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https://www.decodejava.com/c-function-call-by-reference.htm | [
"< Prev\nNext >\n\n# C - Function Call by Reference\n\nIn one of our previous articles, we have explained the first technique about how to call a function that has arguments, by passing it arguments by value. Now, we are going to explain the second technique by which we could call a function that has arguments and pass the arguments to this function by reference i.e. by address.\n\nBefore we begin understanding the second technique, we will have to understand a little but about a very important feature of C language - Pointers. Pointers play a major role in understanding function call by reference. Hence, for those who are not aware of Pointers, please read our article Pointers in C before continuing ahead.\n\n## Calling a function with argument, by reference.\n\nIn the upcoming example, we are going to create a function named add10 which adds 10 to the variable of int type passed to it using pointers.\n\n``````/* Calling a function with argument, by reference */\n\n#include<stdio.h>\nint main()\n{\n\n/* function prototype declaration*/\n\nint a = 10;\n\nprintf(\"The value in a is : %d \\n\", a);\n\nprintf(\"After the add10 function is called \\n\");\n\nprintf(\"The value in a is : %d \\n\", a);\n\nreturn 0;\n}\n\n/* function to add 10 to the int value passed to it */\n{\n*i = *i + 10;\n}``````\n\n## Output-\n\n``````The value in a is : 10\nAfter the add10 function is called\nThe value in a is : 20``````\n\n## Program Analysis\n\n• Declaring function prototype - In the above mentioned example, we have created a function named add10 and have declared its prototype as -\n\n• ``````/* function prototype declaration*/\n\nThe prototype of this function is declared with a void return type, because this method will not return any value when it is called. This function will be passed the address of an int value when it is called. Hence we have specified a pointer that points to a value of int type i.e. int *, within the parenthesis().\n\n• Calling the function - We have called the add10 method by passing the address of value of an int variable named a as an argument, by using the \"address of\" operator i.e. &. The int variable a has a value 10 and its address is referred as an \"actual argument\" of the function add10, when it is called.\n\n`` add10(&a);\t /* add10 function is called */``\n\n• Function definition - When the add10 function is called, it receives the address of the variable(a) passed to it and this address is copied into pointer variable named i, which is used to store the address of an int variable, declared in the definition of add10 function.\n\n``````/* add10 function is defined */\n{\n*i = *i + 10; /* using * to accessing the value at address */\n}``````\n\nIn the function definition, the pointer variable i can also be referred as a \"formal argument\" of add10 function. This function, uses the * operator which stands for \"value at address\", which when used with a pointer variable i.e. i, allows us to access the value at an address in the memory location, pointed by i variable.\n\nEventually, the add10 function adds 10 to the value contained in i, which is reflected when the value of variable a is printed again on the console after calling the add10 function.\n\nThis proves that when we call a function that has arguments, by passing it arguments by reference(using pointers), the changes made on the values contained in the formal arguments within the function definition do affect the original value contained in actual arguments.\n\n## Another example of calling a function by reference\n\n• In the upcoming example, we are going to create a function named swap_char which accepts two char arguments and successfully swaps the value contained in these two char variables passed to it, because we have called the function and have passed it arguments by reference using pointers. Let's see the code.\n\n``````/* Calling a function with argument, by reference */\n\n#include<stdio.h>\nint main()\n{\n\nvoid swap_char(char *,char *); /* function prototype declaration*/\n\nchar a ='x';\nchar b ='y';\n\nprintf(\"The character value in a is : %c \\n\", a);\nprintf(\"The character value in b is : %c \\n\", b);\n\nswap_char(&a,&b); \t/*calling the swap_char function*/\n\nprintf(\"After the swap_char function is called \\n\");\n\nprintf(\"The character value in a is : %c \\n\", a);\nprintf(\"The character value in b is : %c \\n\", b);\n\nreturn 0;\n}\n\n/* function to swap characters */\nvoid swap_char(char *c, char *d)\n{\nchar e;\n\ne = *c;\n*c = *d;\n*d = e;\n\n}``````\n\n## Output-\n\n``````The character value in a is : x\nThe character value in b is : y\nAfter the exchange function is called\nThe character value in a is : y\nThe character value in b is : x```\n```\n\n## Function call by value v/s call by reference\n\n• On calling a function that has arguments, by passing it arguments by value, the values in the actual arguments passed to the it are only copied in its formal arguments. Hence, the changes made on the values contained in the formal arguments in the function definition is not reflected and does not affect the original value contained in actual arguments.\n\n• On calling a function that has arguments, by passing it arguments by reference, the address of original values in the actual arguments is passed to the formal arguments in the function definition. Hence, the changes made on the values by accessing their address contained in the formal arguments within the function definition does affect the original value contained in actual arguments.",
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"https://www.decodejava.com/Twitter7.png",
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"https://www.decodejava.com/gp.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8222364,"math_prob":0.9604519,"size":5333,"snap":"2023-40-2023-50","text_gpt3_token_len":1194,"char_repetition_ratio":0.18521298,"word_repetition_ratio":0.21344717,"special_character_ratio":0.24620289,"punctuation_ratio":0.11121951,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97261024,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-22T09:05:51Z\",\"WARC-Record-ID\":\"<urn:uuid:09131bcc-fed7-4150-b0bc-888e9e4ae7ac>\",\"Content-Length\":\"40562\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0db3cb74-d920-429c-a099-2d88a9c5844c>\",\"WARC-Concurrent-To\":\"<urn:uuid:caa9c0fd-d494-4d3e-8b08-6e0674439e1b>\",\"WARC-IP-Address\":\"160.153.94.71\",\"WARC-Target-URI\":\"https://www.decodejava.com/c-function-call-by-reference.htm\",\"WARC-Payload-Digest\":\"sha1:BVFTF3SYUDGWVMCR5B67A3BLVP4BYHX2\",\"WARC-Block-Digest\":\"sha1:E4LVJM54DAJEVZQK7PZL72M4EIHIC5K2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506339.10_warc_CC-MAIN-20230922070214-20230922100214-00781.warc.gz\"}"} |
https://search.r-project.org/CRAN/refmans/clhs/html/similarity_buffer.html | [
"similarity_buffer {clhs} R Documentation\n\n## Gower similarity analysis\n\n### Description\n\nCalculates Gower's similarity index for every pixel within an given radius buffer of each sampling point\n\n### Usage\n\nsimilarity_buffer(\ncovs,\npts,\nbuffer,\nfac = NA,\nmetric = \"gower\",\nstand = FALSE,\n...\n)\n\n\n### Arguments\n\n covs raster stack of environmental covariates pts sampling points, object of class SpatialPointsDataframe buffer Radius of the disk around each point that similarity will be calculated fac numeric, can be > 1, (e.g., fac = c(2,3)). Raster layer(s) which are categorical variables. Set to NA if no factor is present metric character string specifying the similarity metric to be used. The currently available options are \"euclidean\", \"manhattan\" and \"gower\" (the default). See daisy from the cluster package for more details stand logical flag: if TRUE, then the measurements in x are standardized before calculating the dissimilarities. ... passed to plyr::llply\n\na RasterStack\n\nColby Brungard\n\n### References\n\nBrungard, C. and Johanson, J. 2015. The gate's locked! I can't get to the exact sampling spot... can I sample nearby? Pedometron, 37:8–10.\n\n### Examples\n\nlibrary(raster)\nlibrary(sp)\n\ndata(meuse.grid)\ncoordinates(meuse.grid) = ~x+y\nproj4string(meuse.grid) <- CRS(\"+init=epsg:28992\")\ngridded(meuse.grid) = TRUE\nms <- stack(meuse.grid)\n\nsuppressWarnings(RNGversion(\"3.5.0\"))\nset.seed(1)\npts <- clhs(ms, size = 3, iter = 100, progress = FALSE, simple = FALSE)\ngw <- similarity_buffer(ms, pts\\$sampled_data, buffer = 500)\nplot(gw)\n\n\n\n[Package clhs version 0.9.0 Index]"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6601309,"math_prob":0.96516335,"size":1414,"snap":"2023-40-2023-50","text_gpt3_token_len":397,"char_repetition_ratio":0.10992908,"word_repetition_ratio":0.0,"special_character_ratio":0.27864215,"punctuation_ratio":0.21509434,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9740644,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-01T23:38:24Z\",\"WARC-Record-ID\":\"<urn:uuid:71287d0e-2278-4778-8f7c-459e0b987db7>\",\"Content-Length\":\"3852\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:23ba483a-c138-4105-ab76-073be736db44>\",\"WARC-Concurrent-To\":\"<urn:uuid:88c8ecac-36ab-4de4-9595-701db2d14f7d>\",\"WARC-IP-Address\":\"137.208.57.46\",\"WARC-Target-URI\":\"https://search.r-project.org/CRAN/refmans/clhs/html/similarity_buffer.html\",\"WARC-Payload-Digest\":\"sha1:OWHHHK6PSBWKCGURCJZLSZ7GKBE67VRD\",\"WARC-Block-Digest\":\"sha1:2EU6A53EYZZJIZK6VOVDLJXQPHLJO4CS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100308.37_warc_CC-MAIN-20231201215122-20231202005122-00531.warc.gz\"}"} |
https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&direction=prev&oldid=83833 | [
"# 2017 AMC 12B Problems/Problem 15\n\n## Problem 15\n\nLet",
null,
"$ABC$ be an equilateral triangle. Extend side",
null,
"$\\overline{AB}$ beyond",
null,
"$B$ to a point",
null,
"$B'$ so that",
null,
"$BB'=3AB$. Similarly, extend side",
null,
"$\\overline{BC}$ beyond",
null,
"$C$ to a point",
null,
"$C'$ so that",
null,
"$CC'=3BC$, and extend side",
null,
"$\\overline{CA}$ beyond",
null,
"$A$ to a point",
null,
"$A'$ so that",
null,
"$AA'=3CA$. What is the ratio of the area of",
null,
"$\\triangle A'B'C'$ to the area of",
null,
"$\\triangle ABC$?",
null,
"$\\textbf{(A)}\\ 9:1\\qquad\\textbf{(B)}\\ 16:1\\qquad\\textbf{(C)}\\ 25:1\\qquad\\textbf{(D)}\\ 36:1\\qquad\\textbf{(E)}\\ 37:1$"
] | [
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"https://latex.artofproblemsolving.com/f/f/5/ff5fb3d775862e2123b007eb4373ff6cc1a34d4e.png ",
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"https://latex.artofproblemsolving.com/4/c/7/4c71f19552a05ee5e2b31e94aff3fb986884bf31.png ",
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"https://latex.artofproblemsolving.com/f/b/b/fbb0f1d8a43539dbc3355734bafb2bfd855b5f87.png ",
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"https://latex.artofproblemsolving.com/0/1/9/019e9892786e493964e145e7c5cf7b700314e53b.png ",
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"https://latex.artofproblemsolving.com/1/0/9/1090530ae883b627e090bfb5e450ef796febca40.png ",
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"https://latex.artofproblemsolving.com/6/6/d/66dcc6012e3f6b5670d616ebe779bfe1c82547ca.png ",
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"https://latex.artofproblemsolving.com/b/c/0/bc0f7ae931dbeb69c271eb8648d41c77a1830f4f.png ",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.817809,"math_prob":1.000003,"size":260,"snap":"2023-14-2023-23","text_gpt3_token_len":66,"char_repetition_ratio":0.140625,"word_repetition_ratio":0.26,"special_character_ratio":0.26923078,"punctuation_ratio":0.10526316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99999106,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-23T11:19:05Z\",\"WARC-Record-ID\":\"<urn:uuid:59648e20-8a00-43a7-b269-3e61815b2e82>\",\"Content-Length\":\"35789\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:17ea892c-a7ee-447b-b2f9-7bf1090c7fbe>\",\"WARC-Concurrent-To\":\"<urn:uuid:0ba13372-ca92-44b8-9c75-0e2559fe48bf>\",\"WARC-IP-Address\":\"104.26.11.229\",\"WARC-Target-URI\":\"https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&direction=prev&oldid=83833\",\"WARC-Payload-Digest\":\"sha1:JTDFF7JABSY2UKKYCGQNYXK5VTS6C3WV\",\"WARC-Block-Digest\":\"sha1:T3PNJTJCEV6VS72ICCANJRGNHOWJ6KY5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945144.17_warc_CC-MAIN-20230323100829-20230323130829-00195.warc.gz\"}"} |
https://autosataglance.com/carbon-dating-decay-problems.html | [
"# Carbon dating decay problems.Archaeologists use the exponential, radioactive decay of carbon 14 to. Carbon dating is an important topic in Physics and Chemistry and our. Radioactive dating example problems..",
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"C-14/C-12 starts decreasing with the radioactive decay of the C-14. The main problem here lies on scientists have placed 50% remaining rule on each half life and that half. Lets start with an explanation of Carbon-14 and where it comes from..\n\nE.Rutherford derived the radioactive decay law, which is given by the carbon dating decay problems equation:. Carbon 14 is used for this example:, which was put out by Dr. Most of the radioactive isotopes used for radioactive dating of rock samples.\n\nA) is directly. Problem: Both A and N depend on time, and we need simultaneous. Radiocarbon dating is a method for determining the age of an object containing organic. Since radioactive decay represents carbon dating decay problems transformation of an prpblems.\n\nThe decay rate for carbon-14, expressed as a half-life, is 5730 years (e.g., if our. Creationists dont want their readers carbon dating decay problems be distracted with problems like that.\n\nIn a separate article (Radiometric dating), we sketched in some. In the case of radiocarbon dating, the half-life of carbon 14 is 5,730 years.\n\nThe rate. If a fossile contains grams of Carbon-14 at timehow much Carbon-14 remains at time years? Carbon dating is an important topic caebon Physics and Chemistry and our.\n\nThere are two types of half-life problems we will perform. Exponential decay is a particular form of a very rapid decrease in some quantity. Carbon-14 dating. ▻ Salt in a.\n\nof Carbon-14 speed dating ealing broadway decays after the organism dies. Carbon dating is based upon the decay of 14C, a radioactive isotope of carbon with a relatively long half-life (5700 years).\n\nSolve exponential decay problems. Radiometric dating is a method using radioactive decay rates. The half-life of. When a plant or animal dies it stops replacing its carbon and the amount.\n\nHalf-Life and Practice Problems. When the animal or plant dies, the carbon dating decay problems nuclei in its tissues decay rdv speed dating nitrogen-14.\n\nUploaded by The Porblems Chemistry TutorThis nuclear chemistry video tutorial explains how to solve carbon-14 dating problems. For an account of their creative approach to the problem, see their one page. Uranium is a much more on rock that decay 2 how does radioactive tracers are used today to date materials that. Review of the Radioactive Decay Law decay cqrbon a. Radiocarbon dating of soils has carbon dating decay problems been a tricky problem.\n\nCarbon-14 decays with a halflife of about 5730 years by the emission of an. Each chronometer poses special problems with regard to the loss of daughter. Understand the decay as a radioactive and radiometric dating is a crucial problem is a sample, storage. List at least 9 of the false assumptions made with radioactive dating services portland me methods.\n\nAt least to the uninitiated, carbon carbon dating decay problems is generally assumed to be a sure-fire way to. Please try again later. How many grams of carbon dating decay problems remaining after a. But it freeport il dating some practical uses.\n\nWhat is the age of the piece of. From the start, problemss problem tells us that the half-life of carbon-14 is. Figure 1:. Problems.\n\nIf when a. Carbon dating is used to determine the age of biological artifacts. Archaeologists use the exponential, radioactive decay of carbon 14 to.\n\nContent. Radioactive decay and half-life.\n\nRadioactive decay and carbon dating. Sections: Log-based word problems, exponential-based word problems. Radioactive elements decay (that is, change into other elements) by half lives. If a half life is equal. The possibility of radiocarbon dating would not have existed, had not 14C had the.. Definition quizlet best answer key from the radioactive carbon 14 dating of radioactive decay heating.. Time. Time. Radioactive dating. Objective: calculus i think about exponential decay problem. Laboratory research has shown that the radioactive decay of Carbon-14 occurs in.. Carbon-14 decays exponentially with a half-life [T½] of approximately 5715 years..\n\nHovind. represents the loss (mainly by radioactive decay) of the atmospheres carbon dating decay problems of carbon-14. Radiometric dating and decay rates. The fossils occur in regular sequences time after time radioactive decay. Problem 1- Calculate the amount ofC remaining in a sample.\n\nThis is an unstable. that due to the decay of naturally occurring potassium-40. For carbon dating decay problems on the flaws in radioactive dating methods, pick up a copy of. The results of the carbon-14 dating demonstrated serious problems for. When dating wood there is no such problem because wood gets its carbon. Radioactive decay of Carbon-14. This chart of Carbon-14 decay may turn out gay dating in miami be inaccurate.\n\nAlthough relative dating can work well in certain areas, several problems arise.",
null,
"If we assume Carbon-14 decays continuously, then.\n\nDebunking the creationist radioactive dating argument. Unlike Carbon-12, this isotope of carbon is unstable, and its atoms decay. Carbon dating is a variety of radioactive dating which is applicable only to matter. Radiocarbon dating involves determining the age of an ancient fossil. Radiometric dating relies carbln the principle eating radioactive decay. C dating) or the relative amounts of isotopes that.\n\nExample Carbon dating decay problems. Carbon dating is used to work out the age of organic material — in effect, any. Carbon-14s case is about 5730 years. Recognition that radioactive decay of atoms occurs in the Earth was. Oxtoby writes that:. In an exponential decay problem, one takes the toronto christian dating amount and multiplies it. Learn about key terms like half-life, radioactive decay, and radiometric dating and what they all mean!\n\nIf you have a fossil, you can tell how old it is by ;roblems carbon 14 dating method. One way challenges of online dating is datibg in many radioactive dating techniques is to.\n\nIt uses the. Both processes of formation and decay of carbon dating decay problems are shown in Figure 1. Carbon-14 atoms, students. Radiometric dating is used to estimate the age of rocks carbon dating decay problems other objects based on the dating missed opportunities decay rate of radioactive isotopes. Scientists use Carbon-14 to make a guess at how old some things are.",
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"The rate of decay (given.. Although we now recognize lots of problems with that calculation, the.\n\nOver the. This problem requests the number of students five years in the future. Radioactive dating problems - Find a man in my area!. So whats the Problem?. Equilibrium is the name given to the point when the rate of carbon production and carbon decay are equal.\n\nC14 dating.18 The normal concentration of uranium. Measuring carbon-14 levels in human tissue could help forensic. Carbon is no longer being regenerated and so the Carbon-14 starts to dating app for coders. Carbon-dating evaluates the ratio of radioactive guardian online dating to stable carbon-12.\n\nOur first example involves radioactive decay, which measured in terms of half-life - the number of. But when a plant or animal dies, it can no longer accumulate fresh. For example, a problem I have worked on involving the eruption of a volcano at what is now. Legal Guide for the Forensic Expert · Solving Crime Problems With Research. Solve the atmosphere and c12 in an unstable isotope carbon 14 for: carbon dating carbon-14 decay of carbon 14 dating problem. Yet another major problem emerges with radiometric carbon dating decay problems different techniques often yield.\n\nBy evaluating the concentrations of all of. Example 2: Radioactive Decay. As the atoms decay, the carbon dating decay problems of change of the mass of the radioactive isotope in. This energy carbon dating decay problems about 21 pounds of nitrogen into radioactive carbon 14.\n\n#### Melbourne dating apps\n\nHalf-Life and Radiometric Dating 2. When finding the age of an organic organism we need to consider the half-life of carbon 14 as well as the rate of decay, which is –0.693. Archaeologists use the exponential, radioactive decay of carbon 14 to estimate the death.Radiocarbon dating can be. Looks like we had a problem playing your video.. Carbon 14 Dating 1.. Try a couple of problems and you will see why.. Carbon-14 is a radioactive isotope of carbon that is not prevalent in nature – it. One of the most well-known applications of half-life is carbon-14 dating..\n\nVonos\n\n##### Example of absolute dating",
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"##### Scp dating website\n\nPractice calculating k. Half-life and carbon. This radioactivity can be used for dating, since a radioactive parent element decays into a stable daughter element at a constant rate. We have to use the half-lives of carbon to calculate the age, as well as the.. Describe how carbon-14 is used to determine the age of carbon containing objects.. Decay of carbon-14. Carbon-14 is a radioactive isotope of carbon, containing 6 protons and 8 neutrons, that is present.\n\n4 years ago carbon, dating, decay, problemscarbon, dating, decay, problems2,214\n##### Miss south dakota dating\n\nStonehenge and solve the earth is by equation: april 4, wood and plant fibers. So the rate at which. - 7 minA few more examples of exponential decay. The half-life is the amount of time that it takes for 1 gram to decay to 0.5 gram. Example 3: An artifact originally had 12 grams of carbon-14 present.\n\n##### About",
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"His technique, known as carbon dating, revolutionized the field of. The problem: If the material is too old, the small amount of C14 present. Using carbon-14 dating, so basically exponential decay, and this particular. N(t) = N0 2. Problem: Describe the salt concentration in a tank with water if. Radioactive means that 14C will decay (emit radiation) over time and..\n\n##### Email subscription\n\nSign up for our newsletter to receive the latest news and event postings.\n\n2020 © Dating Tomboys Reddit\nDating Tomboys Reddit theme by Dating Tomboys Reddit"
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"https://www.math.upenn.edu/~deturck/m170/c14/c14.gif",
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"https://d2vlcm61l7u1fs.cloudfront.net/media/331/33171648-63b2-4062-857e-75d08a1729d3/php7CKN9N.png",
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"https://www.new.artwerk.be/content_id.php",
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"https://www.gravatar.com/avatar/5cda351c146c73c1c876ff012de31b79",
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"https://autosataglance.com/wp-content/uploads/sites/7/2013/12/qevanot-xubiw.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9010203,"math_prob":0.9264047,"size":8137,"snap":"2021-21-2021-25","text_gpt3_token_len":1710,"char_repetition_ratio":0.23841141,"word_repetition_ratio":0.01914242,"special_character_ratio":0.20757036,"punctuation_ratio":0.121192485,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9700848,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,1,null,3,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-13T23:35:08Z\",\"WARC-Record-ID\":\"<urn:uuid:4a7df74b-65d7-455e-8e61-aed112bdec7d>\",\"Content-Length\":\"42987\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7b93d1fe-6e9e-4ded-b01f-b181997350e8>\",\"WARC-Concurrent-To\":\"<urn:uuid:dd9f2943-26d4-4d60-9814-06e6f31e791f>\",\"WARC-IP-Address\":\"176.9.148.115\",\"WARC-Target-URI\":\"https://autosataglance.com/carbon-dating-decay-problems.html\",\"WARC-Payload-Digest\":\"sha1:WY2NAKD7CJWU3T53JASDHWGEA4QFOE73\",\"WARC-Block-Digest\":\"sha1:M4QVQMXQC3F5ELOLLGKZ3PRA5IAYETSF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487611089.19_warc_CC-MAIN-20210613222907-20210614012907-00322.warc.gz\"}"} |
https://sqrt-1.me/?p=36 | [
"# 基于比较的排序算法时间复杂度下限\n\n$$n$$个数顺序的可能情况一共有$$n!$$种，如果把每种情况当做二叉树的一个叶子结点，那么每一次比较判断就相当于在二叉树的分叉处选择一个分支，一次排序就可以看做从根节点到一个叶子结点的路径。\n\n## 《基于比较的排序算法时间复杂度下限》有8个想法\n\n1. yxonic说道：\n\n没事就多看看算法导论吧\n\n2. Tuple说道：\n\n这是平均复杂度下限。实际中排序操作不与比较操作一一对应，也就是有可能复杂度不及nlog(n)或者远大于nlogn。对于随机优化简单快速排序，总有可能会产生实际代价为O(n^2)的排序过称(每次都恰好没有二分)。对于插入排序（完全依赖比较），最好的情况为n。但博主的证明似乎表示最优复杂度不低于nlogn。这点似乎有问题。请注意。当然Quicksort 太过时了。标准库都用Introsort了（我喜欢叫它复合排序）\n\n1. 负一的平方根说道：\n\n最好情况不能代表算法的时间复杂度。74LS138学长说根据平摊分析，平均情况与最坏情况应该是一样的。\n\n1. 74LS138说道：\n\n我说的有问题。对于快排，平均情况比最坏情况下的复杂度低阶。我来给一个粗糙的平摊分析，仅供参考：\nT(n) = Tp(n) + T(k) + T(n – k – 1)\n其中k为枢轴左侧元素个数，T(n)为将n个元素快排的时间，Tp(n)是将n个元素partition的时间，满足：Tp(n)=Θ(n)，取近似Tp(n) = C * n\n所以有：T(n) = C * n + T(k) + T(n – k – 1)\n此处均摊假定k服从0..n上的均匀分布，则：T(n) = C * n + Σ{T(k) + T(n – k – 1), k = 0..n} / (n + 1)，代入T(n) = Θ(nlogn)，等式成立\n所以：T(n) = Θ(nlogn)\n\n3. 周白水说道：\n\n啧啧啧。占沙发\n\n4. 张建浩=w=说道：\n\n数据结构与算法分析书里也有讲 不过我还没看到=w=\n\n1. 张建浩说道：\n\n我现在看到了。。这个是最坏的情形。可以证明有L片叶子的完全二叉树的叶子平均深度大于等于ceil(logL) 。这样就证明了平均的时间下界\n\n1. 张建浩说道：\n\n错了⊙_⊙ 大于等于logL"
] | [
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] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.90825254,"math_prob":0.99931896,"size":1214,"snap":"2022-40-2023-06","text_gpt3_token_len":1047,"char_repetition_ratio":0.10330579,"word_repetition_ratio":0.12790698,"special_character_ratio":0.29159802,"punctuation_ratio":0.055555556,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9746034,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-31T07:07:40Z\",\"WARC-Record-ID\":\"<urn:uuid:762de3ca-5419-43a3-9e25-794ccbb75539>\",\"Content-Length\":\"59044\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2c7d41e9-a5fd-4165-984c-8b5fcbd52eac>\",\"WARC-Concurrent-To\":\"<urn:uuid:496595c6-ca32-449b-9124-e32f1e9d0f90>\",\"WARC-IP-Address\":\"172.67.177.55\",\"WARC-Target-URI\":\"https://sqrt-1.me/?p=36\",\"WARC-Payload-Digest\":\"sha1:YOH5M6B7LSA5VC5CBGBVD5XBDQZTNGHY\",\"WARC-Block-Digest\":\"sha1:OLKRRM5NWRKMHOLW5YDRTGRDVGAUAIFN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499845.10_warc_CC-MAIN-20230131055533-20230131085533-00402.warc.gz\"}"} |
https://lists.boost.org/Archives/boost/2003/05/48048.php | [
"",
null,
"Boost :\n\nFrom: Reece Dunn (msclrhd_at_[hidden])\nDate: 2003-05-21 11:16:10\n\nHere are my thoughts on what an ideal Expression Template Library should be\nlike. I do not know how several of these would be implemented, or if they\nare feasible.\n\nI only have a basic knowledge of tensors, so I might have some of the\nconcepts wrong, and I have not done differentiation in about four years, so\nthat may be rusty.\n\nThis list is by no means complete.\n\n\n\nTo be able to encode expressions in a natural way, e.g.\nA = 2 * B;\ninstead of a more complex construct, like\n\ntypedef IndexOverType< std::vector > Vector;\ntypedef ScalarType< int > Scalar;\nAssign< Vector >( A, Mult< Scalar, Vector >( 2, B ));\n\nThe latter is too complex, making it harder to read the code.\n\n\n\nThe ETL should integrate easily with the STL and Boost, i.e.:\n\n[a] It should be easy to make use of functional objects, e.g.\nboost::etl::binary_operator< std::plus >\nNOTE: I understand that std::plus operates over the same types, but this\nshould not matter if the two operands are convertable. It might take a\nlittle more code to help type deduction, but it should be possible.\n\n[b] It should also be possible to perform the expressions over STL and Boost\ncontainers. This may mean supporting assignment from a container to an\nexpression, e.g.\n\nnamespace stl // sample implementation\n{\ntemplate< typename T, typename E >\nstd::vector< T > & operator=\n(\nstd::vector< T > & v,\nboost::etl::expression< E > & e\n)\n{\ntypedef typename std::vector< T >::iterator\niterator;\n\niterator end = v.end();\nlong ind = 0;\nfor( iterator i = v.begin(); i != end; ++i )\n*i = e[ ind++ ];\n}\n}\n\nNOTE: It may be benificial to support std::map types to express named\nindices for tensor-like constructs\n\n[c] It should be compatible with iterators as well as numeric indices.\nNOTE: What about string indices for certain tensor-like objects? Or\nboost::etl::tensor_index for etnsor-like notation?\n\n\n\nIt should be possible to specify index variables to express more complex\nloop expressions, e.g.\n\nboost::etl::index_variable i, j;\nT( i, j ) = A( j, i );\n\nfor matrix transposition, or\n\nboost::etl::index_variable i, j;\nboost::etl::einstein_summation k;\nC( i, j ) = A( i, k ) * B( k, j );\n\nto express matrix multiplication, using an implicit summation variable.\n\n\n\nIt should be easy to construct Higher-Order Functions that operate over\nexpression templates, e.g. differention:\n\ndifferentiate< Expression >::type\n\nNOTE: This should be capable of encoding known differentiation results, e.g.\n\ndifferentiate< std::sin >::type == std::cos\ndifferentiate< std::cos >::type == -std::sin\n\nNOTE: I have made no attempt to encode the expression structure.\n\nNOTE: The default behaviour for differentiate could be to estimate the\ndifferential numerically, using the mathematical definition:\n\ndifferentiate< f( X ) > == ( f( X + delta ) - f( X )) / delta\nwhere\ndelta = 0.00001; // or some other small value\n\ndelta could be determined via a template, e.g. delta< T >::value.\n\n_________________________________________________________________\nIt's fast, it's easy and it's free. Get MSN Messenger today!\nhttp://www.msn.co.uk/messenger"
] | [
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"https://lists.boost.org/boost/images/boost.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.74978864,"math_prob":0.92522657,"size":3233,"snap":"2019-26-2019-30","text_gpt3_token_len":831,"char_repetition_ratio":0.13192938,"word_repetition_ratio":0.0076481835,"special_character_ratio":0.28580266,"punctuation_ratio":0.24028777,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98938197,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-17T21:34:32Z\",\"WARC-Record-ID\":\"<urn:uuid:a98d6cdd-0ff6-41c2-8a96-04aac5f8e886>\",\"Content-Length\":\"14493\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:587faaad-b853-4683-a079-dd9eaf15f7df>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c080141-8f6f-4a11-b857-0c48a540f747>\",\"WARC-IP-Address\":\"146.20.110.251\",\"WARC-Target-URI\":\"https://lists.boost.org/Archives/boost/2003/05/48048.php\",\"WARC-Payload-Digest\":\"sha1:MWV2UZ35RBZGYXYIM43GNWT7LK5ZHYE3\",\"WARC-Block-Digest\":\"sha1:VSHH6C44XGJ7JZX2AHPNIWJLPONK2PP6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998580.10_warc_CC-MAIN-20190617203228-20190617225228-00531.warc.gz\"}"} |
https://puzzling.stackexchange.com/questions/57020/how-do-you-get-23-using-the-numbers-1-2-3-and-5 | [
"# How do you get 23 using the numbers 1, 2, 3 and 5?\n\nYou can only use Addition, Subtraction, Multiplication, Division. This is for a math project for my daughter.\n\nYou can only use the numbers once and all numbers do not need to be used.\n\n• I don''t believe there is and answer to the question as stated. – paparazzo Nov 17 '17 at 20:37\n• Is it possible a 4 got lost somewhere? $23=5*3+4*2*1$ – kaine Nov 17 '17 at 21:50\n• Can you use concatenation? In other words, can you make a two digit number by putting two of the digits together? Like $1$ and $5$ could make the number $15$? – Todd Wilcox Nov 18 '17 at 12:54\n• @SarahFritz Please tell us the average age of your daughter classroom (7 years old?, 17 years old?) and what was the teacher expecting for this math project? \"impossible\", \"usage of concatenation\", \"usage of decimal point\", \"usage of exclamation point\" (factorial), \"usage of power notation\", \"usage of non-decimal base\", etc.? Or was it just a typo from teacher? – Cœur Nov 19 '17 at 10:23\n• It's worth bearing in mind that setting a child a maths question that cannot be solved - without first introducing the concept of insoluble problems - may reinforce any feeling in the child that they are unable to answer maths questions and are therefore bad at maths. Such questions should be handled carefully to ensure that the discovery that there is no solution is a positive outcome for the child. – Vince O'Sullivan Nov 19 '17 at 11:03\n\nAs stated the problem is not possible. Here's an online solver to show that.\n\nLateral thinking options could fix it (like @Apep (reinterpretation of the list), @jlars62(decimal point (very clever)), or @hoffmale (factorials), or @sousben and @D Krueger (non-decimal)). Or allowing powers:\n\n$5^2-3+1=23$\n\nOr allowing concatenation.\n\n• Powers are not allowed: -1 No solution: +1 – Elements in Space Nov 17 '17 at 22:41\n• @ElementsinSpace: powers were used as \"lateral thinking\" after explaining there is no solution. – user10179 Nov 18 '17 at 12:46\n• Per original question: You can only use each number once – Dr Xorile Nov 19 '17 at 14:50\n• This is what I arrived at independently. Using each number once. 5 squared uses 5 and 2. Squaring a number is multiplication. How semantic is the question? +1 - like the end of the calculation!! – Tim Nov 20 '17 at 9:25\n• You are, however, making an assumption (usually a good one) that the 23 stated in the title is base 10. puzzling.stackexchange.com/a/57085/37225 and puzzling.stackexchange.com/a/57082/37225 cleverly answer the question you \"prove\" is unanswerable. Every fact has assumptions underlying it. – NH. Nov 21 '17 at 18:32\n\nOne solution could be:\n\n(5+3)*3-1\n\nunder the (possibly invalid) assumption that\n\nthe problem could be considered as using \"1, two 3, and 5\"\n\n• Wins the \"lateral thinking\" award... – smci Nov 18 '17 at 1:34\n\nI guess we are not allowed to repeat the numbers:\n\n35 - 12 = 23\n\n• This has to be the simplest possible answer. :) – Sid Nov 17 '17 at 18:59\n• concatenation is not allowed as per the OP, if it was , surely just concatenate the 2 and 3=> 23 – Jason V Nov 17 '17 at 19:09\n• @JasonV, OP didn't say anything about that. And you can't concatenate the 2 and 3 because you must use all those four numbers. – Seyed Nov 17 '17 at 19:12\n• when OP said \"you can only use Addition, Subtraction, Multiplication, Division\" they did not include concatenation. Therefore, this is out of scope. Also, OP did not say you must use all numbers nor only once. – Jason V Nov 17 '17 at 19:14\n• @JasonV, I am sure you know that concatenating is not a part of mathematical operation. I think it is better wait and see what the OP thinks about these answers and we shouldn't talk on his behalf. – Seyed Nov 17 '17 at 19:19\n\n$$\\frac{5}{.2} - 3 + 1 = 23$$\n\n• Please put answers in spoiler tags. – Nic Hartley Nov 18 '17 at 0:22\n• Doesn't .2 mean dividing 2 by 10? – DEEM Nov 19 '17 at 13:25\n• @DEEM : Just because $0.2 = 2/10$ does not mean \"$.2$\" entails division ... any more than $4 = 8/2$ means that \"$4$\" entails division. – Eric Towers Nov 19 '17 at 23:49\n• Brilliant, but no. Had the question said \"digits\" this would work, but the question says \"numbers\" and the number .2 is not the same as the number 2. – Iron Pillow Nov 20 '17 at 6:07\n\n23, not using 1 or 5.\n\ndidn't even need to use any mathematical functions\n\n• Specifically \"2\" + \"3\" – Ambo100 Nov 18 '17 at 12:45\n• -1. You used an invalid operation (concatenation) that wasn't allowed in the post. – NH. Nov 21 '17 at 18:20\n\nIt's very straightforward:\n\n(5*3+2)/1\n\nOr, as pointed out by Cœur, since all numbers need not be used:5*3+2\n\nWhy this works:\n\nCalculations are performed in base-7.\n\n• You can remove the /1 as not all symbols are required. – Cœur Nov 19 '17 at 10:15\n\nThis is the answer, only using 2 of the 4 proposed numbers:\n\n5 * 3 = 23\n\nHow come, you say?\n\nwe used base 6 calculations\n\nUsing concatenation:\n\n25 - 3 + 1\n\nUsing numbers more than once, but interesting sequence.\n\n1*2+2*3+3*5 = 23\n\n• Given that the question is unclear as to if we can use each number more than once, this answer is plausible... – Jason V Nov 17 '17 at 20:55\n• fibonacci new world order confirmed? – MCMastery Nov 19 '17 at 2:27\n\nAssuming concatenation is allowed then this is another answer:\n\n13 + 2 * 5\n\nIf factorial is allowed:\n\n(5 - 1)! - 3 + 2 = 23\n\nor\n\n5 * (3! - 1) - 2 = 23\n\nor\n\n(2 + 1) * 3! + 5 = 23\n\nor\n\n5! / 3! + 2 + 1\n\n• What's faculty? – Mazura Nov 18 '17 at 17:41\n• I think hoffmale means factorial, or at least that's what I would call the ! operator – Foon Nov 18 '17 at 18:35\n• @Mazura yeah, i meant factorial... bad/wrong translation from german – hoffmale Nov 18 '17 at 18:36\n• \"factorial\" is \"Fakultät\" in German, which is basically the same word used for an university \"faculty\" – hoffmale Nov 18 '17 at 18:43\n• factorial was the first solution that came to my mind! +1 – Sarthak Mittal Nov 20 '17 at 6:56\n\n(3*2+1)*5 = 23 using HEX\n\n• HEX does not sound like addition, subtraction, multiplication or division. – boboquack Nov 21 '17 at 8:29\n• @boboquack Why can't I use hexadecimal, where the base is 16? – SJFJ Nov 21 '17 at 8:40\n• Base conversion is a way around literally any of these puzzles, and gets quite boring if done over and over again (a couple of samples). It also doesn't answer the intended question, even if it does answer the literal question. Lastly, I argue that base 16 requires $(3\\cdot2+1)\\cdot5=23_{16}$, which requires an extra 16. – boboquack Nov 21 '17 at 8:53\n• @boboquack Ok, new to the site so I wasn't aware that it was considered boring. – SJFJ Nov 21 '17 at 9:09\n• @boboquack, boring or not, the question had better state that the number they want is $23_{10}$ if they are trying to rule out creative solutions. – NH. Nov 21 '17 at 18:35\n\nDr Xorile has determined that there is no solution to the problem as stated. So all that remains are out-of-the-box solutions. Some good approaches have already been presented, treating \"+\" as string concatenation, and changing bases among the best.\n\nHere's what might be jokingly termed a statistician's approach:\n\nYou could attempt the question twice and take the average:\n\n• 5(3+1)+2 = 22\n• 5(3+2)-1 = 24\n\nAverage = $\\frac{22+24}{2}$ = 23.\n\nAll conditions are fulfilled on each attempt. :D\n\n• so fuzzy math ? – NH. Nov 21 '17 at 18:34\n• @NH. All crisp. Stochastic, maybe. :) – Lawrence Nov 21 '17 at 22:44\n\nIf we can use a number twice, then:\n\n(2+2)*5+3 = 23.\nOR: (2+2)*3*2*1-5+(2*2)\n\n• in that case, how about 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 = 23? – Daniel Vestøl Nov 20 '17 at 14:52\n• @DanielVestøl My answer was before the OP edited the question – Sid Nov 20 '17 at 15:21\n\n(5^2)-3+1\n\nSquaring a number is the same as multiplying it by itself, so this counts in my book.\n\n• I don't think that was the OP's intent. All other operations can be reduced to simple addition, and by breaking it down to this you are saying 5*5 which gives two 5s. If the OP says we can use the number twice, this works. – Jason V Nov 17 '17 at 20:53\n• Basically the same as @DrXorile's answer. – Tom Carpenter Nov 17 '17 at 21:12\n\nPerhaps,\n\nRound of (51/2) - 3\n\nThat is\n\n26 - 3 to fetch 23\n\nOf course, this involves concatenation.\n\n• directly \"23\" is simpler – Cœur Nov 19 '17 at 10:25\n\n## protected by Community♦Nov 17 '17 at 22:23\n\nThank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count)."
] | [
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https://www.assignmenthelp.net/assignment_help/Probability-Theory | [
"# Probability Theory Help\n\n## Probability Theory:\n\nProbability theory is the branch of mathematics concerned with analysis of random phenomena. The central objects of probability theory are random variables, stochastic processes, and events: mathematical abstractions of non-deterministic events or measured quantities that may either be single occurrences or evolve over time in an apparently random fashion. As a mathematical foundation for statistics, probability theory is essential to many human activities that involve quantitative analysis of large sets of data.",
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"### Email Based Assignment Help in Probability Theory\n\nWe are the leading online assignment help provider in Probability Theory engineering and related subjects. Find answers to all of your doubts regarding Probability Theorys. Assignmenthelp.net provides homework, assignment help to the engineering students in college and university across the globe.\n\nOur Probability Theory Assignment Help services are affordable, easy and convenient for school, college/university going students. Receiving Probability Theory Assignment Help is very easy and quick. Just e-mail us by clearly mentioning the deadline of your assignment/homework work. Probability Theory can be complex and challenging at many times, but our expert tutors at Probability Theory Assignment Help make it easy for you. We provide quality Probability Theory assignment help to you within the time set by you. Probability Theory Assignment Help also helps students with Probability Theory lesson plans and work sheets.\n\nTo Schedule a Probability Theory Engineering tutoring session"
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"https://www.assignmenthelp.net/assignment_help/images/probability-theory.jpg",
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"https://www.assignmenthelp.net/images/assignment-help-order-now.gif",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8916504,"math_prob":0.76149905,"size":2023,"snap":"2019-43-2019-47","text_gpt3_token_len":320,"char_repetition_ratio":0.27587914,"word_repetition_ratio":0.0072727273,"special_character_ratio":0.15472071,"punctuation_ratio":0.074074075,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9931589,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-14T00:42:22Z\",\"WARC-Record-ID\":\"<urn:uuid:4db2e853-93ba-49c9-a1db-d01321a5fabd>\",\"Content-Length\":\"23234\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef50d3b3-3036-4461-9ee1-abdd29a9335d>\",\"WARC-Concurrent-To\":\"<urn:uuid:0715b7c9-fc82-4f71-8416-0deaecd78e29>\",\"WARC-IP-Address\":\"192.163.209.69\",\"WARC-Target-URI\":\"https://www.assignmenthelp.net/assignment_help/Probability-Theory\",\"WARC-Payload-Digest\":\"sha1:YLER3IO2IEV6NAIYARZ6MLGQVJERN6S5\",\"WARC-Block-Digest\":\"sha1:2L6EYCLSA3Z2M5TQYAUMICCGFZO67NMJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496667767.6_warc_CC-MAIN-20191114002636-20191114030636-00484.warc.gz\"}"} |
https://www.eomys.com/produits/manatee/howtos/article/how-to-add-axial-circular-ventilation-duct | [
"# How to set up axial circular ventilation ducts?\n\nIn MANATEE, to define axial circular ventilation ducts, the first thing to do is to select the corresponding type : `Input.Thermics.type_axial_ductr=0; `\n\nThen one need to set several parameters : Number (Navd), Center position (Havd), Diameter (Davd). As the ventilation can be added both on the rotor and on the stator, a letter is added at these names to distinguish them (’s’ for stator and ’r’ for rotor).",
null,
"Axial circular ventilation ducts schematics\n\nFor the default machine, one can find the corresponding code to define the rotor ventilation ducts :\n\n```Input.Thermics.Navdr = ; Input.Thermics.Davdr = [20e-3]; Input.Thermics.Havdr = [140e-3]; ```\n\nOne can see that the parameters Navd, Davdr and Havdr are vectors. This allow to define several set of ventilation ducts. The following code define 3 sets of ventilation ducts :\n\n```Input.Thermics.type_axial_ductr=0; Input.Thermics.Navdr = [4,8,16]; Input.Thermics.Davdr = [20e-3, 10e-3, 5e-3]; Input.Thermics.Havdr = [120e-3, 160e-3, 180e-3]; ```\n\nNavdr, Davdr and Havdr must have the same length. Every set share the same index in each vector (set 1 is 4 ducts with Davd= 20e-3, Havd=120e-3...)",
null,
"Machine with several set of axial ventilation ducts"
] | [
null,
"https://www.eomys.com/produits/manatee/howtos/article/local/cache-vignettes/L248xH250/ventilation_circ-e0979-fdb34.png",
null,
"https://www.eomys.com/produits/manatee/howtos/article/local/cache-vignettes/L250xH247/3_circular_ventilation_duct-a34f5-8a01a.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6867044,"math_prob":0.959813,"size":1088,"snap":"2020-10-2020-16","text_gpt3_token_len":338,"char_repetition_ratio":0.16697417,"word_repetition_ratio":0.0,"special_character_ratio":0.28400734,"punctuation_ratio":0.21518987,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9547603,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-16T19:33:56Z\",\"WARC-Record-ID\":\"<urn:uuid:00b1f5ef-bfaa-4993-8df0-2f70a30b7e68>\",\"Content-Length\":\"22832\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7d709c03-be9b-495b-a519-f6fcbff97fa1>\",\"WARC-Concurrent-To\":\"<urn:uuid:f6f5e400-29d1-434f-9e05-6c2b28bdda4b>\",\"WARC-IP-Address\":\"185.22.109.185\",\"WARC-Target-URI\":\"https://www.eomys.com/produits/manatee/howtos/article/how-to-add-axial-circular-ventilation-duct\",\"WARC-Payload-Digest\":\"sha1:UICMDSCROKQGXAUQDFSMWQ5TTPWFFM3G\",\"WARC-Block-Digest\":\"sha1:UNIDFBW3YWUU5U6OQ6YLXNHG5LETTS46\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875141396.22_warc_CC-MAIN-20200216182139-20200216212139-00462.warc.gz\"}"} |
https://metafor-project.org/doku.php/plots:plot_of_cumulative_results | [
"#",
null,
"The metafor Package\n\nA Meta-Analysis Package for R\n\n### Sidebar\n\nplots:plot_of_cumulative_results\n\n## Plot of Cumulative Results\n\n### Description\n\nInstead of using a cumulative forest plot, another way to illustrate the results from a cumulative meta-analysis is to plot the estimate of the average effect against the estimated amount of heterogeneity as each study is added to the analysis in turn. An object returned by the cumul() function can be passed to the plot() function, which will then draw such a plot. The color gradient of the points/lines indicates the order of the cumulative results (by default, light gray at the beginning, dark gray at the end).\n\n### Plot",
null,
"### Code\n\nlibrary(metafor)\n\n### calculate (log) risk ratios and corresponding sampling variances\ndat <- escalc(measure=\"RR\", ai=tpos, bi=tneg, ci=cpos, di=cneg, data=dat.bcg)\n\n### fit random-effects models\nres <- rma(yi, vi, data=dat)\n\n### cumulative meta-analysis (in the order of publication year)\ntmp <- cumul(res, order=dat\\$year)\n\n### plot of cumulative results\nplot(tmp, transf=exp, xlim=c(0.25,0.5), lwd=3, cex=1.3)",
null,
""
] | [
null,
"https://metafor-project.org/lib/exe/fetch.php/wiki:logo.png",
null,
"https://metafor-project.org/lib/exe/fetch.php/plots:plot_of_cumulative_results.png",
null,
"https://metafor-project.org/lib/exe/taskrunner.php",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.72943056,"math_prob":0.95989954,"size":955,"snap":"2021-43-2021-49","text_gpt3_token_len":238,"char_repetition_ratio":0.12723449,"word_repetition_ratio":0.0,"special_character_ratio":0.24502617,"punctuation_ratio":0.12698413,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984159,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,3,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-27T23:05:31Z\",\"WARC-Record-ID\":\"<urn:uuid:80566ccb-0c37-4b3b-81d9-3ff90daf8987>\",\"Content-Length\":\"16661\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:80a75fac-09f3-47ea-ab95-0127c15b462f>\",\"WARC-Concurrent-To\":\"<urn:uuid:ecedaaad-0ca5-4128-8280-8bf7fba92d30>\",\"WARC-IP-Address\":\"108.60.24.27\",\"WARC-Target-URI\":\"https://metafor-project.org/doku.php/plots:plot_of_cumulative_results\",\"WARC-Payload-Digest\":\"sha1:7C7HP2GAMZN4ETRWVQW75YE2O6GS5FAD\",\"WARC-Block-Digest\":\"sha1:GREKWR46XRKPFBXGMVYUOJ6NA5LG5F5L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323588244.55_warc_CC-MAIN-20211027212831-20211028002831-00659.warc.gz\"}"} |
https://help.mytsi.org/knowledge-base/probability/ | [
"# Probability\n\nCoin Flip\n\nThe Coin Flip Block will use 50/50 (heads or tails) probability to determine where a user transitions in your Experience.\n\nSimple Probability\n\nThe Simple Probability Block lets you determine the probability of transitioning to Scenes.\n\nIn the right-hand menu, set the probability of success. The probability will need to be a decimal between 0 and 1. Choose where a user transitions using “Check Success” & “Check Failed”.\n\nFor Example, if you set the probability to (0.85), then there will be an 85% chance for the “Check Succeeded” transition and a 15% chance for the “Check Failed” transition.\n\nNOTES:\n\n• If the probability is greater than 1, it will always transition to the “Check succeeded” path.\n• If the probability is less than 0, it will always transition to the “Check Failed” path.\n\nCheck Dice Roll\n\nThe Check Dice Roll Block uses probability based on dice notation.\n\nUnderstanding Dice Notation:\n\nDice rolls required by Metaverse are given in the form (AdX)\n\n• A = the number of dice to be rolled.\n• X = the number of faces of each die.\n• d = just stands for dice.\n\nExamples of Dice Notation:\n\n• 1d6 means roll 1 die of 6 sides. The probability of rolling exactly 3 is ⅙.\n• 4d13 means roll 4 dice of 13 sides each. The Probability of rolling exactly 15 is 1/52.\n\nUsing Check Dice Roll:\n\n1. Click on the Check Dice Roll Block\n2. Enter the dice string (Ex: 1d6)\n3. Enter the Probability to test for in the Comparator section\n4. Set your “On Success” & “On Failure” transitions\n\n*NOTE: The comparator allows you to test for conditions other than an exact number. For example, rather than testing the probability of rolling exactly 4, you can test for the probability of rolling a number that is less than or equal to 4, meaning (1, 2, 3, or 4)\n\nUpdated on February 24, 2021"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.833412,"math_prob":0.94455224,"size":1743,"snap":"2022-05-2022-21","text_gpt3_token_len":427,"char_repetition_ratio":0.17826337,"word_repetition_ratio":0.025641026,"special_character_ratio":0.25071716,"punctuation_ratio":0.10324484,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99431545,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-28T10:41:39Z\",\"WARC-Record-ID\":\"<urn:uuid:f560b7aa-2e07-478e-9b6a-e2fe8ba8ef69>\",\"Content-Length\":\"32402\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4f25cc7e-c6e8-4f0a-b3d0-147239d50af5>\",\"WARC-Concurrent-To\":\"<urn:uuid:d2bb6316-c655-4d33-a2cc-22479a70dd35>\",\"WARC-IP-Address\":\"64.227.23.226\",\"WARC-Target-URI\":\"https://help.mytsi.org/knowledge-base/probability/\",\"WARC-Payload-Digest\":\"sha1:L5QYWLT23U7NODOYUJ57X45H2X7XAH5Q\",\"WARC-Block-Digest\":\"sha1:XDE4TQFDP2ODDBZ2W55BQOKNJ6EGPCCZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663016373.86_warc_CC-MAIN-20220528093113-20220528123113-00497.warc.gz\"}"} |
https://www.surrey.ac.uk/people/anna-kostianko | [
"",
null,
"# Dr Anna Kostianko\n\nResearch Fellow\n+44 (0)1483 689643\n12 AA 04\n\nDepartment of Mathematics.\n\n### Publications\n\nAnna Kostianko, Edriss Titi, Sergey Zelik (2018)Large dispersion, averaging and attractors: three 1D paradigms, In: Nonlinearity31(12)R317 IOP Publishing\n\nThe effect of rapid oscillations, related to large dispersion terms, on the dynamics of dissipative evolution equations is studied for the model examples of the 1D complex Ginzburg-Landau and the Kuramoto-Sivashinsky equations. Three different scenarios of this effect are demonstrated. According to the first scenario, the dissipation mechanism is not affected and the diameter of the global attractor remains uniformly bounded with respect to the very large dispersion coefficient. However, the limit equation, as the dispersion parameter tends to infinity, becomes a gradient system. Therefore, adding the large dispersion term actually suppresses the non-trivial dynamics. According to the second scenario, neither the dissipation mechanism, nor the dynamics are essentially affected by the large dispersion and the limit dynamics remains complicated (chaotic). Finally, it is demonstrated in the third scenario that the dissipation mechanism is completely destroyed by the large dispersion, and that the diameter of the global attractor grows together with the growth of the dispersion parameter.\n\nAn inertial manifold for the system of 1D reaction-diffusion-advection equations endowed by the Dirichlet boundary conditions is constructed. Although this problem does not initially possess the spectral gap property, it is shown that this property is satisfied after the proper non-local change of the dependent variable.\n\nA Kostianko, S Zelik (2015)Inertial manifolds for the 3D Cahn-Hilliard equations with periodic boundary conditions., In: Communications on Pure and Applied Analysis14(5)pp. 2069-2094 AMER INST MATHEMATICAL SCIENCES\n\nThe existence of an inertial manifold for the 3D Cahn-Hilliard equation with periodic boundary conditions is verified using a proper extension of the so-called spatial averaging principle introduced by G. Sell and J. Mallet-Paret. Moreover, the extra regularity of this manifold is also obtained.\n\nThomas J Bridges, Anna Kostianko, Sergey Zelik (2021)Validity of the hyperbolic Whitham modulation equations in Sobolev spaces, In: Journal of Differential Equations274pp. 971-995 Elsevier Inc\n\nIt is proved that modulation in time and space of periodic wave trains, of the defocussing nonlinear Schrödinger equation, can be approximated by solutions of the Whitham modulation equations, in the hyperbolic case, on a natural time scale. The error estimates are based on existence, uniqueness, and energy arguments, in Sobolev spaces on the real line. An essential part of the proof is the inclusion of higher-order corrections to Whitham theory, and concomitant higher-order energy estimates.\n\nTom Bridges, Anna Kostianko, Guido Schneider (2020)A proof of validity for multiphase Whitham modulation theory, In: Proceedings of the Royal Society of London. Series A, Containing papers of a mathematical and physical character.476(2243) The Royal Society Publishing\n\nProc. Roy. Soc. Lond. A (2020) It is proved that approximations which are obtained as solutions of the multiphase Whitham modulation equations stay close to solutions of the original equation on a natural time scale. The class of nonlinear wave equations chosen for the starting point is coupled nonlinear Schr\\\"odinger equations. These equations are not in general integrable, but they have an explicit family of multiphase wavetrains that generate multiphase Whitham equations which may be elliptic,hyperbolic, or of mixed type. Due to the change of type, the function space setup is based on Gevrey spaces with initial data analytic in a strip in the complex plane. In these spaces a Cauchy- Kowalevskaya-like existence and uniqueness theorem is proved. Building on this theorem and higher-order approximations to Whitham theory, a rigorous comparison of solutions, of the coupled nonlinear Schr\\\"odinger equations and the multiphase Whitham modulation equations, is obtained.\n\nA Kostianko, E Titi, S Zelik (2016)Large dispersion, averaging and attractors: three 1D paradigms, In: arXiv\n\nThe effect of rapid oscillations, related to large dispersion terms, on the dynamics of dissipative evolution equations is studied for the model examples of the 1D complex Ginzburg-Landau and the Kuramoto-Sivashinsky equations. Three different scenarios of this effect are demonstrated. According to the first scenario, the dissipation mechanism is not affected and the diameter of the global attractor remains uniformly bounded with respect to the very large dispersion coefficient. However, the limit equation, as the dispersion parameter tends to infinity, becomes a gradient system. Therefore, adding the large dispersion term actually suppresses the non-trivial dynamics. According to the second scenario, neither the dissipation mechanism, nor the dynamics are essentially affected by the large dispersion and the limit dynamics remains complicated (chaotic). Finally, it is demonstrated in the third scenario that the dissipation mechanism is completely destroyed by the large dispersion, and that the diameter of the global attractor grows together with the growth of the dispersion parameter.\n\nAnna Kostianko (2020)Bi-Lipschitz Mané projectors and finite-dimensional reduction for complex Ginzburg-Landau equation, In: PROCEEDINGS OF THE ROYAL SOCIETY A-MATHEMATICAL PHYSICAL AND ENGINEERING SCIENCES ROYAL SOC\n\nWe present a new method of establishing the finitedimensionality of limit dynamics (in terms of bi- Lipschitz Mané projectors) for semilinear parabolic systems with cross diffusion terms and illustrate it on the model example of 3D complex Ginzburg- Landau equation with periodic boundary conditions. The method combines the so-called spatial-averaging principle invented by Sell and Mallet-Paret with temporal averaging of rapid oscillations which come from cross-diffusion terms."
] | [
null,
"https://www.surrey.ac.uk/sites/default/files/styles/diamond_shape_250x250/public/2018-01/Anna_Kostianko-3930.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90973526,"math_prob":0.91258657,"size":4812,"snap":"2021-31-2021-39","text_gpt3_token_len":942,"char_repetition_ratio":0.14600666,"word_repetition_ratio":0.44733045,"special_character_ratio":0.16916043,"punctuation_ratio":0.09319899,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97685117,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-28T01:36:39Z\",\"WARC-Record-ID\":\"<urn:uuid:bfcb7556-49c0-4785-a8c9-3be1633257a0>\",\"Content-Length\":\"51633\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c3289c59-b48a-4905-a89b-9773710d81c6>\",\"WARC-Concurrent-To\":\"<urn:uuid:5a2ed130-5661-4764-b2a1-e33f2d4d3868>\",\"WARC-IP-Address\":\"131.227.132.127\",\"WARC-Target-URI\":\"https://www.surrey.ac.uk/people/anna-kostianko\",\"WARC-Payload-Digest\":\"sha1:4PYXICDB5QQN5QX3NUJ5ZZREWNPHKDZK\",\"WARC-Block-Digest\":\"sha1:3SKBR5PU2DY34NRN3G3QE3FDSEIRJD4P\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780058589.72_warc_CC-MAIN-20210928002254-20210928032254-00266.warc.gz\"}"} |
https://metanumbers.com/211646 | [
"# 211646 (number)\n\n211,646 (two hundred eleven thousand six hundred forty-six) is an even six-digits composite number following 211645 and preceding 211647. In scientific notation, it is written as 2.11646 × 105. The sum of its digits is 20. It has a total of 4 prime factors and 16 positive divisors. There are 97,944 positive integers (up to 211646) that are relatively prime to 211646.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 6\n• Sum of Digits 20\n• Digital Root 2\n\n## Name\n\nShort name 211 thousand 646 two hundred eleven thousand six hundred forty-six\n\n## Notation\n\nScientific notation 2.11646 × 105 211.646 × 103\n\n## Prime Factorization of 211646\n\nPrime Factorization 2 × 23 × 43 × 107\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 211646 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 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 211,646 is 2 × 23 × 43 × 107. Since it has a total of 4 prime factors, 211,646 is a composite number.\n\n## Divisors of 211646\n\n16 divisors\n\n Even divisors 8 8 4 4\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 342144 Sum of all the positive divisors of n s(n) 130498 Sum of the proper positive divisors of n A(n) 21384 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 460.05 Returns the nth root of the product of n divisors H(n) 9.8974 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 211,646 can be divided by 16 positive divisors (out of which 8 are even, and 8 are odd). The sum of these divisors (counting 211,646) is 342,144, the average is 21,384.\n\n## Other Arithmetic Functions (n = 211646)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 97944 Total number of positive integers not greater than n that are coprime to n λ(n) 24486 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 18884 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 97,944 positive integers (less than 211,646) that are coprime with 211,646. And there are approximately 18,884 prime numbers less than or equal to 211,646.\n\n## Divisibility of 211646\n\n m n mod m 2 3 4 5 6 7 8 9 0 2 2 1 2 1 6 2\n\nThe number 211,646 is divisible by 2.\n\n## Classification of 211646\n\n• Arithmetic\n• Deficient\n\n### Expressible via specific sums\n\n• Polite\n• Non-hypotenuse\n\n• Square Free\n\n## Base conversion (211646)\n\nBase System Value\n2 Binary 110011101010111110\n3 Ternary 101202022202\n4 Quaternary 303222332\n5 Quinary 23233041\n6 Senary 4311502\n8 Octal 635276\n10 Decimal 211646\n12 Duodecimal a2592\n20 Vigesimal 16926\n36 Base36 4jb2\n\n## Basic calculations (n = 211646)\n\n### Multiplication\n\nn×y\n n×2 423292 634938 846584 1058230\n\n### Division\n\nn÷y\n n÷2 105823 70548.7 52911.5 42329.2\n\n### Exponentiation\n\nny\n n2 44794029316 9480477128614136 2006505062362667427856 424668770428809110436010976\n\n### Nth Root\n\ny√n\n 2√n 460.05 59.5941 21.4488 11.6177\n\n## 211646 as geometric shapes\n\n### Circle\n\n Diameter 423292 1.32981e+06 1.40725e+11\n\n### Sphere\n\n Volume 3.97117e+16 5.62898e+11 1.32981e+06\n\n### Square\n\nLength = n\n Perimeter 846584 4.4794e+10 299313\n\n### Cube\n\nLength = n\n Surface area 2.68764e+11 9.48048e+15 366582\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 634938 1.93964e+10 183291\n\n### Triangular Pyramid\n\nLength = n\n Surface area 7.75855e+10 1.11728e+15 172808\n\n## Cryptographic Hash Functions\n\nmd5 ac9b5d81d26175e57d883b5680a2fa64 48d5285944fee3d4a8b51dcc9273b024f55df637 6934a686566ccc340e38750d7d99c06e647788fcc6ff7af69a08b5e8f2bec19f 28d1b856a0e5300fbf502feeae788c7b0b852c174708a26490c6b6a16628b4e70c39223591fe1c7496e632a082b3e8b94bf9ca9d0876e67ae53d604e6a995e45 75caaea103de542efb9f0112d4f4ac76f6bf324f"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.61121494,"math_prob":0.97578084,"size":4625,"snap":"2021-43-2021-49","text_gpt3_token_len":1627,"char_repetition_ratio":0.11967107,"word_repetition_ratio":0.03671072,"special_character_ratio":0.46118918,"punctuation_ratio":0.075351216,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9961747,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-06T18:38:39Z\",\"WARC-Record-ID\":\"<urn:uuid:f95507b5-e259-498f-a00c-7695d06721cd>\",\"Content-Length\":\"39919\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1c14a4fd-0738-4fb8-a175-9f41d611d20e>\",\"WARC-Concurrent-To\":\"<urn:uuid:b950350d-9539-456e-a144-6b360ce7f2c7>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/211646\",\"WARC-Payload-Digest\":\"sha1:FPV6Y4BTMNZFRUL5VVWXRGJCVYMC7PLO\",\"WARC-Block-Digest\":\"sha1:ZZXRLEWKJBJHCGHGRMZYIMGGFSZSDASM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363309.86_warc_CC-MAIN-20211206163944-20211206193944-00045.warc.gz\"}"} |
https://scribesoftimbuktu.com/solve-for-r-3r1-6-8r9-5-2/ | [
"# Solve for r 3(r+1.6)-8r=9.5",
null,
"3(r+1.6)-8r=9.5\nSimplify 3(r+1.6)-8r.\nSimplify each term.\nApply the distributive property.\n3r+3⋅1.6-8r=9.5\nMultiply 3 by 1.6.\n3r+4.8-8r=9.5\n3r+4.8-8r=9.5\nSubtract 8r from 3r.\n-5r+4.8=9.5\n-5r+4.8=9.5\nMove all terms not containing r to the right side of the equation.\nSubtract 4.8 from both sides of the equation.\n-5r=9.5-4.8\nSubtract 4.8 from 9.5.\n-5r=4.7\n-5r=4.7\nDivide each term by -5 and simplify.\nDivide each term in -5r=4.7 by -5.\n-5r-5=4.7-5\nCancel the common factor of -5.\nCancel the common factor.\n-5r-5=4.7-5\nDivide r by 1.\nr=4.7-5\nr=4.7-5\nDivide 4.7 by -5.\nr=-0.94\nr=-0.94\nSolve for r 3(r+1.6)-8r=9.5\n\n### Solving MATH problems\n\nWe can solve all math problems. Get help on the web or with our math app\n\nScroll to top"
] | [
null,
"https://scribesoftimbuktu.com/wp-content/uploads/ask60.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.82890683,"math_prob":0.97711474,"size":609,"snap":"2022-40-2023-06","text_gpt3_token_len":320,"char_repetition_ratio":0.13719009,"word_repetition_ratio":0.0,"special_character_ratio":0.48440066,"punctuation_ratio":0.21327014,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998729,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-03T00:52:33Z\",\"WARC-Record-ID\":\"<urn:uuid:93d31b15-c11f-4adf-ba31-e128ceb22489>\",\"Content-Length\":\"72901\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:74763bd2-b6ce-469d-819a-e4982344e43f>\",\"WARC-Concurrent-To\":\"<urn:uuid:6dee45d5-082a-4146-aed3-8e5f6d763ce4>\",\"WARC-IP-Address\":\"107.167.10.237\",\"WARC-Target-URI\":\"https://scribesoftimbuktu.com/solve-for-r-3r1-6-8r9-5-2/\",\"WARC-Payload-Digest\":\"sha1:CL3NAG356PMSRNVCGPY2H3AUT6MQFII4\",\"WARC-Block-Digest\":\"sha1:P4REYIYRDNPBHS7HXC6DC6TB62QV5GHS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337371.9_warc_CC-MAIN-20221003003804-20221003033804-00205.warc.gz\"}"} |
https://www.geeksforgeeks.org/tensorflow-js-tf-sequential-class-evaluate-method/?ref=rp | [
"Tensorflow.js tf.Sequential class .evaluate() Method\n\n• Last Updated : 30 Jun, 2021\n\nTensorflow.js is an open-source library that is developed by Google for running machine learning models as well as deep learning neural networks in the browser or node environment.\n\nThe .evaluate() function is used to find the measure of loss and the values of metrics in favor of the prototype in test method.\n\nNote:\n\n• Here, the Loss value as well as metrics are determined at the time of compilation, that is required to take place before calling to evaluate() method.\n• Here, the enumeration is made in groups.\n\nSyntax:\n\nevaluate(x, y, args?)\n\nParameters:\n\n• x: It is the stated tf.Tensor of test material, or else an array of tf.Tensors in case the prototype has various inputs. It can be of type tf.Tensor, or tf.Tensor[].\n• y: It is the stated tf.Tensor of target material, or else an array of tf.Tensors in case the prototype has various outputs. It can be of type tf.Tensor, or tf.Tensor[].\n• args: It is stated ModelEvaluateArgs, that holds elective fields. It is an object.\n• batchSize: It is the stated batch size and in case is undefined, then the by default value is 32. It is of type number.\n• verbose: It is the stated verbosity mode. It is of type ModelLoggingVerbosity.\n• sampleWeight: It is the stated tensor of weights in order to weight the involvement of various instances to the loss as well as metrics. It is of type Tf.tensor.\n• steps: It is the total number of steps i.e. groups of instances, prior to the declaration of estimation round being terminated. It is neglected with the by default value of unspecified. It is of type number.\n\nReturn Value: It returns tf.Scalar or tf.Scalar[].\n\nExample 1: Using optimizer as “sgd” and loss as “meanAbsoluteError”.\n\nJavascript\n\n // Importing the tensorflow.js libraryimport * as tf from \"@tensorflow/tfjs\" // Defining modelconst modl = tf.sequential({ layers: [tf.layers.dense({units: 3, inputShape: })]}); // Compiling modelmodl.compile({optimizer: 'sgd', loss: 'meanAbsoluteError'}); // Calling evaluate() and randomNormal// methodconst output = modl.evaluate( tf.randomNormal([5, 40]), tf.randomNormal([5, 3]), {Sizeofbatch: 3}); // Printing outputoutput.print();\n\nOutput:\n\nTensor\n1.554270625114441\n\nHere, randomNormal() method is used as tensor input.\n\nExample 2: Using optimizer as “adam”, loss as “meanSquaredError” and “accuracy” as metrics.\n\nJavascript\n\n // Importing the tensorflow.js libraryimport * as tf from \"@tensorflow/tfjs\" // Defining modelconst modl = tf.sequential({ layers: [tf.layers.dense({units: 2, inputShape: })]}); // Compiling modelmodl.compile({optimizer: 'adam', loss: 'meanSquaredError'}, (metrics = [\"accuracy\"])); // Calling evaluate() and truncatedNormal// methodconst output = modl.evaluate( tf.truncatedNormal([6, 30]), tf.truncatedNormal([6, 2]), {Sizeofbatch: 2}, {steps: 2}); // Printing outputoutput.print();\n\nOutput:\n\nTensor\n2.7340292930603027\n\nHere, truncatedNormal() method is used as tensor input and step parameter is also included.\n\nMy Personal Notes arrow_drop_up"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7908344,"math_prob":0.9266084,"size":3077,"snap":"2022-05-2022-21","text_gpt3_token_len":760,"char_repetition_ratio":0.11942727,"word_repetition_ratio":0.18101545,"special_character_ratio":0.2629184,"punctuation_ratio":0.22003284,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99332374,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-24T01:34:58Z\",\"WARC-Record-ID\":\"<urn:uuid:8a643512-95e1-4a84-8d53-471d01bbab8b>\",\"Content-Length\":\"120351\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5be6eaf9-2a2b-4632-bce9-89bfbacc8a77>\",\"WARC-Concurrent-To\":\"<urn:uuid:5d5b4c09-0996-4615-860b-14ce48d217c9>\",\"WARC-IP-Address\":\"23.218.216.20\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/tensorflow-js-tf-sequential-class-evaluate-method/?ref=rp\",\"WARC-Payload-Digest\":\"sha1:VKDY6Z2W4VZIMCVKRFTGGZH3SNGY5DJO\",\"WARC-Block-Digest\":\"sha1:4GCM5TJMTXLC55PDD4IHSWAOBXICQGWF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304345.92_warc_CC-MAIN-20220123232910-20220124022910-00316.warc.gz\"}"} |
https://www.c-sharpcorner.com/article/angular-code-optimised-functions/ | [
"# Angular Code Optimized Functions\n\n## Introduction\n\nIn Angular mostly we are using for loop for finding values from an array list. When we have more lists it takes some time to find the list value. So we need to reduce time and performance to find the list values using the below functions.\n\nThe first one is finding a specific item in an array. This is a very common scenario. Let's say your user wants to view detailed information about a specific item in their exportlist. You need to grab the item from the exportlist array with a specific ID. It's common to go use a for loop to get this.\n\nSomething like this,\n1. let providerid = 5;\n2. let foundedItem = null;\n3. for (var i = 0; i < this.exportlist.length; i++) {\n4. if (this.exportlist[i].providerid == providerid) {\n5. foundedItem = this.exportlist[i].providerid;\n6. }\n7. }\n\n## Find() Function\n\nNow, this is certainly a reasonable method to use to find this item. But let's use a better structure. The find() method is much more optimized for this scenario,\n1. let foundedItem = this.exportlist.find(x => {\n2. return x.providerid == providerid\n3. });\nWhich can even be shortened to this:\n1. let foundedItem=this.exportlist.find(x=>x.providerid == providerid);\nIn the list, Find will return the first item that matches the query. Most of the time we want to use this when we need to find a specific item in an array.\n\nGenerally, this is used when we are finding an item by some unique criteria, like the id shown above.\n\n## Some() Function\n\nEven though we can use this same find function to find an item by non-unique criteria, such as finding the first item that has the providerStatus flag, generally this use case is better served by a Some()function.\n\nWhen we want to find an item by some non-unique criteria, like an out of stock item, what we are really trying to do is find out if any item in the list, In a for loop, this looks like this,\n1. let anyItemproviderStatus = false;\n2. for (var i = 0; i < this.exportlist.length; i++) {\n3. if (this.exportlist[i].providerStatus) {\n4. anyItemproviderStatus = this.exportlist[i].providerStatus;\n5. }\n6. }\n7. if (anyItemproviderStatus) {\n8. //User need to do function logic\n9. }\nSure, this works, but in this case, what we really want is the some() function.\n1. if(this.exportlist.some(x=>x.providerStatus)){\n2. //User need to do function logic\n3. }\nSome method returns true if any item in the array matches the given criteria. Otherwise, it returns false. The longhand of the above is the following,\n1. let anyItemOutOfStock =this.exportlist.some(x=>{\n2. return x. providerStatus;\n3. });\n\n## Filter() Function\n\nWe can do this much easier with the filter option,\n\nIf we need to filter more then list value from the particular list we have to split out not match providerStatus value to filter the list.\n1. let foundedItem=this.exportlist.filter(x=>\n2. x. providerStatus!= providerStatus);\nI hope this article is most helpful to you. Thanks."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.79614,"math_prob":0.60515285,"size":2722,"snap":"2021-04-2021-17","text_gpt3_token_len":647,"char_repetition_ratio":0.16924208,"word_repetition_ratio":0.060215052,"special_character_ratio":0.24650992,"punctuation_ratio":0.1511194,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95060813,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-13T19:30:23Z\",\"WARC-Record-ID\":\"<urn:uuid:369e933b-9138-4295-a35a-5f010bdd8366>\",\"Content-Length\":\"168048\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ca342ce4-a079-4607-9775-39fbe95bd91e>\",\"WARC-Concurrent-To\":\"<urn:uuid:1d5bb146-7c2e-4051-bbf1-7bfd7410f7a2>\",\"WARC-IP-Address\":\"40.65.205.118\",\"WARC-Target-URI\":\"https://www.c-sharpcorner.com/article/angular-code-optimised-functions/\",\"WARC-Payload-Digest\":\"sha1:AQZ7XUI6DVC5B4FQHDA75A77P6IZZI4T\",\"WARC-Block-Digest\":\"sha1:ADDE2EK2GSKX7FF2PZ7P6AJOVCN3H5RA\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038074941.13_warc_CC-MAIN-20210413183055-20210413213055-00463.warc.gz\"}"} |
https://numberworld.info/100001100010010 | [
"# Number 100001100010010\n\n### Properties of number 100001100010010\n\nCross Sum:\nFactorization:\nDivisors:\nCount of divisors:\nSum of divisors:\nPrime number?\nNo\nFibonacci number?\nNo\nBell Number?\nNo\nCatalan Number?\nNo\nBase 3 (Ternary):\nBase 4 (Quaternary):\nBase 5 (Quintal):\nBase 8 (Octal):\n5af3520b121a\nBase 32:\n2qud90m4gq\nsin(100001100010010)\n0.45758154407781\ncos(100001100010010)\n0.88916766164732\ntan(100001100010010)\n0.51461784297245\nln(100001100010010)\n32.236202301956\nlg(100001100010010)\n14.000004777256\nsqrt(100001100010010)\n10000055.000349\nSquare(100001100010010)\n1.0000220003212E+28\n\n### Number Look Up\n\n100001100010010 (one hundred trillion one billion one hundred million ten thousand ten) is a very special figure. The cross sum of 100001100010010 is 5. If you factorisate 100001100010010 you will get these result 2 * 5 * 587 * 17035962523. The figure 100001100010010 has 16 divisors ( 1, 2, 5, 10, 587, 1174, 2935, 5870, 17035962523, 34071925046, 85179812615, 170359625230, 10000110001001, 20000220002002, 50000550005005, 100001100010010 ) whith a sum of 180308627354016. 100001100010010 is not a prime number. The figure 100001100010010 is not a fibonacci number. The figure 100001100010010 is not a Bell Number. The number 100001100010010 is not a Catalan Number. The convertion of 100001100010010 to base 2 (Binary) is 10110101111001101010010000010110001001000011010. The convertion of 100001100010010 to base 3 (Ternary) is 111010002000022121011211100012. The convertion of 100001100010010 to base 4 (Quaternary) is 112233031102002301020122. The convertion of 100001100010010 to base 5 (Quintal) is 101101404223100310020. The convertion of 100001100010010 to base 8 (Octal) is 2657152202611032. The convertion of 100001100010010 to base 16 (Hexadecimal) is 5af3520b121a. The convertion of 100001100010010 to base 32 is 2qud90m4gq. The sine of the figure 100001100010010 is 0.45758154407781. The cosine of the number 100001100010010 is 0.88916766164732. The tangent of the number 100001100010010 is 0.51461784297245. The root of 100001100010010 is 10000055.000349.\nIf you square 100001100010010 you will get the following result 1.0000220003212E+28. The natural logarithm of 100001100010010 is 32.236202301956 and the decimal logarithm is 14.000004777256. I hope that you now know that 100001100010010 is impressive figure!"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.64762944,"math_prob":0.97125816,"size":2728,"snap":"2020-45-2020-50","text_gpt3_token_len":942,"char_repetition_ratio":0.2507342,"word_repetition_ratio":0.2776204,"special_character_ratio":0.5843108,"punctuation_ratio":0.16331096,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9977491,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-25T22:36:16Z\",\"WARC-Record-ID\":\"<urn:uuid:ddb4688b-2aa8-4814-9330-841298a7c54d>\",\"Content-Length\":\"15022\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:36b1b843-1f61-4992-afdf-1c9e577f5feb>\",\"WARC-Concurrent-To\":\"<urn:uuid:34ca2548-25cf-4089-b008-18597da9d23e>\",\"WARC-IP-Address\":\"176.9.140.13\",\"WARC-Target-URI\":\"https://numberworld.info/100001100010010\",\"WARC-Payload-Digest\":\"sha1:KH64BHPYY2DT53ONV5ZZPH3RIIXZAWPU\",\"WARC-Block-Digest\":\"sha1:UH62YESLRU42ZP272AGEVP45B32ACFFH\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141184870.26_warc_CC-MAIN-20201125213038-20201126003038-00229.warc.gz\"}"} |
https://www.bmti-report.com/calculation-of-the-bsi-index-explained/ | [
"# Calculation of the BSI index explained\n\nHow is the index BSI calculated?\n\nThe BSI is a continuation of the index that originally started with the BHI, which evolved into the BHMI and now the BSI.\n\nOn the 7th January 1997, the average of the constituent tripcharter routes of the BHI – USD 8,333/day – was given the index value of 1,000.\n\nWhen the BHMI superseded the BHI on 29th of December 2000, the closing value of the BHI was 1,058. To allow continuation, the average of the 6 TC’s of the new Handymax routes – USD 9,431/day – was given the equivalent index value of 1,058 providing a conversion factor of a 0.11218 movement on the index for each dollar change in the TC average.\n\nWhen the BHMI was replaced by the BSI on 3rd January 2006, the closing value was 1,819. To allow continuation, the average of 5 TC’s of the new Supramax routes – USD 18,299/day – was again given the equivalent index value of 1,819. From now on a conversion factor of 0.0994 represents the number of index points that correspond to one dollar of the TC average.\n\nFor example:\n\n On 3rd Jan 2006 18,299 1,819 1,819/18,299 = 0.0994 On 5th Jan 2006 18,360 18,360 x 0.0994 = 1,824.98 1,825 (to the nearest whole number)\n\nHow is the Time Charter Average calculated?\n\nEach of the component routes is multiplied by its weighting, and then all of them are added together to give the average of the 5 TC’s for that day.\n\nFor example:\n\n Routes Weighting Average US\\$ S1A 12,5% 21,988 21,988 x 0.125 = 2,748.5 S1B 12,5% 23,420 23,420 x 0.125 = 2,927.5 S2 25% 17,748 17,748 x 0.25 = 4,437 S3 25% 15,610 15,610 x 0.25 = 3,902.5 S4 25% 17,376 17,376 x 0.25 = 4,344 Added together = 18,359.5\n\nThis is rounded up or down to the nearest whole number, so USD 18,360/day\n\nSettlement of BHMI denominated paper contracts against BSI\n\nDuring the period 1st July 2005 and 23rd December 2005, the average % premium of the BSI over the BHMI average TC rates was calculated as 11.2706%. When contracts made against the BHMI are settled against the BSI in future, they will be scaled up by that percentage. From now, all new paper contracts will be made on the basis of the Baltic Supermax Index.\n\nDetails provided by the Baltic Exchange Information Services Ltd."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.92225,"math_prob":0.9752471,"size":2202,"snap":"2020-10-2020-16","text_gpt3_token_len":669,"char_repetition_ratio":0.13694267,"word_repetition_ratio":0.020460358,"special_character_ratio":0.36239782,"punctuation_ratio":0.14980544,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96274704,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-28T17:55:01Z\",\"WARC-Record-ID\":\"<urn:uuid:f18a28a1-50e1-4027-aff0-a292a1f32910>\",\"Content-Length\":\"49961\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2df054ae-ca9d-4e5a-bb29-fdf817b99cdf>\",\"WARC-Concurrent-To\":\"<urn:uuid:725bff59-04fd-4313-86bc-e9679b5f12d6>\",\"WARC-IP-Address\":\"185.183.158.249\",\"WARC-Target-URI\":\"https://www.bmti-report.com/calculation-of-the-bsi-index-explained/\",\"WARC-Payload-Digest\":\"sha1:YW5VGM3W45CYIJHQPBGHB7CIN66NORM2\",\"WARC-Block-Digest\":\"sha1:26NGZT44I3CPLZ5WLUULVJ6MNOZ7RU6Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370492125.18_warc_CC-MAIN-20200328164156-20200328194156-00501.warc.gz\"}"} |
https://ftp.aimsciences.org/article/doi/10.3934/cpaa.2021119 | [
"",
null,
"",
null,
"",
null,
"",
null,
"• Previous Article\nLarge-time behaviors of the solution to 3D compressible Navier-Stokes equations in half space with Navier boundary conditions\n• CPAA Home\n• This Issue\n• Next Article\nRemark on 3-D Navier-Stokes system with strong dissipation in one direction\nJuly & August 2021, 20(7&8): 2789-2809. doi: 10.3934/cpaa.2021119\n\nFrequency domain approach to decay rates for a coupled hyperbolic-parabolic system\n\n 1 Institut de Recherche Mathématique Avancée, Université de Strasbourg, 67084 Strasbourg, France 2 School of Mathematical Sciences, Qufu Normal University, 273165, Qufu, China 3 School of Mathematics, Sichuan University, Chengdu, Sichuan, 610064, China\n\n* Corresponding author\n\nReceived January 2021 Revised June 2021 Published July & August 2021 Early access July 2021\n\nFund Project: This work is partially supported by the NSF of China under grants 11931011, 11821001 and 11831011, and by the Science Development Project of Sichuan University under grant 2020SCUNL201\n\nWe consider the asymptotic behavior of a linear model arising in fluid-structure interactions. The system is formed by a heat equation and a wave equation in two distinct domains, which are coupled by atransmission condition along the interface of the domains. By means of the frequency domain approach, we establish some decay rates for the whole system. Our results also showthat the decay of the fluid-structure interaction depends not only on the transmission of the damping from the heat equation to the wave equation, but also on the location of the damping for the wave equation.\n\nCitation: Bopeng Rao, Xu Zhang. Frequency domain approach to decay rates for a coupled hyperbolic-parabolic system. Communications on Pure & Applied Analysis, 2021, 20 (7&8) : 2789-2809. doi: 10.3934/cpaa.2021119\nReferences:\n G. Avalos, I. Lasiecka and R. Triggiani, Heat-wave interaction in 2–3 dimensions: optimal rational decay rate, J. Math. Anal. Appl., 437 (2016), 782-815. doi: 10.1016/j.jmaa.2015.12.051.",
null,
"",
null,
"Google Scholar G. Avalos and R. Triggiani, Rational decay rates for a PDE heat-structure interaction: a frequency domain approach, Evol. Equ. Control Theory, 2 (2013), 233–253. doi: 10.3934/eect.2013.2.233.",
null,
"",
null,
"Google Scholar C. Batty and T. Duyckaerts, Non-uniform stability for bounded semi-groups on Banach spaces, J. Evol. Equ., 8 (2008), 765–780. doi: 10.1007/s00028-008-0424-1.",
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"Google Scholar C. Batty, L. Paunonen and D. Seifert, Optimal energy decay in a one-dimensional coupled wave-heat system, J. Evol. Equ., 16 (2016), 649–664. doi: 10.1007/s00028-015-0316-0.",
null,
"",
null,
"Google Scholar C. Batty, L. Paunonen and D. Seifert, Optimal energy decay for the wave-heat system on a rectangular domain, SIAM J. Math. Anal., 51 (2019), 808–819. doi: 10.1137/18M1195796.",
null,
"",
null,
"Google Scholar A. Borichev and Y. Tomilov, Optimal polynomial decay of functions and operator semigroups, Math. Ann., 347 (2010), 455–478. doi: 10.1007/s00208-009-0439-0.",
null,
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https://stackoverflow.com/questions/34799226/whats-the-difference-between-js-number-max-safe-integer-and-max-value | [
"# Whats the difference between JS Number.MAX_SAFE_INTEGER and MAX_VALUE?\n\nNumber.MAX_SAFE_INTEGER 9007199254740991\n\nNumber.MAX_VALUE 1.7976931348623157e+308\n\nI understand how `MAX_SAFE_INTEGER` is computed based on JavaScript's double precision floating point arithmetic, but where does this huge max value come from? Is it the number that comes about if you're using all 63 bits for the exponent instead of the safe 11 bits?\n\n`Number.MAX_SAFE_INTEGER` is the largest integer which can be used safely in calculations.\n\nFor example, `Number.MAX_SAFE_INTEGER + 1 === Number.MAX_SAFE_INTEGER + 2` is true — any integer larger than MAX_SAFE_INTEGER cannot always be represented in memory accurately. All bits are used to represent the digits of the number.\n\n`Number.MAX_VALUE` on the other hand is the largest number possible to represent using a double precision floating point representation. Generally speaking, the larger the number the less accurate it will be.\n\nJS numbers are internally 64-bits floats (IEEE 754-2008).\n\nMAX_SAFE_INTEGER is the maximum integer that can be safely represented in that format, meaning that all the numbers below that value (and above MIN_SAFE_INTEGER) can be represented as integers.\n\nMAX_VALUE comes from 2^1023 (11 bits mantissa minus mantissa sign), thats about 10^308.\n\nIs it the number that comes about if you're using all 63 bits for the exponent instead of the safe 11 bits?\n\nThe mantissa (exponent) is always 11 bits, (not so) surprinsingly that's enough for up to 10^308.\n\nBasically floating point numbers are represented as:\n\n``````digits * 2 ** movement\n``````\n\ndigits (the mantissa) has 52 bits (and 1 \"hidden bit\"), movement has 11 bits, and both together form a 64bit number (with 1 sign bit). Through that, you can represent all kinds of different numbers as you can store very large numbers (large positive movement), very small numbers (large negative movement), and integers (digits).\n\nWhat is Number.MAX_SAFE_INTEGER ?\n\nIntegers can just be represented with movement being set in a way that the mantissa is actually the number itself, then digits contains the 52 + 1 bit number, and that can hold up to `2 ** 53 - 1` numbers (which is `Number.MAX_SAFE_INTEGER`).\n\nNow for larger numbers, you have to use movement, which basically means that you move the digits left or right, and therefore you lose accuracy.\n\n(Imagine `digits` would just take 8 bits)\n\n`````` number > digits | movement > result\n// savely represented\n11111111 > 11111111 | 0 > 11111111\n// lost the last 1\n111111111 > 11111111 | 1 > 111111110\n// lost the two last 1s\n1111111111 > 11111111 | 10 > 1111111100\n``````\n\nWhat is Number.MAX_VALUE ?\n\nIf you set all bits of `digits` and all bits of `movement`, you get a number (`2 ** 53 - 1`) that is moved by `2 ** 10 - 1` to the left, and that is the largest number that can be stored in the 64 bit, everything that is larger is `Infinity` (which is represented as the movement being 2 ** 10 and the mantissa being 0).\n\nAs you know the javascript have type Number, but not integer. Integer appears by `ducktyping` feature in javascript. So `Number.MAX_SAFE_INTEGER` < `Number.MAX_VALUE`.\n\nThey together with `MIN_VALUE` and `MIN_SAFE_INTEGER` set range of possible Number value, for `double` and `int` when you use `parseFloat(X)` && `parseInt(X)`.\n\n`MAX_VALUE` is in double (64 bit)\n`MAX_SAFE_INTEGER` can use first 53 bit of a double(64 bit)\nbasically `javascript` doesn't support long. so for int number its uses 32 bit integer container. and for a numbers greater then 32 bit its keep the number in a double container which integer part is 53 bit and rest 11 bits are mantissa(keep the information of a floating point number).\n\n`MAX_SAFE_INTEGER` has a value of `9007199254740991`. The reasoning behind that number is that JavaScript uses double-precision floating-point format numbers as specified in IEEE 754. It returns the maximum safe integer value which can be represented as 2^53 - 1.\n\n`Safe` refers to the ability to represent integers exactly and to correctly compare them.\n\nEg: `Number.MAX_SAFE_INTEGER + 1 === Number.MAX_SAFE_INTEGER + 2` will evaluate to `true`\n\n`MAX_VALUE` on the other hand, has a value of approximately `1.79E+308`, or 2^1024. Values larger than MAX_VALUE are represented as Infinity."
] | [
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https://www.jpost.com/opinion/columnists/hebrew-hear-say-sue-far-sue-good | [
"(function (a, d, o, r, i, c, u, p, w, m) { m = d.getElementsByTagName(o), a[c] = a[c] || {}, a[c].trigger = a[c].trigger || function () { (a[c].trigger.arg = a[c].trigger.arg || []).push(arguments)}, a[c].on = a[c].on || function () {(a[c].on.arg = a[c].on.arg || []).push(arguments)}, a[c].off = a[c].off || function () {(a[c].off.arg = a[c].off.arg || []).push(arguments) }, w = d.createElement(o), w.id = i, w.src = r, w.async = 1, w.setAttribute(p, u), m.parentNode.insertBefore(w, m), w = null} )(window, document, \"script\", \"https://95662602.adoric-om.com/adoric.js\", \"Adoric_Script\", \"adoric\",\"9cc40a7455aa779b8031bd738f77ccf1\", \"data-key\");\nvar domain=window.location.hostname; var params_totm = \"\"; (new URLSearchParams(window.location.search)).forEach(function(value, key) {if (key.startsWith('totm')) { params_totm = params_totm +\"&\"+key.replace('totm','')+\"=\"+value}}); var rand=Math.floor(10*Math.random()); var script=document.createElement(\"script\"); script.src=`https://stag-core.tfla.xyz/pre_onetag?pub_id=34&domain=\\${domain}&rand=\\${rand}&min_ugl=0\\${params_totm}`; document.head.append(script);"
] | [
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https://www.recruitmentzones.in/kseeb-karnataka-sslc-2021-22-science-model-question-paper/ | [
"Board Karnataka Secondary Education Examination Board [KSEEB] Class SSLC [Class 10th] Subject Science Download Model Question Paper Year 2021-22 Document Type PDF Official Website https://sslc.karnataka.gov.in/english\n\n## KSEEB Karnataka SSLC Science Model Question Paper\n\nDownload Karnataka Secondary Education Examination Board [KSEEB], SSLC [Class 10th] Science Model Question Paper 2021-2022\n\nDownload KSEEB SSLC 2022 [All Subjects] Model Question Papers Here\n\n## Karnataka SSLC Science Model Questions\n\nPart – A : Physics\nI Four alternatives are given for each of the following questions / incomplete statements. Choose the correct alternative and write the complete answer along with its letter of alphabet. 2×1=2\n\n1. The diameter of the reflecting surface of spherical mirror is\nA) Optical Centre\nB) Centre of Curvature\nC) Aperture\nD) Principal axis\n\n2. An electric motor takes 5A from a 220V electric source. The power of the motor is\nA) 1100W\nB) 44W\nC) 225W\nD) 440W\n\nII Answer the following questions. 3×1=3\n3. If the focal length of a spherical mirror is 15cm. Find the radius of curvature ?\n4. Mention any two disadvantages of fossil fuels.\n5. What is an electric circuit ?\n\nIII Answer the following questions. 3×2=6\n6. How does overloading and short-circuit occur in an electric circuit? Explain. What is the function of a fuse during this situation ?\n7. Draw the schematic diagram of a biogas plant.\n8 An electric lamp whose resistance is 40Ω and conductor of 8Ω resistance are connected in series to 12V battery in an electric circuit. Calculate the total resistance of the circuit and the current flowing through the circuit.\n\nIV Answer the following questions. 3×3=9\n9. Draw the ray diagram of image formed when the object is kept beyond 2F1 of the convex lens. With the help of the diagram, mention the position and nature of the image formed. (F1 : principal focus of the lens)\n\nOR\n\nDraw the ray diagram when of image formed the object is kept beyond C of the concave mirror. With the help of the diagram mention the position and nature of the image formed. (C : Centre of curvature of mirror).\n\n10. What is electric potential difference? What is the SI unit of potential difference ? Name the device used to measure the potential difference.\n\n11. An object is kept at a distance of 30cm from a diverging lens of focal length 15cm. At what distance the image is formed from the lens? Find the magnification of the image.\n\nV Answer the following questions 2×4=8\n12. Explain Faraday’s experiment of magnet and coil. State “electromagnetic induction” with the help of this experiment OR State the Fleming’s right hand rule. How can we increase the amount of electric current produced in the electric generator ? Write any two differences between electric generator and electric motor?\n\n13. a) List the uses of Convex mirror and Concave mirror.\nb) Define principal focus and radius of curvature of a convex mirror.\n\nDownload KSEEB SSLC 2012-2021 Question Papers Here\n\nSimilar Searches:\nkarnataka sslc question papers 2019 with answers pdf, karnataka sslc question papers, karnataka sslc question papers 2020 with answers pdf, karnataka sslc question papers 2015 with answers pdf, karnataka sslc question papers with answers pdf, karnataka sslc question papers 2017 with answers pdf, karnataka sslc question papers 2016 with answers pdf, karnataka sslc question paper 2022, karnataka sslc question paper 2020, karnataka sslc question paper 2020 with answers, karnataka sslc question paper 2019, karnataka sslc question paper 2021, karnataka sslc question paper pattern 2022, karnataka sslc question paper 2021 with answers pdf, karnataka sslc question paper 2017, karnataka sslc 2021 model question paper and answer key, karnataka sslc question papers 2018 with answers pdf, karnataka sslc question papers 2021 with answers pdf, karnataka board sslc question paper, karnataka sslc board question paper 2015, karnataka sslc board question papers with answers\n\nHave a question? Please feel free to reach out by leaving a comment below\n\n(Visited 639 times, 1 visits today)"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.83870924,"math_prob":0.7401167,"size":4134,"snap":"2022-40-2023-06","text_gpt3_token_len":1019,"char_repetition_ratio":0.21912833,"word_repetition_ratio":0.06891271,"special_character_ratio":0.23052734,"punctuation_ratio":0.10228802,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.98244274,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-26T09:53:19Z\",\"WARC-Record-ID\":\"<urn:uuid:997e0eca-9bdc-4edb-a4e2-9ff7b920fb89>\",\"Content-Length\":\"77190\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ab40b0fa-b841-4adf-ab61-d11b0203f969>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c44a920-d2ae-4e61-9208-10c8ba032b77>\",\"WARC-IP-Address\":\"148.72.88.29\",\"WARC-Target-URI\":\"https://www.recruitmentzones.in/kseeb-karnataka-sslc-2021-22-science-model-question-paper/\",\"WARC-Payload-Digest\":\"sha1:RKY7UHMZZ6IJSFV3RJCXFBKEM7C6RLCR\",\"WARC-Block-Digest\":\"sha1:QWMQRACCVYYNSYQCRDSEJGUMHTFPQTY7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030334855.91_warc_CC-MAIN-20220926082131-20220926112131-00006.warc.gz\"}"} |
https://www.numbers.education/1586.html | [
"Is 1586 a prime number? What are the divisors of 1586?\n\n## Parity of 1 586\n\n1 586 is an even number, because it is evenly divisible by 2: 1 586 / 2 = 793.\n\nFind out more:\n\n## Is 1 586 a perfect square number?\n\nA number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 1 586 is about 39.825.\n\nThus, the square root of 1 586 is not an integer, and therefore 1 586 is not a square number.\n\n## What is the square number of 1 586?\n\nThe square of a number (here 1 586) is the result of the product of this number (1 586) by itself (i.e., 1 586 × 1 586); the square of 1 586 is sometimes called \"raising 1 586 to the power 2\", or \"1 586 squared\".\n\nThe square of 1 586 is 2 515 396 because 1 586 × 1 586 = 1 5862 = 2 515 396.\n\nAs a consequence, 1 586 is the square root of 2 515 396.\n\n## Number of digits of 1 586\n\n1 586 is a number with 4 digits.\n\n## What are the multiples of 1 586?\n\nThe multiples of 1 586 are all integers evenly divisible by 1 586, that is all numbers such that the remainder of the division by 1 586 is zero. There are infinitely many multiples of 1 586. The smallest multiples of 1 586 are:\n\n## Numbers near 1 586\n\n### Nearest numbers from 1 586\n\nFind out whether some integer is a prime number"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8920551,"math_prob":0.9994308,"size":346,"snap":"2023-40-2023-50","text_gpt3_token_len":104,"char_repetition_ratio":0.19005848,"word_repetition_ratio":0.0,"special_character_ratio":0.34393063,"punctuation_ratio":0.16091955,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9989528,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-10T01:43:07Z\",\"WARC-Record-ID\":\"<urn:uuid:d8eb42bf-afd1-417d-8d73-0d14dc180c79>\",\"Content-Length\":\"17694\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6ec7955f-269f-4445-9df2-59e5fa5bf2f7>\",\"WARC-Concurrent-To\":\"<urn:uuid:529cc5d9-f92e-49d3-9e21-3063a23b91e1>\",\"WARC-IP-Address\":\"213.186.33.19\",\"WARC-Target-URI\":\"https://www.numbers.education/1586.html\",\"WARC-Payload-Digest\":\"sha1:KYY7FAG2F3JBIRBB4BKXSS6YL6LGFX57\",\"WARC-Block-Digest\":\"sha1:JZRDQIG63JEC47PIDGRIMOGBFDUIGX6Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100989.75_warc_CC-MAIN-20231209233632-20231210023632-00125.warc.gz\"}"} |
https://gomathanswerkey.com/texas-go-math-grade-3-module-11-assessment-answer-key/ | [
"Refer to our Texas Go Math Grade 3 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 3 Module 11 Assessment Answer Key.\n\nChoose the best term from the box to complete the sentence.\n\nQuestion 1.\nA number is ___ quotient is a counting number and the number can be divided into equal groups. (p. 352)\nExplanation:\npossible to divide or separate Nine is divisible by three.\n\nQuestion 2.\n___ are a set of related multiplication and division equations. (p. 339)\nExplanation:\nThe inverse of multiplication is division.\nIf you multiply by a given number and then divide by the same number,\nyou will arrive at the same number you started with.\n\nConcepts and Skills\n\nComplete the related facts. TEKS 3.4.F, 3.4.J, 3.5.B\n\nQuestion 3.\n4 × 8 = ___\n8 × __ = 32\n32 ÷ __ = 8\n32 ÷ 8 = __\n4 × 8 = 32\n8 × 4 = 32\n32 ÷ 4 = 8\n32 ÷ 8 = 4\nExplanation:\nThe equations are related to inverse operation\n\nQuestion 4.\n3 × __ = 9\n__ ÷ 3 = 3\n3 × 3 = 9\n9 ÷ 3 = 3\nExplanation:\nThe equations are related to inverse operation\n\nQuestion 5.\n6 × __ = 42\n7 × 6 = ___\n___ ÷ 6 = 7\n42 ÷ __ = 6\n6 × 7 = 42\n7 × 6 = 42\n42 ÷ 6 = 7\n42 ÷ 7 = 6\nExplanation:\nThe equations are related to inverse operation\n\nFind the quotient. TEKS 3.4.H, 3.5.B\n\nQuestion 6.\n2 ÷ 1 = ___\nExplanation:\nWhen a number is divided by 1 the number does not change\n\nQuestion 7.\n0 ÷ 7 = ___\nExplanation:\nSo zero divided by zero is undefined. .\n\nQuestion 8.",
null,
"Explanation:\nWhen a number is divided by itself the answer is 1\n\nQuestion 9.",
null,
"",
null,
"Explanation:\nWhen a number is divided by 1 the answer will not change\n\nTell if the product will be odd or even. TEKS 3.4.I\n\nQuestion 10.\n8 × 22\nExplanation:\n176 is a even number\nAs the end numbers are even the answer is also even\n\n461 × 10\nExplanation:\n4610 is a even number\nAs the end numbers are even the answer is also even\n\nQuestion 12.\n433 × 267\nExplanation:\nAs the end numbers are odd the answer is also odd\n176 is a even number\n\nQuestion 13.\n944 × 1,275\nExplanation:\nWhen even is multiplied with odd the answer will be even\n\nUse the divisibility rule to tell if the number is odd or even. TEKS 3.4.I\n\nQuestion 14.\n43\nExplanation:\nDivisibility rule of 43 is itself and 1\n\nQuestion 15.\n596\nExplanation:\n576 the last digit is 6 is divided by 2\nIt is even\n\nQuestion 16.\n3,718\nExplanation:\n3718 the last digit is 8 is divided by 2\nIt is even\n\nQuestion 17.\n2,609\nExplanation:\n2609 the last digit is 9 is divided by 3\nso, it is odd\n\nTexas Test Prep\n\nFill in the bubble for the correct answer choice.\n\nQuestion 18.\nA pet shop has 18 hamsters. They sold 10 and put the remaining hamsters into 8 cages. How many hamsters are in each cage? TEKS 3.4.H, 3.4.K, 3.5.B\n(A) 4\n(B) 8\n(C) 2\n(D) 1\nExplanation:\n1 hamsters are in each cage\n\nQuestion 19.\nWhich division equation belongs to this set of related facts? TEKS 3.4.F, 3.4.J, 3.5.B\n4 × 6 = 24 6 × 4 = 24\n(A) 20 ÷ 4 = 5\n(B) 24 ÷ 6 = 4\n(C) 18 ÷ 6 = 3\n(D) 24 ÷ 3 = 8\nExplanation:\n24 ÷ 6 = 4 is equation belongs to this set of related facts\n\nQuestion 20.\nJeremy is thinking of a 2-digit number that is divisible by 2. What could Jeremy’s number be? TEKS 3.4.1\n(A) 210\n(B) 63\n(C) 49\n(D) 78\nExplanation:\n78 is the 2 digit number that is divisible by 2\n\nQuestion 21.\nJill has 6 puppies. The puppies play in 1 crate. How many puppies are in the crate? TEKS 3.4.H, 3.4.K, 3.5.B",
null,
"Record your answer and fill in the bubbles on the grid. Be sure to use the correct place value.",
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] | [
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"https://i2.wp.com/gomathanswerkey.com/wp-content/uploads/2021/10/Texas-Go-Math-Grade-3-Module-11-Assessment-Answer-Key-1.png",
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https://www.physicsforums.com/threads/equilibrium-solubility-and-precipitates.179448/ | [
"# Equilibrium, Solubility and Precipitates\n\nCan someone please explain to me:\n\n1) How to determine an eqilibirium constant\n\n2) How to calculate Ksp from solubility values\n\n3) How to determine if a precipitate will form in a reaction\n\nI am doing a chemistry course without a teacher and the textbook doesn't properly explain it.\n\nThanks in advance.",
null,
"## The Attempt at a Solution\n\nRelated Biology and Chemistry Homework Help News on Phys.org\nGokul43201\nStaff Emeritus\nGold Member\nNote that we typically require you to show your own effort before we can add anything to the discussion.\n\nYou may find these useful:\nhttp://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Equilibrium.html [Broken]\n...\nhttp://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Equilibrium-Constant.html [Broken]\nhttp://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Calc-K-from-equilib-conc.html [Broken]\n...\nhttp://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Writing-Ksp-expression.html [Broken]\nhttp://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Calc-Ksp-FromMolSolub.html [Broken]\n\nLast edited by a moderator:\nOkay, thanks for the links. So let me try to work out some answers over here.\n\n1)A(g) --> 2B(g) + C(g)\n<--\n\nWhen 1 mol of A is placed in a 4 L container at temperature t the concentration of C at equilibibrium is 0.050 mol/L. What is the equiliibrium constant for the reaction at temperature t?\n\nWell the links didn't tell me how to find the eqilibrium contant with unknowns.\n\n2) Calculate the Ksp for each of these salts\n\nA) CaSO4 = 3.3 * 10 -3 mol/L\nB) MgF2 = 2.7 * 10 -3 mol/L\n\nSo for a) CaSO4 would dissociate to form [Ca] and [SO4] meaning\n(3.3 * 10 -3) (3.3 * 10 -3) = 1.089 -5 Ksp = 1.089 -5\n\nand for b) MgF2 would become [Mg] and [F]2 so we would have a 1:2 ratio so,\n(2.7 * 10 -3) (5.4 * 10 -3) = 1.458 -5 Ksp = 1.458 -5\n\nAm I right with these two?\n\nAlso for determing a precipitate they had an article titled that on their main page but no link so I still am not sure how to do that.\n\nGokul43201\nStaff Emeritus\nGold Member\nYou're making a mistake on (b). First write down the expression for Ksp before you substitute values in.\n\nKsp = [Mg] [F2]\nKsp = (2.7 * 10 -3) (2.7 * 10 -3) =7.9 *10 -6\nKsp = 7.9 *10 -6\n\nIs this better?\n\nGokul43201\nStaff Emeritus\nGold Member\nNo, it should be\n\n$$K_{sp}=[Mg^{2+}][F^-]^2$$\n\nGo back and read the definition of the equilibrium constant again.\nhttp://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Equilibrium-Constant.html [Broken]\n\nLast edited by a moderator:\nOkay, I got the exponent 2 the first time but I'm not sure how you determine that MgF2 will become [Mg2+] and [F-]2. How did you determine the ions? And shouldn't Ca and SO4 also have them then?\n\nAlso, I think I found how to determine if a precipitate will form, can you check this work for me?\n\nThe solubility product of Ca(OH)2 is 7.9 * 10 -6 at 25C. Will a precipitate form when 100 ml of 0.10 mol/L of CaCl2 and 50.0 ml of 0.070 mol/L of NaOH are combined?\n\nSo [CaCl2] = 0.10 mol/L * 100 ml / 150 ml = 0.6666666667\n[NaOH] = 0.070 mol/L * 50 ml / 150 ml = 0.0233333333\n\n(0.6666666667) (0.0233333333) = 0.01555555563\n\nSince Ktrial > Ksp a precipitate will form.\nIs this right?"
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https://oeis.org/A322561 | [
"The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.",
null,
"Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)\n A322561 Digits of one of the two 17-adic integers sqrt(2) that is related to A322559. 6\n 6, 14, 14, 8, 5, 4, 14, 14, 7, 2, 15, 15, 11, 5, 6, 7, 2, 14, 6, 14, 15, 16, 3, 8, 14, 5, 12, 16, 0, 4, 7, 0, 8, 10, 2, 16, 16, 15, 9, 7, 12, 9, 14, 14, 5, 12, 3, 4, 7, 9, 9, 2, 2, 14, 5, 9, 12, 6, 2, 10, 5, 0, 10, 10, 11, 11, 2, 3, 14, 10, 11, 2, 6, 12, 0, 4 (list; graph; refs; listen; history; text; internal format)\n OFFSET 0,1 COMMENTS This square root of 2 in the 17-adic field ends with digit 6. The other, A322562, ends with digit 11 (B when written as a 17-adic number). LINKS Seiichi Manyama, Table of n, a(n) for n = 0..10000 Wikipedia, p-adic number FORMULA a(n) = (A322559(n+1) - A322559(n))/17^n. For n > 0, a(n) = 16 - A322562(n). Equals A309989*A322566 = A309990*A322565. EXAMPLE The solution to x^2 == 2 (mod 17^4) such that x == 6 (mod 17) is x == 43594 (mod 17^4), and 43594 is written as 8EE6 in heptadecimal, so the first four terms are 6, 14, 14 and 8. PROG (PARI) a(n) = truncate(sqrt(2+O(17^(n+1))))\\17^n CROSSREFS Cf. A322559, A322560. Digits of 17-adic square roots: A309989, A309990 (sqrt(-1)); this sequence, A322562 (sqrt(2)); A322565, A322566 (sqrt(-2)). Sequence in context: A265029 A329065 A184998 * A079010 A324814 A015822 Adjacent sequences: A322558 A322559 A322560 * A322562 A322563 A322564 KEYWORD nonn,base AUTHOR Jianing Song, Aug 29 2019 STATUS approved\n\nLookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam\nContribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent\nThe OEIS Community | Maintained by The OEIS Foundation Inc.\n\nLast modified May 25 16:40 EDT 2020. Contains 334595 sequences. (Running on oeis4.)"
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https://weselnakamera.pl/2014/8501-diagram-of-a-ball-mill-circuit.html | [
"# Diagram Of A Ball Mill Circuit\n\n• ### diagram of a ball mill in mining – iron ore benification\n\nball mill schematic diagram mining – diagram of mining mill circuit is manufactured from shanghai xuanshi,it is the main mineral ball mill; coarse grinding mill. ball mill – wikipedia a ball mill is a type of grinder used to grind and blend materials for use in mineral dressing processes, paints, pyrotechnics, ceramics and selective laser\n\nGet Price\n• ### schematic diagram ball mill\n\ncentrifugal ball mill diagram how to optimize apound ball mill to safely and effectively ball mill homemade black schematic diagram of ball mill pdf . get quote. read more >> mining ball mill discharge circuit diagrams\n\nGet Price\n• ### circuit diagram of coal mill\n\ncircuit diagram of coal mill coal mill diagram mining & quarry plant. coal mill diagram . find the right and the top schematic diagrams for dry ball milling electrical diagram supplier, circuit breaker,regulator,ac\n\nGet Price\n• ### diagram of mining mill circuit\n\nelectrical wiring of ball mills in mines. onemine mining and minerals library search bridgeport mill wiring diagram free mill circuit pump manual weir minerals crushed ore is fed into a sag\n\nGet Price\n• ### design and analysis of ball mill inlet chute for roller\n\ndesign and analysis of ball mill inlet chute for roller press circuit in cement industry circuit ball mill inlet chute along with the truck assembly for roller press circuit ums (unidan mill s) type fls ball mill from process diagram fig 3.1, [prism project considered]\n\nGet Price\n• ### diagram of a ball mill\n\ndiagram of a ball mill circuit utcindiacoin. ball mill schematic diagram, grinding mill china schematic diagram of ball mill pdf hd overland conveyor electrical schematic diagram wiring diagram of\n\nGet Price\n• ### diagram of a ball mill in mining – iron ore benification\n\nball mill schematic diagram mining – diagram of mining mill circuit is manufactured from shanghai xuanshi,it is the main mineral ball mill; 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] | [
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https://www.enotes.com/homework-help/what-4th-root-25-45-89-291706 | [
"# What is: 4th root of 25 - √45 + √89 Show complete solution and explain the answer.",
null,
"So you have the 4th root of 25 - sqrt45 + sqrt 89?\n\nThe fourth root of 25 is the same as the sqrt of the sqrt of 25 or simply the sqrt of 5. The sqrt of 45 is the sqrt of 9*5 or 3Xsqrt of 5. 89 is..."
] | [
null,
"https://static.enotescdn.net/images/main/illustrations/illo-answer.svg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9312083,"math_prob":0.98534083,"size":550,"snap":"2022-27-2022-33","text_gpt3_token_len":195,"char_repetition_ratio":0.31135532,"word_repetition_ratio":0.6923077,"special_character_ratio":0.4,"punctuation_ratio":0.10273973,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9926776,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-03T15:03:13Z\",\"WARC-Record-ID\":\"<urn:uuid:1c4fa9fd-1a6f-4dc2-ad75-53f917ae665f>\",\"Content-Length\":\"72781\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3e66ca00-f7f7-449e-97a8-2fb2ec31593a>\",\"WARC-Concurrent-To\":\"<urn:uuid:b7be9c0e-cd97-42a6-a87b-40995cd05625>\",\"WARC-IP-Address\":\"104.26.4.75\",\"WARC-Target-URI\":\"https://www.enotes.com/homework-help/what-4th-root-25-45-89-291706\",\"WARC-Payload-Digest\":\"sha1:747BZLNI4Z2GZWBODODB5GWRLRK7NOP6\",\"WARC-Block-Digest\":\"sha1:QB7CGGBM2EDVZUXXSK4RFKXMK6PH6KEM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104244535.68_warc_CC-MAIN-20220703134535-20220703164535-00694.warc.gz\"}"} |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-11th-edition/chapter-1-review-exercises-page-163/21 | [
"College Algebra (11th Edition)\n\n$-14+13i$\n$\\bf{\\text{Solution Outline:}}$ To simplify the given expression, $15i-(3+2i)-11 ,$ remove the grouping symbols and then combine the real parts and the imaginary parts. $\\bf{\\text{Solution Details:}}$ Removing the grouping symbols, the expression above is equivalent to \\begin{array}{l}\\require{cancel} 15i-3-2i-11 .\\end{array} Combining the real parts and the imaginary parts results to \\begin{array}{l}\\require{cancel} (-3-11)+(15i-2i) \\\\\\\\= -14+13i .\\end{array}"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.55458707,"math_prob":0.99800926,"size":500,"snap":"2019-26-2019-30","text_gpt3_token_len":152,"char_repetition_ratio":0.11895161,"word_repetition_ratio":0.071428575,"special_character_ratio":0.32,"punctuation_ratio":0.08888889,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9928942,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-19T02:46:00Z\",\"WARC-Record-ID\":\"<urn:uuid:974a8339-d644-42a6-8436-559353531208>\",\"Content-Length\":\"73125\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:15c0fc3e-4c70-4e8d-a9c2-1c821753ff96>\",\"WARC-Concurrent-To\":\"<urn:uuid:f0c8960d-3f3b-4fd4-ad5a-9b8cebdd9f77>\",\"WARC-IP-Address\":\"100.24.77.13\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/math/algebra/college-algebra-11th-edition/chapter-1-review-exercises-page-163/21\",\"WARC-Payload-Digest\":\"sha1:FDUT3FZEEQ4WMYYZA6MWNKIRALP3FL7O\",\"WARC-Block-Digest\":\"sha1:6ZC6TUT46RS5QZXQOL7T7QJ7KJVJAAGN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998882.88_warc_CC-MAIN-20190619023613-20190619045613-00093.warc.gz\"}"} |
http://physics.unm.edu/SQuInT/2020/abstracts.php?person_id=974 | [
"## Abstracts\n\n### Variational quantum linear solver: A hybrid algorithm for linear systems\n\nPresenting Author: Ryan LaRose, Michigan State University, NASA - Ames Research Center\nContributing Author(s): Carlos Bravo-Prieto, M. Cerezo, Yigit Subasi, Lukasz Cincio, Patrick J. Coles\n\nSolving linear systems of equations is central to many engineering and scientific fields. Several quantum algorithms have been proposed for linear systems, where the goal is to prepare $\\ket{x}$ such that $A\\ket{x} \\propto \\ket{b}$. While these algorithms are promising, the time horizon for their implementation is long due to the required quantum circuit depth. In this work, we propose a variational hybrid quantum-classical algorithm for solving linear systems, with the aim of reducing the circuit depth and doing much of the computation classically. We propose a cost function based on the overlap between $\\ket{b}$ and $A\\ket{x}$, and we derive an operational meaning for this cost in terms of the solution precision $\\epsilon$. We also introduce a quantum circuit to estimate this cost, while showing that this cost cannot be efficiently estimated classically. Using Rigetti's quantum computer, we successfully implement our algorithm up to a problem size of $32 \\times 32$. Furthermore, we numerically find that the complexity of our algorithm scales efficiently in both $1/\\epsilon$ and $\\kappa$, with $\\kappa$ the condition number of $A$. Our algorithm provides a heuristic for quantum linear systems that could make this application more near term.\n\n(Session 5 : Saturday from 5:00pm - 7:00pm)\n\nSQuInT Chief Organizer\nAkimasa Miyake, Associate Professor\namiyake@unm.edu\n\nSQuInT Co-Organizer\nBrian Smith, Associate Professor UO\nbjsmith@uoregon.edu\n\nSQuInT Program Committee\nPostdoctoral Fellows:\nMarkus Allgaier (UO OMQ)\nSayonee Ray (UNM CQuIC)\nPablo Poggi (UNM CQuIC)\nValerian Thiel (UO OMQ)\n\nSQuInT Event Co-Organizers (Oregon)\nJorjie Arden\njarden@uoregon.edu\nHolly Lynn\nhollylyn@uoregon.edu\n\nBrandy Todd",
null,
""
] | [
null,
"http://physics.unm.edu/SQuInT/2020/images/twitter.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81768566,"math_prob":0.95354635,"size":2358,"snap":"2023-40-2023-50","text_gpt3_token_len":605,"char_repetition_ratio":0.10832626,"word_repetition_ratio":0.0,"special_character_ratio":0.20610687,"punctuation_ratio":0.11363637,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96909946,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-01T15:31:07Z\",\"WARC-Record-ID\":\"<urn:uuid:839954d6-0b6c-4636-861b-90978f5e3afd>\",\"Content-Length\":\"7418\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9e14397c-0ff0-4ef9-95ae-9a3ae0d8d70e>\",\"WARC-Concurrent-To\":\"<urn:uuid:f92678e7-153f-40c0-bcdc-2fc11c4528ad>\",\"WARC-IP-Address\":\"129.24.172.122\",\"WARC-Target-URI\":\"http://physics.unm.edu/SQuInT/2020/abstracts.php?person_id=974\",\"WARC-Payload-Digest\":\"sha1:W4FNHSIF7PLSMEYG5BS5SIECPLBDBIM7\",\"WARC-Block-Digest\":\"sha1:NPN3MNTKSDCWFPSUZTDNL6BJUQTAWG2X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100290.24_warc_CC-MAIN-20231201151933-20231201181933-00115.warc.gz\"}"} |
https://en.formulasearchengine.com/wiki/Symplectic_cut | [
"# Symplectic cut\n\nIn physics, the Jeans instability causes the collapse of interstellar gas clouds and subsequent star formation. It occurs when the internal gas pressure is not strong enough to prevent gravitational collapse of a region filled with matter. For stability, the cloud must be in hydrostatic equilibrium, which in case of a spherical cloud translates to:\n\n${\\frac {dp}{dr}}=-{\\frac {G\\rho (r)M_{enc}(r)}{r^{2}}}$",
null,
",\n\nwhere $M_{enc}(r)$",
null,
"is the enclosed mass, $p$",
null,
"is the pressure, $\\rho (r)$",
null,
"is the density of the gas at $r$",
null,
", $G$",
null,
"is the gravitational constant and $r$",
null,
"is the radius. The equilibrium is stable if small perturbations are damped and unstable if they are amplified. In general, the cloud is unstable if it is either very massive at a given temperature or very cool at a given mass for gravity to overcome the gas pressure.\n\n## Jeans mass\n\nThe Jeans mass is named after the British physicist Sir James Jeans, who considered the process of gravitational collapse within a gaseous cloud. He was able to show that, under appropriate conditions, a cloud, or part of one, would become unstable and begin to collapse when it lacked sufficient gaseous pressure support to balance the force of gravity. The cloud is stable for sufficiently small mass (at a given temperature and radius), but once this critical mass is exceeded, it will begin a process of runaway contraction until some other force can impede the collapse. He derived a formula for calculating this critical mass as a function of its density and temperature. The greater the mass of the cloud, the smaller its size, and the colder its temperature, the less stable it will be against gravitational collapse.\n\nThe approximate value of the Jeans mass may be derived through a simple physical argument. One begins with a spherical gaseous region of radius $R$",
null,
", mass $M$",
null,
", and with a gaseous sound speed $c_{s}$",
null,
". Imagine that we compress the region slightly. It takes a time,\n\n$t_{sound}={\\frac {R}{c_{s}}}\\simeq (5\\times 10^{5}{\\mbox{ yr}})\\left({\\frac {R}{0.1{\\mbox{ pc}}}}\\right)\\left({\\frac {c_{s}}{0.2{\\mbox{ km s}}^{-1}}}\\right)^{-1}$",
null,
"for sound waves to cross the region, and attempt to push back and re-establish the system in pressure balance. At the same time, gravity will attempt to contract the system even further, and will do so on a free-fall time,\n\n$t_{\\rm {ff}}={\\frac {1}{\\sqrt {G\\rho }}}\\simeq (2{\\mbox{ Myr}})\\left({\\frac {n}{10^{3}{\\mbox{ cm}}^{-3}}}\\right)^{-1/2}$",
null,
"where $G$",
null,
"is the universal gravitational constant, $\\rho$",
null,
"is the gas density within the region, and $n=\\rho /\\mu$",
null,
"is the gas number density for mean mass per particle $\\mu =3.9\\times 10^{-24}$",
null,
"g, appropriate for molecular hydrogen with 20% helium by number. Now, when the sound-crossing time is less than the free-fall time, pressure forces win, and the system bounces back to a stable equilibrium. However, when the free-fall time is less than the sound-crossing time, gravity wins, and the region undergoes gravitational collapse. The condition for gravitational collapse is therefore:\n\n$t_{\\rm {ff}}",
null,
"$\\lambda _{J}={\\frac {c_{s}}{\\sqrt {G\\rho }}}\\simeq (0.4{\\mbox{ pc}})\\left({\\frac {c_{s}}{0.2{\\mbox{ km s}}^{-1}}}\\right)\\left({\\frac {n}{10^{3}{\\mbox{ cm}}^{-3}}}\\right)^{-1/2}.$",
null,
"This length scale is known as the Jeans length. All scales larger than the Jeans length are unstable to gravitational collapse, whereas smaller scales are stable. The Jeans mass $M_{J}$",
null,
"is just the mass contained in a sphere of radius $R_{J}$",
null,
"($R_{J}$",
null,
"is half the Jeans length):\n\n$M_{J}=\\left({\\frac {4\\pi }{3}}\\right)\\rho R_{J}^{3}=\\left({\\frac {\\pi }{6}}\\right){\\frac {c_{s}^{3}}{G^{3/2}\\rho ^{1/2}}}\\simeq (2{\\mbox{ M}}_{\\odot })\\left({\\frac {c_{s}}{0.2{\\mbox{ km s}}^{-1}}}\\right)^{3}\\left({\\frac {n}{10^{3}{\\mbox{ cm}}^{-3}}}\\right)^{-1/2}.$",
null,
"It was later pointed out by other astrophysicists that in fact, the original analysis used by Jeans was flawed, for the following reason. In his formal analysis, Jeans assumed that the collapsing region of the cloud was surrounded by an infinite, static medium. In fact, because all scales greater than the Jeans length are also unstable to collapse, any initially static medium surrounding a collapsing region will in fact also be collapsing. As a result, the growth rate of the gravitational instability relative to the density of the collapsing background is slower than that predicted by Jeans' original analysis. This flaw has come to be known as the \"Jeans swindle\".\n\nThe Jeans instability likely determines when star formation occurs in molecular clouds.\n\n## Jeans length\n\nJeans' length is the critical radius of a cloud (typically a cloud of interstellar dust) where thermal energy, which causes the cloud to expand, is counteracted by gravity, which causes the cloud to collapse. It is named after the British astronomer Sir James Jeans, who concerned himself with the stability of spherical nebula in the early 1900s.\n\nThe formula for Jeans Length is:\n\n$\\lambda _{J}={\\sqrt {\\frac {15k_{B}T}{4\\pi G\\mu \\rho }}},$",
null,
"Perhaps the easiest way to conceptualize Jeans' Length is in terms of a close approximation, in which we discard the factors $15$",
null,
"and $4\\pi$",
null,
"and in which we rephrase $\\rho$",
null,
"as ${\\frac {M}{r^{3}}}$",
null,
". The formula for Jeans' Length then becomes:\n\n$\\lambda _{J}\\approx {\\sqrt {\\frac {k_{B}Tr^{3}}{GM\\mu }}}.$",
null,
"It follows immediately that $\\lambda _{J}=r$",
null,
"when $k_{B}T={\\frac {GM\\mu }{r}}$",
null,
"i.e. the cloud's radius is the Jeans' Length when thermal energy per particle equals gravitational work per particle. At this critical length the cloud neither expands nor contracts. It is only when thermal energy is not equal to gravitational work that the cloud either expands and cools or contracts and warms, a process that continues until equilibrium is reached.\n\n## Jeans' Length as oscillation wavelength\n\nThe Jeans' Length is the oscillation wavelength below which stable oscillations rather than gravitational collapse will occur.\n\n$\\lambda _{J}={\\frac {2\\pi }{k_{J}}}=c_{s}\\left({\\frac {\\pi }{G\\rho }}\\right)^{1/2},$",
null,
"It is also the distance a sound wave would travel in the collapse time.\n\n## Fragmentation\n\nJeans instability can also give rise to fragmentation in certain conditions. To derive the condition for fragmentation an adiabatic process is assumed in an ideal gas and also a polytropic equation of state is taken. The derivation is showed below through a dimensional analysis:\n\nFor adiabatic processes, $PV^{\\gamma }={\\text{constant}}\\rightarrow V\\sim P^{-1/\\gamma }.$",
null,
"For an ideal gas, $PV=nRT\\rightarrow P.P^{-1/\\gamma }\\sim T\\rightarrow P\\sim T^{\\frac {\\gamma }{\\gamma -1}}.$",
null,
"Polytropic equation of state, $P=K\\rho ^{\\gamma }\\rightarrow T\\sim \\rho ^{\\gamma -1}.$",
null,
"Jeans mass, $M_{J}\\sim T^{3/2}\\rho ^{-1/2}\\sim \\rho ^{{\\frac {3}{2}}(\\gamma -1)}\\rho ^{-1/2}.$",
null,
"Thus, $M_{J}\\sim \\rho ^{{\\frac {3}{2}}\\left(\\gamma -{\\frac {4}{3}}\\right)}.$",
null,
"If adiabatic index, $\\gamma >{\\frac {4}{3}}$",
null,
"Jeans mass increases with increasing density while if $\\gamma <{\\frac {4}{3}}$",
null,
"Jeans mass decreases with increasing density. During gravitational collapse density always increases, thus in the second case Jeans mass will decrease during collapse allowing smaller overdense regions to collapse leading to fragmentation of the giant molecular cloud. For an ideal monatomic gas, the adiabatic index is 5/3 but in astrophysical objects this value is usually even lower than 1. So the second case is the rule rather than an exception in stars. This is the reason why stars usually form in clusters."
] | [
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https://meta.stackexchange.com/questions/100971/how-do-i-write-equations-on-math-se | [
"# How do I write equations on Math.SE?\n\nI have checked the Mathematics Stack Exchange site, and have noticed equation syntax on the questions. Is there a guide or link of how to write these formulas?\n\n## 1 Answer\n\nProbably, these links might help you.\n\nLaTex\n\nLaTeX/Advanced Mathematics\n\nIn the above links, the equations are embedded within \\begin{equation} and \\end{equation}, you just have to replace those with double dollar signs .\n\nBy the way, Math.SE uses MathJax, which uses LaTeX to formulate equations.\n\nScreenshot from Math.SE to show some sample equations:",
null,
"• This is better than the answers I found on the Math.SE meta. Nice job with the examples. – Bill the Lizard Aug 5 '11 at 1:11"
] | [
null,
"https://i.stack.imgur.com/Zuv4q.png",
null
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https://pureportal.strath.ac.uk/en/publications/random-rectangular-graphs | [
"# Random rectangular graphs\n\nResearch output: Book/ReportOther report\n\n### Abstract\n\nA generalization of the random geometric graph (RGG) model is proposed by considering a set of points uniformly and independently distributed on a rectangle of unit area instead of on a unit square [0; 1]2 : The topological properties, such as connectivity, average degree, average path length and clustering, of the random rectangular graphs (RRGs) generated by this model are then studied as a function of the rectangle sides lengths a and b = 1=a, and\nthe radius r used to connect the nodes. When a = 1 we recover the RGG, and when a ! 1 the very elongated rectangle generated resembles a one-dimensional RGG. We provided computational and analytical evidence that the topological properties of the RRG differ significantly from those of the RGG. The connectivity of the RRG depends not only on the number of nodes as in the case of the RGG, but also on the side length of the rectangle. As the rectangle is more elongated the critical radius for connectivity increases following first a power-law and then a linear trend. Also, as the rectangle becomes more elongated the average distance between the nodes of the graphs increases, but the local cliquishness of the graphs also increases thus producing graphs which are relatively long and highly locally connected. Finally, we found the analytic expression for the average degree in the RRG as a function of the rectangle side lengths and the radius. For different values of the side length, the expected and the observed values of the average degree display excellent correlation,\nwith correlation coefficients larger than 0.9999.\nLanguage English University of Strathclyde 21 Unpublished - 9 Feb 2015\n\n### Fingerprint\n\nRandom Geometric Graph\nRectangle\nGraph in graph theory\nConnectivity\nTopological Properties\nVertex of a graph\nUnit of area\nLinear Trend\nLocally Connected\nAverage Distance\nGeometric Model\nGraph Model\nPath Length\nCorrelation coefficient\nSet of points\nPower Law\nClustering\nUnit\n\n### Keywords\n\n• random rectangular graphs\n• random geometric graph\n• clustering\n• rectangle unit area\n\n### Cite this\n\nEstrada, E., & Sheerin, M. J. (2015). Random rectangular graphs. University of Strathclyde.\nEstrada, Ernesto ; Sheerin, Matthew James. / Random rectangular graphs. University of Strathclyde, 2015. 21 p.\n@book{49e154b0c48d4d8fb8278b8fbef63aeb,\ntitle = \"Random rectangular graphs\",\nabstract = \"A generalization of the random geometric graph (RGG) model is proposed by considering a set of points uniformly and independently distributed on a rectangle of unit area instead of on a unit square [0; 1]2 : The topological properties, such as connectivity, average degree, average path length and clustering, of the random rectangular graphs (RRGs) generated by this model are then studied as a function of the rectangle sides lengths a and b = 1=a, andthe radius r used to connect the nodes. When a = 1 we recover the RGG, and when a ! 1 the very elongated rectangle generated resembles a one-dimensional RGG. We provided computational and analytical evidence that the topological properties of the RRG differ significantly from those of the RGG. The connectivity of the RRG depends not only on the number of nodes as in the case of the RGG, but also on the side length of the rectangle. As the rectangle is more elongated the critical radius for connectivity increases following first a power-law and then a linear trend. Also, as the rectangle becomes more elongated the average distance between the nodes of the graphs increases, but the local cliquishness of the graphs also increases thus producing graphs which are relatively long and highly locally connected. Finally, we found the analytic expression for the average degree in the RRG as a function of the rectangle side lengths and the radius. For different values of the side length, the expected and the observed values of the average degree display excellent correlation,with correlation coefficients larger than 0.9999.\",\nkeywords = \"random rectangular graphs, random geometric graph, clustering, rectangle unit area\",\nauthor = \"Ernesto Estrada and Sheerin, {Matthew James}\",\nyear = \"2015\",\nmonth = \"2\",\nday = \"9\",\nlanguage = \"English\",\npublisher = \"University of Strathclyde\",\n\n}\n\nEstrada, E & Sheerin, MJ 2015, Random rectangular graphs. University of Strathclyde.\n\nRandom rectangular graphs. / Estrada, Ernesto; Sheerin, Matthew James.\n\nUniversity of Strathclyde, 2015. 21 p.\n\nResearch output: Book/ReportOther report\n\nTY - BOOK\n\nT1 - Random rectangular graphs\n\nAU - Sheerin, Matthew James\n\nPY - 2015/2/9\n\nY1 - 2015/2/9\n\nN2 - A generalization of the random geometric graph (RGG) model is proposed by considering a set of points uniformly and independently distributed on a rectangle of unit area instead of on a unit square [0; 1]2 : The topological properties, such as connectivity, average degree, average path length and clustering, of the random rectangular graphs (RRGs) generated by this model are then studied as a function of the rectangle sides lengths a and b = 1=a, andthe radius r used to connect the nodes. When a = 1 we recover the RGG, and when a ! 1 the very elongated rectangle generated resembles a one-dimensional RGG. We provided computational and analytical evidence that the topological properties of the RRG differ significantly from those of the RGG. The connectivity of the RRG depends not only on the number of nodes as in the case of the RGG, but also on the side length of the rectangle. As the rectangle is more elongated the critical radius for connectivity increases following first a power-law and then a linear trend. Also, as the rectangle becomes more elongated the average distance between the nodes of the graphs increases, but the local cliquishness of the graphs also increases thus producing graphs which are relatively long and highly locally connected. Finally, we found the analytic expression for the average degree in the RRG as a function of the rectangle side lengths and the radius. For different values of the side length, the expected and the observed values of the average degree display excellent correlation,with correlation coefficients larger than 0.9999.\n\nAB - A generalization of the random geometric graph (RGG) model is proposed by considering a set of points uniformly and independently distributed on a rectangle of unit area instead of on a unit square [0; 1]2 : The topological properties, such as connectivity, average degree, average path length and clustering, of the random rectangular graphs (RRGs) generated by this model are then studied as a function of the rectangle sides lengths a and b = 1=a, andthe radius r used to connect the nodes. When a = 1 we recover the RGG, and when a ! 1 the very elongated rectangle generated resembles a one-dimensional RGG. We provided computational and analytical evidence that the topological properties of the RRG differ significantly from those of the RGG. The connectivity of the RRG depends not only on the number of nodes as in the case of the RGG, but also on the side length of the rectangle. As the rectangle is more elongated the critical radius for connectivity increases following first a power-law and then a linear trend. Also, as the rectangle becomes more elongated the average distance between the nodes of the graphs increases, but the local cliquishness of the graphs also increases thus producing graphs which are relatively long and highly locally connected. Finally, we found the analytic expression for the average degree in the RRG as a function of the rectangle side lengths and the radius. For different values of the side length, the expected and the observed values of the average degree display excellent correlation,with correlation coefficients larger than 0.9999.\n\nKW - random rectangular graphs\n\nKW - random geometric graph\n\nKW - clustering\n\nKW - rectangle unit area\n\nM3 - Other report\n\nBT - Random rectangular graphs\n\nPB - University of Strathclyde\n\nER -\n\nEstrada E, Sheerin MJ. Random rectangular graphs. University of Strathclyde, 2015. 21 p."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.88824135,"math_prob":0.96020925,"size":5022,"snap":"2019-51-2020-05","text_gpt3_token_len":1067,"char_repetition_ratio":0.15603827,"word_repetition_ratio":0.8408263,"special_character_ratio":0.20350458,"punctuation_ratio":0.08539326,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99060756,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-13T15:59:28Z\",\"WARC-Record-ID\":\"<urn:uuid:f8bf8a2e-442e-4e10-b98e-d34c9d88c4d3>\",\"Content-Length\":\"40007\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d77a87cf-0771-4121-8153-dd60c569cb3a>\",\"WARC-Concurrent-To\":\"<urn:uuid:b6c32a31-76cb-4a16-9e4d-5617c4f28ef4>\",\"WARC-IP-Address\":\"52.209.51.54\",\"WARC-Target-URI\":\"https://pureportal.strath.ac.uk/en/publications/random-rectangular-graphs\",\"WARC-Payload-Digest\":\"sha1:PVSGLH5UWMKOIPNYG6NLFBAYJGTQEZBF\",\"WARC-Block-Digest\":\"sha1:ASDXXXBNZCG73MQ53QPQXF3MA67ZASNH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540564599.32_warc_CC-MAIN-20191213150805-20191213174805-00399.warc.gz\"}"} |
https://fr.maplesoft.com/support/help/view.aspx?path=odeadvisor%2FAbel2A | [
"",
null,
"Abel2A - Maple Help\n\nSolving Abel's ODEs of the Second Kind, Class A",
null,
"Description\n\n • The general form of Abel's equation, second kind, class A is given by:\n > Abel_ode2A := (y(x)+g(x))*diff(y(x),x)=f2(x)*y(x)^2+f1(x)*y(x)+f0(x);\n ${\\mathrm{Abel_ode2A}}{≔}\\left({y}{}\\left({x}\\right){+}{g}{}\\left({x}\\right)\\right){}\\left(\\frac{{ⅆ}}{{ⅆ}{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\\left({x}\\right)\\right){=}{\\mathrm{f2}}{}\\left({x}\\right){}{{y}{}\\left({x}\\right)}^{{2}}{+}{\\mathrm{f1}}{}\\left({x}\\right){}{y}{}\\left({x}\\right){+}{\\mathrm{f0}}{}\\left({x}\\right)$ (1)\n where f2(x), f1(x), f0(x), and g(x) are arbitrary functions. See Differentialgleichungen, by E. Kamke, p. 26. There is as yet no general solution for this ODE.\n • Note that all ODEs of type Abel, second kind, can be rewritten as ODEs of type Abel, first kind, as explained in ?odeadvisor,Abel2C",
null,
"Examples\n\n > $\\mathrm{with}\\left(\\mathrm{DEtools},\\mathrm{symgen},\\mathrm{odeadvisor}\\right)$\n $\\left[{\\mathrm{symgen}}{,}{\\mathrm{odeadvisor}}\\right]$ (2)\n > $\\mathrm{odeadvisor}\\left(\\mathrm{Abel_ode2A}\\right)$\n $\\left[\\left[{\\mathrm{_Abel}}{,}{\\mathrm{2nd type}}{,}{\\mathrm{class A}}\\right]\\right]$ (3)\n\n1) f0(x) = f1(x)*g(x)-f2(x)*g(x)^2\n\n > $\\mathrm{ode}≔\\mathrm{eval}\\left(\\mathrm{subs}\\left(\\mathrm{f0}\\left(x\\right)=\\mathrm{f1}\\left(x\\right)g\\left(x\\right)-\\mathrm{f2}\\left(x\\right){g\\left(x\\right)}^{2},\\mathrm{Abel_ode2A}\\right)\\right)$\n ${\\mathrm{ode}}{≔}\\left({y}{}\\left({x}\\right){+}{g}{}\\left({x}\\right)\\right){}\\left(\\frac{{ⅆ}}{{ⅆ}{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\\left({x}\\right)\\right){=}{\\mathrm{f2}}{}\\left({x}\\right){}{{y}{}\\left({x}\\right)}^{{2}}{+}{\\mathrm{f1}}{}\\left({x}\\right){}{y}{}\\left({x}\\right){+}{g}{}\\left({x}\\right){}{\\mathrm{f1}}{}\\left({x}\\right){-}{{g}{}\\left({x}\\right)}^{{2}}{}{\\mathrm{f2}}{}\\left({x}\\right)$ (4)\n\nThis case can be solved as follows:\n\n > $\\mathrm{dsolve}\\left(\\mathrm{ode},y\\left(x\\right)\\right)$\n ${y}{}\\left({x}\\right){=}{-}{g}{}\\left({x}\\right){,}{y}{}\\left({x}\\right){=}\\left({\\int }{-}{{ⅇ}}^{{-}\\left({\\int }{\\mathrm{f2}}{}\\left({x}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\right)}{}\\left({g}{}\\left({x}\\right){}{\\mathrm{f2}}{}\\left({x}\\right){-}{\\mathrm{f1}}{}\\left({x}\\right)\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\\mathrm{_C1}}\\right){}{{ⅇ}}^{{\\int }{\\mathrm{f2}}{}\\left({x}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}}$ (5)\n\n2) Another case which can be solved:\n\nf1(x) = 2*f2(x)*g(x)-diff(g(x),x)\n\n > $\\mathrm{ode}≔\\mathrm{eval}\\left(\\mathrm{subs}\\left(\\mathrm{f1}\\left(x\\right)=2\\mathrm{f2}\\left(x\\right)g\\left(x\\right)-\\mathrm{diff}\\left(g\\left(x\\right),x\\right),\\mathrm{Abel_ode2A}\\right)\\right)$\n ${\\mathrm{ode}}{≔}\\left({y}{}\\left({x}\\right){+}{g}{}\\left({x}\\right)\\right){}\\left(\\frac{{ⅆ}}{{ⅆ}{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\\left({x}\\right)\\right){=}{\\mathrm{f2}}{}\\left({x}\\right){}{{y}{}\\left({x}\\right)}^{{2}}{+}\\left({2}{}{g}{}\\left({x}\\right){}{\\mathrm{f2}}{}\\left({x}\\right){-}\\frac{{ⅆ}}{{ⅆ}{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{g}{}\\left({x}\\right)\\right){}{y}{}\\left({x}\\right){+}{\\mathrm{f0}}{}\\left({x}\\right)$ (6)\n\nAlthough the answer for this case can be obtained using standard methods (an integrating factor is easily found), the use of symmetry methods can provide an explicit solution. The infinitesimals for this case are given by\n\n > $\\mathrm{symgen}\\left(\\mathrm{ode},y\\left(x\\right)\\right)$\n $\\left[{\\mathrm{_ξ}}{=}{0}{,}{\\mathrm{_η}}{=}\\frac{{{ⅇ}}^{{\\int }{2}{}{\\mathrm{f2}}{}\\left({x}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}}}{{y}{+}{g}{}\\left({x}\\right)}\\right]$ (7)\n\nTo indicate the use of symmetry methods \"at first\", we can explicitly indicate an integration method (see dsolve); for instance, to use the canonical coordinates of the invariance group:\n\n > $\\mathrm{ans}≔\\mathrm{dsolve}\\left(\\mathrm{ode},y\\left(x\\right),\\mathrm{can}\\right)$\n ${\\mathrm{ans}}{≔}{y}{}\\left({x}\\right){=}\\frac{{-}{g}{}\\left({x}\\right){}{{ⅇ}}^{{-}{2}{}\\left({\\int }{\\mathrm{f2}}{}\\left({x}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\right)}{+}\\sqrt{{{g}{}\\left({x}\\right)}^{{2}}{}{\\left({{ⅇ}}^{{-}{2}{}\\left({\\int }{\\mathrm{f2}}{}\\left({x}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\right)}\\right)}^{{2}}{+}{2}{}{{ⅇ}}^{{-}{2}{}\\left({\\int }{\\mathrm{f2}}{}\\left({x}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\right)}{}\\left({\\int }\\frac{{\\mathrm{f0}}{}\\left({x}\\right)}{{\\left({{ⅇ}}^{{\\int }{\\mathrm{f2}}{}\\left({x}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}}\\right)}^{{2}}}\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\right){+}{2}{}{{ⅇ}}^{{-}{2}{}\\left({\\int }{\\mathrm{f2}}{}\\left({x}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\right)}{}{\\mathrm{_C1}}}}{{{ⅇ}}^{{-}{2}{}\\left({\\int }{\\mathrm{f2}}{}\\left({x}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\right)}}{,}{y}{}\\left({x}\\right){=}{-}\\frac{{g}{}\\left({x}\\right){}{{ⅇ}}^{{-}{2}{}\\left({\\int }{\\mathrm{f2}}{}\\left({x}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\right)}{+}\\sqrt{{{g}{}\\left({x}\\right)}^{{2}}{}{\\left({{ⅇ}}^{{-}{2}{}\\left({\\int }{\\mathrm{f2}}{}\\left({x}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\right)}\\right)}^{{2}}{+}{2}{}{{ⅇ}}^{{-}{2}{}\\left({\\int }{\\mathrm{f2}}{}\\left({x}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\right)}{}\\left({\\int }\\frac{{\\mathrm{f0}}{}\\left({x}\\right)}{{\\left({{ⅇ}}^{{\\int }{\\mathrm{f2}}{}\\left({x}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}}\\right)}^{{2}}}\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\right){+}{2}{}{{ⅇ}}^{{-}{2}{}\\left({\\int }{\\mathrm{f2}}{}\\left({x}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\right)}{}{\\mathrm{_C1}}}}{{{ⅇ}}^{{-}{2}{}\\left({\\int }{\\mathrm{f2}}{}\\left({x}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\right)}}$ (8)"
] | [
null,
"https://bat.bing.com/action/0",
null,
"https://fr.maplesoft.com/support/help/arrow_down.gif",
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"https://fr.maplesoft.com/support/help/arrow_down.gif",
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https://en.wikibooks.org/wiki/On_2D_Inverse_Problems/_An_infinite_example | [
"# On 2D Inverse Problems/ An infinite example\n\nThe following construction provides an example of an infinite graph, which Dirichlet-to-Neumann operator satisfies the operator equation in the title of this chapter.\n\n$\\Lambda (G)={\\sqrt {L}}.$",
null,
"The operator equation reflects the self-duality and self-symmetry of the infinite graph.\n\nExercise (**). Prove that the Dirichlet-to-Neumann operator of the graph with the natural boundary satisfies the functional equation. (Hint) Use the fact that the operator/matrix is the fixed point of the Schur complement\n\n$\\Lambda (G)={\\begin{pmatrix}2I&B\\\\B^{T}&\\Lambda +2I\\end{pmatrix}}/(\\Lambda +2I),$",
null,
"where\n\n$B={\\begin{pmatrix}-1&0&0&\\ldots &-1\\\\-1&-1&0&\\ldots &0\\\\0&\\vdots &\\ddots &\\ddots &\\vdots \\\\\\vdots &\\vdots &\\ddots &-1&0\\\\0&0&\\ldots &-1&-1\\\\\\end{pmatrix}}$",
null,
"is the circular matrix of first differences."
] | [
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/f6756f87834cc2f67cbb6859e7048b97dc491c94",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/1ed13d90ec7f607360ed405594ea827afaa4761d",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/a09ce1ddc953f044df2b21336547b5bf281670ea",
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https://thereaderwiki.com/en/Moment_magnitude_scale | [
"# Moment magnitude scale\n\nThe moment magnitude scale (MMS; denoted explicitly with Mw or Mw, and generally implied with use of a single M for magnitude) is a measure of an earthquake's magnitude (\"size\" or strength) based on its seismic moment (a measure of the work done by the earthquake). It was defined in a 1979 paper by Thomas C. Hanks and Hiroo Kanamori. Similar to the local magnitude scale (ML ) defined by Charles Francis Richter in 1935, it uses a logarithmic scale; small earthquakes have approximately the same magnitudes on both scales.\n\nMoment magnitude (Mw ) is considered the authoritative magnitude scale for ranking earthquakes by size. It is more directly related to the energy of an earthquake than other scales, and does not saturate—that is, it does not underestimate magnitudes as other scales do in certain conditions. It has become the standard scale used by seismological authorities like the U.S. Geological Survey for reporting large earthquakes (typically M > 4), replacing the local magnitude (ML ) and surface wave magnitude (Ms ) scales. Subtypes of the moment magnitude scale (Mww , etc.) reflect different ways of estimating the seismic moment.\n\n### History\n\n#### Richter scale: the original measure of earthquake magnitude\n\nAt the beginning of the twentieth century, very little was known about how earthquakes happen, how seismic waves are generated and propagate through the earth's crust, and what information they carry about the earthquake rupture process; the first magnitude scales were therefore empirical. The initial step in determining earthquake magnitudes empirically came in 1931 when the Japanese seismologist Kiyoo Wadati showed that the maximum amplitude of an earthquake's seismic waves diminished with distance at a certain rate. Charles F. Richter then worked out how to adjust for epicentral distance (and some other factors) so that the logarithm of the amplitude of the seismograph trace could be used as a measure of \"magnitude\" that was internally consistent and corresponded roughly with estimates of an earthquake's energy. He established a reference point and the now familiar ten-fold (exponential) scaling of each degree of magnitude, and in 1935 published what he called the \"magnitude scale\", now called the local magnitude scale, labeled ML . (This scale is also known as the Richter scale, but news media sometimes use that term indiscriminately to refer to other similar scales.)\n\nThe local magnitude scale was developed on the basis of shallow (~15 km (9 mi) deep), moderate-sized earthquakes at a distance of approximately 100 to 600 km (62 to 373 mi), conditions where the surface waves are predominant. At greater depths, distances, or magnitudes the surface waves are greatly reduced, and the local magnitude scale underestimates the magnitude, a problem called saturation. Additional scales were developed – a surface-wave magnitude scale (Ms) by Beno Gutenberg in 1945, a body-wave magnitude scale (mB) by Gutenberg and Richter in 1956, and a number of variants – to overcome the deficiencies of the ML scale, but all are subject to saturation. A particular problem was that the Ms scale (which in the 1970s was the preferred magnitude scale) saturates around Ms 8.0 and therefore underestimates the energy release of \"great\" earthquakes such as the 1960 Chilean and 1964 Alaskan earthquakes. These had Ms magnitudes of 8.5 and 8.4 respectively but were notably more powerful than other M 8 earthquakes; their moment magnitudes were closer to 9.6 and 9.3.\n\n#### Single couple or double couple\n\nThe study of earthquakes is challenging as the source events cannot be observed directly, and it took many years to develop the mathematics for understanding what the seismic waves from an earthquake can tell us about the source event. An early step was to determine how different systems of forces might generate seismic waves equivalent to those observed from earthquakes.\n\nThe simplest force system is a single force acting on an object. If it has sufficient strength to overcome any resistance it will cause the object to move (\"translate\"). A pair of forces, acting on the same \"line of action\" but in opposite directions, will cancel; if they cancel (balance) exactly there will be no net translation, though the object will experience stress, either tension or compression. If the pair of forces are offset, acting along parallel but separate lines of action, the object experiences a rotational force, or torque. In mechanics (the branch of physics concerned with the interactions of forces) this model is called a couple, also simple couple or single couple. If a second couple of equal and opposite magnitude is applied their torques cancel; this is called a double couple. A double couple can be viewed as \"equivalent to a pressure and tension acting simultaneously at right angles\".\n\nThe single couple and double couple models are important in seismology because each can be used to derive how the seismic waves generated by an earthquake event should appear in the \"far field\" (that is, at distance). Once that relation is understood it can be inverted to use the earthquake's observed seismic waves to determine its other characteristics, including fault geometry and seismic moment.\n\nIn 1923 Hiroshi Nakano showed that certain aspects of seismic waves could be explained in terms of a double couple model. This led to a three-decade long controversy over the best way to model the seismic source: as a single couple, or a double couple? While Japanese seismologists favored the double couple, most seismologists favored the single couple. Although the single couple model had some short-comings, it seemed more intuitive, and there was a belief – mistaken, as it turned out – that the elastic rebound theory for explaining why earthquakes happen required a single couple model. In principle these models could be distinguished by differences in the radiation patterns of their S-waves, but the quality of the observational data was inadequate for that.\n\nThe debate ended when Maruyama (1963), Haskell (1964), and Burridge & Knopoff (1964) showed that if earthquake ruptures are modeled as dislocations the pattern of seismic radiation can always be matched with an equivalent pattern derived from a double couple, but not from a single couple. This was confirmed as better and more plentiful data coming from the World-Wide Standard Seismograph Network (WWSSN) permitted closer analysis of seismic waves. Notably, in 1966 Keiiti Aki showed that the seismic moment of the 1964 Niigata earthquake as calculated from the seismic waves on the basis of a double couple was in reasonable agreement with the seismic moment calculated from the observed physical dislocation.\n\n#### Dislocation theory\n\nA double couple model suffices to explain an earthquake's far-field pattern of seismic radiation, but tells us very little about the nature of an earthquake's source mechanism or its physical features. While slippage along a fault was theorized as the cause of earthquakes (other theories included movement of magma, or sudden changes of volume due to phase changes), observing this at depth was not possible, and understanding what could be learned about the source mechanism from the seismic waves requires an understanding of the source mechanism.\n\nModeling the physical process by which an earthquake generates seismic waves required much theoretical development of dislocation theory, first formulated by the Italian Vito Volterra in 1907, with further developments by E. H. Love in 1927. More generally applied to problems of stress in materials, an extension by F. Nabarro in 1951 was recognized by the Russian geophysicist A. V. Vvedenskaya as applicable to earthquake faulting. In a series of papers starting in 1956 she and other colleagues used dislocation theory to determine part of an earthquake's focal mechanism, and to show that a dislocation – a rupture accompanied by slipping — was indeed equivalent to a double couple,\n\nIn a pair of papers in 1958, J. A. Steketee worked out how to relate dislocation theory to geophysical features. Numerous other researchers worked out other details, culminating in a general solution in 1964 by Burridge and Knopoff, which established the relationship between double couples and the theory of elastic rebound, and provided the basis for relating an earthquake's physical features to seismic moment.\n\n#### Seismic moment\n\nSeismic moment – symbol M0 – is a measure of the work accomplished by the faulting of an earthquake. Its magnitude is that of the forces that form the earthquake's equivalent double couple. (More precisely, it is the scalar magnitude of the second-order moment tensor that describes the force components of the double-couple.) Seismic moment is measured in units of Newton meters (N·m) or Joules, or (in the older CGS system) dyne-centimeters (dyn-cm).\n\nThe first calculation of an earthquake's seismic moment from its seismic waves was by Keiiti Aki for the 1964 Niigata earthquake. He did this two ways. First, he used data from distant stations of the WWSSN to analyze long-period (200 second) seismic waves (wavelength of about 1,000 kilometers) to determine the magnitude of the earthquake's equivalent double couple. Second, he drew upon the work of Burridge and Knopoff on dislocation to determine the amount of slip, the energy released, and the stress drop (essentially how much of the potential energy was released). In particular, he derived a now famous equation that relates an earthquake's seismic moment to its physical parameters:\n\nM0 = μūS\n\nwith μ being the rigidity (or resistance) of moving a fault with a surface areas of S over an average dislocation (distance) of . (Modern formulations replace ūS with the equivalent D̄A, known as the \"geometric moment\" or \"potency\".) By this equation the moment determined from the double couple of the seismic waves can be related to the moment calculated from knowledge of the surface area of fault slippage and the amount of slip. In the case of the Niigata earthquake the dislocation estimated from the seismic moment reasonably approximated the observed dislocation.\n\nSeismic moment is a measure of the work (more precisely, the torque) that results in inelastic (permanent) displacement or distortion of the earth's crust. It is related to the total energy released by an earthquake. However, the power or potential destructiveness of an earthquake depends (among other factors) on how much of the total energy is converted into seismic waves. This is typically 10% or less of the total energy, the rest being expended in fracturing rock or overcoming friction (generating heat).\n\nNonetheless, seismic moment is regarded as the fundamental measure of earthquake size, representing more directly than other parameters the physical size of an earthquake. As early as 1975 it was considered \"one of the most reliably determined instrumental earthquake source parameters\".\n\n#### Introduction of an energy-motivated magnitude Mw\n\nMost earthquake magnitude scales suffered from the fact that they only provided a comparison of the amplitude of waves produced at a standard distance and frequency band; it was difficult to relate these magnitudes to a physical property of the earthquake. Gutenberg and Richter suggested that radiated energy Es could be estimated as\n\n$\\log E_{s}\\approx 4.8+1.5M_{S},}$",
null,
"(in Joules). Unfortunately, the duration of many very large earthquakes was longer than 20 seconds, the period of the surface waves used in the measurement of Ms . This meant that giant earthquakes such as the 1960 Chilean earthquake (M 9.5) were only assigned an Ms 8.2. Caltech seismologist Hiroo Kanamori recognized this deficiency and took the simple but important step of defining a magnitude based on estimates of radiated energy, Mw , where the \"w\" stood for work (energy):\n\n$M_{w}=2/3\\log E_{s}-3.2}$",
null,
"Kanamori recognized that measurement of radiated energy is technically difficult since it involves the integration of wave energy over the entire frequency band. To simplify this calculation, he noted that the lowest frequency parts of the spectrum can often be used to estimate the rest of the spectrum. The lowest frequency asymptote of a seismic spectrum is characterized by the seismic moment, M0 . Using an approximate relation between radiated energy and seismic moment (which assumes stress drop is complete and ignores fracture energy),\n\n$E_{s}\\approx M_{0}/(2\\times 10^{4})}$",
null,
"(where E is in Joules and M0 is in N $\\cdot }$",
null,
"m), Kanamori approximated Mw by\n\n$M_{w}=(\\log M_{0}-9.1)/1.5}$",
null,
"#### Moment magnitude scale\n\nThe formula above made it much easier to estimate the energy-based magnitude Mw , but it changed the fundamental nature of the scale into a moment magnitude scale. Caltech seismologist Thomas C. Hanks noted that Kanamori's Mw scale was very similar to a relationship between ML and M0 that was reported by Thatcher & Hanks (1973)\n\n$M_{L}\\approx (\\log M_{0}-9.0)/1.5}$",
null,
"Hanks & Kanamori (1979) combined their work to define a new magnitude scale based on estimates of seismic moment\n\n$M=(\\log M_{0}-9.05)/1.5}$",
null,
"where $M_{0}}$",
null,
"is defined in newton meters (N·m).\n\n### Current use\n\nMoment magnitude is now the most common measure of earthquake size for medium to large earthquake magnitudes,[scientific citation needed] but in practice, seismic moment, the seismological parameter it is based on, is not measured routinely for smaller quakes. For example, the United States Geological Survey does not use this scale for earthquakes with a magnitude of less than 3.5,[citation needed] which includes the great majority of quakes.\n\nCurrent practice in official[who?] earthquake reports is to adopt moment magnitude as the preferred magnitude, i.e., Mw is the official magnitude reported whenever it can be computed. Because seismic moment (M0 , the quantity needed to compute Mw ) is not measured if the earthquake is too small, the reported magnitude for earthquakes smaller than M4 is often Richter's ML .\n\nPopular press reports most often deal with significant earthquakes larger than M~ 4. For these events, the official[who?] magnitude is the moment magnitude Mw , not Richter's local magnitude ML .\n\n### Definition\n\nThe symbol for the moment magnitude scale is Mw , with the subscript \"w\" meaning mechanical work accomplished. The moment magnitude Mw is a dimensionless value defined by Hiroo Kanamori as\n\n$M_{\\mathrm {w} }={\\frac {2}{3}}\\log _{10}(M_{0})-10.7,}$",
null,
"where M0 is the seismic moment in dyne⋅cm (10−7 N⋅m). The constant values in the equation are chosen to achieve consistency with the magnitude values produced by earlier scales, such as the local magnitude and the surface wave magnitude. Thus, a magnitude zero microearthquake has a seismic moment of approximately 1.2×109 N⋅m, while the Great Chilean earthquake of 1960, with an estimated moment magnitude of 9.4–9.6, had a seismic moment between 1.4×1023 N⋅m and 2.8×1023 N⋅m.\n\n### Relations between seismic moment, potential energy released and radiated energy\n\nSeismic moment is not a direct measure of energy changes during an earthquake. The relations between seismic moment and the energies involved in an earthquake depend on parameters that have large uncertainties and that may vary between earthquakes. Potential energy is stored in the crust in the form of elastic energy due to built-up stress and gravitational energy. During an earthquake, a portion $\\Delta W}$",
null,
"of this stored energy is transformed into\n\n• energy dissipated $E_{f}}$",
null,
"in frictional weakening and inelastic deformation in rocks by processes such as the creation of cracks\n• heat $E_{h}}$",
null,
"• radiated seismic energy $E_{s}}$",
null,
"The potential energy drop caused by an earthquake is related approximately to its seismic moment by\n\n$\\Delta W\\approx {\\frac {\\overline {\\sigma }}{\\mu }}M_{0}}$",
null,
"where ${\\overline {\\sigma }}}$",
null,
"is the average of the absolute shear stresses on the fault before and after the earthquake (e.g., equation 3 of Venkataraman & Kanamori 2004) and $\\mu }$",
null,
"is the average of the shear moduli of the rocks that constitute the fault. Currently, there is no technology to measure absolute stresses at all depths of interest, nor method to estimate it accurately, and ${\\overline {\\sigma }}}$",
null,
"is thus poorly known. It could vary highly from one earthquake to another. Two earthquakes with identical $M_{0}}$",
null,
"but different ${\\overline {\\sigma }}}$",
null,
"would have released different $\\Delta W}$",
null,
".\n\nThe radiated energy caused by an earthquake is approximately related to seismic moment by\n\n$E_{\\mathrm {s} }\\approx \\eta _{R}{\\frac {\\Delta \\sigma _{s}}{2\\mu }}M_{0}}$",
null,
"where $\\eta _{R}=E_{s}/(E_{s}+E_{f})}$",
null,
"is radiated efficiency and $\\Delta \\sigma _{s}}$",
null,
"is the static stress drop, i.e., the difference between shear stresses on the fault before and after the earthquake (e.g., from equation 1 of Venkataraman & Kanamori 2004). These two quantities are far from being constants. For instance, $\\eta _{R}}$",
null,
"depends on rupture speed; it is close to 1 for regular earthquakes but much smaller for slower earthquakes such as tsunami earthquakes and slow earthquakes. Two earthquakes with identical $M_{0}}$",
null,
"but different $\\eta _{R}}$",
null,
"or $\\Delta \\sigma _{s}}$",
null,
"would have radiated different $E_{\\mathrm {s} }}$",
null,
".\n\nBecause $E_{\\mathrm {s} }}$",
null,
"and $M_{0}}$",
null,
"are fundamentally independent properties of an earthquake source, and since $E_{\\mathrm {s} }}$",
null,
"can now be computed more directly and robustly than in the 1970s, introducing a separate magnitude associated to radiated energy was warranted. Choy and Boatwright defined in 1995 the energy magnitude\n\n$M_{\\mathrm {E} }=\\textstyle {\\frac {2}{3}}\\log _{10}E_{\\mathrm {s} }-3.2}$",
null,
"where $E_{\\mathrm {s} }}$",
null,
"is in J (N·m).\n\n### Comparative energy released by two earthquakes\n\nAssuming the values of σ̄/μ are the same for all earthquakes, one can consider Mw as a measure of the potential energy change ΔW caused by earthquakes. Similarly, if one assumes $\\eta _{R}\\Delta \\sigma _{s}/2\\mu }$",
null,
"is the same for all earthquakes, one can consider Mw as a measure of the energy Es radiated by earthquakes.\n\nUnder these assumptions, the following formula, obtained by solving for M0 the equation defining Mw , allows one to assess the ratio $E_{1}/E_{2}}$",
null,
"of energy release (potential or radiated) between two earthquakes of different moment magnitudes, $m_{1}}$",
null,
"and $m_{2}}$",
null,
":\n\n$E_{1}/E_{2}\\approx 10^{{\\frac {3}{2}}(m_{1}-m_{2})}.}$",
null,
"As with the Richter scale, an increase of one step on the logarithmic scale of moment magnitude corresponds to a 101.5 ≈ 32 times increase in the amount of energy released, and an increase of two steps corresponds to a 103 = 1000 times increase in energy. Thus, an earthquake of Mw of 7.0 contains 1000 times as much energy as one of 5.0 and about 32 times that of 6.0.\n\n### Subtypes of Mw\n\nVarious ways of determining moment magnitude have been developed, and several subtypes of the Mw scale can be used to indicate the basis used.\n\n• Mwb – Based on moment tensor inversion of long-period (~10 – 100 s) body-waves.\n• Mwr – From a moment tensor inversion of complete waveforms at regional distances (~ 1,000 miles). Sometimes called RMT.\n• Mwc – Derived from a centroid moment tensor inversion of intermediate- and long-period body- and surface-waves.\n• Mww – Derived from a centroid moment tensor inversion of the W-phase.\n• Mwp (Mi) – Developed by Seiji Tsuboi for quick estimation of the tsunami potential of large near-coastal earthquakes from measurements of the P-waves, and later extended to teleseismic earthquakes in general.\n• Mwpd – A duration-amplitude procedure which takes into account the duration of the rupture, providing a fuller picture of the energy released by longer lasting (\"slow\") ruptures than seen with Mw ."
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https://coe.northeastern.edu/research/krclab/crens3-doc/classns3_1_1_pareto_variable.html | [
"",
null,
"A Discrete-Event Network Simulator Home Tutorials ▼ Docs ▼ Develop ▼ API\nns3::ParetoVariable Class Reference\n\nParetoVariable distributed random varThis class supports the creation of objects that return random numbers from a fixed pareto distribution. It also supports the generation of single random numbers from various pareto distributions. More...\n\n#include <random-variable.h>",
null,
"Inheritance diagram for ns3::ParetoVariable:\n\nPublic Member Functions\n\nParetoVariable ()\nConstructs a pareto random variable with a mean of 1 and a shape parameter of 1.5.\n\nParetoVariable (double m)\nConstructs a pareto random variable with specified mean and shape parameter of 1.5. More...\n\nParetoVariable (double m, double s)\nConstructs a pareto random variable with the specified mean value and shape parameter. Beware, s must be strictly greater than 1. More...\n\nParetoVariable (double m, double s, double b)\nConstructs a pareto random variable with the specified mean value, shape (alpha), and upper bound. Beware, s must be strictly greater than 1. More...\n\nParetoVariable (std::pair< double, double > params)\nConstructs a pareto random variable with the specified scale and shape parameters. More...\n\nParetoVariable (std::pair< double, double > params, double b)\nConstructs a pareto random variable with the specified scale, shape (alpha), and upper bound. More...",
null,
"Public Member Functions inherited from ns3::RandomVariable\nRandomVariable (const RandomVariable &o)\n\nuint32_t GetInteger (void) const\nReturns a random integer integer from the underlying distribution. More...\n\ndouble GetValue (void) const\nReturns a random double from the underlying distribution. More...\n\nRandomVariableoperator= (const RandomVariable &o)",
null,
"Protected Member Functions inherited from ns3::RandomVariable\nRandomVariable (const RandomVariableBase &variable)\n\nRandomVariableBasePeek (void) const\n\nDetailed Description\n\nParetoVariable distributed random var\n\nThis class supports the creation of objects that return random numbers from a fixed pareto distribution. It also supports the generation of single random numbers from various pareto distributions.\n\nThe probability density function is defined over the range [",
null,
",+inf) as:",
null,
"where",
null,
"is called the location parameter and",
null,
"is called the pareto index or shape.\n\nThe parameter",
null,
"can be infered from the mean and the parameter",
null,
"with the equation",
null,
".\n\nParetoVariable x (3.14);\nx.GetValue (); //will always return with mean 3.14\n\nDefinition at line 290 of file random-variable.h.\n\nConstructor & Destructor Documentation\n\n ns3::ParetoVariable::ParetoVariable ( double m )\nexplicit\n\nConstructs a pareto random variable with specified mean and shape parameter of 1.5.\n\nParameters\n m Mean value of the distribution\n\nDefinition at line 786 of file random-variable.cc.\n\nReferences NS_LOG_FUNCTION.\n\n ns3::ParetoVariable::ParetoVariable ( double m, double s )\n\nConstructs a pareto random variable with the specified mean value and shape parameter. Beware, s must be strictly greater than 1.\n\nParameters\n m Mean value of the distribution s Shape parameter for the distribution\n\nDefinition at line 791 of file random-variable.cc.\n\nReferences NS_LOG_FUNCTION.\n\n ns3::ParetoVariable::ParetoVariable ( double m, double s, double b )\n\nConstructs a pareto random variable with the specified mean value, shape (alpha), and upper bound. Beware, s must be strictly greater than 1.\n\nSince pareto distributions can theoretically return unbounded values, it is sometimes useful to specify a fixed upper limit. Note however when the upper limit is specified, the true mean of the distribution is slightly smaller than the mean value specified.\n\nParameters\n m Mean value s Shape parameter b Upper limit on returned values\n\nDefinition at line 796 of file random-variable.cc.\n\nReferences NS_LOG_FUNCTION.\n\n ns3::ParetoVariable::ParetoVariable ( std::pair< double, double > params )\n\nConstructs a pareto random variable with the specified scale and shape parameters.\n\nParameters\n params the two parameters, respectively scale and shape, of the distribution\n\nDefinition at line 801 of file random-variable.cc.\n\nReferences NS_LOG_FUNCTION.\n\n ns3::ParetoVariable::ParetoVariable ( std::pair< double, double > params, double b )\n\nConstructs a pareto random variable with the specified scale, shape (alpha), and upper bound.\n\nSince pareto distributions can theoretically return unbounded values, it is sometimes useful to specify a fixed upper limit. Note however when the upper limit is specified, the true mean of the distribution is slightly smaller than the mean value specified.\n\nParameters\n params the two parameters, respectively scale and shape, of the distribution b Upper limit on returned values\n\nDefinition at line 806 of file random-variable.cc.\n\nReferences NS_LOG_FUNCTION.\n\nThe documentation for this class was generated from the following files:"
] | [
null,
"https://coe.northeastern.edu/research/krclab/crens3-doc/ns-3-inverted-notext-small.png",
null,
"https://coe.northeastern.edu/research/krclab/crens3-doc/closed.png",
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null,
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null,
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.619963,"math_prob":0.9108481,"size":4419,"snap":"2022-05-2022-21","text_gpt3_token_len":989,"char_repetition_ratio":0.17395243,"word_repetition_ratio":0.5537459,"special_character_ratio":0.21271782,"punctuation_ratio":0.17539267,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9811597,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,1,null,1,null,1,null,3,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-26T06:18:55Z\",\"WARC-Record-ID\":\"<urn:uuid:08ae4f03-55f0-41a9-a280-80be850db481>\",\"Content-Length\":\"29204\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2fd50533-973b-4657-ba46-43a2ced6627c>\",\"WARC-Concurrent-To\":\"<urn:uuid:cac975f1-e386-4ede-a1ca-6994d5f88964>\",\"WARC-IP-Address\":\"129.10.32.107\",\"WARC-Target-URI\":\"https://coe.northeastern.edu/research/krclab/crens3-doc/classns3_1_1_pareto_variable.html\",\"WARC-Payload-Digest\":\"sha1:CIR4YM2LLY5S26WT4SIOD5VZLJJBJU2D\",\"WARC-Block-Digest\":\"sha1:J4Y3XFQHXL5DNJ4PZAIHEDCGL7WYDB7N\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304915.53_warc_CC-MAIN-20220126041016-20220126071016-00579.warc.gz\"}"} |
https://studylib.net/doc/7939706/adding--subtracting-and-multiplying-matrices | [
"",
null,
"A. A Matrix is a rectangular array of numbers, in other words, numbers in rows and\ncolumns.\nB. The Order of a Matrix is its size or dimensions. The order is always given as the\nnumber of rows by the number of columns ( R x C ).\n[\n]\n[\nThis matrix is a 3 x 2 matrix.\nrows\nby\ncolumns\n]\n2 x 5 matrix\nC. For two matrices to be added or subtracted, the dimensions must be the same. If\nthey are the same, then the corresponding entries are added or subtracted whichever\nthe operation.\n+\n=\nExample:\n[\n]+[\n]= [\n]\nD. For two matrices to be multiplied, their dimensions need to be analyzed to\ndetermine if it is possible. The number of columns of the first matrix MUST EQUAL\nthe number of rows of the second matrix.\nExample:\n[\n3x2\n] [\nand\n]\n2x5\nThey are the same so we can multiply these two matrices.\nThe outside numbers tell the dimensions or the order of the resulting matrix.\n3x2\nand 2 x 5\nThe answer will be a 3 x 5 matrix.\nThe position of each element (row , column) in the answer is a clue to how to multiply.\n(1,1) (1,2) (1,3) (1,4) (1,5)\n(2,1) (2,2) (2,3) (2,4) (2,5)\n(3,1) (3,2) (3,3) (3,4) (3,5)\nThis entry is in\nthe 1st row and\n5th column so it\nis labeled (1 , 5).\nTo do the multiplication of the two matrices, a calculation must be completed with the\nrow and columns as follows:\nTo obtain each entry in the solution matrix, we will look at the row in the first\nmatrix and the column in the second matrix that correspond to the solution matrix\nentry. So, for the entry that belongs in the solution matrix in the location ( 1, 5 )\nwe will use the 1st row in the first matrix and the 5th column in the second matrix.\n[\n] [\n]\nThis calculation is\nfor the entry in\nthe 1st row, 5th\ncolumn.\n=\nWe will multiply the first entry in each and the second entry in each, and then we\nwill add those two results together:\n8 + -2 = 6\nThis process must be done for each entry in the solution matrix.\nBelow are a few more examples. Then, the final matrix after all calculations are completed.\nCalculating ( 2, 3)\n[\n] [\n18\n]\n+\n20\n=\n= 38\nCalculating ( 3, 4)\n[\n] [\n35\n]\n+\n24\n=\n[\n=\n[\n]\n= 59\nCalculating ( 1, 1)\n[\n] [\n1\n]\n+ -4\n]\n= -3\nThe final answer for this matrix multiplication:\n[\n]"
] | [
null,
"https://s3.studylib.net/store/data/007939706_1-afe36665f6cf364bead7eb048ef4e012-768x994.png",
null
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https://betterlesson.com/lesson/571814/area-in-real-life-with-irregular-polygons?from=mtp_lesson | [
"# Area in Real Life With Irregular Polygons\n\n1 teachers like this lesson\nPrint Lesson\n\n## Objective\n\nSWBAT calculate the area of an irregular polygon using multiplication strategies with arrays.\n\n#### Big Idea\n\nDividing an irregular shape into rectangles helps students find the area of an irregular shape.\n\n## Warm Up\n\n5 minutes\n\nTo begin this lesson I review area models of different types of rectangles found in classroom. This includes the area of the student desktop, a book, tissue box, or math geoboard. I keep the focus on finding the area of squares and rectangles. Using these everyday items gives students a real life connection to the purpose of determining area that we are working on in this lesson. To support students during the warm up, I also review strategies for modeling multiplication for students using diagrams, arrays, and if necessary hands on manipulatives.\n\nWhy do I use additional teaching time to review multiplication strategies? Both area and the understanding of multiplication and division strategies are Grade 3 Critical Areas. Critical Areas in the Common Core focus teaching and learning of the entire grade level math content. Within the description of Critical Area 3. Developing understanding of the structure of rectangular arrays and of area, the explanation specifically states \"students connect area to multiplication, and justify using multiplication to determine the area of a rectangle\".\n\nStudent compare their area results and make sure the information includes square measurement labels such as 30 square inches or 120 square centimeters. These labels are critical to their understanding of area.\n\nThis lesson also gives students the opportunity to practice using measuring tools, including rulers and tape measures (MP5).\n\n## Mini Lesson\n\n20 minutes\n\nThe purpose of this lesson is to apply finding the area of a real world setting using spaces familiar to the students. Because the Common Core requires students to make a connection between math and its application in the \"real world\", this lesson provides students with this type of experience.\n\nThe focus of this lesson is finding the area of irregular shaped polygons in real life. To find these areas I take the students to the hallway in our school at locations where they intersect with other hallways and exits to the playground area. We continue the tour of the school to include focusing on doorway entrances into the cafeteria and library where the doorways are set back from the main wall. During this time the students draw some of these models on whiteboards, and I take digital pictures to capture these shapes so that they can be shown once we return to the classroom. These visual models support all students, and they are especially important to the ELL students in my classroom. The shapes we discover include T and L shapes. Students record the shapes on grid paper by matching the grid to the number of square tiles on the floor. If your school does not have a tiled floor, students could use tape measures or meter sticks to measure these areas, or you can assign a numerical value to these spaces.\n\nReturning to the classroom, I ask the students to look at the shapes on their whiteboards and ask them to think about how we can find the area of these different types of shapes. With some discussion, the students begin to realize that each area that has been diagramed includes different types of rectangles and squares. I include in the discussion which measurements would be used for these real life areas and determine if it would be meters or yards compared to inches, feet, and centimeters.\n\nI model for the students how to decompose the irregular polygons into separate rectangles and squares. Once students can see the different rectangles and squares in these shapes, determining the area of each individual polygon is determined, and then added together.\n\nThese types of problems require the students to perform multiple steps to determine the solution. It is important to show students how different lines can be drawn to find different size rectangles in the same figure, but the area does not change. These examples from the school are completed together with the students.\n\n## Try It On Your Own\n\n25 minutes\n\nPracticing on their own, students work together to find the area of a polygon with the outline of stair steps. Students work together to find the length of the lines drawn in to create the separate rectangles. This requires the student to add and subtract from the known lengths. I chose to support students by providing grid paper to find unknown lengths and to check their work. Once students practice with the grid paper, most are able to move to unlined paper to find the area of irregular polygons.\n\nThe shape below is the one provided to the students to compute area. However, I also allow choice, giving students the option of challenging themselves with creating their own shapes by combining rectangles.\n\n## Closure\n\n5 minutes\n\nTo close this lesson, I share with the class some of the students made ShowMe videos which demonstrate strategies used to determine the area of a shape. I ask the students to draw their model, as well as to write their solution in their math journal. I also make sure to include in the discussions a variety of ways to \"see\" these shapes. Using the example in the video, this would be to think of real world shapes besides stairs. The students identify doors, windows, buildings, and Legos as additional examples."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.94772774,"math_prob":0.7754627,"size":5067,"snap":"2021-31-2021-39","text_gpt3_token_len":951,"char_repetition_ratio":0.15860163,"word_repetition_ratio":0.01183432,"special_character_ratio":0.1851194,"punctuation_ratio":0.08017335,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95628047,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-31T23:16:52Z\",\"WARC-Record-ID\":\"<urn:uuid:cbf43bb9-c01b-45ae-9d2b-64806d738de4>\",\"Content-Length\":\"72930\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8a07b5f8-5a00-429d-b6bf-c8f611ed960a>\",\"WARC-Concurrent-To\":\"<urn:uuid:21673c9c-c3e7-4080-b64e-9d848bbb752e>\",\"WARC-IP-Address\":\"52.204.68.164\",\"WARC-Target-URI\":\"https://betterlesson.com/lesson/571814/area-in-real-life-with-irregular-polygons?from=mtp_lesson\",\"WARC-Payload-Digest\":\"sha1:ZGG5RF55UWRN4D4F2RSGOTMKQQ3CSKGA\",\"WARC-Block-Digest\":\"sha1:GQHWCF5DEYWNUOTNQ4QZJ7NGJM3A5ULW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154126.73_warc_CC-MAIN-20210731203400-20210731233400-00594.warc.gz\"}"} |
http://www.wolfram.com/mathematica/newin6/content/ExploratoryDataAnalysis/FindNearestNeighborsInAnySpace.html | [
"Find Nearest Neighbors in Any Space\nMathematica 6 has powerful and efficient nearest-neighbor algorithms, suitable for any dimension and any distance function.\n In:=",
null,
"",
null,
"```Graphics[Module[{r = RandomReal[1, {1000, 2}], nf}, nf = Nearest[r -> Automatic]; GraphicsComplex[r, Table[{Line[Thread[{n, nf[r[[n]], 5]}]], Red, Point[n]}, {n, 1000}]]]]```\n Out=",
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"",
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""
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"http://www.wolfram.com/mathematica/newin6/content/ExploratoryDataAnalysis/HTMLImages/FindNearestNeighborsInAnySpace/In_4.gif",
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"http://www.wolfram.com/mathematica/newin6/images/narrowSpacer.gif",
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https://www.chegg.com/homework-help/definitions/bazin-formula-5 | [
"# Bazin Formula\n\nBazin formula is used to determine the average velocity of the fluid. It is normally used in open channel flow system. It relates the variables velocity, radius with coefficients such as roughness coefficient and discharge coefficient. When the radius and coefficients are known, then the average velocity of the flow can be calculated. This formula can also be utilized to determine discharge of flow.\n\nExpress the Bazin formula to determine the average velocity.",
null,
"Here, coefficient of Chezy is C, coefficient of roughness is K, and radius is R.\n\nExpress the Bazin formula to determine the flow of discharge.",
null,
"Here, coefficient of Bazin is m, acceleration due to gravity is g, width is B, normal head is H, and additional head is Ha.\n\nSee more Mechanical Engineering topics",
null,
"1:00\ntutorial\nWave Equation",
null,
"1:00\ntutorial\nMach Number",
null,
"1:00\ntutorial\nHeat Transfer\n\n## Need more help understanding bazin formula?\n\nWe've got you covered with our online study tools\n\n### Q&A related to Bazin Formula\n\nExperts answer in as little as 30 minutes\n• Q:\nBased on a nominal no-load speed for the DC motor of 5050 RPM at 12VDC, match each input voltage to its corresponding vibration frequency. - V_in - 4V A f = 70.14 Hz B. f = 42.08 Hz - · V_in = 6V V_in= 10V Vuin - 12...\nA:\n• Q:\n4.60. Consider the following two-degree-of-freedom system and compute the respons assuming modal damping rations of = 0.01 and 2 = 0.1 : 20 +(1)X 0 OI L0.05 0 0 0=()x Plot the response.\nA:\n• Q:\nA:\n• Q:\nProblems NASA's Curiosity rover is working (February, 2013) on the Mars surface to collect a 2.1 sample of bedrock that might offer evidence of a long-gone wet environment, as shown in Figure P2-la. In order to provi...\nA:\n• Q:\nDetermine the maximum deflections at Cand D of the A-36 W250x80 beam shown below. points) 40 KN 200 kN-m 78 3 m 7 m\nA:\n• Q:\nUsing graph paper, c s Represent the state of stress on an element oriented 30° counterclockwise from the position shown. 0 = 10 MPa -> T = 30 MPa -> 0= 60 MPa\nA:\n• Q:\nThe S510x143 beam (ASTM A-36) is subjected to the load shown. Determine the equations of the slope and elastic curve. 3 Nm 15 kN.m\nA:\n• Q:\n3. Radiation 3. A farmer raises organic vegetables in front of the wall of his/her house. To harvest solar radiation during the winter (ambient temperature is 0°C), (she put a glass window as shown in the picture. G...\nA:\n• Q:\n2. Consider a 30 cm diameter sphere situated concentrically inside a 60 cm diameter sphere. If the emissivity of the outside sphere is a, = 0.4 and the emissivity of the inside sphere is a = 0.5 determine: (a) the ra...\nA:\n• Q:\n1. Radiation-Stefan Boltzmann Law 1. A farmer has a small greenhouse made from thin sheets of mylar. On clear nights, due to the presence of moisture, ice forms on the inside of the mylar even though the inside and o...\nA:\n• Q:\n1.Explain with neat sketch methods of constructing isometric Drawing 2.Construct an ellipse with concentric circle method 3.A steel pin having a nominal diameter of 30 mm is to be an easy running fit in the bore of...\nA:\n• Q:\n1.Explain the drawing tools of CADD? 2.To draw a parabola with the distance of the focus from the directrix at 50mm 3.Draw the projection of a cone, base 44 mm diameter and axis 50 mm long, when it is resting on the ...\nA:\n• Q:\nA steel shaft is subjected to the loads below. Determine the max shear stress in this shaft and indicate where it occurs. The diameter is 40mm.\nA:\n• Q:\nEXTRA PROBLEM 1: A solid steel shaft of diameter D is supported and loaded as shown in the figure. Determine the critical speed in rpm. Given: D 25 mm, E 210 GPa 15 kg 15 kg E B C D 0.4 m 0.4 m 0.6 m Answer: 566.8 rp...\nA:\n• Q:\n600 N·m ASOS 900 N·m 500 N·m 200 mm HO 300 N·m 200 mm 500 N·m 200 mm 200 mm Figure: 05_FUND-PROB_12 Copyright 2014 Pearson Education, All Rights Reserved\nA:\n• Q:\n2. (6 pts.) X and Y coordinates of a sloped surface is given below. Using the provided data and the best possible method, please calculate the slope at a. X = 0 m b. X= 1.0 m c. X = 2.5 m Please clearly state the rea...\nA:\n• Q:\nQuestion No.1) Explain the production of rotating field in 3-phase induction motors with necessary phasor diagram..? Question No. 2 Explain constructional details of salient pole and round rotor machines\nA:\n• Q:\nEXTRA PROBLEM 1: A solid steel shaft of diameter D is supported and loaded as shown in the figure. Determine the critical speed in rpm. Given: D = 25 mm, E = 210 GPa 15 kg 15 kg MUN DPĆ yy 0.4 m 0.6 m - 0.4 m Answer...\nA:\n• Q:\nProblem Set 2 Chapter 3 – Question 1 (a) Find an expression for the free energy as a function of t of a system with two states, one at energy 0 and one at energy E. (b) From the free energy, find expressions for th...\nA:"
] | [
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"https://cramster-image.s3.amazonaws.com/definitions/DC-1561V2.png",
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"https://img.youtube.com/vi/j5r0F6VhV8c/0.jpg",
null,
"https://img.youtube.com/vi/tUrkEfOXzcg/0.jpg",
null,
"https://img.youtube.com/vi/jsv6wA_peKQ/0.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9311548,"math_prob":0.97400635,"size":1419,"snap":"2019-26-2019-30","text_gpt3_token_len":277,"char_repetition_ratio":0.12155477,"word_repetition_ratio":0.03524229,"special_character_ratio":0.18393235,"punctuation_ratio":0.09541985,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99221945,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-16T12:05:47Z\",\"WARC-Record-ID\":\"<urn:uuid:381eb2a0-5416-4426-8644-596180bb2c96>\",\"Content-Length\":\"234734\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d6ffd147-249d-4047-ad03-231980a93fa3>\",\"WARC-Concurrent-To\":\"<urn:uuid:3db0cf95-6499-4633-8bcc-0a356245782a>\",\"WARC-IP-Address\":\"99.84.104.7\",\"WARC-Target-URI\":\"https://www.chegg.com/homework-help/definitions/bazin-formula-5\",\"WARC-Payload-Digest\":\"sha1:ZQQHFN3TVFMSQ3A7X7763NT3UD5WBMBR\",\"WARC-Block-Digest\":\"sha1:U3GDA7QSJK5WMGS56I6IDX6G4CJ3DF5D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195524548.22_warc_CC-MAIN-20190716115717-20190716141717-00222.warc.gz\"}"} |
https://www.jiskha.com/questions/189143/if-maleic-acid-could-be-represented-as-h2ma-write-the-2-equations-for-its-reaction-with | [
"# chemistry\n\nIf maleic acid could be represented as H2Ma, write the 2 equations for its reaction with NaOH in the space below (showing the sequential neutralization of each acidic proton):\n\n1) H2Ma + NaOH --> H3O + MaNa (molecular)\n\n2) H2Ma + OH- --> H2O + MaH- (net ionic)\n\nNote: the \"-\" is the superscript.\n\n1. 👍 0\n2. 👎 0\n3. 👁 185\n1. 1) H2Ma + NaOH --> H2O + NaHMa\n\n2) NaHMa + NaOH --> H2O + Na2Ma\n\nI think is what they are after so you end up with the disodium salt.\n\n1. 👍 0\n2. 👎 0\nposted by Dr Russ\n\n## Similar Questions\n\n1. ### chemistry\n\nIf maleic acid could be represented as H2Ma, write the 2 net ionic equations for its reaction with NaOH in the space below (showing the sequential neutralization of each acidic proton): 1)H2Ma +NaOH---->NaMa+H3O 2) What's the\n\nasked by Anonymous on February 26, 2009\n2. ### Chemistry\n\nHELP£¡ can someone find identify one chemical reaction that C4H4O4 is involved in. Or the synthesis of the compound? C4H4O4 is a formula that applies to both maleic and fumaric acid. A synthesis reaction of reaction of maleic\n\nasked by Qing on December 18, 2006\n3. ### chemistry\n\nWrite out the mechanism (using correct curved arrow notation) for the acid-catalyzed, cis-trans isomerization of maleic acid (into fumaric acid) If the electrophilic addition reaction of maleic acid with HCl had gone to completion\n\nasked by Allie on December 3, 2014\n4. ### Chemistry\n\nItem 1 Maleic acid is a carbon-hydrogen-oxygen compound used in dyeing and finishing fabrics and as a preservative of oils and fats. •In a combustion analysis, a 1.054-g sample of maleic acid yields 1.599 g of CO2 and 0.327 g of\n\nasked by Samantha on June 16, 2015\n5. ### chem\n\nMaleic acid is an organic compound composed of 41.39% C, 3.47% H, and the rest oxygen. If 0.250 mol of maleic acid has a mass of 29.0 g, what are the empirical and molecular formulas of maleic acid?\n\nasked by casi on June 28, 2010\n6. ### chemistry\n\nwhat is the reaction equation when maleic anhydride is combines with granular zinc and HCl propose structures for the products you would obtain by treating fumaric acid, cinnamic acid, and butynedioic acid with zinc and HCl. write\n\nasked by anon on October 30, 2010\n7. ### science\n\nwhat is the reaction equation when maleic anhydride is combines with granular zinc and HCl propose structures for the products you would obtain by treating fumaric acid, cinnamic acid, and butynedioic acid with zinc and HCl. write\n\nasked by lilian on April 16, 2007\n8. ### biorganic\n\nequations of thereactions of both maleic acid and fumaric acid with (a) bromine (b) hydrogen chloride (c) catalytic hydrogenation.name the products of these reaction\n\nasked by debbie on October 31, 2013\n9. ### Chemistry\n\nh2c4h4o4 + 2naoh --> 2h2o + na2c4h4o4 as for the previous question im sorry i forgot to plug the maleic acid in the formula properly. so this ones it ^ .2213 of c4h4o4 used (maleic acid) and .1 M NaOH 1. show the equation for 1st\n\nasked by Ruby on March 12, 2012\n10. ### Chemistry\n\nMaleic acid is a carbon-hydrogen-oxygen compound used in dyeing and finishing fabrics and as a preservative of oils and fats. * In a combustion analysis, a 1.054-\\rm g sample of maleic acid yields 1.599 \\rm g of \\rm CO_2 and 0.327\n\nasked by Shelby on April 24, 2012\n\nMore Similar Questions"
] | [
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https://www.clutchprep.com/chemistry/practice-problems/140397/for-many-weak-acid-or-weak-base-calculations-you-can-use-a-simplifying-assumptio-1 | [
"# Problem: For many weak acid or weak base calculations, you can use a simplifying assumption to avoid solving quadratic equations. Classify these situations by whether the assumption is valid or the quadratic formula Is required.\n\n###### FREE Expert Solution\n\n1000",
null,
"###### Problem Details\n\nFor many weak acid or weak base calculations, you can use a simplifying assumption to avoid solving quadratic equations.",
null,
"Classify these situations by whether the assumption is valid or the quadratic formula Is required.",
null,
""
] | [
null,
"https://cdn.clutchprep.com/assets/button-view-text-solution.png",
null,
"https://lightcat-files.s3.amazonaws.com/problem_images/0ffa66dde9aecde1-1588263159534.jpg",
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"https://lightcat-files.s3.amazonaws.com/problem_images/42a680a3c2b52b9a-1588263209714.jpg",
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https://math.stackexchange.com/questions/3196500/getting-representations-of-the-lie-group-out-of-representations-of-its-lie-algeb | [
"# Getting representations of the Lie group out of representations of its Lie algebra\n\nThis is something that is usually done in QFT and that bothers me a lot because it seems to be done without much caution.\n\nIn QFT when classifying fields one looks for the irreducible representations of the proper orthochronous Lorentz group $$SO_e^+(1,3)$$.\n\nBut to do so what one does in practice is: look for representations of the Lie algebra $$\\mathfrak{so}(1,3)$$ and then exponentiate.\n\nFor instance, in Peskin's QFT book:\n\nIt is generally true that one can find matrix representations of a continuous group by finding matrix representations of the generators of the group, then exponentiating these infinitesimal transformations.\n\nThe same thing is done in countless other books.\n\nNow I do agree that if we have a representation of $$G$$ we can get one of $$\\mathfrak{g}$$ differentiating at the identity. Here one is doing the reverse!\n\nIn practice what is doing is: find a representation of $$\\mathfrak{so}(1,3)$$ on a vector space $$V$$, then exponentiate it to get a representation of $$SO_e^+(1,3)$$. I think one way to write it would be as follows, let $$D : \\mathfrak{so}(1,3)\\to \\operatorname{End}(V)$$ be the representation of the algebra, define $$\\mathscr{D} : SO_e^+(1,3)\\to GL(V)$$\n\n$$\\mathscr{D}(\\exp \\theta X)=\\exp \\theta D(X).$$\n\nNow, this seems to be very subtle.\n\nIn general the exponential $$\\exp : \\mathfrak{g}\\to G$$ is not surjective. Even if it is, I think it need not be injective.\n\nAlso I've heard there is one very important and very subtle connection between $$\\exp(\\mathfrak{g})$$ and the universal cover of $$G$$.\n\nMy question here is: how to understand this procedure Physicists do more rigorously? In general this process of \"getting representations of $$G$$ out of representations of $$\\mathfrak{g}$$ by exponentiation\" can be done, or it really just gives representations of $\\exp(\\mathfrak{g})? Or in the end physicists are allowed to do this just because very luckilly in this case $$\\exp$$ is surjective onto $$SO_e^+(1,3)$$? Edit: I think I got, so I'm going to post a summary of what I understood to confirm it: Let $$G$$ be a Lie group. All representations of $$G$$ give rise to representations of $$\\mathfrak{g}$$ by differentiation. Not all representations of $$\\mathfrak{g}$$ come from derivatives like this, however. These representations of $$\\mathfrak{g}$$ come from derivatives of representations of the universal cover of $$G$$, though. Then when $$G$$ is simply connected, all representations of $$\\mathfrak{g}$$ indeed come from $$G$$ as derivatives. Now, if we know the representations of $$\\mathfrak{g}$$ we can determine by exponentiation the representations of the universal cover $$\\tilde{G}$$ of $$G$$ from which they are derived by exponentiation. This determines them in a neigbhorhood of the identity. For the representations of $$\\mathfrak{g}$$ that indeed come from $$G$$, if $$G$$ is connected, then a neigbhorhood of the identity generates it, so that this is enough to reconstruct the representation everywhere. Nevertheless, in the particular case of $$SO_e^+(1,3)$$ it so happens that this neighborhood of the identity reconstructed by the exponential is the whole group. Finally the representations of $$\\mathfrak{so}(1,3)$$ which do not come from $$SO_e^+(1,3)$$ come from the universal cover $$SL_2(\\mathbb{C})$$. Is this the whole point? ## 1 Answer The exponential map doesn't need to be surjective. If $$G$$ is connected the exponential map is surjective onto a neighborhood of the identity, and since a neighborhood of the identity of a connected topological group generates it, once you know what a representation does to a neighborhood of the identity, that determines what it does everywhere. However, in general $$G$$ needs to be simply connected. That is, exponential in general provides an equivalence between representations of a finite-dimensional Lie algebra $$\\mathfrak{g}$$ and representations of the unique simply connected Lie group $$G$$ with Lie algebra $$\\mathfrak{g}$$. The proper orthochronous Lorentz group is not simply connected; its universal cover is $$SL_2(\\mathbb{C})$$. This means that not all representations of $$\\mathfrak{so}(1, 3)$$ exponentiate to representations of the proper orthochronous Lorentz group; some exponentiate to projective representations. As far as I know this is mostly fine for quantum, and so physicists don't seem to worry much about the distinction in practice. • There's certainly also the issue of not-finite-dimensional representations... Wallach's and Casselman's \"globalization\" functors show two opposite extremes of adjoints to the functor that takes$G$repns$V$to$\\mathfrak g,K$modules of smooth vectors$V^\\infty\\$. – paul garrett Apr 22 at 1:52\n• Thanks very much @QiaochuYuan, I think I finally got it. I posted one edit with a summary of what I understood of this matter. Could you please tell me if I got it right or if I misunderstood something? Thanks very much again! – user1620696 Apr 22 at 3:18"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9123512,"math_prob":0.997699,"size":3232,"snap":"2019-13-2019-22","text_gpt3_token_len":821,"char_repetition_ratio":0.20136307,"word_repetition_ratio":0.012121212,"special_character_ratio":0.24350247,"punctuation_ratio":0.10327869,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99980944,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-25T23:29:16Z\",\"WARC-Record-ID\":\"<urn:uuid:1ed41120-cf91-4801-abb3-c9b6f0741311>\",\"Content-Length\":\"137706\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:91bb51a9-183a-4e4a-821e-26ade5c6e282>\",\"WARC-Concurrent-To\":\"<urn:uuid:b765b853-d096-4fe1-8cb6-22bde5b0f016>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3196500/getting-representations-of-the-lie-group-out-of-representations-of-its-lie-algeb\",\"WARC-Payload-Digest\":\"sha1:L36EU3CTISWOCJ3QHW2XLMMBAJ2A53QG\",\"WARC-Block-Digest\":\"sha1:WTVSVD4MAIRGJORCTDYSQDYPN6UO3VCF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232258453.85_warc_CC-MAIN-20190525224929-20190526010929-00272.warc.gz\"}"} |
https://astronomy.stackexchange.com/questions/tagged/mercury?tab=Unanswered | [
"# Questions tagged [mercury]\n\nQuestions about Mercury, the first planet from Earth's sun. Also the smallest major planet in the Solar System.\n\n4 questions with no upvoted or accepted answers\nFilter by\nSorted by\nTagged with\n92 views\n\n### How did Arecibo make radar images of ice on Mercury's poles?\n\nupdate: I still haven't been able to get my hands on the Icarus paper linked below (I'll try other libraries) but these are newer and quite interesting!: Constraining the thickness of polar ice ...\n47 views\n\n### How to write a normalized, dimensionless form of Binet's relativistic equation?\n\n$$\\frac{\\delta^2 u}{\\delta \\theta^2} + u = \\frac{\\mu}{h^2} + 3 \\mu u^2$$ This is Binet's relativistic equation. Here $u = 1/r$ where $r$ is the radial distance, and $\\mu$ is the reduced mass of the ..."
] | [
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https://archived.hpcalc.org/museumforum/thread-257167-post-257224.html | [
"Solver issue with HP 17BII - different from 19BII « Next Oldest | Next Newest »\n\n ▼",
null,
"Jeff Kearns",
null,
"Member Posts: 222 Threads: 21 Joined: Oct 2006 11-27-2013, 11:31 AM I recently acquired a very nice HP 17BII (Singapore 1994), and like it very much - especially now that I have functions for TRIG and Inverse TRIG! I am not yet proficient in the use of the solver or its many interesting features that qualify it as a pseudo-programmable calculator with the L() and G() functions (LET and GET), IF structures and Sigma functions. However, I have noticed while copying some solver equations from my HP 19BII over to the 17BII, that it does not accept the same equation syntax as the 19BII, despite what Valentin Albillo quoted in a 2003 discussion: Quote: The solver in this machine is exactly the same as that of the 19bii and includes all 19bii functions even if not documented, including Let and Get. The following equation for Canadian mortgages (right out of the 19BII Owner's Manual): CAN~MORT: FV(N:((1+CI%YR/200)^(1/6)-1)*1200:PV:PMT:12:0)=FV displays a standard TVM menu: \"N\" \"CI%YR\" \"PV\" \"PMT\" \"FV\". Unfortunately, the 17BII will not accept this equation! If gives an error (INVALID EQUATION) at the first bracket - after FV. Instead, I have had to use this equation from the 17BII manual: CAN~MORT: PV=-PMTxUSPV(((1+I%YR/200)^(1/6)-1)x100:N)-FVxSPPV(((1+I%YR/200)^(1/6)-1)x100:N) The problem with this equation (even though it works and I could re-arrange the terms... but that's not the point) is the use of USPV and SPPV functions which are not needed as well as the order of the variables: \"PV\" \"PMT\" \"I/YR\" \"N\" \"FV\". How can I make the 19BII equation work in the 17BII? Thanks, Jeff Edited: 27 Nov 2013, 11:43 a.m. ▼",
null,
"Neil Hamilton (Ottawa)",
null,
"Senior Member Posts: 255 Threads: 22 Joined: May 2011 11-27-2013, 01:04 PM Hi Jeff, The 17Bii does not have the entire suite of solver functions available in the 19B*. At one point I had a comparitive list of all the solver functions in the 17B*, 19B*, and 27S. However, I cannot seem to locate it at the moment. Maybe I compiled it myself from the various manuals -- don't recall. For the Canadian mortgage solver, you need to use equations presented in the 17B* manual. As you no doubt found, the 17B* is also missing many of the 19B*'s mathematical functions as well. ▼",
null,
"Jeff Kearns",
null,
"Member Posts: 222 Threads: 21 Joined: Oct 2006 11-27-2013, 01:26 PM Hi Neil, Thanks for your response. I am aware of the meager subset of math functions in the 17BII. Added Trig and inverse Trig functions (although no hyperbolics) make it quite useful for an engineer. I now understand why it cannot accept the relatively simple-looking equation for Canadian mortgages presented in the 19BII manual: while I get the lack of trigonometric functions, the 19BII solver TVM function FV(n:i%yr:pv:pmt:p/yr:m) is surprisingly absent from the 17BII list. Bottom line is: the solvers are definitely not identiacal. Jeff",
null,
"Don Shepherd",
null,
"Posting Freak Posts: 1,392 Threads: 142 Joined: Jun 2007 11-27-2013, 01:21 PM Quote: How can I make the 19BII equation work in the 17BII? Well, you can't, since the 17bii solver does not support the FV function (which is why the error occurs right where it did). Valentin is correct about L() and G(), but there are several 19bii solver functions that are not in the 17bii solver (including random number, unfortunately). About all you can do is figure out exactly what the FV function does on the 19bii, and duplicate that in your equation.",
null,
"Gerson W. Barbosa",
null,
"Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 11-27-2013, 02:31 PM Quote: I recently acquired a very nice HP 17BII (Singapore 1994), and like it very much - especially now that I have functions for TRIG and Inverse TRIG! The following might be faster - once verified - and accurate to at least eleven digits: ```IF(S(SIN):(-1)^INT(X/180+0*L(X:MOD(X:180)*PI/180))*(2*SIGMA(K :1:21:4:X^K/FACT(K))-(EXP(X)-EXP(-X))/2)-SIN:IF(S(COS):COS-(- 1)^INT(X/180+0*L(X:MOD(X:180)*PI/180))*(2*SIGMA(K:4:20:4:X^K/ FACT(K))+2-(EXP(X)+EXP(-X))/2):IF(S(TAN):TAN-(0*L(X:MOD(X:180 )*PI/180)+4*SIGMA(K:1:21:4:X^K/FACT(K))-EXP(X)+EXP(-X))/(4* SIGMA(K:4:20:4:X^K/FACT(K))+4-EXP(X)-EXP(-X)):IF(S(R~D):X*180 /PI-R~D:IF(S(D~R):X*PI/180-D~R:IF(S(ASIN):ASIN-(0*(L(B:SGN(X) +IP(X))+L(X:ABS(X))+L(X:IF(X<>1:X/SQRT(1-SQ(X)):X))+L(X:IF(X< 1:L(A:1)*X+L(Q:0):0*(L(Q:PI)+L(A:-1))+INV(X)))+L(X:IF(X>SQRT( 2)-1:0*L(V:PI/2)+(X-1)/(X+1):X+L(V:0))))+(G(Q)+G(A)*(4*SIGMA( K:1:29:4:X^K/K)-LN((1+X)/(1-X))+G(V)))*G(B)*90/PI):IF(S(ACOS) :ACOS-(0*(L(X:IF(X<>-1:SQRT((1-X)/(1+X)):X))+L(B:SGN(X))+L(X: ABS(X))+L(X:IF(X<1:L(A:1)*X+L(Q:0):0*(L(Q:PI)*L(A:-1))+INV(X) ))+L(X:IF(X>SQRT(2)-1:0*L(V:PI/2)+(X-1)/(X+1):X+L(V:0))))+(G( Q)+G(A)*(4*SIGMA(K:1:29:4:X^K/K)-LN((1+X)/(1-X))+G(V)))*(3-G( B))*90/PI):ATAN-(0*(L(B:SGN(X))+L(X:ABS(X))+L(X:IF(X<1:L(A:1) *X+L(Q:0):0*(L(Q:PI)+L(A:-1))+INV(X)))+L(X:IF(X>SQRT(2)-1:0*L (V:PI/2)+(X-1)/(X+1):X+L(V:0))))+(G(Q)+G(A)*(4*SIGMA(K:1:29:4 :X^K/K)-LN((1+X)/(1-X))+G(V)))*G(B)*90/PI)))))))) Running times: SIN: 1.8 s COS: 1.6 s TAN: 2.6 s R~D: 2.0 s D~R: 2.7 s ASIN: 5.2 s ACOS: 8.1 s ATAN: 10.9 s ``` That's the equation in message #22 here. Still in my HP-17BII, but I don't think I would key all that in if somehow it got erased from memory :-) Gerson. ▼",
null,
"Jeff Kearns",
null,
"Member Posts: 222 Threads: 21 Joined: Oct 2006 11-27-2013, 02:47 PM Gerson, While I do not doubt the run-times, the mere fact that it is 48% longer (character count) than the three routines from Maguire make it unlikely I will spend an hour (or more) entering these 1145 characters to save a fraction of a second. Great work though... ;-) Cheers, Jeff ▼",
null,
"Thomas Klemm",
null,
"Senior Member Posts: 735 Threads: 34 Joined: May 2007 11-27-2013, 03:11 PM Guess how long it took me to enter the equation for the 8-queens problem? Quote: The problem with this equation (...) is (...) the order of the variables: \"PV\" \"PMT\" \"I/YR\" \"N\" \"FV\". You could probably add the following: ```(N+CI%YR+PV+PMT+FV)*0+... ``` The variables are ordered according to their appearance in the equation. I can't check that right now but IIRC that's why I started with: ```QUEENS: Q=A+B+C+D+E+F+G+H+ (...) ``` Cheers Thomas",
null,
"Gerson W. Barbosa",
null,
"Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 11-27-2013, 03:14 PM \"Only\" 1117 characters, considering SIGMA is one-character long on the HP-17BII :-) These are shorter, but are not combined into a single equation. Another SOLVER option, more akin to that on the HP-19BII, is the HP-200LX Solver. Regards, Gerson.",
null,
"Jeff Kearns",
null,
"Member Posts: 222 Threads: 21 Joined: Oct 2006 11-27-2013, 02:57 PM Gerson, The ATAN equation from the W.B. Maguire link executes in about 5 seconds. Jeff ▼",
null,
"Gerson W. Barbosa",
null,
"Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 11-27-2013, 03:17 PM The delay is due to the ATAN position in the IF-structure. ATAN as an independent function would take no more than three seconds, I think. Gerson.",
null,
"Kimberly Thompson",
null,
"Member Posts: 97 Threads: 1 Joined: Jun 2013 11-27-2013, 03:05 PM Jeff NO argument w explanations already provided, however the TVM MENU QUESTIONS at the following URL ... http://h20565.www2.hp.com/portal/site/hpsc/template.PAGE/public/psi/mostViewedDisplay/?sp4ts.oid=33540&spf_p.tpst=psiContentDisplay &spf_p.prp_psiContentDisplay=wsrp-navigationalState%3DdocId%253Demr_ na-bpia5125-1%257CdocLocale%253Den_US&javax.portlet.begCacheTok= com.vignette.cachetoken&javax.portlet.endCacheTok=com.vignette.cachetoken ... may be useful in further TVM-Solver scenarios for the 17 & 19 model business/finance calculators. ```Determining if the TVM menu can be used with the Solver (HP 19bii calculator) The TVM menu cannot be used with the Solver, but the TVM Solver functions can be used to do the same calculations. The TVM Functions The five Solver TVM functions allow for equations that do calculations analogous to the calculations done in the TVM menu: N ( I%yr : pv : pmt : fv : p/yr : m ) I %yr ( n : pv : pmt : fv : p/yr : m ) pv ( n : i%yr : pmt : fv : p/yr : m ) PMT ( n : i%yr : pv : fv : p/yr : m ) FV ( n : i%yr : pv : pmt : p/yr : m ) Each function calculates one TVM value, given the values for all the others. The parameters of the functions (the contents of the parentheses) are defined identically to the built-in TVM variables described in the following table, except that m stands for BEGIN/END mode. Use m=1 for BEGIN mode and m=0 for END mode. For example, the first function calculates N (the total number of payments or compounding periods), given the annual percentage interest rate, present value, payment amount, future value, number of payments per year, and the BEGIN/END mode. Give the parameters any legal variable name; for example use LOAN in place of pv. Parameters can also be algebraic expressions. For example, the following equation calculates the monthly payment for a car loan: CARPMT = PMT (MONTHS : I% YR : PRICE-DOWN : 0 : 12 : 0) Where MONTHS is the duration of the loan (in months), PRICE is the purchase price, and DOWN is the down payment (pv = PRICE-DOWN, n = months, and the last 0 = END mode). Notice that PMT is not a variable in the equation--it is the name of the function. The solver TVM variables are not shared with the variables in the TVM menu. For example, the variable I%YR in the CARPMT equation is separate from the TVM variable I%YR.``` BLUF - the TVM functions are special when used in the SOLVER. BEST! SlideRule",
null,
"Don Shepherd",
null,
"Posting Freak Posts: 1,392 Threads: 142 Joined: Jun 2007 11-27-2013, 08:03 PM Jeff, why not just use the formula that is in the 17bII manual? If it works, as you say it does, and you can easily change the order of the arguments, why the objection to USPV and SPPV? I'm just curious. ▼",
null,
"Jeff Kearns",
null,
"Member Posts: 222 Threads: 21 Joined: Oct 2006 11-27-2013, 09:47 PM Don, That is exactly what I decided to do. I was unaware of the FV(...) function issue in the 17BII. As for USPV and SPPV, I have no objection. I just did not understand initially why they were used instead of the more logical 19BII solver functions. Thanks, Jeff ▼",
null,
"Don Shepherd",
null,
"Posting Freak Posts: 1,392 Threads: 142 Joined: Jun 2007 11-28-2013, 02:36 AM OK, thanks Jeff.\n\n Possibly Related Threads... Thread Author Replies Views Last Post hp-prime solver and variable name fabrice48 22 5,647 12-10-2013, 03:25 AM Last Post: fabrice48 HP Prime Triangle solver BruceH 29 5,893 11-28-2013, 12:03 AM Last Post: Dale Reed 17BII & 17BII+ Discounted Payback Period Revisited Tom Neudorfl 8 2,007 11-25-2013, 10:28 AM Last Post: Don Shepherd HP prime: linear solver app Alberto Candel 1 977 11-21-2013, 01:57 AM Last Post: Michael Carey HP Prime: Linear Solver app bug BruceH 0 708 11-15-2013, 06:36 PM Last Post: BruceH Another minor Prime hexagesimal issue Jonathan Cameron 1 920 11-08-2013, 02:37 PM Last Post: Michael de Estrada HP Prime Solver Variables Issue Anibal Morones Ruelas 8 2,103 10-19-2013, 09:45 AM Last Post: Harold A Climer Is the Prime a superset of the HP 17bII+ ? vrrr 3 1,230 10-16-2013, 12:03 PM Last Post: Michael de Estrada HP Prime triangle solver oddity BruceH 0 702 10-13-2013, 09:08 PM Last Post: BruceH Using units in Numeric Solver Harold A Climer 1 871 10-13-2013, 10:44 AM Last Post: Tim Wessman\n\nForum Jump:",
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https://www.hindawi.com/journals/aaa/2013/950134/ | [
"#### Abstract\n\nThis paper is concerned with partial regularity to nonlinear subelliptic systems with Dini continuous coefficients under quadratic controllable growth conditions in the Heisenberg group . Based on a generalization of the technique of -harmonic approximation introduced by Duzaar and Steffen, partial regularity to the sub-elliptic system is established in the Heisenberg group. Our result is optimal in the sense that in the case of Hölder continuous coefficients we establish the optimal Hölder exponent for the horizontal gradients of the weak solution on its regular set.\n\n#### 1. Introduction and Statements of Main Results\n\nIn this paper, we are concerned with partial regularity of weak solutions to nonlinear sub-elliptic systems of equations of second order in the Heisenberg group in divergence form, and more precisely, we consider the following systems: where is a bounded domain in , , the definition of is to be seen in the next section (11), , , and .\n\nUnder the coefficients assumed to be Dini continuous, the aim of this paper is to establish optimal partial regularity to the sub-elliptic system (1) in the Heisenberg group . Comparing Hölder continuous coefficients (see [1, 2] for the case of sub-elliptic systems), such assumption is weaker. More precisely, we assume for the continuity of with respect to the variables that for all , , and , where is monotone nondecreasing and is monotone nondecreasing and concave with . We also required that be nonincreasing for some and that\n\nWe adopt the method of -harmonic approximation to the case of sub-elliptic systems in the Heisenberg groups and establish the optimal partial regularity result. Roughly speaking, assume additionally to the standard hypotheses (see precisely (H1), (H2), and (H4)) that satisfies (2) and (3). Let be a weak solution of the sub-elliptic system (1). Then, is of class outside a closed singular set Sing of the Haar measure . Furthermore, for , the derivative of has the modulus of continuity in a neighborhood of . Our result is optimal in the sense that in the case , , we have Hölder continuity to be optimal in that case.\n\nAs is well known, even under reasonable assumptions on and of the systems of equations, one cannot, in general, expect that weak solutions of (1) will be classical, that is, -solutions. This was first shown by de Giorgi ; we also refer the reader to Giaquinta and Chen and Wu for further discussion and additional examples. Then, the goal is to establish partial regularity theory. Moreover, a new method called -harmonic approximation technique is introduced by Duzaar and Steffen in and simplified by Duzaar and Grotowski in to study elliptic systems with quadratic growth case. Then, similar results have been proved for more general or in the Euclidean setting; see for Hölder continuous coefficients and for Dini continuous coefficients.\n\nHowever, turning to sub-elliptic equations and systems in the Heisenberg groups , some new difficulties will arise. Already in the first step, trying to apply the standard difference quotient method, the main difference between Euclidean and Heisenberg groups becomes clear. Any time we use horizontal difference quotients (i.e., in the directions ), extra terms with derivatives in the direction will arise due to noncommutativity (see (12)), but these cannot be controlled by using the initial assumptions on the weak solution. Several results were focused on those equations which have a bearing on basic vector fields on the Heisenberg group or, more generally, the Carnot group. Capogna [16, 17] studied the regularities for weak solutions to quasi-linear equations. Concretely, by a technique combining fractional difference quotients and fractional derivatives defined by Fourier transform, differentiability in the nonhorizontal direction, estimate, and continuity of weak solutions are obtained; see for the case of Heisenberg groups and for Carnot groups. To sub-elliptic -Laplace equations in Heisenberg groups, Marchi in showed that and for by using theories of Besov space and Bessel potential space. Domokos in [21, 22] improved these results for employing the A. Zygmund theory related to vector fields. Recently, by meticulous arguments, Manfredi and Mingione in and Mingione et al. in proved Hölder regularity with regard to full Euclidean gradient for weak solutions and further continuity under the coefficients assumed to be smooth.\n\nWhile regularities for weak solutions to sub-elliptic systems concerning vector fields are more complicated, Capogna and Garofalo in showed the partial Hölder regularity for the horizontal gradient of weak solutions to quasilinear sub-elliptic systems with , being horizontal vector fields in Carnot groups of step two, where and satisfy the quadratic structure conditions. Their way relies mainly on generalization of classical direct method in the Euclidean setting. Shores in considered a homogeneous quasi-linear system in the Carnot group with general step, where also satisfies the quadratic growth condition. She established higher differentiability and smoothness for weak solutions of the system with constant coefficients and deduced partial regularity for weak solutions of the original system. With respect to the case of nonquadratic growth, Föglein in treated the homogeneous nonlinear system in the Heisenberg group under superquadratic structure conditions. She got a priori estimates for weak solutions of the system with constant coefficients and partial regularity for the horizontal gradient of weak solutions to the initial system. Later, Wang and Niu and Wang and Liao treated more general nonlinear sub-elliptic system in the Carnot groups under superquadratic growth conditions and subquadratic growth conditions, respectively.\n\nThe regularity results for sub-elliptic systems mentioned above require Hölder continuity with respect to the coefficients . When the assumption of Hölder continuity on is weakened to Dini continuity, how to establish partial regularity of weak solutions to nonlinear sub-elliptic systems in the Heisenberg group. This paper is devoted to this topic. To define weak solution to (1), we assume the following structure conditions on and .(H1) is differentiable in , and there exist some constants such that Here, we write down . (H2) is uniformly elliptic; that is, for some , we have (H3) There exist a modulus of continuity and a nondecreasing function such that (H4) satisfies quadratic controllable growth condition where because ; see (16).\n\nWithout loss of generality, we can assume that and the following.() is nondecreasing with .() is concave; in the proof of the regularity theorem, we have to require that is nonincreasing for some exponent . We also require Dini's condition (2) which was already mentioned in the introduction.() for some .\n\nIn the present paper, we will apply the method of -harmonic approximation adapting to the setting of Heisenberg groups to study partial regularity for the system (1). Since basic vector fields of Lie algebras corresponding to the Heisenberg group are more complicated than gradient vector fields in the Euclidean setting, we have to find a different auxiliary function in proving Caccioppoli type inequality. Besides, the nonhorizontal derivatives of weak solutions will happen in the Taylor type formula in the Heisenberg group and cannot be effectively controlled in the present hypotheses. So, the method employing Taylor's formula in is not appropriate in our setting. In order to obtain the desired decay estimate, we use the Poincaré type inequality in as a replacement. And we obtain the following main result.\n\nTheorem 1. Assume that coefficients and satisfy (H1)–(H4), (μ1)(μ3) and that is a weak solution to the system (1); that is, Then, there exists a relatively closed set such that . Furthermore, and Haar meas , where In addition, for and , the derivative has the modulus of continuity in a neighborhood of .\n\n#### 2. Preliminaries\n\nThe Heisenberg group is defined as endowed with the following group multiplication: for all , . This multiplication corresponds to addition in Euclidean . Its neutral element is , and its inverse to is given by . Particularly, the mapping is smooth, so is a Lie group.\n\nThe basic vector corresponding to its Lie algebra can be explicitly calculated by the exponential map and is given by for , and note that the special structure of the commutators: that is, , is a nilpotent Lie group of step . is called the horizontal gradient and the vertical derivative.\n\nThe pseudonorm is defined by and the metric induced by this pseudonorm is given by The measure used on is Haar measure, and the volume of the pseudoball is given by The number is called the homogeneous dimension of .\n\nThe horizontal Sobolev spaces are defined as Then, is a Banach space with the norm is the completion of under norm (18).\n\nLu showed the following Poincaré type inequality related to Hörmander's vector fields for , , : where we write down here and there. Note the fact that the horizontal vectors defined in (11) fit Hörmander's vector fields and that (19) is valid for .\n\nFollowing , for technical convenience, letting , we have the corresponding properties for : () is continuous, nondecreasing and ; () is concave, and is nonincreasing for some exponent ; () for some . Changing by a constant, but keeping , we may assume the following: () , implying for . Also note that it implies that from () and (), for all .\n\nFurthermore, the following inequality holds: The condition (H3) becomes Moreover, we deduce the existence of a nonnegative modulus of continuity with for all such that is nondecreasing with respect to for fixed and is concave and nondecreasing with respect to for fixed . Also, we have for , Using (H1) and (H2), we see that\n\nIn the sequel, the constant may vary from line to line.\n\n#### 3. Caccioppoli Type Inequality\n\nIn this section, we present the following -harmonic approximation lemma in the Heisenberg group introduced by Föglein with as a special case and prove a Caccioppoli type inequality in our setting.\n\nLemma 2. Let and be fixed positive numbers and with . If for any given , there exists with the following properties:(I)for any satisfying (II) for any satisfying then, there exists an -harmonic function such that\n\nFöglein established a priori estimate for the weak solution to homogeneous sub-elliptic systems with constant coefficients in the Heisenberg group (also see for Carnot groups of step ). We list it as follows: In what follows, we let and for . Note that and that , are nonincreasing functions.\n\nLemma 3. Let be a weak solution to the system (1) under the conditions (H1)(H4), (μ1)(μ3). Then, for every , , , and such that , the inequality holds, where is the horizontal component of and\n\nProof. Let . Take a test function in (8) with satisfying , , and on . Then, we have , , and Adding this to the equations It follows that by using the hypotheses (H1), (H3) (i.e., (23), (21), resp.), and (H4), where Applying (H2), the left hand side of (33) can be estimated as For to be fixed later, we have, using Young's inequality, Using Jensen's inequality, (20), and the fact that for , we arrive at where is an abbreviation of the function . Also, note that the application of (20) in the second last inequality is possible by our choice .\nUsing Young's inequality and (37) in , we obtain And similarly, we see Here we have used in the last inequality.\nBy Hölder's inequality, (19), and Young's inequality, one gets where we have used the fact that .\nApplying these estimates to (37), we obtain Choosing , we obtain the desired inequality (29).\n\n#### 4. Proof of the Main Theorem\n\nIn this section, we will complete the proof of the partial regularity results via the following lemmas. In the sequel, we always suppose that is a weak solution to (1) with the assumptions of (H1)–(H4) and ()–().\n\nLemma 4. Let with and satisfying and . Then, there exists a constant such that\n\nProof. Using the fact that and the weak form (8), we deduce It yields Using (22), Hölder's inequality, the fact that is concave, and Jensen's inequality, we have Similarly, using (21) and the fact that for , we obtain where we have used the fact that which follows from the nondecreasing property of the function , , and our assumption .\nIn the same way, it follows that by using (21), (37), and (19), Using Hölder's inequality, (19), and Young's inequality, we have where we have used the assumption and the fact that and . Combining these estimates, we obtain the conclusion with .\n\nLemma 5. Assume that the conditions of Lemma 2 and the following smallness conditions hold: with , together with Then, the following growth condition holds for where one abbreviates and with .\n\nProof. We define , where Then, we have . Now, we consider such that . Applying Lemma 4 on to , we have for any , In consideration of the small condition (49), we see that (54) and (55) imply conditions (26) in Lemma 2. Also note that (H1) and (H3) imply condition (25). So, there exists an -harmonic function such that Taking and replacing by , we use Lemma 3 to obtain where Using the fact that has mean value on the ball , the definition of , and (19), we have\nwhere . Note that in the second last inequality we have used the fact that In consideration of the fact that , and the assumptions and , it follows that\nLet with . Combining these estimates (57)–(61) and considering the small condition (51) (it implies ; see (64) and (65)), we deduce that We now specify , such that . Note that the small condition (50) implies with , and it yields where we have used the a priori estimate (28) for the -harmonic function . Furthermore, using (19) and recalling the definition of and , we have Combining these estimates with (62), we have"
] | [
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http://www.global-sci.com/intro/article_detail/cicp/11160.html | [
"Volume 20, Issue 2\nAn Improved Second-Order Finite-Volume Algorithm for Detached-Eddy Simulation Based on Hybrid Grids\n\nCommun. Comput. Phys., 20 (2016), pp. 459-485.\n\nPublished online: 2018-04\n\nPreview Purchase PDF 90 1100\nExport citation\n\nCited by\n\n• Abstract\n\nA hybrid grid based second-order finite volume algorithm has been developed for Detached-Eddy Simulation (DES) of turbulent flows. To alleviate the effect caused by the numerical dissipation of the commonly used second order upwind schemes in implementing DES with unstructured computational fluid dynamics (CFD) algorithms, an improved second-order hybrid scheme is established through modifying the dissipation term of the standard Roe's flux-difference splitting scheme and the numerical dissipation of the scheme can be self-adapted according to the DES flow field information. By Fourier analysis, the dissipative and dispersive features of the new scheme are discussed. To validate the numerical method, DES formulations based on the two most popular background turbulence models, namely, the one equation Spalart-Allmaras (SA) turbulence model and the two equation k−ω Shear Stress Transport model (SST), have been calibrated and tested with three typical numerical examples (decay of isotropic turbulence, NACA0021 airfoil at 60◦ incidence and 65◦ swept delta wing). Computational results indicate that the issue of numerical dissipation in implementing DES can be alleviated with the hybrid scheme, the resolution for turbulence structures is significantly improved and the corresponding solutions match the experimental data better. The results demonstrate the potentiality of the present DES solver for complex geometries.\n\n• Keywords\n\n• BibTex\n• RIS\n• TXT\n@Article{CiCP-20-459, author = {}, title = {An Improved Second-Order Finite-Volume Algorithm for Detached-Eddy Simulation Based on Hybrid Grids}, journal = {Communications in Computational Physics}, year = {2018}, volume = {20}, number = {2}, pages = {459--485}, abstract = {\n\nA hybrid grid based second-order finite volume algorithm has been developed for Detached-Eddy Simulation (DES) of turbulent flows. To alleviate the effect caused by the numerical dissipation of the commonly used second order upwind schemes in implementing DES with unstructured computational fluid dynamics (CFD) algorithms, an improved second-order hybrid scheme is established through modifying the dissipation term of the standard Roe's flux-difference splitting scheme and the numerical dissipation of the scheme can be self-adapted according to the DES flow field information. By Fourier analysis, the dissipative and dispersive features of the new scheme are discussed. To validate the numerical method, DES formulations based on the two most popular background turbulence models, namely, the one equation Spalart-Allmaras (SA) turbulence model and the two equation k−ω Shear Stress Transport model (SST), have been calibrated and tested with three typical numerical examples (decay of isotropic turbulence, NACA0021 airfoil at 60◦ incidence and 65◦ swept delta wing). Computational results indicate that the issue of numerical dissipation in implementing DES can be alleviated with the hybrid scheme, the resolution for turbulence structures is significantly improved and the corresponding solutions match the experimental data better. The results demonstrate the potentiality of the present DES solver for complex geometries.\n\n}, issn = {1991-7120}, doi = {https://doi.org/10.4208/cicp.190915.240216a}, url = {http://global-sci.org/intro/article_detail/cicp/11160.html} }\nTY - JOUR T1 - An Improved Second-Order Finite-Volume Algorithm for Detached-Eddy Simulation Based on Hybrid Grids JO - Communications in Computational Physics VL - 2 SP - 459 EP - 485 PY - 2018 DA - 2018/04 SN - 20 DO - http://doi.org/10.4208/cicp.190915.240216a UR - https://global-sci.org/intro/article_detail/cicp/11160.html KW - AB -\n\nA hybrid grid based second-order finite volume algorithm has been developed for Detached-Eddy Simulation (DES) of turbulent flows. To alleviate the effect caused by the numerical dissipation of the commonly used second order upwind schemes in implementing DES with unstructured computational fluid dynamics (CFD) algorithms, an improved second-order hybrid scheme is established through modifying the dissipation term of the standard Roe's flux-difference splitting scheme and the numerical dissipation of the scheme can be self-adapted according to the DES flow field information. By Fourier analysis, the dissipative and dispersive features of the new scheme are discussed. To validate the numerical method, DES formulations based on the two most popular background turbulence models, namely, the one equation Spalart-Allmaras (SA) turbulence model and the two equation k−ω Shear Stress Transport model (SST), have been calibrated and tested with three typical numerical examples (decay of isotropic turbulence, NACA0021 airfoil at 60◦ incidence and 65◦ swept delta wing). Computational results indicate that the issue of numerical dissipation in implementing DES can be alleviated with the hybrid scheme, the resolution for turbulence structures is significantly improved and the corresponding solutions match the experimental data better. The results demonstrate the potentiality of the present DES solver for complex geometries.\n\nYang Zhang, Laiping Zhang, Xin He & Xiaogang Deng. (2020). An Improved Second-Order Finite-Volume Algorithm for Detached-Eddy Simulation Based on Hybrid Grids. Communications in Computational Physics. 20 (2). 459-485. doi:10.4208/cicp.190915.240216a\nCopy to clipboard\nThe citation has been copied to your clipboard"
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https://www.scribd.com/document/76008357/Thms-and-Postulates | [
"You are on page 1of 3\n\n# Geometry Postulates and Theorems (ch.\n\n1-3)\nUnique Line Postulate: Through any two points, there is exactly one line. Line Intersection Theorem: Two different lines intersect in at most one point. Distance Formula: If two points on a line have coordinates x and y, the distance between them is |x - y|. Distance Addition Property: If B is on\n\nAC , then AB+BC=AC\n\nTriangle Inequality Postulate: The sum of the lengths of any two sides of a triangle is greater than the length of the third side. Angle Addition Property: If (except for point V) is in the interior of mA V C+ mC V B= mA V B.\n\nVC\n\nAVB, then\n\nLPT Linear Pair Theorem: If two angles form a linear pair, then they are supplementary. VAT Vertical Angles Theorem: If two angles are vertical angles, then they have equal measures. Postulates of Equality and Operations: For any real numbers a, b and c: a. RPE Reflexive property of equality: a=a. b. SPE Symmetric property of equality: If a=b, then b=a. c. TPE Transitive property of equality: If a=b, and b=c, then a=c. d. APE Addition (Subtraction) property of equality: If a=b, then a+c=b+c. e. MPE Multiplication (Division) property of equality: If a=b, then ac=bc. Postulates of Inequality and Operations: For any real numbers a, b and c: a. TPI Transitive property of inequality: If a < b, and b < c, then a < c. b. API Addition property of inequality: If a < b, then a+c < b+c. c. MPI Multiplication property of inequality: If a < b and c > 0, then ac < bc. (c is positive) If a < b and c < 0, then ac > bc. (c is negative) Equation to Inequality Property (Part to Whole): If a and b are positive numbers and a+b=c, then c > a and c > b. Substitution property: If a=b, then a may be substituted for b in any expression. CAP Corresponding Angles Postulate: Corresponding angles have the measure if and only if the lines are parallel. Parallel Lines and Slope Theorem: Two nonvertical lines are parallel if and only if they have the same slope. Transitive Property of Parallel Lines: In a plane, if line l is parallel to line m and line m is parallel to line n, then line l is parallel to line n. Two Perpendiculars Theorem: If two coplanar lines l and m are each perpendicular to the same line, then they are parallel to each other. Perpendicular to Parallels Theorem: In a plane, if a line is perpendicular to one of two parallel lines, then it is also perpendicular to the other. Perpendicular Lines and Slopes Theorem: Two nonvertical lines are perpendicular if and only if the product of their slopes is -1 (they are opposite reciprocals).\n\n## Geometry Postulates and Theorems (ch. 4-6)\n\nFigure Reflection Theorem: If certain points determine a figure, then its reflection image is the corresponding figure determined by the reflection images of those points. CPCF Corresponding Parts of Congruent Figures Theorem: If two figures (triangles, quads, etc.) are congruent, then any pair of corresponding parts (segments and angles) is congruent. ABCD Theorem: Every Isometry (reflection, translation, rotation or glide reflection) preserves angle measure, betweenness, collinearity (lines), and distance (lengths of segments). Postulates of Congruence: For any figures F, G and H: a. RPC Reflexive property of congruence: b. SPC Symmetric property of congruence: If c. TPC Transitive property of congruence: If\n\n## F @F . F @G , then G @F . F @G and G @H , then F @H .\n\nSegment Congruence Theorem: Two segments are congruent if and only if they have the same length. Angle Congruence Theorem: Two angles are congruent if and only if they have the same measure. AIA Alternate Interior Angles Theorem: Two lines cut by a transversal are parallel if and only if alternate interior angles are congruent. AEA Alternate Exterior Angles Theorem: Two lines cut by a transversal are parallel if and only if alternate exterior angles are congruent. Perpendicular Bisector Theorem: If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. Uniqueness of Parallels Theorem: Through a point not on a line, there is exactly one line parallel to the given line. Triangle-Sum Theorem: The sum of the measures of the angles of a triangle is 180o. Quadrilateral-Sum Theorem: The sum of the measure of the angles of a convex quadrilateral is 360. Polygon-Sum Theorem: The sum of the measures of the angles of a convex n-gon is (n-2)*180. Flip-Flop Theorem: If F and G are points or figures and rm(F) = G, then rm(G) = F. Segment Symmetry Theorem: Every segment has exactly two symmetry lines: its perpendicular bisector, and the line containing the segment. Side-Switching Theorem: If one side of an angle is reflected over the line containing the angle bisector, its image is the other side of the angle. Angle Symmetry Theorem: The line containing the bisector of an angle is a symmetry line of the angle. Circle Symmetry Theorem: A circle is reflection-symmetric to any line through its center. Symmetric Figures Theorem: If a figure is symmetric, then any pair of corresponding parts under the symmetry is congruent. Isosceles Triangle Symmetry Theorem: The line containing the bisector of the vertex angle of an isosceles triangle is a symmetry line for the triangle. Isosceles Triangle Base Angles Theorem: If a triangle has two congruent sides, then the angles opposite them are congruent.\n\nEquilateral Triangle Symmetry Theorem: Every equilateral triangle has three symmetry lines, which are bisectors of its angles or perpendicular bisectors of its sides. Equilateral Triangle Angle Theorem: If a triangle is equilateral then it is equiangular. Kite Symmetry Theorem: The line containing the ends of a kite is a symmetry line for the kite. Kite Diagonal Theorem: The symmetry diagonal of a kite is the perpendicular bisector of the other diagonal and bisects the two angles at the ends of the kite. Rhombus Diagonal Theorem: Each diagonal of a rhombus is the perpendicular bisector of the other diagonal. Trapezoid Angle Theorem: In a trapezoid, consecutive angles between a pair of parallel sides are supplementary. Isosceles Trapezoid Symmetry Theorem: The perpendicular bisector of one base of an isosceles trapezoid is the perpendicular bisector of the other base and a symmetry line for the trapezoid. Isosceles Trapezoid Theorem: In an isosceles trapezoid, the non-base sides are congruent. Rectangle Symmetry Theorem: The perpendicular bisectors of the sides of a rectangle are symmetry lines for the rectangle. Center of a Regular Polygon Theorem: In any regular polygon there is a point (its center) which is equidistant from all of its vertices. Regular Polygon Symmetry Theorem: Every regular n-gon possesses n symmetry lines, which are the perpendicular bisectors of each of its sides and the bisectors of each of its angles. Every regular n-gon possesses n-fold rotational symmetry."
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https://www.arxiv-vanity.com/papers/1908.00905/ | [
"###### Abstract\n\nWe explain the setup for using the pde2path libraries for Hopf bifurcation and continuation of branches of periodic orbits and give implementation details of the associated demo directories. See [Uec19a] for a description of the basic algorithms and the mathematical background of the examples. Additionally we explain the treatment of Hopf bifurcations in systems with continuous symmetries, including the continuation of traveling waves and rotating waves in equivariant systems as relative equilibria, the continuation of Hopf bifurcation points via extended systems, and some simple setups for the bifurcation from periodic orbits associated to critical Floquet multipliers going through .\n\nUser guide on Hopf bifurcation and time periodic orbits with pde2path\n\n[4mm] Hannes Uecker\n\n[2mm] Institut für Mathematik, Universität Oldenburg, D26111 Oldenburg,\n\n[2mm] November 25, 2020\n\nMSC: 35J47, 35B22, 37M20\nKeywords: Hopf bifurcation, periodic orbit continuation, Floquet multipliers, partial differential equations, finite element method\n\n## 1 Introduction",
null,
"Figure 1: Results for (3.1), from cmdsHPc.m (a), evalplot.m (b), and bru1dcmds.m (c). (a) Bifurcation lines as obtained from branch point (Turing–line) and Hopf point (Hopf and wave line) continuation. Compare [Uec19a, Fig.7a] or [YDZE02]. (b) Spectrum of the linearization of (3.1) around Us, a=0.95 fixed. The dots on the real part show the admissible wave numbers on the subsequently used 1D domain. (c) Bifurcation diagram of Hopf and Turing branches; see also [Uec19a, Fig.7] for the associated solution plots.\n\nIn [Uec19a] we describe the basic algorithms in pde2path to study Hopf bifurcations111i.e.: the bifurcation of (branches of) time periodic orbits (in short Hopf orbits) from steady states; accordingly, we shall call these branches Hopf branches; in PDEs of the form\n\n ∂tu=−G(u,λ),u=u(x,t), x∈Ω, t∈R, (1.1)\n\nwhere . Here , with some bounded domain, , is a parameter (vector), and the diffusion, advection and linear tensors , and the nonlinearity , can depend on , and parameters. The boundary conditions (BC) for (1.1) are of “generalized Neumann” form, i.e.,\n\n n⋅(c⊗∇u)+qu=g, (1.2)\n\nwhere is the outer normal and again and may depend on , , and parameters, and over rectangular domains there additionally is the possibility of periodic BC in one or more directions. See [Uec19a], and the steady state tutorials at [Uec19c], for more details on .\n\npde2path spatially discretizes the PDE (1.1), (1.2) via piecewise linear finite elements, leading to the high–dimensional ODE problem (with a slight misuse of notation)\n\n M˙u=−G(u,λ),u=u(t)∈Rnu,G(u)=Ku−Mf(u), (1.3)\n\nwhere is the number of unknowns, with the number of grid points. In (1.3), is the mass matrix, is the stiffness matrix, which typically corresponds to the diffusion term , and contains the rest, which we often also call the ’nonlinearity’. However, (1.3) is really a sort of symbolic notation to express the spatially discretized version of (1.1), and in (1.3) can also involve nonlinear terms and first order derivatives coming from in (1.1). See, e.g., [RU18, RU17] for more details.\n\nHere we first present implementation details for the four Hopf bifurcation test problems considered in [Uec19a], thus giving a tutorial on how to treat Hopf bifurcations in pde2path. See [Uec19c] for download of the package, including the demo directories, and various documentation and tutorials. In particular, since the Hopf examples are somewhat more involved than the steady case we recommend to new users of pde2path to first look into [RU18], which starts with some simple steady state problems.\n\nThe first Hopf demo problem (demo cgl, subdir of hopfdemos) is a cubic–quintic complex Ginzburg–Landau (cGL) equation, which we consider over 1D, 2D, and 3D cuboids with homogeneous Neumann and Dirichlet BC. Next we consider a Brusselator system (demo brussel) from [YDZE02], which shows interesting interactions between Turing branches and Turing–Hopf branches. As a non–dissipative example we treat the canonical system for an optimal control problem (see also [Uec17]) of “optimal pollution” (demo pollution). This is still of the form (1.1), but is ill–posed as an initial value problem, since it features “backward diffusion”. Nevertheless, we continue steady states and obtain Hopf bifurcations and branches of periodic orbits.\n\nIn §5 we give three tutorial examples for Hopf bifurcations in systems with continuous symmetries, namely a reaction-diffusion problem with mass conservation (demo mass-cons), a Kuramoto-Sivashinsky equation with translational and boost invariance (demos kspbc2 and kspbc4), and a FHN type system with (breathing) pulses featuring an approximate translational symmetry (demo symtut/breathe). Such symmetries were not considered systematically in [Uec19a]. They require phase conditions, first for the reliable continuation of steady states and detection of (Hopf or steady) bifurcations, which typically lead to the coupling of (1.1) with algebraic equations. To compute Hopf branches we then also need to set up suitable phase conditions for the Hopf orbits. See also [RU17], where preliminary results for the breathing pulses are discussed, and one more example of Hopf orbits for systems with symmetries is considered, namely modulated traveling fronts in a combustion model. Moreover, additional to [Uec19a, RU17] we explain some further routines such as continuation of Hopf bifurcation points, and some simple setups for branch switching from Hopf orbits, i.e., for Hopf pitchforks (critical multiplier ) and period doubling (critical multiplier ). See also [Uec19b, §8] for further examples of period–doubling bifurcations in a classical two–component Brusselator system, following [YZE04].\n\nIn §6 we consider Hopf bifurcation in equivariant systems, which generically leads to coexistence of standing waves (SWs) and traveling waves (TWs). We start with the cGL equation over an interval with periodic BC (pBC) (demo cglpbc) and use an appropriate additional phase condition to continue TWs, periodic in the lab frame, but steady in the co–moving frame) as relative equilibria, from which we obtain modulated TWs via Hopf bifurcation in the co–moving frame. A similar approach for the cGL equation in a disk with Neumann BC (demo cgldisk) yields spiral waves as rotating waves (RWs), and meandering spirals as modulated RWs. Then we review and extend an example from [Uec19a, §3.2], namely a reaction diffusion system in a disk and with Robin BC (demo gksspirals), following [GKS00]. In all these examples, the ’interesting’ Hopf bifurcation points have multiplicity at least two, leading to SWs vs TWs (or RWs). Our default branch switching so far only deals with simple Hopf bifurcation points systematically, but we use and explain an ad hoc modification for Hopf points of higher multiplicity, allowing to select, e.g., TW vs SW branches.\n\nThe user interfaces for Hopf bifurcations reuse the standard pde2path setup and no new user functions are necessary, except for the case of symmetries, which requires one additional user function. For the basic ideas of continuation and bifurcation in steady problems we refer to [UWR14] (and the references therein), for pde2path installation hints and review of data structures to [dWDR18], for a general soft introduction to [RU18], and for the basic algorithms implemented in the hopf library of pde2path to [Uec19a]. Thus, here we concentrate on how to use these routines, and on recent additions.\n\nThe basic setup of all demos is similar. Each demo directory contains:\n\n• Function files named *init.m for setting up the geometry and the basic pde2path data, where * stands for the problem, e.g., cgl (and later brussel, …).\n\n• Main script files, such as cmds*d.m where * stands for the space dimension.\n\n• Function files sG.m and sGjac.m for setting up the rhs of the equation and its Jacobian. Most examples are 2nd order semilinear, i.e., of the form with diffusion matrix , and we put the ’nonlinearity’ (i.e., everything except diffusion) into a function nodalf.m, which is then called in sG.m, but also in mesh-adaption and in normal form computations. An exception is the KS equation, see §5.2.\n\n• a function oosetfemops.m for setting up the system matrices.\n\n• A few auxiliary functions, for instance small modifications of the basic plotting routine hoplot.m from the hopf library, which we found convenient to have problem adjusted plots.\n\n• Some auxiliary scripts auxcmds.m, which contain commands, for instance for creating movies or for mesh–refinement, which are not genuinely related to the Hopf computations, but which we find convenient for illustrating either some mathematical aspects of the models, or some further pde2path capabilities, and which we hope the user will find useful as well.\n\n• For the demo pollution (an optimal control problem) we additionally have the functions polljcf.m, which implements the objective function, and pollbra.m, which combines output from the standard Hopf output hobra.m and the standard OC output ocbra.m.\n\nIn all examples, the meshes are chosen rather coarse, to quickly get familiar with the software. We did check for all examples that these coarse meshes give reliable results by running the same simulations on finer meshes, without qualitative changes. We give hints about the timing and indications of convergence, but we refrain from a genuine convergence analysis. In some cases (demos cgl in 3D and brussel in 2D, and cgldisk, gksspirals) we additionally switch off the on the fly computation of Floquet multipliers and instead compute the multipliers a posteriori at selected points on branches. Computing the multipliers simultaneously is possible as well, but may be slow. Nevertheless, even with the coarse meshes some commands, e.g., the continuation of Hopf branches in 3+1D (with about 120000 total degrees of freedom), may take several minutes. All computational times given in the following are from a 16GB i7 laptop with Linux Mint 18 and Matlab 2016b.\n\nIn §2 to §6 we explain the implementations for the four demos from [Uec19a] and the extensions, and thus in passing also the main data-structures and functions from the hopf library. In particular, in §3 we extend some results from [Uec19a] by Hopf point continuation, and in §5 explain the setup for the systems with symmetries and hence constraints. For the unconstrained theory, and background on the first three example problems, we refer to [Uec19a], but for easier reference and to explain the setup with constraints, we also give the pertinent formulas in Appendix A. This also helps understanding a number of new functions, for instance for continuation of TWs and RWs, and for bifurcation from periodic orbits, and Appendix B contains an overview of the functions in the hopf library and of the relevant data structures for quick reference. For comments, questions, and bugs, please mail to .\n\nAcknowledgment. Many thanks to Francesca Mazzia for providing TOM [MT04], which was essential help for setting up the hopf library; to Uwe Prüfert for providing OOPDE; to Tomas Dohnal, Jens Rademacher and Daniel Wetzel for some testing of the Hopf examples; to Daniel Kressner for pqzschur; and to Dieter Grass for the cooperation on distributed optimal control problems, which was one of my main motivations to implement the hopf library.\n\n## 2 The cGL equation as an introductory example: Demo cgl\n\nWe consider the cubic-quintic complex Ginzburg–Landau equation\n\n ∂tu=Δu+(r+iν)u−(c3+iμ)|u|2u−c5|u|4u,u=u(t,x)∈C, (2.1)\n\nwith real parameters . In real variables with , (2.1) can be written as the 2–component system\n\n (2.2)\n\nWe set , and use as the main bifurcation parameter. Considering (2.2) on, e.g., a (generalized) rectangle\n\n Ω=(−l1π,l1π)×⋯×(−ldπ,ldπ) (2.3)\n\nwith homogeneous Dirichlet BC or Neumann BC, or with periodic BC, we can explicitly calculate all Hopf bifurcation points from the trivial branch , located at , with wave vector .\n\nIn particular from the bifurcation point of view, an important feature of the cGL equation (2.1) are its various symmetries, as also discussed in [RU17]. For homogeneous Neumann and Dirichlet BCs, (2.1) has the gauge (or rotational) symmetry , i.e., (2.1) is equivariant wrt the action of the special orthogonal group . Periodic boundary conditions imply a translation invariance as an in general additional equivariance, as on the real line. However, for wavetrains (or traveling waves) , where and are the amplitude, frequency and wave number, respectively, the rotation and translation have the same group orbits. Therefore, in contrast to the steady case [RU17], for our purposes here the gauge symmetry will not play a role. Additional, (2.1) has the reflection symmetry (1D, and analogously for suitable BC over higher dimensional boxes with suitable BC). Thus, in summary, for pBC (and also for the cGL in a disk with e.g., Neumann BC where the role of spatial translation will be played by spatial rotation, the pertinent symmetry group is . Consequently, many Hopf bifurcation points from the trivial solution will be at least double, and in §6 we discuss this case and the associated questions of the bifurcation of standing vs traveling waves, and their numerical treatment in pde2path.\n\nHere we shall first focus on (2.1) over boxes with BC that break the translational invariance, such that only discrete symmetries remain, since, as noted above, for Hopf bifurcation the gauge invariance is equivalent to time shifts, and hence is automatically factored out by the phase condition for time shifts. Moreover, we shall choose boxes such that all HBPs are simple.\n\n### 2.1 General setup\n\nThe cGL demo directory consists, as noted above, of some function files to set up and describe (2.2), some script files to run the computations, and a few auxiliary functions and scripts to explain additional features, or, e.g., to produce customized plots. An overview is given in Table 1, and for this ’first’ demo we discuss the main files in some detail, while in later demos we will mainly focus on differences to this basic template.\n\nAs main functions we have\n\n• cGLinit.m, which (depending on the spatial dimension) sets up the domain, mesh, boundary conditions, and sets the relevant function handles [email protected] and [email protected] to encode the rhs of (2.2);\n\n• sG.m and sGjac, which encode (2.2) and the associated Jacobian of ;\n\nThen we have three script files, cmds*d.m, where *=1,2,3 stands for the spatial dimension. These are all very similar, i.e., only differ in file names for output and some plotting commands, but the basic procedure is always the same:\n\n1. call cGLinit, then cont to find the HBPs from the trivial branch , ;\n\n2. compute branches of periodic orbits (including Floquet multipliers) by calling hoswibra and cont again, then plotting.\n\nListings 1-5 discuss the dimension independent (function) files. We use the OOPDE setup, and thus we refer to [RU18] for a general introduction concerning the meaning of the stiffness matrix , the mass matrix and the BC matrices (and , not used here), and the initialization methods stanpdeo*D, , setting up an OOPDE object which contains the geometry and FEM space, and the methods to assemble the system matrices.\n\n### 2.2 1d\n\nIn 1D we use Neumann BC, and spatial, and (without mesh-refinement) temporal discretization points. Listings 6 and 8 shows the main script file cmds1d.m for 1D computations (with some omissions wrt to plotting), while Fig. 2 shows some output generated by running cmds1d. The norm in (a) is\n\n ∥u∥∗:=∥u∥L2(Ω×(0,T),RN)/√T|Ω|, (2.4)\n\nwhich is our default for plotting of Hopf branches. During the continuation the default plotting routine hoplot also plots the time–series for some mesh point , selected by the index p.hopf.x0i, which is set in cGLinit (see also Fig. 2(b))). The simulations run in less than 10 seconds per branch, but the rather coarse meshes lead to some inaccuracies. For instance, the first three HBPs, which analytically are at , are obtained at , and (b) also shows some visible errors in the period . However, these numerical errors quickly decay if we increase and , and runtimes stay small.",
null,
"Figure 2: Selected outputs from cmds1d.m, i.e., numerical bifurcation diagrams, example plots and (leading 20) Floquet multipliers for (2.2) on the domain Ω=(−π,π) with Neumann BC, 30 grid–points in x. Parameters (ν,μ,c3,c5)=(1,0.1,−1,1), hence bifurcations at (restricting to the first three branches) r=0 (k=0, spatially homogeneous branch, black), r=1/4 (k=1/2, blue) and r=1 (k=1, red). The thick part of the black line in (a) indicates the only stable periodic solutions. The black dots in (a) and (d) are from the (spatially homogeneous) analytical solution, see [Uec19a]. For m=20 there is a visible error in T. The right panel of (d) shows the numerical T for different m (m=20 black, m=40 red-dashed, m=60 blue-dotted), which illustrates the convergence of the numerical solution towards the analytical solution (6.8). Similarly, the periods also converge on the other branches (see cmds1dconv.m). The second plot in (b) shows a time series at the point p.hopf.x0i from b1/pt27. See also Fig. 3(b) for a plot of b3/pt17.\n\nAs usual we recommend to run cmds1d cell-by-cell to see the effect(s) of the individual cells.\n\nSwitching to continuation in another parameter works just as for stationary problems by calling p=hoswiparf(…). See Cells 1 and 2 of cgl/auxcmds1.m for an example, and Fig. 3(a) for illustration. Cells 3 and 4 of auxcmds1.m then contain examples for mesh-refinement in , for which there are essentially two options. The first is to use p.sw.para=3 and the mesh-adaption of TOM, the second is hopftref, see Listing 9.",
null,
"Figure 3: Example outputs from auxcmds1.m. (a) Continuing the solution b1/pt28 from Fig. 2(a,b) in c5, with comparison to the analytical formula [Uec19a, §3.1]. (b), (c) Solution at b3/pt17 before and after mesh-refinement in t via hopftref, here near t∗=4.\n\n### 2.3 Remarks on Floquet multipliers and time integration\n\nFor the Floquet multipliers , ( with the number of spatial discretization points, see (1.3)) we recall from [Uec19a, §2.4] that we have two algorithms for their computation:222in the software we typically call the Floquet multipliers instead of\n\n• FA1 (encoded in the function floq) computes multipliers as eigenvalues of the monodromy matrix .\n\n• FA2(encoded in floqps) uses a periodic Schur decomposition of the matrices building to compute all multipliers.\n\nFA2 is generally much more accurate and robust, but may be slow.333 For floqps one needs to mex percomplex.f(F) in the directory pqzschur, see the README file there. See also line 15 in Listing 6. There always is the trivial Floquet multiplier associated to translational in , and we use with the numerical as a measure for the accuracy of the multiplier computation. Furthermore we define the index of a periodic orbit as\n\n ind(uH)=number of multipliers γ with |γ|>1 (numerically: |γ|>1+p.hopf.% fltol), (2.5)\n\nsuch that indicates instability.\n\nOn b1 in Fig. 2, initially there is one unstable multiplier , i.e., , which passes through 1 to enter the unit circle at the fold. On b2 we start with , and have after the fold. Near another multiplier moves through 1 into the unit circle, such that afterwards we have , with, for instance at . Thus, we may expect a bifurcation near , and similarly we can identify a number of possible bifurcation on b3 and other branches. The trivial multiplier is close to in all these computations, using floq.\n\nIn cgl/auxcmds2.m we revisit these multiplier computations, and complement them with time-integration. For the latter, the idea is to start time integration from some point on the periodic orbit, e.g. , and to monitor, inter-alia, , where by default . Without approximation error for the computation of (including the period ) and of we would have . In general, even if is stable we cannot expect that, in particular due to errors in which will accumulate with , but nevertheless we usually can detect instability of if at some there is a qualitative change in the time–series of .444 The time integration hotintxs takes inter alia the number npp of time steps per period as argument. Time integration is much faster than the BVP solver used to compute the periodic orbits, and thus npp can be chosen significantly larger than the number of time-discretization points in the BVP solver. Thus, choosing or appears a reasonable practice. In Fig. 4(a), where we use the smaller amplitude periodic solution at for the IC, this happens right from the start. Panel (b) illustrates the stability of the larger amplitude periodic solution at , while in (c) the instability of the solution on h2 at manifests around , with subsequent convergence to the (stable) spatially homogeneous periodic orbit",
null,
"Figure 4: Selected output from auxcmds2.m, i.e., stability experiments for (2.2) in 1D. (a) IC h1/pt8, time series of ∥u(⋅,t)−u0∥∞ and u1(x,t), showing the convergence to the larger amplitude solution at the same r. (b) IC h1/pt27 from Fig. 2, where we plot ∥u(⋅,t)−u0∥∞ for t∈[0,4T], which shows stability of the periodic orbit, and a good agreement for the temporal period under time integration. (c) instability of b2/pt19 from Fig. 2, and again convergence to the solution on the b1 branch. Note that the time–stepping is much finer than the appearance of the solution plots, but we only save the solution (and hence plot) every 100th step, cf. footnote 4.\n\n### 2.4 2d\n\nIn 2D we choose homogeneous Dirichlet BC for , see lines 8,9 in cGLinit, and oosetfemops.m. Then the first two HBPs are at (, and (). The script file cmds2d.m follows the same principles as cmds1d.m, and includes some time integration as well, and in the last cell an example for creating a movie of a periodic orbits.\n\nFigure 5 shows some results from cmds2d.m, obtained on a coarse mesh of points, hence spatial unknowns, yielding the numerical values and . With temporal discretization points, the computation of each Hopf branch then takes about a minute. Again, the numerical HBPs converge to the exact values when decreasing the mesh width, but at the prize of longer computations for the Hopf branches. For the Floquet multipliers we obtain a similar picture as in 1D. The first branch has up to the fold, and afterwards, and on b2 decreases from 3 to 2 at the fold and to 1 near . Panel (c) illustrates the 2D analogue of Fig. 4(c), i.e., the instability of the second Hopf branch and stability of the first.",
null,
"Figure 5: Example plots from cmds2d.m. (a) Bifurcation diagrams of the first 2 Hopf branches for (2.2) in 2D. (b) Solution snapshot from b2/pt10, at t=0,310T,610T,910T. (c) Time integration starting from (b) (t=0), with convergence to the first Hopf branch.\n\n### 2.5 3d\n\nTo illustrate that exactly the same setup also works in 3D, in cmds3d.m and Fig. 6 we consider (2.2) over . Here we use a very coarse tetrahedral mesh of points, thus DoF in space. Analytically, the first 2 HBPs are () and (, but with the coarse mesh we numerically obtain and . Again, this can be greatly improved by, e.g., halving the spatial mesh width, but then the Hopf branches become very expensive. Using , the computation of the branches (with 15 continuation steps each) in Fig. 6 takes about 10 minutes, and a call of floqap to a posteriori compute the Floquet multipliers about 50 seconds. Again, on b1, up to fold and afterwards, while on b2 decreases from 3 to 2 at the fold and to 1 at the end of the branch, and time integration from an IC from b2 yields convergence to a periodic solution from b1.\n\nThe script cmds2d.m follows the same principles as the 1D and 2D scripts. In 3D, the “slice plot” in Fig. 6(b), indicated by p.plot.pstyle=1 should be used as a default setting, while the isolevels in (c) (via p.plot.pstyle=2) often require some fine tuning. Additionally we provide a “face plot” option p.plot.pstyle=3, which however is useless for Dirichlet BC.",
null,
"Figure 6: Example plots from cmds3d.m. (a) Bifurcation diagram of first 2 Hopf branches for (2.2) in 3D. (b,c) Solution snapshots at t=0 and t=T/2 for the blue dot in (a); slice-plot in (b), and isolevel plot in (c) with levels 0.525m1+0.475m2 and 0.475m1+0.525m2, where m1=minx,tu1(x,t) and m2=maxx,tu1(x,t).\n\n## 3 An extended Brusselator: Demo brussel\n\nIn [Uec19a, §3.2] we consider an example with an interesting interplay between stationary patterns and Hopf bifurcations, and where there are typically many eigenvalues with small real parts, such detecting HBPs with bifcheck=2 without first using initeig for setting a guess for a shift is problematic. The model, following [YDZE02] is an ’extended Brusselator’, namely the three component reaction–diffusion system\n\n ∂tu=DuΔu+f(u,v)−cu+dw,∂tv=DvΔv+g(u,v),∂tw=DwΔw+cu−dw, (3.1)\n\nwhere , , with kinetic parameters and diffusion constants . We consider (3.1) on rectangular domains in 1D and 2D, with homogeneous Neumann BC for all three components. The system has the trivial spatially homogeneous steady state\n\n Us=(u,v,w):=(a,b/a,ac/d),\n\nand in suitable parameter regimes it shows co-dimension 2 points between Hopf, Turing–Hopf (aka wave), and (stationary) Turing bifurcations from . A discussion of these instabilities of in the plane is given in [YDZE02] for fixed parameters\n\n (c,d,Du,Dv,Dw)=(1,1,0.01,0.1,1). (3.2)\n\nIn our simulations we additionally fix , and take as the primary bifurcation parameter.\n\nFor the quite rich bifurcation results, which include primary spatially homogeneous and patterned Hopf bifurcations from , and Turing bifurcations from followed by secondary Hopf bifurcations, we refer to [Uec19a, §3.2]. Regarding the implementation, Table 2 lists the scripts and functions in brussel. Except for the additional component ( instead of ) this is quite similar to cgl, with one crucial difference, in particular in 2D, on which we focus in §3.2. First, however, we shall focus on 1D and additional to [Uec19a, §3.2] compute bifurcation lines in the parameter plane by branch point continuation and Hopf point continuation, and compute secondary bifurcations from Hopf orbits.\n\n### 3.1 1d\n\nListing 11 shows the startup in 1D. We fix and choose as the continuation parameter, starting at , over the domain , , where is chosen to have the first bifurcation from to a Turing-Hopf (or wave) branch. This follows from, e.g., [YDZE02], see also [Uec19a], but also from the bifurcation lines and spectral plots in Fig. 1(a,b), which we explain below. The ’standard’ files such as bruinit.m, oosetfemops.m, sG.m, and sGjac.m are really standard and thus we refer to their sources. Moreover, regarding initeig in line 5 of bru1dcmds.m we refer to §3.2, where this becomes crucial. The continuation in line 6 of bru1dcmds.m then yields the first three bifurcations as predicted from Fig. 1(a,b), and subsequently further steady and Hopf bifurcations. See Fig. 1(c) for the BD of these branches, and [Uec19a, §3.4] for further discussion. Here we first want to explain how Fig. 1(a) can be computed by Hopf point continuation (HPC) and branch point continuation BPC, see Listing 12.\n\n#### 3.1.1 Hopf point continuation\n\nSimilar to fold continuation and branch point continuation ([DRUW14] and [Uec19b, §3.4]), Hopf point continuation (HPC) can be done via suitable extended systems. Here we use [Gov00, §4.3.2]\n\n H(U):=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝GGuϕr+ωMϕiGuϕi−ωMϕrcTϕr−1cTϕi⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠=0∈R3nu+2,U=(u,ϕr,ϕi,ω,λ), (3.3)\n\nwhere is the desired eigenvalue of , an associated eigenvector, and is a normalization vector. We thus have equations for the real unknowns , and in [Gov00, Proposition 4.3.3] it is shown that (3.3) is regular at a simple Hopf bifurcation point.\n\nThus, (3.3) can be used for localization of (simple) Hopf points (implemented in the pde2path function hploc) if a sufficiently good initial guess is given, and, moreover, freeing a second parameter we can use the extended system\n\n H(U,w)=(H(U,w)p(U,w,ds))=(00)∈R(3nu+2)+1 (3.4)\n\nfor HPC, where is the standard arclength condition, with suitable weights for the parameters . This requires the Jacobian\n\n ∂UH(U)=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝Gu000Gλ∂u(Guϕr)GuωMMϕi∂λ(Guϕr)∂u(Guϕi)−ωMGu−Mϕr∂λ(Guϕi)0cT00000cT00⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠∈R(3nu+2)×(3nu+2). (3.5)\n\nHere, is already available, the expressions are cheap from finite differences, as well as the derivatives needed in the arclength continuation, and expressions such as are easy. Thus the main task is to compute the directional 2nd derivatives\n\n (∂u(Guϕr)∂u(Guϕr))∈R2nu×nu. (3.6)\n\nThis can be done numerically, but this may be expensive, and for semilinear problems it is typically easy to write a function hpjac (or with some other problem specific name name) which returns (3.6), and which should be registered as [email protected] (or [email protected]). For instance, for components and we have\n\n ∂u(Guϕr)=−M∂u(f1,u1ϕ1+f1,u2ϕ2f2,u1ϕ1+f2,u2ϕ2)=−M(f1,u1u1ϕ1+f1,u2u1ϕ2f1,u1u2ϕ1+f1,u2u2ϕ2f2,u1u1ϕ1+f2,u2u1ϕ2f2,u1u2ϕ1+f2,u2u2ϕ2), (3.7)\n\nand we obtain the same expression for with . Accordingly, brussel/hpjac.m returns for the three component semilinear system (3.1). For general testing we also provide the function hpjaccheck, which checks p.fuha.hpjac against finite differences.\n\nTo initialize HPC, the user can call p=hpcontini(’hom1d’,’hpt1’,1,’hpc1’), see line 2 of Listing 12, where the third argument gives the new free parameter . Here , which is at position 1 in the parameter vector. This triples p.nu and sets a number of further switches, for instance for automatically taking care of the structure of in (3.5). For convenience hpcontini also directly sets [email protected], which of course the user can reset afterwards. Then calling cont will continue (3.4) in , and thus we produce the ’wave’ and ’Hopf’ lines in Fig. 1(a). Use hpcontexit to return to ’normal’ continuation (in the original primary parameter).\n\nSimilarly, BPC is based on the extended system [Mei00, §3.3.2]\n\n H(U)=⎛⎜ ⎜ ⎜ ⎜⎝G(u,λ)+μMψGTu(u,w)ψ∥ψ∥22−1⟨ψ,Gλ(u,w)⟩⎞⎟ ⎟ ⎟ ⎟⎠=0∈R2nu+2,U=(u,ψ,w), (3.8)\n\nwhere is a (simple) BP (for the continuation in ), is an adjoint kernel vector, with the primary active parameter and as additional active parameter. The BPC requires the Jacobian of which is potentially difficult to implement. However, again for semilinear problems has a similar structure as (3.7), see [Uec19b, §3.4] for further details. In particular, for (3.1) it can be implemented rather easily, see bpjac.m. The actual BPC is then initialized by calling bpcontini, see line 11 of Listing 12, and the BPC produces the ’Turing line’ in Fig. 1(a). Using bpcontexit returns to ’normal’ continuation.\n\n#### 3.1.2 Bifurcations from the first Hopf branch\n\nAs a second extension of what is presented for (3.1) in [Uec19a, §3.3] we give some results on bifurcations from the first Hopf branch h1 in Fig. 1(c), associated to Floquet multipliers going through .(Of course, a critical multiplier also goes through 1 for a periodic orbit fold as, e.g., for the cGL in §2, but here we are interested in genuine bifurcations.) The used method is described in Appendix A.2, together with the case of period doubling bifurcations associated to Floquet multipliers going through .\n\n###### Remark 3.1.\n\nOur methods are somewhat preliminary in the following sense:\n(a) The localization of the branch points uses a simple bisection based on the change of ind, cf. (2.5), as a multiplier crosses the unit circle. Such bisections work well and robustly for bifurcations from steady branches (and can always be improved to high accuracy using the above extended systems for BPs and HPs), but the bisection for critical multipliers is often more difficult (i.e., less accurate) due to many multipliers close to the unit circle, and also often due to sensitive dependence of the on the continuation parameter , and, moreover, on the numerical time discretization fineness. Typically, some trial and error is needed here.\n(b) The computation of predictors for branch switching is currently based on the classical monodromy matrix , see (A.16) and (A.17). As explained in, e.g., [Lus01], see also [Uec19a] and §4, this may be unstable numerically, in particular for non–dissipative problems. The multiplier computations should then be based on the periodic QZ-Schur algorithm (FA2 algorithm in pde2path, see also §4), but for the branch switching predictor computations this has not been implemented yet.\n\nDespite Remark 3.1, for ’nice’ problems the branch–switching seems to be robust enough. Figure 1 shows some results from bru1dcmds_b, and Listing 13 shows the basic commands. In line 2 we reload a point from the first Hopf branch and set p.hopf.bisec which determines how many bisections are done to localize a BP after the detection of an index change. Here (and in other problems) p.hopf.bisec=5 seems a reasonable value. We then continue further and find the two BPs on 1dh1 indicated in Fig. 1 where the red and magenta branches bifurcate. The branch switching is done in lines 4-8. Typically this requires a rather large ds (all this of course depends on scaling), and often one step with a large residual tolerance is needed to get on the bifurcating branch (which indicates that the predictor is not very accurate). However, once on the bifurcating branch, p.nc.tol can and should be decreased again. The same strategy is used for the second (magenta) bifurcating branch.",
null,
"Figure 7: Results for (3.1) from bru1dcmds_b.m. (a) Bifurcation diagram, extenting Fig. 1(c) by the secondary pitchfork bifurcations from the primary Hopf branch (orange); first PD branch t1 in red, second PD branch t2 in magenta. (b,c) solution and Floquet plots. The magenta branch intermediately has index 3, before for increasing b first two unstable multipliers come back inside the unit circle via a Neimark-Sacker scenario, and then the last unstable γ goes through 1 and the orbit gains stability near b=3.3.\n\n### 3.2 2d\n\nWe close the discussion of (3.1) with some comments on the continuation of solution branches in 2D. See [Uec19a, Fig.10] for example results, where we consider (3.1) on . Already on this rather small domain the linearization of (3.1) around has many small real eigenvalues. Therefore, the Hopf eigenvalues (with imaginary parts near ) are impossible to detect by computing just a few eigenvalues close to , as illustrated in Fig. 8(a,b). Thus, the preparatory step initeig already used in 1D in line 5 of Listing 11 becomes vital. This uses a Schur complement algorithm to compute a guess for the spectral shift near which we expect Hopf eigenvalues during the continuation of a steady branch, see [Uec19a, §2.1 and Fig.8] for illustration.",
null,
"Figure 8: (a,b) neig smallest eigenvalues of the linearization of (3.1) around Us at b=2.75, remaining parameters from (3.2); HD1 with neig=200 will not detect any Hopf points. (c) (LABEL:reso) yields a guess ω1=0.9375 for the ω value at Hopf bifurcation, and then HD2 with neig=(3,3) is reliable and fast: (d) shows the three eigenvalues closest to 0 in blue, and the three eigenvalues closest to iω1 in red.\n\nAs indicated in the caption of Listing 14, at the start of the (1D and 2D) Turing branches we do some adaptive mesh–refinement. The used error estimator is given in Listing 15. The further BPs and HBPs then obtained are very close to the BPs and HBPs on the coarser mesh, but the resolution of the bifurcating Hopf branches becomes considerably better, with a moderate increase of computation time, which in any case is faster than starting with a uniform spatial mesh yielding a comparable accuracy.\n\n## 4 A canonical system from optimal control: Demo pollution\n\nIn [Uec16, GU17], pde2path has been used to study infinite time horizon distributed optimal control (OC) problems, see also [Uec17] for a tutorial on OC computations with pde2path. As an example for such problems with Hopf bifurcations555which so far could not be found in the systems studied in [Uec16, GU17] we consider, following [Wir00], a model in which the states and are the emissions of some firms and the pollution stock, and the control models the abatement policy of the firms. The objective is to maximize\n\n J(v0(⋅),k(⋅,⋅)):=∫∞0e−ρtJca(v(t),k(t))dt, (4.1a)\n\nwhere is the spatially averaged current value function, with local current value , , where is the discount rate. Using Pontryagin’s Maximum Principle, the so called canonical system for the states and co-states (or Lagrange multipliers or shadow prices) can be formally derived as a first order necessary optimality condition, using the intertemporal transversality condition\n\n limt→∞e−ρt∫Ω⟨v,λ⟩dx=0. (4.2)\n\n ∂tv =DΔv+f1(v,k),v|t=0=v0, (4.3a) ∂tλ =−DΔλ+f2(v,k), (4.3b) where f1(v,k)=(−k,v1−α(v2))T, f2(v,k)=(ρλ1−p−λ2,(ρ+α′(v2))λ2+β)T, ∂nλ=0 on ∂Ω, and where the control k is given by k=k(λ1)=−(1+λ1)/γ. (4.3c)\n\nFor convenience we set , and write (4.3) as\n\n ∂tu=−G(u):=DΔu+f(u), (4.4)\n\nwhere diag, . Besides the boundary condition on we have the initial condition (only) for the states. A solution of the canonical system (4.4) is called a canonical path, and a steady state of (4.4) (which automatically fulfills (4.2)) is called a canonical steady state (CSS). Due to the backward diffusion in , and since we only have initial data for half the variables, (4.4) is not well posed as an initial value problem. Thus, one method for OC problems of type (4.1) is to first study CSS, and then canonical paths connecting some initial states to some CSS . This requires the so-called saddle-point property for , and if this holds, then canonical paths to can often be obtained from a continuation process in the initial states, see [Uec17].\n\nA natural next step is to search for time–periodic solutions of canonical systems, which obviously also fulfill (4.2). The natural generalization of the saddle point property is that\n\n d(uH):=ind(uH)−nu2=0, (4.5)\n\ni.e., that exactly half of the Floquet multipliers are in the unit circle. In the (low–dimensional) ODE case, there then exist methods to compute connecting orbits to (saddle type) periodic orbits with , see [BPS01, GCF08], which require comprehensive information on the Floquet multipliers and the associated eigenspace of . A future aim is to extend these methods to periodic orbits of PDE OC systems.\n\nHowever, in [Uec19a, §3.4] we only illustrate that Hopf orbits can appear as candidates for optimal solutions in OC problems of the form (4.1), and that the computation of Floquet multipliers via the periodic Schur decomposition floqps can yield reasonable results, even when computation via floq completely fails.\n\nFor all parameter values, (4.4) has an explicit spatially homogeneous CSS, see [Uec19a], and by a suitable choice of parameters we obtain Hopf bifurcations to spatially homogeneous and spatially patterned time periodic orbits. Concerning the implementation, Table 3 gives an overview of the involved scripts and functions. Since we again use the OOPDE setting, and since we restrict to 1D, although (4.4) is a four component system, much of this is very similar to the cgl demo in 1D, with the exceptions that: (a) we also need to implement the objective value and other OC related features; (b) similar to brussel it is useful to prepare the detection of HBPs via initeig; (c) we need to use flcheck=2 throughout. Thus, in Listings 16-18 we comment on these points, and for plots illustrating the results of running pollcmds.m refer to [Uec19a, §3.4].\n\n## 5 Hopf bifurcation with symmetries\n\nIf the PDE (1.1) has (continuous) symmetries, then already for the reliable continuation of steady states it is often necessary to augment (1.1) by suitable phase conditions, in the form\n\n Q(u,λ,w)=0∈RnQ (5.1)\n\nwhere stands for the required additional active parameters, see [RU17] for a review. For instance, if (1.1) is spatially homogeneous and we consider periodic BC, then we have a translational invariance, and (in 1D) typically augment (1.1) by the phase condition\n\n ⟨∂xu∗,u⟩=0∈R, (5.2)\n\nwhere (for scalar ) , and where is either a fixed reference profile or the solution from the previous continuation step. We thus have additional equations, and consequently must free 1 additional parameter.\n\nSimilarly, we must add phase conditions to the computation of Hopf orbits (additional to the phase condition (A.6) fixing the translational invariance in ). This is in general not straightforward, since (5.1), with (5.2) as an example, is not of the form and thus cannot simply be appended to (1.1). Instead, the steady phase conditions (5.1) must be suitably modified and explicitly appended to the Hopf system, see (A.9). Examples for the case (5.2) have been discussed in [RU17, §4], namely the cases of modulated fronts, and of breathers.\n\nHere we give two more examples, and extend the breather example to compute period doubling bifurcations. The first example deals with Hopf orbits in a reaction diffusion system with mass conservation, and the second with Hopf orbits in the Kuramoto-Sivashinsky (KS) equation, where we need two phase conditions, one for mass conservation and one to fix the translational invariance. For both problems we restrict to 1D; like, e.g., the cGL equation, they both can immediately be transferred to 2D (where for the KS equation we need a third phase condition , cf. (5.9c)), but the solution spaces and bifurcations then quickly become “too rich”, such that – as often – 2D setups only make sense if there are specific questions to be asked.\n\n### 5.1 Mass conservation: Demo mass-cons\n\nAs a toy problem for mass conservation in a reaction diffusion system we consider\n\n ∂tu1=Δu1+d2Δ"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8891233,"math_prob":0.96189165,"size":40068,"snap":"2021-04-2021-17","text_gpt3_token_len":9894,"char_repetition_ratio":0.14996007,"word_repetition_ratio":0.015137329,"special_character_ratio":0.22896077,"punctuation_ratio":0.1400985,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9832375,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-18T03:19:54Z\",\"WARC-Record-ID\":\"<urn:uuid:d507e400-98d0-485e-848b-caf58641bd74>\",\"Content-Length\":\"1049569\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7e54be18-cbb0-4415-aa4a-130f1d558ff2>\",\"WARC-Concurrent-To\":\"<urn:uuid:39160eea-5730-4444-9b2e-12574f1a5c3a>\",\"WARC-IP-Address\":\"104.21.14.110\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/1908.00905/\",\"WARC-Payload-Digest\":\"sha1:TB6G3L32ICNUEEWARLY3EGXZMVHCEO2N\",\"WARC-Block-Digest\":\"sha1:SCJNAR2SNZICSGU6JKUDHAWCSLOD6XBT\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703514121.8_warc_CC-MAIN-20210118030549-20210118060549-00024.warc.gz\"}"} |
https://lykyfubuqani.redoakpta.com/investigation-of-resistivity-of-nichrome-wire-essay-17556vg.html | [
"# Investigation of resistivity of nichrome wire essay\n\nVariables There are expected variables that can be implemented in this experiment; these are the offending variable. The power growing was set to 2 presentations and the current and why was measured using the voltmeter and evaluation respectively.\n\nI think that the most of my results was waiting enough for me to go a valid conclusion about how the marker of the idea affected the resistance. Get Colleague Essay Get access to this skill to get all help you need with your summary and educational issues.\n\nLap, I think my resistance was sufficient to obtain rank results. Precaution was bothered to make sure there were no time hazards due to do-circuit. The wires will have been premeasured out to the ideas mm with each other of 50mm. This would have done the area of the academic from remaining constant and would have strayed my results.\n\nRarely is no possible way to tell the key electrical resistivity without having the exact composition of the winner and this is because time is a property that is specific to the reader of the conductor and slight changes in the intended can produce significant changes in the thesis for the resistivity of the reader.\n\nI will also necessary, using the text pack, how many volts spirit through the world. If the distance eared, I predict that so will the admissions assuming that the reader is the same diameter all the way along and not round.\n\nI will pay up during the entire to ensure that I do not mean myself if something strikes. This was easily due to find error somewhere along the basic. Another limitation to the topic of the experiment is the other of the electrical resistivity of the Writer wire. The actual range is 1.\n\nThe substantive value for resistivity, given by the reader the wire was obtained from is 1. One equation can often represented by l on the y-axis and R on the x-axis.\n\nI will be best overall for the resistivity of the supernatural over different distances of wire. I contribute that as the institution distance increases so will the time.\n\nThe length of the wire partners the resistance as of the library the current must travel. I have much to base my investigation on the language of the wire. The Ohms Law institutions that the current rate through a wire or resistor at every temperature is proportional to the time difference, a high resistance wire passes a large current and a low income passes a large quantity.\n\nIn fact, the wires were staring around several bobs and the writing between the bobs were trying to give an accurate estimate of the analysis of the wire. To determine the written-sectional area of the para, the diameter was first key using a micrometer screw gauge. In this country I will only think 1 factor, which is the end of the wire.\n\nThe platforms I will control will be the relative of wire resistivity and the cross-sectional destination of the wire.\n\nParas were recorded in the facts table. The pass of the wire. Exotic up the circuit as shown 2. A rather wire would have more resistance then that which is limitless. The room temperature — since if the universe is increased the particles in the time will move faster and this would not clear the experiment to be significant.\n\nThe four lines to this equation are: In dresser, I will make sure I upset the power pack off after each main. This is Potential difference P. The endeavor of the circuit for very lengths were collected as raw data and are needed below: For accuracy, and also to think time, I put all the raw jump into a spreadsheet, plucked the correct formulas and hence engaging all the necessary values see results were.\n\nThis way, bowling on secondary data is referenced and a higher level of behaviour can be convinced in the emotions.\n\nNichrome wire, over 50cm compensation. To precede the accuracy of my work, I could have only three separate ideas of wire, suddenly from different coils, which would have forsworn me a better average.\n\nI will give off the power growing, move the crocodile clip that was at 5cm up to 10cm, and scholarship on the power concentrate. There are a few moments points that are farther away from the reader of best fit than the others, but they are still likely with the general trend. Nichrome, as the name suggests, is an alloy of Multiple, Chromium and Iron.\n\nI will also point that there is a chance indication that the essay is isolated by means of a rarity and an L. Autobiography, this would affect the audience and the challenge would be unfair.",
null,
"Investigating the Resistance of a Wire Over Set Distances Essay Sample. For this investigation I looked at all the factors that could affect the results of my experiment.\n\nResearch The equation for finding resistance is GET EVEN A BETTER ESSAY WE WILL WRITE A CUSTOM ESSAY SAMPLE ON An investigation into the relationship between TOPICS SPECIFICALLY FOR YOU Order now R= V/I Where V = potential difference (volts) and I = current (amps) Current is the rate of flow of charge.",
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"An amp [ ]. Resistance of a Wire Investigation Essay - Resistance of a Wire Investigation Aim: To investigate the effects of resistance on a nichrome wire Planning My planning will begin with providing a background understanding on the theory of resistance and how it participates in the overall process of the electric current.\n\nRead this essay on Resistivity of Nichrome.",
null,
"Come browse our large digital warehouse of free sample essays. Investigation of the Resistance of a Wire The Nichrome wire has electrical resistivity of ρe = 10 -6 Ωm, thermal conductivity of k = 25 Wm-1K-1, and surface emissivity of ε = a.\n\nAssuming negligible temperature variations. The resistance of a wire is given by the equation R = ρl / A where ρ is the resistivity of the metal from which the wire is made, l is the length of the wire and A is its cross-sectional area.\n\nUsing the circuit below, you will make measurements of current and voltage for different lengths of wire.\n\nThe housing that encloses the coiled wire is at Tsur = 50˚C. The Nichrome wire has electrical resistivity of ρe = 10 -6 Ωm, thermal conductivity of k = 25 Wm-1K-1, and surface emissivity of ε = a.\n\nInvestigation of resistivity of nichrome wire essay\nRated 3/5 based on 89 review\nInvestigation of the resistance of a wire Free Essay"
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https://dsp.stackexchange.com/questions/15708/determining-the-uncertainty-of-an-autocorrelation/15788#15788 | [
"# Determining the uncertainty of an autocorrelation\n\nMy problem should probably be built up from the beginning, so lets start there. I performed a certain experiment 25 times. Every time, the experiment consists of 5000 measurements, and each measurement returns either 1 ('yes') or 0 ('no'). I know that each time I measure 1, the probability that this measurement is correct is $F_1$, while if I measure 0 I know that it's correct with probability $F_0$.\n\nNow, subsequently I want to compute the autocorrelation function of these measurements I use the following formula",
null,
"with a slightly different normalization, but in principle that should not matter, because what I want to do is fit this autocorrelation to an exponential decay and find the decay time. To do so, I thought I had two options: Either I fit the 25 datasets separately, and then find the mean of the decay time $t_1$ and maybe say something about its variance, or I can find the mean autocorrelation of the 25 datasets, and fit that one. I've decided that that is probably the best option, as the individual datasets can be quite noisy with strange fluctuations, while the mean looks much cleaner.\n\nBut now I'm stuck wondering what I should be doing to find the uncertainty in the fit. Should I have introduced some sort of uncertainty due to the imperfect measurements ($F_0$ and $F_1$), or should I have introduced some sort of standard error of the mean for the 'mean autocorrelation' data? That last one seems plausible, as of course there is also a standard deviation in the 25 datasets, but then I get a little confused as to how I should do that. The formula for the standard error of the mean seems to be $\\frac{\\sigma}{\\sqrt{n}}$, but this is a bit unfair as in my 5000 measurements, the autocorrelation drops to 0 after around 50 measurements. For the fit I therefore also only use the first ~100 points of the autocorrelation, as there's no real need to fit the next 4000 points that are all pretty much equal to 0. So should I instead only calculate the mean and standard error of the first 100 points? Would that be a fair way of finding the uncertainty? The subsequent fitting method will already give error bars for the $t_1$, but this of course heavily depends on the uncertainty in the data.\n\nA final sort of sidetrail, what kind of quantitative method of finding how good I fit should I use for exponential decay? Reduced Chi square, Adjusted R squared, or something else entirely? I suppose this is not intended for dsp though, more for a statistics forum, so feel free to ignore it.\n\n• Regarding your final question, look up \"coefficient of determination\". It is a way of quantifying goodness of fit.\n– John\nApr 17, 2014 at 17:17\n\nAs far as I can see the main question is: How to account for the uncertainty in the measurements of $X_t$ (i.e. $F$).\n\nOne approach to do this would be to run some sort of Monte Carlo simulation on the data. I.e. for each experiment you can resample a large number of times (say 1000 or 10000) where each point has $1-F$ probability of being flipped. By calculating the auto-correlation of each resample you get a distribution of the correlation for each $k$. Then take the 95% (or another value) confidence interval to get an uncertainty in $R(k)$.\n\nIt may also be possible to do this analytically but I'm not sure for binary data (my instinct says not or at least not simply).\n\nThis uncertainty in $R$ should be accounted for in your fitting of the data when calculating the decay time. This can definitely be done analytically and may be implemented in whatever you are using to do the fit (I think matlab does). Alternatively a similar Monte Carlo simulation could be done but is probably more effort.\n\nIf you think all the data follows the same trend (not unreasonable if it is a repeated experiment) you could calculate a mean for each auto-correlation distance as you suggest. This is effectively taking a manual Monte Carlo with 25 resamples. Then you can calculate the standard error as you suggest (note: $n=25$, not $5000$). I would shy away from this approach as it ignores the uncertainty in you initial measurement which you presumably have a good reason for giving.\n\n• Hm, thank you for your answer, it's very useful. The question indeed boils down to how the uncertainty in the measurements propagates all the way down to the decay time of the correlation functions. I'm not sure I completely understand the Monte Carlo part, it sounds a little like a bootstrap method I guess. Do you perhaps have a reference that describes how one would go about this? What you describe later, with the 25 samples, also makes a lot of sense. The 25 data sets are indeed repeated experiments, so I think this might be a valid approach. Apr 22, 2014 at 15:39\n• Yes, bootstrap and monte carlo are very similar. As far as I understand the difference is that bootstrap resamples the empirical data (e.g. the X values at a point across the 25 tests) whereas Monte Carlo resamples based on the probability distribution. Apr 23, 2014 at 9:21\n• For references for anything uncertainty related I tend to look at the GUM. It is primarily about continuous data but some things are applicable to binary info. In particular Supplement 1 is all about Monte Carlo methods. Although it's not the most practical guide. Here is a more practical guide. Apr 23, 2014 at 9:22"
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https://www.hitlershollywood.de/demand/aggregate-demand-and-suply-model-and-its-assumptions.html | [
"HOT NEWS\n1. Home\n2. /\n3. aggregate demand and suply model and its assumptions\n\n# aggregate demand and suply model and its assumptions",
null,
"aggregate demand and suply model and its assumptions\n\nThe AD–AS or aggregate demand–aggregate supply model is a macroeconomic model that explains price level and output through the relationship of aggregate demand and aggregate supply It is based on the theory of John Maynard Keynes presented in his work The General Theory of Employment Interest and Money\n\nGet a Quote Send Message\n\n## MORE DETAILS:\n\n•",
null,
"### What are the assumptions of aggregate demand and aggregate\n\nWhat are the assumptions of aggregate demand and aggregate supply models Aggregate demand and Aggregate supply The aggregate demand curve is\n\nMore Detail\n•",
null,
"### Aggregate Supply Definition\n\nAggregate supply also known as total output is the total supply of goods and services produced within an economy at a given overall price in a given period It is represented by the aggregate\n\nMore Detail\n•",
null,
"### The Aggregate Demand and Aggregate Supply Model\n\nSimilarly in macro economic model of aggregate demand and aggregate supply we study the determination of general price level and does not explain the relative prices of various products We explain below in detail the concepts of aggregate demand AD and aggregate supply AS curves and their likely shape and factors determining them\n\nMore Detail\n•",
null,
"### Economics Flashcards Quizlet\n\naccording to the aggregate demand and aggregate supply model in the long run what is the impact of an increase in the money supply a its simplicity b its assumptions c its predictions d its mathematical structure c Subjects Arts and Humanities Math Social Science Other Languages Science Features\n\nMore Detail\n•",
null,
"### Chapter 11 Aggregate SupplyAggregate Demand Model\n\nStart studying Chapter 11 Aggregate SupplyAggregate Demand Model Learn vocabulary terms and more with flashcards games and other study tools\n\nMore Detail\n•",
null,
"### LECTURE NOTES ON MACROECONOMIC PRINCIPLES\n\nModel of Aggregate Demand and Supply The model of aggregate demand and aggregate supplyis used by economists to explain short‐run fluctuations in economic activity around its long‐run trend The model focuses on the behavior of two variables ‐ The\n\nMore Detail\n•",
null,
"### National income and price determination Macroeconomics\n\nIn this unit youll learn how the aggregate supply and aggregate demand model helps explain the determination of equilibrium national output and the general price level as well as to analyze and evaluate the effects of fiscal policy Youll also learn about the impact of economic fluctuations on the economy’s output and price level both in the short run and in the long run\n\nMore Detail\n•",
null,
"### Aggregate Supply and Aggregate Demand ASAD Model\n\nSupply and demand models are useful for examining the behavior of one good or market but what about looking at a whole economy Luckily the aggregate supply and aggregate demand model lets us\n\nMore Detail\n•",
null,
"### Top 4 Models of Aggregate Supply of Wages With Diagram\n\nADVERTISEMENTS The following points highlight the top four models of Aggregate Supply of Wages The Models are 1 StickyWage Model 2 The Worker Misperception Model 3 The Imperfect Information Model 4 The StickyPrice Model Aggregate Supple Model 1 StickyWage Model The proximate reason for the upward slope of the AS curve is slow sluggish\n\nMore Detail\n•",
null,
"### Aggregate demand and aggregate supply\n\nEconomists use the model of aggregate demand and aggregate supply to analyse economic fluctuations On the vertical axis is the overall level of prices On the horizontal axis is the economy’s total output of goods and services Output and the price level adjust to the point at which the aggregatesupply and aggregatedemand curves intersect\n\nMore Detail\n•",
null,
"### The Classical Economic Model » Economics Tutorials\n\nAn increase in money supply from M1 to M2 leads to a shift in the aggregate demand curve from AD to AD’ This is because the classical model employs the Quantity Theory of Money MV PY where M is the money supply V is the velocity of money in circulation P is the level of price and Y is the output\n\nMore Detail\n•",
null,
"### How the ADAS model incorporates growth unemployment\n\nShifts in aggregate demand Demandpull inflation under Johnson Real GDP driving price Costpush inflation How the ADAS model incorporates growth unemployment and inflation Shifts in aggregate supply Lesson summary Changes in the ADAS model in the short run\n\nMore Detail\n•",
null,
"### SparkNotes Aggregate Supply Aggregate Supply and\n\ndepicts the ASAD model The intersection of the shortrun aggregate supply curve the longrun aggregate supply curve and the aggregate demand curve gives the equilibrium price level and the equilibrium level of output This is the starting point for all problems dealing with the AS AD model Shifts in Aggregate Demand in the ASAD Model\n\nMore Detail\n•",
null,
"### The Aggregate Demand and Aggregate Supply Model\n\nSimilarly in macro economic model of aggregate demand and aggregate supply we study the determination of general price level and does not explain the relative prices of various products We explain below in detail the concepts of aggregate demand AD and aggregate supply AS curves and their likely shape and factors determining them\n\nMore Detail\n•",
null,
"### Aggregate Demand Supply Analysis Bizfluent\n\nThe aggregate supply curve is a curve showing the relationship between a nations price level and the quantity of goods supplied by its producers The Short Run Aggregate Supply SRAS curve is an upwardsloping curve and represents how firms will respond to what they perceive as changing demand\n\nMore Detail\n•",
null,
"Aug 21 2017 · The problem with the ISLM model The starting point of the Aggregate DemandAggregate Supply or ADAS model is an assumption in the ISLM model and in the cross model that limits its usefulness This is the assumption that if firms where to choose the profit maximizing quantity of L LOPT they would produce more than the aggregate demand\n\nMore Detail\n•",
null,
"### Aggregate Demand Idle Time and Unemployment The\n\nIII A Model of Aggregate Demand Idle Time and Unemployment This section develops the main model of the article The model keeps the architecture of the Barro and Grossman 1971 model but takes a matching approach to the labor and product markets instead of a disequilibrium approach IIIA Assumptions\n\nMore Detail\n•",
null,
"### Aggregate Supply AS Curve CliffsNotes\n\nSo there is some uncertainty as to whether the economy will supply more real GDP as the price level rises In order to address this issue it has become customary to distinguish between two types of aggregate supply curves the short‐run aggregate supply curve and the long‐run aggregate supply\n\nMore Detail\n•",
null,
"### Aggregate demand Wikipedia\n\nIn macroeconomics Aggregate Demand AD or Domestic Final Demand DFD is the total demand for final goods and services in an economy at a given time It is often called effective demand though at other times this term is is the demand for the gross domestic product of a country It specifies the amount of goods and services that will be purchased at all possible price levels\n\nMore Detail\n•",
null,
"### Solved Question 4 10 Marks Employ The Aggregate Demand\n\nQuestion Question 4 10 Marks Employ The Aggregate Demand And Supply Model For The Australian Economy To Analyse The Consequences For Real GDP And The General Price Level Of The Following Scenarios Assume That The Economy Operates In The Intermediate Range Of The Aggregate Supply Curve In Your Response Clearly State Your Assumptions And Illustrate Your\n\nMore Detail\n•",
null,
"### The Classical Economic Model » Economics Tutorials\n\nAn increase in money supply from M1 to M2 leads to a shift in the aggregate demand curve from AD to AD’ This is because the classical model employs the Quantity Theory of Money MV PY where M is the money supply V is the velocity of money in circulation P is the level of price and Y is the output\n\nMore Detail\n•",
null,
"### Imperfect Information and Aggregate Supply\n\nThe maximand of each firm is its perceived real profits as given by 1 ˆ Xit it it it t it it t EPYPWHP 4 2 Blanchard and Kiyotaki 1987 present an early example Gali 2008 gives a recent textbook presentation on these models in the context of aggregate supply\n\nMore Detail\n•",
null,
"### The Neoclassical Perspective and Potential GDP\n\nIn the neoclassical model the aggregate supply curve is drawn as a vertical line at the level of potential GDP If AS is vertical then it determines the level of real output no matter where the aggregate demand curve is drawn Over time the LRAS curve shifts to the right as\n\nMore Detail\n•",
null,
"### SparkNotes Aggregate Supply Aggregate Supply and\n\ndepicts the ASAD model The intersection of the shortrun aggregate supply curve the longrun aggregate supply curve and the aggregate demand curve gives the equilibrium price level and the equilibrium level of output This is the starting point for all problems dealing with the AS AD model Shifts in Aggregate Demand in the ASAD Model\n\nMore Detail\n•",
null,
"### Aggregate Demand And Aggregate Supply Intelligent\n\nApr 10 2019 · The ‘natural rate of unemployment’ is the rate of unemployment at equilibrium at this rate wages are in equilibrium and aggregate demand and aggregate supply are also in balance If the demand for labor decreases then wages will fall and labor employed falls This logic follows that at the given wage rate those who want to work will work\n\nMore Detail\n•",
null,
"### 242 Building a Model of Aggregate Demand and Aggregate Supply\n\nConfusion sometimes arises between the aggregate supply and aggregate demand model and the microeconomic analysis of demand and supply in particular markets for goods services labor and capital Read the following Clear It Up feature to gain an understanding of\n\nMore Detail\n•",
null,
"### Aggregate Demand Curve A Close View Economics\n\nADVERTISEMENTS The below mentioned article provides a close view on aggregate demand curve Aggregate demand is the relationship between then quantity of output and the aggregate price level The Quantity Equation as Aggregate Demand The quantity theory tells us that MV PY where M is the money supply V is the velocity of money\n\nMore Detail\n•",
null,
"### Lecture Notes Aggregate Demand and Aggregate Supply\n\nAggregate Demand Aggregate Supply and the Business Cycle Having explained the theoretical framework we are now ready to explain business cycle behavior using the Aggregate DemandAggregate Supply model Generally economic expansions and contractions are driven by shifts in the Aggregate Demand or Aggregate Supply curves\n\nMore Detail\n•",
null,
"### Classical Theory of Employment Assumptions Equation\n\nIn Figure1 A D L is the aggregate demand for labour curve and S L is the aggregate supply of labour curve Intersection of these two curves at point E determines the equilibrium real wage rate WP at full employment level ON\n\nMore Detail\n•",
null,
"### Keynesian vs Classical models and policies Economics Help\n\nBecause of the different opinions about the shape of the aggregate supply and the role of aggregate demand in influencing economic growth there are different views about the cause of unemployment 25 thoughts on “ Keynesian vs Classical models and policies\n\nMore Detail\n•",
null,
"### Aggregate demand shocks and economic growth\n\nWe have argued that the two main assumptions behind this traditional view are first that in the wake of aggregate demand shocks the economy converges to its “normal” growth path and second that this “normal” growth path is unaffected by the aggregate demand shocks\n\nMore Detail\n•",
null,
"### QUIZ 11 12 Consider the assumptions ol the Classical\n\nUnformatted text preview Consider the assumptions ol the Classical Model 1Using the line diawing toot draw the longnin aggregate supply curve such that real GDP is 10 trillion 2 Using the line drawing toot draw the aggregate demand curve\n\nMore Detail\n•",
null,
"### Imperfect Information and Aggregate Supply\n\nThe maximand of each firm is its perceived real profits as given by 1 ˆ Xit it it it t it it t EPYPWHP 4 2 Blanchard and Kiyotaki 1987 present an early example Gali 2008 gives a recent textbook presentation on these models in the context of aggregate supply\n\nMore Detail\n•",
null,
"### Classicalneoclassical model UITS\n\nA Simple Neoclassical Model Assumptions zMarket economy with private property zMarkets are fully competitive zAll variables in the model are either endogenous or exogenous and supplied zInitially there is no government zExcept when indicated the general equilibrium assumptions obtain zTwo kinds of individual agents exist in this economy firms and households\n\nMore Detail\n•",
null,
"### Aggregate Demand AD Curve CliffsNotes\n\nThe aggregate demand curve represents the total quantity of all goods and services demanded by the economy at different price example of an aggregate demand curve is given in Figure The vertical axis represents the price level of all final goods and services The aggregate price level is measured by either the GDP deflator or the CPI\n\nMore Detail\n\n## Our Latest News\n\n• ### best america stone crusher plants",
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https://prime-numbers.fandom.com/wiki/1,123?diff=18089&oldid=7695 | [
"## FANDOM\n\n857 Pages\n\n1,123 is a prime number from 1001-2000. 1,123 has 2 factors, 1 and 1,123. It is the 188th prime number, and the 20th prime number from 1001-2000.\n\n## Relationship with other odd numbers",
null,
"Edit\n\n### The numbers before",
null,
"Edit\n\n• The previous prime number is 1,117.\n• 1,117 and 1,123 are sexy primes.\n• 1,121 is the product of 19 and 59.\n• The previous prime number before 1,117 is 1,109.\n• 1,111 is divisible by 11.\n\n### The numbers after",
null,
"Edit\n\n• The next prime number is 1,129.\n• 1,123 and 1,129 are sexy primes.\n• 1,127 is divisible by 7.\n• 1,151 is the next prime number after 1129. The prime gap is 22 numbers apart.\n• 1,131 is divisible by 3, along with the next three numbers, 1,137, 1,143, and 1,149.\n• 1,133 is divisible by 11.\n• 1,135 and 1,145 are divisible by 5.\n• 1,139 is the product of 17 and 67.\n• 1,141 is divisible by 7.\n• 1,147 is the product of 31 and 37.\n\n## Trivia",
null,
"Edit\n\nCommunity content is available under CC-BY-SA unless otherwise noted."
] | [
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"data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D",
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"data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D",
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"data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D",
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http://nothings.org/gamedev/post_mix_3d_panning.html | [
"# A Failed Idea: Post-Mix Panning For 3D View Direction Changes\n\n## Abstract\n\nThis article describes a technique that doesn't quite work for applying \"listening direction\" changes separately from other 3D mixing, specifically after mixing together multiple audio sounds. Even if this worked, it would be a solution in search of a problem. But it doesn't actually work.\n\n## Overview\n\nIn this article, I present a close-but-no-cigar failed technique that attempts to allow mixing 3d-positioned audio voices together, then performing additional 3D rotations of the mixed buffer.\n\nIt is not immediately obvious that this is actually useful. One scenario where it could be useful would be one in which all of the following are true:\n\n• The cost of mixing is significant (e.g. you are mixing thousands of 3D-spatialized sounds together)\n• Most mixed sounds' 3D doppler and distance attenuation effects are approximately constant or at least predictable (i.e. can ignore user input)\n• Mixed sounds' 3D rotation (pan) effect must be updated dynamically (e.g. due to mouselook, but only if failing to do so is audible)\n• You need to mix \"ahead\" (e.g. on a platform where the OS might disrupt the mixing thread, so you have to mix more than one tick's worth of audio, even though you're respatializing every tick, thus \"wasting\" lots of mixing work most of the time)\n\n## Mixing model\n\nI assume there are three kinds of sounds: \"2D\" sounds which are not spatialized relative to a moving listener; \"modifiable\" sounds which must have their distance attenuation, doppler, or other non-pan effects updated regularly, or which can stop playing under interactive control; and \"predictable\" sounds, which have their distance attenuation, doppler, etc. set once when the sound begins playing, and which once started, cannot be stopped/canceled or otherwise modified.\n\nI assume mixing occurs in floating-point or otherwise with wide range, with a final clamping to the output range after mixing is performed.\n\nIn the mixing model I propose, each class of sounds is mixed into separate buffers (actually, 2D and modifiable can be combined). A final mix step mixes these buffers together before the final clamp.\n\nIn this article I describe a mechanism that allows mixing all of the \"predictable\" sounds together, then applying a \"rotation\" step during the final mixing of the buffers before clamping. In the mix-ahead low-latency-rotation scenario, the predictable sounds are mixed significantly far ahead, but are never remixed. Each new predictable sound that is created is added into the existing predictable mix buffer. The \"final mix\" step must be executed for the entire premixed buffer every tick, but the final mix step is independent of the number of sounds being played.\n\nThe remainder of this article ignores the existence of the 2D and modifiable sounds.\n\nWe also ignore the need issues with avoiding clicks when changing parameter values at discrete points in time, since these are easily avoided by using fixups (crossfades or parameter sweeps) during the final rotation step.\n\n## Spatialization model\n\nI assume that the non-directional aspects of spatialization (such as distance attenuation, doppler, and reverb) are independent of listener orientation, and so can be applied while mixing sounds together. I assume the effect of the viewpoint tilting up/down is ignored: or, equivalently, after mixing, all sounds are assumed to lie in a single plane around the viewer.\n\n### Panning model\n\nGiven a variable t which ranges from -1 (left) to 1 (right), one classic model is:\n\n```Left attenuation = clamp( 1-t, 0,1)\nRight attenuation = clamp(-1+t, 0,1)```\n\nwhich means that when t=0, Left = Right = 1.\n\nThe model I use here assumes t ranges from 0 (left) to 1 (right):\n\n```Left = cos(t*pi/2)\nRight = sin(t*pi/2)```\n\nThis is the \"equal power\" model of panning (left^2 + right^2 is constant).\n\n### 3D spatialization\n\nGiven a sound in direction D expressed in radians, we can then map that direction to the t value of the above equation. Assuming D=0 is centered, then we want D=-pi/2 to be t=0, and D=pi/2 to be t=1, so t = (D+pi/2)/pi.\n\nSubstituting this into the equal power model gives us:\n\n``` L = cos(D/2+pi/4)\nR = sin(D/2+pi/4)```\n\nHowever, attenuation values should always be positive, and the above will result in a negative attenuation values for directions behind the viewer. To avoid this, we take the absolute value:\n\n``` L = |cos(D/2+pi/4)|\nR = |sin(D/2+pi/4)|```\n\nFinally, we want to explicitly separate out view direction so we can alter it on the fly. Let the direction to the sound be D, and let V be direction of the viewer (listener), then:\n\n``` L = |cos(pi/4 + D/2 - V/2)|\nR = |sin(pi/4 + D/2 - V/2)|```\n\nThis model applies only panning as orientation spatialization. There are other possible orientation spatialization effects, such as having sounds behind the listener be further attenuated or lowpass filtered. These effects can only be applied in a very coarsely approximate way, discussed later.\n\n## Simplified problem\n\n#### With an unacceptable approximation\n\nSuppose we consider the phase inversions produced by negative attenuation values to be acceptable (they're not).\n\nWe could then remove the absolute-value computations from our attenuation calculation:\n\n``` L = cos(pi/4 + D/2 - V/2)\nR = sin(pi/4 + D/2 - V/2)```\n\nWe can then apply trigonometric sum identities:\n\n``` L = cos(pi/4 + D/2) * cos(-V/2) - sin(pi/4 + D/2) * sin(-V/2)\nR = sin(pi/4 + D/2) * cos(-V/2) + cos(pi/4 + D/2) * sin(-V/2)```\n\nThis is just a simple 2D rotation by -V/2.\n\nThis makes the overall algorithm simple: mix each sound using into the \"predictable mix buffer\" using the weights:\n\n``` L = cos(pi/4 + D/2)\nR = sin(pi/4 + D/2)```\n\nThen, during final mix, apply the rotation by -V/2 based on the current listening direction. Because the rotation is linear, and because the sounds mixing is linear, this will correctly position all of the already-mixed sounds to match the viewer listening direction, except with incorrect negative attenuations.\n\n## Full problem\n\n#### For which there does not appear to be a solution\n\nThe above solution does not work correctly because sometimes the attenutations are negative, and the phases of the sounds invert.\n\nTo correct this, we must force attenuations to be positive:\n\n``` L = |cos(pi/4 + D/2 - V/2)|\nR = |sin(pi/4 + D/2 - V/2)|```\n\nWe can't achieve this with a simple rotation.\n\nTo address this, we'll attempt to split these into discrete cases. First, let's define the absolute values of trig functions assuming angles in the range 0 .. 2pi.\n\n``` |cos(a)| = a > pi/2 && a < 3pi/2 ? -cos(a) : cos(a)\n|sin(a)| = a > pi ? -sin(a) : sin(a)```\n\nTp perform the substitution, we first imagine defining a substituted variable:\n\n` E' = pi/4 - D/2 - V/2`\n\nHowever, to simplify notation, we'll use 2x that:\n\n` E = pi/2 + D - V`\n\nHowever, we must clamp E/2 to 2pi, so E must clamp to 4pi:\n\n` E = (pi/2 + D - V) mod 4pi`\n\n``` L = |cos(E/2)| = E/2 > pi/2 && E < 3pi/2 ? -cos(E/2) : cos(E/2)\nR = |sin(E/2)| = E/2 > pi ? -sin(E/2) : sin(E/2)```\n\nWe now express the conditions in terms of quadrants so we can understand it geometrically:\n\n``` Q=0 if 0 <= E/2 < pi/2\nQ=1 if pi/2 <= E/2 < pi\nQ=2 if pi <= E/2 < 3*pi/2\nQ=3 if 3*pi/2 <= E/2 < 2*pi```\n\n``` L = Q==1 || Q == 2 ? -cos(E/2) : cos(E/2)\nR = Q==2 || Q == 3 ? -sin(E/2) : sin(E/2)```\n\nThe reason this is interesting is because we can potentially separately classify each sound into which quadrant it is, and then create separate mix buffers, one per quadrant. This would then allow the rotation step to correctly compute a \"rotation\" which produces a non-negative attenuation for the given mix buffer, since all the sounds in the mix buffer are in the same quadrant, and thus their sin()s and cos()s are consistently positive or negative (and thus so is their sum).\n\nWe can't actually do this, though, because the quadrants of the final mixed sound depend on both the source direction and the listening direction.\n\nSo now we need to simplify the equation so we can classify based on the sound and listening direction as simply as possible.\n\nWe modify E as follows: combine the pi/2 into D:\n\n``` D' = D + pi/2\nE = (D - V) mod 4pi```\n\nSubstitute D' with D' mod 2pi, which won't affect the sound:\n\n` D'' = (D + pi/2) mod 2pi`\n\nAdd another 2pi to D'', which won't affect the sound:\n\n` F = ((D + pi/2) mod 2pi) + 2pi`\n\nAnd now define E as:\n\n` E = (F - V) mod 4pi`\n\nSince F now ranges from 2pi..4pi, and V ranges from 0pi..2pi, we can drop the mod:\n\n` E = F - V`\n\n``` Q=0 if 0 <= F-V < pi\nQ=1 if pi <= F-V < 2*pi\nQ=2 if 2*pi <= F-V < 3*pi\nQ=3 if 3*pi <= F-V < 4*pi```\n\nHere F is a value that depends only on D, the source direction, which we assume is constant.\n\nThe problem here is that there is no further we can go. For any given value of F, there is some V for which a small V epsilon changes which quadrant (F-V)/2 is in, thus needing to introduce an extra negation. For two F's very near each other, which have been mixed into a shared buffer, the V at which point this transition occurs is different, but since they're already mixed and we must choose a specific negation, there's no way to avoid one or the other ending up with a negative phase.\n\nWhat I was hoping was that you could classify the sounds into quadrants or octants or some such, even more finely. Although storing a separate mix buffer for each quadrant/octant would use more memory and require more mixing work, it would still be independent of the total number of sounds. It would also allows you to coarsely apply other effects like attenuation and low-pass filtering, but only on a per-quadrant/octant basis.\n\nIt's possible that if you divided sounds into octants, the amount of \"inverted phase\" you would get would be small enough that it wouldn't be objectionable, but I have no reason to think this is true and so I assume instead this technique doesn't work. (Clearly, if you divided directions into small enough sectors, eventually there'd be some division where it was unobjectionable, but that might involve so many mix buffers that it is slower than the naive implementation.)\n\nActually, I guess it's easier to set aside the math and understand it this way: there's a plane through the listener. Sounds that transition from in front of the listener, in the front halfspace to being behind the listener, in the back halfspace, must flip the sign of one of the two attenuations. Thus when the listener rotates, any objects which cross that plane cannot work with this mechanism. It is possible you could \"unmix\" them from the premixed buffer and remix them on the new side, but then you still pay a big cost on rapid listener orientation changes (but not as big as the naive implementation, since you only need to update sounds every 180 degrees).\n\n## Conclusion\n\nOh well, this doesn't work.\n\nThe plain rotation for the \"simple problem\" introduces phase inversions which are unacceptable for 3D spatialization. However, it is an interesting effect in the environment of music effects processing.\n\nFor example, given a mono siginal, a stereo delay line can be set up with a rotation in the feedback path, and the echoes generated by the delay will rotate around in stereo. As the echoes go through the quadrants, some of the time the echoes will have one channel with inverted phase, producing a distinctive but not objectionable stereo effect. Alternatively, only one channel of the delay can be output, producing a delay whose echoes fade in and out over time (some will be inverted, but this is inaudible in mono)."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.911556,"math_prob":0.96648204,"size":8810,"snap":"2019-51-2020-05","text_gpt3_token_len":2270,"char_repetition_ratio":0.11208267,"word_repetition_ratio":0.019011406,"special_character_ratio":0.2644722,"punctuation_ratio":0.09156493,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99613184,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-10T10:47:17Z\",\"WARC-Record-ID\":\"<urn:uuid:8bbb8861-137a-4c09-ab78-3d1074b56c25>\",\"Content-Length\":\"25247\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:38843d9b-5e86-475d-ae1f-30ddfa674278>\",\"WARC-Concurrent-To\":\"<urn:uuid:ace269a0-f5ea-4ab3-bc49-7632720877a8>\",\"WARC-IP-Address\":\"69.163.218.157\",\"WARC-Target-URI\":\"http://nothings.org/gamedev/post_mix_3d_panning.html\",\"WARC-Payload-Digest\":\"sha1:AHD5FSLDYOEZTPKZE6BW2O726FMCDN7N\",\"WARC-Block-Digest\":\"sha1:KJGQLAIFBEMLFM5QRJD7SNCO2CZCBT6B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540527205.81_warc_CC-MAIN-20191210095118-20191210123118-00034.warc.gz\"}"} |
https://mathexamination.com/class/lindstr%C3%B6m-quantifier.php | [
"## Do My LindströM Quantifier Class",
null,
"A \"LindströM Quantifier Class\" QE\" is a basic mathematical term for a generalized constant expression which is used to fix differential formulas and has services which are periodic. In differential Class resolving, a LindströM Quantifier function, or \"quad\" is utilized.\n\nThe LindströM Quantifier Class in Class kind can be expressed as: Q( x) = -kx2, where Q( x) are the LindströM Quantifier Class and it is a crucial term. The q part of the Class is the LindströM Quantifier constant, whereas the x part is the LindströM Quantifier function.\n\nThere are 4 LindströM Quantifier functions with appropriate service: K4, K7, K3, and L4. We will now look at these LindströM Quantifier functions and how they are resolved.\n\nK4 - The K part of a LindströM Quantifier Class is the LindströM Quantifier function. This LindströM Quantifier function can also be written in partial fractions such as: (x2 - y2)/( x+ y). To resolve for K4 we increase it by the appropriate LindströM Quantifier function: k( x) = x2, y2, or x-y.\n\nK7 - The K7 LindströM Quantifier Class has a solution of the kind: x4y2 - y4x3 = 0. The LindströM Quantifier function is then increased by x to get: x2 + y2 = 0. We then need to increase the LindströM Quantifier function with k to get: k( x) = x2 and y2.\n\nK3 - The LindströM Quantifier function Class is K3 + K2 = 0. We then increase by k for K3.\n\nK3( t) - The LindströM Quantifier function equationis K3( t) + K2( t). We increase by k for K3( t). Now we increase by the LindströM Quantifier function which gives: K2( t) = K( t) times k.\n\nThe LindströM Quantifier function is also referred to as \"K4\" because of the initials of the letters K and 4. K indicates LindströM Quantifier, and the word \"quad\" is pronounced as \"kah-rab\".\n\nThe LindströM Quantifier Class is among the primary approaches of resolving differential formulas. In the LindströM Quantifier function Class, the LindströM Quantifier function is first multiplied by the appropriate LindströM Quantifier function, which will offer the LindströM Quantifier function.\n\nThe LindströM Quantifier function is then divided by the LindströM Quantifier function which will divide the LindströM Quantifier function into a genuine part and a fictional part. This provides the LindströM Quantifier term.\n\nLastly, the LindströM Quantifier term will be divided by the numerator and the denominator to get the ratio. We are entrusted to the right-hand man side and the term \"q\".\n\nThe LindströM Quantifier Class is a crucial concept to understand when fixing a differential Class. The LindströM Quantifier function is just one technique to fix a LindströM Quantifier Class. The methods for resolving LindströM Quantifier equations include: particular value decay, factorization, optimal algorithm, numerical service or the LindströM Quantifier function approximation.\n\n## Hire Someone To Do Your LindströM Quantifier Class\n\nIf you want to become familiar with the Quartic Class, then you require to first begin by looking through the online Quartic page. This page will show you how to utilize the Class by utilizing your keyboard. The explanation will also show you how to develop your own algebra formulas to help you study for your classes.\n\nBefore you can comprehend how to study for a LindströM Quantifier Class, you must first understand using your keyboard. You will discover how to click on the function keys on your keyboard, in addition to how to type the letters. There are 3 rows of function keys on your keyboard. Each row has 4 functions: Alt, F1, F2, and F3.\n\nBy pressing Alt and F2, you can multiply and divide the worth by another number, such as the number 6. By pushing Alt and F3, you can use the 3rd power.\n\nWhen you press Alt and F3, you will key in the number you are attempting to multiply and divide. To increase a number by itself, you will press Alt and X, where X is the number you want to multiply. When you press Alt and F3, you will enter the number you are attempting to divide.\n\nThis works the very same with the number 6, other than you will just key in the two digits that are six apart. Lastly, when you press Alt and F3, you will use the 4th power. Nevertheless, when you press Alt and F4, you will use the real power that you have discovered to be the most suitable for your problem.\n\nBy using the Alt and F function keys, you can multiply, divide, and then utilize the formula for the 3rd power. If you need to multiply an odd number of x's, then you will require to enter an even number.\n\nThis is not the case if you are attempting to do something complex, such as multiplying 2 even numbers. For instance, if you want to increase an odd number of x's, then you will require to enter odd numbers. This is especially real if you are trying to figure out the response of a LindströM Quantifier Class.\n\nIf you wish to transform an odd number into an even number, then you will require to push Alt and F4. If you do not know how to multiply by numbers by themselves, then you will need to use the letters x, a b, c, and d.\n\nWhile you can increase and divide by utilize of the numbers, they are much easier to use when you can look at the power tables for the numbers. You will need to do some research study when you initially begin to utilize the numbers, however after a while, it will be force of habit. After you have developed your own algebra equations, you will have the ability to develop your own multiplication tables.\n\nThe LindströM Quantifier Solution is not the only method to resolve LindströM Quantifier equations. It is important to discover trigonometry, which uses the Pythagorean theorem, and after that use LindströM Quantifier formulas to solve issues. With this technique, you can know about angles and how to resolve issues without having to take another algebra class.\n\nIt is important to attempt and type as quickly as possible, because typing will assist you understand about the speed you are typing. This will assist you compose your answers much faster.\n\n## Pay Someone To Take My LindströM Quantifier Class",
null,
"A LindströM Quantifier Class is a generalization of a direct Class. For instance, when you plug in x=a+b for a given Class, you obtain the worth of x. When you plug in x=a for the Class y=c, you obtain the worths of x and y, which give you a result of c. By using this fundamental principle to all the equations that we have attempted, we can now solve LindströM Quantifier equations for all the worths of x, and we can do it quickly and effectively.\n\nThere are many online resources offered that supply complimentary or cost effective LindströM Quantifier formulas to resolve for all the worths of x, including the expense of time for you to be able to make the most of their LindströM Quantifier Class project assistance service. These resources typically do not require a membership charge or any type of investment.\n\nThe answers provided are the outcome of complex-variable LindströM Quantifier formulas that have been solved. This is likewise the case when the variable used is an unidentified number.\n\nThe LindströM Quantifier Class is a term that is an extension of a linear Class. One advantage of using LindströM Quantifier formulas is that they are more basic than the linear formulas. They are much easier to resolve for all the values of x.\n\nWhen the variable utilized in the LindströM Quantifier Class is of the type x=a+b, it is easier to fix the LindströM Quantifier Class due to the fact that there are no unknowns. As a result, there are fewer points on the line defined by x and a continuous variable.\n\nFor a right-angle triangle whose base points to the right and whose hypotenuse points to the left, the right-angle tangent and curve chart will form a LindströM Quantifier Class. This Class has one unknown that can be found with the LindströM Quantifier formula. For a LindströM Quantifier Class, the point on the line specified by the x variable and a constant term are called the axis.\n\nThe presence of such an axis is called the vertex. Since the axis, vertex, and tangent, in a LindströM Quantifier Class, are an offered, we can find all the worths of x and they will sum to the provided values. This is accomplished when we use the LindströM Quantifier formula.\n\nThe aspect of being a consistent factor is called the system of equations in LindströM Quantifier equations. This is in some cases called the central Class.\n\nLindströM Quantifier equations can be solved for other values of x. One way to solve LindströM Quantifier formulas for other worths of x is to divide the x variable into its factor part.\n\nIf the variable is provided as a positive number, it can be divided into its factor parts to get the typical part of the variable. This variable has a magnitude that amounts to the part of the x variable that is a continuous. In such a case, the formula is a third-order LindströM Quantifier Class.\n\nIf the variable x is unfavorable, it can be divided into the same part of the x variable to get the part of the x variable that is multiplied by the denominator. In such a case, the formula is a second-order LindströM Quantifier Class.\n\nSolution help service in solving LindströM Quantifier formulas. When using an online service for solving LindströM Quantifier equations, the Class will be fixed quickly."
] | [
null,
"https://mathexamination.com/Do-My-Math-Class.webp",
null,
"https://mathexamination.com/Take-My-Math-Class.webp",
null
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https://qubytes.org/2021/01/19/will-quantum-computers-without-error-correction-be-useful-for-optimization/ | [
"# Will Quantum Computers without Error Correction be Useful for Optimization?\n\nAuthors: Daniel Stilck Franca, Raul Garcia-Patron\n\nFirst Author’s Institution: University of Copenhagen\n\nStatus: Pre-print on arXiv\n\n## The Big Idea\n\nThe authors develop a theoretical technique to identify situations where a noisy quantum computer without error correction loses to current classical optimization methods. The authors use their technique to provide estimates for many popular quantum algorithms running on near-term devices, including variational quantum eigensolvers,quantum approximate optimization (both gate-based), and quantum annealing (think D-Wave). The authors found that for quantum computers without error correction “substantial quantum advantages are unlikely for optimization unless current noise rates are decreased by orders of magnitude or the topology of the problem matches that of the device. This is the case even if the number of qubits increases.”\n\nThe author’s conclusion allows others researchers to identify and focus on the slim regime of experiments where quantum advantage without error correction is still possible, and shift more time into development of error-corrected quantum computers. Even seemingly “negative” results advance the field in meaningful ways.\n\n## Currently in Quantum Computing\n\nCurrent quantum computers are noisy (errors frequently occur), and of intermediate size: large enough to compete with the best classical computers on (almost) useless problems, but not large enough to be fault-tolerant (~50-100 qubits). A fault tolerant quantum computer has enough error correction so that errors mid-calculation do not affect the final result. Fault tolerance requires more and better qubits than are available today. The community is fairly confident fault tolerant quantum computers will outperform classical computers on many useful problems, but it is unclear if noisy, intermediate scale quantum (NISQ) computers can do the same.\n\nOptimization problems are a very useful, profitable, and ubiquitous class of problem where the goal is to minimize or maximize something: cost, energy, path length, etc. Optimization problems occur everywhere, from financial portfolios to self-driving cars, and often belong to the NP complexity class, which is widely accepted as extremely difficult for classical computers to solve. Comparing computers based on ability to solve optimization problems has two benefits. First, it is easy to see which computer’s solution is better. Second, if quantum computers have an advantage, there are immediate applications.\n\n## How to Tell if a NISQ Computer is a Poor Optimizer",
null,
"Quantum state diagram. In red is the region of classical optimization superiority. Figure 1 from Limitations of optimization algorithms on noisy quantum devices\n\nGet familiar with the state diagram- I’ll be using it to explain the entire technique!\n\nThe optimization task is to minimize a “cost function,” which takes an input and assigns a “cost”(the function’s output). Here, the input is the quantum state of the device",
null,
"$\\rho$, and the “cost” is the energy of the state. The function is characterized by a linear operator",
null,
"$H$ (usually a Hamiltonian). Every",
null,
"$H$ has a family of thermal equilibrium states (inputs) that do not change with time. The device has a unique thermal equilibrium state (Gibbs state for short) labelled by",
null,
"$\\sigma_\\beta$ for every temperature.",
null,
"$\\beta$ represents the inverse of temperature, so",
null,
"$\\beta = 0$ is “burning hot,” though for such a tiny device “hot” just means randomly scrambled (i.e. noisy). Likewise,",
null,
"$\\beta = \\infty$ is absolute zero temperature, meaning everything is perfectly in order and functioning as intended (i.e. noiseless).\n\nFigure 1 from the paper shows device states (labelled black dots) at various points in a quantum computation. The noisy quantum computer with",
null,
"$n$ qubits initialized at state",
null,
"$\\rho=|0\\rangle^{\\otimes n}$ attempts to follow the absolute zero temperature path (orange arrow) through the space of “noiseless” quantum states (black Bloch sphere) and arrive at the true answer",
null,
"$\\sigma_{\\beta=\\infty}$. However, the real computation takes a noisy path (black arrow) that after enough time, leads to the steady state of the noise at",
null,
"$\\sigma_{\\beta=0}$.\n\nThe quantum device state at an intermediate point in the real computation",
null,
"$\\Phi(\\rho)$ is too difficult to simulate classically (NP complexity), but at each intermediate state",
null,
"$\\Phi(\\rho)$, there is a set of states(located on the blue line) with the same cost function value (i.e. energy) to within some error",
null,
"$\\epsilon$. One of those equal-energy states is the Gibbs state at temperature",
null,
"$\\beta$, denoted in the diagram by",
null,
"$e^{-\\beta H} / \\mathcal{Z}$. In fact, we can “mirror” the real path of the quantum device (black arrow) as it progresses in time by matching all intermediate states with Gibbs states, producing a “mirror descent” (green line) from the correct answer to the steady state of the noise",
null,
"$\\sigma_{\\beta} = 0$.\n\nFor any class of optimization problems, there is a critical threshold",
null,
"$\\beta_C$ below which we are guaranteed to efficiently classically compute the Gibbs state, providing a cost function estimate better than the current quantum state",
null,
"$\\Phi(\\rho)$: this threshold is depicted in red and labelled",
null,
"$\\mathcal{C}$. Once a quantum computation passes the threshold( i.e. has an equivalent Gibbs state at with",
null,
"$\\beta<\\beta_C$), one has certified classical superiority. All that remains to use this technique is a method for quantifying how far along in",
null,
"$\\beta$ the associated quantum state",
null,
"$\\Phi(\\rho)$ is.\n\nThe authors quantify the distance between the quantum state",
null,
"$\\Phi(\\rho)$ and the steady state of the noise",
null,
"$\\sigma_{\\beta=0}$ by the relative entropy (for our purposes, relative entropy is simply a way to measure the distance between quantum states). The relative entropy only decreases throughout the computation (meaning",
null,
"$\\Phi(\\rho)$ is always moving towards",
null,
"$\\sigma_{\\beta=0}$, never away). Sparing the relative entropy proof (see the paper), it is possible to identify an upper bound on the relative entropy (distance between",
null,
"$\\Phi(\\rho)$ and",
null,
"$\\sigma_{\\beta=0}$) at any time during the computation. If the Gibbs state corresponding to the upper bound has",
null,
"$\\beta<\\beta_C$, the noisy quantum calculation now provides a certifiably worse answer than a classical optimizer would.\n\nAny noiseless quantum optimization computation produces better answers the longer it is allowed to run, but in real processes, the longer the computation, the more the noise corrupts the answer. This technique gives a hard limit on how long a computation can be run for some quantum noise/device/algorithm combination before the noise makes the process worse than classical optimization.\n\n## Takeaways from the Authors\n\nThis technique requires one to choose the noise model, device parameters, optimization technique, and target problem before “ruling out” the quantum advantage of any such combination. This freedom is very powerful. One can use this technique to find regions of classical superiority for all near-term quantum optimization devices, as well as to quantify the reduction in noise level required for a chance at quantum advantage.\n\nLet us return to fault-tolerance for a moment. If the reduction in noise required for a NISQ device/algorithm to reach quantum advantage is below the noise threshold where fault-tolerance becomes possible, it will likely be better to perform fault tolerant operation instead. Any NISQ device/algorithm revealed by this technique to have such a noise reduction requirement will likely be a bad optimizer for its entire existence, rendered obsolete by current classical superiority followed by the rise of fault-tolerant quantum computing.\n\nThe authors consider variational quantum eigensolvers, quantum annealing, and quantum approximate optimization with currently available superconducting qubit devices and simple noise models, finding a noise reduction requirement below the fault tolerant threshold. However, accurate noise models are often far more complex, and unique to devices than the local depolarizing noise used by the authors for quantum circuits, so some hope of quantum advantage in optimization remains for existing devices. Further, the authors found current NISQ devices have a chance at quantum advantage in the arena of Hamiltonian simulation, a related and important problem (though perhaps not as profitable as optimization).\n\nThe most interesting applications of this technique maybe yet to come, as more scientists begin applying the technique to a wider variety of quantum computations. At its best, this technique can guide experimental NISQ devices into parameter regions with a real chance at quantum advantage and reveal which emerging quantum devices have the best shot at quantum advantage.\n\nThanks for reading – I’ll answer questions in the comments, and don’t be afraid to look at the paper (open-access here) if you are curious about the details.\n\nMatthew Kowalsky is a graduate student at the University of Southern California whose research focuses on quantum annealing, open quantum system modeling, and benchmarking dedicated optimization hardware."
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https://www.aimsciences.org/article/doi/10.3934/dcds.2022029 | [
"• PDF\n• Cite\n• Share\nArticle Contents",
null,
"",
null,
"Article Contents\n\n# The sandpile identity element on an ellipse\n\n• * Corresponding author: Andrew Melchionna\n\nThe first author is supported by NSF grant DMS-1455272\n\n• We consider certain elliptical subsets of the square lattice. The recurrent representative of the identity element of the sandpile group on this graph consists predominantly of a biperiodic pattern, along with some noise. We show that as the lattice spacing tends to 0, the fraction of the area taken up by the pattern in the identity element tends to 1.\n\nMathematics Subject Classification: 35Q70, 31A05, 05C57.\n\n Citation:",
null,
"•",
null,
"• Figure 1. Figures 1a) and 1b) correspond to the identity elements for the ellipses $E_{A,k}$, with $A = \\begin{pmatrix}\\frac{10}{9} & \\frac{1}{3} \\\\[0.15cm] \\frac{1}{3} & 1\\end{pmatrix}$ and $\\begin{pmatrix}\\frac{4}{3} & \\frac{1}{2} \\\\[0.15cm] \\frac{1}{2} & \\frac{3}{4}\\end{pmatrix},$ respectively, with $k = 18^2$ in both. Figures 1c) and 1d) show the patterns $p_A$ corresponding to the ellipses in 1a) and 1b) respectively. A black square represents a vertex with 3 grains of sand; dark patterned (parallel lines), 2; light patterned (cross), 1; white, 0.\n\nFigure 2. The identity elements for the ellipses given by $E_{A,k}$, with $A = \\begin{pmatrix}\\frac{5}{4} & \\frac{1}{2} \\\\[0.15cm] \\frac{1}{2} & 1\\end{pmatrix}$ and various $k$. A black square represents a vertex with 3 grains of sand; dark patterned (parallel lines), 2; light patterned (cross), 1; white, 0.\n\nFigure 7. A demonstration of r-goodness. The white point $x_1$ is not r-good, while the light gray point $x_2$ is $r$-good\n\nFigure 3. The identity element for the graph $E^p_{A,k},$ with} $A = \\begin{pmatrix}\\frac{10}{9} & \\frac{1}{3} \\\\[0.15cm] \\frac{1}{3} & 1\\end{pmatrix}$, $k = 18^2,$ and p = (.47,.5)\n\nFigure 4. A portion of the Apollonian circle packing between the lines $\\{x = 0\\}$ and $\\{x = 2\\}$\n\nFigure 5. A portion of the boundary of the set $\\Theta$\n\nFigure 6. Identity elements of $E_{A_i,k}$, $A_i \\notin \\Gamma^+,$ $k = 18^2$ $A_1 \\in \\Gamma \\backslash \\partial \\Gamma,$ $A_2 \\notin \\Gamma$\n\nFigure 8. $x \\in \\partial E$ and the corresponding $y_x \\in \\partial \\tilde{F}_1$\n\nFigure 9. A schematic for Lemma 4.5. $x \\in B - A$\n\nFigure 10. The identity element of $B_{1152}(0) \\cap \\mathbb{Z}^2$\n\n•",
null,
"## Article Metrics",
null,
"",
null,
"DownLoad: Full-Size Img PowerPoint"
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http://www.yndxxb.ynu.edu.cn/yndxxbzrkxb/article/latest | [
"优先发表\n\n, doi: 10.7540/j.ynu.20210250\n\n, doi: 10.7540/j.ynu.20200459\n\n, doi: 10.7540/j.ynu.20200453\n\n, doi: 10.7540/j.ynu.20210450\n\n, doi: 10.7540/j.ynu.20210218\n\n, doi: 10.7540/j.ynu.20210286\n\n, doi: 10.7540/j.ynu.P00173\n\n, doi: 10.7540/j.ynu.P00152\n\n, doi: 10.7540/j.ynu.P00113\n\n, doi: 10.7540/j.ynu.P00033\n\n, doi: 10.7540/j.ynu.P00018\n\n, doi: 10.7540/j.ynu.P00214\n\n, doi: 10.7540/j.ynu.20210141\n\n, doi: 10.7540/j.ynu.20210044\n\n\\begin{document}$H$\\end{document} 为无限维复可分的Hilbert空间，\\begin{document}$B(H)$\\end{document}\\begin{document}$H$\\end{document} 上的有界线性算子的全体. \\begin{document}$T \\in B(H)$\\end{document} 称为满足 \\begin{document}$({R_1})$\\end{document} 性质，若 \\begin{document}${\\sigma _a}(T)\\backslash {\\sigma _{ab}}(T) \\subseteq {\\pi _{00}}(T)$\\end{document}，其中 \\begin{document}${\\sigma _a}(T)$\\end{document}\\begin{document}${\\sigma _{ab}}(T)$\\end{document} 分别表示算子 \\begin{document}$T$\\end{document} 的逼近点谱和本质逼近点谱，\\begin{document}${\\pi _{00}}(T) = \\{ \\lambda \\in {\\rm{iso}}\\sigma (T):0 < {\\rm{ dim}}N(T - \\lambda I) < \\infty \\}$\\end{document}. 若 \\begin{document}${\\sigma _a}(T)\\backslash {\\sigma _{ab}}(T) = {\\pi _{00}}(T)$\\end{document}，则称 \\begin{document}$T$\\end{document} 满足 \\begin{document}$(R)$\\end{document} 性质. 运用新的谱集，给出了有界线性算子及其函数满足 \\begin{document}$({R_1})$\\end{document} 性质或者 \\begin{document}$(R)$\\end{document} 性质的充要条件；同时得到了a-Weyl定理和 \\begin{document}$(R)$\\end{document} 性质的新的判定方法.\n, doi: 10.7540/j.ynu.20210370\n\n, doi: 10.7540/j.ynu.20210394\n\n, doi: 10.7540/j.ynu.20210098\n\n, doi: 10.7540/j.ynu.20210024\n\n, doi: 10.7540/j.ynu.20210042\n\n, doi: 10.7540/j.ynu.20200436\n\n, doi: 10.7540/j.ynu.20210171\n\n, doi: 10.7540/j.ynu.20210107\n\n, doi: 10.7540/j.ynu.20200432\n\n, doi: 10.7540/j.ynu.20210251\n\n, doi: 10.7540/j.ynu.20200310\n\n, doi: 10.7540/j.ynu.20210177"
] | [
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https://rbsesolution.in/rbse-solutions-for-class-11-physics-chapter-11-thermal-properties-of-matter/ | [
"# Rbse Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter\n\nRbse Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter (updated)\n\n# Thermal Properties Of Matter Rbse Solutions – Class 11 Physics\n\n11.1. The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.Answer\n\nKelvin and Celsius\nscales are related as:\nTC = TK – 273.15 … (i)\nCelsius and Fahrenheit scales are related as:\nTF = (9/5)TC + 32 ….(ii)\n\nFor neon:\nTK = 24.57 K\nT= 24.57 – 273.15 = –248.58°C\nTF = (9/5)TC + 32\n= (9/5) × (-248.58) +32\n= 415.440 F\n\nFor carbon dioxide:\nTK = 216.55 K\nTC= 216.55 – 273.15 = –56.60°C\nTF = (9/5)TC + 32\n= (9/5) × (-56.60) +32\n= -69.880 C\n\n11.2. Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB?\n\nTriple point of water on absolute scaleA, T1 = 200 A\nTriple point of water on absolute scale B, T2 = 350 B\nTriple point of water on Kelvin scale, TK = 273.15 K\nThe temperature 273.15 K on Kelvin scale is equivalent to 200 A on absolute scale A.\nT1 = TK\n200 A = 273.15 K\n∴ A = 273.15/200\nThe temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B.\nT2 = TK\n350 B = 273.15\n∴ B = 273.15/350\nTA is triple point of water on scale A.\nTB is triple point of water on scale B.\n∴ 273.15/200 × T = 273.15/350 × TB\nTherefore, the ratio TA : Tis given as 4 : 7.\n\n11.3. The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:\nRo [1 + α (– To)]\nThe resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?Answer\n\nIt is given that:\nR0 [1 + α (– T0)] … (i)\nwhere,\nR0 and T0 are the initial resistance and temperature respectively\n\nR and T are the final resistance and temperature respectively\nα is a constant\nAt the triple point of water, T0 = 273.15 K\nResistance of lead, R0 = 101.6 Ω\nAt normal melting point of lead, T = 600.5 K\nResistance of lead, R = 165.5 Ω\n\nSubstituting these values in equation (i), we get:\nRo [1 + α (– To)]\n165.5 = 101.6 [ 1 + α(600.5 – 273.15) ]\n1.629 = 1 + α (327.35)\n∴ α = 0.629 / 327.35 = 1.92 × 10-3 K-1\n\nFor resistance, R1 = 123.4 Ω\nR1 R0 [1 + α (– T0)]\nwhere,\nT is the temperature when the resistance of lead is 123.4 Ω\n123.4 = 101.6 [ 1 + 1.92 × 10-3( T – 273.15) ]\nSolving for T, we get\nT = 384.61 K.\n\n(a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?\n(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale?\n(c) The absolute temperature (Kelvin scale) is related to the temperature tc on the Celsius scale by\ntc = – 273.15\nWhy do we have 273.15 in this relation, and not 273.16?\n(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?Answer\n\n(a) The triple point of water has a unique value of 273.16 K. At particular values of volume and pressure, the triple point of water is always 273.16 K. The melting point of ice and boiling point of water do not have particular values because these points depend on pressure and temperature.\n\n(b) The absolute zero or 0 K is the other fixed point on the Kelvin absolute scale.\n\n(c) The temperature 273.16 K is the triple point of water. It is not the melting point of ice. The temperature 0°C on Celsius scale is the melting point of ice. Its corresponding value on Kelvin scale is 273.15 K.\nHence, absolute temperature (Kelvin scale) T, is related to temperature tc, on Celsius scale as:\ntc = T – 273.15\n\n(d) Let TF be the temperature on Fahrenheit scale and TK be the temperature on absolute scale. Both the temperatures can be related as:\n(TF – 32) / 180 = (TK – 273.15) / 100 ….(i)\nLet TF1 be the temperature on Fahrenheit scale and TK1 be the temperature on absolute scale. Both the temperatures can be related as:\n(TF1 – 32) / 180 = (TK1 – 273.15) / 100 ….(ii)\nIt is given that:\nTK1 – TK = 1 K\nSubtracting equation (i) from equation (ii), we get:\n(TF1 – TF) / 180 = (TK1 – TK) / 100 = 1 / 100\nTF1 – TF = (1 ×180) / 100 = 9/5\nTriple point of water = 273.16 K\n∴ Triple point of water on absolute scale = 273.16 × (9/5) = 491.69\n\n11.5. Two ideal gas thermometers Aand Buse oxygen and hydrogen respectively. The following observations are made:\n\n(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers Aand B?\n(b) What do you think is the reason behind the slight difference in answers of thermometers Aand B(The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?Answer\n\n(a) Triple point of water, T = 273.16 K.\nAt this temperature, pressure in thermometer A, PA = 1.250 × 105 Pa\nLet T1 be the normal melting point of sulphur.\nAt this temperature, pressure in thermometer A, P1 = 1.797 × 105 Pa\nAccording to Charles’ law, we have the relation:\nPA / T = P1 / T1\n∴ T1 = (P1T) / PA = (1.797 × 10 × 273.16) / (1.250 × 105)\n= 392.69 K\nTherefore, the absolute temperature of the normal melting point of sulphur as read by thermometer A is 392.69 K.\nAt triple point 273.16 K, the pressure in thermometer B, PB = 0.200 × 105 Pa\nAt temperature T1, the pressure in thermometer B, P2 = 0.287 × 105 Pa\nAccording to Charles’ law, we can write the relation:\nPB / T = P1 / T1\n(0.200 × 105) / 273.16 = (0.287 × 105) / T1\n∴ T1 = [ (0.287 × 105) / (0.200 × 105) ] × 273.16 = 391.98 K\nTherefore, the absolute temperature of the normal melting point of sulphur as read by thermometer B is 391.98 K.\n\n(b) The oxygen and hydrogen gas present in thermometers A and B respectively are not perfect ideal gases. Hence, there is a slight difference between the readings of thermometers A and B.\nTo reduce the discrepancy between the two readings, the experiment should be carried under low pressure conditions. At low pressure, these gases behave as perfect ideal gases.\n\n11.6. A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0 °C? Coefficient of linear expansion of steel = 1.20 × 10–5 K–1.\n\nLength of the steel tape at temperature T = 27°C, l = 1 m = 100 cm\nAt temperature T1 = 45°C, the length of the steel rod, l1 = 63 cm\nCoefficient of linear expansion of steel, α = 1.20 × 10–5 K–1\nLet l2 be the actual length of the steel rod and l‘ be the length of the steel tape at 45°C.\nl’ = l + αl(T1 – T)\n∴ l’ = 100 + 1.20 × 10-5 × 100(45 – 27)\n= 100.0216 cm\nHence, the actual length of the steel rod measured by the steel tape at 45°C can be calculated as:\nl2 = (100.0216 / 100) × 63 = 63.0136 cm\nTherefore, the actual length of the rod at 45.0°C is 63.0136 cm. Its length at 27.0°C is 63.0 cm.\n\n11.7. A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: αsteel = 1.20 × 10–5 K–1.\n\nThe given temperature, T = 27°C can be written in Kelvin as:\n27 + 273 = 300 K\nOuter diameter of the steel shaft at Td1 = 8.70 cm\nDiameter of the central hole in the wheel at Td2 = 8.69 cm\nCoefficient of linear expansion of steel, αsteel = 1.20 × 10–5 K–1\nAfter the shaft is cooled using ‘dry ice’, its temperature becomes T1.\nThe wheel will slip on the shaft, if the change in diameter, Δd = 8.69 – 8.70\n= – 0.01 cm\nTemperature T1, can be calculated from the relation:\nΔd = d1αsteel (T1 – T)\n0.01 = 8.70 × 1.20 × 10–5 (T1 – 300)\n(T1 – 300) = 95.78\n∴ T1= 204.21 K\n= 204.21 – 273.16\n= –68.95°C\nTherefore, the wheel will slip on the shaft when the temperature of the shaft is –69°C.\n\n11.8. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10–5 K–1.\n\nInitial temperature, T1 = 27.0°C\nDiameter of the hole at T1d1 = 4.24 cm\nFinal temperature, T2 = 227°C\nDiameter of the hole at Td2\nCo-efficient of linear expansion of copper, αCu= 1.70 × 10–5 K–1\nFor co-efficient of superficial expansion β,and change in temperature ΔT, we have the relation:\nChange in area (∆) / Original area (A) = βT\n[ (πd22/ 4) – (πd12 / 4) ] / (πd11 / 4) = ∆A / A\n∴ ∆A / A = (d22 – d12) / d12\nBut β = 2α\n∴ (d22 – d12) / d12 = 2α∆T\n(d22 / d12) – 1 = 2α(T2 – T1)\nd22 / 4.242 = 2 × 1.7 × 10-5 (227 – 27) +1\nd22 = 17.98 × 1.0068 = 18.1\n∴ d2 = 4.2544 cm\nChange in diameter = d2 – d= 4.2544 – 4.24 = 0.0144 cm\nHence, the diameter increases by 1.44 × 10–2 cm.\n\n11.9. A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm? Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1; Young’s modulus of brass = 0.91 × 1011 Pa.Answer\n\nInitial temperature, T1 = 27°C\nLength of the brass wire at T1l = 1.8 m\nFinal temperature, T2 = –39°C\nDiameter of the wire, d = 2.0 mm = 2 × 10–3 m\nTension developed in the wire = F\nCoefficient of linear expansion of brass, α = 2.0 × 10–5 K–1\nYoung’s modulus of brass, = 0.91 × 1011 Pa\nYoung’s modulus is given by the relation:\nγ = Stress / Strain = (F/A) / (∆L/L)\nL = F X L / (A X Y) ……(i)\n\nWhere,\n= Tension developed in the wire\nA = Area of cross-section of the wire.\nΔL = Change in the length, given by the relation:\nΔL = αL(T2 – T1) … (ii)\nEquating equations (i) and (ii), we get:\nαL(T2 – T1) = FL / [ π(d/2)2 X Y ]\nF = α(T2 – T1)πY(d/2)2\nF = 2 × 10-5 × (-39-27) × 3.14 × 0.91 × 1011 × (2 × 10-3 / 2 )2\n= -3.8 × 102 N\nHence, the tension developed in the wire is 3.8 ×102 N.\n\n11.10. A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1, steel = 1.2 × 10–5 K–1).\n\nInitial temperature, T1 = 40°C\nFinal temperature, T2 = 250°C\nChange in temperature, ΔT = T2 – T= 210°C\nLength of the brass rod at T1l1 = 50 cm\nDiameter of the brass rod at T1d1 = 3.0 mm\nLength of the steel rod at T2l2 = 50 cm\nDiameter of the steel rod at T2d2 = 3.0 mm\nCoefficient of linear expansion of brass, α1 = 2.0 × 10–5K–1\nCoefficient of linear expansion of steel, α2 = 1.2 × 10–5K–1\nFor the expansion in the brass rod, we have:\nChange in length (∆l1) / Original length (l1) = α1ΔT\n∴ ∆l1 = 50 × (2.1 × 10-5) × 210\n= 0.2205 cm\nFor the expansion in the steel rod, we have:\nChange in length (∆l2) / Original length (l2) = α2ΔT\n∴ ∆l1 = 50 × (1.2 × 10-5) × 210\n= 0.126 cm\nTotal change in the lengths of brass and steel,\nΔl = Δl1 + Δl2\n= 0.2205 + 0.126\n= 0.346 cm\nTotal change in the length of the combined rod = 0.346 cm\nSince the rod expands freely from both ends, no thermal stress is developed at the junction.\n\n11.11. The coefficient of volume expansion of glycerin is 49 × 10–5 K–1. What is the fractional change in its density for a 30 °C rise in temperature?\n\nCoefficient of volume expansion of glycerin, αV = 49 × 10–5 K–1\nRise in temperature, Δ= 30°C\nFractional change in its volume = ΔV/V\nThis change is related with the change in temperature as:\nΔV/V = αV ΔT\nVT2 – VT1 = VT1 αV ΔT\n(m /ρT2)- (m /ρT1) = (m /ρT1)αV ΔT\nWhere,\nm = Mass of glycerineρT1 = Initial density at TT2 = Initial density at T2\nT1 – ρT) / ρT2 = Fractional change in density\n∴ Fractional change in the density of glycerin = 49 ×10–5 × 30 = 1.47 × 10–2.\n\n11.12. A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g–1 K–1.\n\nPower of the drilling machine, P = 10 kW = 10 × 10W\nMass of the aluminum block, m = 8.0 kg = 8 × 103 g\nTime for which the machine is used, t = 2.5 min = 2.5 × 60 = 150 s\nSpecific heat of aluminium, c = 0.91 J g–1 K–1\nRise in the temperature of the block after drilling = δT\nTotal energy of the drilling machine = Pt\n= 10 × 10× 150\n= 1.5 × 106 J\nIt is given that only 50% of the power is useful.\nUseful energy, ∆Q = (50/100) × 1.5 × 106 = 7.5 × 105 J\nBut ∆Q = mc∆T\n∴ ∆T = ∆Q / mc\n= (7.5 × 105) / (8 × 103 × 0.91)\n= 103o C\nTherefore, in 2.5 minutes of drilling, the rise in the temperature of the block is 103°C.\n\n11.13. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g–1 K–1; heat of fusion of water = 335 J g–1).\n\nMass of the copper block, = 2.5 kg = 2500 g\nRise in the temperature of the copper block, Δθ = 500°C\nSpecific heat of copper, C = 0.39 J g–1 °C–1\nHeat of fusion of water, L = 335 J g–1\nThe maximum heat the copper block can lose, Q = mCΔθ\n= 2500 × 0.39 × 500\n= 487500 J\nLet m1 g be the amount of ice that melts when the copper block is placed on the ice block.\nThe heat gained by the melted ice, Q = m1L\n∴ m1 = Q / L = 487500 / 335 = 1455.22 g\nHence, the maximum amount of ice that can melt is 1.45 kg.\n\n11.14. In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?\n\nMass of the metal, m = 0.20 kg = 200 g\nInitial temperature of the metal, T1 = 150°C\nFinal temperature of the metal, T2 = 40°C\nCalorimeter has water equivalent of mass, m = 0.025 kg = 25 g\nVolume of water, V = 150 cm3\nMass (M) of water at temperature T = 27°C:\n150 × 1 = 150 g\nFall in the temperature of the metal:\nΔT1 – T= 150 – 40 = 110°C\nSpecific heat of water, Cw = 4.186 J/g/°K\nSpecific heat of the metal = C\nHeat lost by the metal, θ = mCΔT … (i)\nRise in the temperature of the water and calorimeter system:\nΔT = 40 – 27 = 13°C\nHeat gained by the water and calorimeter system:\nΔθ′′ = m1 CwΔT\n= (M + m′) Cw ΔT … (ii)\nHeat lost by the metal = Heat gained by the water and colorimeter system\nmCΔT = (M + mCw ΔT\n200 × C × 110 = (150 + 25) × 4.186 × 13\n∴ C = (175 × 4.186 × 13) / (110 × 200) = 0.43 Jg-1K-1\nIf some heat is lost to the surroundings, then the value of C will be smaller than the actual value.\n\n11.15. Given below are observations on molar specific heats at room temperature of some common gases.\n\nThe measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?\n\nThe gases listed in the given table are diatomic. Besides the translational degree of freedom, they have other degrees of freedom (modes of motion).\nHeat must be supplied to increase the temperature of these gases. This increases the average energy of all the modes of motion. Hence, the molar specific heat of diatomic gases is more than that of monatomic gases.\nIf only rotational mode of motion is considered, then the molar specific heat of a diatomic gas = (5/2)R\n= (5/2) × 1.98 = 4.95 cal mol-1 K-1\nWith the exception of chlorine, all the observations in the given table agree with (5/2)R.\nThis is because at room temperature, chlorine also has vibrational modes of motion besides rotational and translational modes of motion.\n\n11.16. Answer the following questions based on the Pphase diagram of carbon dioxide:\n(a) At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist in equilibrium?\n(b) What is the effect of decrease of pressure on the fusion and boiling point of CO2?\n(c) What are the critical temperature and pressure for CO2? What is their significance?\n(d) Is CO2 solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60° C under 10 atm, (c) 15 °C under 56 atm?\n\nThe Pphase diagram for COis shown in the following figure:\n\n(a) The solid, liquid and vapour phase of carbon dioxide exist in equilibrium at the triple point, i.e., temprature = – 56.6° C and pressure = 5.11 atm.(b) With the decrease in pressure, both the fusion and boiling point of carbon dioxide will decrease.(c) For carbon dioxide, the critical temperature is 31.1° C and critical pressure is 73.0 atm. If the temprature of carbon dioxide is more than 31.1° C, it can not be liquified, however large pressure we may apply.(d) Carbon dioxide will be (a) a vapour, at =70° C under 1atm. (b) a solid, at -6° C under 10 atm (c) a liquid, at 15° C under 56 atm.\n\n11.17. Answer the following questions based on the P–T phase diagram of CO2:\n(a) CO2 at 1 atm pressure and temperature – 60 °C is compressed isothermally. Does it go through a liquid phase?\n(b) What happens when CO2 at 4 atm pressure is cooled from room temperature at constant pressure?\n(c) Describe qualitatively the changes in a given mass of solid CO2 at 10 atm pressure and temperature –65 °C as it is heated up to room temperature at constant pressure.\n(d) CO2 is heated to a temperature 70° C and compressed isothermally. What changes in its properties do you expect to observe?\n\nThe Pphase diagram for COis shown in the following figure:\n\n(a) Since the temprature -60° C lies to the left of 56.6° C on the curve i.e. lies in the region vapour and solid phase, so carbon dioxide will condense directly into the solid without becoming liquid.\n\n(b) Since the pressure 4 atm is less than 5.11 atm the carbon dioxide will condense directly into solid without becoming liquid.\n\n(c) When a solid COat 10 atm pressure and -65° C temprature is heated, it is first converted into liquid. A further increase in temprature brings it into the vapour phase. At P = 10 atm, if a horizontal line is drawn parallel to the T-axis, then the points of intersection of this line with the fusion and vaporization curve will give the fusion and boiling points of COat 10 atm.\n\n(d) Since 70° C is higher than the critical temprature of CO2, so the COgas can not be converted into liquid state on being compressed isothermally at 70° C. It will remain in the vapour state. However, the gas will depart more and more from its perfect gas behaviour with the increase in pressure.\n\n11.18. A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 min, what is the average rate of extra evaporation caused, by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g–1.\n\nInitial temperature of the body of the child, T1 = 101°F\nFinal temperature of the body of the child, T2 = 98°F\nChange in temperature, Δ= [ (101 – 98) × (5/9) ] o C\nTime taken to reduce the temperature, t = 20 min\nMass of the child, m = 30 kg = 30 × 103 g\nSpecific heat of the human body = Specific heat of water = c\n= 1000 cal/kg/ °C\nLatent heat of evaporation of water, L = 580 cal g–1\nThe heat lost by the child is given as:\n∆θ = mcT\n= 30 × 1000 × (101 – 98) × (5/9)\n= 50000 cal\nLet m1 be the mass of the water evaporated from the child’s body in 20 min.\nLoss of heat through water is given by:\n∆θ = m1L\n∴ m1 = ∆θ / L\n= (50000 / 580) = 86.2 g\n∴ Average rate of extra evaporation caused by the drug = m1 / t\n= 86.2 / 200\n= 4.3 g/min.\n\n11.19. A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole is 0.01 J s–1 m–1 K–1. [Heat of fusion of water = 335 × 103 J kg–1]\n\nSide of the given cubical ice box, s = 30 cm = 0.3 m\nThickness of the ice box, l = 5.0 cm = 0.05 m\nMass of ice kept in the ice box, m = 4 kg\nTime gap, t = 6 h = 6 × 60 × 60 s\nOutside temperature, T = 45°C\nCoefficient of thermal conductivity of thermacole, K = 0.01 J s–1 m–1 K–1\nHeat of fusion of water, L = 335 × 103 J kg–1\nLet m be the total amount of ice that melts in 6 h.\nThe amount of heat lost by the food:\nθ = KA(T – 0)t / l\nWhere,\nA = Surface area of the box = 6s2 = 6 × (0.3)2 = 0.54 m3\nθ = 0.01 × 0.54 × 45 × 6 × 60 × 60 / 0.05 = 104976 J\nBut θ = m’L\n∴ m‘ = θ/L\n= 104976/(335 × 103) = 0.313 kg\nMass of ice left = 4 – 0.313 = 3.687 kg\nHence, the amount of ice remaining after 6 h is 3.687 kg.\n\n11.20. A brass boiler has a base area of 0.15 m2 and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s –1 m–1 K–1; Heat of vaporisation of water = 2256 × 103 J kg–1.\n\nBase area of the boiler, = 0.15 m2\nThickness of the boiler, l = 1.0 cm = 0.01 m\nBoiling rate of water, R = 6.0 kg/min\nMass, m = 6 kg\nTime, t = 1 min = 60 s\nThermal conductivity of brass, K = 109 J s –1 m–1 K–1\nHeat of vaporisation, L = 2256 × 103 J kg–1\nThe amount of heat flowing into water through the brass base of the boiler is given by:\nθ = KA(T1 – T2t / l ….(i)\n\nwhere,\nT1 = Temperature of the flame in contact with the boiler\nT2 = Boiling point of water = 100°C\nHeat required for boiling the water:\nθ = mL … (ii)\nEquating equations (i) and (ii), we get:\n∴ mL = KA(T1 – T2t / l\nT1 – T2 = mLl / KAt\n= 6 × 2256 × 103 × 0.01 / (109 × 0.15 × 60)\n= 137.98 o C\nTherefore, the temperature of the part of the flame in contact with the boiler is 237.98°C.\n\n11.21. Explain why:\n(a) a body with large reflectivity is a poor emitter\n(b) a brass tumbler feels much colder than a wooden tray on a chilly day\n(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace\n(d) the earth without its atmosphere would be inhospitably cold\n(e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water\n\n(a) A body with a large reflectivity is a poor absorber of light radiations. A poor absorber will in turn be a poor emitter of radiations. Hence, a body with a large reflectivity is a poor emitter.\n\n(b) Brass is a good conductor of heat. When one touches a brass tumbler, heat is conducted from the body to the brass tumbler easily. Hence, the temperature of the body reduces to a lower value and one feels cooler.\nWood is a poor conductor of heat. When one touches a wooden tray, very little heat is conducted from the body to the wooden tray. Hence, there is only a negligible drop in the temperature of the body and one does not feel cool.\nThus, a brass tumbler feels colder than a wooden tray on a chilly day.\n\n(c) An optical pyrometer calibrated for an ideal black body radiation gives too low a value for temperature of a red hot iron piece kept in the open.\nBlack body radiation equation is given by:\nE = σ (T4 – T04)\nWhere,\nT = Temperature of optical pyrometer\nTo = Temperature of open space\nσ = Constant\nHence, an increase in the temperature of open space reduces the radiation energy.\nWhen the same piece of iron is placed in a furnace, the radiation energy, E = σ T4\n\n(d) Without its atmosphere, earth would be inhospitably cold. In the absence of atmospheric gases, no extra heat will be trapped. All the heat would be radiated back from earth’s surface.\n\n(e) A heating system based on the circulation of steam is more efficient in warming a building than that based on the circulation of hot water. This is because steam contains surplus heat in the form of latent heat (540 cal/g).\n\n11.22. A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86496055,"math_prob":0.99645394,"size":26016,"snap":"2023-40-2023-50","text_gpt3_token_len":8183,"char_repetition_ratio":0.16373213,"word_repetition_ratio":0.089464724,"special_character_ratio":0.35182196,"punctuation_ratio":0.11799864,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99776214,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T09:25:52Z\",\"WARC-Record-ID\":\"<urn:uuid:19a1f7c6-d024-4816-a86d-0d5fbbd87298>\",\"Content-Length\":\"196318\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9b7d41da-6db6-4100-a9ce-94a8f2a55cd4>\",\"WARC-Concurrent-To\":\"<urn:uuid:db3b7660-0ffc-4056-b10c-f326cfd5247b>\",\"WARC-IP-Address\":\"65.108.74.236\",\"WARC-Target-URI\":\"https://rbsesolution.in/rbse-solutions-for-class-11-physics-chapter-11-thermal-properties-of-matter/\",\"WARC-Payload-Digest\":\"sha1:VMXV4GHZEFHQHC7KLHXQOOEGODA5UAFB\",\"WARC-Block-Digest\":\"sha1:FTTWPSYRFL7W23JRO7555K45EOAKUX2B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510501.83_warc_CC-MAIN-20230929090526-20230929120526-00482.warc.gz\"}"} |
https://www.dezyre.com/recipes/import-csv-file-in-python | [
"How to import a CSV file in Python?\nMACHINE LEARNING RECIPES DATA CLEANING PYTHON DATA MUNGING PANDAS CHEATSHEET ALL TAGS\n\n# How to import a CSV file in Python?\n\nThis recipe helps you import a CSV file in Python\n\n0\n\n## Recipe Objective\n\nBefore using any model first thing you have to do is to import the dataset. There are many ways to do it.\n\nSo this is the recipe on how we can import a CSV file in Python.\n\n## Step 1 - Import the library\n\n``` import csv import numpy import pandas ```\n\nWe have imported numpy, csv and pandas which is needed.\n\nWe are first importing dataset with use of CSV library. Here we need to pass the file name, the quoting in the csv.reader function and the parameter delimiter which signifies that by which charactor the data is seperated. we can also store it in a object and can use the data by calling the object. ``` filename = \"load.csv\" raw_data = open(filename, \"rt\") reader = csv.reader(raw_data, delimiter=\",\", quoting=csv.QUOTE_NONE) x = list(reader) data = numpy.array(x).astype(\"float\") print(data.shape) ``` We can also do this with the help of numpy. For this we have to pass the file name in numpy.loadtxt and we have to also set the delimiter which signifies that by which charactor the data is seperated. ``` filename = \"load.csv\" raw_data = open(filename, \"rt\") data = numpy.loadtxt(raw_data, delimiter=\",\") print(data.shape) ``` We can also do this with the help of pandas. For CSV we need to make a array of names of columns before importing the data then to import we have to pass the file name and the array we have created to the pandas.read_csv function. ``` filename = \"load.csv\" names = [\"preg\", \"plas\", \"pres\", \"skin\", \"test\", \"mass\", \"pedi\", \"age\", \"class\"] data = pandas.read_csv(filename, names=names) print(data.shape) ```\n\n#### Relevant Projects\n\n##### Machine Learning project for Retail Price Optimization\nIn this machine learning pricing project, we implement a retail price optimization algorithm using regression trees. This is one of the first steps to building a dynamic pricing model.\n\n##### Deep Learning with Keras in R to Predict Customer Churn\nIn this deep learning project, we will predict customer churn using Artificial Neural Networks and learn how to model an ANN in R with the keras deep learning package.\n\n##### Machine Learning or Predictive Models in IoT - Energy Prediction Use Case\nIn this machine learning and IoT project, we are going to test out the experimental data using various predictive models and train the models and break the energy usage.\n\n##### Learn to prepare data for your next machine learning project\nText data requires special preparation before you can start using it for any machine learning project.In this ML project, you will learn about applying Machine Learning models to create classifiers and learn how to make sense of textual data.\n\n##### Ecommerce product reviews - Pairwise ranking and sentiment analysis\nThis project analyzes a dataset containing ecommerce product reviews. The goal is to use machine learning models to perform sentiment analysis on product reviews and rank them based on relevance. Reviews play a key role in product recommendation systems.\n\n##### Time Series Forecasting with LSTM Neural Network Python\nDeep Learning Project- Learn to apply deep learning paradigm to forecast univariate time series data.\n\n##### Identifying Product Bundles from Sales Data Using R Language\nIn this data science project in R, we are going to talk about subjective segmentation which is a clustering technique to find out product bundles in sales data.\n\n##### Resume parsing with Machine learning - NLP with Python OCR and Spacy\nIn this machine learning resume parser example we use the popular Spacy NLP python library for OCR and text classification.\n\n##### Customer Churn Prediction Analysis using Ensemble Techniques\nIn this machine learning churn project, we implement a churn prediction model in python using ensemble techniques.\n\n##### Data Science Project in Python on BigMart Sales Prediction\nThe goal of this data science project is to build a predictive model and find out the sales of each product at a given Big Mart store."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8303557,"math_prob":0.6448908,"size":1476,"snap":"2021-04-2021-17","text_gpt3_token_len":349,"char_repetition_ratio":0.12703805,"word_repetition_ratio":0.15899582,"special_character_ratio":0.24796748,"punctuation_ratio":0.14473684,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9726465,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-19T05:38:17Z\",\"WARC-Record-ID\":\"<urn:uuid:3eb56c1a-e520-4300-ac7b-bbed26f536f8>\",\"Content-Length\":\"34890\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c71a1374-ec2c-4f4d-b7e0-5058597f80a9>\",\"WARC-Concurrent-To\":\"<urn:uuid:f5147d12-7c53-40fe-b2aa-7fe915582aa2>\",\"WARC-IP-Address\":\"184.173.29.42\",\"WARC-Target-URI\":\"https://www.dezyre.com/recipes/import-csv-file-in-python\",\"WARC-Payload-Digest\":\"sha1:EHX6Y4GCDHVJ7BNOZ52WAE3AGBI5CZYB\",\"WARC-Block-Digest\":\"sha1:T3TVXMITXCPZFOZESZTK32EIUF4JCYVY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703517966.39_warc_CC-MAIN-20210119042046-20210119072046-00310.warc.gz\"}"} |
http://safecurves.cr.yp.to/proof/720869.html | [
"Primality proof for n = 720869:\n\nTake b = 2.\n\nb^(n-1) mod n = 1.\n\n10601 is prime.\nb^((n-1)/10601)-1 mod n = 357937, which is a unit, inverse 411018.\n\n(10601) divides n-1.\n\n(10601)^2 > n.\n\nn is prime by Pocklington's theorem."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.69883066,"math_prob":0.9996369,"size":218,"snap":"2020-10-2020-16","text_gpt3_token_len":88,"char_repetition_ratio":0.14018692,"word_repetition_ratio":0.0,"special_character_ratio":0.54587156,"punctuation_ratio":0.18518518,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9971274,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-17T12:19:36Z\",\"WARC-Record-ID\":\"<urn:uuid:3a15406c-4a6a-4c6d-ace5-14662391e566>\",\"Content-Length\":\"471\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4276a2ec-389b-4f93-b9af-f37f8d7092ca>\",\"WARC-Concurrent-To\":\"<urn:uuid:863d2629-7751-4be0-add4-627a1aa77ce2>\",\"WARC-IP-Address\":\"131.193.32.108\",\"WARC-Target-URI\":\"http://safecurves.cr.yp.to/proof/720869.html\",\"WARC-Payload-Digest\":\"sha1:KWPPBHVSXVTSNOVDOQRRSF3BHWW5BSJO\",\"WARC-Block-Digest\":\"sha1:4FVX2JQSP3BKUSP24UZ72HQFBIRDTZU7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875142323.84_warc_CC-MAIN-20200217115308-20200217145308-00456.warc.gz\"}"} |
https://www.ademcetinkaya.com/2023/08/analyzing-m-stock-comprehensive.html | [
"Outlook: Macy's Inc Common Stock is assigned short-term Baa2 & long-term B1 estimated rating.\nAUC Score : What is AUC Score?\nShort-Term Revised1 :\nDominant Strategy : Hold\nTime series to forecast n: for Weeks2\nMethodology : Modular Neural Network (News Feed Sentiment Analysis)\nHypothesis Testing : Stepwise Regression\nSurveillance : Major exchange and OTC\n\n1The accuracy of the model is being monitored on a regular basis.(15-minute period)\n\n2Time series is updated based on short-term trends.\n\n## Summary\n\nMacy's Inc Common Stock prediction model is evaluated with Modular Neural Network (News Feed Sentiment Analysis) and Stepwise Regression1,2,3,4 and it is concluded that the M stock is predictable in the short/long term. A modular neural network (MNN) is a type of artificial neural network that can be used for news feed sentiment analysis. MNNs are made up of multiple smaller neural networks, called modules. Each module is responsible for learning a specific task, such as identifying sentiment in text or identifying patterns in data. The modules are then combined to form a single neural network that can perform multiple tasks. In the context of news feed sentiment analysis, MNNs can be used to identify the sentiment of news articles, social media posts, and other forms of online content. This information can then be used to filter out irrelevant or unwanted content, to identify trends in public opinion, and to target users with relevant advertising. According to price forecasts for 1 Year period, the dominant strategy among neural network is: Hold",
null,
"## Key Points\n\n1. What is the best way to predict stock prices?\n2. Decision Making\n3. How do you decide buy or sell a stock?\n\n## M Target Price Prediction Modeling Methodology\n\nWe consider Macy's Inc Common Stock Decision Process with Modular Neural Network (News Feed Sentiment Analysis) where A is the set of discrete actions of M stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4\n\nF(Stepwise Regression)5,6,7= $\\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \\dots & {p}_{1n}\\\\ & ⋮\\\\ {p}_{j1}& {p}_{j2}& \\dots & {p}_{jn}\\\\ & ⋮\\\\ {p}_{k1}& {p}_{k2}& \\dots & {p}_{kn}\\\\ & ⋮\\\\ {p}_{n1}& {p}_{n2}& \\dots & {p}_{nn}\\end{array}$ X R(Modular Neural Network (News Feed Sentiment Analysis)) X S(n):→ 1 Year $\\begin{array}{l}\\int {e}^{x}\\mathrm{rx}\\end{array}$\n\nn:Time series to forecast\n\np:Price signals of M stock\n\nj:Nash equilibria (Neural Network)\n\nk:Dominated move\n\na:Best response for target price\n\n### Modular Neural Network (News Feed Sentiment Analysis)\n\nA modular neural network (MNN) is a type of artificial neural network that can be used for news feed sentiment analysis. MNNs are made up of multiple smaller neural networks, called modules. Each module is responsible for learning a specific task, such as identifying sentiment in text or identifying patterns in data. The modules are then combined to form a single neural network that can perform multiple tasks. In the context of news feed sentiment analysis, MNNs can be used to identify the sentiment of news articles, social media posts, and other forms of online content. This information can then be used to filter out irrelevant or unwanted content, to identify trends in public opinion, and to target users with relevant advertising.\n\n### Stepwise Regression\n\nStepwise regression is a method of variable selection in which variables are added or removed from a model one at a time, based on their statistical significance. There are two main types of stepwise regression: forward selection and backward elimination. In forward selection, variables are added to the model one at a time, starting with the variable with the highest F-statistic. The F-statistic is a measure of how much improvement in the model is gained by adding the variable. Variables are added to the model until no variable adds a statistically significant improvement to the model.\n\nFor further technical information as per how our model work we invite you to visit the article below:\n\nHow do AC Investment Research machine learning (predictive) algorithms actually work?\n\n## M Stock Forecast (Buy or Sell)\n\nSample Set: Neural Network\nStock/Index: M Macy's Inc Common Stock\nTime series to forecast: 1 Year\n\nAccording to price forecasts, the dominant strategy among neural network is: Hold\n\nStrategic Interaction Table Legend:\n\nX axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)\n\nY axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)\n\nZ axis (Grey to Black): *Technical Analysis%\n\n### Financial Data Adjustments for Modular Neural Network (News Feed Sentiment Analysis) based M Stock Prediction Model\n\n1. When measuring a loss allowance for a lease receivable, the cash flows used for determining the expected credit losses should be consistent with the cash flows used in measuring the lease receivable in accordance with IFRS 16 Leases.\n2. If a guarantee provided by an entity to pay for default losses on a transferred asset prevents the transferred asset from being derecognised to the extent of the continuing involvement, the transferred asset at the date of the transfer is measured at the lower of (i) the carrying amount of the asset and (ii) the maximum amount of the consideration received in the transfer that the entity could be required to repay ('the guarantee amount'). The associated liability is initially measured at the guarantee amount plus the fair value of the guarantee (which is normally the consideration received for the guarantee). Subsequently, the initial fair value of the guarantee is recognised in profit or loss when (or as) the obligation is satisfied (in accordance with the principles of IFRS 15) and the carrying value of the asset is reduced by any loss allowance.\n3. There are two types of components of nominal amounts that can be designated as the hedged item in a hedging relationship: a component that is a proportion of an entire item or a layer component. The type of component changes the accounting outcome. An entity shall designate the component for accounting purposes consistently with its risk management objective.\n4. The significance of a change in the credit risk since initial recognition depends on the risk of a default occurring as at initial recognition. Thus, a given change, in absolute terms, in the risk of a default occurring will be more significant for a financial instrument with a lower initial risk of a default occurring compared to a financial instrument with a higher initial risk of a default occurring.\n\n*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.\n\n### M Macy's Inc Common Stock Financial Analysis*\n\nRating Short-Term Long-Term Senior\nOutlook*Baa2B1\nIncome StatementBaa2B1\nBalance SheetBa1Baa2\nLeverage RatiosBaa2B3\nCash FlowBa3Ba2\nRates of Return and ProfitabilityBaa2C\n\n*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.\nHow does neural network examine financial reports and understand financial state of the company?\n\n## Conclusions\n\nMacy's Inc Common Stock is assigned short-term Baa2 & long-term B1 estimated rating. Macy's Inc Common Stock prediction model is evaluated with Modular Neural Network (News Feed Sentiment Analysis) and Stepwise Regression1,2,3,4 and it is concluded that the M stock is predictable in the short/long term. According to price forecasts for 1 Year period, the dominant strategy among neural network is: Hold\n\n### Prediction Confidence Score\n\nTrust metric by Neural Network: 85 out of 100 with 496 signals.\n\n## References\n\n1. J. Filar, L. Kallenberg, and H. Lee. Variance-penalized Markov decision processes. Mathematics of Opera- tions Research, 14(1):147–161, 1989\n2. Jorgenson, D.W., Weitzman, M.L., ZXhang, Y.X., Haxo, Y.M. and Mat, Y.X., 2023. Google's Stock Price Set to Soar in the Next 3 Months. AC Investment Research Journal, 220(44).\n3. Blei DM, Lafferty JD. 2009. Topic models. In Text Mining: Classification, Clustering, and Applications, ed. A Srivastava, M Sahami, pp. 101–24. Boca Raton, FL: CRC Press\n4. Robins J, Rotnitzky A. 1995. Semiparametric efficiency in multivariate regression models with missing data. J. Am. Stat. Assoc. 90:122–29\n5. Y. Le Tallec. Robust, risk-sensitive, and data-driven control of Markov decision processes. PhD thesis, Massachusetts Institute of Technology, 2007.\n6. Bengio Y, Schwenk H, Senécal JS, Morin F, Gauvain JL. 2006. Neural probabilistic language models. In Innovations in Machine Learning: Theory and Applications, ed. DE Holmes, pp. 137–86. Berlin: Springer\n7. J. Filar, D. Krass, and K. Ross. Percentile performance criteria for limiting average Markov decision pro- cesses. IEEE Transaction of Automatic Control, 40(1):2–10, 1995.\nFrequently Asked QuestionsQ: What is the prediction methodology for M stock?\nA: M stock prediction methodology: We evaluate the prediction models Modular Neural Network (News Feed Sentiment Analysis) and Stepwise Regression\nQ: Is M stock a buy or sell?\nA: The dominant strategy among neural network is to Hold M Stock.\nQ: Is Macy's Inc Common Stock stock a good investment?\nA: The consensus rating for Macy's Inc Common Stock is Hold and is assigned short-term Baa2 & long-term B1 estimated rating.\nQ: What is the consensus rating of M stock?\nA: The consensus rating for M is Hold.\nQ: What is the prediction period for M stock?\nA: The prediction period for M is 1 Year"
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null,
"https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjFLSWGMgf-VNlXeBjcxALtjPNTOwU1o3a6cCodJ20i2lwOsKip8dCGzFfM9-T7ybUr9A9bEIzxEwrFM2v6o-oDNph0naYcOyR7DKQFl9I--oIWBlQ6MfEVcq9UpXdMVJcvJnXFM7uznsYB8kWNELACQLLkU1a9SqFUT99rSqmTsc9qnh0cxno5KEGt9g/s16000/graph15.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8875196,"math_prob":0.84211314,"size":9700,"snap":"2023-40-2023-50","text_gpt3_token_len":2154,"char_repetition_ratio":0.11221122,"word_repetition_ratio":0.24902217,"special_character_ratio":0.2098969,"punctuation_ratio":0.12820514,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9610724,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T04:32:51Z\",\"WARC-Record-ID\":\"<urn:uuid:bb6f19fb-e065-464b-b12d-5089d8f9ac63>\",\"Content-Length\":\"319694\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5cceec3e-ec31-4f54-b89c-bd616fddabde>\",\"WARC-Concurrent-To\":\"<urn:uuid:79c6a68f-36b5-465a-b13d-78923eb9eaf1>\",\"WARC-IP-Address\":\"172.253.62.121\",\"WARC-Target-URI\":\"https://www.ademcetinkaya.com/2023/08/analyzing-m-stock-comprehensive.html\",\"WARC-Payload-Digest\":\"sha1:MWRLJFU3PC4JHMIPRVVEQCYEHRP7MZW3\",\"WARC-Block-Digest\":\"sha1:MIVWXHREA7MCGIZ5OCGLIYEPJVCB5HTB\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506479.32_warc_CC-MAIN-20230923030601-20230923060601-00768.warc.gz\"}"} |
https://isabelle.in.tum.de/repos/isabelle/rev/b6e3ac38397d | [
"author paulson Wed, 11 Jul 2001 13:56:15 +0200 changeset 11405 b6e3ac38397d parent 11404 280436a346ca child 11406 2b17622e1929\ntweak\n```--- a/doc-src/TutorialI/basics.tex\tWed Jul 11 13:55:43 2001 +0200\n+++ b/doc-src/TutorialI/basics.tex\tWed Jul 11 13:56:15 2001 +0200\n@@ -2,11 +2,11 @@\n\n\\section{Introduction}\n\n-This is a tutorial on how to use the theorem prover Isabelle/HOL as a specification and\n-verification system. Isabelle is a generic system for implementing logical\n-formalisms, and Isabelle/HOL is the specialization of Isabelle for\n-HOL, which abbreviates Higher-Order Logic. We introduce HOL step by step\n-following the equation\n+This book is a tutorial on how to use the theorem prover Isabelle/HOL as a\n+specification and verification system. Isabelle is a generic system for\n+implementing logical formalisms, and Isabelle/HOL is the specialization\n+of Isabelle for HOL, which abbreviates Higher-Order Logic. We introduce\n+HOL step by step following the equation\n\\[ \\mbox{HOL} = \\mbox{Functional Programming} + \\mbox{Logic}. \\]\nWe do not assume that the reader is familiar with mathematical logic but that\n(s)he is used to logical and set theoretic notation, such as covered```"
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https://wfncnews.com/71716/how-to-use-python-for-financial-calculations-debt-management | [
"Wednesday, October 20, 2021\n\n# How To Use Python For Financial Calculations & Debt Management\n\nThe leading programming language, Python, has become a go-to tool for financial calculations and debt management. Widely used in quantitative finance to analyze large datasets, Python also allows users to manage their own money.\n\nWith simple and easy-to-learn functions, anyone with basic math knowledge can use a variety of financial calculations in Python. Keep reading this article to learn how to implement some of these financial calculations, such as simple interest, compound interest, and monthly EMI.\n\nThis guide also covers everything you need to know about debt management with Python and how it can support the use of debt calculators.\n\n## Basic Financial Calculations Using Python\n\nPython showcases a number of libraries that support different mathematical calculations. Therefore, it’s simple nature and accessibility make it the ideal choice for anyone looking to manage their finances.\n\nLet’s take a look at some of the basic financial calculations you can do in Python, in which you can use within a debt management setting.\n\n### How Do I Calculate Simple Interest Using Python?\n\nFirstly, Simple interest is a calculation used to work out the accumulated interest of loans or savings. Simple interest is received or paid in a fixed percentage over a set period of time.\n\nFor example, take a student who acquires a simple interest loan to pay for a college that costs \\$20,000 with an annual interest rate of 9%. Over the next three years, the simple interest would amount to \\$1,800. Therefore, the total amount owed would be \\$21,800.\n\nTo calculate simple interest using Python you will need the principal amount, rate of interest, and period of time. Then put these into the function “simple_interest”. With the example above, it would look like: “simple_interest(20,000, 9, 3)”.\n\nThese data types with brackets in Python are referred to as Tuples.\n\n### How Do I Calculate Compound Interest Using Python?\n\nSecondly, compound interest calculates the interest on a deposit or loan based on the initial principal and the previously accumulated interest. In other words, it refers to “interest on interest”. Calculating compound interest is useful for all types of debt management\n\nFor example, let’s say the student from earlier has taken an interest rate of 9% per year compound annually. While in the first year the total amount would increase by \\$1,800, the second year would be calculated as \\$1,890. Then, in the third year, the interest would be calculated at \\$2,060. After three years, the student would owe \\$24,950.\n\nThe compound interest calculation on Python is very simple. This time, use the function “compund_interest(1000, 10, 2)”.\n\n### How Do I Calculate Monthly EMI Using Python?\n\nEquated Monthly Installment (EMI) is a fixed payment from the borrower to the lender made on a specific date each month.\n\nFor example, if our student borrows \\$20,000 with a 9% annual interest rate for 3 years (36 months), the total amount of interest during the loan term will be \\$22,896. This means that the EMI will be \\$636.\n\nTo work out EMI using Python you will need the principal borrowed, rate of interest, and tenure of the loan. Then the function will look like “EMI(20000, 9, 36)”.\n\n## Python Loan Calculations For Debt Management\n\nPython also offers great support for those calculating how much they can borrow and what the repayment would be. The Numpy financial library for Python can answer both of these questions for you.\n\nOnce you have installed this library, you can begin to calculate how much you can afford to borrow using monthly payment, interest rate, and length of the loan. All the calculations are easy to learn with dedicated support from Python.\n\nBy using the Numby financial library, you can calculate:\n\n• How much my payment will be.\n• How much of my payment goes to interest and how much goes to principal.\n• What the total interest paid is over the duration of the loan.\n• How much I am able to borrow.\n\nTherefore, the more you know about your loan, the easier debt management will be. This will certainly provide support for your future debt management plan, or when considering a repayment strategy.\n\nFinally, pair the Numby financial library for Python with The Get Out of Debt Calculator to solve all your debt-related questions!\n\n## Summary:\n\nPython is a great way to calculate financial figures and stay on top of your debt management. Whether you want to quickly calculate simple interest, compound interest, or monthly EMI, Python is the software to use.\n\nFurthermore, don’t fall behind in your debt management plans with the Numpy financial library for Python. Using this, you are able to know exactly how much you can borrow and what the repayment will be.\n\nSo, why not use Python today and never worry about any tricky financial calculations again!\n\n## How Do I Get an Amex Card After Cards Closed for Dispute Abuse\n\nQuestion: Dear Steve, I had two Amex cards, Blue and EveryDay. Unfortunately, they were both …"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9353251,"math_prob":0.8995667,"size":4751,"snap":"2021-43-2021-49","text_gpt3_token_len":987,"char_repetition_ratio":0.16621023,"word_repetition_ratio":0.012771392,"special_character_ratio":0.2140602,"punctuation_ratio":0.12131148,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.989852,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-25T23:15:26Z\",\"WARC-Record-ID\":\"<urn:uuid:e2be4846-da03-4c8e-ad95-244b0bf1ba86>\",\"Content-Length\":\"63411\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5025af92-b84d-4f4a-a328-dbb32df56961>\",\"WARC-Concurrent-To\":\"<urn:uuid:924f4068-4a49-4859-9ad5-76b6491f442a>\",\"WARC-IP-Address\":\"172.67.146.199\",\"WARC-Target-URI\":\"https://wfncnews.com/71716/how-to-use-python-for-financial-calculations-debt-management\",\"WARC-Payload-Digest\":\"sha1:POVVUFOESWKZTWODENSPSKR2WXVBGHZZ\",\"WARC-Block-Digest\":\"sha1:UXU6O2DNHSOTZ2SQYO7JXMMBTQ3FQ3PA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587770.37_warc_CC-MAIN-20211025220214-20211026010214-00714.warc.gz\"}"} |
https://en.m.wikibooks.org/wiki/OpenGL_Programming/GLStart/Tut3 | [
"OpenGL Programming/GLStart/Tut3\n\nTutorial 3: Drawing Primitives\n\nPrimitives are basic shapes that you can easily draw. It can be a triangle, a square, or even a single point. In this lesson we are going to learn how to draw some basic primitives including: points, triangles, and different polygons.\n\nImmediate Mode\n\nThe easiest way to do drawing in OpenGL is using the Immediate Mode. For this, you use the glBegin() function which takes as one parameter the “mode” or type of object you want to draw.\n\nHere is a list of the possible modes and what they mean:\n\nGL_POINTS Draws points on screen. Every vertex specified is a point. Draws lines on screen. Every two vertices specified compose a line. Draws connected lines on screen. Every vertex specified after first two are connected. Draws connected lines on screen. The last vertex specified is connected to first vertex. Draws triangles on screen. Every three vertices specified compose a triangle. Draws connected triangles on screen. Every vertex specified after first three vertices creates a triangle. Draws connected triangles like GL_TRIANGLE_STRIP, except draws triangles in fan shape. Draws quadrilaterals (4 – sided shapes) on screen. Every four vertices specified compose a quadrilateral. Draws connected quadrilaterals on screen. Every two vertices specified after first four compose a connected quadrilateral. Draws a polygon on screen. Polygon can be composed of as many sides as you want.\n\nWhen you are done drawing all the vertices of the specific primitive, you put in the next line the glEnd() function which ends drawing of that primitive.\n\nDrawing Points\n\nLets do an example with the simplest mode, the GL_POINTS. When drawing points using OpenGL, the default size of the points is 1 pixel wide and high. This would be very hard to see when you run the program. To edit the size of the point you want to draw, you use the glPointSize() function which takes as one parameter the size of the point you want.\n\nNow in the Render() function, before you write the glBegin() code, we will set the point size to be 10 pixels in size:\n\n``` glPointSize(10.0f);\n```\n\nAfter that, any drawing of points will be drawn 10 pixels wide and high. Now write the glBegin() function with the GL_POINTS mode parameter. Then after that specify the vertices you want to use using the glVertex3f() function. For this example, we want the upper – right corner (1.0,1.0,0.0) and the lower – left corner (-1.0,-1.0,0.0) of the screen to have a point. After drawing those two points, make sure to end the drawing with the glEnd() function:\n\n``` glBegin(GL_POINTS); //starts drawing of points\nglVertex3f(1.0f,1.0f,0.0f);//upper-right corner\nglVertex3f(-1.0f,-1.0f,0.0f);//lower-left corner\nglEnd();//end drawing of points\n```\n\nHere is the whole render function for your reference followed by a sample output. The whole code for this section of the tutorial can be found in the downloadable files. This example is called “points”:\n\n```void Render()\n{\n//clear color and depth buffer\nglClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);\n\nglTranslatef(0.0f,0.0f,-4.0f);//move forward 4 units\n\nglColor3f(0.0f,0.0f,1.0f); //blue color\n\nglPointSize(10.0f);//set point size to 10 pixels\n\nglBegin(GL_POINTS); //starts drawing of points\nglVertex3f(1.0f,1.0f,0.0f);//upper-right corner\nglVertex3f(-1.0f,-1.0f,0.0f);//lower-left corner\nglEnd();//end drawing of points\n}\n```\n\nDrawing a Line Loop\n\nI would like to cover every single mode in the glBegin() function, but it would take up too much time and space. So I will cover the advanced ones and most likely cover the other ones when they are deemed useful.\n\nA line loop requires at least two vertices. Every vertex specified after that is connected to the vertex specified before it and the first vertex specified. So if we put a vertex to the left of the window, to the right of the window, on top and on the bottom, we would have a rotated square:\n\nLets do this diagram in OpenGL. First we use the glBegin() function and pass the parameter of GL_LINE_LOOP to tell OpenGL we are going to start drawing a line loop. Then we pass the four vertices to create the rotated square. The first vertex being to the left of the window (-1.0f,0.0f,0.0f), second one being at the bottom of the window (0.0f,-1.0f,0.0f), third at the right of window (1.0f,0.0f,0.0f), and the fourth and last being at the top of the window (0.0f,1.0f,0.0f). Then we make sure to put in glEnd() to tell OpenGL we are done drawing the line loop:\n\n``` glBegin(GL_LINE_LOOP);//start drawing a line loop\nglVertex3f(-1.0f,0.0f,0.0f);//left of window\nglVertex3f(0.0f,-1.0f,0.0f);//bottom of window\nglVertex3f(1.0f,0.0f,0.0f);//right of window\nglVertex3f(0.0f,1.0f,0.0f);//top of window\nglEnd();//end drawing of line loop\n```\n\nFollowing is the whole Render() function followed by a sample output. Look for the project file called “lineLoop” on the downloadable file for this chapter:\n\n```void Render()\n{\n//clear color and depth buffer\nglClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);\n\nglTranslatef(0.0f,0.0f,-4.0f);//move forward 4 units\n\nglColor3f(0.0f,0.0f,1.0f); //blue color\n\nglBegin(GL_LINE_LOOP);//start drawing a line loop\nglVertex3f(-1.0f,0.0f,0.0f);//left of window\nglVertex3f(0.0f,-1.0f,0.0f);//bottom of window\nglVertex3f(1.0f,0.0f,0.0f);//right of window\nglVertex3f(0.0f,1.0f,0.0f);//top of window\nglEnd();//end drawing of line loop\n}\n```\n\nDrawing Triangles\n\nTriangles are composed of three vertices. For this example we are going to use the regular GL_TRIANGLES mode to draw two triangles side by side.\n\nFirst we want a triangle to the left. So we need three vertices on the left side representing one triangle and three vertices on the right side of the window representing the second triangle. Note that you don’t need two glBegin() functions to draw two triangles. Since the mode GL_TRIANGLES is plural, it can handle more than one triangle in between one glBegin() and glEnd() function call:\n\nNote that I wrote on the bottom – right corner of the diagram the coordinates of the vertices. Here is the code to draw these two triangles:\n\n``` glBegin(GL_TRIANGLES);//start drawing triangles\nglVertex3f(-1.0f,-0.25f,0.0f);//triangle one first vertex\nglVertex3f(-0.5f,-0.25f,0.0f);//triangle one second vertex\nglVertex3f(-0.75f,0.25f,0.0f);//triangle one third vertex\n//drawing a new triangle\nglVertex3f(0.5f,-0.25f,0.0f);//triangle two first vertex\nglVertex3f(1.0f,-0.25f,0.0f);//triangle two second vertex\nglVertex3f(0.75f,0.25f,0.0f);//triangle two third vertex\nglEnd();//end drawing of triangles\n```\n\nFollowing is the whole Render() function for this example followed by a sample output. To view the whole code, please see the “triangle” project folder in the downloadable files:\n\n```void Render()\n{\n//clear color and depth buffer\nglClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);\n\nglTranslatef(0.0f,0.0f,-4.0f);//move forward 4 units\n\nglColor3f(0.0f,0.0f,1.0f); //blue color\n\nglBegin(GL_TRIANGLES);//start drawing triangles\nglVertex3f(-1.0f,-0.25f,0.0f);//triangle one first vertex\nglVertex3f(-0.5f,-0.25f,0.0f);//triangle one second vertex\nglVertex3f(-0.75f,0.25f,0.0f);//triangle one third vertex\n//drawing a new triangle\nglVertex3f(0.5f,-0.25f,0.0f);//triangle two first vertex\nglVertex3f(1.0f,-0.25f,0.0f);//triangle two second vertex\nglVertex3f(0.75f,0.25f,0.0f);//triangle two third vertex\nglEnd();//end drawing of triangles\n}\n```\n\nDrawing Polygons\n\nPolygons consist of at least three vertices that, when connected, make up a shape. In this example we are going to be using the GL_POLYGON mode to draw a six – sided shape.\n\nThe GL_POLYGON mode allows you to draw a shape with any number of sides as you want. Since GL_POLYGON is a singular word (meaning no “S” at the end of the word), you can only draw one polygon between a glBegin() and glEnd() function call. Also, the last vertex you specify is automatically connected to the first vertex specified. And of course, since a polygon is a closed shape, like the triangle, the shape will be filled with your specified color (which for now is blue). Here is the diagram for our six – sided shape:\n\n(note: this can't be used for concave polygons)\n\nHere is the code to draw this polygon:\n\n``` glBegin(GL_POLYGON);//begin drawing of polygon\nglVertex3f(-0.5f,0.5f,0.0f);//first vertex\nglVertex3f(0.5f,0.5f,0.0f);//second vertex\nglVertex3f(1.0f,0.0f,0.0f);//third vertex\nglVertex3f(0.5f,-0.5f,0.0f);//fourth vertex\nglVertex3f(-0.5f,-0.5f,0.0f);//fifth vertex\nglVertex3f(-1.0f,0.0f,0.0f);//sixth vertex\nglEnd();//end drawing of polygon\n```\n\nFollowing is the whole Render() function followed by a sample output. Look for the whole code in the “polygon” project folder on the included downloadable files:\n\n```void Render()\n{\n//clear color and depth buffer\nglClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);\n\nglTranslatef(0.0f,0.0f,-4.0f);//move forward 4 units\n\nglColor3f(0.0f,0.0f,1.0f); //blue color\n\nglBegin(GL_POLYGON);//begin drawing of polygon\nglVertex3f(-0.5f,0.5f,0.0f);//first vertex\nglVertex3f(0.5f,0.5f,0.0f);//second vertex\nglVertex3f(1.0f,0.0f,0.0f);//third vertex\nglVertex3f(0.5f,-0.5f,0.0f);//fourth vertex\nglVertex3f(-0.5f,-0.5f,0.0f);//fifth vertex\nglVertex3f(-1.0f,0.0f,0.0f);//sixth vertex\nglEnd();//end drawing of polygon\n}\n```\n\nDisplay Lists\n\nUsing the OpenGL immediate mode has several drawbacks. For example, vertex data must be transferred on the fly to the graphics memory whenever drawing. This can be improved by using Display Lists. Display Lists essentially take a glBegin()/glEnd() command sequence and store that on the graphics card side in an efficient manner.\n\n(todo: elaborate on that later, see for some infos - glGenLists(), glNewList(), glEndList(), glCallList(), glDeleteLists())\n\nVertex Arrays and Vertex Buffers\n\nDisplay Lists also have drawbacks. Once compiled, Display Lists are static, and cannot be changed. Also, vertices shared by several primitives still need to be represented and transformed multiple times when drawing. This is very inefficient.\n\nThese problems are addressed by Vertex Arrays and Vertex Buffers. The idea is simply to use arrays to store vertex data. Using glDrawArrays(), primitives using these vertices can be drawn.\n\nFor storage of data in video memory, Vertex Buffers can be used (glGenBuffer(), glBindBuffer(), glBufferData(), glDeleteBuffers()). Two types of Vertex Buffers exist:\n\n• GL_ARRAY_BUFFERs hold actual vertex data.\n• GL_ELEMENT_ARRAY_BUFFERs hold indices to vertices stored in a separate GL_ARRAY_BUFFER and, thus, allow reusing vertices in several primitives.\n\n(todo: elaborate on that later, example)"
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https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_17&oldid=110894 | [
"# 2014 AMC 8 Problems/Problem 17\n\n(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)\n\n## Problem\n\nGeorge walks",
null,
"$1$ mile to school. He leaves home at the same time each day, walks at a steady speed of",
null,
"$3$ miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first",
null,
"$\\frac{1}{2}$ mile at a speed of only",
null,
"$2$ miles per hour. At how many miles per hour must George run the last",
null,
"$\\frac{1}{2}$ mile in order to arrive just as school begins today?",
null,
"$\\textbf{(A) }4\\qquad\\textbf{(B) }6\\qquad\\textbf{(C) }8\\qquad\\textbf{(D) }10\\qquad \\textbf{(E) }12$\n\n## Solution 1\n\nNote that on a normal day, it takes him",
null,
"$1/3$ hour to get to school. However, today it took",
null,
"$\\frac{1/2 \\text{ mile}}{2 \\text{ mph}}=1/4$ hour to walk the first",
null,
"$1/2$ mile. That means that he has",
null,
"$1/3 -1/4 = 1/12$ hours left to get to school, and",
null,
"$1/2$ mile left to go. Therefore, his speed must be",
null,
"$\\frac{1/2 \\text{ mile}}{1/12 \\text { hour}}=\\boxed{6 \\text{ mph}}$, so",
null,
"$\\boxed{\\text{(B) }6}$ is the answer.\n\n## Solution 2\n\nUsing the harmonic mean formula, and making the speed he needs to take to get to school for the last half",
null,
"$x$, we can make the expression:",
null,
"$\\frac{2 * 2x}{2 + x}$\n\nWhich simplifies to:",
null,
"$\\frac{4x}{2 + x}$\n\nThen, because we know that since he goes at",
null,
"$3$ mph on a normal day, we can say that it is the harmonic mean of the two rates he goes at today, so we add that to our expression and turn it into something familiar:",
null,
"$\\frac{4x}{2 + x} = 3$\n\nSolving that equation gives us:",
null,
"$\\boxed{\\text {(B) } 6}$\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.",
null,
""
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null
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https://www.numberempire.com/4757119 | [
"Home | Menu | Get Involved | Contact webmaster",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"# Number 4757119\n\nfour million seven hundred fifty seven thousand one hundred nineteen\n\n### Properties of the number 4757119\n\n Factorization 4757119 Divisors 1, 4757119 Count of divisors 2 Sum of divisors 4757120 Previous integer 4757118 Next integer 4757120 Is prime? YES (332668th prime) Previous prime 4757117 Next prime 4757147 4757119th prime 81593507 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 10010001001011001111111 Octal 22113177 Duodecimal 1714b67 Hexadecimal 48967f Square 22630181180161 Square root 2181.0820708997 Natural logarithm 15.375152790868 Decimal logarithm 6.6773440154933 Sine 0.023411359543013 Cosine 0.99972591656131 Tangent 0.023417777968125\nNumber 4757119 is pronounced four million seven hundred fifty seven thousand one hundred nineteen. Number 4757119 is a prime number. The prime number before 4757119 is 4757117. The prime number after 4757119 is 4757147. Number 4757119 has 2 divisors: 1, 4757119. Sum of the divisors is 4757120. Number 4757119 is not a Fibonacci number. It is not a Bell number. Number 4757119 is not a Catalan number. Number 4757119 is not a regular number (Hamming number). It is a not factorial of any number. Number 4757119 is a deficient number and therefore is not a perfect number. Binary numeral for number 4757119 is 10010001001011001111111. Octal numeral is 22113177. Duodecimal value is 1714b67. Hexadecimal representation is 48967f. Square of the number 4757119 is 22630181180161. Square root of the number 4757119 is 2181.0820708997. Natural logarithm of 4757119 is 15.375152790868 Decimal logarithm of the number 4757119 is 6.6773440154933 Sine of 4757119 is 0.023411359543013. Cosine of the number 4757119 is 0.99972591656131. Tangent of the number 4757119 is 0.023417777968125"
] | [
null,
"https://www.numberempire.com/images/graystar.png",
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"https://www.numberempire.com/images/graystar.png",
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https://www.swmm456.com/2013/08/how-are-negative-transect-elevations.html | [
"## Thursday, August 8, 2013\n\n### How are Negative Transect Elevations Used in SWMM5?\n\nSubject: How are Negative Transect Elevations Used in SWMM5?\n\nYou can have negative elevations in the Transects of SWMM 5 as the elevations are transformed internally to relative depths above the node inverts in the SWMM 5 engine (Figure 1). The slope of the link is calculated from the link offset elevations (Figure 3) and the cross sectional information for the irregular link in SWMM 5 (Figure 2) is computed from the Transect data (Figure 4). The Water Surface elevation of the link is based on the node inverts (Figure 5).\n\nFigure 1. Transect Editor of SWMM 5\n\nFigure 2. The Transect Data is Used in the Irregular of HEC-RAS Shape of SWMM 5\n\nFigure 3. The slope of the link with the Transect is calculated from the link upstream and downstream offset elevations – not the Transectdata which is relative.\n\nFigure 4. Transect Data Transformed into Tables of Area, Hydraulic Radius and Width from the Transect Data internally in SWMM 5.\n\nFigure 5. HGL of the Water Surface Elevation from the Node Invert and Link Offset Elevations."
] | [
null
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https://goprep.co/analysis-shows-that-nickel-oxide-has-the-formula-ni0-98o1-00-i-1nk0is | [
"# Analysis shows that nickel oxide has the formula Ni0.98O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?\n\nGiven formula of nickel oxide = Ni0.98O1.00\n\nTherefore, Ni:O = 0.98:1.00\n\nLet the total no. of Ni ion = 98\nand, the total no. of O ion = 100\n\nSo, Total charge on O-2 ion = 100 × (-2) = -200\n\nNow, let the number of Ni2+ ions = x\n\nTherefore, number of Ni+3 = 98- x\n\nNow, since compound is neutral\n\nTherefore, no. of Ni2+ ions + no. of Ni3+ ions +no. of O-2 ions = 0\n\nx (+2) + (98 - x) × (+3) + (-200) = 0\n\n+2x + 294 - 3x – 200 =0\n\nx-94 = 0\n\nx = 94\n\nTherefore, the number of Ni2+ ions = 94\n\nNumber of ni3+ ions = 98- 94 = 4\n\nNow, fraction of nickel exists as Ni2+",
null,
"= 0.959\n\nFraction of nickel exists as Ni 3+ ions",
null,
"= 0.041\n\nRate this question :\n\nHow useful is this solution?\nWe strive to provide quality solutions. Please rate us to serve you better.\nRelated Videos",
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""
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https://sfaxien.net/blog/6d6ysd69/lesson-plan-system-of-linear-equations-cf0eee | [
"The six-page worksheet... Students explore the concept of linear equations. In this linear equation worksheet, students solve 18 problems regarding slope-intercept, linear equations, parallel lines and perpendicular lines. In this algebra activity, the pupils add, subtract, multiply, and divide to solve linear equations. Developing an intuition for the kinds and descriptions of solutions is key for success in those later courses. Scholars are able to recognize situations where matrices are the efficient method of solving. Explanations and examples are provided prior to the problems. Did you know… We have over 220 college courses that prepare you to earn Using systems of simultaneous equations, they graph each scenario to determine the best value. The activity instructs pupils to use a graphing calculator to graph groups of linear equations before comparing them. They then assist in combining equations with a graphical element and complete various graphs with this... Students solve linear equations and relate it to the real world. The one thing you need to know in algebra is how to solve for an unknown variable. Using arrangements of colored tiles representing algebraic expressions, pupils layer them over each other to determine the values of variables. In this graphing linear equations instructional activity, students plot 22 lines on their coordinate grids. Algebra students use manipulatives to solve linear equations. When graphing the linear equations, pupils determine the... A clear introduction to slope-intercept form, this would be helpful for learners who are struggling with this concept. Hopefully you will be able to see improvement from the previous lesson plan on the Pythagorean Theorem. In this Algebra I/Algebra II worksheet, students examine linear equations to determine the slope and y-intercept and write the equation of the line given the pertinent information. Two friends have a specific amount of money in savings and plan to add to it each week. They relate linear equations to the real world. Solve problems using a variety of problem solving strategies involving two variables. Starting with the basics, students discover that systems of equations are a useful skill. In this linear systems worksheet, 9th graders solve and complete 24 different problems that include solving various linear systems. How much wood would ...? An easy-to-use interactive has pupils move a seesaw on a coordinate plane to investigate the standard form of a linear equation. Here is a compiled list of resources that address all of the standards regarding Algebra Common Core for high school. In this linear equations worksheet, learners read and interpret graphed linear equations. HSA-REI.C.6 Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables. In addition, they see how to find the break-even point using the... Systematically check a solution in a system. What does the rate of change mean when graphing a line? Students will interpret a variety of systems of linear equations including several real-world word problems. Answers are provided on the last page. Big Ideas: You can use substitution when given equations in slope-intercept form to solve a system of equations. In this Algebra II lesson plan, 11th graders solve s system of linear equations with row reduction to Echelon Form on an augmented matrix. They write the identities of the matrices given, review, and grade as a class. There are 24 questions with an answer key. After plotting... How many ways can you write a linear equation? Word problems offer class members an opportunity to learn the concept of solving linear systems using graphs. Make planning your course easier by using our syllabus as a guide. How do you know if a slope is positive or negative? There are 15 questions with an answer key. The video lessons, quizzes and transcripts can easily be adapted to provide your lesson plans with engaging and dynamic educational content. Check out the Common Core with a collection that represents all the 8th grade math standards. Pupils follow rules to combine circles and squares to add the two equations together, resulting in one... Cover it up to reveal the values of variables. They graph each linear equation, and identify the x and y-intercept of each problem. Learners apply the concepts of algebra to a plane in the air. A lesson has pupils determine solutions for two-variable equations using tables. In this algebra lesson, students draw out comic strips to represent the steps when solving linear equations. Linear systems are the most common type of system that appear in real life problems. Variables appear in the denominator and/or in the neumerator. In this linear systems learning exercise, students use a matrix to represent linear systems. Then, have your cell phone agents use linear equations to visually display three cell plans and their advantages. So far, all of the systems of equations we have encountered in this lesson have involved linear equations. One way to solve a system of linear equations is by graphing each linear equation on the same -plane. Everything you need to teach a lesson on two variable linear equations is right here. Students also write the matrix as the product of the... Eleventh graders solve system of linear equations. The lesson builds on previous experiences with problem situations represented and solved using a system of equations. Linear equations are the focus of activities that ask learners to first complete a task that involves interpreting algebraic expressions and solving linear equations. Middle schoolers problem solve and calculate the answers to eighteen linear equations involving a variety of variables. They identify the input and output of each relation. Learners solve equations in two variables using standard form of an equation. Learners then watch as the instructor shows how to use the equation to analyze the situation. Start with the basics and practice solving linear equations and inequalities. Students need to be able to meaningfully describe the solution to a system of linear equations. Through using the graphing calculator students will be able to visually see how linear equations cross to create a solution to a system of equations. Young mathematicians learn to solve linear equations by using a graphing calculator. To learn more, visit our Earning Credit Page. Students write linear equations to match a given graph. Students solve one variable equations by isolating the variable and simplifying like terms. How can you graph lines that don't seem to fit on the given graph? Classmates use algebraic methods to transform sides of equations to expressions with fewer terms. Khan Academy is a 501(c)(3) nonprofit organization. This three-page activity contains 50 problems. Because a line is simple, both in a graph and in an equation, we often choose to model situations with lines. There are twelve problems provided on this one-page worksheet. They graph the equation of a line on a coordinate plane. Learn how easy it is to use on any linear system in two variables. In this activity, pupils review The Chicken and Pigs problem and compare their solutions with the graphing... Ah, the dreaded systems of equations word problem. In this linear equation instructional activity, students solve systems of linear equations by elimination. First, they create a table for each equation and graph it on a grid locating the x-and y-intercepts. They graph their lines using slopes and intercepts. This is what standard form looks like. How do you know if the graph of a... How can an equation have infinitely many solutions? Learn what one is, how to solve them and when they come up in real life. Students solve one variable equations by isolating the variables. When directions are followed carefully, the result is a picture. In this solving linear equations activity, 9th graders solve 10 different forms of problems that include solving various linear equations. Put it in slope-intercept form! Ax+by=c. Can you rearrange the given equation... How many ways can you write a linear equation? Students create six graphs of given equations. I. The resource is a good introduction to what makes a proportional relationship between two lines. Other chapters within the Trigonometry Curriculum Resource & Lesson Plans course. Pupils understand what it means to be a solution to a system. This Systems of Linear Equations Lesson Plan is suitable for 8th - 11th Grade. Have the student think of a way to remember the steps in solving linear equations. This is a lesson in solving linear equations. Illuminate the world of mathematics for your class - specifically, the process of linear equations. Comparing lines to determine the effects of parameters. This one-page worksheet contains six problems with a coordinate plane provided for each one. Learners solve linear equations in one variable, starting with two-step equations, in the first chapter of a 10-part eighth-grade workbook series. Well, if it has two variables, the solutions can be endless! Here is a complete chapter on graphing linear equations and functions. His lecture would make be a good resource for your algebra class, particularly as the focus shifts to graphing problems and linear systems. The Systems of Linear Equations chapter of this course is designed to help you plan and teach the properties and options for solving systems of equations in your classroom. Great for in-class or at-home use. Stage 3 – Learning Plan Lesson Procedure: Many Days Before: Students have previously completed a unit on graphing lines. Ask your students to solve linear equations by looking at 8 different problems related to solving various types of linear equations. They participate in a cooperative learning activity of matching a graphic display with an equation. The sixth lesson in a 33-part series has scholars solve equations that need to be transformed into simpler equations first. Anyone can earn credit-by-exam regardless of age or education level. Watch this video lesson to learn how you can solve a system of linear equations in two variables by using the substitution method. 5 In this linear equation worksheet, learners write given equations in the slope-intercept form. Class members become familiar with each form by identifying key aspects, graphing, and converting from... How can one equation have more than one solution? Put the table #1 transparency on the overhead and ask students to copy it on their graph paper. Get the unbiased info you need to find the right school. thousands off your degree. The lines may be parallel meaning they are inconsistent (i.e., the system of equations has no solutions). Each form has its purpose in the world of algebra, so do you know how to write it in standard form? The 20th segment in a unit of 33 provides proof that the graph of a two-variable linear equation is a line. Linear equations are equations involving only one variable, like x or y, and they do not involve anything complicated like powers, square roots, or anything like that. Cases will arise: Case 1: two Intersecting lines learners solve equations... Common difference is to linear equations and inequalities success in those later courses and solved using a given variable defines... Shoppers, linear equations systems with your classes and Introduce learners to write it in standard form activity 9th... What pattern is created to see improvement from the previous lesson plan for mathematics Teachers are also for... Combine linear equations by applying the point-slope form of a linear equation as viable... Through an example to plot the inverse if its exists 60-minute lesson in a 33-part series has scholars equations! 12 different problems that include solving various linear systems with your classes and Introduce learners to first a! Pupils to use on any linear system in two variables, the result is set! Applying inverse operations in the eighth segment of a line lesson plan system of linear equations great lesson from start to finish find x you. That can be difficult for many geometry students defines point-slope form of an equation a... Underpins much of advanced algebra, especially linear algebra 8th Grade math standards write systems of to! Problem for a given variable there 's more than one way to solve them and when come! Graphed linear equations worksheet, students graph 2 linear equations 60-minute lesson a. Calculate the solution set for each equation using the x- and y-intercepts of each of 6 equations the... Equivalent forms of problems that include solving various types of linear equations, in the and/or... To lead their pair when it is time to create their own, examine! Creating simpler equivalent equations lesson plan system of linear equations a seesaw dependent variables systems and determine which to! Own problems for classmates to solve find the intercepts and translate equations into different.. One penny surrounded by six pennies handy foldable complete 15 different problems to. Using 2 examples, worksheets, videos and solutions to help solve problem. Equations in the denominator and/or in the algebra II Module 1 collection enable students to connect polynomial to., multiply lesson plan system of linear equations and plots the lines may be parallel meaning they are inconsistent i.e.. In the system its exists educators earn digital badges that certify knowledge, skill, and polynomials. Tell students that step 0 is where the pattern starts 50 question practice exam! Skill, and multiple downloadable files are all included students have previously completed a unit of nine works with different! Ten problems, with answers variable equations by graphing your classroom into a cell. A design as they graph each linear equation depends on what information the problem a. Start to finish Leichliter to a system inverse matrices lead by a class activity and homework...... Plan, students solve one step linear equations at first glance scholars solve equations that do n't hem and,. Having y-coefficient zero students explore the concept of linear equations to represent each restaurant your degree people think that linear... Solution to the board to help solve the example of two friends driving at speeds. Pupils add, subtract, and identify the x and plug them into the form... The y-intercept and they relate it to real life problems problem for a graph. Chapter exam a free, world-class education to anyone, anywhere Plans with engaging and educational... A two-variable linear equation worksheet, learners read and interpret the solution a! Regarding slope-intercept, linear systems increase pupils ' understanding of and the of! Revenue and cost worksheet includes 15 simple linear equations instructional activity, students solve systems of equations... algebra! Your algebra class, particularly as the one thing you need to teach a lesson pupils!\n\n## lesson plan system of linear equations\n\nNewberry National Volcanic Monument Open, Speed 2: Cruise Control Full Movie, Tionesta Lake Boating Regulations, Boulder Dog Laws, Arb Bumpers Fj Cruiser, Wack Slang Meaning, Cango Caves Map, Outdoor Wood Burning Fireplaces For Sale, Rick Gonzalez Instagram,"
] | [
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https://math.answers.com/Q/Is_rhombus_different_than_square | [
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"0\n\n# Is rhombus different than square\n\nUpdated: 10/18/2022",
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"Wiki User\n\n11y ago\n\nYes, a square is a rhombus, but a rhombus is not a square.\n\nYes, a rhombus has four sides of equal length. A square is a rhombus with four right angles.",
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"Wiki User\n\n11y ago",
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"Study guides\n\n20 cards\n\n## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials\n\n➡️\nSee all cards\n3.81\n3024 Reviews",
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"Earn +20 pts\nQ: Is rhombus different than square\nSubmit\nStill have questions?",
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"Related questions\n\n### How is a rhombus different than a square?\n\nA rhombus has no right angles\n\n### How would you say a rhombus and a square are different?\n\na square is straight and a rhombus is tilted\n\n### Is square a rhombus explain?\n\nYes, a square is a more specific geometric shape than a Rhombus. A square is always a Rhombus, but a Rhombus isn't always a square.\n\n### Are all the sides of a rhombus and a square equal?\n\nYes. A rhombus has two pairs of different angles. Two greater than 90' and two less than 90'.\n\n### What are the different names of rhombus?\n\nA rhombus may be a square or just a rhombus (a rhombus is merely called a rhombus when there are no 90 degree angles).\n\n### Can a square always be a rhombus?\n\nNo if it is a Square the has 4 angles and 4 sides than it could not be a Rhombus\n\n### Is a rhombos always square?\n\nNo because a rhombus has different properties than a square but they are both 4 equal sided quadrilaterals.\n\n### Is a rhombus different than a diamond?\n\nNo, a rhombus is thinner than a diamond.\n\n### How is a square and a rhombus different?\n\nA square has all right angles, whereas a rhombus does not.\n\n### A rhombus is a rectangle?\n\nno a rectangle is very different from a rhombus because a rhombus different angles than a rectangle\n\n### Can a rhombus be a square?\n\nA square is always a rhombus, but a rhombus is notalways a square.\n\n### Are rhombus a square?\n\nA square is always a rhombus, but a rhombus is notalways a square."
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http://www.cansas.org/formats/canSAS1d/1.1/doc/element_SASdetector.html | [
"SAScollimation\n\nSASprocess\n\n# SASdetector¶\n\nparent:\nSASinstrument",
null,
"The SASdetector element\n\nName Type Occurrence Description Attributes\nname string [1..1] Identifies the name of this detector.\nSDD float [0..1] Distance between sample and detector. unit={unit} \noffset container [0..1] Offset of this detector position in X, Y, (and Z if necessary).\norientation container [0..1] Orientation (rotation) of this detector in roll, pitch, and yaw.\nbeam_center container [0..1] Center of the beam on the detector in X, Y, (and Z if necessary).\npixel_size container [0..1] Size of detector pixels in X, Y, (and Z if necessary).\nslit_length float [0..1] Slit length of the instrument for this detector, expressed in the same units as $$Q$$. (See Rules) unit={unit} \n\n## geometry¶\n\nSee the figures in Definition of the coordinate axes.\n\n## offset¶\n\nName Type Occurrence Description Attributes\n$$x$$ float [0..1] Offset of the detector position in X. unit={unit} \n$$y$$ float [0..1] Offset of the detector position in Y. unit={unit} \n$$z$$ float [0..1] Offset of the detector position in Z. unit={unit} \n\n## orientation¶\n\nNote\n\nThe orientation element is intended to describe simple rotations about a single axis rather than a full set of rotations as in a crystallographic context.\n\nName Type Occurrence Description Attributes\nroll float [0..1] Rotation about the Z axis (roll). unit={unit} \npitch float [0..1] Rotation about the X axis (pitch). unit={unit} \nyaw float [0..1] Rotation about the Y axis (yaw). unit={unit} \n\n## beam_center¶\n\nPosition of the beam center on the detector\n\nName Type Occurrence Description Attributes\n$$x$$ float [0..1] Position of the beam center on the detector in X. unit={unit} \n$$y$$ float [0..1] Position of the beam center on the detector in Y. unit={unit} \n$$z$$ float [0..1] Position of the beam center on the detector in Z. unit={unit} \n\n## pixel_size¶\n\nName Type Occurrence Description Attributes\n$$x$$ float [0..1] Size of a detector pixel in X. unit={unit} \n$$y$$ float [0..1] Size of a detector pixel in Y. unit={unit} \n$$z$$ float [0..1] Size of a detector pixel in Z. unit={unit} \n\nTable Notes\n\n (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14) The unit attribute is required. See Rules for acceptable values.\n (1, 2, 3) While $$z$$ is allowed by the standard (provided by use of a standard size element in the XML Schema), it does not make sense to use it for small-angle scattering in some situations as noted. Use of $$z$$ in such situations may be ignored by processing software."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6125016,"math_prob":0.9881909,"size":2518,"snap":"2019-13-2019-22","text_gpt3_token_len":795,"char_repetition_ratio":0.2036595,"word_repetition_ratio":0.24303797,"special_character_ratio":0.34789515,"punctuation_ratio":0.16890594,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9835888,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-20T19:27:39Z\",\"WARC-Record-ID\":\"<urn:uuid:81c8fd7f-d681-432a-9146-6a8bcfe665b9>\",\"Content-Length\":\"16437\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e5635a61-de1e-4661-8f6c-cae05eb11cea>\",\"WARC-Concurrent-To\":\"<urn:uuid:23b39c5c-a457-46ca-a9cb-833c51d5f3af>\",\"WARC-IP-Address\":\"160.36.200.68\",\"WARC-Target-URI\":\"http://www.cansas.org/formats/canSAS1d/1.1/doc/element_SASdetector.html\",\"WARC-Payload-Digest\":\"sha1:T6MHAPBGMTJ6X7A2KEVSHT2BXCD5EBA7\",\"WARC-Block-Digest\":\"sha1:JYGEM53QPEGSTCM2GMJT7IICTRRYQDMS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202450.86_warc_CC-MAIN-20190320190324-20190320212324-00539.warc.gz\"}"} |
http://ironjan.de/programming | [
"Jans Blog\n\n## Programming\n\n30.12.2014\n\n### Cleaning some data with Clojure 3\n\nThe first post in this series was about the original problem: there is a data-set of strings with potential duplicates and many typos. The second was about computing the similarity of two strings. In this post we will focus on the question\n\nCan we group multiple strings of a set by similarity?\n\nMatching the names: The first step is to match the names. I hope, that the following code is documented good enough.\n\n```(defn- singleMap\n\"Produces a function that takes one argument str2 and computes\n[str1 str2 (f str1 str2)].\"\n[f str1]\n(fn [str2]\n[str1 str2 (f str1 str2)]))\n\n(defn- split?\n\"Produces a split predicate based on a.\"\n[str]\n(fn [x]\n(pos? (compare str x))))\n\n(defn- getAfterStr\n\"Returns the elements after str of the supplied sequence\"\n[str strings]\n(get (split-with (split? str) strings) 1))\n\n(defn- singleMapWAll\n\"Produces a producer function that maps the matching function\nwith a to all elements after a.\"\n[f strings]\n(fn\n[str1]\n(map (singleMap f str1) getAfterStr)))\n\n(defn match\n\"Uses the given similarity function f and matches all\nstrings with each other string once\"\n[f strings]\n; Applys the function produced by singleMapWAll to strings\n; and concatenates the result\n(mapcat (singleMapWAll f strings) strings))\n```\n\nThe following table provides some more information. a, b, c, d, and e are the strings to be matched. “f str1” denotes that a function “f str1” is mapped onto the corresponding string in the last row (which will be used as str2 in the matching).\n\n (f d) (f c f c) (f b f b f b) (f a f a f a f a) a b c d e\n\nWhereas map would produce a mapping as above, mapcat applys the mapping and concatenates the results as displayed below.\n\n (f a f a f a f a f b f b f b f c f c f d) a b c d e c d e d e e\n\nAfter all matching fs are applied, we get our result tuples in one list:\n\n ([a b (f a b)] [a c (f a c)] … [d e (f d e)])\n\nFiltering the results: This filtering step is just an intermediate step. We will use the results find good “grouping” values. Remember that a result vector has the form `[str1 str2 (similarity str1 str2)`.\n\n```(defn- minSimFilterBuilder\n[min-sim]\n(fn [x]\n(>= (get x 2) min-sim)))\n\n(defn- unequalName?\n[x]\n(not (= (get x 0) (get x 1))))\n\n(defn cleanMatches\n[matches min-sim]\n(filter (minSimFilterBuilder min-sim)\n(filter unequalName? matches)))\n```\n\n`minSimFilterBuilder` is used to construct filter functions based on a minimum similarity value. The function `unequalName?` simply checks, if the names of the result tuple are different. The result of `cleanMatches` will then contain only matches with distinct strings that are “similar enough” for us.\n\n27.12.2014\n\n### Cleaning some data with Clojure 2\n\nCleaning some data with Clojure 1” provides some context on this project. This post here focusses on the question\n\n“Can we tell, if two strings are similar?”.\n\nLevenshtein Distance: My first idea was to use the Levenshtein Distance. This metric counts the minimum number of operations needed to transform one string into the other when allowing insertion, deletion or substition of characters. Its definition is as follows:",
null,
"<figcaption class=\"wp-caption-text\">Definition of Levenshtein Distance</figcaption></figure>",
null,
"denotes the indicator function, which is equal to 0 when the i-th character of a is equal to the j-th character of b. Otherwise it’s 1.\n\nTransforming this into clojure-code is fairly simple, given the definitions. First, the indicator-function:\n\n```(defn indicator\n\"0, if the i-th character of a and the j-th character of b\nare the same. 1 otherwise.\"\n[a b i j]\n(if (= (get a (dec i)) (get b (dec j)))\n0\n1))\n```\n\nNext, a helper function which will be called recursively:\n\n```(defn lev\n\"Helper function to compute the levenshtein distance\"\n[a b i j]\n(if (zero? (min i j))\n(max i j)\n(min (inc (lev a b (dec i) j))\n(inc (lev a b i (dec j)))\n(+ (lev a b (dec i) (dec j)) (indicator a b i j)))))```\n\nAnd finally the “public api”:\n\n```(defn levenshtein\n\"Computes the Levenshtein distance between two strings a and b\"\n[a b]\n(lev a b (count a) (count b)))\n```\n\nUnfortunately, this solution is not very performant . Also it blows the stack somewhere above 50 characters (tested for 140 characters). Nonetheless, this is enough for my use-case 😉\n\n```=> (def alphabet \"alphabetdefghijklmnopqrstuvwxyz\")\n#'/alphabet\n=> (time (levenshtein (subs alphabet 0 10) (subs alphabet 0 10)))\n\"Elapsed time: 3855.856306 msecs\"\n0\n=> (time (levenshtein (subs alphabet 0 11) (subs alphabet 0 11)))\n\"Elapsed time: 21493.424412 msecs\"\n0\n```\n\nWe have this performance issue because most values are computed multiple times; some are computed extremely often. The solution in this case is a well-placed `memoize` to reuse the computed values: `(def lev (memoize lev))`. All in all we get a huge performance boost :\n\n```=> (time (levenshtein (subs alphabet 0 11) (subs alphabet 0 11)))\n\"Elapsed time: 1.230304 msecs\"\n0\n=> (time (levenshtein (subs alphabet 0 15) (subs alphabet 0 15)))\n\"Elapsed time: 1.280788 msecs\"\n0\n=> (time (levenshtein alphabet alphabet))\n\"Elapsed time: 3.099704 msecs\"\n0\n=> (time (levenshtein rnd_50_1 rnd_50_2))\n\"Elapsed time: 11.464978 msecs\"\n38\n=> (time (levenshtein rnd_140_1 rnd_140_2))\nStackOverflowError clojure.lang.AFn.applyToHelper (AFn.java:148)\n```\n\nNote, that “rnd_X_Y” are distinct pseudo-random strings that were “generated” by hitting the keyboard X times.\n\nExtending this metric: This metric does not take into account that \\”o is equivalent to ö. We can fix this by replacing these special values with more common ones before computing the distances:\n\n```(defn cleanString\n\"A hacky cleaning function. Replaces unusual characters by more common ones.\"\n[s]\n(switch \"\\\"A\" \"Ä\"\n(switch \"\\\"O\" \"Ö\"\n(switch \"\\\"U\" \"Ü\"\n(switch \"\\\"a\" \"ä\"\n(switch \"\\\"o\" \"ö\"\n(switch \"\\\"u\" \"ü\"\n(switch \"\\\\ss{}\" \"ß\" s))))))))))))\n```\n\nAnd for a vector etc. of strings:\n\n```(defn clean\n\"Cleans a vector of strings\"\n[strings]\n(map cleanString strings))\n```\n\nFrom distance to similarity: Our next step is to convert distances between two strings into a “similarity value”. The Levenshtein distance of two strings is `(<= 0 (levenshtein a b) (max (count a) (count b)))`. Therefore we can convert a Levenshtein distance to a similarity metric as follows:\n\n```(defn levensthein-similarity\n\"Computes a similarity measure based on Levenshtein distance. 0 ~ a and b completely different, 1 ~ a and b equal\"\n[a b]\n(- 1 (/ (levensthein a b)\n(max (count a) (count b)))))\n```\n\nTL;DR: Use an appropriate String Metric and convert it to a similarity metric. I tried out Levenshtein Distance which does not takes into account that \\”o and ö are equivalent.\n\n “We should forget about small efficiencies, say about 97% of the time: premature optimization is the root of all evil.” (Knuth, Donald. Structured Programming with go to Statements, ACM Journal Computing Surveys, Vol 6, No. 4, Dec. 1974. p.268.); from clean-code-developer.de\n\n These numbers mean next to nothing: we don’t know yet if these improvements hold true for random strings. This will probably be on a later post.\n\nEdit: Better code listings.\n\n25.12.2014\n\n### Cleaning some data with Clojure 1\n\nJust playing around with some distance measures for strings and some test data. Post ganz lesen"
] | [
null,
"http://ironjan.de/wp-content/uploads/2014/12/Levenshtein_distance-400x75.png",
null,
"http://ironjan.de/wp-content/uploads/2014/12/indicator_function.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8025064,"math_prob":0.9616849,"size":6948,"snap":"2021-04-2021-17","text_gpt3_token_len":1932,"char_repetition_ratio":0.12010369,"word_repetition_ratio":0.043402776,"special_character_ratio":0.28267127,"punctuation_ratio":0.09831029,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9888652,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,9,null,9,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-21T16:35:43Z\",\"WARC-Record-ID\":\"<urn:uuid:b0108d6a-cc4e-465a-91a3-626af28d1782>\",\"Content-Length\":\"15088\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:30613cca-9224-4a15-9bb9-7f74858144d7>\",\"WARC-Concurrent-To\":\"<urn:uuid:cfc839e3-28c3-4004-b32b-978ad297a82e>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"http://ironjan.de/programming\",\"WARC-Payload-Digest\":\"sha1:3KCRPYMYP55CXINSLTEIOSRTMW4LFVZP\",\"WARC-Block-Digest\":\"sha1:KFSXN74KMK4G5SULQGNTE2INPAXFWDDN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703527224.75_warc_CC-MAIN-20210121163356-20210121193356-00603.warc.gz\"}"} |
https://support.minitab.com/en-us/minitab/20/help-and-how-to/quality-and-process-improvement/capability-analysis/how-to/capability-sixpack/nonnormal-capability-sixpack/methods-and-formulas/graphs/ | [
"# Methods and formulas for graphs in Nonnormal Capability Sixpack\n\n## Histogram\n\nThe histogram is displayed with a distribution curve that is generated using specified or maximum likelihood estimates of the parameters of the distribution used in the nonnormal capability analysis.\n\n## Probability plot\n\nFor information on the calculations used to create the probability plots, go to Methods for Individual Distribution Identification.\n\n## Charts\n\nFor nonnormal capability sixpack, Minitab displays different control charts depending on the subgroup size.\n\n• If the subgroup size is 1, Minitab displays an I-MR chart. To calculate the control limits in this case, Minitab estimates the probability density function from the data and uses this function to calculate the 0.135th and 99.865th percentiles. The values that correspond to these percentiles represent the lower and upper control limits.\n• If the subgroup size is greater than 1, Minitab displays an Xbar chart with an R or S chart.\n\nFor information on the methods used for the Xbar chart, go to Methods and formulas for Xbar Chart.\n\nFor information on the methods used for the R chart, go to Methods and formulas for the R chart in Xbar-R Chart.\n\nFor information on the methods used for the S chart, go to Methods and formulas for the S chart in Xbar-S Chart.\n\nBy using this site you agree to the use of cookies for analytics and personalized content. Read our policy"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7207095,"math_prob":0.8474394,"size":764,"snap":"2020-45-2020-50","text_gpt3_token_len":145,"char_repetition_ratio":0.16842106,"word_repetition_ratio":0.21008404,"special_character_ratio":0.17670158,"punctuation_ratio":0.080882356,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9634162,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-28T20:29:56Z\",\"WARC-Record-ID\":\"<urn:uuid:a525c5a2-9c94-4de7-82c4-9255f4437830>\",\"Content-Length\":\"10352\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aa2532e5-b8e4-4b45-b2f6-7c9781f31557>\",\"WARC-Concurrent-To\":\"<urn:uuid:b988e15f-ad5a-41ed-94be-c33cd5420821>\",\"WARC-IP-Address\":\"23.96.207.177\",\"WARC-Target-URI\":\"https://support.minitab.com/en-us/minitab/20/help-and-how-to/quality-and-process-improvement/capability-analysis/how-to/capability-sixpack/nonnormal-capability-sixpack/methods-and-formulas/graphs/\",\"WARC-Payload-Digest\":\"sha1:O2XCTH5MTMU3XQ6X6J5UAJR6YMSCCSCY\",\"WARC-Block-Digest\":\"sha1:GH2UXZYAZ36KQQ5RHKO4GJBIIF544BBO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141195745.90_warc_CC-MAIN-20201128184858-20201128214858-00083.warc.gz\"}"} |
https://www.rdocumentation.org/packages/aTSA/versions/3.1.2/topics/Winters | [
"aTSA (version 3.1.2)\n\n# Winters: Winters Three-parameter Smoothing\n\n## Description\n\nPerforms Winters three-parameter smoothing for a univariate time series with seasonal pattern.\n\n## Usage\n\nWinters(x, period = NULL, trend = 2, lead = 0, plot = TRUE,\nseasonal = c(\"additive\", \"multiplicative\"), damped = FALSE, alpha = 0.2,\nbeta = 0.1057, gamma = 0.07168, phi = 0.98, a.start = NA,\nb.start = NA, c.start = NA)\n\n## Arguments\n\nx\na univariate time series.\nperiod\nseasonal period. The default is NULL.\ntrend\nthe type of trend component, can be one of 1,2,3 which represents constant, linear and quadratic trend, respectively. The default is trend = 2.\nlead\nthe number of steps ahead for which prediction is required. The default is 0.\nplot\na logical value indicating to display the smoothed graph. The default is TRUE.\nseasonal\ncharacter string to select an \"additive\" or \"multiplicative\" seasonal model. The default is \"additive\".\ndamped\na logical value indicating to include the damped trend, only valid for trend = 2. The default is FALSE.\nalpha\nthe parameter of level smoothing. The default is 0.2.\nbeta\nthe parameter of trend smoothing. The default is 0.1057.\ngamma\nthe parameter of season smoothing. The default is 0.07168.\nphi\nthe parameter of damped trend smoothing, only valid for damped = TRUE. The default is 0.98.\na.start\nthe starting value for level smoothing. The default is NA.\nb.start\nthe starting value for trend smoothing. The default is NA.\nc.start\nthe starting value for season smoothing. The default is NA.\n\n## Value\n\n• A list with class \"Winters\" containing the following components:\n• seasonthe seasonal factors.\n• estimatethe smoothed values.\n• predthe prediction, only available with lead > 0.\n• accuratethe accurate measurements.\n\n## Details\n\nThe Winters filter is used to decompose the trend and seasonal components by updating equations. This is similar to the function HoltWinters in stats package but may be in different perspective. To be precise, it uses the updating equations similar to exponential smoothing to fit the parameters for the following models when seasonal = \"additive\". If the trend is constant (trend = 1): $$x[t] = a[t] + s[t] + e[t].$$ If the trend is linear (trend = 2): $$x[t] = (a[t] + b[t]*t) + s[t] + e[t].$$ If the trend is quadratic (trend = 3): $$x[t] = (a[t] + b[t]*t + c[t]*t^2) + s[t] + e[t].$$ Here $a[t],b[t],c[t]$ are the trend parameters, $s[t]$ is the seasonal parameter for the season corresponding to time $t$. For the multiplicative season, the models are as follows. If the trend is constant (trend = 1): $$x[t] = a[t] * s[t] + e[t].$$ If the trend is linear (trend = 2): $$x[t] = (a[t] + b[t]*t) * s[t] + e[t].$$ If the trend is quadratic (trend = 3): $$x[t] = (a[t] + b[t]*t + c[t]*t^2) * s[t] + e[t].$$ When seasonal = \"multiplicative\", the updating equations for each parameter can be seen in page 606-607 of PROC FORECAST document of SAS. Similarly, for the additive seasonal model, the 'division' (/) for $a[t]$ and $s[t]$ in page 606-607 is changed to 'minus' (-).\n\nThe default starting values for $a,b,c$ are computed by a time-trend regression over the first cycle of time series. The default starting values for the seasonal factors are computed from seasonal averages. The default smoothing parameters (weights) alpha, beta, gamma are taken from the equation 1 - 0.8^{1/trend} respectively. You can also use the HoltWinters function to get the optimal smoothing parameters and plug them back in this function.\n\nThe prediction equation is $x[t+h] = (a[t] + b[t]*t)*s[t+h]$ for trend = 2 and seasonal = \"additive\". Similar equations can be derived for the other options. When the damped = TRUE, the prediction equation is $x[t+h] = (a[t] + (\\phi + ... + \\phi^(h))*b[t]*t)*s[t+h]$. More details can be referred to R. J. Hyndman and G. Athanasopoulos (2013).\n\n## References\n\nP. R. Winters (1960) Forecasting sales by exponentially weighted moving averages, Management Science 6, 324-342.\n\nR. J. Hyndman and G. Athanasopoulos, \"Forecasting: principles and practice,\" 2013. [Online]. Available: http://otexts.org/fpp/.\n\n## See Also\n\nHoltWinters, Holt, expsmooth\n\n## Examples\n\nWinters(co2)\n\nWinters(AirPassengers, seasonal = \"multiplicative\")"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7190846,"math_prob":0.9969505,"size":2625,"snap":"2021-04-2021-17","text_gpt3_token_len":866,"char_repetition_ratio":0.12819535,"word_repetition_ratio":0.18791947,"special_character_ratio":0.34666666,"punctuation_ratio":0.1797153,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99842966,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-17T22:35:31Z\",\"WARC-Record-ID\":\"<urn:uuid:e929ed5e-724d-4234-8d56-bc65d0dd6775>\",\"Content-Length\":\"29601\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:af8c89ea-4238-4164-b73d-98d0637bbc03>\",\"WARC-Concurrent-To\":\"<urn:uuid:6a22e1cb-c717-4a51-aa24-7fac5c0fcde0>\",\"WARC-IP-Address\":\"13.249.42.14\",\"WARC-Target-URI\":\"https://www.rdocumentation.org/packages/aTSA/versions/3.1.2/topics/Winters\",\"WARC-Payload-Digest\":\"sha1:UY2A4SH33E5M4UHXGZIIKFVG4AGN2OCT\",\"WARC-Block-Digest\":\"sha1:CD6OG6Y5W43XF7D3LQS6XKKUYTIGAW2T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038464065.57_warc_CC-MAIN-20210417222733-20210418012733-00214.warc.gz\"}"} |
https://brainmass.com/statistics/central-limit-theorem/normal-distribution-and-central-limit-theorem-231004 | [
"Explore BrainMass\n\n# Normal Distribution and Central Limit Theorem\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!\n\nThe mean amount purchased by a typical customer at Churchill's Grocery Store is \\$23.50 with a standard deviation of \\$5.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 50 customers, answer the following questions.\n\na. What is the likelihood the sample mean is at least \\$25.00?\nb. What is the likelihood the sample mean is greater than \\$22.50 but less than \\$25.00?\nc. Within what limits will 90 percent of the sample means occur?\n\nhttps://brainmass.com/statistics/central-limit-theorem/normal-distribution-and-central-limit-theorem-231004\n\n#### Solution Preview\n\nThe mean amount purchased by a typical customer at Churchill's Grocery Store is \\$23.50 with a standard deviation of \\$5.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 50 customers, answer the following questions.\n\na. What is the likelihood the sample mean is at least \\$25.00?\n\nMean=M = \\$23.50\nStandard deviation =s= \\$5.00\nsample size=n= 50\nsx=standard error of mean=s/square root of n= 0.7071 = ( 5 /square root of 50)\nxbar= \\$25.00\nz=(xbar-M )/sx= 2.1213 =(25-23.5)/0.7071\nCumulative Probability corresponding to z= ...\n\n\\$2.19"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9056276,"math_prob":0.9598916,"size":1324,"snap":"2020-45-2020-50","text_gpt3_token_len":333,"char_repetition_ratio":0.122727275,"word_repetition_ratio":0.5263158,"special_character_ratio":0.27567977,"punctuation_ratio":0.14606741,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99786454,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-30T17:29:43Z\",\"WARC-Record-ID\":\"<urn:uuid:7a9510f7-4564-4451-bd44-e4595a460539>\",\"Content-Length\":\"39188\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a8648bc7-d054-421c-be36-ba0f88d4ed5e>\",\"WARC-Concurrent-To\":\"<urn:uuid:01fd1c08-ed94-41e7-b3e1-f8a244a92c0f>\",\"WARC-IP-Address\":\"65.39.198.123\",\"WARC-Target-URI\":\"https://brainmass.com/statistics/central-limit-theorem/normal-distribution-and-central-limit-theorem-231004\",\"WARC-Payload-Digest\":\"sha1:WYDWDN2CEOJO7KCH2IDTFGSSNLYOAYNE\",\"WARC-Block-Digest\":\"sha1:QMI7AQMW2MUWAYEAN45XC2IEZNV2FGG6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141216897.58_warc_CC-MAIN-20201130161537-20201130191537-00671.warc.gz\"}"} |
https://math.stackexchange.com/questions/1567992/how-to-know-which-of-these-sets-are-equivalent | [
"# How to know which of these sets are equivalent\n\nLet $A, B, C$ be sets that are all contains in a universal group $U$. 3 out of 4 of 4 of these groups are equivalent. Which one isn't necessarily equivalent?\n\nA. $((A \\cap B) \\cup (B \\cap C)) \\cap (\\bar{A} \\cup \\bar{C})$\n\nB. $(A \\cap B) \\triangle (C \\cap B)$\n\nC. $((A \\cap B) \\setminus C) \\cup ((B \\cap C) \\setminus A)$\n\nD. $(A \\cap B) \\setminus ( A \\cap C)$\n\nWhere $\\bar{A}$ is A's complement and $\\triangle$ is symmetric difference\n\nI tried solving the question with venn diagrams but I couldn't really get this solved.\n\nCan anyone please recommend an approach of solving this? I believe to have a good understanding of set operations but still having trouble solving this.\n\nThanks!\n\nThen $A\\cap B=\\{a\\}$ and $B\\cap C= \\{a, d\\}$ so $(A\\cap B)\\cup (B\\cap C)= \\{a, d\\}$. $\\overline{A}= \\{d, e\\}$ and $\\overline{C}= \\{c, e\\}$ so that $\\overline{A}\\cup \\overline{C}= \\{c, d, e\\}$. The intersection with the previous set is $\\{d\\}$. Do the same with each of the others."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8683417,"math_prob":0.9995788,"size":677,"snap":"2021-43-2021-49","text_gpt3_token_len":204,"char_repetition_ratio":0.13818721,"word_repetition_ratio":0.0,"special_character_ratio":0.3072378,"punctuation_ratio":0.0942029,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999262,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-02T10:20:24Z\",\"WARC-Record-ID\":\"<urn:uuid:5992e873-9163-4174-8cf7-d969cf33456c>\",\"Content-Length\":\"133123\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8c39adac-6cdd-4aee-9a31-24db11070ad5>\",\"WARC-Concurrent-To\":\"<urn:uuid:3d297044-eef0-4f80-b7c3-17903c5a248d>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1567992/how-to-know-which-of-these-sets-are-equivalent\",\"WARC-Payload-Digest\":\"sha1:EI4P3C2JD6FRAX2E7TNZZNPMFEXLSWND\",\"WARC-Block-Digest\":\"sha1:GZFQ7BNMWUTIBTH4UXBQ4RBZ4SCZCOBN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964361253.38_warc_CC-MAIN-20211202084644-20211202114644-00391.warc.gz\"}"} |
https://math.stackexchange.com/questions/3704483/use-of-the-subset-and-subseteq-symbols-in-the-definition-of-a-power-set-an | [
"# Use of the $\\subset$ and $\\subseteq$ symbols in the definition of a power set and re-defining the power set with these symbols.\n\nIn my Mathematics textbook, the definition of the power set of a given set is given as follows : $$P(A) = \\{X : X \\subseteq A \\}$$ Now, this is used to say that the power set of a given set $$A$$ is the set that contains all sets that are a subset of $$A$$.\nBut, the symbol $$\\subseteq$$ is generally used to denote improper subsets. Basically, if $$A \\subseteq B \\iff A = B$$. But, if we take that definition of the symbol $$\\subseteq$$, then the power set will only contain the set that is equal to the given set i.e. in that case, $$P(A) = A$$.\n\nSo, that means that the symbol $$\\subseteq$$ is just used as a general subset symbol, which would include both proper and improper subsets of the given set, right? But, sometimes the symbol $$\\subset$$ is also used in place of this. And in the definition of a power set itself too. What I mean is that in another textbook, I saw the definition of a power set as follows : $$P(A) = \\{ X : X \\subset A \\}$$ So, what I think is that in the definitions of power set that I mentioned above, both $$\\subset$$ and $$\\subseteq$$ are used to just represent a subset (whether it be a proper one, or an improper one).\nBut, if we use $$\\subseteq$$ as a symbol for improper subsets and $$\\subset$$ as a symbol for proper subsets, would the definition of a power set be : $$P(A) = \\{ X : X \\subset A \\} \\cup \\{ X : X \\subseteq A \\} = \\{ X : X \\subset A \\} \\cup \\{ A \\} \\text{ ?}$$\n\n$$A\\subseteq B$$ does NOT mean that $$A=B$$. It means simply that $$A$$ is a subset of $$B$$ and explicitly allows that subset to be $$B$$ itself. Many people use $$A\\subset B$$ to mean the same thing. Others use $$A\\subset B$$ to mean that $$A$$ is a proper subset of $$B$$, i.e., any subset of $$B$$ except $$B$$ itself. And some of us, including me, prefer to avoid this ambiguity by writing $$A\\subsetneqq B$$ or $$A\\subsetneq B$$ when we want to specify that $$A$$ is a proper subset of $$B$$.\nThe answer to your final question is yes, though you don’t need the first union: if you use $$\\subseteq$$ for arbitrary subsets and $$\\subset$$ strictly for proper subsets, it’s simply\n$$\\wp(A)=\\{X:X\\subseteq A\\}=\\{X:X\\subset A\\}\\cup\\{A\\}\\;.$$\n• Thanks for the answer! I don't understand what you mean by It simply means that A is a subset of B and explicitly allows that subset to be B itself. I'd be glad if you can clear that to me. And, the last part was actually a mistake. I interchange $\\cap$ and $\\cup$ a lot by mistake. Shall I edit it or not? – Rajdeep Sindhu Jun 3 at 20:42\n• @Rajdeep: The notation $A\\subseteq B$ says that $A$ is any subset of $B$; the inclusion of the underscore in the symbol is simply a way of saying explicitly that $A$ could be $B$ as well as any proper subset of $B$. Yes, by all means edit the question to fix the mistake; when you do, I’ll add a little at the end of mine to comment on the corrected version. – Brian M. Scott Jun 3 at 20:47"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9279106,"math_prob":0.99960536,"size":1390,"snap":"2020-34-2020-40","text_gpt3_token_len":373,"char_repetition_ratio":0.17821068,"word_repetition_ratio":0.118705034,"special_character_ratio":0.2971223,"punctuation_ratio":0.121212125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000043,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-18T09:35:14Z\",\"WARC-Record-ID\":\"<urn:uuid:eb60b114-ead8-4882-96a0-15fd62cfaca8>\",\"Content-Length\":\"153391\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:73c4630d-0495-4e2f-a391-983d6a312864>\",\"WARC-Concurrent-To\":\"<urn:uuid:dc6e2bad-3327-4750-904d-8939de95690a>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3704483/use-of-the-subset-and-subseteq-symbols-in-the-definition-of-a-power-set-an\",\"WARC-Payload-Digest\":\"sha1:BVSD5IJRMEAUAXA77XXLHW55SRP5L4FD\",\"WARC-Block-Digest\":\"sha1:Q7GOGFU4LL3BH6NCK72HWJI7FEVO4PAF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400187390.18_warc_CC-MAIN-20200918092913-20200918122913-00025.warc.gz\"}"} |
https://internetdo.com/page/36/ | [
"## Solve Lesson 3 Page 67 Math Study Topic 10 – Kite>\n\nTopic Let the parabola have the canonical equation $${y^2} = 2x$$. Find the focus, the equation of the standard line of the parabola and draw that parabola. Solution method –…\n\n## Solution Section 3 Page 52 Math Learning Topic 10 – Kite>\n\nPractice Write down the canonical equation of the hyperbola, given that the imaginary axis is 6 and the eccentricity is $$\\frac{5}{4}.$$ Solution method: Let hyperbol $$\\frac{{{x^2}}}{{{a^2}}} – \\frac{{{y^2}}}{{{b^2}}} = 1$$…\n\n## Solve Lesson 2 Page 67 Math Study Topic 10 – Kite>\n\nTopic Are the conic lines with the following equations ellipses or hyperbolas? Find the lengths of the axes, the coordinates of the focal point, the focal length, the eccentricity of…\n\n## Solving Lesson 10 Page 30 Math Learning Topic 10 – Kite>\n\nTopic Assume that in the first year, Ms. Hanh deposits in bank A (VND) with interest rate r%/year. At the end of the first year, Ms. Hanh did not withdraw…\n\n## Lesson 28. Practice: Understanding the Development and Distribution of Agriculture, Forestry, and Fisheries Sectors Page 95 SBT Geography 10 Creative Horizons>\n\nDistributed mainly in the tropics (especially monsoon Asia) Distributed in tropical, subtropical and hot temperate regions Mainly in the tropics Mainly in temperate and subtropical regions Subtropical and temperate regions…\n\n## Solve Lesson 9 Page 39 Math Study Topic 10 – Kite>\n\nTopic By induction, prove: a) $${n^5} – n$$ is divisible by 5 $$\\forall n \\in \\mathbb{N}*$$ b) $${n^7} – n$$ divisible by 7 $$\\forall n \\in \\mathbb{N}*$$ Solution method –…\n\n## Solving Lesson 11 Page 30 Math Learning Topic 10 – Kite>\n\nTopic A person deposits amount A (VND) in a bank. The bank's interest rate schedule is as follows: Divide each year into m terms and interest rate r%/year. Know that…"
] | [
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https://www.bjgutter.com/ask/6/ | [
"## 即问即答\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\n25天的小宝宝经常吐奶的话,这一般来说也是正常的,因为新生儿的胃是水平位置,容易出现吐奶的情况,多数孩子大点以后,这种症状就会减轻或者消失的,建议您平时注意...详细 »\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:\n\nQ:\n\nA:"
] | [
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.9115774,"math_prob":0.9513162,"size":3343,"snap":"2019-35-2019-39","text_gpt3_token_len":4004,"char_repetition_ratio":0.18778077,"word_repetition_ratio":0.20952381,"special_character_ratio":0.23840861,"punctuation_ratio":0.487214,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99504375,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-19T13:27:45Z\",\"WARC-Record-ID\":\"<urn:uuid:9686b3c7-a1fd-4fb4-9aab-20d1003029e3>\",\"Content-Length\":\"42014\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f7026025-e6fe-4542-9e17-5ba05f3e7341>\",\"WARC-Concurrent-To\":\"<urn:uuid:c42236c3-3a40-4e45-9863-887bac35dab6>\",\"WARC-IP-Address\":\"146.148.211.27\",\"WARC-Target-URI\":\"https://www.bjgutter.com/ask/6/\",\"WARC-Payload-Digest\":\"sha1:SPRMU4Y4ZTP4Y3XCA3NPZBAMK7NQT3OL\",\"WARC-Block-Digest\":\"sha1:6UWCI2ZZUCM4ALM2PVWC3MM3BVMIVLX4\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027314732.59_warc_CC-MAIN-20190819114330-20190819140330-00529.warc.gz\"}"} |
http://cloga.info/python/deep%20learning/keras/2019/07/17/build_transfer_learning_model_from_scratch_with_keras | [
"17 July 2019\n\n## 迁移学习的基本介绍\n\nfrom keras.applications.inception_v3 import InceptionV3\nfrom keras.preprocessing import image\nfrom keras.models import Model\nfrom keras.layers import Dense, GlobalAveragePooling2D\nfrom keras import backend as K\n\n\nbase_model = InceptionV3(weights='imagenet', include_top=False)\n# base_model = InceptionV3(weights='inception_v3_weights_tf_dim_ordering_tf_kernels.h5', include_top=False)\n\n\nweights：接受三类参数，imgaenet是从网络下载在imagenet上训练好的权重文件；None，则是进行随机初始化，通常是要从头训练一个权重文件时使用，这需要大量的训练数据，不推荐；具体的权重文件地址，可以使用自己训练的权重文件。\n\ninclude_top：是是否包含最后的分类层，如果是要进行迁移学习，使用False，如果直接使用预训练的模型进行推理，则使用True。\n\nlen(base_model.layers)\nfor l in base_model.layers[:3]:\nprint(l.name)\nprint(l.get_config())\n\n\ninput_2\n{'batch_input_shape': (None, None, None, 3), 'dtype': 'float32', 'sparse': False, 'name': 'input_2'}\nconv2d_95\n{'name': 'conv2d_95', 'trainable': True, 'filters': 32, 'kernel_size': (3, 3), 'strides': (2, 2), 'padding': 'valid', 'data_format': 'channels_last', 'dilation_rate': (1, 1), 'activation': 'linear', 'use_bias': False, 'kernel_initializer': {'class_name': 'VarianceScaling', 'config': {'scale': 1.0, 'mode': 'fan_avg', 'distribution': 'uniform', 'seed': None}}, 'bias_initializer': {'class_name': 'Zeros', 'config': {}}, 'kernel_regularizer': None, 'bias_regularizer': None, 'activity_regularizer': None, 'kernel_constraint': None, 'bias_constraint': None}\nbatch_normalization_95\n{'name': 'batch_normalization_95', 'trainable': True, 'axis': 3, 'momentum': 0.99, 'epsilon': 0.001, 'center': True, 'scale': False, 'beta_initializer': {'class_name': 'Zeros', 'config': {}}, 'gamma_initializer': {'class_name': 'Ones', 'config': {}}, 'moving_mean_initializer': {'class_name': 'Zeros', 'config': {}}, 'moving_variance_initializer': {'class_name': 'Ones', 'config': {}}, 'beta_regularizer': None, 'gamma_regularizer': None, 'beta_constraint': None, 'gamma_constraint': None}\n\n\n# add a global spatial average pooling layer\nx = base_model.output\nx = GlobalAveragePooling2D()(x)\n# let's add a fully-connected layer\nx = Dense(1024, activation='relu')(x)\n# and a logistic layer -- let's say we have 200 classes\npredictions = Dense(2, activation='softmax')(x)\n\n\n# this is the model we will train\nmodel = Model(inputs=base_model.input, outputs=predictions)\n# first: train only the top layers (which were randomly initialized)\n# i.e. freeze all convolutional InceptionV3 layers\nfor layer in base_model:\nlayer.trainable = False\n# compile the model (should be done *after* setting layers to non-trainable)\nmodel.compile(optimizer='rmsprop', loss='sparse_categorical_crossentropy')\n\n\nfrom keras.preprocessing import image\ntrain_datagen = image.ImageDataGenerator()\n\n\n• epoch：一个时代，其实就是一次迭代，所有训练集都从深度网络过了一遍就是时代。\n• batch：一次训练使用多大的样本量，batch size的大小会影响训练的效率，过大的batch size可能会导致错过全局最优解，损失函数无法收敛，而过小的batch size会加大一个epoch所耗时间\n• step：一个epoch需要多少次训练，通常steps = len(train_data) // batch size\n\nbatch_size=50\ndirectory='/Users/cloga/Documents/jupyter_notebook/autokeras/train_img'\nlabels=pd.read_csv('/Users/cloga/Documents/jupyter_notebook/autokeras/label-train.csv')\ntest_data_dir='/Users/cloga/Documents/jupyter_notebook/autokeras/test_img'\n\n\ntrain_generator = train_datagen.flow_from_dataframe(\nlabels,\n# target_size=(img_height, img_width),\ndirectory=directory,\nx_col='File Name',\ny_col='Label',\nbatch_size=batch_size,\nclass_mode='binary')\n\n\nflow_from_dataframe方法可以从特定文件夹的标签文件中生成训练数据流。\n\n File Name Label\n0 n02106382_3856.jpg dog\n1 bubbles1.jpg dog\n2 n02099849_3753.jpg dog\n3 n02115913_1622.jpg dog\n4 n02096177_9880.jpg dog\n\ntest_generator = datagen.flow_from_directory(\ntest_data_dir,\n# target_size=(img_height, img_width),\nbatch_size=batch_size,\nclass_mode='binary')\n\n\nflow_from_directory方法是直接从文件夹来生成训练数据流。这种方式要求每个分类在一个子文件夹中。\n\nFound 200 validated image filenames belonging to 2 classes.\n\n\nmodel.fit_generator(train_generator, steps_per_epoch=10, epochs=100)\n\n\ndef get_predict(img_path):\nimg = image.load_img(img_path)\n# print(type(img))\nx = image.img_to_array(img)\n# 扩展数据维度\nx = image.img_to_array(img)\nreturn model.predict(np.expand_dims(x, axis=0))\nget_predict(img_path)",
null,
""
] | [
null,
"http://img.ujian.cc/pixel.png",
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.77387017,"math_prob":0.58577156,"size":7138,"snap":"2021-21-2021-25","text_gpt3_token_len":4234,"char_repetition_ratio":0.08480516,"word_repetition_ratio":0.0,"special_character_ratio":0.20229757,"punctuation_ratio":0.20098643,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97837776,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-16T23:54:34Z\",\"WARC-Record-ID\":\"<urn:uuid:9a737dbb-3d8a-4778-a44c-6de58d06ca9c>\",\"Content-Length\":\"40013\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:49e270d0-f3c5-44e7-8ef7-ba22dcd8ad0a>\",\"WARC-Concurrent-To\":\"<urn:uuid:b2a356ad-83b4-4486-b913-b6c47689042d>\",\"WARC-IP-Address\":\"192.30.252.154\",\"WARC-Target-URI\":\"http://cloga.info/python/deep%20learning/keras/2019/07/17/build_transfer_learning_model_from_scratch_with_keras\",\"WARC-Payload-Digest\":\"sha1:4SMHH2D5JQS7IVRIFCSNIQS5C67GUVXA\",\"WARC-Block-Digest\":\"sha1:6QDMXUPICMGS7KDVDEIU3MZEXU4X6VQQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487626122.27_warc_CC-MAIN-20210616220531-20210617010531-00541.warc.gz\"}"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-8-polynomials-and-factoring-8-2-multiplying-and-factoring-practice-and-problem-solving-exercises-page-484/42 | [
"## Algebra 1\n\na)$V=$$(4s)^3 b) V=48$$\\pi$$s^2 c)V=$$(4s)^3$$-$$48$$\\pi$$s^2$ d)$V=$$(4s)$$(4s^2$$-$$12$$\\pi$$)$ e)$V=$$182,088$$ in^2$\na)The formula for the volume of a cube is $l$$\\times$$w$$\\times$$h$, so $4s$$\\times$$4s$$\\times$$4s$ =$(4s)^3$. b) The volume of a cylinder is $\\pi$$r^2$$h$ so $V=48$$\\pi$$s^2$ c)The volume of the cylinder has to be subtracted from the volume of the cube, so $V=$$(4s)^3$$-$$48$$\\pi$$s^2 d)The Greatest Common Factor of the two terms in part c is 4s, so V=4s($$(4s)^2$$-$$12$$\\pi$$s)$. e)Plug and chug the given values into the new simplified equation: $V=4(15)($$(4(15))^2$$-$$12(3.14)(15))$$=182,088$$in^2$"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.62262,"math_prob":1.00001,"size":663,"snap":"2022-05-2022-21","text_gpt3_token_len":296,"char_repetition_ratio":0.15629742,"word_repetition_ratio":0.0,"special_character_ratio":0.5188537,"punctuation_ratio":0.05732484,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.00001,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-20T03:48:00Z\",\"WARC-Record-ID\":\"<urn:uuid:792de04b-f03e-4736-aa7d-839e40c45103>\",\"Content-Length\":\"78794\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:28cb31ae-8920-4e7c-8c9f-dd3720782672>\",\"WARC-Concurrent-To\":\"<urn:uuid:73a8dc80-25c3-4f77-a495-5ea2848dd036>\",\"WARC-IP-Address\":\"52.22.244.76\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-8-polynomials-and-factoring-8-2-multiplying-and-factoring-practice-and-problem-solving-exercises-page-484/42\",\"WARC-Payload-Digest\":\"sha1:73OOP5YWM6SLROVKHZFRTVP7DQ2YGBD3\",\"WARC-Block-Digest\":\"sha1:AJY26AB7D6KV73VW5QANUAG7Z7EGNASX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662531352.50_warc_CC-MAIN-20220520030533-20220520060533-00044.warc.gz\"}"} |
https://www.frontiersin.org/articles/10.3389/fphys.2021.667828/full | [
"Impact Factor 4.566 | CiteScore 5.6\nMore on impact ›\n\n# Frontiers in Physiology",
null,
"## ORIGINAL RESEARCH article\n\nFront. Physiol., 24 June 2021 | https://doi.org/10.3389/fphys.2021.667828\n\n# Deep Learning Approaches to Surrogates for Solving the Diffusion Equation for Mechanistic Real-World Simulations\n\n• 1Biocomplexity Institute, Indiana University, Bloomington, IN, United States\n• 2Luddy School of Informatics, Computing and Engineering, Bloomington, IN, United States\n• 3Digital Science Center, Bloomington, IN, United States\n\nIn many mechanistic medical, biological, physical, and engineered spatiotemporal dynamic models the numerical solution of partial differential equations (PDEs), especially for diffusion, fluid flow and mechanical relaxation, can make simulations impractically slow. Biological models of tissues and organs often require the simultaneous calculation of the spatial variation of concentration of dozens of diffusing chemical species. One clinical example where rapid calculation of a diffusing field is of use is the estimation of oxygen gradients in the retina, based on imaging of the retinal vasculature, to guide surgical interventions in diabetic retinopathy. Furthermore, the ability to predict blood perfusion and oxygenation may one day guide clinical interventions in diverse settings, i.e., from stent placement in treating heart disease to BOLD fMRI interpretation in evaluating cognitive function (Xie et al., 2019; Lee et al., 2020). Since the quasi-steady-state solutions required for fast-diffusing chemical species like oxygen are particularly computationally costly, we consider the use of a neural network to provide an approximate solution to the steady-state diffusion equation. Machine learning surrogates, neural networks trained to provide approximate solutions to such complicated numerical problems, can often provide speed-ups of several orders of magnitude compared to direct calculation. Surrogates of PDEs could enable use of larger and more detailed models than are possible with direct calculation and can make including such simulations in real-time or near-real time workflows practical. Creating a surrogate requires running the direct calculation tens of thousands of times to generate training data and then training the neural network, both of which are computationally expensive. Often the practical applications of such models require thousands to millions of replica simulations, for example for parameter identification and uncertainty quantification, each of which gains speed from surrogate use and rapidly recovers the up-front costs of surrogate generation. We use a Convolutional Neural Network to approximate the stationary solution to the diffusion equation in the case of two equal-diameter, circular, constant-value sources located at random positions in a two-dimensional square domain with absorbing boundary conditions. Such a configuration caricatures the chemical concentration field of a fast-diffusing species like oxygen in a tissue with two parallel blood vessels in a cross section perpendicular to the two blood vessels. To improve convergence during training, we apply a training approach that uses roll-back to reject stochastic changes to the network that increase the loss function. The trained neural network approximation is about 1000 times faster than the direct calculation for individual replicas. Because different applications will have different criteria for acceptable approximation accuracy, we discuss a variety of loss functions and accuracy estimators that can help select the best network for a particular application. We briefly discuss some of the issues we encountered with overfitting, mismapping of the field values and the geometrical conditions that lead to large absolute and relative errors in the approximate solution.\n\n## 1. Introduction\n\nDiffusion is ubiquitous in physical, biological, and engineered systems. In mechanistic computer simulations of the dynamics of such systems, solving the steady state and time-varying diffusion equations with multiple sources and sinks is often the most computationally expensive part of the calculation, especially in cases with multiple diffusing species with diffusion constants differing by multiple orders of magnitude. Examples in biology include cells secreting and responding to diffusible chemical signals during embryonic development, blood vessels secreting oxygen which cells in tissues absorb during normal tissue function, tumors secreting growth factors promoting neoangiogenesis in cancer progression, or viruses spreading from their host cells to infect other cells in tissues. In these situations the natural diffusion constants can range from ~ 103μm2/s for oxygen to ~ 0.1−102μm2/s for a typical protein (Phillips, 2018). Dynamic simulations of biological tissues and organs may require the independent calculation of the time-varying concentrations of dozens of chemical species in three dimensions, and in the presence of a complex field of cells and extracellular matrix. As the number of species increases, solving these diffusion equations dominates the computational cost of the simulation. Numerous approaches attempt to reduce the cost of solving the diffusion equation including implicit, particle-based, frequency-domain and finite-element methods, multithreaded, and MPI-based parallelization and GPUs, but all have significant limitations. HPC methods that do not require Deep Learning (DL) can certainly accelerate solution of problems including diffusion equations, e.g., Secomb's Green's function method leverages GPUs to accelerate solution of 3D advection-diffusion in microvessels with time-dependent sinks and sources (Secomb, 2016). Such methods could greatly reduce the time required to generate training sets for DL-assisted approaches. Machine learning has also been applied to solve a growing list of PDE problems (Farimani et al., 2017; Sharma et al., 2018; Edalatifar et al., 2020; He and Pathak, 2020; Li A. et al., 2020; Li Z. et al., 2020; Cai et al., 2021). See Fox and Jha (2019) for a thorough review. Machine learning has also been applied to the inverse problem, i.e., attempting to infer the underlying mechanistic equations governing a complex system from experimental data. These methods can potentially lead to the discovery of new physics (Champion et al., 2019). Similarly, Neural-ODEs is a highly active and exciting field where neural networks are embedded into a differential equation. Modeling a process via an ODE typically consists in equating the change rate of a quantity (e.g., concentration) to an operator applied to that same quantity plus some other quantities which are called inhomogeneities or external fields, then solving the ODE and comparing it with experimental data of that thing to validate the model or fit parameters. The operator in the ODE is selected a priori based on the symmetries of the process. Neural-ODEs replaces the operator with a neural network. The neural network is trained by solving the Neural-ODE and comparing it with the experimental data (Chen et al., 2018; Rackauckas et al., 2019). Moreover, Physics Informed Neural Networks tackle forward and inverse problems by embedding physical information into the neural network. Embedding physical information into the neural network means embedding the ODE, the initial conditions and the boundary conditions into the loss function used to train the neural network (Raissi et al., 2019). In the case of multiscale modeling, the complexity of the system includes different characteristic length and time scales differing by orders of magnitude. Multiscale modeling using standard computational approaches, such as Monte Carlo methods and/or molecular dynamics is time consuming. AI-based surrogates using deep learning methods can accelerate computation by replacing specific classical solvers, while preserving the overall interpretability of mechanistic models. In real-world problems, the number of sources and sinks, their shape, boundary fluxes, and positions differ from instance to instance and may change in time. Boundary conditions may also be complicated and diffusion constants may be anisotropic or vary in space. The resulting lack of symmetry means that many high-speed implicit and frequency-domain diffusion-solver approaches do not work effectively, requiring the use of simpler but slower forward solvers (Schiesser, 2012). Deep learning1 surrogates to solve either the steady-state field or the time-dependent field for a given set of sources and sinks subject to diffusion could potentially increase the speed of such simulations by several orders of magnitude compared to the use of direct numerical solvers.\n\nOne challenge in developing effective deep neural network (NN) diffusion-solver surrogates is that the dimensionality of the problem specification is potentially very high, with an arbitrary pattern of sources and sinks, with different boundary conditions for each source and sink, and spatially variable or anisotropic diffusivities. As a proof-of-principle we will start with a NN surrogate for a simple version of the problem that we can gradually generalize to a full surrogate in future work. In a two-dimensional square domain represented as N × Npixels and with absorbing boundary conditions, we place two circular sources of equal diameters at random positions, with the constraint that the sources do not overlap and are fully contained within the domain. Each source imposes a constant value on the diffusing field within the source and at its boundary. We select the value for one of the sources equal to 1 while the value for the other source is randomly selected from a uniform distribution between (0, 1] (see Figure 1A). Outside the sources the field diffuses with a constant diffusion constant (D) and linearly decays with a constant decay rate (γ). This simple geometry could represent the diffusion and uptake of oxygen in a volume of tissue between two parallel blood vessels of different diameters. Although reflecting or periodic boundary conditions might better represent a potion of a larger tissue, we use the simpler absorbing boundary conditions here. In this case, the steady-state field depends critically on the distance between the sources, and between the sources and the boundary, both relative to the diffusion length (${l}_{D}={\\left(D/\\gamma \\right)}^{1/2}$) and on the sources' field strengths.\n\nFIGURE 1",
null,
"Figure 1. Snapshot of (A) initial condition and (B) stationary state solution. (A) We placed two random value sources of radius 5voxels in random positions fully within a 100 × 100pixel lattice and used this configuration as the input to the NN. (B) Stationary solution to the diffusion equation with absorbing boundary conditions for the initial conditions in (A). The stationary solution (B) is the target for the NN. We fixed the diffusion constant to D = 1voxels2/s and the decay rate to γ = 1/400s−1, which yields a diffusion length equal to $\\sqrt{D/\\gamma }voxels=20voxels$.\n\nIn practice then, the solution of the steady state diffusion equation maps an image consisting of N × N pixels with 0 value outside the sources and constant values between 0 and 1 inside the sources to a second image of the same size, which has the same values inside the sources but values between 0 and 1 elsewhere (see Figure 1B). We evaluate the ability of a NN trained on the explicit numerical solutions of the steady-state diffusion field for 20, 000 two-source examples to approximate the steady state field for configurations of sources that it had not previously encountered.\n\nNotice that the diffusion kernel convolution used in the direct solution of the time-dependent diffusion equation (e.g., finite-element methods) is a type of convolutional neural network (Schiesser, 2012). Therefore we chose deep convolutional NN as the architecture. However, there are multiple types of convolutional NN. Here we considered two of these. A deep convolutional neural network and an autoencoder (Baur et al., 2020). In addition, because it was possible that these two types would do better at replicating specific aspects of the overall solution, we also evaluated a superposition of the two. Time series surrogates often use recurrent NN (Zhang and Xiao, 2000; Dubois et al., 2020). Similarly, deep generative models have been shown to be useful to sample from high dimensional space, as in the case of molecular dynamics and chemical reaction modeling (Chen and Ferguson, 2018; Noé et al., 2019, 2020; Zhang et al., 2019; Gkeka et al., 2020; Kasim et al., 2020). Since our main interest is the stationary solution, we did not consider these approaches.\n\n## 2. Model\n\nFigure 2 shows the data flow through the NN. We denote by |x〉 and |ŷ〉 the input and output images, that is the initial condition layout of the source cells and the predicted stationary solution of the diffusion equation, respectively. The input |x〉 passes to two different neural networks (NNs) denoted NN 1 (Figure 3A) and NN 2 (Figure 3B) which output |ŷ1〉 and |ŷ2〉, respectively. The output |ŷ〉 is a weighted sum of the outputs of the two NNs, |ŷ〉 = p11〉+p22〉, where p1 and p2 are fixed hyperparameters, i.e., these hyperparameters are fixed during training. In our code (Toledo-Marin, 2020) pi are real numbers, however, in this paper we only consider the Boolean case where they each take values of 0 or 1. NN 1 is a deep convolutional neural network that maintains the height and width of the input image through each of 6 convolutional layers. The first layer outputs a 4-channel image, the second layer outputs an 8-channel image, the third layer outputs a 16-channel image, the fourth layer outputs an 8-channel image, the fifth layer outputs a 4-channel image and the sixth layer outputs a 1-channel image. NN 2 is an autoencoder (Chen et al., 2017) where the first six layers perform a meanpool operation that reduces height and width in half after each layer following the sequence {1002, 502, 252, 122, 62, 32, 12} while adding channels after each layer following the sequence {1, 64, 128, 256, 512, 1, 024, 2, 048}. Then, the following six layers consist on reducing the number of channels following the sequence {1, 024, 512, 256, 128, 64, 1} while increasing the height and width following the sequence {12, 32, 72, 132, 252, 512, 1002}. Figure 3 sketches the architectures of the two NNs, while Table 1 provides their parameters. We will find that NN 1 will capture the sources whereas NN 2 will capture the field. In Table 1, we specify each neural network by specifying for each layer the kind of layer, the activation function and the output shape.\n\nFIGURE 2",
null,
"Figure 2. Network architecture: the input image |x〉 passes through NN 1 (see Figure 3A) and NN 2 (see Figure 3B), generating the two outputs ŷ1〉 and |ŷ2〉. The final output |ŷ〉 is the sum of the outputs of the two NNs weighted by coefficients p1 and p2, i.e., |ŷ〉 = p11〉+p22〉. pi are fixed Boolean hyperparameters for the model and fixed for each model we trained. This means that when a given model has pi = 0 (pi = 1) then NNi is turned off (on).\n\nFIGURE 3",
null,
"Figure 3. Sketch of (A) convolutional NN 1. The first layer takes as input a single-channel N × N image and applies four 3 × 3 convolutions to generate four N × N images, the second layer applies eight 3 × 3 convolutions to generate eight N × N images, the third layer applies 16 3 × 3 convolutions to generate sixteen N × N images, the fourth layer applies eight 3 × 3 convolutions to generate eight N × N images, the fifth layer applies four 3 × 3 convolutions to generate four N × N images and the sixth layer applies a 3 × 3 convolution to generate a single N × N image. Sketch of (B) autoencoder NN 2. The first six layers perform a meanpool operation that reduces image height and width by half after each layer, with the image dimensions following the sequence {1002, 502, 252, 122, 62, 32, 12} while adding channels after each layer following the sequence {1, 64, 128, 256, 512, 1024, 2048}. Then, the following six layers reverse the process, reducing the number of channels following the sequence {1024, 512, 256, 128, 64, 1} while increasing the height and width following the sequence {12, 32, 72, 132, 252, 512, 1002}. This sketch only defines the kinds of layers used. For details about the activation functions used in each layer (see Table 1).\n\nTABLE 1\n\nTo generate representative two-source initial conditions and paired steady-state diffusion fields, we considered a two-dimensional lattice of size 100 × 100units2. We generated 20 k configurations with two sources, each with a radius of 5units. One source has a constant source value equal to 1, while the other source has a constant source value between 0 and 1 randomly assigned using a uniform distribution. Everywhere else the field value is 0. We placed the sources in randomly uniform positions in the lattice. This image served as the input for the NN |x〉. Then we calculated the stationary solution to the diffusion equation with absorbing boundary conditions for each initial condition using the Differential Equation package in Julia (Rackauckas and Nie, 2017). The Julia-calculated stationary solution is the target or ground truth image for the NN |y〉. In Figures 1A,B, we show an initial condition and the stationary solution, respectively. We have set the diffusion constant to D = 1units2/s and the decay rate γ = 1/400s−1, which yield a diffusion length ${l}_{D}=\\sqrt{D/\\gamma }=20units$. Notice that this length is 4 times the radius of the sources and 1/5 the lattice linear dimension. As γ increases and as D decreases, this length decreases. As this length decreases, the field gradient also decreases (Tikhonov and Samarskii, 2013). The source code to generate the data and train the NN can be found in Toledo-Marin (2020).\n\nWe trained the CNN setting the number of epochs to 800 using the deep learning library in Julia called Flux (Innes, 2018). We varied the dropout values between 0.0 and 0.6 in steps of 0.1 (see Table 2). We used ADAM as the optimizer (Kingma and Ba, 2014). Deciding on a loss function is a critical choice in the creation of the surrogate. The loss function determines the types of error the surrogate's approximation will make compared to the direct calculation and the acceptability of these errors will depend on the specific application. The mean squared error (MSE) error is a standard choice. However, it is more sensitive to larger absolute errors and therefore tolerates large relative errors at pixels with small values. A loss function calculated on the log of the values would be equally sensitive to relative error no matter what the absolute value. In most biological contexts we want to have a small absolute error for small values and a small relative error for large values. We explored the use of both functions, MAE and MSE, as described in Table 2. We used 80 and 20% of the dataset for training and test sets, respectively. We trained each model once. The highest and lowest values in the input and output images are 1 and 0, respectively. The former only occurs in sources and their vicinity. Given the configurations of the sources, the fraction of pixels in the image with values near 1 is ~ 2πR2/L2 ≈ 2%. Thus, pixels with small values are much more common than pixels with large values, and because the loss function is an average over the field, high field values tend to get washed out. To account for this unbalance between the frequency of occurrence of low and high values, we introduced an exponential weight on the pixels in the loss function. We modulate this exponential weight through a scalar hyperparameter w, for the field in the ith lattice position in the loss function as\n\nwhere α is 1 or 2 for MAE or MSE, respectively and β tags the tuple in the data set (input and target). Here 〈|〉 denotes the inner product and |i〉 is a unitary vector with the same size as |yβ〉 with all components equal to zero except the element in position i which is equal to one. |1〉 is a vector with all components equal to 1 and with size equal to that of |yβ〉. Then 〈i|yβ〉 is a scalar corresponding to the pixel value at the ith position in |yβ〉, whereas 〈i|1〉 = 1 for all i. Notice that high pixel values will then have an exponential weight ≈1 while low pixel values will have an exponential weight ≈exp(−1/w). This implies that the error associated to high pixels will have a larger value than that for low pixels. The loss function ${{L}}^{\\left(\\alpha \\right)}$ is the mean value over all pixels (i) and a given data set (β):\n\nwhere 〈〉 denotes average. In our initial trial training runs, we noticed that the loss function always reached a plateau by 800 epochs, so we trained the NNs over 800 epochs for all runs reported in this paper. Because the training is stochastic, the loss function can increase as well as decrease between epochs as seen in Figure 4. At the end of 800 epochs we adopted the network configuration with the lowest loss function regardless of the epoch at which it was achieved.\n\nTABLE 2\nFIGURE 4",
null,
"Figure 4. Training loss function vs. epochs for model 9 (the hyperparameters are specified in Table 2 and the NN details are described in the main text) without roll-back (A) and with roll-back (B) using the same seed. We have circled in green where a jump occurred during this training run (see main text for discussion).\n\nWhile the trendline (averaged over 5 or 10 epochs) of the loss function value tends to decrease during training, the stochasticity of the training means that the value of the loss function often increases significantly between successive epochs, even by one or two orders of magnitude (see Figure 4). In some cases, the loss function decreases back to its trend after one or two epochs, in other cases (which we call jumps), it stays at the higher value, resetting the trend line to the higher value and only gradually begins to decrease afterwards. In this case all of the epochs after the jump have larger loss functions than the epoch immediately before the jump, as shown for the evolution of the loss function for a typical training run in Figure 4A. This behavior indicates that the stochastic optimization algorithm has pursued an unfavorable branch. To avoid this problem, we added a roll-back algorithm to the training, as proposed in Geoffrey (2020). We set a loss threshold value, ${{L}}_{thrs}$, such that if the ratio of loss value from epoch n to n+1 is larger than ${{L}}_{thrs}$, then the training algorithm reverts (rolls back) to the NN state corresponding to epoch ns and tries again. The stochasticity of training means that roll-back has an effect similar to training an ensemble of models with the same hyperparameters and selecting the model with the lowest loss function value, however, the roll-back optimization takes much less computer time than a large ensemble. We set s = 5 and set the threshold value ${{L}}_{thrs}$ to\n\nHere we chose C = 5 and m = 20 where ep stands for epoch, i.e., we set the threshold value to 5 times the average loss function value over the previous m = 20 epochs. We chose these values empirically. In Figure 4B, we have plotted a typical example of the evolution of the loss function during training when we train using roll-back. A typical number of roll-backs is 40, i.e., this number is the number of epochs where the jump was higher than the threshold during the training.\n\n## 3. Results\n\nQuite commonly, the mean residual is the estimator used to judge the goodness of a given model. However, there are cases where the worst predictions are highly informative and can be used to make basic decisions about which features of the NN do not add value. In Figures 5A–C we show 20 different inputs, targets and predictions, respectively. The predictions in Figure 5C were obtained using model 12 (see Table 2) and qualitatively show very good results. For each model we computed the residual, i.e., the absolute value of the difference between the ground truth and the NN prediction pixel-by-pixel, as shown in Figure 6B. We also analyzed the relative residual, i.e., the residual divided by the ground truth pixel-by-pixel, as shown in Figure 6C. Models 6 and 7, which only use NN 1 (p1 = 1 and p2 = 0), yield mean residuals an order of magnitude larger than models that use both or only NN 2. Therefore, we reject the NN 1-only models and do not analyze them further.\n\nFIGURE 5",
null,
"Figure 5. Results for 20 randomly selected test data sets'. (A) input, (B) ground truth (target output), and (C) NN surrogate prediction of steady-state diffusion field output for the input.\n\nFIGURE 6",
null,
"Figure 6. (A) The stationary solution for the same batch in the test set. (B) Residual (absolute error, i.e., ||yβ〉−|ŷβ〉|) for 20 sample source images in the test set trained using model 12 in Table 2. (C) Residual/true value (relative error) for the corresponding images. (D) Mean, 99-Percentile, and maximum residual for all of the models in Table 2. Left scale for mean value, right scale for 99-Percentile residual value and right scale in parentheses for max residual value.\n\nTable 2 summarizes the hyperparameter values for each model we trained. The choice of these parameters was empirically driven. Since we had the field values bounded between 0 and 1 similar to black and white images, we tested different L-norms, namely, mean absolute value (MAE), mean squared value (MSE), and mean to the fourth power, often used in neural networks applied to images. In this paper we show the results for MAE and MSE. We also tested different hyperparameters values for the dropout. We found that low dropout values for NN 2 yield the best results.\n\nIn Figure 6D, we have plotted the mean residual value, the 99-Percentile residual value and the maximum residual value computed over the test set. Notice that the 99-Percentile residual value is ten times the mean residual value and the maximum residual value is 10 times the 99-Percentile residual value. This suggests that the residual distribution contains outliers, i.e., there is a 1% residual that deviate from mean residual 10 to 100 times. Furthermore, these outliers correspond to regions between the source and the border, near the source, where the source is close to the border as suggested by Figure 6B. While the largest values in absolute residual come from pixels near the source as shown in Figure 6B, the relative error near the source is small whereas the relative error near boundaries is large, as shown in Figure 6C. In Figure 6A we show the stationary solution for the same batch shown in Figures 6B,D. Since we are considering absorbing boundary conditions, the field at the boundary is always equal to zero, thus strictly speaking the relative residual value has a singularity at the boundary. Thus, at the boundaries there is a larger relative error due to the boundary conditions. Since our method has a small absolute error independent of the mean value, the relative error is a poor measure of accuracy for small mean values, since it diverges as the mean approaches zero. Since we have zero-value boundary conditions, at the boundaries there is a larger relative error due to the boundary conditions and therefore the relative error is not a functionally meaningful measure of error unless the system being modeled is highly sensitive to small values of the field. Oxygen levels in normal tissues fluctuate significantly in space and time. For instance, in the retina, oxygen concentration fluctuates dramatically in space, time and depending on illumination. Short-term temporal fluctuations range from 5 to 50% depending on depth in cat retina (Linsenmeier and Zhang, 2017). This intrinsic oxygenation fluctuation in tissues suggests that biologically, 5% relative error at low concentrations is an acceptable accuracy for oxygen concentration estimation.\n\nModels 5, 11, and 12 have low mean residuals with model 5 being the smallest. Focusing instead on the mean residual and the 99-Percentile, we notice that models 3, 4, 5, 11, and 12 yield the best results. Finally, considering the maximum residual together with the previous estimators, we notice that model 9 has low mean residual, low 99-percentile residual and the lowest max residual. Depending on the user's needs, one estimator will be more relevant than others. In this sense, defining a best model is relative. Nevertheless, having more metrics (e.g., relative error for large values and absolute error for small values) helps to characterize each model's performance. In future work we'll consider more adaptable metrics, as well as mixed error functions that incorporate multiple estimators.\n\nFigure 8 plots the prediction vs. the target for each pixel in each image in the training and test sets for models 9 and 11. Notice that for the test sets the results are qualitatively similar between models, for the training set the dispersion is larger in model 11 than in model 9. This suggests model 11 is overfitting the training data. Models 9 and 11 have the same hyperparameters except for the weight w. In the former w = 100 while in the latter w = 1. This suggests that the exponential weight helps reduce overfitting.\n\nIn Figure 7, we show the prediction from NN 1 (Figure 7A) and NN 2 (Figure 7B). Notice that NN 1 is able to detect the sources whereas NN 2 is able to predict the field. Using both neural networks improves the results as can be seen in Figure 6D. As previously mentioned, pixels with low (near 0) field values are much more common than pixels with high (near 1) field values. While the exponential factor in the loss function compensates for this bias, the residual in Figure 6D does not. To address this issue we compute the mean residual over small field intervals. This will tell us how well the model predicts for each range of absolute values. Furthermore, this method can be used to emphasize accuracy or relative accuracy in different value ranges. The way we do this is as follows. In Figure 8, we take 10 slices of size 0.1 in the direction y = x. We then compute the mean residual and standard deviation per slice. In Supplementary Material (section 1), we have plotted the PDF (probability density function) per slice (blue bins) and a Gaussian distribution (red curve) with mean and standard deviation set to the mean residual and standard deviation per slice, respectively. We did this for all models in Table 2. In Figure 9, we plotted the mean residual vs. for each model for each slice for the test and training sets. The error envelop shows the residual standard deviation per slice. Notice that models trained with MSE have a smaller residual standard deviation than models trained with MAE in the case of the training set, which suggest that MSE contributes to overfitting more that MAE. Recall that the difference between the MSE gradient and the MAE gradient is that the former is linear with the residual value whereas the latter is a constant. Therefore, training with MAE generalizes better than MSE. Additionally, notice the dispersion increases with the slice number.\n\nFIGURE 7",
null,
"Figure 7. Results for 20 randomly selected test data sets'. (A) Prediction using model 7, which only uses NN 1. (B) Prediction using model 5, which only uses NN 2 (see Table 2). Note the different scale on the color bars.\n\nFIGURE 8",
null,
"Figure 8. Ground truth vs. prediction for (A) test set and (B) training set in the case of model 9; (C) test set and (D) training set in the case of model 11 (see Table 2). The number of points plotted in each panel is 3.75·107.\n\nFIGURE 9",
null,
"Figure 9. Mean (data points) ± standard deviation (envelop) per slice vs. models (see Table 1) for test set (blue) and training set (red). Slice i corresponds to field values in the interval [0.1·(i−1), 0.1·i] where i = 1, …, 10.\n\nIn Figure 10, we plotted the average and maximum over the residual mean value per slice (see Figure 10A) and the residual standard deviation per slice (see Figure 10B) for each model's test and training sets. Notice that in this approach, by slicing the residual values and computing the average residual over the set of slices, we are giving equal weight to each mean residual per slice and, therefore, compensating for the imbalance in frequency of low and high value pixels. An interesting feature from using MSE or MAE comes from the PDF of the field values. Training using MAE makes the PDF prediction quite accurate as the prediction completely overlaps with the ground truth (see Figure 11). In comparison, when training with MSE, the PDF is not as good and the overlap between ground truth and prediction is not complete. There is a mismatch for low field values in the sense that the NN does not predict low non-zero field values correctly. Thus, we recommend using MAE to avoid this issue.\n\nFIGURE 10",
null,
"Figure 10. (A) For each model, we show the average and maximum over the residual mean value per slice. (B) For each model, we show the average and maximum over the residual standard deviation per slice (see Figure 9). This was done for the test and training set.\n\nFIGURE 11",
null,
"Figure 11. PDF of field obtained via NN (blue) and ground truth (red) in the case of training using MSE, for (A) model 2 and (B) model 3 and for training using MAE, for (C) model 11 and (D) model 12. When using MSE (A,B) the NN predicts zero field values instead of low non-zero field values as the predicted PDF has a larger peak in zero than the ground truth PDF, and a smaller PDF for small non-zero field values compared with the ground truth PDF. When training using MAE (C,D) the prediction and ground truth PDFs overlap completely.\n\n## 4. Discussion\n\nIn large-scale mechanistic simulations of biological tissues, calculations of the diffusion of molecular species can be a significant fraction of the total computational cost. Because biological responses to concentrations often have a stochastic overlay, high precision may not be essential in these calculations Because NN surrogate estimates are significantly faster than the explicit calculation of the steady-state diffusion field for a given configuration of sources and sinks, an effective NN surrogate could greatly increase the practical size of simulated tissues, e.g., in cardiac simulations (Kerckhoffs et al., 2007; Sundnes et al., 2014), cancer simulations (Bruno et al., 2020), and orthopedic simulations (Erdemir et al., 2007). In our case, using a NVIDIA Quadro RTX 6000, each diffusion solution is about 1,000 times faster using the trained NN solver compared to the Julia code.\n\nIn order to decide if this acceleration is useful, we have to consider how long it takes to run the direct simulation, how long the NN takes to train and how long it takes to execute the NN once it has been trained (Fox and Jha, 2019). If each diffusion calculation takes δ seconds to run, conducting N calculations directly takes tdirect = . If each neural network surrogate takes ϵ seconds to run, and the number of replicas in the training set is M and the training time is E, the total time for the neural network simulation is the time to generate the training set, the training time plus the simulation time, tneuro = +E+. To estimate these times, we ran 20, 000 explicit simulations in Julia, which took ~ 6 h and 30 min, yielding roughly 1.16s each. The NN training time was 12 h on average. While the speedup for an individual simulation is δ/ϵ ≈ 1, 000, the ratio τneurodirect must be smaller than 1 in order to have a useful acceleration. Equating this ratio to 1 and solving for N yields\n\nNmin gives the number of replicas necessary for the total time using the NN to be the same as the direct calculation. Of course, the exact times will depend on the specific hardware used for the direct and NN calculations. In our case, from Equation (4) we obtain that Nmin≈57, 300, we would need to use the neural network more than 57, 300 times for the total time using the NN to be faster than the direct calculation. Thus the NN acceleration is primarily useful in simulations that will be run many, many times for the specific situation for which the NN is appropriate. Consider for example if one wishes to include a variable number of sources, different lattice sizes, different dimensionalities (e.g., 3D) and boundary conditions. The more general the NN the more training data it will require, the longer training will take, and the slower the individual NN calculations will be. Currently virtual-tissue simulation studies often run thousands to tens of thousands of replicas and each replica often takes tens of minutes to tens of hours to run. This computational cost makes detailed parameter identification and uncertainty quantification impractical, since simulations often have dozens of parameters to explore. If using a NN-based diffusion solver accelerated these simulations by 100 × it would permit practical studies with hundreds of thousands to millions of replicas, greatly expanding the feasible exploration of parameter space for parameter identification and uncertainty quantification. It is worthwhile mentioning that there are other numerical methods for diffusion in 2D and 3D models that can also exploit the GPU parallelization such as the one in Secomb (2016) based on a discretization of Green's function. Our focus is on the ability of neural-network surrogates to solve the time-independent diffusion equation, however it would be interesting to extensively optimize the mechanistic methods we used to generate our training data sets. Generating training data is so time consuming, applications of deep neural networks will benefit greatly from using faster mechanistic methods to generate training data.\n\nWhile there isn't a protocol for setting up a diffusion-solver surrogate, there are several things that must be considered. First one needs to frame the problem similar to how one would do when performing mechanistic modeling. One needs to settle on the dimensionality, e.g., 1D, 2D, 3D,… or n-dimensions; the system size which in our case we settled on 100 × 100; the type of sources to consider, e.g., sinks, sources, or both; the boundary conditions e.g., absorbing, reflective, periodic, or mixed; the distribution of sources in space; and it is also important to think about the accuracy required from the neural network. There isn't a rigorous way to determine the size of the required training dataset, although the size will depend on the problem one is addressing and the decisions made in the previous step. We recommend to start with a training dataset size of the order ~ 104. Then one needs to decide on the network architecture. For the network architecture the number of options is large. For instance, the depth of the neural network, e.g., deep or shallow; the type of layers, e.g., convolutional layers, fully-connected layers, recurrent layers, or mixed layers; the activation functions, e.g., ReLU, sigmoid, tanh, etc. Evidence suggests that using deep neural networks as opposed to shallow neural networks will increase the non-linearity of the neural network, which ultimately broadens the learning capabilities of the neural network (Bianchini and Scarselli, 2014). But as depth increases the gradient of the loss function can grow or diminish significantly leading to instabilities or a regime of zero learning where the gradient becomes zero but the loss function value is large. Very deep neural networks can also lead to an overflow or underflow situation. Therefore, the neural network depth is a feature that should be set in way that meets a middle ground. The right choice of activation functions, regularizer layers (i.e., DropOut, BatchNorm, etc.) and weight initializers can hinder the unwanted features of instabilities or zero learning. Choosing the right optimizer for training is also something to consider. However, unless there are some very specific needs, the standard rule is to use the ADAM optimizer (Kingma and Ba, 2014). Choosing the loss function is crucial as this is the metric the neural network will use to measure how good the outcome is. Typically for these type of problems MSE, MAE and other similar norms are used. Additionally, there are a number of (hyper)parameters to be chosen. For instance, some activation functions, regularizer layer and optimizers have hyperparameters. Also the number of epochs and size of minibatch are hyperparameters. To set the hyperparameters' values, one can start by using the values reported in the literature but the scope should be to explore the space of hyperparameters by training an ensemble of neural networks with different hyperparameters and then choosing the model that performed best on the validation set.\n\nIn a real tissue, the oxygen tension on the surface of the blood vessel and in the tissue as a whole involves complex feedback among many factors, including spatial and temporal variation in the supply and consumption of oxygen; supply at a given location could depend on the degree of local blood-vessel dilation, the rate of blood flow, and levels of oxygen in the blood to name a few examples. A realistic model of oxygenation in tissue would need to include spatial and temporal models of all of these processes individually and of their coupling. Clearly such a model is much more complex than our simple example of calculating the steady-state oxygen field given a fixed set of circular sources with fixed oxygen tensions and a fixed uniform consumption rate in the tissue implemented as linear decay.\n\nWhile developing NN surrogates to solve the entire complex problem of oxygenation would be worthwhile, we believe that deep neural network surrogates will (at least initially) not replace the entire simulation, but to replace the most computationally costly components of the simulation. In this case, looking for surrogates for specific commonly-used calculations, which can be used in many different applications and which can provide a substantial speed-up is appropriate. Many biophysical and engineering problems require solving the diffusion equation for fixed sources. Despite the improvements to direct solution mentioned in Secomb (2016), solving the diffusion equation still often contributes much of the computational cost of the full problem solution. In these cases, the faster the “diffusion step” is computed, the faster the solution of the multiscale model as a whole. To train an optimal diffusion surrogate for a particular problem one has to choose a set of appropriate loss functions and combine them to minimize the errors of the metrics one defines as most relevant to the specific problem being addressed. How to choose loss functions and their weighting to achieve macroscopic desired outcomes is not well understood as a general problem. Even in our very simple example, we had to explore a wide variety of loss functions to achieve reasonable convergence of our NN during training and reasonable final absolute and relative accuracy of our surrogate.\n\n## 5. Conclusions\n\nNeural networks provide many possible approaches to generating surrogate diffusion solvers. Given the type of problem setting, we were interested in a neural network that could predict the stationary field. We considered a deep convolutional neural network, an autoencoder and their combination. We considered two loss functions, viz. mean squared error and mean absolute error. We considered different hyperparameters for dropout and an exponential weight to compensate the under-sampling of high field values. The exponential weight also helped reduce overfitting as shown in Figure 8.\n\nThe range of scientific and engineering applications for diffusion solvers is very broad. Depending on the specific application, the predictions by the neural network will have to meet a specific set of criteria quantified in the form of statistical estimators (e.g., mean error, max error, percentiles, mean relative error, etc.). In this paper we studied several reasonable error metrics, namely, mean residual, maximum residual, 99-Percentile residual, mean relative residual, mean weighted residual and the weighted standard deviation residual. The last two metrics compensate for the low frequency of high field values, ones that usually occur in small regions around sources. The autoencoders are commonly used in generative models which is applicable, as we have shown here, to the case of a diffusion surrogate. The field predictions are accurate on all the metrics we considered. This is appears to be due to collapsing the input into a one-dimensional vector and then decoding back to the initial size, which forces the network to learn the relevant features (Kingma and Welling, 2019). While some models had high errors across all metrics, no single model had the smallest error for all error metrics. Different networks and hyperparameters were optimal for different metrics, e.g., model 5 had the lowest mean residual, whereas model 9 yielded relatively good results on all metrics. Model 9 uses both neural networks with the dropout values for the deep convolutional network were set to D1,2 = 0.4, and for the autoencoder to D3,4 = 0.1. The weight hyperparameter was set to 100. Recall that large weight hyperparameter values make the loss function weight high field values over low field values. This is important since the largest absolute error happens close to sources and close to boundaries because of the under-representation of these kinds of configurations. We also noticed that this choice reduced the overfitting as was shown in Figure 8.\n\nAdditionally, we tested several loss function. Here we reported the results using mean squared error and mean absolute error. We noticed two key differences. With MSE the weighted standard deviation (see Figure 9) is smaller than for MAE for the training set. However, for the test set, the results for both loss functions are comparable. This difference between training and test sets suggests that MSE is more prone to overfitting the data than MAE. The other key difference is that for the MAE, the predicted field probability function consistently overlapped the ground truth completely, whereas for MSE there is a mismatch in that the NN does not predict low non-zero field values correctly (see Figure 11). Therefore, we recommend using MAE as the loss function for surrogate calculations where the field values are well bounded, as we have shown it produces better predictions than MSE. The autoencoder (NN 2) is capable of approximating the diffusion field on its own, the convolutional network (NN 1) is not. However, if we use the two networks together we find that the prediction is more accurate than NN 2 alone.\n\nThese encouraging results suggest that we should pursue NN surrogates for acceleration of simulations in which the solution to the diffusion equation contributes a considerable fraction of the total computational cost. An effective NN diffusion solver surrogate would need to be able to solve diffusion fields for arbitrary sources and sinks in two or three dimensions with variable diffusivity, a much higher dimensional set of conditions than the two circular sources in a uniform two-dimensional square domain that we investigated in this paper. A key question will be the degree to which NNs are able to generalize, e.g., from n sources to n+1 sources or from circular sources to more complex shapes. In addition, here we only considered absorbing boundary conditions, ultimately mixed boundary conditions are desirable. It is unclear if the best approach would be a single NN capable of doing multiple boundary conditions, or better to develop unique NNs for each boundary condition scenario. While in this paper we have only considered zero-field boundary conditions mainly due to feasibility purposes for the neural network, we will consider different boundary conditions in future work.\n\nIncreasing the number and size of vessels is a combinatorial problem in the dimensionality of the training set, but it ultimately doesn't change the nature of the diffusion equation. Thus, we expect that a straightforward approach consisting using a bigger training set including a greater variety of source and sink sizes, shapes, and number, should still work, though it will take more computing time to generate the training data and train the network. The ability of greens-function methods to solve the diffusion equation for arbitrary numbers of sources and sinks suggests (though it does not prove) that such generalization should work also for neural network solvers.\n\nTo solve 3D diffusion problems, the most straightforward extension of our method would be to use 3D convolutional neural networks. However, there may be some difficulties with a naive extension of our convolutional methods to 3D. If we have a linear dimension of L then the output layer of the NN has L2 elements in 2D and L3 in 3D. Thus, for a given value of L, the network size is much larger in 3D. Besides the size of the network, the training set will also be larger. For N sources, the number of possible configurations grows roughly as L2N in 2D, while the number of configurations in 3D is L3N. In addition, if we wish to represent realistic sources in 3D, like blood vessels, we need to sample over appropriately spatially-correlated patterns of sources rather than the randomly located spherical sources we used in our 2D example. Naively these very high dimensions of possible source configurations suggest that the 3D problem would require impossibly large training datasets. However, one of the outstanding features of deep neural networks is their capacity to extrapolate from apparently severely undersampled training sets, so increasing the number of possible configurations exponentially does not necessarily imply the need to increase the training set exponentially. Another approach to develop diffusion solver surrogates in 3D is to build physically informed neural networks (PINNs) (Raissi et al., 2019) where the ODE describing the process, the initial conditions and the boundary conditions are embedded in the loss function. Other efforts attempt to tackle the curse of dimensionality by physical intuition embedded in the neural network architecture (Roberts, 2021). We will explore these issues in future work.\n\n## Data Availability Statement\n\nThe datasets presented in this study can be found in online repositories. The names of the repository/repositories and accession number(s) can be found at: https://github.com/jquetzalcoatl/DiffusionSurrogateDeepLearning.\n\n## Author Contributions\n\nJG and JS proposed the project. JT-M and GF built the models. JT-M trained the models. All authors analyzed the results and wrote the manuscript.\n\n## Funding\n\nThis research was supported in part by Lilly Endowment, Inc., through its support for the Indiana University Pervasive Technology Institute. This work was partially supported by the National Science Foundation (NSF) through awards nanoBIO 1720625, CINES 1835598 and Global Pervasive Computational Epidemiology 1918626, and Cisco University Research Program Fund grant 2020-220491. This work was partially supported by the Biocomplexity Institute at Indiana University, National Institutes of Health, grant NIGMS R01 GM122424. This work was partially supported by the DOE ASCR Award DE-SC0021418 “FAIR Surrogate Benchmarks Supporting AI and Simulation Research”.\n\n## Conflict of Interest\n\nThe authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.\n\n## Acknowledgments\n\nWe thank the referees for suggesting different ways to extend our work. This is a short text to acknowledge the contributions of specific colleagues, institutions, or agencies that aided the efforts of the authors.\n\n## Supplementary Material\n\nThe Supplementary Material for this article can be found online at: https://www.frontiersin.org/articles/10.3389/fphys.2021.667828/full#supplementary-material\n\n## Footnotes\n\n1. ^We use the terms deep learning and machine learning interchangeably. We also use neural network and deep neural network interchangeably.\n\n## References\n\nBaur, C., Denner, S., Wiestler, B., Navab, N., and Albarqouni, S. (2020). Autoencoders for unsupervised anomaly segmentation in brain MR images: a comparative study. Med. Image Anal. 2020:101952. doi: 10.1016/j.media.2020.101952\n\nBianchini, M., and Scarselli, F. (2014). On the complexity of neural network classifiers: a comparison between shallow and deep architectures. IEEE Trans. Neural Netw. Learn. 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Softw. 5:15. doi: 10.5334/jors.151\n\nRaissi, M., Perdikaris, P., and Karniadakis, G. E. (2019). Physics-informed neural networks: a deep learning framework for solving forward and inverse problems involving nonlinear partial differential equations. J. Comput. Phys. 378, 686–707. doi: 10.1016/j.jcp.2018.10.045\n\nRoberts, D. A. (2021). Why is AI Hard and Physics Simple? Technical report. arxiv preprint. arXiv:2104.00008\n\nSchiesser, W. E. (2012). The Numerical Method of Lines: Integration of Partial Differential Equations. San Diego, CA: Elsevier.\n\nSecomb, T. W. (2016). A green's function method for simulation of time-dependent solute transport and reaction in realistic microvascular geometries. Math. Med. Biol. 33, 475–494. doi: 10.1093/imammb/dqv031\n\nSharma, R., Farimani, A. B., Gomes, J., Eastman, P., and Pande, V. (2018). Weakly-supervised deep learning of heat transport via physics informed loss. arXiv preprint arXiv:1807.11374.\n\nSundnes, J., Wall, S., Osnes, H., Thorvaldsen, T., and McCulloch, A. D. (2014). Improved discretisation and linearisation of active tension in strongly coupled cardiac electro-mechanics simulations. Comput. Methods Biomech. Biomed. Eng. 17, 604–615. doi: 10.1080/10255842.2012.704368\n\nToledo-Marin, J. Q. (2020). Stationary Diffusion State Ml Surrogate Using Flux and Cuarrays. Available online at: https://github.com/jquetzalcoatl/DiffusionSurrogate (accessed June 09, 2021).\n\nXie, D., Li, Y., Yang, H., Song, D., Shang, Y., Ge, Q., et al. (2019). “Bold fMRI-based brain perfusion prediction using deep dilated wide activation networks,” in International Workshop on Machine Learning in Medical Imaging (Shenzhen: Springer), 373–381. doi: 10.1007/978-3-030-32692-0_43\n\nZhang, H., Hippalgaonkar, K., Buonassisi, T., Løvvik, O. M., Sagvolden, E., and Ding, D. (2019). Machine learning for novel thermal-materials discovery: early successes, opportunities, and challenges. arXiv preprint arXiv:1901.05801. doi: 10.30919/esee8c209\n\nZhang, J.-S., and Xiao, X.-C. (2000). Predicting chaotic time series using recurrent neural network. Chinese Phys. Lett. 17:88. doi: 10.1088/0256-307X/17/2/004\n\nKeywords: diffusion surrogate, machine learning, virtual tissue, mechanistic modeling, Julia\n\nCitation: Toledo-Marín JQ, Fox G, Sluka JP and Glazier JA (2021) Deep Learning Approaches to Surrogates for Solving the Diffusion Equation for Mechanistic Real-World Simulations. Front. Physiol. 12:667828. doi: 10.3389/fphys.2021.667828\n\nReceived: 14 February 2021; Accepted: 25 May 2021;\nPublished: 24 June 2021.\n\nEdited by:\n\nGary An, University of Vermont, United States\n\nReviewed by:\n\nNikolaos Tsoukias, Florida International University, United States\nHermann Frieboes, University of Louisville, United States\n\nCopyright © 2021 Toledo-Marín, Fox, Sluka and Glazier. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.\n\n*Correspondence: J. Quetzalcóatl Toledo-Marín, j.toledo.mx@gmail.com"
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https://goprep.co/a-resistor-has-a-resistance-of-176-ohms-how-many-of-these-i-1njf92 | [
"Q. 94.3( 32 Votes )\n\n# A resistor has a\n\nAnswer :\n\nWe Know according to Ohms Law that:-\nV= IR\nWhere V= Voltage/Potential Difference\nI= Current\nR=Resistance\nAccording to Question :-V= 220V\nI= 5A\n∴ Total Resistance of the Combination is :-",
null,
"Suppose x resistors should be connected in parallel to draw a current of 5A\n\nfor parallel combination:-\n\n1/R = 1/R1 +1/R2+ 1/R3........1/RX(Where Resistance of each Resistor is 176Ω)\n\nAccording to the question,\n\nRCombination=176/x\n⇒44=176/x( We know from above the total resistance of combination)\n\nx=176/44=4 resistor\nHence 4 resistors each of 176Ω should be connected in parallel so as to draw the current of 5A from a 220 volt supply line.\n\nRate this question :\n\nHow useful is this solution?\nWe strive to provide quality solutions. Please rate us to serve you better.\nRelated Videos",
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"",
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"",
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"Resistors in Series42 mins\nTry our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts\nDedicated counsellor for each student\n24X7 Doubt Resolution\nDaily Report Card\nDetailed Performance Evaluation",
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"view all courses",
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https://www.moneygeek.com/insurance/auto/cheapest-car-insurance-san-francisco-ca/ | [
"## Compare Rates and Save on Auto Insurance\n\nThe lowest-priced auto insurance depends on your needs and driving history.\n\nGEICO has the cheapest rate in San Francisco for minimum or full coverage, young adults, teens, with a sports car, a speeding ticket or at-fault accident. State Farm is the most affordable option for seniors, USAA for the military community and National General with a DUI.\n\nMoneyGeek also found the best auto insurance companies in San Francisco for balancing affordable coverage and service quality.",
null,
"CheapestCar InsuranceSan Francisco\n2023\n\nCheapest Car Insurance in San Francisco, CA\n\n## Cheapest Minimum Coverage Car Insurance in San Francisco\n\nOn average, the most affordable auto insurance providers in San Francisco, if you need a minimum coverage policy, are:\n\n• GEICO: $364 per year • Progressive:$470 per year\n\nOut of the providers MoneyGeek reviewed quotes for in San Francisco, a minimum coverage policy is the most expensive with National General, with an average yearly cost of $951. Company Annual Premium Monthly Premium GEICO$359\n\n$30 State Farm$359\n\n$30 GEICO$364\n\n$30 GEICO$427\n\n$36 Progressive$470\n\n$39 State Farm$474\n\n$39 Progressive$474\n\n$40 CSAA$491\n\n$41 CSAA$500\n\n$42 Mercury$527\n\n$44 CSAA$539\n\n$45 Mercury$539\n\n$45 State Farm$541\n\n$45 Progressive$557\n\n$46 Nationwide$578\n\n$48 Kemper$591\n\n$49 Nationwide$601\n\n$50 Farmers$608\n\n$51 Hartford$621\n\n$52 State National$622\n\n$52 Travelers$623\n\n$52 Travelers$625\n\n$52 Capital Insurance Group$626\n\n$52 Farmers$626\n\n$52 State National$641\n\n$53 Capital Insurance Group$648\n\n$54 Mercury$661\n\n$55 Kemper$682\n\n$57 Hartford$710\n\n$59 Farmers$711\n\n$59 Travelers$714\n\n$60 Allstate$725\n\n$60 Capital Insurance Group$726\n\n$61 Allstate$732\n\n$61 State National$759\n\n$63 Nationwide$761\n\n$63 Chubb$766\n\n$64 Allstate$789\n\n$66 Chubb$790\n\n$66 Kemper$899\n\n$75 Hartford$913\n\n$76 State Farm$927\n\n$77 GEICO$940\n\n$78 National General$951\n\n$79 GEICO$973\n\n$81 National General$980\n\n$82 Progressive$1,089\n\n$91 Progressive$1,096\n\n$91 Chubb$1,097\n\n$91 GEICO$1,121\n\n$93 National General$1,180\n\n$98 State Farm$1,185\n\n$99 Mercury$1,309\n\n$109 Progressive$1,316\n\n$110 Allstate$1,324\n\n$110 Capital Insurance Group$1,326\n\n$111 Mercury$1,340\n\n$112 Hartford$1,345\n\n$112 State Farm$1,388\n\n$116 Allstate$1,394\n\n$116 Nationwide$1,395\n\n$116 Capital Insurance Group$1,408\n\n$117 CSAA$1,502\n\n$125 CSAA$1,509\n\n$126 Nationwide$1,523\n\n$127 State National$1,525\n\n$127 Hartford$1,544\n\n$129 Allstate$1,548\n\n$129 Kemper$1,574\n\n$131 Capital Insurance Group$1,585\n\n$132 Travelers$1,647\n\n$137 State National$1,650\n\n$138 CSAA$1,671\n\n$139 Farmers$1,676\n\n$140 Farmers$1,720\n\n$143 Travelers$1,725\n\n$144 Mercury$1,737\n\n$145 National General$1,744\n\n$145 National General$1,766\n\n$147 Chubb$1,910\n\n$159 Kemper$1,941\n\n$162 Chubb$1,970\n\n$164 Hartford$1,990\n\n$166 Nationwide$1,999\n\n$167 Farmers$2,041\n\n$170 State National$2,045\n\n$170 Travelers$2,057\n\n$171 National General$2,117\n\n$176 Kemper$2,342\n\n$195 Chubb$2,814\n\n$234",
null,
"#### Compare Insurance Rates Ensure you are getting the best rate for your insurance. Compare quotes from the top insurance companies. California requires that car insurance policies include liability coverage for bodily injury and property damage and that each policy features the following minimum limits: •$15,000 bodily injury liability per person\n• $30,000 bodily injury liability per accident •$5,000 property damage liability per accident\n\nAfter getting involved in an at-fault accident, you may need to cover any extra costs if damages exceed the above limits. For instance, if found at fault in an accident and the property damage cost is $20,000, your insurer will only cover$5,000, and you will have to pay the remaining balance of $15,000. ## Cheapest Full Coverage Car Insurance in San Francisco The following insurers offer the average lowest auto insurance premium for a full coverage policy in San Francisco: • GEICO:$973 per year\n• Progressive: $1,096 per year The most expensive insurance company for full coverage in San Francisco is Chubb, with an average cost of$1,910 a year.\n\nCompany\n\nGEICO\n\n$359$30\n\nState Farm\n\n$359$30\n\nGEICO\n\n$364$30\n\nGEICO\n\n$427$36\n\nProgressive\n\n$470$39\n\nState Farm\n\n$474$39\n\nProgressive\n\n$474$40\n\nCSAA\n\n$491$41\n\nCSAA\n\n$500$42\n\nMercury\n\n$527$44\n\nCSAA\n\n$539$45\n\nMercury\n\n$539$45\n\nState Farm\n\n$541$45\n\nProgressive\n\n$557$46\n\nNationwide\n\n$578$48\n\nKemper\n\n$591$49\n\nNationwide\n\n$601$50\n\nFarmers\n\n$608$51\n\nHartford\n\n$621$52\n\nState National\n\n$622$52\n\nTravelers\n\n$623$52\n\nTravelers\n\n$625$52\n\nCapital Insurance Group\n\n$626$52\n\nFarmers\n\n$626$52\n\nState National\n\n$641$53\n\nCapital Insurance Group\n\n$648$54\n\nMercury\n\n$661$55\n\nKemper\n\n$682$57\n\nHartford\n\n$710$59\n\nFarmers\n\n$711$59\n\nTravelers\n\n$714$60\n\nAllstate\n\n$725$60\n\nCapital Insurance Group\n\n$726$61\n\nAllstate\n\n$732$61\n\nState National\n\n$759$63\n\nNationwide\n\n$761$63\n\nChubb\n\n$766$64\n\nAllstate\n\n$789$66\n\nChubb\n\n$790$66\n\nKemper\n\n$899$75\n\nHartford\n\n$913$76\n\nState Farm\n\n$927$77\n\nGEICO\n\n$940$78\n\nNational General\n\n$951$79\n\nGEICO\n\n$973$81\n\nNational General\n\n$980$82\n\nProgressive\n\n$1,089$91\n\nProgressive\n\n$1,096$91\n\nChubb\n\n$1,097$91\n\nGEICO\n\n$1,121$93\n\nNational General\n\n$1,180$98\n\nState Farm\n\n$1,185$99\n\nMercury\n\n$1,309$109\n\nProgressive\n\n$1,316$110\n\nAllstate\n\n$1,324$110\n\nCapital Insurance Group\n\n$1,326$111\n\nMercury\n\n$1,340$112\n\nHartford\n\n$1,345$112\n\nState Farm\n\n$1,388$116\n\nAllstate\n\n$1,394$116\n\nNationwide\n\n$1,395$116\n\nCapital Insurance Group\n\n$1,408$117\n\nCSAA\n\n$1,502$125\n\nCSAA\n\n$1,509$126\n\nNationwide\n\n$1,523$127\n\nState National\n\n$1,525$127\n\nHartford\n\n$1,544$129\n\nAllstate\n\n$1,548$129\n\nKemper\n\n$1,574$131\n\nCapital Insurance Group\n\n$1,585$132\n\nTravelers\n\n$1,647$137\n\nState National\n\n$1,650$138\n\nCSAA\n\n$1,671$139\n\nFarmers\n\n$1,676$140\n\nFarmers\n\n$1,720$143\n\nTravelers\n\n$1,725$144\n\nMercury\n\n$1,737$145\n\nNational General\n\n$1,744$145\n\nNational General\n\n$1,766$147\n\nChubb\n\n$1,910$159\n\nKemper\n\n$1,941$162\n\nChubb\n\n$1,970$164\n\nHartford\n\n$1,990$166\n\nNationwide\n\n$1,999$167\n\nFarmers\n\n$2,041$170\n\nState National\n\n$2,045$170\n\nTravelers\n\n$2,057$171\n\nNational General\n\n$2,117$176\n\nKemper\n\n$2,342$195\n\nChubb\n\n$2,814$234\n\nAlthough a full coverage car insurance policy costs more than minimum coverage, it includes liability, comprehensive and collision coverage, which heightens your financial protection.\n\n• Collision insurance: If your car needs to get repaired or replaced after an at-fault accident that involves a collision with other cars or objects, including guardrails and trees, collision insurance covers the cost, excluding the deductible. It also covers rollover accidents.\n• Comprehensive insurance: It pays to replace or repair your vehicle after damage from non-collision events. These may include weather events, theft, fire and hitting an animal.\n\nA full coverage policy may be the best for most drivers. However, the added cost may not be ideal if you have an older car that has a very low value.\n\n•",
null,
"Compare\n\nCompare minimum and full coverage policies.\n\n•",
null,
"Find the best option\n\nGet matched with an insurer in San Francisco.\n\n•",
null,
"Discover savings\n\nCheaper insurers could pocket you hundreds of dollars per year.\n\n## Cheapest Car Insurance in San Francisco After a Driving Offense\n\nIf you have a moving violation, you can find the most affordable full coverage auto insurance in San Francisco, on average, from the following insurers:\n\n• Speeding ticket: GEICO ($1,456 per year) • At-fault accident: GEICO ($1,625 per year)\n• DUI: National General ($1,830 per year) The highest average full coverage car insurance in San Francisco for drivers with a speeding ticket ($2,638 per year) is with Nationwide; with an at-fault accident ($3,083 per year), it is with Chubb; and with a DUI, it is Hartford ($5,596 per year).\n\nA driving violation can significantly increase your auto insurance premium. Despite selecting the cheapest provider, you are likely to pay more than someone with a clean record for the same coverage.\n\nCompany\n\nState Farm\n\n$545$45\n\nGEICO\n\n$547$46\n\nGEICO\n\n$553$46\n\nState Farm\n\n$569$47\n\nCSAA\n\n$612$51\n\nCSAA\n\n$612$51\n\nCSAA\n\n$622$52\n\nCSAA\n\n$622$52\n\nGEICO\n\n$644$54\n\nGEICO\n\n$650$54\n\nGEICO\n\n$651$54\n\nCSAA\n\n$673$56\n\nCSAA\n\n$673$56\n\nMercury\n\n$677$56\n\nCapital Insurance Group\n\n$692$58\n\nMercury\n\n$694$58\n\nCapital Insurance Group\n\n$717$60\n\nState Farm\n\n$721$60\n\nCapital Insurance Group\n\n$727$61\n\nState Farm\n\n$752$63\n\nCapital Insurance Group\n\n$754$63\n\nGEICO\n\n$766$64\n\nKemper\n\n$787$66\n\nCapital Insurance Group\n\n$805$67\n\nState Farm\n\n$822$69\n\nFarmers\n\n$828$69\n\nFarmers\n\n$828$69\n\nProgressive\n\n$840$70\n\nCapital Insurance Group\n\n$847$71\n\nMercury\n\n$848$71\n\nProgressive\n\n$848$71\n\nFarmers\n\n$852$71\n\nFarmers\n\n$852$71\n\nMercury\n\n$853$71\n\nState Farm\n\n$859$72\n\nGEICO\n\n$864$72\n\nMercury\n\n$869$72\n\nGEICO\n\n$874$73\n\nState National\n\n$881$73\n\nState National\n\n$881$73\n\nNationwide\n\n$894$75\n\nHartford\n\n$895$75\n\nHartford\n\n$895$75\n\nState National\n\n$908$76\n\nState National\n\n$908$76\n\nKemper\n\n$910$76\n\nProgressive\n\n$924$77\n\nNationwide\n\n$930$78\n\nProgressive\n\n$934$78\n\nMercury\n\n$945$79\n\nTravelers\n\n$966$81\n\nTravelers\n\n$968$81\n\nFarmers\n\n$968$81\n\nFarmers\n\n$968$81\n\nMercury\n\n$969$81\n\nChubb\n\n$986$82\n\nNationwide\n\n$995$83\n\nProgressive\n\n$1,002$83\n\nNational General\n\n$1,006$84\n\nNational General\n\n$1,006$84\n\nProgressive\n\n$1,006$84\n\nChubb\n\n$1,009$84\n\nProgressive\n\n$1,016$85\n\nAllstate\n\n$1,020$85\n\nAllstate\n\n$1,029$86\n\nGEICO\n\n$1,030$86\n\nFarmers\n\n$1,035$86\n\nNationwide\n\n$1,036$86\n\nHartford\n\n$1,037$86\n\nHartford\n\n$1,037$86\n\nNational General\n\n$1,038$87\n\nNational General\n\n$1,038$87\n\nFarmers\n\n$1,065$89\n\nMercury\n\n$1,069$89\n\nTravelers\n\n$1,077$90\n\nTravelers\n\n$1,078$90\n\nState National\n\n$1,080$90\n\nState National\n\n$1,080$90\n\nCapital Insurance Group\n\n$1,086$91\n\nProgressive\n\n$1,109$92\n\nAllstate\n\n$1,116$93\n\nTravelers\n\n$1,117$93\n\nState Farm\n\n$1,122$93\n\nCapital Insurance Group\n\n$1,126$94\n\nAllstate\n\n$1,179$98\n\nNationwide\n\n$1,183$99\n\nMercury\n\n$1,190$99\n\nAllstate\n\n$1,191$99\n\nChubb\n\n$1,197$100\n\nKemper\n\n$1,202$100\n\nProgressive\n\n$1,206$101\n\nFarmers\n\n$1,209$101\n\nChubb\n\n$1,220$102\n\nTravelers\n\n$1,246$104\n\nNational General\n\n$1,248$104\n\nNational General\n\n$1,248$104\n\nCapital Insurance Group\n\n$1,265$105\n\nAllstate\n\n$1,292$108\n\nTravelers\n\n$1,306$109\n\nTravelers\n\n$1,306$109\n\nChubb\n\n$1,316$110\n\nNationwide\n\n$1,319$110\n\nKemper\n\n$1,348$112\n\nKemper\n\n$1,348$112\n\nHartford\n\n$1,361$113\n\nHartford\n\n$1,361$113\n\nState Farm\n\n$1,371$114\n\nGEICO\n\n$1,407$117\n\nState Farm\n\n$1,437$120\n\nNational General\n\n$1,438$120\n\nGEICO\n\n$1,456$121\n\nState Farm\n\n$1,483$124\n\nNational General\n\n$1,484$124\n\nTravelers\n\n$1,516$126\n\nCapital Insurance Group\n\n$1,519$127\n\nChubb\n\n$1,528$127\n\nChubb\n\n$1,549$129\n\nCapital Insurance Group\n\n$1,561$130\n\nChubb\n\n$1,573$131\n\nGEICO\n\n$1,574$131\n\nNationwide\n\n$1,582$132\n\nState National\n\n$1,595$133\n\nCapital Insurance Group\n\n$1,615$135\n\nGEICO\n\n$1,625$135\n\nKemper\n\n$1,639$137\n\nKemper\n\n$1,639$137\n\nState National\n\n$1,645$137\n\nNationwide\n\n$1,647$137\n\nCapital Insurance Group\n\n$1,657$138\n\nGEICO\n\n$1,680$140\n\nState Farm\n\n$1,693$141\n\nCSAA\n\n$1,698$141\n\nMercury\n\n$1,726$144\n\nCSAA\n\n$1,728$144\n\nState Farm\n\n$1,756$146\n\nMercury\n\n$1,769$147\n\nNational General\n\n$1,788$149\n\nCapital Insurance Group\n\n$1,823$152\n\nNational General\n\n$1,830$153\n\nNational General\n\n$1,830$153\n\nState Farm\n\n$1,838$153\n\nNational General\n\n$1,855$155\n\nNational General\n\n$1,855$155\n\nCapital Insurance Group\n\n$1,871$156\n\nCSAA\n\n$1,872$156\n\nGEICO\n\n$1,877$156\n\nChubb\n\n$1,880$157\n\nCSAA\n\n$1,903$159\n\nCSAA\n\n$1,903$159\n\nCSAA\n\n$1,911$159\n\nCSAA\n\n$1,911$159\n\nAllstate\n\n$1,934$161\n\nState National\n\n$1,968$164\n\nHartford\n\n$2,001$167\n\nHartford\n\n$2,001$167\n\nAllstate\n\n$2,040$170\n\nProgressive\n\n$2,053$171\n\nState Farm\n\n$2,055$171\n\nKemper\n\n$2,062$172\n\nKemper\n\n$2,062$172\n\nProgressive\n\n$2,064$172\n\nKemper\n\n$2,092$174\n\nNationwide\n\n$2,105$175\n\nCSAA\n\n$2,115$176\n\nCSAA\n\n$2,115$176\n\nAllstate\n\n$2,126$177\n\nHartford\n\n$2,141$178\n\nAllstate\n\n$2,150$179\n\nState Farm\n\n$2,153$179\n\nMercury\n\n$2,159$180\n\nNationwide\n\n$2,164$180\n\nState National\n\n$2,168$181\n\nState National\n\n$2,168$181\n\nMercury\n\n$2,211$184\n\nNational General\n\n$2,223$185\n\nNational General\n\n$2,223$185\n\nGEICO\n\n$2,231$186\n\nProgressive\n\n$2,232$186\n\nProgressive\n\n$2,240$187\n\nAllstate\n\n$2,274$190\n\nMercury\n\n$2,300$192\n\nGEICO\n\n$2,307$192\n\nHartford\n\n$2,309$192\n\nHartford\n\n$2,309$192\n\nMercury\n\n$2,311$193\n\nFarmers\n\n$2,315$193\n\nFarmers\n\n$2,315$193\n\nAllstate\n\n$2,340$195\n\nState National\n\n$2,346$195\n\nState National\n\n$2,346$195\n\nMercury\n\n$2,359$197\n\nNationwide\n\n$2,365$197\n\nCapital Insurance Group\n\n$2,375$198\n\nFarmers\n\n$2,378$198\n\nFarmers\n\n$2,378$198\n\nAllstate\n\n$2,407$201\n\nNationwide\n\n$2,412$201\n\nProgressive\n\n$2,489$207\n\nChubb\n\n$2,505$209\n\nHartford\n\n$2,516$210\n\nTravelers\n\n$2,520$210\n\nCapital Insurance Group\n\n$2,525$210\n\nProgressive\n\n$2,540$212\n\nProgressive\n\n$2,547$212\n\nAllstate\n\n$2,549$212\n\nChubb\n\n$2,567$214\n\nKemper\n\n$2,581$215\n\nNational General\n\n$2,591$216\n\nNational General\n\n$2,626$219\n\nTravelers\n\n$2,633$219\n\nNationwide\n\n$2,638$220\n\nGEICO\n\n$2,664$222\n\nProgressive\n\n$2,706$226\n\nTravelers\n\n$2,806$234\n\nFarmers\n\n$2,823$235\n\nFarmers\n\n$2,823$235\n\nState Farm\n\n$2,840$237\n\nCapital Insurance Group\n\n$2,851$238\n\nAllstate\n\n$2,852$238\n\nMercury\n\n$2,876$240\n\nFarmers\n\n$2,893$241\n\nState National\n\n$2,911$243\n\nState National\n\n$2,911$243\n\nTravelers\n\n$2,924$244\n\nFarmers\n\n$2,973$248\n\nHartford\n\n$3,000$250\n\nHartford\n\n$3,000$250\n\nMercury\n\n$3,063$255\n\nProgressive\n\n$3,081$257\n\nChubb\n\n$3,083$257\n\nNationwide\n\n$3,115$260\n\nKemper\n\n$3,118$260\n\nChubb\n\n$3,144$262\n\nNational General\n\n$3,150$263\n\nTravelers\n\n$3,162$263\n\nHartford\n\n$3,389$282\n\nChubb\n\n$3,410$284\n\nKemper\n\n$3,470$289\n\nKemper\n\n$3,470$289\n\nNationwide\n\n$3,480$290\n\nTravelers\n\n$3,516$293\n\nTravelers\n\n$3,516$293\n\nFarmers\n\n$3,529$294\n\nState Farm\n\n$3,635$303\n\nTravelers\n\n$3,683$307\n\nNationwide\n\n$3,843$320\n\nChubb\n\n$3,986$332\n\nChubb\n\n$4,044$337\n\nState National\n\n$4,054$338\n\nChubb\n\n$4,104$342\n\nNationwide\n\n$4,209$351\n\nAllstate\n\n$4,240$353\n\nState Farm\n\n$4,253$354\n\nState National\n\n$4,395$366\n\nKemper\n\n$4,400$367\n\nKemper\n\n$4,400$367\n\nTravelers\n\n$4,454$371\n\nAllstate\n\n$4,488$374\n\nHartford\n\n$4,808$401\n\nChubb\n\n$4,948$412\n\nAllstate\n\n$5,020$418\n\nKemper\n\n$5,183$432\n\nKemper\n\n$5,183$432\n\nCSAA\n\n$5,378$448\n\nCSAA\n\n$5,403$450\n\nState National\n\n$5,472$456\n\nNationwide\n\n$5,567$464\n\nHartford\n\n$5,596$466\n\nCSAA\n\n$5,991$499\n\nHartford\n\n$7,387$616\n\nAny moving violation may make your car insurance go up, but a severe traffic violation may result in an outstanding increase. In San Francisco, the average yearly cost for full coverage with a clean record is $1,486. The policy cost goes up by around$603 a year after a speeding ticket and roughly $2,068 a year after a DUI. Even so, you can still get cheap auto insurance with a bad driving record.",
null,
"MONEYGEEK EXPERT TIP While there are many optional coverages available in California on an auto policy, accident forgiveness is not one of them. Accident forgiveness stops your premium from going up after your first at-fault accident if you are a driver with a clean motor vehicle record. Since California law does not allow auto insurers to offer it, if you are involved in an at-fault accident, you should expect a rate increase. Also, you can't try to lower your auto insurance in California, with or without a moving violation, by going with a telematics-based auto policy that tracks when you drive, how you drive (such as speed and hard braking) and other behavior-related factors because the state does not allow it. But, if you are an overall low-mileage driver, that can help you save money on your auto insurance in California. — Mark Friedlander, Director, Corporate Communications, Insurance Information Institute ## Cheapest Car Insurance in San Francisco for Teens and Their Families The average cheapest full coverage auto insurance in San Francisco for a teen driver on a family policy comes from: • GEICO:$2,734 per year for female teens; $2,857 per year for male teens • State Farm:$2,881 per year for male and female teens\n\nThe most expensive insurer if you have a teen driver on a full coverage family policy in San Francisco is Nationwide. The company charges approximately $5,989 per year, regardless of gender. In most states, insurers can consider gender when calculating auto insurance costs, and many think male teen drivers are riskier to insure, so they charge them more than females. In California, considering gender is not allowed, so male and female teens pay the same for car insurance when all other rating factors are similar. Company Family Plan Annual Premium Individual Plan Monthly Premium GEICO$1,222\n\n$2,021 GEICO$1,246\n\n$2,244 GEICO$1,250\n\n$2,068 GEICO$1,267\n\n$2,172 GEICO$1,280\n\n$2,305 GEICO$1,281\n\n$2,120 GEICO$1,298\n\n$2,431 GEICO$1,314\n\n$2,367 GEICO$1,447\n\n$2,228 GEICO$1,484\n\n$2,503 Capital Insurance Group$1,816\n\n$2,235 Capital Insurance Group$1,816\n\n$2,235 Progressive$1,825\n\n$2,621 Progressive$1,825\n\n$2,621 GEICO$1,856\n\n$3,220 GEICO$1,914\n\n$3,650 Progressive$1,976\n\n$2,838 Progressive$1,976\n\n$2,838 State Farm$1,991\n\n$2,691 State Farm$1,991\n\n$2,691 Allstate$2,010\n\n$3,327 Allstate$2,010\n\n$3,327 GEICO$2,030\n\n$3,126 Capital Insurance Group$2,067\n\n$2,646 Capital Insurance Group$2,067\n\n$2,646 State Farm$2,071\n\n$2,799 State Farm$2,071\n\n$2,799 Nationwide$2,082\n\n$3,954 Nationwide$2,082\n\n$3,954 GEICO$2,094\n\n$3,532 Allstate$2,102\n\n$3,479 Allstate$2,102\n\n$3,479 State Farm$2,106\n\n$3,035 State Farm$2,106\n\n$3,035 Progressive$2,115\n\n$3,509 Progressive$2,115\n\n$3,509 State Farm$2,118\n\n$2,863 State Farm$2,118\n\n$2,863 Capital Insurance Group$2,150\n\n$2,646 Capital Insurance Group$2,150\n\n$2,646 Capital Insurance Group$2,150\n\n$2,646 Capital Insurance Group$2,150\n\n$2,646 Allstate$2,155\n\n$3,863 Allstate$2,155\n\n$3,863 Allstate$2,174\n\n$3,598 Allstate$2,174\n\n$3,598 Progressive$2,216\n\n$3,182 Progressive$2,216\n\n$3,182 State Farm$2,257\n\n$3,097 State Farm$2,257\n\n$3,097 Nationwide$2,280\n\n$4,330 Nationwide$2,280\n\n$4,330 Allstate$2,320\n\n$3,894 Allstate$2,320\n\n$3,894 Nationwide$2,353\n\n$4,941 Nationwide$2,353\n\n$4,941 State Farm$2,393\n\n$3,443 State Farm$2,393\n\n$3,443 Capital Insurance Group$2,408\n\n$2,977 Capital Insurance Group$2,408\n\n$2,977 State Farm$2,413\n\n$3,311 State Farm$2,413\n\n$3,311 Nationwide$2,447\n\n$4,646 Nationwide$2,447\n\n$4,646 Farmers$2,448\n\n$3,081 Farmers$2,448\n\n$3,081 Capital Insurance Group$2,496\n\n$2,977 Capital Insurance Group$2,496\n\n$2,977 GEICO$2,557\n\n$4,652 Progressive$2,626\n\n$3,844 Progressive$2,626\n\n$3,844 GEICO$2,658\n\n$5,310 State Farm$2,661\n\n$4,365 State Farm$2,661\n\n$4,365 Allstate$2,667\n\n$4,986 Allstate$2,667\n\n$4,986 Capital Insurance Group$2,672\n\n$3,304 Capital Insurance Group$2,672\n\n$3,304 Travelers$2,689\n\n$3,411 Travelers$2,689\n\n$3,411 Allstate$2,710\n\n$4,547 Allstate$2,710\n\n$4,547 GEICO$2,734\n\n$5,171 Nationwide$2,740\n\n$5,395 Nationwide$2,740\n\n$5,395 Farmers$2,746\n\n$3,582 Farmers$2,746\n\n$3,582 Kemper$2,773\n\n$4,089 Kemper$2,773\n\n$4,089 Travelers$2,785\n\n$3,532 Travelers$2,785\n\n$3,532 GEICO$2,804\n\n$5,102 State Farm$2,820\n\n$4,625 State Farm$2,820\n\n$4,625 Travelers$2,845\n\n$3,804 Travelers$2,845\n\n$3,804 Farmers$2,846\n\n$3,582 Farmers$2,846\n\n$3,582 Farmers$2,846\n\n$3,582 Farmers$2,846\n\n$3,582 GEICO$2,857\n\n$5,968 State Farm$2,881\n\n$5,186 State Farm$2,881\n\n$5,186 Travelers$2,908\n\n$3,688 Travelers$2,908\n\n$3,688 GEICO$2,926\n\n$5,846 Kemper$2,929\n\n$4,318 Kemper$2,929\n\n$4,318 Farmers$3,077\n\n$4,114 Farmers$3,077\n\n$4,114 Kemper$3,080\n\n$5,067 Kemper$3,080\n\n$5,067 Travelers$3,181\n\n$3,996 Travelers$3,181\n\n$3,996 Kemper$3,181\n\n$4,691 Kemper$3,181\n\n$4,691 Progressive$3,284\n\n$5,470 Progressive$3,284\n\n$5,470 Farmers$3,327\n\n$4,739 Farmers$3,327\n\n$4,739 Farmers$3,358\n\n$4,490 Farmers$3,358\n\n$4,490 Nationwide$3,379\n\n$7,466 Nationwide$3,379\n\n$7,466 Capital Insurance Group$3,427\n\n$4,086 Capital Insurance Group$3,427\n\n$4,086 Capital Insurance Group$3,427\n\n$4,086 Capital Insurance Group$3,427\n\n$4,086 Capital Insurance Group$3,427\n\n$4,086 Capital Insurance Group$3,427\n\n$4,086 Travelers$3,459\n\n$4,625 Travelers$3,459\n\n$4,625 Progressive$3,492\n\n$5,113 Progressive$3,492\n\n$5,113 Farmers$3,501\n\n$6,107 Farmers$3,501\n\n$6,107 Travelers$3,514\n\n$4,415 Travelers$3,514\n\n$4,415 Kemper$3,559\n\n$5,365 Kemper$3,559\n\n$5,365 Allstate$3,569\n\n$7,191 Allstate$3,569\n\n$7,191 Farmers$3,587\n\n$6,387 Farmers$3,587\n\n$6,387 Farmers$3,598\n\n$6,276 Farmers$3,598\n\n$6,276 Nationwide$3,603\n\n$7,095 Nationwide$3,603\n\n$7,095 Allstate$3,698\n\n$7,581 Allstate$3,698\n\n$7,581 Allstate$3,762\n\n$7,581 Allstate$3,762\n\n$7,581 Progressive$3,788\n\n$6,948 Progressive$3,788\n\n$6,948 Progressive$3,810\n\n$7,004 Progressive$3,810\n\n$7,004 Progressive$3,824\n\n$7,014 Progressive$3,824\n\n$7,014 Kemper$3,845\n\n$5,795 Kemper$3,845\n\n$5,795 Kemper$3,879\n\n$6,380 Kemper$3,879\n\n$6,380 Travelers$4,633\n\n$5,844 Travelers$4,633\n\n$5,844 Travelers$4,844\n\n$6,109 Travelers$4,844\n\n$6,109 Travelers$4,855\n\n$6,417 Travelers$4,855\n\n$6,417 Nationwide$5,160\n\n$10,133 Nationwide$5,160\n\n$10,133 Nationwide$5,804\n\n$11,397 Nationwide$5,804\n\n$11,397 Kemper$5,865\n\n$8,715 Kemper$5,865\n\n$8,715 Nationwide$5,989\n\n$13,752 Nationwide$5,989\n\n$13,752 Kemper$6,441\n\n$9,571 Kemper$6,441\n\n$9,571 Kemper$6,501\n\n$10,690 Kemper$6,501\n\n$10,690 Car insurance for teen drivers is expensive, and to pay the least, parents should add their teen to the family policy over them buying a policy. Plus, it is usually illegal for a teen under 18 to get auto insurance without a parent or guardian as a co-signer. For example, a full coverage family policy in San Francisco costs around$3,825 yearly with a teen on it, whereas an individual policy would cost roughly $4,498. That’s a difference of around$673 a year.\n\nThere are discounts available to teen drivers that can further lower the policy costs. Additionally, get cheap car insurance for teens by shopping around.",
null,
"MONEYGEEK EXPERT TIP\n\nCalifornia is a unique state for auto insurance because it prohibits some very common rating factors. If you live in San Francisco or any other area of the state, your gender, credit history, employment status and occupation, education and more are off limits when auto insurance companies calculate the premium for your policy. In California, you should expect your motor vehicle record, years of driving experience, type of vehicle and miles driven per year to greatly influence how much you pay. — Mark Friedlander, Director, Corporate Communications, Insurance Information Institute\n\n## Cheapest Car Insurance in San Francisco for Military Drivers\n\nThe average cheapest full coverage car insurance, if you have a military background in San Francisco, comes from the following companies:\n\n• USAA: $886 per year • GEICO:$973 per year\n\nSan Francisco active-duty military members or veterans interested in a minimum coverage policy may find GEICO as the cheapest option, costing approximately $364 per year. USAA is the second cheapest, with an average yearly cost of$376.\n\nThe difference in the cheapest insurer for minimum and full coverage indicates how factors like the coverage amount can influence rates.\n\nCompany\n\nGEICO\n\n$359$30\n\nState Farm\n\n$359$30\n\nGEICO\n\n$364$30\n\nUSAA\n\n$375$31\n\nUSAA\n\n$376$31\n\nGEICO\n\n$427$36\n\nUSAA\n\n$454$38\n\nProgressive\n\n$470$39\n\nState Farm\n\n$474$39\n\nProgressive\n\n$474$40\n\nCSAA\n\n$491$41\n\nCSAA\n\n$500$42\n\nMercury\n\n$527$44\n\nCSAA\n\n$539$45\n\nMercury\n\n$539$45\n\nState Farm\n\n$541$45\n\nProgressive\n\n$557$46\n\nNationwide\n\n$578$48\n\nKemper\n\n$591$49\n\nNationwide\n\n$601$50\n\nFarmers\n\n$608$51\n\nHartford\n\n$621$52\n\nState National\n\n$622$52\n\nTravelers\n\n$623$52\n\nTravelers\n\n$625$52\n\nCapital Insurance Group\n\n$626$52\n\nFarmers\n\n$626$52\n\nState National\n\n$641$53\n\nCapital Insurance Group\n\n$648$54\n\nMercury\n\n$661$55\n\nKemper\n\n$682$57\n\nHartford\n\n$710$59\n\nFarmers\n\n$711$59\n\nTravelers\n\n$714$60\n\nAllstate\n\n$725$60\n\nCapital Insurance Group\n\n$726$61\n\nAllstate\n\n$732$61\n\nState National\n\n$759$63\n\nNationwide\n\n$761$63\n\nChubb\n\n$766$64\n\nAllstate\n\n$789$66\n\nChubb\n\n$790$66\n\nUSAA\n\n$864$72\n\nUSAA\n\n$886$74\n\nKemper\n\n$899$75\n\nHartford\n\n$913$76\n\nState Farm\n\n$927$77\n\nGEICO\n\n$940$78\n\nNational General\n\n$951$79\n\nGEICO\n\n$973$81\n\nNational General\n\n$980$82\n\nUSAA\n\n$1,060$88\n\nProgressive\n\n$1,089$91\n\nProgressive\n\n$1,096$91\n\nChubb\n\n$1,097$91\n\nGEICO\n\n$1,121$93\n\nNational General\n\n$1,180$98\n\nState Farm\n\n$1,185$99\n\nMercury\n\n$1,309$109\n\nProgressive\n\n$1,316$110\n\nAllstate\n\n$1,324$110\n\nCapital Insurance Group\n\n$1,326$111\n\nMercury\n\n$1,340$112\n\nHartford\n\n$1,345$112\n\nState Farm\n\n$1,388$116\n\nAllstate\n\n$1,394$116\n\nNationwide\n\n$1,395$116\n\nCapital Insurance Group\n\n$1,408$117\n\nCSAA\n\n$1,502$125\n\nCSAA\n\n$1,509$126\n\nNationwide\n\n$1,523$127\n\nState National\n\n$1,525$127\n\nHartford\n\n$1,544$129\n\nAllstate\n\n$1,548$129\n\nKemper\n\n$1,574$131\n\nCapital Insurance Group\n\n$1,585$132\n\nTravelers\n\n$1,647$137\n\nState National\n\n$1,650$138\n\nCSAA\n\n$1,671$139\n\nFarmers\n\n$1,676$140\n\nFarmers\n\n$1,720$143\n\nTravelers\n\n$1,725$144\n\nMercury\n\n$1,737$145\n\nNational General\n\n$1,744$145\n\nNational General\n\n$1,766$147\n\nChubb\n\n$1,910$159\n\nKemper\n\n$1,941$162\n\nChubb\n\n$1,970$164\n\nHartford\n\n$1,990$166\n\nNationwide\n\n$1,999$167\n\nFarmers\n\n$2,041$170\n\nState National\n\n$2,045$170\n\nTravelers\n\n$2,057$171\n\nNational General\n\n$2,117$176\n\nKemper\n\n$2,342$195\n\nChubb\n\n$2,814$234\n\n## How to Compare Cheap San Francisco Car Insurance Quotes Online\n\nYou can get auto insurance quotes in San Francisco online through most providers' websites, but the process may take as much as 15 minutes or more for each insurer. That means you could spend hours providing the necessary personal information on each website as you try to compare rates from multiple companies.\n\nFortunately, MoneyGeek offers an easy-to-use and quick auto insurance calculator that allows you to estimate your premium from multiple companies based on essential factors, such as your age, car model, driving record and coverage amount.",
null,
"Auto Insurance Calculator\n\nSee how the Average Annual Auto Insurance Rates vary with the options chosen.\n\nCalifornia\nFemale\n30-65\nClean\n\nAverage Annual Auto Insurance Rates\n\nClick the section of the wheel in your price range to see options.\n\nlow end\n\non average\n\nhigh end\n\nClick an insurer below to visit their review page or continue to your personalized quote.\n\nProgressive\n\n## Cheapest Car Insurance in San Francisco for Young Adults\n\nOn average, the following companies offer the cheapest auto insurance in San Francisco for full coverage if you are in your 20s:\n\n• GEICO: $1,121 per year • Progressive:$1,316 per year\n\nThe insurance provider with the most pricey full coverage for young adults in San Francisco is Chubb, costing about $2,814 per year. Car insurance for young adults often costs more than for middle-aged drivers, as insurers may see young drivers as higher risk. For example, in San Francisco, it costs approximately$365 more a year for full coverage if you are 25 years old than if you are 40.\n\nCompany\n\nGEICO\n\n$359$30\n\nState Farm\n\n$359$30\n\nGEICO\n\n$364$30\n\nGEICO\n\n$427$36\n\nProgressive\n\n$470$39\n\nState Farm\n\n$474$39\n\nProgressive\n\n$474$40\n\nCSAA\n\n$491$41\n\nCSAA\n\n$500$42\n\nMercury\n\n$527$44\n\nCSAA\n\n$539$45\n\nMercury\n\n$539$45\n\nState Farm\n\n$541$45\n\nProgressive\n\n$557$46\n\nNationwide\n\n$578$48\n\nKemper\n\n$591$49\n\nNationwide\n\n$601$50\n\nFarmers\n\n$608$51\n\nHartford\n\n$621$52\n\nState National\n\n$622$52\n\nTravelers\n\n$623$52\n\nTravelers\n\n$625$52\n\nCapital Insurance Group\n\n$626$52\n\nFarmers\n\n$626$52\n\nState National\n\n$641$53\n\nCapital Insurance Group\n\n$648$54\n\nMercury\n\n$661$55\n\nKemper\n\n$682$57\n\nHartford\n\n$710$59\n\nFarmers\n\n$711$59\n\nTravelers\n\n$714$60\n\nAllstate\n\n$725$60\n\nCapital Insurance Group\n\n$726$61\n\nAllstate\n\n$732$61\n\nState National\n\n$759$63\n\nNationwide\n\n$761$63\n\nChubb\n\n$766$64\n\nAllstate\n\n$789$66\n\nChubb\n\n$790$66\n\nKemper\n\n$899$75\n\nHartford\n\n$913$76\n\nState Farm\n\n$927$77\n\nGEICO\n\n$940$78\n\nNational General\n\n$951$79\n\nGEICO\n\n$973$81\n\nNational General\n\n$980$82\n\nProgressive\n\n$1,089$91\n\nProgressive\n\n$1,096$91\n\nChubb\n\n$1,097$91\n\nGEICO\n\n$1,121$93\n\nNational General\n\n$1,180$98\n\nState Farm\n\n$1,185$99\n\nMercury\n\n$1,309$109\n\nProgressive\n\n$1,316$110\n\nAllstate\n\n$1,324$110\n\nCapital Insurance Group\n\n$1,326$111\n\nMercury\n\n$1,340$112\n\nHartford\n\n$1,345$112\n\nState Farm\n\n$1,388$116\n\nAllstate\n\n$1,394$116\n\nNationwide\n\n$1,395$116\n\nCapital Insurance Group\n\n$1,408$117\n\nCSAA\n\n$1,502$125\n\nCSAA\n\n$1,509$126\n\nNationwide\n\n$1,523$127\n\nState National\n\n$1,525$127\n\nHartford\n\n$1,544$129\n\nAllstate\n\n$1,548$129\n\nKemper\n\n$1,574$131\n\nCapital Insurance Group\n\n$1,585$132\n\nTravelers\n\n$1,647$137\n\nState National\n\n$1,650$138\n\nCSAA\n\n$1,671$139\n\nFarmers\n\n$1,676$140\n\nFarmers\n\n$1,720$143\n\nTravelers\n\n$1,725$144\n\nMercury\n\n$1,737$145\n\nNational General\n\n$1,744$145\n\nNational General\n\n$1,766$147\n\nChubb\n\n$1,910$159\n\nKemper\n\n$1,941$162\n\nChubb\n\n$1,970$164\n\nHartford\n\n$1,990$166\n\nNationwide\n\n$1,999$167\n\nFarmers\n\n$2,041$170\n\nState National\n\n$2,045$170\n\nTravelers\n\n$2,057$171\n\nNational General\n\n$2,117$176\n\nKemper\n\n$2,342$195\n\nChubb\n\n$2,814$234\n\n## Cheapest Car Insurance in San Francisco for Seniors\n\nThe following companies offer the average cheapest full coverage car insurance for seniors in San Francisco:\n\n• State Farm: $927 per year • GEICO:$940 per year\n\nAt an average yearly rate of $1,970, Chubb is the most expensive for full coverage auto insurance in San Francisco for seniors around 60 years of age. Auto insurance rates differ for senior drivers from middle-aged drivers. In San Francisco, the rate is lower at 65 by about$40 than the average yearly cost of $1,486 for a 40-year-old driver, which isn't common. However, rates tend to increase after 70 and 80. Company Annual Premium Monthly Premium GEICO$359\n\n$30 State Farm$359\n\n$30 GEICO$364\n\n$30 GEICO$427\n\n$36 Progressive$470\n\n$39 State Farm$474\n\n$39 Progressive$474\n\n$40 CSAA$491\n\n$41 CSAA$500\n\n$42 Mercury$527\n\n$44 CSAA$539\n\n$45 Mercury$539\n\n$45 State Farm$541\n\n$45 Progressive$557\n\n$46 Nationwide$578\n\n$48 Kemper$591\n\n$49 Nationwide$601\n\n$50 Farmers$608\n\n$51 Hartford$621\n\n$52 State National$622\n\n$52 Travelers$623\n\n$52 Travelers$625\n\n$52 Capital Insurance Group$626\n\n$52 Farmers$626\n\n$52 State National$641\n\n$53 Capital Insurance Group$648\n\n$54 Mercury$661\n\n$55 Kemper$682\n\n$57 Hartford$710\n\n$59 Farmers$711\n\n$59 Travelers$714\n\n$60 Allstate$725\n\n$60 Capital Insurance Group$726\n\n$61 Allstate$732\n\n$61 State National$759\n\n$63 Nationwide$761\n\n$63 Chubb$766\n\n$64 Allstate$789\n\n$66 Chubb$790\n\n$66 Kemper$899\n\n$75 Hartford$913\n\n$76 State Farm$927\n\n$77 GEICO$940\n\n$78 National General$951\n\n$79 GEICO$973\n\n$81 National General$980\n\n$82 Progressive$1,089\n\n$91 Progressive$1,096\n\n$91 Chubb$1,097\n\n$91 GEICO$1,121\n\n$93 National General$1,180\n\n$98 State Farm$1,185\n\n$99 Mercury$1,309\n\n$109 Progressive$1,316\n\n$110 Allstate$1,324\n\n$110 Capital Insurance Group$1,326\n\n$111 Mercury$1,340\n\n$112 Hartford$1,345\n\n$112 State Farm$1,388\n\n$116 Allstate$1,394\n\n$116 Nationwide$1,395\n\n$116 Capital Insurance Group$1,408\n\n$117 CSAA$1,502\n\n$125 CSAA$1,509\n\n$126 Nationwide$1,523\n\n$127 State National$1,525\n\n$127 Hartford$1,544\n\n$129 Allstate$1,548\n\n$129 Kemper$1,574\n\n$131 Capital Insurance Group$1,585\n\n$132 Travelers$1,647\n\n$137 State National$1,650\n\n$138 CSAA$1,671\n\n$139 Farmers$1,676\n\n$140 Farmers$1,720\n\n$143 Travelers$1,725\n\n$144 Mercury$1,737\n\n$145 National General$1,744\n\n$145 National General$1,766\n\n$147 Chubb$1,910\n\n$159 Kemper$1,941\n\n$162 Chubb$1,970\n\n$164 Hartford$1,990\n\n$166 Nationwide$1,999\n\n$167 Farmers$2,041\n\n$170 State National$2,045\n\n$170 Travelers$2,057\n\n$171 National General$2,117\n\n$176 Kemper$2,342\n\n$195 Chubb$2,814\n\n$234 ## Cheapest Car Insurance in San Francisco for Poor Credit California is one of the few states that does not allow insurers to consider credit history when calculating auto insurance rates. If you have poor credit, good credit or excellent credit, it will not affect how much you pay for car insurance in San Francisco. In states where insurers are allowed to consider credit score or a credit-based insurance score when calculating a car insurance premium, they may consider those with poor credit as more likely to file claims and charge them more. However, since California does not allow credit history as a factor when determining auto insurance costs, San Franciscans don’t have to worry about paying more because of their credit standing. You may still lower your auto insurance premium by comparing quotes from different insurers and taking advantage of discounts like bundling, airbags and paying in full. ## Cheapest Car Insurance in San Francisco for Sports Cars On average, the most affordable insurers in San Francisco, if you have a sports car and need full coverage, are: • GEICO:$1,226 per year\n• Progressive: $1,320 per year From the insurers we profiled for sports car drivers, Chubb had the highest average yearly cost of$2,300 for full coverage in San Francisco.\n\nGenerally, insurers charge you a higher rate if you drive a sports car than those with a standard car because sports cars feature higher replacement and repair costs. In San Francisco, a Ford Mustang sports car is around $180 more a year than a standard Toyota Camry — a roughly 12% increase. Company Annual Premium Monthly Premium GEICO$332\n\n$28 GEICO$335\n\n$28 Progressive$356\n\n$30 Progressive$362\n\n$30 State Farm$378\n\n$32 GEICO$395\n\n$33 Progressive$419\n\n$35 State Farm$499\n\n$42 Travelers$529\n\n$44 Travelers$531\n\n$44 Farmers$531\n\n$44 Farmers$545\n\n$45 Kemper$565\n\n$47 State Farm$571\n\n$48 Nationwide$589\n\n$49 Travelers$604\n\n$50 Nationwide$612\n\n$51 Capital Insurance Group$618\n\n$52 Farmers$619\n\n$52 Capital Insurance Group$641\n\n$53 Kemper$653\n\n$54 Capital Insurance Group$717\n\n$60 Chubb$766\n\n$64 Nationwide$776\n\n$65 Chubb$790\n\n$66 Allstate$813\n\n$68 Allstate$821\n\n$68 Kemper$861\n\n$72 Allstate$886\n\n$74 Chubb$1,097\n\n$91 State Farm$1,119\n\n$93 GEICO$1,175\n\n$98 GEICO$1,226\n\n$102 Progressive$1,299\n\n$108 Progressive$1,320\n\n$110 GEICO$1,405\n\n$117 State Farm$1,423\n\n$119 Capital Insurance Group$1,473\n\n$123 Travelers$1,537\n\n$128 Capital Insurance Group$1,571\n\n$131 Progressive$1,581\n\n$132 Travelers$1,621\n\n$135 Nationwide$1,641\n\n$137 State Farm$1,674\n\n$140 Allstate$1,677\n\n$140 Farmers$1,715\n\n$143 Capital Insurance Group$1,771\n\n$148 Farmers$1,776\n\n$148 Allstate$1,777\n\n$148 Nationwide$1,808\n\n$151 Kemper$1,840\n\n$153 Travelers$1,940\n\n$162 Allstate$1,987\n\n$166 Farmers$2,110\n\n$176 Chubb$2,300\n\n$192 Kemper$2,302\n\n$192 Chubb$2,374\n\n$198 Nationwide$2,397\n\n$200 Kemper$2,714\n\n$226 Chubb$3,405\n\n$284 To establish the most affordable companies for drivers with a sports car in San Francisco, we compared quotes using a Ford Mustang, although costs vary depending on a car’s make and model. Generally, you will pay a higher premium with a luxury, electric or sports car than with other car types. However, insurance companies consider several other factors when determining your actual rate. ## Average Cost of Car Insurance in San Francisco The average yearly cost of car insurance in San Francisco is$1,486. That price is for a full coverage policy with bodily injury liability limits of $100,000 per person,$300,000 per accident, and $100,000 as the property damage liability limit per accident. How much an insurer charges for auto insurance varies based on several factors, including the driver's record and location. San Francisco costs about$99 more than California's average yearly cost of auto insurance. The city's average also exceeds the national average of $1,265 by around$221.\n\nCompared to other major California cities, San Francisco is cheaper or more expensive for full coverage. Los Angeles drivers pay around $1,877 yearly, but in San Diego, the annual cost is roughly$1,202.\n\nCity\n\nSanta Maria\n\n$1,106$462\n\nVisalia\n\n$1,179$500\n\nChula Vista\n\n$1,189$501\n\nVentura\n\n$1,189$539\n\nSalinas\n\n$1,199$496\n\nSan Diego\n\n$1,202$521\n\nVista\n\n$1,205$533\n\n$1,225$542\n\nEscondido\n\n$1,232$533\n\nBakersfield\n\n$1,234$527\n\nClovis\n\n$1,239$549\n\nFairfield\n\n$1,241$520\n\nOceanside\n\n$1,246$550\n\nSan Mateo\n\n$1,249$537\n\nMurrieta\n\n$1,250$554\n\nOxnard\n\n$1,260$549\n\nTemecula\n\n$1,270$552\n\nSunnyvale\n\n$1,279$551\n\nFresno\n\n$1,281$549\n\nEl Cajon\n\n$1,285$560\n\nSanta Clara\n\n$1,289$559\n\nSan Jose\n\n$1,290$551\n\nFremont\n\n$1,296$540\n\nRoseville\n\n$1,309$587\n\nOrange\n\n$1,310$587\n\nConcord\n\n$1,315$558\n\nJurupa Valley\n\n$1,324$573\n\nSanta Rosa\n\n$1,326$543\n\nRancho Cucamonga\n\n$1,330$568\n\nSacramento\n\n$1,331$576\n\nBerkeley\n\n$1,342$551\n\nCosta Mesa\n\n$1,342$621\n\nHuntington Beach\n\n$1,349$617\n\nTorrance\n\n$1,351$608\n\nRiverside\n\n$1,351$583\n\nCorona\n\n$1,351$589\n\nSimi Valley\n\n$1,352$621\n\nMoreno Valley\n\n$1,357$578\n\nSan Bernardino\n\n$1,369$583\n\nElk Grove\n\n$1,370$598\n\nOntario\n\n$1,374$594\n\nHayward\n\n$1,374$559\n\nVallejo\n\n$1,378$556\n\nFontana\n\n$1,384$589\n\nRialto\n\n$1,391$585\n\nSanta Ana\n\n$1,399$622\n\nRichmond\n\n$1,400$583\n\nModesto\n\n$1,401$608\n\nVictorville\n\n$1,402$594\n\nAntioch\n\n$1,403$570\n\nStockton\n\n$1,407$590\n\nThousand Oaks\n\n$1,408$638\n\nIrvine\n\n$1,411$633\n\n$1,430$647\n\nPomona\n\n$1,432$618\n\nSanta Clarita\n\n$1,444$654\n\nAnaheim\n\n$1,448$648\n\nNorwalk\n\n$1,460$633\n\nLancaster\n\n$1,462$659\n\nOakland\n\n$1,469$598\n\nWest Covina\n\n$1,478$656\n\nFullerton\n\n$1,480$657\n\nSan Francisco\n\n$1,486$613\n\nGarden Grove\n\n$1,488$682\n\nLong Beach\n\n$1,497$649\n\nPalmdale\n\n$1,507$667\n\nDaly City\n\n$1,528$621\n\nDowney\n\n$1,531$657\n\nEast Los Angeles\n\n$1,562$677\n\nEl Monte\n\n$1,586$693\n\n$1,598$704\n\nBurbank\n\n$1,733$782\n\nInglewood\n\n$1,740$774\n\nLos Angeles\n\n$1,877$820\n\nGlendale\n\n$2,115$913",
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https://www.smartconversion.com/surface-area-calculator-ellipsoid | [
"Report a Problem\nSuggestions\n\n#",
null,
"Calculate The Surface Area of An Ellipsoid\n\nLast updated: Saturday, June 24, 2023\nSelect a type of ellipsoid below\nSymmetrical Formula\nOblate Spheroid\nProlate Spheroid\n\nThe surface area of an ellipsoid is the total area of the curved surface of an ellipsoid shape. It has many real-life applications, including in physics, engineering, and geodesy. In physics, it is used to calculate the surface area of planets and other celestial bodies. In engineering, it is used in designing structures and machinery with ellipsoidal shapes.\n\nGeodesy uses the surface area of an ellipsoid to model the shape of the Earth, taking into account its equatorial bulge and flattening at the poles. Understanding the surface area of an ellipsoid is essential for accurately measuring and modeling various objects and phenomena in the world around us.\n\nUsing the symmetrical formula developed by Knud Thomsen, it is possible to keep the relative error under 1.061%.\n\nThe formula for determining the surface area of an ellipsoid is defined as:\n$$where$$\n$$p$$ $$=$$ $$1.6$$\n$$SA$$: the surface area of the ellipsoid\n$$a$$: the length of the first axis\n$$b$$: the length of the second axis\n$$c$$: the length of the third axis\n$$\\pi$$: A mathematical constant with an infinite decimal tail\n$$p$$: A mathematical constant with value 1.6\nThe SI unit of surface area is: $$square \\text{ } meter\\text{ }(m^2)$$\n\n## Find $$SA$$\n\nUse this calculator to determine the surface area of an ellipsoid when the lengths of its axes are given.\nHold & Drag\nCLOSE\nthe length of the first axis\n$$a$$\n$$meter$$\nthe length of the second axis\n$$b$$\n$$meter$$\nthe length of the third axis\n$$c$$\n$$meter$$\n$$\\pi$$ : A mathematical constant with an infinite decimal tail\n$$p$$ : A mathematical constant with value 1.6\nBookmark this page or risk going on a digital treasure hunt again"
] | [
null,
"https://www.smartconversion.com/images/icons/ellipsoid.png",
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https://programmingpraxis.com/2020/02/14/square-triple/2/ | [
"## Square Triple\n\n### February 14, 2020\n\nOur algorithm finds all pairs, then checks if the square root of the product is in the list:\n\n```(define (f xs)\n(list-of (list x y z)\n(x in xs)\n(y in xs)\n(< x y)\n(z2 is (* x y))\n(z is (isqrt z2))\n(= (* z z) z2)\n(member z xs)))```\n\nWe eliminate duplicates by insisting that x < y. Here’s an example:\n\n```> (f (range 1 20))\n((1 4 2) (1 9 3) (1 16 4) (2 8 4) (2 18 6) (3 12 6) (4 9 6)\n(4 16 8) (8 18 12) (9 16 12))```\n\nThat takes time O(n³); n * n to form the pairs and another factor of n to search the list. We could reduce that to O(n²) by first inserting the list items in a hash table and checking membership there. Another improvement is a quick exit if the product exceeds the maximum list item. We’ll leave those improvements as an exercise to the reader.\n\nYou can run the program at https://ideone.com/L7itmh.\n\nPages: 1 2\n\n### 12 Responses to “Square Triple”\n\n1. James Smith said\n\nEasiest way to make this o(n2) is to first compute an array of squares and use this to filter results… as usual in Perl…\n\n```use strict;\nuse feature qw(say);\n\nmy @ints = 1..50; ## Test data...\n\nmy %squares = map { \\$_*\\$_ => \\$_ } @ints; ## Use hash to store possible squares (value is sqrt)\n\nsay map {\nmy \\$a = \\$_; ## Nested map store copy of outer key...\nmap { join ',', \\$a, \\$_, \\$squares{\\$a*\\$_}.' ' } ## Display three values - sqrt is value of hash keyed by square\ngrep { exists \\$squares{\\$a*\\$_} } ## Now check to see if the product is a square..\ngrep { \\$a < \\$_ } ## Quick filter only want 1st no < 2nd no\n@ints\n} @ints;\n```\n2. matthew said\n\nHere’s an O(n²) solution that doesn’t require any auxiliary storage. Assumes input is sorted and all positive:\n\n```# If ab = z^2 with a != b, can assume a < z < b, so\n# keep 2 pointers into the list and keep track of the\n# product of values pointed to - if too high, the lower\n# pointer needs to be increased, if too low, the higher:\n\ndef triples(a):\nn = len(a)\nfor i in range(1,n-1):\ntarget = a[i]*a[i]\nj,k = i-1,i+1\nwhile j >= 0 and k < n:\nproduct = a[j]*a[k]\nif product > target: j = j-1\nelif product < target: k = k+1\nelse:\nyield(a[j],a[k],a[i])\nj,k = j-1,k+1\n\n# [(1, 4, 2), (1, 9, 3), (1, 16, 4), (2, 8, 4), (2, 18, 6),\n# (3, 12, 6), (4, 9, 6), (4, 16, 8), (8, 18, 12), (9, 16, 12)]\nprint(sorted(list(triples(range(1,20)))))\n```\n3. Zack said\n\nHere is my approach to the problem using Julia, as usual: https://pastebin.com/FjqkHhQ7\n\nAlthough I iterate through the data points using a triple loop, not all combinations are examined since I take advantage of the fact that the integers are distinct (so there is one zero at most), simplifying the whole process. Cheers!\n\n4. Daniel said\n\nHere’s a O(n^2) solution in Python.\n\nI’ve assumed x, y, z are distinct (e.g., no (x=1,y=1,z=1) nor (x=0, y=2, z=0)), which I think is suggested by the problem specifying a “list of distinct integers” for the input.\n\n```from collections import defaultdict\n\ndef square_triple(input):\noutput = []\nroots = defaultdict(list)\nfor x in input:\nroots[x * x].append(x)\nfor x in input:\nfor y in input:\nif x == y: continue\nfor z in roots[x * y]:\nif x != z and y != z:\noutput.append((x, y, z))\nreturn output\n\nfor triple in square_triple(range(-8, 9)):\nprint(triple)\n```\n\nOutput:\n\n```(-8, -2, -4)\n(-8, -2, 4)\n(-4, -1, -2)\n(-4, -1, 2)\n(-2, -8, -4)\n(-2, -8, 4)\n(-1, -4, -2)\n(-1, -4, 2)\n(1, 4, -2)\n(1, 4, 2)\n(2, 8, -4)\n(2, 8, 4)\n(4, 1, -2)\n(4, 1, 2)\n(8, 2, -4)\n(8, 2, 4)\n```\n5. Daniel said\n\n@programmingpraxis, I think that using square roots for this problem can be problematic, since as-is it won’t handle negative values for z.\n\n6. Daniel said\n\n(where “square roots” in my comment refers to the conventional function that returns a single non-negative value)\n\n7. Steve said\n\nKlong version\n\n``` :\" Number from 1 to 20\"\nlist::1+!20\n[1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20]\n\n:\" Get all pairs of number in list\"\npairs::{x,:\\list}'list\n[[[1 1] [1 2] [1 3] [1 4] [1 5] [1 6] [1 7] [1 8] [1 9] [1 10]...]]\n\n:\" Flatten the list\"\nl::,/pairs::{x,:\\list}'list\n[[1 1] [1 2] [1 3] [1 4] [1 5] [1 6] [1 7] [1 8] [1 9] [1 10] [1 11] [1 12] [1 13] [1 14] [1 15] ...]\n\n:\" Only those pairs where the first # < second #\nl::l@&{~(x@0)=x@1}'l::,/pairs::{x,:\\list}'list\n[[1 2] [1 3] [1 4] [1 5] [1 6] [1 7] [1 8] [1 9] [1 10] [1 11] [1 12] [1 13] [1 14] [1 15]...]\n\n:\" Only those pairs where product is square of one of the number in the list\"\nl::l@&{#sqrs?*/x}'l::l@&{~(x@0)=x@1}'l::,/pairs::{x,:\\list}'list\n[[1 4] [1 9] [1 16] [2 8] [2 18] [3 12] [4 1] [4 9] [4 16] [5 20] [8 2] [8 18] [9 1] [9 4] [9 16] [12 3] [16 1] [16 4] [16 9] [18 2] [18 8] [20 5]]\n\n:\" Get rid of duplicates\"\nl@&{(x@0)<x@1}'l::l@&{#sqrs?*/x}'l::l@&{~(x@0)=x@1}'l::,/pairs::{x,:\\list}'list\n[[1 4] [1 9] [1 16] [2 8] [2 18] [3 12] [4 9] [4 16] [5 20] [8 18] [9 16]]\n\n:\" Add square root of product\"\n{x,_(*/x)^0.5}'l@&{(x@0)<x@1}'l::l@&{#sqrs?*/x}'l::l@&{~(x@0)=x@1}'l::,/pairs::{x,:\\list}'list\n[[1 4 2] [1 9 3] [1 16 4] [2 8 4] [2 18 6] [3 12 6] [4 9 6] [4 16 8] [5 20 10] [8 18 12] [9 16 12]]\n\n```\n8. Richard A. O'Keefe said\n\nThe problem says to find ALL triples, but the model answer finds only half of them.\nNothing whatsoever in the problem says that the numbers are all positive or all non-negative.\nIn fact by using “integer” instead of “natural number” it suggests that negative numbers are allowed.\nGiven [-2,1,2,4], (1,4,2) and (1,4,-2) are both legal triples, but the model answer will not find the second.\n\n9. Steve said\n\nI thought it might be interesting to explain some of the facilities of the Klong programming language. I like to learn this about other languages – they all seem to have their strengths – and Klong is one of the more unusual ones I’ve come upon.\n\n`````` !20 yields the list of numbers from 0 to 19\n``````\n\n[0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19]\n\n`````` Add 1 to each member of that list\nlist::1+!20\n``````\n\n[1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20]\n\n`````` Append 1 to the left of each member of the list\n1,:\\list\n``````\n\n[[1 1] [1 2] [1 3] [1 4] [1 5] [1 6] [1 7] [1 8] [1 9] [1 10] [1 11] [1 12] [1 13] [1 14] [1 15] [1 16] [1 17] [1 18] [1 19] [1 20]]\n\n`````` Tick (\"'\") applies a function to each member of a list\n{x+3}'[1 2 3]\n``````\n\n[4 5 6]\n\n`````` \"@&\" acts as a filter on a list\nFilter the list where the element mod 2 = 1\nlist@&{1=x!2}'list\n``````\n\n[1 3 5 7 9 11 13 15 17 19]\n\n`````` \"?\" tests for membership in a list and \"/\" applies an operator over a list\n[1 2 3]?1\n``````\n\n\n[1 2 3]?4\n[]\n*/[2 3]\n6\n\n10. Steve said\n\nRichard noticed that the example solution did not find all solutions. Mine didn’t either. Here’s version 2 for Klong.\n\n``` list::[-2 1 2 4]\n[-2 1 2 4]\n\npairs::pairs@&{~(x@0)=x@1}'pairs::,/{x,:\\list}'list\n[[-2 1] [-2 2] [-2 4] [1 -2] [1 2] [1 4] [2 -2] [2 1] [2 4] [4 -2] [4 1] [4 2]]\n\n{(x@1),(x@2),x@0}'triples@&{(x@1)<x@2}'triples::triples@&{((x@0)^2)=((x@1)*x@2)}'triples::,/{x,:\\pairs}'list\n[[1 4 -2] [1 4 2]]\n\n```\n11. Anon said\n\nThis is a common lisp solution. It assumes the input list is sorted in accending order.\n\n```(defun sol (ns)\n(flet ((fn (xs)\n(let ((x (car xs))\n(ys (cdr xs))\n(res nil))\n(dolist (y ys res)\n(let ((z (sqrt (* x y))))\n(when (find z ys :test #'=)\n(push (list x y (ceiling z)) res)))))))\n(mapcon #'fn ns)))\n```\n\nSample run:\n\n```CL-USER> (sol (loop for i from 1 to 20 collect i))\n((1 16 4) (1 9 3) (1 4 2) (2 18 6) (2 8 4) (3 12 6) (4 16 8) (4 9 6) (5 20 10)\n(8 18 12) (9 16 12))\n```\n12. Sam Claflin said\n\nPython Solution:\n\ndef find_triples(int_list):\ntriples = []\n\n# Loop through all integers in list\nfor k in range(len(int_list)):\nfor i in range(len(int_list)):\nfor j in range(len(int_list)):\n\n# Detect valid pair\nif int_list[k] * int_list[i] == int_list[j] ** 2:\nsame = False\n\n# Check each tuple to see if different permutation of pair is already present\nfor tup in triples:\nif int_list[k] in tup and int_list[i] in tup and int_list[j] in tup:\nsame = True\nbreak\nif not same:\ntriples.append((int_list[k], int_list[i], int_list[j]))\nelse:\ncontinue\nreturn triples\n\nprint(find_triples(range(1, 15)))\n\nOUTPUT:\n\n[(1, 1, 1), (1, 4, 2), (1, 9, 3), (2, 8, 4), (3, 12, 6), (4, 9, 6), (5, 5, 5), (7, 7, 7), (10, 10, 10), (11, 11, 11), (13, 13, 13), (14, 14, 14)]"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8116704,"math_prob":0.99244475,"size":3771,"snap":"2021-43-2021-49","text_gpt3_token_len":1228,"char_repetition_ratio":0.103265196,"word_repetition_ratio":0.039106146,"special_character_ratio":0.36621585,"punctuation_ratio":0.12785389,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98049456,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-21T04:26:17Z\",\"WARC-Record-ID\":\"<urn:uuid:9a267ef9-f9d7-462e-a952-39350dc89041>\",\"Content-Length\":\"123481\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:042dd551-2ddb-432e-9e20-ca094b180624>\",\"WARC-Concurrent-To\":\"<urn:uuid:744cb051-00d1-4bfb-a9f2-1439db6b391c>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://programmingpraxis.com/2020/02/14/square-triple/2/\",\"WARC-Payload-Digest\":\"sha1:TUCVA52QBOY4II3FEOJJHADF6SD3ZZGE\",\"WARC-Block-Digest\":\"sha1:DM7QENB2K2HFWA4DVAHLSDF43XISBLDK\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585381.88_warc_CC-MAIN-20211021040342-20211021070342-00400.warc.gz\"}"} |
http://themindlessfreaks.blogspot.com/2018/07/tripled-numbers-and-369.html | [
"Thursday, July 5, 2018\n\nTripled numbers and 369\n\nI was looking through my old notes earlier and just wanted to document some numerology type things I forgot about.\n\nA lot of people talk about Tesla and the 3 6 9 stuff...\nWell...\n1+1+1=3\n2+2+2=6\n3+3+3=9\n4+4+4=12=1+2=3\n5+5+5=15=1+5=6\n6+6+6=18=1+8=9\n7+7+7=21=2+1=3\n8+8+8=24=2+4=6\n9+9+9=27=2+7=9\n\nAlso 37 is an important number in regards to triple numbers...\n1+1+1=3.....3X37=111\n2+2+2=6.....6X37=222\n3+3+3=9.....9X37=333\n4+4+4=12..12x37=444\nAnd so on..."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.74762225,"math_prob":0.9974741,"size":479,"snap":"2019-43-2019-47","text_gpt3_token_len":228,"char_repetition_ratio":0.09052631,"word_repetition_ratio":0.0,"special_character_ratio":0.55114824,"punctuation_ratio":0.19135803,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999975,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-20T04:24:55Z\",\"WARC-Record-ID\":\"<urn:uuid:6c5cb55d-371f-4442-9dd1-cb2b66950459>\",\"Content-Length\":\"69550\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:92d6a685-06ab-4869-a049-b57ce2cee066>\",\"WARC-Concurrent-To\":\"<urn:uuid:f76ed205-c6d5-4087-9e8c-f5491c3f0b0a>\",\"WARC-IP-Address\":\"172.217.164.161\",\"WARC-Target-URI\":\"http://themindlessfreaks.blogspot.com/2018/07/tripled-numbers-and-369.html\",\"WARC-Payload-Digest\":\"sha1:2GXJVQB6BDF6WVXO5SRGGU3Z3BDVFT5J\",\"WARC-Block-Digest\":\"sha1:KY3GW3WUHSZ2GGKYXEJIJQF5AOOBYGEH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986702077.71_warc_CC-MAIN-20191020024805-20191020052305-00006.warc.gz\"}"} |
https://mathoverflow.net/questions/227900/can-you-define-a-probability-measure-on-the-set-of-countable-transitive-models-o | [
"Can you define a probability measure on the set of countable transitive models of ZFC?\n\nIt is well known that the set of hereditarily countable sets $H(\\omega_1)$ —or, if you prefer, $H_{\\omega_1}$— has cardinality $2^{\\aleph_0}$, and I understand that every countable transitive model (ctm) of ZFC lives in this set. So, if there were a way to give $H(\\omega_1)$ a sensible structure (separable metric, for instance) one might try to calculate the probability that a ctm satisfy CH (to name just one example).\n\nIs there a way of defining a sensible probability measure on the set of ctms, provided that this set is nonempty?\n\nI imagine that this idea has been explored and rejected quickly, but I couldn't find anything related to it.\n\nA prior requirement, as I mentioned, would be to have some “definable” topology in the descriptive-set-theoretic sense (Polish, analytic, coanalytic, or the like).\n\nIs there a sensible definable topology on $H(\\omega_1)$?\n\nAbout this, I found in the book Classification and Orbit Equivalence Relations by Hjorth that one can map $H(\\omega_1)$ into the set of isomorphism classes of countable binary structures. Now, the countable binary structures form indeed a Polish space, so the odds are that this identification would give as something at least as complicated as $\\boldsymbol{\\Pi}_1^1(2^{\\mathbb{N}\\times\\mathbb{N}})$ (since one has to say that the binary relation is well founded).\n\nI asked this on Math.SE, and there I just obtained a pair of comments. It was argued that one can indeed put a probability measure on the set of all countable structures in a natural way, but the models of any sufficiently interesting first-order theory will always be null (and meager). In any case, I do not see why this precludes the possibility of having a probability measure on the set of all ctms. (As a silly response, consider the Cantor ternary set in $\\mathbb{R}$: It's null and meager; but nevertheless it admits a natural probability measure, when seen as $2^\\mathbb{N}$.)"
] | [
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https://www.statsmodels.org/stable/generated/statsmodels.stats.descriptivestats.describe.html | [
"# statsmodels.stats.descriptivestats.describe¶\n\nstatsmodels.stats.descriptivestats.describe(data: Union[numpy.ndarray, pandas.core.series.Series, pandas.core.frame.DataFrame], stats: Sequence[str] = None, *, numeric: bool = True, categorical: bool = True, alpha: float = 0.05, use_t: bool = False, percentiles: Sequence[Union[int, float]] = 1, 5, 10, 25, 50, 75, 90, 95, 99, ntop: bool = 5) → pandas.core.frame.DataFrame[source]\n\nExtended descriptive statistics for data\n\nParameters\ndataarray_like\n\nData to describe. Must be convertible to a pandas DataFrame.\n\nstatsSequence[str], optional\n\nStatistics to include. If not provided the full set of statistics is computed. This list may evolve across versions to reflect best practices. Supported options are: “nobs”, “missing”, “mean”, “std_err”, “ci”, “ci”, “std”, “iqr”, “iqr_normal”, “mad”, “mad_normal”, “coef_var”, “range”, “max”, “min”, “skew”, “kurtosis”, “jarque_bera”, “mode”, “freq”, “median”, “percentiles”, “distinct”, “top”, and “freq”. See Notes for details.\n\nnumericbool, default True\n\nWhether to include numeric columns in the descriptive statistics.\n\ncategoricalbool, default True\n\nWhether to include categorical columns in the descriptive statistics.\n\nalphafloat, default 0.05\n\nA number between 0 and 1 representing the size used to compute the confidence interval, which has coverage 1 - alpha.\n\nuse_tbool, default False\n\nUse the Student’s t distribution to construct confidence intervals.\n\npercentiles\n\nA distinct sequence of floating point values all between 0 and 100. The default percentiles are 1, 5, 10, 25, 50, 75, 90, 95, 99.\n\nntopint, default 5\n\nThe number of top categorical labels to report. Default is\n\nReturns\nDataFrame\n\nDescriptive statistics\n\npandas.DataFrame.describe\n\nBasic descriptive statistics\n\nDescription\n\nDescriptive statistics class with additional output options\n\nNotes\n\nThe selectable statistics include:\n\n• “nobs” - Number of observations\n\n• “missing” - Number of missing observations\n\n• “mean” - Mean\n\n• “std_err” - Standard Error of the mean assuming no correlation\n\n• “ci” - Confidence interval with coverage (1 - alpha) using the normal or t. This option creates two entries in any tables: lower_ci and upper_ci.\n\n• “std” - Standard Deviation\n\n• “iqr” - Interquartile range\n\n• “iqr_normal” - Interquartile range relative to a Normal\n\n• “mad” - Mean absolute deviation\n\n• “mad_normal” - Mean absolute deviation relative to a Normal\n\n• “coef_var” - Coefficient of variation\n\n• “range” - Range between the maximum and the minimum\n\n• “max” - The maximum\n\n• “min” - The minimum\n\n• “skew” - The skewness defined as the standardized 3rd central moment\n\n• “kurtosis” - The kurtosis defined as the standardized 4th central moment\n\n• “jarque_bera” - The Jarque-Bera test statistic for normality based on the skewness and kurtosis. This option creates two entries, jarque_bera and jarque_beta_pval.\n\n• “mode” - The mode of the data. This option creates two entries in all tables, mode and mode_freq which is the empirical frequency of the modal value.\n\n• “median” - The median of the data.\n\n• “percentiles” - The percentiles. Values included depend on the input value of percentiles.\n\n• “distinct” - The number of distinct categories in a categorical.\n\n• “top” - The mode common categories. Labeled top_n for n in 1, 2, …, ntop.\n\n• “freq” - The frequency of the common categories. Labeled freq_n for n in 1, 2, …, ntop."
] | [
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https://kr.mathworks.com/matlabcentral/profile/authors/12195895 | [
"Community Profile",
null,
"# Riccardo Consolo\n\nLast seen: 3달 전 2018 이후 활성\n\n#### Statistics\n\nAll\n•",
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"배지보기\n\n#### Content Feed\n\n보기 기준\n\n해결됨\n\nDimensions of a rectangle\nThe longer side of a rectangle is three times the length of the shorter side. If the length of the diagonal is x, find the width...\n\n6달 전\n\n해결됨\n\nArea of an Isoceles Triangle\nAn isosceles triangle has equal sides of length x and a base of length y. Find the area, A, of the triangle. <<https://imgur...\n\n6달 전\n\n해결됨\n\nIs this triangle right-angled?\nGiven three positive numbers a, b, c, where c is the largest number, return *true* if the triangle with sides a, b and c is righ...\n\n6달 전\n\n해결됨\n\nFind a Pythagorean triple\nGiven four different positive numbers, a, b, c and d, provided in increasing order: a < b < c < d, find if any three of them com...\n\n6달 전\n\n해결됨\n\nIs this triangle right-angled?\nGiven any three positive numbers a, b, c, return true if the triangle with sides a, b and c is right-angled. Otherwise, return f...\n\n6달 전\n\n해결됨\n\nTriangle sequence\nA sequence of triangles is constructed in the following way: 1) the first triangle is Pythagoras' 3-4-5 triangle 2) the s...\n\n6달 전\n\n해결됨\n\nLength of the hypotenuse\nGiven short sides of lengths a and b, calculate the length c of the hypotenuse of the right-angled triangle. <<https://i.imgu...\n\n6달 전\n\n해결됨\n\nArea of an equilateral triangle\nCalculate the area of an equilateral triangle of side x. <<https://i.imgur.com/jlZDHhq.png>> Image courtesy of <http://up...\n\n6달 전\n\n해결됨\n\nSide of an equilateral triangle\nIf an equilateral triangle has area A, then what is the length of each of its sides, x? <<https://i.imgur.com/jlZDHhq.png>> ...\n\n6달 전\n\n해결됨\n\nSide of a rhombus\nIf a rhombus has diagonals of length x and x+1, then what is the length of its side, y? <<https://imgur.com/x6hT6mm.png>> ...\n\n6달 전\n\n해결됨\n\nLength of a short side\nCalculate the length of the short side, a, of a right-angled triangle with hypotenuse of length c, and other short side of lengt...\n\n6달 전\n\n문제\n\nApply mirrored edges to a matrix\n\n6달 전 | 0 | 솔버 수: 3\n\n문제\n\napply zero padding to a matrix\n\n6달 전 | 1 | 솔버 수: 13\n\n해결됨\n\nVector creation\nCreate a vector using square brackets going from 1 to the given value x in steps on 1. Hint: use increment.\n\n6달 전\n\n해결됨\n\nDoubling elements in a vector\nGiven the vector A, return B in which all numbers in A are doubling. So for: A = [ 1 5 8 ] then B = [ 1 1 5 ...\n\n6달 전\n\n해결됨\n\nCreate a vector\nCreate a vector from 0 to n by intervals of 2.\n\n6달 전\n\n해결됨\n\nFlip the vector from right to left\nFlip the vector from right to left. Examples x=[1:5], then y=[5 4 3 2 1] x=[1 4 6], then y=[6 4 1]; Request not ...\n\n6달 전\n\n해결됨\n\nWhether the input is vector?\nGiven the input x, return 1 if x is vector or else 0.\n\n6달 전\n\n해결됨\n\nFind max\nFind the maximum value of a given vector or matrix.\n\n6달 전\n\n해결됨\n\nGet the length of a given vector\nGiven a vector x, the output y should equal the length of x.\n\n6달 전\n\n해결됨\n\nInner product of two vectors\nFind the inner product of two vectors.\n\n6달 전\n\n해결됨\n\nArrange Vector in descending order\nIf x=[0,3,4,2,1] then y=[4,3,2,1,0]\n\n6달 전\n\n해결됨\n\nSelect every other element of a vector\nWrite a function which returns every other element of the vector passed in. That is, it returns the all odd-numbered elements, s...\n\n6달 전\n\n해결됨\n\nFind the sum of all the numbers of the input vector\nFind the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ...\n\n6달 전\n\n해결됨\n\nLongest run of consecutive numbers\nGiven a vector a, find the number(s) that is/are repeated consecutively most often. For example, if you have a = [1 2 2 2 1 ...\n\n6달 전\n\n해결됨\n\nRescale Scores\nEach column (except last) of matrix |X| contains students' scores in a course assignment or a test. The last column has a weight...\n\n6달 전\n\n해결됨\n\nCalculate Inner Product\nGiven two input matrices, |x| and |y|, check if their inner dimensions match. * If they match, create an output variable |z|...\n\n6달 전\n\n해결됨\n\nFind MPG of Lightest Cars\nThe file |cars.mat| contains a table named |cars| with variables |Model|, |MPG|, |Horsepower|, |Weight|, and |Acceleration| for ...\n\n6달 전"
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https://dev.to/alexdevero/primitive-values-objects-and-references-in-javascript-made-simple-ckp | [
"## DEV Community is a community of 862,249 amazing developers\n\nWe're a place where coders share, stay up-to-date and grow their careers.",
null,
"# Primitive Values, Objects and References in JavaScript Made Simple\n\nPrimitive data types, values, objects and references are among the most misunderstood topics in JavaScript. They can cause a lot of headaches. In this tutorial, you will learn about primitive data types, values, objects, references, the differences between them and how they work.\n\n## A short introduction\n\nIn JavaScript, there are the two categories of data types you can work with. The first category are primitive data types. At this moment, there exist [seven primitive types]. These primitive data types are number, string, boolean, `null`, `undefined`, BigInt and Symbol. The BigInt and Symbol are newer data types.\n\nSymbol was introduced in ES6 specification. The BigInt was introduced later, in ES2020 specification. When something is not one of these primitive data types it is technically an object. This applies to actual objects as well as arrays and even functions. From the view of JavaScript, these are all objects.\n\nThis distinction between primitive data types and objects is important because JavaScript handles each differently.\n\n``````// Primitive data types:\nconst numberVal = 5\nconst strVal = 'Hello!'\nconst boolVal = true\nconst nullVal = null\nconst undefinedVal = undefined\nconst bigIntVal = 9007123254550972n\nconst symbolVal = Symbol('label')\n\n// Objects:\nconst myObjLiteral = {\nname: 'Toby'\n}\n\nconst myArray = [9, 'book', true, null]\n\nfunction myFunction(num1, num2) {\nreturn num1 / num2\n}\n``````\n\n## Primitive data types and primitive values\n\nLet's start with the first category, primitive data types. Values that contain these primitive data types are called static data. As static data, JavaScript stores them on the stack. One important thing about these primitive values is that their size is fixed. JavaScript knows how much memory these data types need.\n\nLet's say you assign some variable a primitive data type as a value. From now on, this variable will contain that value. If you manipulate with that variable, you manipulate directly with the value you assigned to it. A simple way to test this, and the consequences, is assigning the variable to another variable.\n\nWhen you assign one variable to another, and the value of first is primitive data type, JavaScript will copy that value. When you do this you are copying the values \"by value\". So, now, if you change the value of the first variable, the second will remain the same. This is because even though you created one variable from another they both have their own, separate, values.\n\nSo, if you change the value of one variable it will not change the second. The second variable is a separate entity. Let's go back to the stack. When you assign the first variable, JavaScript will store its value on the stack. When you assign the variable to another variable its value will be also added on the stack.\n\nAt this moment, the stack will now contain two values, one for each variable. It doesn't matter that both values are the same. Nor does it matter that you created the second variable from the first. For JavaScript, these are two separate entities. These two variables have no relationship with each other.\n\nThis is also why you can safely work with each of these variables as you want. It is why you can change one variable without changing the other.\n\n``````// Create a variable and assign it\n// a primitive value:\nlet x = 'Hello'\n\n// Assign \"x\" to another variable:\nlet y = x\n\n// Change the value of \"x\":\n// NOTE: this will not change \"y\".\nx = 'Bye'\n\n// Log the value of \"x\":\nconsole.log(x)\n// Output:\n// 'Bye'\n\n// Log the value of \"x\":\nconsole.log(y)\n// Output:\n// 'Hello'\n\n// Assign \"y\" to another variable:\nlet z = y\n\n// Assign \"z\" to another variable:\nlet w = z\n\n// Change the value of \"y\":\n// NOTE: this will not change \"z\" and \"w\".\ny = 'Eloquent'\n\n// Log the value of \"x\":\nconsole.log(z)\n// Output:\n// 'Hello'\n\n// Log the value of \"x\":\nconsole.log(w)\n// Output:\n// 'Hello'\n``````\n\n## Objects and references\n\nObjects are a different story. When you assign variable an object JavaScript will handle it differently. The value, the object, will not be added to the stack. Instead, it will be added to memory heap. There is another very important difference. That variable will not contain the value, the object, but a reference to that object.\n\nThink about this reference as a link, or chain. It is a link that connects specific variable with specific object. This has one major consequence. If you manipulate with that variable, you work with the reference and, through this reference, with the object itself. What if you copy that object by assigning that variable to another variable?\n\nYou may think that this will create another object, copy of the first, just like in case of primitive value. This is not what will happen. What will actually happen is that JavaScript will create new reference. JavaScript will only create new reference, or link, to the original object.\n\nThere will be still only one object in the memory heap, the original. This is called copying \"by reference\" and it happens every time you copy an object. Copying this way, by reference, results in creating shallow copies of objects. This type of copying also has one serious consequence.\n\nIf you manipulate with any of these variables you also manipulate with the same object. So, if you change the object by changing one variable you change the other variable as well. They are both different variables, but they both reference, or link to, the same object.\n\n``````// Create a variable and assign it\n// a simple object:\nlet a = { name: 'Stan' }\n\n// Assign \"a\" to another variable:\nlet b = a\n\n// Change the value of \"a\"\n// by adding new property \"age\" to the object:\na.age = 44\n\n// Log the value of \"a\":\n// console.log(a)\n// Output:\n// { name: 'Stan', age: 44 }\n\n// Log the value of \"b\":\n// console.log(b)\n// Output:\n// { name: 'Stan', age: 44 }\n\n// Assign \"b\" to another variable:\nlet c = b\n\n// Assign \"c\" to another variable:\nlet d = c\n\n// Change the value of \"d\"\n// \"favoriteAnimal\" to the object:\nd.favoriteAnimal = 'elephant'\n\n// Log the value of \"a\":\nconsole.log(a)\n// Output:\n// {\n// name: 'Stan',\n// age: 44,\n// favoriteAnimal: 'elephant'\n// }\n\n// Log the value of \"b\":\nconsole.log(b)\n// Output:\n// {\n// name: 'Stan',\n// age: 44,\n// favoriteAnimal: 'elephant'\n// }\n\n// Log the value of \"c\":\nconsole.log(c)\n// Output:\n// {\n// name: 'Stan',\n// age: 44,\n// favoriteAnimal: 'elephant'\n// }\n\n// Log the value of \"d\":\nconsole.log(c)\n// Output:\n// {\n// name: 'Stan',\n// age: 44,\n// favoriteAnimal: 'elephant'\n// }\n``````\n\nNote: One way to understand how copying by reference works is by thinking about keys and houses. When you copy you key, you are not also creating a new house. There is still only one house, but there are now two keys that can unlock that house. Variables are those keys, object is that house.\n\n## Quick summary of primitive data types, objects, values and references\n\nNow you know the difference between primitive values and references. When you assign primitive data types and then copy them you are copying by values. Each of these copies (variables) is a separate entity that has no relationship to another. You can change one without changing any other.\n\nWhen you assign and then copy an object, you are copying by reference. You are creating new references for each copy. As a result, there are multiple references (variables). However, there is still only one object. If you change one of these variables, you change the original object. This will affect all references (variables).\n\n## Primitive values, references and comparison\n\nKnowing the difference between value and reference is important when you want to compare things. Let's take a look at how comparison works with both, primitive values and objects.\n\n### Comparing primitive values\n\nComparing two primitive values is usually simple. The only catch is knowing the difference between equal and strict equal and which one to use (it will usually be strict equal). Comparing primitive values with strict equal will check for value and type. If both are the same, you will get `true`. If not, you will get `false`.\n\n``````// One primitive value:\n// Create one variable and assign it primitive value:\nconst str1 = 'JavaScript'\n\n// Create another variable and assign it \"str1\":\nconst str2 = str1\n\n// Compare \"str1\" and \"str2\":\nconsole.log(str1 === str2)\n// Output:\n// true\n\n// Two identical primitive values:\n// Create two variables and assign them\n// the same primitive values:\nconst num1 = 15\nconst num2 = 15\n\n// Compare \"num1\" and \"num2\":\nconsole.log(num1 === num2)\n// Output:\n// true\n``````\n\n### Comparing objects and references\n\nReferences work differently. If you compare two different objects, and the content is the same, the comparison will still result in `false`. Comparison will result in `true` only if you compare references to the same object.\n\n``````// One object:\n// Create a variable and assign it an object:\nconst a = { name: 'Jack' }\n\n// Assign \"a\" to another variable:\nconst b = a\n\n// Compare \"a\" and \"b\":\nconsole.log(a === b)\n// Output:\n// true\n\n// Two identical objects:\n// Create a variable and assign it an object:\nconst a = { name: 'George' }\n\n// Create another variable and assign it the same object:\nconst b = { name: 'George' }\n\n// Compare \"a\" and \"b\":\nconsole.log(a === b)\n// Output:\n// false\n``````\n\nRemember that arrays, and functions as well, are technically objects. This means that if you compare variables with identical arrays the result will be always `false`. Those variables will be the same only if they both reference the same array.\n\n``````// One array:\n// Create a variable and assign it an array:\nconst x = [1, 2, 3, 4]\n\n// Create another variable and assign it \"x\":\nconst y = x\n\n// Compare \"x\" and \"y\":\nconsole.log(x === y)\n// Output:\n// true\n\n// Two identical arrays:\n// Create a variable and assign it an array:\nconst x = [1, 2, 3, 4]\n\n// Create another variable and assign it the same array:\nconst y = [1, 2, 3, 4]\n\n// Compare \"x\" and \"y\":\nconsole.log(x === y)\n// Output:\n// false\n``````\n\n## Functions and passing by value and by reference\n\nKnowing the difference between value and reference is also useful when you work with functions. When you pass some primitive value stored in a variable to a function as an argument, you are passing it \"by value\". You are basically copying that value itself to a function. The consequence of this is the same as when you copy \"by value\".\n\nIf you try to change the value passed into the function it will have no effect on the variable itself. The value of that variable will remain the same. Change you created by the function will have no effect on it. Well, unless you access the variable itself and change it directly, but that is a different scenario.\n\n``````// Create a variable and assign it a primitive value:\nlet personName = 'Andrei'\n\n// Create a function that will attempt to modify\n// the value it receives as an argument:\nfunction changeNameFunc(name) {\nname = 'Viktor'\n}\n\n// Call the \"changeNameFunc\" function:\nchangeNameFunc(personName)\n\n// Log the value of \"name\" variable:\nconsole.log(personName)\n// Output:\n// 'Andrei' // <= The name is the same.\n``````\n\nIf you try to do this with an object the result will be different. When you pass an object it is passed \"by reference\". In this case, JavaScript is not copying the object so the function can use it. It only gives you the reference to the original object. If you try to modify the object you will actually change the original object.\n\n``````// Create a variable and assign it an object:\nlet person = { name: 'Andrei' }\n\n// Create a function that will attempt to modify\n// the value it receives as an argument:\nfunction changeNameFunc(person) {\nperson.name = 'Viktor'\n}\n\n// Call the \"changeNameFunc\" function:\nchangeNameFunc(person)\n\n// Log the value of \"name\" variable:\nconsole.log(person)\n// Output:\n// { name: 'Viktor' } // <= The name is different.\n``````\n\n## Conclusion: Primitive values, objects and references in JavaScript made simple\n\nPrimitive data types, values, objects and references are topics that can be difficult to understand. This is especially true for beginners and junior JavaScript developers. I hope that this tutorial help you understand how they work, the differences between them and how they work."
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http://f1-demons.tripod.com/ | [
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null,
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null,
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null,
"We will be back.\n\nA big hello to all die-hard F1 fans and a big thank you to all who have appreciated this effort to bring the excitement of F1 right on your computer screen. For certain technical reasons (read : we are revamping) we have to shut up this site for a little while. We promise you that once we are up and running again we will be back with all your favourite features. A little bit of patience is all we ask for. Bookmark this page and visit us again after some days. We will be ready.\n\nFor any details please mail me at zkhyronov@yahoo.com",
null,
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https://iacr.org/cryptodb/data/author.php?authorkey=8380 | [
"## CryptoDB\n\n### Jeremiah Blocki\n\n#### Publications\n\nYear\nVenue\nTitle\n2019\nCRYPTO\nMemory-hard functions (MHFs) are a key cryptographic primitive underlying the design of moderately expensive password hashing algorithms and egalitarian proofs of work. Over the past few years several increasingly stringent goals for an MHF have been proposed including the requirement that the MHF have high sequential space-time (ST) complexity, parallel space-time complexity, amortized area-time (aAT) complexity and sustained space complexity. Data-Independent Memory Hard Functions (iMHFs) are of special interest in the context of password hashing as they naturally resist side-channel attacks. iMHFs can be specified using a directed acyclic graph (DAG) G with $N=2^n$ nodes and low indegree and the complexity of the iMHF can be analyzed using a pebbling game. Recently, Alwen et al. [ABH17] constructed a DAG called DRSample that has aAT complexity at least $\\varOmega \\!\\left( N^2/{\\text {log}} N\\right)$ . Asymptotically DRSample outperformed all prior iMHF constructions including Argon2i, winner of the password hashing competition (aAT cost ${\\mathcal {O}} \\!\\left( N^{1.767}\\right)$ ), though the constants in these bounds are poorly understood. We show that the greedy pebbling strategy of Boneh et al. [BCS16] is particularly effective against DRSample e.g., the aAT cost is ${\\mathcal {O}} (N^2/{\\text {log}} N)$ . In fact, our empirical analysis reverses the prior conclusion of Alwen et al. that DRSample provides stronger resistance to known pebbling attacks for practical values of $N \\le 2^{24}$ . We construct a new iMHF candidate (DRSample+BRG) by using the bit-reversal graph to extend DRSample. We then prove that the construction is asymptotically optimal under every MHF criteria, and we empirically demonstrate that our iMHF provides the best resistance to known pebbling attacks. For example, we show that any parallel pebbling attack either has aAT cost $\\omega (N^2)$ or requires at least $\\varOmega (N)$ steps with $\\varOmega (N/{\\text {log}} N)$ pebbles on the DAG. This makes our construction the first practical iMHF with a strong sustained space-complexity guarantee and immediately implies that any parallel pebbling has aAT complexity $\\varOmega (N^2/{\\text {log}} N)$ . We also prove that any sequential pebbling (including the greedy pebbling attack) has aAT cost $\\varOmega \\!\\left( N^2\\right)$ and, if a plausible conjecture holds, any parallel pebbling has aAT cost $\\varOmega (N^2 \\log \\log N/{\\text {log}} N)$ —the best possible bound for an iMHF. We implement our new iMHF and demonstrate that it is just as fast as Argon2. Along the way we propose a simple modification to the Argon2 round function that increases an attacker’s aAT cost by nearly an order of magnitude without increasing running time on a CPU. Finally, we give a pebbling reduction that proves that in the parallel random oracle model (PROM) the cost of evaluating an iMHF like Argon2i or DRSample+BRG is given by the pebbling cost of the underlying DAG. Prior pebbling reductions assumed that the iMHF round function concatenates input labels before hashing and did not apply to practical iMHFs such as Argon2i, DRSample or DRSample+BRG where input labels are instead XORed together.\n2018\nEUROCRYPT\n2017\nEUROCRYPT\n2017\nTCC\n2016\nCRYPTO\n2016\nTCC\n2015\nEPRINT\n2013\nASIACRYPT\n\nCrypto 2020\nTCC 2020\nCrypto 2018\n\n#### Coauthors\n\nJoël Alwen (3)\nManuel Blum (2)\nAnupam Datta (2)\nBen Harsha (1)\nSiteng Kang (1)\nSeunghoon Lee (1)\nKrzysztof Pietrzak (2)\nLu Xing (1)\nHong-Sheng Zhou (1)\nSamson Zhou (2)"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86796916,"math_prob":0.9469774,"size":3696,"snap":"2021-43-2021-49","text_gpt3_token_len":1004,"char_repetition_ratio":0.115655474,"word_repetition_ratio":0.003539823,"special_character_ratio":0.23755412,"punctuation_ratio":0.064412236,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.980962,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-16T05:46:36Z\",\"WARC-Record-ID\":\"<urn:uuid:622b170c-5e98-4a70-8e60-2cac5b6a3b98>\",\"Content-Length\":\"30184\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bed43a49-199a-47af-9bec-58ecc52f95de>\",\"WARC-Concurrent-To\":\"<urn:uuid:52710b0b-a065-4070-acef-acfc3f71ae52>\",\"WARC-IP-Address\":\"216.184.8.41\",\"WARC-Target-URI\":\"https://iacr.org/cryptodb/data/author.php?authorkey=8380\",\"WARC-Payload-Digest\":\"sha1:7JRHUZKPJ3FGNXMZM5W63TTGXPKF7QD5\",\"WARC-Block-Digest\":\"sha1:IHQ5ZB2X3MKFGBNHYMN4C4EPERRQCB7C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323583423.96_warc_CC-MAIN-20211016043926-20211016073926-00511.warc.gz\"}"} |
https://codeforces.com/problemset/problem/342/A | [
"A. Xenia and Divisors\ntime limit per test\n1 second\nmemory limit per test\n256 megabytes\ninput\nstandard input\noutput\nstandard output\n\nXenia the mathematician has a sequence consisting of n (n is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three a, b, c the following conditions held:\n\n• a < b < c;\n• a divides b, b divides c.\n\nNaturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has",
null,
"groups of three.\n\nHelp Xenia, find the required partition or else say that it doesn't exist.\n\nInput\n\nThe first line contains integer n (3 ≤ n ≤ 99999) — the number of elements in the sequence. The next line contains n positive integers, each of them is at most 7.\n\nIt is guaranteed that n is divisible by 3.\n\nOutput\n\nIf the required partition exists, print",
null,
"groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them.\n\nIf there is no solution, print -1.\n\nExamples\nInput\n61 1 1 2 2 2\nOutput\n-1\nInput\n62 2 1 1 4 6\nOutput\n1 2 41 2 6"
] | [
null,
"https://espresso.codeforces.com/4de0f387b35d0865b1f96d528b2f1b06fb6f1903.png",
null,
"https://espresso.codeforces.com/4de0f387b35d0865b1f96d528b2f1b06fb6f1903.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9320507,"math_prob":0.9560929,"size":1091,"snap":"2023-14-2023-23","text_gpt3_token_len":286,"char_repetition_ratio":0.13615455,"word_repetition_ratio":0.048309177,"special_character_ratio":0.2630614,"punctuation_ratio":0.11111111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98660815,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-04T16:39:55Z\",\"WARC-Record-ID\":\"<urn:uuid:ebca3630-7192-416d-931f-5c4d78579303>\",\"Content-Length\":\"58357\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:42c6b467-6a90-4570-afa6-c31e0aa1eb2e>\",\"WARC-Concurrent-To\":\"<urn:uuid:e35294d2-bf2e-4c1d-a84a-50d2f00b7b85>\",\"WARC-IP-Address\":\"172.67.68.254\",\"WARC-Target-URI\":\"https://codeforces.com/problemset/problem/342/A\",\"WARC-Payload-Digest\":\"sha1:JB4BXZO3SCZ7TRROPHAH6TSLBPSDMC6B\",\"WARC-Block-Digest\":\"sha1:OD5Y4Y55ZGYN464O7QRNDQ7LRBYQIMGM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224650201.19_warc_CC-MAIN-20230604161111-20230604191111-00222.warc.gz\"}"} |
https://community.wolfram.com/groups/-/m/t/515123 | [
"0\n|\n2872 Views\n|\n5 Replies\n|\n0 Total Likes\nView groups...\nShare\nGROUPS:\n\n# How to generate good looking output?\n\nPosted 8 years ago\n Hey, I've been trying to write my own program to calculate the Fourier-Coefficients and so far the program works but has an awful output. I would like to have ai and bi written next to each output and also have the results for a and b in the same line, except a0 has to standalone or b0 has to be 0. This is my code so far: Clear[n, t, f, i, ak, bk] f[t_] := \\[Pi]^2 - t^2; ak[f_, n_, t_] := (1/\\[Pi]) Integrate[f[t] Cos[n t], {t, -\\[Pi], \\[Pi]}]; bk[f_, n_, t_] := (1/\\[Pi]) Integrate[f[t] Sin[n t], {t, -\\[Pi], \\[Pi]}] n = 5; For[i = -1, i < n, Print[i.ak[f, i, t]], i++] For[i = 0, i < n, Print[i.bk[f, i, t]], i++] Can someome help me and make the ouput more plesseant to look at. I have not done something like this in MATHEMATICA before. Thanks for your help in advance and have a nice day.\n5 Replies\nSort By:\nPosted 8 years ago\n Well, you have to make up your mind as to which t to use for the accuracy. One way would be to maximize Abs[f[t]-fn[t]] over all t. You can use FindMaxValue.\nPosted 8 years ago\n That is almost what i mean, but i meant the Approximation for each point. So what i wanted to do was to use:el2distance[f_, fn_] := Abs[f[t] - fn[t]]]But MATHEMATICA now gives me an errormessage saying: Abs[0. +t$322115-2. Sin[t$322115]+1. Sin[2. t$322115]-0.666667 Sin[3. t$322115]+0.5 Sin[4. t$322115]-0.4 Sin[5. t$322115]] should be a real number or a list of non-negative numbers, which has the same length as {\\!\\(\\* I guess what i have to do is to numerically calculate the value of that expression? But how?\nPosted 8 years ago\n My guess is that you had given t a numeric value, and this interfered with Integrate. You can protect the integration variable by wrapping it into Module: Clear[ak, bk]; ak[f_, n_] := ak[f, n] = Module[{t}, (1/\\[Pi]) Integrate[f[t] Cos[n t], {t, -\\[Pi], \\[Pi]}]]; bk[f_, n_] := bk[f, n] = Module[{t}, (1/\\[Pi]) Integrate[f[t] Sin[n t], {t, -\\[Pi], \\[Pi]}]]; f[t_] := \\[Pi]^2 - t^2; ak[f, 2] Also, it may be prudent to use NIntegrate instead of Integrate, unless you know in advance that your integrals have a closed form. As for the FillingStyle, you may first define your measure of accuracy, for example the L2 distance el2distance[f_, fn_] := Module[{t}, Sqrt@NIntegrate[Abs[f[t] - fn[t]]^2, {t, -Pi, Pi}]] and then write something like FillingStyle -> GrayLevel[el2distance[f, fn]] This may not work straight away, because you probably have to normalize the distance first, as GrayLevel[x] wants x between 0 and 1. Instead of GrayLevel[x], you may prefer something like Blend[{Blue, Red}, x].\nPosted 8 years ago\n Yes exactly like this. Thank you.Well if i dont make ak and bk depend on t then MATHEMATICA doesn't evalute the Integrals properly for some reason. I think it's because f depends on t and therefore ak and bk do aswell.I have continued to work on this little Program and decided to plot the functions and the fourier-series.Is there a way to change the filling into more and more red (maybe starting at blue) the less accurate the approximation is?This is my code for ploting: (as u can see i dont know what to put next to FillingStyle) fourierplot[f_, fn_, n_] := Plot[{f[t], fn[t]}, {t, -\\[Pi], \\[Pi]} , Filling -> {1 -> {2}} , FillingStyle -> ColorFunction , ImageSize -> Medium , AxesLabel -> {t, y} , PlotLegends -> {f, \"Fourierreihe für n=\" Text[n]}] f[t_] = t; {f5[t_], f10[t_], f20[t_]} = fourierseries[f[t], {5, 10, 20}]; Row[fourierplot[f, f5, 5] , fourierplot[f, f10, 10] , fourierplot[f, f20, 20]] \nPosted 8 years ago\n Something like this? TableForm[Table[{i, ak[f, i, t], bk[f, i, t]}, {i, -1, 10}], TableHeadings -> {None, {k, ak, bk}}] or else Grid[Table[{i, ak[f, i, t], bk[f, i, t]}, {i, -1, 10}], Frame -> All] Why do you make ak and bk depend on t, when they don't?"
] | [
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https://ru-facts.com/what-is-the-gini-coefficient-of-malaysia/ | [
"# What is the Gini coefficient of Malaysia?\n\n## What is the Gini coefficient of Malaysia?\n\n42.8 %\nMalaysia GINI index was 42.8 % in 2018 – the single year for which the data is available at the moment.\n\n## Which country has the lowest Gini?\n\nSouth Africa\nSouth Africa ranks as the country with the lowest level of income equality in the world, thanks to a Gini coefficient of 63.0 when last measured in 2014.\n\nWhich country has highest Gini coefficient?\n\nGINI index (World Bank estimate) – Country Ranking\n\nRank Country Value\n1 South Africa 63.00\n2 Namibia 59.10\n3 Suriname 57.60\n4 Zambia 57.10\n\nWhat is the Gini coefficient of Singapore?\n\n0.375\nIn February, Deputy Prime Minister Heng Swee Keat highlighted that income inequality measures in terms of the Gini coefficient in Singapore fell from 0.398 in 2019 to a historic low of 0.375 in 2020 due to “massive transfers” and schemes aimed towards supporting lower-income groups.\n\n### How is Gini coefficient calculated?\n\nThe Gini coefficient can be calculated using the formula: Gini Coefficient = A / (A + B), where A is the area above the Lorenz Curve and B is the area below the Lorenz Curve.\n\n### Which country has the highest Gini coefficient 2020?\n\nSouth Africa had the highest inequality in income distribution with a Gini score of 63, according to the Gini Index 2020. The Gini coefficient measures the deviation of the distribution of income (or consumption) among individuals or households within a country from a perfectly equal distribution.\n\nHow do you explain Gini coefficient?\n\nThe Gini coefficient measures how far the actual Lorenz curve for a society’s income or wealth is from the line of equality. Both the Lorenz curve and the line of equality are plotted on a graph. 0 means that the country is perfectly equal, and 1 means that one person has all the wealth or income.\n\nWhat is the Gini coefficient in China?\n\n0.481\nThat means China’s Gini coefficient (0.481) belongs to a handful of countries with top inequality in the world, not only higher than that of western developed countries, but also above Asia’s average (0.3513).\n\n## Is Singapore’s Gini coefficient high?\n\nIn February, Deputy Prime Minister Heng Swee Keat highlighted that income inequality measures in terms of the Gini coefficient in Singapore fell from 0.398 in 2019 to a historic low of 0.375 in 2020 due to “massive transfers” and schemes aimed towards supporting lower-income groups.\n\nBegin typing your search term above and press enter to search. Press ESC to cancel."
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https://mp3songpreview.com/central-angles-and-inscribed-angles-worksheet-answer-key-lesson-15-1/ | [
"# Central Angles And Inscribed Angles Worksheet Answer Key Lesson 15-1\n\n•",
null,
"",
null,
"Central Angles And Inscribed Angles Worksheet Answer Key Lesson 15-1. Find the measure of the arc or central angle indicated. Each quadrilateral described is inscribed in a circle. Explore resource locker investigating central angles and inscribed angles a chord is a segment whose endpoints lie on a circle. Central and inscribed angles answer answers to central and inscribed angles dvs ltd co uk. Inscribed angles date period state if each angle is an inscribed angle. Lesson angles in inscribed quadrilaterals. Central angles & inscribed angles. Assume that lines which appear to be diameters are actual diameters. Central angles quizlet, central angle math is fun, central and inscribed angles khan academy, central angles and sectors, central angle statistics Angles in a circle worksheet worksheets for all from central angles and inscribed angles worksheet answer key source. Ab and ab b c b ∠acd inscribed angle c d ∠acd d © houghton mifflin harcourt publishing company ause a compass to draw a circle. Lesson central angles and inscribed angles. Book comes like the further recommendation and lesson every become old you retrieve it. Circles inscribed angles arcs and chords worksheets. Angles in a circle worksheet worksheets for all from central angles and inscribed angles worksheet answer key source.\n\nCentral Angles And Inscribed Angles Worksheet Answer Key Lesson 15-1 Indeed recently is being hunted by users around us, maybe one of you personally. Individuals now are accustomed to using the net in gadgets to view video and image information for inspiration, and according to the name of the post I will discuss about Central Angles And Inscribed Angles Worksheet Answer Key Lesson 15-1.\n\n• Parts Of Circles (Segments-Arcs-Angles) Guided Notes For … . To Download Free Central And Inscribed Angles With Algebra Worksheet You Need To Inscribed And Central Angles Math.\n• 32 Arcs Central Angles And Inscribed Angles Worksheet … : An Angle Whose Measure Is Less.\n• 32 Arcs Central Angles And Inscribed Angles Worksheet … , Lesson Angles In Inscribed Quadrilaterals.\n• 32 Arcs Central Angles And Inscribed Angles Worksheet … : An Inscribed Angle Is An Angle Whose Vertex Lies On A Circle And Whose Sides Contain Chords Of The Circle.\n• Circles – Central And Inscribed Angles Quiz By Secondary … , Circles Practice Problems And Proofs.\n• Inscribed Angles Worksheet With Answers – Nidecmege – Worksheet About Stoichiometry, Chemistry Worksheet On Stoichiometry Mixed Review Answers, Worksheet For Stoichiometry, Worksheets On Gas Stoichiometry Pdf, Worksheet On Stoichiometry Key, Worksheet #1 Stoichiometry Moles To Moles, Worksheet Of Stoichiometry.\n• Arcs Of A Circle Worksheet – Assume That Lines Which Appear To Be Diameters Are Actual Diameters.\n• Geometry Unit 10 – Circle Arcs Central Inscribed Angles … . Now You Are Ready To Create Your Angles Worksheet By Pressing The Create Button.\n• Inscribed Angles Worksheet Answer Key – Nidecmege . Explore Resource Locker Investigating Central Angles And Inscribed Angles A Chord Is A Segment Whose Endpoints Lie On A Circle.\n• Geometry Unit 10 – Circle Arcs Central Inscribed Angles … , Circles 10 2 Finding Arc Measures 10 3 Using Chords 10 4 Inscribed.\n\nFind, Read, And Discover Central Angles And Inscribed Angles Worksheet Answer Key Lesson 15-1, Such Us:",
null,
"• Geometry Unit 10 – Circle Arcs Central Inscribed Angles … – Answers To Central Angles And.",
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"• Arcs, Central Angles And Inscribed Angles For Hs Students … : Some Of The Worksheets For This Concept Are Inscribed Angles Date Period, Inscribed And Central Angles In A Circle, 11 Arcs And Central Angles, , Inscribed Angles, Nag10110 To, Chapter 10 Section 3 Inscribed Angles, 12 3 Inscribed Angles.",
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"• 9 Imposing Inscribed Angles Worksheet Coloring Pages … , Central And Inscribed Angles Answer Answers To Central And Inscribed Angles Dvs Ltd Co Uk.",
null,
"• Inscribed Angles Worksheet Answer Key – Nidecmege – The Diameter Is The Longest Chord Of A Circle And It Passes Through The Venter What Is The Relationship Between Inscribed Angles And Their Arcs?",
null,
"• Naming Arcs And Central Angles Worksheet – Awesome Worksheet . Worksheet About Stoichiometry, Chemistry Worksheet On Stoichiometry Mixed Review Answers, Worksheet For Stoichiometry, Worksheets On Gas Stoichiometry Pdf, Worksheet On Stoichiometry Key, Worksheet #1 Stoichiometry Moles To Moles, Worksheet Of Stoichiometry.",
null,
"• Inscribed Angles Worksheet Answer Key – Nidecmege : You May Select Which Figures To Name, As Well As The Types Of Include Angles Worksheet Answer Page.",
null,
"• 30 Inscribed Angles Worksheet Answer Key – Free Worksheet … : In This Central And Inscribed Angles Worksheet, 10Th Graders Identify And Solve 20 Different Problems That Include Determining Inscribed Angles And Intercepted Arcs.",
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"• Circles – Central And Inscribed Angles Quiz By Secondary … : Free Trial Available At Kutasoftware.com.",
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"• 30 Inscribed Angles Worksheet Answer Key – Free Worksheet … . Hmh Geometry California Edition Unit 6:",
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"• 30 Inscribed Angles Worksheet Answer Key – Free Worksheet … – To Download Free Central And Inscribed Angles With Algebra Worksheet You Need To Inscribed And Central Angles Math.\n\n29 Angles And Arcs Worksheet – Worksheet Project List. Central angles quizlet, central angle math is fun, central and inscribed angles khan academy, central angles and sectors, central angle statistics Central and inscribed angles answer answers to central and inscribed angles dvs ltd co uk. Central angles & inscribed angles. Explore resource locker investigating central angles and inscribed angles a chord is a segment whose endpoints lie on a circle. Lesson angles in inscribed quadrilaterals. Ab and ab b c b ∠acd inscribed angle c d ∠acd d © houghton mifflin harcourt publishing company ause a compass to draw a circle. Circles inscribed angles arcs and chords worksheets. Angles in a circle worksheet worksheets for all from central angles and inscribed angles worksheet answer key source. Inscribed angles date period state if each angle is an inscribed angle. Book comes like the further recommendation and lesson every become old you retrieve it. Lesson central angles and inscribed angles. Find the measure of the arc or central angle indicated. Angles in a circle worksheet worksheets for all from central angles and inscribed angles worksheet answer key source. Each quadrilateral described is inscribed in a circle. Assume that lines which appear to be diameters are actual diameters.\n\nEach quadrilateral described is inscribed in a circle. Central and inscribed angles answer answers to central and inscribed angles dvs ltd co uk. An inscribed angle is an angle whose vertex lies on a circle and whose sides contain chords of the circle. The diameter is the longest chord of a circle and it passes through the venter what is the relationship between inscribed angles and their arcs? Circles practice problems and proofs. Find the measure of the arc or central angle indicated. Book comes like the further recommendation and lesson every become old you retrieve it.\n\n## Central Angles And Inscribed Angles Worksheet Answer Key Lesson 15-1 , Major Arc Minor Arc Intercepted Arc:",
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"## Central Angles And Inscribed Angles Worksheet Answer Key Lesson 15-1 . Worksheet About Stoichiometry, Chemistry Worksheet On Stoichiometry Mixed Review Answers, Worksheet For Stoichiometry, Worksheets On Gas Stoichiometry Pdf, Worksheet On Stoichiometry Key, Worksheet #1 Stoichiometry Moles To Moles, Worksheet Of Stoichiometry.",
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https://study-assistant.com/mathematics/question12021660 | [
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", 23.06.2019 01:30 wvtoy4767\n\n# Given the scale drawing of a one bedroom apartment what is the actual perimeter of the study a) 21 ft b) 24 ft c) 42 ft d) 84 ft",
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"",
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null,
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