll often be in or near the commonly accepted range at selective colleges. The exception may be the highly selective colleges.\n\n## How do I calculate my salary percentile?\n\nTo determine the salary range percentile, you must first calculate the difference between the maximum and minimum salary figures. For example, if the salary range for a particular position is between \\$45,000 and \\$75,000, the difference between those two figures would be \\$30,000.\n\n## How to find the 25th percentile in statistics?\n\nGiven here is a simple online 25 th percentile calculator to find the 25th rank of percentile. 25 th Percentile is also referred to as the first quartile in statistics. For example, a test score that is greater than or equal to 25% of the scores of people taking the test is said to be 25th percentile, where 25 is the percentile rank.\n\nHow to calculate 10th percentile step by step?\n\nFollow the steps above to calculate the 10th percentile. 1 (.1 x 8)=.8 (round to 1) 2 K=33 (greater than) and k=30 (greater than or equal to) 3 Average. (33 + 30) / 2 = 31.5\n\nWhat does the 90th percentile mean on a test?\n\nFor instance, if you take a standardized test and your score is greater than or equal to 90% of all other scores, your percentile rank is the 90th percentile. It’s also important to note that the percentile rank may not denote an actual test score or other assessment score.\n\n### How to calculate the percentile rank of an item?\n\nThe value 25 represents the n variable in the formula: Percentile rank = 80 / [100 x (25 + 1)] Add one to the total number of values in the data set to get this: Percentile rank = 80 / [100 x (26)] 3. Multiply the sum of the number of items and one by 100. Once you add one to your n value, multiply this sum by 100." ]

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[ "Question: Genetic correlations from LDHub - how to test if two datasets have significantly different rg with a trait?\n0\nkettchy0 wrote:\n\nHello all,\n\nThis is probably (hopefully!) a very simple question, but I've reached a bit of a mental block about it.\n\nI have been using Alkes Price's LDHub website (http://ldsc.broadinstitute.org/). It takes GWAS summary statistics as input and calculates the genetic correlations with other traits in the database. It outputs for each trait:\n\n• rg, the genetic correlation\n• se, the standard error of rg\n• a z score\n• a p-value for rg\n\nThis is all fine. What I'm struggling on is comparing rgs. I'm theorising that for a given trait, the value of rg might vary between two groups, say European ancestry vs African ancestry. However, I'm struggling on a formal test for that.\n\nUsing R I've muddled something together, but I'm not sure if it's valid. rg.A and rg.B are the genetic correlations in the two groups A and B between my trait of interest and a trait in the database (likewise for se.A and se.B).\n\n``````dat\\$diffZ <- (dat\\$rg.A - dat\\$rg.B) / sqrt(dat\\$se.A^2 + dat\\$se.B^2)\ndat\\$diffP <- 2*pnorm(abs(dat\\$diffZ), lower.tail=F)\n``````\n\nThanks!" ]

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[ "• # Elementary Mathematics Grade 3 Unit 5\n\nSubject: Mathematics\nTimeline: 16 days\nUnit 5 Title: Introduction to Fractions\n\nUnit Overview:\n\nThis unit will give the opportunity for students to build a conceptual understanding of fractions.  There is focus on fractions as part of a whole and part of a collection (set).  The unit will emphasize identifying fractions using shapes and ordering fractions on a number line.\n\nUnit Objectives:\n\nAt the end of this unit, all students must be able to build and draw fractions on a shape or figure, find the fractional units on a number line or ruler, and explain why fractional units are not always equal.\n\nFocus Standards:\n\nPA.CCSS.Math.Content.CC.2.1.3.C.1 Explore and develop an understanding of fractions as numbers.(3.NF.1, 3.NF.2)\n\nMathematical Practice Standards:\n\n#1 Make sense of problems and persevere in solving them.\n\nMathematically proficient students start by explaining to themselves the meaning of a problem and look for an entry point to its solution. They then plan a solution pathway rather than jumping into a solution attempt.  They monitor and evaluate their progress and change course, if needed.  Students might rely on using concrete objects or pictures to help conceptualize and solve a problem.  They then check their answers to problems using a different method, and continually ask themselves, “Does this make sense?”  These students can also understand the approaches of others to solving complex problems and identify correspondences between different approaches.\n\n#2 Reason abstractly and quantitatively.\n\nMathematically proficient students make sense of quantities and their relationships in problem situations.  They bring two complementary abilities to bear on problems, the ability to decontextualize-to abstract the given situation and represent it symbolically and manipulate the representing symbols as if they have a life of their own, without necessarily attending to their referents and contextualize-to pause as needed to during the manipulation process in order to probe into the referents for the symbols involved.\n\n#3 Construct viable arguments and critique the reasoning of others.\n\nMathematically proficient students understand and use states assumptions, definitions, and previously established results in constructing arguments.  They make a statement that they believe to be true but not yet proved and then build a progression of statements to explore its truth.  They justify their conclusions, communicate them to others and respond to the arguments of others.  Elementary students can construct arguments using concrete referents such as objects, drawings, diagrams, and actions.\n\n#4 Model with mathematics.\n\nMathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. In early grades, this might be as simple as writing an addition equation to describe a situation.\n\n#5 Use appropriate tools strategically.\n\nMathematically proficient students consider the available tools when solving a mathematical problem. These tools might include pencil and paper, concrete models, a ruler, a protractor, or a calculator. Proficient students are sufficiently familiar with tools appropriate for their grade to make sound decisions about when each of these tools might be helpful, recognizing both the insight to be gained and their limitations.\n\n#6 Attend to Precision.\n\nMathematically proficient students try to communicate precisely to others.  They use clear definitions in discussions with others and in their own reasoning.  They state the meaning of the symbols they choose, including the equals sign, consistently and appropriately.  They calculate accurately and efficiently and give carefully formulated explanations to each other.\n\n#7  Look for and make use of structure.\n\nMathematically proficient students look closely to discern a pattern or structure. Young students, for example, might notice that three and seven more is the same amount as seven and three more, or they may sort a collection of shapes according to how many sides the shapes have.\n\n# 8 Look for and express regularity in repeated reasoning.\n\nMathematically proficient students notice if calculations are repeated, and look both for general methods and for shortcuts... They continually evaluate the reasonableness of their intermediate results.\n\nConcepts - Students will know:\n• Fractions are wholes partitioned into equal parts.\n• The numerator of a fraction tells the selected number of parts.\n• The denominator of a fraction tells the total number of parts.\n• In equal wholes, the smaller the denominator, the larger the fractional parts.\nCompetencies -Students will be able to:\n• Identify fractions as equal groups\n• Name the numerator and denominator of fractions\n• Identify fractions of a whole and fractions of a collection\n• Place fractions on a number line\n\nAssessments:\n• Unit 5 Assessment\n• Daily RSA\n\nElements of Instruction:\n\nStudents will have just completed a unit on measurement which will help their understanding of fractions as parts of parts of a whole.\n\nDifferentiation:\n\nEach lesson has differentiation options for each portion of the lesson.  Additional differentiation options are listed with directions and student masters in the Teacher’s Guide to Games.\n\nInterdisciplinary Connections:\n\nNone" ]

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[ "#### Stoichiometry Problems\n\n1.      How many atoms are there in one mole of oxygen gas?\n\n2.      What is the mass of\na)      238 moles of iron metal\nb)      0.0035 moles of Na2HPO4\nc)      7.70 x 10−6 moles of helium gas\nd)      1.206 x 1024 atoms of N2 gas\n\n3.      How many moles are contained in the following substances\na)      1.50 g of silver metal\nb)      a 127 g spool of copper wire\nc)      4.00 g sample of argon gas\n\n4.      If potassium chlorate is heated gently, the crystals will melt.  Further heating will decompose it to give oxygen gas and potassium chloride.\na)      Write the formula for the reactants and products.\nb)      Write the balanced equation for the decomposition of potassium chlorate.\nc)      How many grams of potassium chloride are produced when 3.5 moles of potassium chlorate are decomposed?\nd)      How many moles of potassium chlorate are needed to give 1.5 moles of oxygen gas.\ne)      How many liters of oxygen gas at STP will be produced by decomposing 122.6 g of potassium chlorate?\n\n5.      Xenon is a noble gas. One of the first compounds made from xenon was xenon tetrafluoride, XeF4. When 10.0 g of xenon gas react completely  with fluorine to give xenon tetrafluoide, 39.8 kJ of energy are released. The equation for the reaction is:\n\nXe(g)  +  2F2(g)  →  XeF4(s)\n\nUsing this information calculate how much energy is released when 1 mole of xenon gas reacts.\n\n6a) How would you go about preparing 150 mL of 1.5 M barium nitrate solution from solid barium nitrate, Ba(NO3)2?\nb) Determine the molarity of a solution containing 0.15 g of KOH in 25 mL of solution.\n\n6.      Solutions of FeBr3 and KOH are mixed and a reddish solid, called a precipitate, Fe(OH)3, is formed.\na)      Write the molecular equation for the above reaction. It must be balanced.\nb)      Write the total ionic equation for the reaction.\nc)      Write the net ionic equation for the reaction." ]

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[ "# Create numpy array with random elements from list\n\nIs there an efficient way to create an arbitrary long numpy array where each dimension consists of n elements drawn from a list of length >= n? Each element in the list can be drawn only once for each dimension.\n\nFor instance, if I have the list `l = ['cat', 'mescaline', 'popcorn']`, I want to be able to, for instance by typing something like `np.random.pick_random(l, (3, 2), replace=false)`, create an array `array([['cat', 'popcorn'], ['cat', 'popcorn'], ['mescaline', 'cat']])`.\n\nThank you.\n\n• Is there something wrong with the dirt simple and obvious import random; random.shuffle()? – Jim Dennis Nov 11 '12 at 21:04\n• I wonder why does it have to be `numpy`? In general numpy is for numerical type of calculations, hence its name is short for numerical python, granted it does support other types ... pythons own `random.sample` might be better for this `[random.sample(['cat', 'mescaline', 'popcorn'], number_of_members) for index in xrange(number_of_arrays)]`... – Samy Vilar Nov 11 '12 at 21:20\n• @samy-vilar The reason is that I want to avoid slow loops. I will use this for Monte Carlo simulation, so I will need quite large arrays. – Aae Nov 12 '12 at 0:52\n• @jim-dennis The difference in performance when I create large arrays. – Aae Nov 12 '12 at 0:53\n\nTheres a couple of ways of doing this, each has their pros/cons, the following four where just from the top of my head ...\n\n• pythons own `random.sample`, is simple and built in, though it may not be the fastest...\n• `numpy.random.permutation` again simple but it creates a copy of which we have to slice, ouch!\n• `numpy.random.shuffle` is faster since it shuffles in place, but we still have to slice.\n• `numpy.random.sample` is the fastest but it only works on the interval 0 to 1 so we have to normalize it, and convert it to ints to get the random indices, at the end we still have to slice, note normalizing to the size we want does not generate a uniform random distribution.\n\nHere are some benchmarks.\n\n``````import timeit\nfrom matplotlib import pyplot as plt\n\nsetup = \\\n\"\"\"\nimport numpy\nimport random\n\nnumber_of_members = 20\nvalues = range(50)\n\"\"\"\n\nnumber_of_repetitions = 20\narray_sizes = (10, 200)\n\npython_random_times = [timeit.timeit(stmt = \"[random.sample(values, number_of_members) for index in xrange({0})]\".format(array_size),\nsetup = setup,\nnumber = number_of_repetitions)\nfor array_size in xrange(*array_sizes)]\n\nnumpy_permutation_times = [timeit.timeit(stmt = \"[numpy.random.permutation(values)[:number_of_members] for index in xrange({0})]\".format(array_size),\nsetup = setup,\nnumber = number_of_repetitions)\nfor array_size in xrange(*array_sizes)]\n\nnumpy_shuffle_times = [timeit.timeit(stmt = \\\n\"\"\"\nrandom_arrays = []\nfor index in xrange({0}):\nnumpy.random.shuffle(values)\nrandom_arrays.append(values[:number_of_members])\n\"\"\".format(array_size),\nsetup = setup,\nnumber = number_of_repetitions)\nfor array_size in xrange(*array_sizes)]\n\nnumpy_sample_times = [timeit.timeit(stmt = \\\n\"\"\"\nvalues = numpy.asarray(values)\nrandom_arrays = [values[indices][:number_of_members]\nfor indices in (numpy.random.sample(({0}, len(values))) * len(values)).astype(int)]\n\"\"\".format(array_size),\nsetup = setup,\nnumber = number_of_repetitions)\nfor array_size in xrange(*array_sizes)]\n\nline_0 = plt.plot(xrange(*array_sizes),\npython_random_times,\ncolor = 'black',\nlabel = 'random.sample')\n\nline_1 = plt.plot(xrange(*array_sizes),\nnumpy_permutation_times,\ncolor = 'red',\nlabel = 'numpy.random.permutations'\n)\n\nline_2 = plt.plot(xrange(*array_sizes),\nnumpy_shuffle_times,\ncolor = 'yellow',\nlabel = 'numpy.shuffle')\n\nline_3 = plt.plot(xrange(*array_sizes),\nnumpy_sample_times,\ncolor = 'green',\nlabel = 'numpy.random.sample')\n\nplt.xlabel('Number of Arrays')\nplt.ylabel('Time in (s) for %i rep' % number_of_repetitions)\nplt.title('Different ways to sample.')\nplt.legend()\n\nplt.show()\n``````\n\nand the result:", null, "So it looks like `numpy.random.permutation` is the worst, not surprising, pythons own `random.sample` is holding it own, so it looks like its a close race between `numpy.random.shuffle` and `numpy.random.sample` with `numpy.random.sample` edging out, so either should suffice, even though `numpy.random.sample` has a higher memory footprint I still prefer it since I really don't need to build the arrays I just need the random indices ...\n\n``````\\$ uname -a\nDarwin Kernel Version 10.8.0: Tue Jun 7 16:33:36 PDT 2011; root:xnu-1504.15.3~1/RELEASE_I386 i386\n\n\\$ python --version\nPython 2.6.1\n\n\\$ python -c \"import numpy; print numpy.__version__\"\n1.6.1\n``````\n\nUPDATE\n\nUnfortunately `numpy.random.sample` doesn't draw unique elements from a population so you'll get repitation, so just stick with shuffle is just as fast.\n\nUPDATE 2\n\nIf you want to remain within numpy to leverage some of its built in functionality just convert the values into numpy arrays.\n\n``````import numpy as np\nvalues = ['cat', 'popcorn', 'mescaline']\nnumber_of_members = 2\nN = 1000000\nrandom_arrays = np.asarray([values] * N)\n_ = [np.random.shuffle(array) for array in random_arrays]\nsubset = random_arrays[:, :number_of_members]\n``````\n\nNote that N here is quite large as such you are going to get repeated number of permutations, by permutations I mean order of values not repeated values within a permutation, since fundamentally theres a finite number of permutations on any giving finite set, if just calculating the whole set then its n!, if only selecting k elements its n!/(n - k)! and even if this wasn't the case, meaning our set was much larger, we might still get repetitions depending on the random functions implementation, since shuffle/permutation/... and so on only work with the current set and have no idea of the population, this may or may not be acceptable, depends on what you are trying to achieve, if you want a set of unique permutations, then you are going to generate that set and subsample it.\n\n• Thanks for the effort. The efficiency of the numpy.shuffle method is okay. However it doesn't save me from slow loops when doing calculations on the array. For instance I would like to do sum(random_arrays, axis=1). Sorry I am so unclear in what I'm looking for. – Aae Nov 13 '12 at 13:02\n• umm random_arrays.sum(axis = 1)? random_arrays should be a numpy type. Also note that shuffle may generate non-unique permutations depending on the number of random arrays you need, if you truly want unique permutations than you are going to have to generate them manually and sub sample them, also note that `numpy.random.choice` was added in 1.7 Im currently at 1.6.1, docs.scipy.org/doc/numpy-dev/reference/generated/… Im not sure about its performance need to test it, but it may be slower since it gens new arrays ... – Samy Vilar Nov 18 '12 at 1:54\n• Maybe I misunderstood, but the way I did it generates a 'list': pastee.org/d76bb The permutations shouldn't be unique. – Aae Nov 19 '12 at 18:39\n• @Aae I've updated to work with numpy, also sum can only be applied to numeric values, here you have string values, if you want to use indices, simply replace values by `range(leng(values))` and it should work. – Samy Vilar Nov 19 '12 at 22:05\n\nHere's a way to do it using numpy's `np.random.randint`:\n\n``````In : l = np.array(['cat', 'mescaline', 'popcorn'])\n\nIn : l[np.random.randint(len(l), size=(3,2))]\nOut:\narray([['cat', 'popcorn'],\n['popcorn', 'popcorn'],\n['mescaline', 'cat']],\ndtype='|S9')\n``````\n\nEDIT: after the additional details that each element should appear at most once in each row\n\nthis is not very space efficient, do you need something better?\n\n``````In : l = np.array(['cat', 'mescaline', 'popcorn'])\n\nIn : array([np.random.choice(l, 3, replace=False) for i in xrange(5)])\nOut:\narray([['mescaline', 'popcorn', 'cat'],\n['mescaline', 'popcorn', 'cat'],\n['popcorn', 'mescaline', 'cat'],\n['mescaline', 'cat', 'popcorn'],\n['mescaline', 'cat', 'popcorn']],\ndtype='|S9')\n``````\n• Thank you for this. However, there is one detail I forgot to mention. The new array shouldn't consist of dimensions which contain the same element more than once (if it hasn't been listed more than once in the list). – Aae Nov 12 '12 at 0:56\n• The update gives the desired result, but it's not very efficient. And efficiency is really what I'm requesting. Sorry if I have been unclear. – Aae Nov 14 '12 at 20:20\n• @Aae then you should specify what type of efficiency is important to you. speed? memory? – davidbrai Nov 15 '12 at 11:55\n• Speed is the important. I mentioned it in a comment above (‘avoid slow loops’) but I guess I could have made it clearer. – Aae Nov 16 '12 at 1:02\n``````>>> import numpy\n>>> l = numpy.array(['cat', 'mescaline', 'popcorn'])\n>>> l[numpy.random.randint(0, len(l), (3, 2))]\narray([['popcorn', 'mescaline'],\n['mescaline', 'popcorn'],\n['cat', 'cat']],\ndtype='|S9')\n``````\n• Thank you. But as I said to the other person here: there is one detail I forgot to mention. The new array shouldn't consist of dimensions which contain the same element more than once (if it hasn't been listed more than once in the list). – Aae Nov 12 '12 at 0:57" ]

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[ "## Events\n\n MUGI 2023Seville, Spain 16-21 April, 2023 MUG 2023Park City, Utah 15-20 October, 2023\nMore\n\n## News\n\nMore\nUsing multiple operation modes to represent a non-linear process unit operation is very common in LP.  For example, the reformate yield and qualities are non-linear functions of a reformer severity.  We could use a set of modes for operations at different severities in order to represent this.    Below is an example where we use three modes at 90, 95 and 100 severity to represent a reformer.", null, "An LP can find the optimal solution that uses some combination of the modes to represent the best severity.  To have this structure work properly, however, we need to ensure that only one or two adjacent modes can be active at the same time, so that a severity of 98, for example, is interpolated between the 95 and 100 data, not 90 and 100.  This can be done using MIP (mixed integer programming), with a  type 2 Special Ordered Set (SOS).  The drawback is that in large models, MIP may cause long and unpredictable run times.  In GRTMPS, we can use non-linear equations directly using Adherent Recursion.  Below we’ll present a simple way to fit the above three modes representation into polynomial equations.  We can then use just one mode with Adherent Recursion PUPs to model the non-linear behavior, completely avoiding MIP.\n\nTake a look at the Hydrogen (HYD) yield, the three modes correspond to three data points : (Severity,Yield) = (90,0.900), (95,0.925) and (100,0.960).  Assume that the yield (y) can be expressed as a function of the severity (x) in the polynomial form: y =ax2  +bx +c.\n\nPlugging in values from the three data points, gives three simultaneous equations and three unknowns:  a, b and c.", null, "", null, "We can find the coefficients a, b, and c by multiply the inverse of A from the left on both sides:", null, "Excel provides array formulas for finding matrix inverse (MINVERSE) and matrix multiplication (MMULT).  We can use them to easily find the polynomial coefficients a, b and c, as shown below:", null, "Here are the steps in the example:\n1.  Create matrix A in cells B1:D3.\n2.  Select cells of the same size in F1:H3.\n3.  Type in the formula =MINVERSE(\\$B\\$1:\\$D\\$3) and click Ctrl+Shift+Enter key (to indicate an array formula where the result appears in multiple cells).  This gives the inverse of the matrix,  A-1 in cells F1:H3.\n4.  Create the RHS matrix in cells D5:D7.\n5.  Select cells G5:G7.  Type formula =MMULT(\\$F\\$1:\\$H\\$3,D5:D7) and click Ctrl+Shift+Enter.\nThe coefficients a, b, and c  are now in cells G5:G7.  (Note: In matrix multiplication, the order matters.  Make sure the cells for the inverse A matrix are given before the RHS values.)\n\nAfter getting a, b and c for polynomials for all the yields, yield qualities, utilities, losses etc., we can replace the three operations with a single mode connected to these formulae via PUP codes.", null, "The equations for the PUPs could be put in a Process Simulator Interface workbook or you can write them directly in the Adherent Recursion panel [TABLE 206.9] using Type 7 equations.   Type 7 is a 6th power polynomial of the generic form Ax6 + Bx5 + Cx4 + Dx3+ Ex2  + Fx + K1.   So  the calculated value for a goes in E, b in F and c in K1, as shown below.", null, "Severity, our x,  is indicated under PUP1 as RFS, an operating parameter constraint controlled to the allowed range in the LP model.  Where the relationship is to severity is actually linear, such as for reformate yield, code YYR#, then E is effectively zero.\n\nThe original modal representation only allowed severities in the range 90 to 100, but with a function based model, there is the possibility of extrapolating to higher and lower values - although you should plot the curve and check how far the equation looks trustworthy.    Generally speaking, if we have n modes which vary with respect to the input variable, we can use this method to fit the data points into a polynomial of the (n-1) power, creating a curvilinear description of the relationship to drive the SLP optimization model.\n\nFrom Richard's Desk, 22nd October 2018." ]

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[ "go to:   myTestBook.com\n print help! Print:   Use this to print without Ads and Toolbar (taks a few seconds). dotted fields in the header are editable. Report an error\n The following text/image is for 1 through 10", null, "Free Worksheets From myTestBook.com ------ © myTestBook.com, Inc.\n-\n Question 1 What action will bring points U and V of the line UV into second and fourth quadrant, respectively? A. rotating 90o counter-clockwise B. rotating 180o counter-clockwise C. rotating 90o clockwise D. rotating 180o clockwise\n Question 2 Line UV is dilated using a scale factor 2 (increased twice). Center of dilation is point (0,0). Find new coordinates of point U. A. (−6,−6) B. (−3,−3) C. (6,6) D. (−9,−9)\n Question 3 Circle with center K is dilated using a scale factor 0.5 (decreased half in its size). Center of dilation is the center of the circle. Find the coordinates of points where the circle with cut on the x-axis now. A. (12,0) and (15,0) B. (9,0) and (12,0) C. (6,0) and (18,0) D. (9,0) and (15,0)\n Question 4 Circle with center K is reflected across the line XY. Find new coordinates of its center point K. A. (12,6) B. (12,0) C. (0,0) D. (−12,0)\n Question 5 Triangle LMN is dilated using a scale factor 0.5 (decreased half). Center of dilation is the point L. Find new coordinates of points M and N. A. M(23,7) & N(19,10) B. M(21,5) & N(19,13) C. M(21,7) & N(19,10) D. M(21,7) & N(19,13)\n Question 6 Triangle LMN is reflected across the line LS. Find new coordinates point N. A. (17,13) B. (14,13) C. (−14,13) D. (14,−13)\n Question 7 Square ABCD is dilated using a scale factor 0.5 (decreased half). Center of dilation is the point D. Find new coordinates of points A, B and C. A. A(0,21) & B(4,21) & C(4,17) B. A(0,21) & B(4,21) & C(6,17) C. A(0,25) & B(4,25) & C(4,17) D. A(0,21) & B(5,21) & C(5,17)\n Question 8 Square FGHI is rotated 90o counter-clockwise around the point F. Find new coordinates of the point H. A. (3,15) B. (3,12) C. (−3,18) D. (3,18)\n Question 9 Square FGHI is dilated using a scale factor 0.5 (decreased half). Center of dilation is the point H. Find new coordinates of points F and G. A. F(−1,10) & G(3,10) B. F(0,6) & G(3,9) C. F(0,9) & G(3,9) D. F(−3,9) & G(3,9)\n Question 10 Square OPQR is dilated using a scale factor 2 (increased twice). Center of dilation is point O. Find new coordinates of point Q. A. (25,−8) B. (29,−8) C. (29,−4) D. (27,−6)\nFree Worksheets From myTestBook.com ------ © myTestBook.com, Inc.\n-" ]

[ null, "http://mytestbook.com/images/Grade8/Math/cr1337_272_inline_test2.gif", null ]

[ "# Solve the following: The ratio of Boys to Girls in a college is 3:2 and 3 girls out of 500 and 2 boys out of 50 of that college are good singers. A good singer is chosen what is the probability that - Mathematics and Statistics\n\nSum\n\nSolve the following:\n\nThe ratio of Boys to Girls in a college is 3:2 and 3 girls out of 500 and 2 boys out of 50 of that college are good singers. A good singer is chosen what is the probability that the chosen singer is a girl?\n\n#### Solution\n\nLet event S: Student is a good singer,\n\nevent B: Student is a boy,\n\nevent G: Student is a girl.\n\nSince the ratio of boys to girls is 3:2 and\n3 girls out of 500 and 2 boys out of 50 are good singers.\n\n∴ P(B) = 3/5, P(G) = 2/5, \"P\"(\"S\"/\"G\") = 3/500, \"P\"(\"S\"/\"B\") = 2/50.\n\n\"P\"(\"S\") = \"P\"(\"G\") xx \"P\"(\"S\"/\"G\") + \"P\"(\"B\") xx \"P\"(\"S\"/\"B\")\n\n= 2/5 xx 3/500 + 3/5 xx 2/50\n\n= (2xx3)/5(1/500+1/50)\n\n= 6/5 xx 11/500\n\n= 33/1250\n\nRequired probability = \"P\"(\"G\"/\"S\")\n\nBy Bayes’ theorem,\n\n\"P\"(\"G\"/\"S\") = (\"P\"(\"G\") \"P\"(\"S\"/\"G\"))/(\"P\"(\"S\")\n\n= (2/5 xx 3/500)/(33/1250)\n\n= 1/11\n\nConcept: Baye'S Theorem\nIs there an error in this question or solution?\n\n#### APPEARS IN\n\nBalbharati Mathematics and Statistics 1 (Arts and Science) 11th Standard Maharashtra State Board\nChapter 9 Probability\nMiscellaneous Exercise 9 | Q II. (17) | Page 214" ]

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