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https://ncatlab.org/nlab/show/kernel+functor | [
"# nLab kernel functor\n\nIn noncommutative ring theory, particularly in the subject of noncommutative localization of rings, a kernel functor is any left exact additive subfunctor of the identity functor on the category ${}_R Mod$ of left modules over a ring $R$. There is a bijective correspondence between kernel functors and uniform filters of ideals in $R$. A functor $\\sigma: {}_R Mod\\to {}_R Mod$ is idempotent if $\\sigma\\sigma = \\sigma$ and a preradical if it is additive subfunctor of the identity and $\\sigma(M/\\sigma(M))=0$ for all $M$ in ${}_R Mod$. A kernel functor $\\sigma: {}_R Mod\\to {}_R Mod$ is said to be an idempotent kernel functor if $\\sigma(M/\\sigma(M))=0$ for all $M$ in ${}_R Mod$; it is idempotent as we see by calculating\n\n$\\sigma \\sigma M = \\sigma Ker(M\\to M/\\sigma M) = Ker (\\sigma M\\to \\sigma(M/\\sigma M)) = Ker(\\sigma M\\to M/\\sigma M) = \\sigma M$\n\nIn the last step, we used that $\\sigma$ is a subfunctor of the identity, hence the compositions $\\sigma M\\hookrightarrow M\\to M/\\sigma M$ and $\\sigma M\\to \\sigma(M/\\sigma M)\\to M/\\sigma M$ coincide.\n\nThe basic reference is\n\n• O. Goldman, Rings and modules of quotients, J. Algebra 13, 1969 10–47, MR245608, doi\n\nwhich is clearly written from the point of view of a ring theorist. Unfortunately, it just creates another formalism in localization theory of the categories of modules over a ring for basically the same results as P. Gabriel succeeded by more categorical formulations in his thesis published 7 years earlier. Some of the methods from Goldman, and even more from Gabriel apply for more general Grothendieck categories.\n\n• Pascual Jara, Alain Verschoren, Conchi Vidal, Localization and sheaves: a relative point of view, Pitman Research Notes in Mathematics Series, 339. Longman, Harlow, 1995. xiv+235 pp.\n• J. L. Bueso, P. Jara, A. Verschoren, Compatibility, stability, and sheaves, Monographs and Textbooks in Pure and Applied Mathematics, 185. Marcel Dekker, Inc., New York, 1995. xiv+265 pp.\n\nLast revised on June 8, 2011 at 18:07:06. See the history of this page for a list of all contributions to it."
] | [
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https://lup.lub.lu.se/search/publication/30d892fa-0ade-495a-b193-62e6711efeaa | [
"# Lund University Publications\n\n## LUND UNIVERSITY LIBRARIES\n\nAdvanced\n\n### Trimmed moebius inversion and graphs of bounded degree\n\n(2008) 25th International Symposium on Theoretical Aspects of Computer Science (STACS 2008) p.85-96\nAbstract\nWe study ways to expedite Yates's algorithm for computing the zeta and Moebius transforms of a function defined on the subset lattice. We develop a trimmed variant of Moebius inversion that proceeds point by point, finishing the calculation at a subset before considering its supersets. For an n-element universe U and a family F of its subsets, trimmed Moebius inversion allows us to compute the number of parkings, coverings, and partitions of U with k sets from F in time within a polynomial factor (in n) of the number of supersets of the members of F. Relying on an intersection theorem of Chung et al. (1986) to bound the sizes of set families, we apply these ideas to well-studied combinatorial optimisation problems on graphs of maximum... (More)\nWe study ways to expedite Yates's algorithm for computing the zeta and Moebius transforms of a function defined on the subset lattice. We develop a trimmed variant of Moebius inversion that proceeds point by point, finishing the calculation at a subset before considering its supersets. For an n-element universe U and a family F of its subsets, trimmed Moebius inversion allows us to compute the number of parkings, coverings, and partitions of U with k sets from F in time within a polynomial factor (in n) of the number of supersets of the members of F. Relying on an intersection theorem of Chung et al. (1986) to bound the sizes of set families, we apply these ideas to well-studied combinatorial optimisation problems on graphs of maximum degree A. In particular, we show how to compute the Domatic Number in time within a polynomial factor of (2(Delta+1) - 2)(n/(Delta+1)) and the Chromatic Number in time within a polynomial factor of (2(Delta+1) - Delta - 1)(n/(Delta+1)) For any constant A, these bounds are 0 ((2 - epsilon)(n)) for epsilon > 0 independent of the number of vertices n. (Less)\nPlease use this url to cite or link to this publication:\nauthor\norganization\npublishing date\ntype\nChapter in Book/Report/Conference proceeding\npublication status\npublished\nsubject\nhost publication\nSTACS 2008: Proceedings of the 25th Annual Symposium on Theoretical Aspects of Computer Science\npages\n85 - 96\npublisher\nLABRI - Laboratoire Bordelais de Recherche en Informatique\nconference name\n25th International Symposium on Theoretical Aspects of Computer Science (STACS 2008)\nconference location\nBordeaux, France\nconference dates\n2008-02-21 - 2008-02-23\nexternal identifiers\n• wos:000254982900007\n• scopus:45749123453\nproject\nExact algorithms\nlanguage\nEnglish\nLU publication?\nyes\nid\n30d892fa-0ade-495a-b193-62e6711efeaa (old id 1407345)\nalternative location\nhttp://arxiv.org/pdf/0802.2834v1\ndate added to LUP\n2009-06-02 14:42:22\ndate last changed\n2019-02-20 09:21:21\n```@inproceedings{30d892fa-0ade-495a-b193-62e6711efeaa,\nabstract = {We study ways to expedite Yates's algorithm for computing the zeta and Moebius transforms of a function defined on the subset lattice. We develop a trimmed variant of Moebius inversion that proceeds point by point, finishing the calculation at a subset before considering its supersets. For an n-element universe U and a family F of its subsets, trimmed Moebius inversion allows us to compute the number of parkings, coverings, and partitions of U with k sets from F in time within a polynomial factor (in n) of the number of supersets of the members of F. Relying on an intersection theorem of Chung et al. (1986) to bound the sizes of set families, we apply these ideas to well-studied combinatorial optimisation problems on graphs of maximum degree A. In particular, we show how to compute the Domatic Number in time within a polynomial factor of (2(Delta+1) - 2)(n/(Delta+1)) and the Chromatic Number in time within a polynomial factor of (2(Delta+1) - Delta - 1)(n/(Delta+1)) For any constant A, these bounds are 0 ((2 - epsilon)(n)) for epsilon > 0 independent of the number of vertices n.},\nauthor = {Björklund, Andreas and Husfeldt, Thore and Kaski, Petteri and Koivisto, Mikko},\nlanguage = {eng},\nlocation = {Bordeaux, France},\npages = {85--96},\npublisher = {LABRI - Laboratoire Bordelais de Recherche en Informatique},\ntitle = {Trimmed moebius inversion and graphs of bounded degree},\nyear = {2008},\n}\n\n```"
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https://doc.sagemath.org/html/en/reference/combinat/sage/combinat/crystals/bkk_crystals.html | [
"# Benkart-Kang-Kashiwara crystals for the general-linear Lie superalgebra#\n\nclass sage.combinat.crystals.bkk_crystals.CrystalOfBKKTableaux(ct, shape)#\n\nBases: CrystalOfWords\n\nCrystal of tableaux for type $$A(m|n)$$.\n\nThis is an implementation of the tableaux model of the Benkart-Kang-Kashiwara crystal [BKK2000] for the Lie superalgebra $$\\mathfrak{gl}(m+1,n+1)$$.\n\nINPUT:\n\n• ct – a super Lie Cartan type of type $$A(m|n)$$\n\n• shape – shape specifying the highest weight; this should be a partition contained in a hook of height $$n+1$$ and width $$m+1$$\n\nEXAMPLES:\n\nsage: T = crystals.Tableaux(['A', [1,1]], shape = [2,1])\nsage: T.cardinality()\n20\n\nclass Element#\ngenuine_highest_weight_vectors(index_set=None)#\n\nReturn a tuple of genuine highest weight elements.\n\nA fake highest weight vector is one which is annihilated by $$e_i$$ for all $$i$$ in the index set, but whose weight is not bigger in dominance order than all other elements in the crystal. A genuine highest weight vector is a highest weight element that is not fake.\n\nEXAMPLES:\n\nsage: B = crystals.Tableaux(['A', [1,1]], shape=[3,2,1])\nsage: B.genuine_highest_weight_vectors()\n([[-2, -2, -2], [-1, -1], ],)\nsage: B.highest_weight_vectors()\n([[-2, -2, -2], [-1, -1], ],\n[[-2, -2, -2], [-1, 2], ],\n[[-2, -2, 2], [-1, -1], ])\n\nshape()#\n\nReturn the shape of self.\n\nEXAMPLES:\n\nsage: T = crystals.Tableaux(['A', [1, 2]], shape=[2,1])\nsage: T.shape()\n[2, 1]"
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https://calcforme.com/percentage-calculator/395-is-289-percent-of-what | [
"# 395 is 289 Percent of what?\n\n## 395 is 289 Percent of 136.68\n\n%\n\n395 is 289% of 136.68\n\nCalculation steps:\n\n395 ÷ ( 289 ÷ 100 ) = 136.68\n\n### Calculate 395 is 289 Percent of what?\n\n• F\n\nFormula\n\n395 ÷ ( 289 ÷ 100 )\n\n• 1\n\nPercent to decimal\n\n289 ÷ 100 = 2.89\n\n• 2\n\n395 ÷ 2.89 = 136.68 So 395 is 289% of 136.68\n\nExample"
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https://www.physicsforums.com/threads/dumb-slope-question.4971/ | [
"# Dumb slope question\n\n## Main Question or Discussion Point\n\nHi,\n\nI was looking at a problem from an old test and I got confused about something.\n\nQ. Find the slope of the tangent line to the given curve at a given point:\n\ny = x(x^2 + 3)^1/2 ; P(1,2)\n\nS. First I found the derivative:\n[(x^2 + 3)^1/2] + [(x^2)(x^2 + 3)^-1/2]\n\nPlugged in the values for x and y at the given point:\n[(4)^1/2] + [(4)^-1/2]\n\na) 2 b) -5/2 c) 3/2 d) 5/2 e) none of these\n\nSquare root of 4 can be + OR - 2, no?\n\nThat would mean I can have 4 possible solutions:\n(4/2) + (1/2) = 5/2\n(4/2) + (-1/2) = 3/2\n(-4/2) + (1/2) = -3/2\n(-4/2) + (-1/2) = -5/2\n\nSince 3 of these are listed as choices, how did I pick the right one? (I chose d)5/2 and it was marked correct)\n\nForgive me if this is shockingly stupid...I am trying to review all my calculus and physics to start classes up again in the winter.\n\nNEVERMIND!!!!\n\nI feel dumb...\n\nOf course, the original equation must also hold true, which means that (4)^1/2 can only be +2\n\nHallsofIvy\nHomework Helper\nIn addition, \"square root\" is a FUNCTION which means it can have only one value. The square root of any positive number, a, is, by definition, the POSITIVE number, x, such that x2= a.\n\nOf course, x2= a has TWO solutions: they are [sqrt](a) and -[sqrt](a).\n\nOriginally posted by HallsofIvy\nIn addition, \"square root\" is a FUNCTION which means it can have only one value. The square root of any positive number, a, is, by definition, the POSITIVE number, x, such that x2= a.\n\nOf course, x2= a has TWO solutions: they are [sqrt](a) and -[sqrt](a).\n\nWhen you say FUNCTION, do you actually mean CONTINOUS ONE-TO-ONE FUNCTION?\n\nHallsofIvy"
] | [
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http://blogsee.cn/qspevdu_t1001001010 | [
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null,
"省直辖县级行政单位海南标签制作|荐_海南万佳达优良的海南标签供应\n\n品牌:万佳达,万佳达包装,海南万佳达包装\n\n出厂地:南丹县(城关镇)\n\n报价：面议\n\n海南万佳达包装乐天堂fun88备用\n\n黄金会员：",
null,
"经营模式：生产型\n\n主营：海南纸箱包装,海南精美礼盒,海南手提袋,海南水果箱包装,海南水印箱\n\n•",
null,
"报价：面议\n\n甘肃博琳国际文化发展乐天堂fun88备用\n\n黄金会员：",
null,
"经营模式：服务型\n\n主营：甘肃旅游规划设计,兰州旅游宣传片,甘肃文化旅游标识导示系统,甘肃旅游策划,兰州旅...\n\n•",
null,
"防伪渠道-贵阳信誉好的卓霖防伪信息密码锁项目\n\n品牌:品牌防伪,智慧溯源,数字营销\n\n出厂地:柳北区\n\n报价：面议\n\n贵州卓霖科技乐天堂fun88备用\n\n黄金会员：",
null,
"经营模式：生产型\n\n主营：卓霖密码锁,防伪信息化系统,小马追溯系统,一物一码系统,市场稽查系统\n\n•",
null,
"服务周到的防伪|贵阳专业的卓霖防伪信息密码锁推荐\n\n品牌:品牌防伪,智慧溯源,数字营销\n\n出厂地:柳北区\n\n报价：面议\n\n贵州卓霖科技乐天堂fun88备用\n\n黄金会员：",
null,
"经营模式：生产型\n\n主营：卓霖密码锁,防伪信息化系统,小马追溯系统,一物一码系统,市场稽查系统\n\n•",
null,
"报价：面议\n\n肇庆市中天印务乐天堂fun88备用\n\n黄金会员：",
null,
"经营模式：贸易型\n\n主营：可变数据不干胶,透明不干胶,PP合成纸,铜版纸不干胶,热敏纸不干胶\n\n•",
null,
"寻找防伪_专业可靠的卓霖防伪信息密码锁哪里有\n\n品牌:品牌防伪,智慧溯源,数字营销\n\n出厂地:柳北区\n\n报价：面议\n\n贵州卓霖科技乐天堂fun88备用\n\n黄金会员：",
null,
"经营模式：生产型\n\n主营：卓霖密码锁,防伪信息化系统,小马追溯系统,一物一码系统,市场稽查系统\n\n•",
null,
"防伪怎么样_贵州专业的卓霖防伪信息密码锁公司\n\n品牌:品牌防伪,智慧溯源,数字营销\n\n出厂地:柳北区\n\n报价：面议\n\n贵州卓霖科技乐天堂fun88备用\n\n黄金会员：",
null,
"经营模式：生产型\n\n主营：卓霖密码锁,防伪信息化系统,小马追溯系统,一物一码系统,市场稽查系统\n\n•",
null,
"创意标签-哪里能买到有品质的透明不干胶\n\n品牌:中天,,\n\n出厂地:合山市\n\n报价：面议\n\n肇庆市中天印务乐天堂fun88备用\n\n黄金会员：",
null,
"经营模式：贸易型\n\n主营：可变数据不干胶,透明不干胶,PP合成纸,铜版纸不干胶,热敏纸不干胶\n\n•",
null,
"防伪信息_实力可靠的卓霖密码锁服务商_卓霖科技，防伪信息\n\n品牌:品牌防伪,智慧溯源,数字营销\n\n出厂地:柳北区\n\n报价：面议\n\n贵州卓霖科技乐天堂fun88备用\n\n黄金会员：",
null,
"经营模式：生产型\n\n主营：卓霖密码锁,防伪信息化系统,小马追溯系统,一物一码系统,市场稽查系统\n\n•",
null,
"选择防伪|卓霖科技提供专业的卓霖防伪信息密码锁\n\n品牌:品牌防伪,智慧溯源,数字营销\n\n出厂地:柳北区\n\n报价：面议\n\n贵州卓霖科技乐天堂fun88备用\n\n黄金会员：",
null,
"经营模式：生产型\n\n主营：卓霖密码锁,防伪信息化系统,小马追溯系统,一物一码系统,市场稽查系统\n\n• 没有找到合适的供应商？您可以发布采购信息\n\n没有找到满足要求的供应商？您可以搜索 标签标牌批发 标签标牌公司 标签标牌厂\n\n### 最新入驻厂家\n\n相关产品:\n省直辖县级行政单位海南标签制作 兰州景区指示牌厂家 防伪渠道 服务周到的防伪 广东可变数据不干胶 寻找防伪 防伪怎么样 创意标签 防伪信息 选择防伪"
] | [
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] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.7897998,"math_prob":0.5147303,"size":815,"snap":"2020-10-2020-16","text_gpt3_token_len":988,"char_repetition_ratio":0.18372379,"word_repetition_ratio":0.074074075,"special_character_ratio":0.20368098,"punctuation_ratio":0.30769232,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9702989,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40],"im_url_duplicate_count":[null,2,null,null,null,2,null,null,null,2,null,null,null,5,null,null,null,2,null,null,null,5,null,null,null,6,null,null,null,1,null,null,null,5,null,null,null,6,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-06T03:55:43Z\",\"WARC-Record-ID\":\"<urn:uuid:c7bd9af2-3ebb-4b0a-bf9a-afc197053207>\",\"Content-Length\":\"104433\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:561e5223-b7be-4545-b683-66c3e2e9ccd7>\",\"WARC-Concurrent-To\":\"<urn:uuid:c6270afb-eeba-4154-aa4a-33bd64b0c34e>\",\"WARC-IP-Address\":\"154.81.91.146\",\"WARC-Target-URI\":\"http://blogsee.cn/qspevdu_t1001001010\",\"WARC-Payload-Digest\":\"sha1:THGU7Y2DAS232VT4CJ7OMP2G7V2DGG4Y\",\"WARC-Block-Digest\":\"sha1:HYPYTTUPY2FGSY32DUIQLJOZMRCOHSCB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371618784.58_warc_CC-MAIN-20200406035448-20200406065948-00093.warc.gz\"}"} |
https://www.numbersaplenty.com/2331 | [
"Search a number\nBaseRepresentation\nbin100100011011\n310012100\n4210123\n533311\n614443\n76540\noct4433\n93170\n102331\n11182a\n121423\n1310a4\n14bc7\n15a56\nhex91b\n\n2331 has 12 divisors (see below), whose sum is σ = 3952. Its totient is φ = 1296.\n\nThe previous prime is 2311. The next prime is 2333. The reversal of 2331 is 1332.\n\n2331 = 103 + 113.\n\nIt is a happy number.\n\n2331 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.\n\n2331 is an esthetic number in base 4, because in such base its adjacent digits differ by 1.\n\nIt is not a de Polignac number, because 2331 - 26 = 2267 is a prime.\n\nIt is a super-2 number, since 2×23312 = 10867122, which contains 22 as substring.\n\nIt is a Harshad number since it is a multiple of its sum of digits (9).\n\nIt is a Duffinian number.\n\nIt is a nialpdrome in base 5, base 7 and base 8.\n\nIt is a zygodrome in base 5 and base 8.\n\nIt is not an unprimeable number, because it can be changed into a prime (2333) by changing a digit.\n\nIt is a polite number, since it can be written in 11 ways as a sum of consecutive naturals, for example, 45 + ... + 81.\n\n22331 is an apocalyptic number.\n\n2331 is a gapful number since it is divisible by the number (21) formed by its first and last digit.\n\n2331 is a deficient number, since it is larger than the sum of its proper divisors (1621).\n\n2331 is a wasteful number, since it uses less digits than its factorization.\n\n2331 is an evil number, because the sum of its binary digits is even.\n\nThe sum of its prime factors is 50 (or 47 counting only the distinct ones).\n\nThe product of its digits is 18, while the sum is 9.\n\nThe square root of 2331 is about 48.2804308183. The cubic root of 2331 is about 13.2591013771.\n\nAdding to 2331 its reverse (1332), we get a palindrome (3663).\n\nSubtracting from 2331 its reverse (1332), we obtain a palindrome (999).\n\nIt can be divided in two parts, 2 and 331, that added together give a palindrome (333).\n\nThe spelling of 2331 in words is \"two thousand, three hundred thirty-one\".\n\nDivisors: 1 3 7 9 21 37 63 111 259 333 777 2331"
] | [
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https://answerswallet.com/in-which-electric-circuit-would-the-voltmeter-read-10-volts-g17k/ | [
"# In which electric circuit would the voltmeter read 10 volts\n\nAsked by wiki @ in Physics viewed by 855 People\n\nIn which electric circuit would the voltmeter read 10 volts ?\n\nGiven that,\n\nVoltage = 10 volt\n\nSuppose, The three resistance is connected in parallel and each resistance is 12 Ω. find the current in the electric circuit.\n\nWe need to calculate the equivalent resistance\n\nUsing formula of parallel",
null,
"Put the value into the formula",
null,
"",
null,
"",
null,
"We need to calculate the current in the circuit\n\nUsing ohm's law",
null,
"",
null,
"Where, V = voltage\n\nR = resistance\n\nPut the value into the formula",
null,
"",
null,
"Hence, The current in the circuit is 2.5 A\n\n## Which would cause an electric circuit to lack a current\n\nAsked by wiki @ in Chemistry viewed by 162 persons\n\nWhich would cause an electric circuit to lack a current? open switch closed switch lighted bulbs unlighted bulbs\n\n## The electric potential at a given location in a circuit\n\nAsked by wiki @ in Physics viewed by 241 persons\n\nWhich of the following statements are true about the electric potential or the electric potential difference? (a) The SI unit of the electric potential is volts (V). (b) The electric …\n\n## Which unit of electricity does the work in the circuit\n\nAsked by wiki @ in Physics viewed by 231 persons\n\n5. Which unit of electricity does the work in a circuit? A. Ohm B. Ampere C. Coulomb D. Volt\n\n## In an operating electrical circuit the source of potential difference\n\nAsked by wiki @ in Physics viewed by 160 persons\n\nIn an operating electrical circuit the source of potential difference could be\n\n## Which device provides electrical energy to run an electric circuit\n\nAsked by wiki @ in Chemistry viewed by 250 persons\n\nWhich device provides electrical energy to run an electric circuit? the wire the switch the battery the light bulb\n\n## If the resistance of an electric circuit is 12 ohms\n\nAsked by wiki @ in Physics viewed by 237 persons\n\nIf the resistance of an electric circuit is 12 ohms and the voltage in the circuit is 60 V, the current flowing through the circuit is A. 0.2 A. B. …\n\n## Which reading strategy would most help in comprehending the sentence\n\nAsked by wiki @ in English viewed by 441 persons\n\nRead the excerpt from \"How the Internet and Other Technologies Came About.\" Because most cities would no longer exist, messages would have to be broken up into pieces, scattered throughout …\n\n## A voltmeter should be connected to the circuit being tested\n\nAsked by wiki @ in Engineering viewed by 618 persons\n\nVoltage drop testing is being discussed. Technician A connects the red voltmeter lead to the most positive side of the circuit being tested and the black lead to the most …\n\n## Which of the following would be the best electrical insulator\n\nAsked by wiki @ in Physics viewed by 329 persons\n\nWhich of the following would be best used as an electrical insulator? A. copper B. silver C. salt water D. wood\n\n## Which strategy would most likely improve a student's reading comprehension\n\nAsked by wiki @ in Computers and Technology viewed by 417 persons\n\n## Which item s would be sufficient to make a circuit\n\nAsked by wiki @ in Physics viewed by 236 persons\n\nWhich item(s) would be sufficient to make a circuit? a. Conductor b. Insulator c. Conductor and Battery d. Insulator and Battery e. Battery\n\n## What association would you expect if graphing height and weight\n\nAsked by wiki @ in Mathematics viewed by 186 persons\n\nWhat is the term for data that are grouped closely together? (1 point outlier linear positive clustering 2 what association would you expect if graphing height and weight? (1 point …\n\n## Read the excerpt from the lovesong of j. alfred prufrock.\n\nAsked by wiki @ in English viewed by 165 persons\n\nRead the excerpt from \"The Love Song of J. Alfred Prufrock.\" Shall I say, I have gone at dusk through narrow streets And watched the smoke that rises from the …\n\n## In what circumstances would a property insurance claim be rejected\n\nAsked by wiki @ in Mathematics viewed by 147 persons\n\nIn what circumstance would a property insurance claim be rejected? The insurance company finds that a homeowner intentionally caused damage. The property damage is caused by a natural disaster, such …\n\n## How would you describe eckels in a sound of thunder\n\nAsked by wiki @ in English viewed by 126 persons\n\nIn Ray Bradbury’s short story \"A Sound of Thunder,\" which of Mr. Eckels’s character traits leads to drastic consequences for the rest of the world? (1 point) his lack of …\n\n## A 0.25 kg arrow with a velocity of 12m s\n\nAsked by wiki @ in Physics viewed by 1035 persons\n\n## Uranium 235 undergoes an alpha decay to produce thorium 231\n\nAsked by wiki @ in Physics viewed by 997 persons\n\n## A golf club hits a stationary 0.050 kilogram golf ball\n\nAsked by wiki @ in Physics viewed by 928 persons\n\n## Water waves in a small tank are 06 m long\n\nAsked by wiki @ in Physics viewed by 923 persons\n\n## A tank of water in the shape of a cone\n\nAsked by wiki @ in Physics viewed by 864 persons\n\n## In which electric circuit would the voltmeter read 10 volts\n\nAsked by wiki @ in Physics viewed by 855 persons\n\nAsked by wiki @ in Physics viewed by 852 persons\n\n## The initial speed of a cannonball is 0.20 km s\n\nAsked by wiki @ in Physics viewed by 839 persons\n\n## A 60 kilogram student jumps down from a laboratory counter\n\nAsked by wiki @ in Physics viewed by 836 persons\n\n## A 90 kg person rides a spinning amusement park ride\n\nAsked by wiki @ in Physics viewed by 826 persons\n\n## A golfer hits a ball at an angle of 25\n\nAsked by wiki @ in Physics viewed by 780 persons\n\n## A 5.8 x 10 4 watt elevator motor can lift\n\nAsked by wiki @ in Physics viewed by 777 persons\n\n## All of the following are advantages of outsourcing except it\n\nAsked by wiki @ in Physics viewed by 774 persons\n\n## 7 types of electromagnetic waves from lowest to highest frequency\n\nAsked by wiki @ in Physics viewed by 672 persons\n\n## Which of the following statements about diffusion is true\n\nAsked by wiki @ in Physics viewed by 665 persons"
] | [
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"https://tex.z-dn.net/",
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"https://tex.z-dn.net/",
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"https://tex.z-dn.net/",
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https://www.py.cn/jishu/jichu/20871.html | [
"• 写文章\n• 提问\n\n# python中loc函数的用法是什么?",
null,
"loc()：基于标签索引\n\niloc()：基于整数索引\n\ndf.loc[ 行索引, 列索引]",
null,
"",
null,
"```>>> df = pd.DataFrame(np.random.randn(8, 4),\nindex = ['a','b','c','d','e','f','g','h'],\ncolumns = ['A', 'B', 'C', 'D'])\n>>> df\nA B C D\na -0.484976 1.958562 -0.073555 0.524286\nb 1.681393 1.041901 -0.109796 0.836486\nc 0.352229 0.656365 0.590963 0.908981\nd 1.325258 1.199558 0.953455 -0.192507\ne 0.573300 -0.202530 -0.699603 1.504382\n-1.423372 -0.311816 0.680950 -1.619343\n0.771233 -0.101350 -0.207373 1.242127\n0.084874 -0.655007 -0.834754 0.072229\n>>> df.loc['a', ['A', 'B']]\nA -0.484976\nB 1.958562```",
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"https://oss.py.cn/pycn/upload/image/622/184/532/1605512665335502.png",
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https://es.mathworks.com/matlabcentral/fileexchange/33608-snake-evolution-gp?s_tid=prof_contriblnk | [
"## snake evolution, GP\n\nversión 1.0.0.0 (935 KB) por\nGenetic programing example at snake evolution simulation.\n\n547 descargas\n\nVer licencia\n\nhttp://simulations.narod.ru/\n\nSee this anitmation here:\n\nTo fast test the code run\nzz_fast.m to make evolution\nthen run zz_motion_terst_1_snakes.m to see history of this evolution as animation of best individuals.\n\nSnake evolution is made as Genenetic programming. Gene Expression Programming (GEP) was used. GEP is developed by FERREIRA. FERREIRA's articles is incuded in pdf files. It is parameterization of binary tree of sequence of mathematics operations.\n\nSnake shape is represented as function f=f(d,t), where t is time. Snake is links where angles between links is controeld with angles phi=phi(t), angles phi is differences of f function. For example:\nphi(1)=f(L,t)-f(0,t)\nphi(2)=f(2*L,t)-f(L,t)\nphi(3)=f(4*L,t)-f(2*L,t)\nect.\nSo if phi not too big then shape of snake is f(d,t) function.\n\nIt is classic GP with crossover and mutations. Two point crossover was used. Elitism was used. In gene head size is 24, that means that formula can have 24 operations max. + - * / sin cos sign operations was used. phi angle cannot be more then pi/2 at absolute value.\n\n### Citar como\n\nMaxim Vedenyov (2022). snake evolution, GP (https://www.mathworks.com/matlabcentral/fileexchange/33608-snake-evolution-gp), MATLAB Central File Exchange. Recuperado .\n\n##### Compatibilidad con la versión de MATLAB\nSe creó con R2010b\nCompatible con cualquier versión"
] | [
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https://oro.open.ac.uk/28349/ | [
"# ORTHOMAX rotation problem. a differential equation approach\n\nTrendafilov, Nickolay T. and Chu, Moody T. (1998). ORTHOMAX rotation problem. a differential equation approach. Behaviormetrika, 25(1) pp. 13–23.\n\n## Abstract\n\nIn the present paper the ORTHOMAX rotation problem is reconsidered. It is shown that its solution can be presented as a steepest ascent flow on the manifold of orthogonal matrices. A matrix formulation of the ORTHOMAX problem is given as an initial value problem for matrix differential equation of first order. The solution can be found by any available ODE numerical integrator. Thus the paper proposes a convergent method for direct matrix solution of the ORTHOMAX problem. The well-known first order necessary condition for the VARIMAX maximizer is reestablished for the ORTHOMAX case without using Lagrange multipliers. Additionally new second order optimality conditions are derived and as a consequence an explicit second order necessary condition for further classification of the ORTHOMAX maximizer is obtained.\n\n## Metrics\n\n### Public Attention\n\nAltmetrics from Altmetric\n\n### Number of Citations\n\nCitations from Dimensions"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7740965,"math_prob":0.96934474,"size":1561,"snap":"2021-21-2021-25","text_gpt3_token_len":359,"char_repetition_ratio":0.11046885,"word_repetition_ratio":0.03539823,"special_character_ratio":0.2005125,"punctuation_ratio":0.10080645,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9760626,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-14T14:51:32Z\",\"WARC-Record-ID\":\"<urn:uuid:30e15bed-a3b8-4973-a8be-56bcee9c9de4>\",\"Content-Length\":\"52600\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d7e4307c-fce5-4397-bf6c-5de328b1c1d0>\",\"WARC-Concurrent-To\":\"<urn:uuid:cdf83061-b8cd-4801-b9eb-232a695fdfe4>\",\"WARC-IP-Address\":\"137.108.200.18\",\"WARC-Target-URI\":\"https://oro.open.ac.uk/28349/\",\"WARC-Payload-Digest\":\"sha1:FIB3EJBQHQ552ZZ3X5IXHDAY4FCYVT3X\",\"WARC-Block-Digest\":\"sha1:EIR5ZSGVUCTPKTZYTNT7EIAXANJOU6YE\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989526.42_warc_CC-MAIN-20210514121902-20210514151902-00028.warc.gz\"}"} |
https://www.allmath.com/blogs/ratio-and-proportion | [
"",
null,
"# Ratio and Proportion: Defined and Explained with Calculations\n\nPublish Date: Jun 29, 2021\n\n## Table of Content\n\nRatios and proportions are widely practiced concepts. We have to deal with them on a daily basis. This is why it is necessary to learn about them.\n\nIn math, both of these concepts are greatly related to fractions. It is because the ratio is fundamentally a fraction. And proportion is the comparison of two fractions.\n\nIt is no wonder if you are vaguely familiar with ratios. For example, you would have come across situations where the ratio of one group is higher than the other.\n\nSimilarly, You might have an idea about proportions already. For instance, you know that a motorcycle is faster than a bicycle in the same way a sports car is faster than a normal car.\n\nIn this blog post, we will discuss ratios and propositions with examples, so that by the end of this post you will have no confusion.\n\n## What are Ratios?\n\nRatio can be defined as,\n\nA ratio compares the quantity of one kind to the quantity of another kind. For ratios to exist, the quantities should have the same units.\n\nRatios are represented as:\n\na:b or a/b\n\nThe symbols of both slash (/) and colon (:) are used to represent a ratio. The first value, here a, is known as antecedent. The second value, b in this case, is called the consequent.\n\n## Types of Ratios\n\nThere are two ways ratios exist.\n\n• ### Part to part ratios\n\nA ratio in which two different quantities are compared. For instance, in a cake mixture, the ratio of flour to oil or milk. At a party, the ratio of bottles of orange juice to the bottles of mango or strawberry juice.\n\n• ### Part to whole ratios\n\nA ratio in which a quantity is compared to the whole amount. For example, a ratio of males in a city to the total population of the city.\n\n## What are Proportions?\n\nA proportion can be defined as,\n\nA proportion compares two ratios. Through a proportion, we get to know that in two ratios a:b and c:d, a is related to b in the same way as c is related to d.\n\nThe proportion of two ratios is represented as:\n\na : b : : c : d or a : b = c : d\n\nThe double semicolon (: :) or equal to (=) tells that the two ratios mentioned are in equal proportion. The values a and d are known as extremes and the values b and c are known as means.\n\n### Parts of Proportion\n\nIn a proportion “a : b : : c : d\n\n• The mean between b and c is ab.\n• c is third proportional to a and b.\n• d is fourth proportional to a, b and c.\n\n## What do Ratios and Proportions tell us?\n\nIn a ratio, one quantity is compared in terms of “times” with the other quantity. This means if A and B are in a ratio of 3:1, then A is 3 times the quantity of B.\n\nFor example, In a cake recipe, flour is 2 times the quantity of oil. It can be represented in ratio as\n\n2 : 1\n\nA proportion tells a person that two ratios in proportion increase or decrease in a similar way.\n\nWe can see this in an example. Previously we discussed that in a cake recipe, flour is two times the quantity of oil. This recipe is to make a one-pound cake.\n\nTo make a 2-pound cake, increase the quantity of both in the same way, we will multiply two on both sides.\n\n= 2 x 2 : 2 x 1\n\n= 4 : 2\n\nThis ratio is in proportion to our previous ratio 2:1. It can be written as,\n\n2 : 1 :: 4 : 2\n\n## How to find Ratios and Proportions?\n\nIt will be easy to learn about ratios with the help of an example.\n\n### Example - Ratio\n\nIn a fruit basket, there are 15 fruits. Out of these 15 fruits, 6 are bananas and 3 are apples. Find the ratio of\n\n• Bananas to rest of the fruits\n• Apples to total fruits\n• Apples to bananas\n\nSolution:\n\nStep 1: For Bananas:\n\nTo find in which ratio bananas are present in the fruit basket, we will have to subtract bananas from the total number of fruits. It is a “part to part” ratio.\n\nTotal number of bananas = 6\n\nTotal fruits in the basket = 15\n\nFruits - bananas = 15 - 6 = 9\n\nThis means there are 9 more fruits in the basket. The ratio of bananas to the rest of the fruits is:\n\n6 : 9\n\nAfter dividing by 3:\n\n2 : 3\n\nStep 2: For apples to the total fruits.\n\nWe will not need to subtract no. of apples from no. of fruits in this part. This is because we have to find the “part to the whole” ratio.\n\nNo. of apples = 3\n\nNo. of total fruits = 15\n\nThe ratio is:\n\n3 : 15\n\nAfter simplifying,\n\n1 : 5\n\nStep 3: Apples to bananas.\n\nIt is also a “part to part” ratio.\n\nNo. of apples = 3\n\nNo. of bananas = 6\n\nThe ratio is,\n\n3 : 6\n\nSimplifying:\n\n1 : 2\n\nA proportion can be formed by multiplying or dividing the ratio by the same number on both sides. After that writing the new ratio in proportion (with double semicolon i.e ::) the old ratio.\n\nWe usually determine whether a proportion exists or not. We can prove this in 3 ways. Let’s see how to do that with the help of an example.\n\n### Example - Proportion\n\nProve that the ratios 3:6 and 6:12 are in proportion.\n\nSolution:\n\nStep 1: By simplifying.\n\n3 : 6 : : 6 : 12\n\nDividing the first ratio by 3 and the second ratio by 6.\n\n1 : 2 : : 1 : 2\n\nBoth ratios are equal, hence proportion exists.\n\nStep 2: Cross multiplication.\n\nWrite the ratios in the fraction form and cross multiply the fractions.\n\n3 x 12 = 6 x 6\n\n36 = 36\n\nBoth sides are the same. Hence, the proportion is true.\n\nStep 3: Decimals.\n\nSolve the fractions for decimal numbers.\n\n3/6 = 0.2\n\n6/12 = 0.2\n\nBoth of the decimals are the same. Proportion exists.\n\nUnder this heading, we will solve the problems related to ratios and proportions.\n\n### Problem No. 1\n\nThere are two black paint cans. In the first can we mix white paint into the black paint in a ratio of 6:9. In the second can we mix white in a ratio of 4:5. Which ratio is bigger?\n\nSolution:\n\nBy bigger ratio we mean, In which fraction (can) antecedent (white paint) is in more quantity. It is very easy to find. Just solve the fractions and compare.\n\nStep 1: Solve fractions.\n\n6:9 = 6/9 = 0.6\n\n4:5 = 4 /5 = 0.8\n\nStep 2: Compare.\n\nAs 0.8 > 0.6, so the ratio 4:5 is greater.\n\nYou can also find this through the ratio calculator.\n\n### Problem No. 2\n\nFind a in proportion, 4 : a : : 25 : 100.\n\nSolution:\n\nStep 1: Write the ratios in fractions and cross multiply.\n\n4 x 100 = a x 25\n\na = (4 x 100) / 25\n\na = 400 / 25\n\na = 16\n\nTry solving this problem using the proportional calculator"
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https://themathblogger.blog/2009/09/27/how-many-part-ii/ | [
"# How many… part II\n\nQ: How many strings of four decimal digits\n(a) Do not contains the same digit twice?\n(b) Begin with an odd digit?\n(c) Have exactly 3 digits that are 9?\n\nA: Back to the slot method… Four digits: _ _ _ _\n\n(a) Do not contain the same digit twice (basically asking for no repeats).. So, how many choices are there for the first slot? This questions is actually up in the air. Can 0 occupy the first digit or not? Usually, for example 0456 = 456 (which is a three-digit-number, not a four-digit-number)… This all depends on your interpretation, or your teacher’s interpretation!\n\nI believe due to the wording of this problem that 0 can occupy the first slot. Since we are looking for “strings” and not necessarily four-digit numbers. The first slot may be 0-9 (10 choices):\n\n10 _ _ _\n\nNow, the second digit can be 0-9 (10 choices).. However, it cannot be the same as the first choice, so minus 1 choice = 9 choices.\n\n10 9 _ _\n\nThe third digit can be 0-9 (10 choices), but minus 2 choices (cannot match slot 1 or slot 2) = 8 choices\n\n10 9 8 _\n\nFourth slot only has 7 choices (same logic as above)\n\n10 9 8 7\n\nThere are 10*9*8*7 total 4-digit numbers with distinct digits.\n\n(b) Begin with an odd digit?\n\nAgain, four digits: _ _ _ _\n\nThe first slot must be odd: 1, 3, 5, 7, 9 = 5 choices\n\n5 _ _ _\n\nThe rest of the digits can be any choice 0-9 (10 choices)\n\n5 10 10 10\n\n(c) Have exactly three digits that are 9?\n\nMany ways to solve this… But, we will do it my way.. Ha, we always do it my way on here…\n\nSo, the number must look like one of the follow:\n\n(1) 9 9 9 _ or\n\n(2) 9 9 _ 9 or\n\n(3) 9_ 9 9 or\n\n(4) _ 9 9 9\n\nThe 9s above do not represent choices, but rather that the number 9 is locked in and must be 9, no other number. So, in option (1) there are 9 choices for the final digit (the number can be 0-8). In option (2) there are 9 choices for the blank (0-8 are your options). Again for option (3): there are 9 choices for the blank. And, option (4) has 9 choices for the blank too.\n\nNow, this is an “or” problem. You have option (1) or (2) or (3) or (4). In counting, or means addition.\n\nSo, there are 9 + 9 + 9 + 9 = 36 total four-digit strings that have exactly 3 nines."
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https://opus.bibliothek.uni-wuerzburg.de/frontdoor/index/index/year/2021/docId/17246 | [
"## Search for new high-mass phenomena in the dilepton final state using 36 fb$$^{−1}$$ of proton-proton collision data at $$\\sqrt{s}=13$$ TeV with the ATLAS detector\n\nPlease always quote using this URN: urn:nbn:de:bvb:20-opus-172462\n\n• A search is conducted for new resonant and non-resonant high-mass phenomena in dielectron and dimuon final states. The search uses 36.1 fb$$^{−1}$$ of proton-proton collision data, collected at $$\\sqrt{s}=13$$ TeV by the ATLAS experiment at the LHC in 2015 and 2016. No significant deviation from the Standard Model prediction is observed. Upper limits at 95% credibility level are set on the cross-section times branching ratio for resonances decaying into dileptons, which are converted to lower limits on the resonance mass, up to 4.1 TeV for theA search is conducted for new resonant and non-resonant high-mass phenomena in dielectron and dimuon final states. The search uses 36.1 fb$$^{−1}$$ of proton-proton collision data, collected at $$\\sqrt{s}=13$$ TeV by the ATLAS experiment at the LHC in 2015 and 2016. No significant deviation from the Standard Model prediction is observed. Upper limits at 95% credibility level are set on the cross-section times branching ratio for resonances decaying into dileptons, which are converted to lower limits on the resonance mass, up to 4.1 TeV for the E$$_6$$-motivated $$Z^′_χ$$. Lower limits on the $${qqℓℓ}$$ contact interaction scale are set between 2.4 TeV and 40 TeV, depending on the model.",
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"",
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"• Dokument_1.pdf",
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"URN: urn:nbn:de:bvb:20-opus-172462 Journal article Fakultät für Physik und Astronomie English Journal of High Energy Physics 2017 2017 182 Journal of High Energy Physics (2017) 182. https://doi.org/10.1007/JHEP10(2017)182 https://doi.org/10.1007/JHEP10(2017)182 5 Naturwissenschaften und Mathematik / 53 Physik / 539 Moderne Physik Beyond Standard Model; Hadron-Hadron scattering (experiments); High energy physics 2021/03/17 The ATLAS Collaboration",
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https://math.stackexchange.com/questions/tagged/markov-process?sort=frequent&pageSize=50 | [
"# Questions tagged [markov-process]\n\nA stochastic process satisfying the Markov property: the distribution of the future states given the value of the current state does not depend on the past states. Use this tag for general state space processes (both discrete and continuous times); use (markov-chains) for countable state space processes.\n\n101 questions\n3answers\n8k views\n\n### What is the importance of the infinitesimal generator of Brownian motion?\n\nI have read that the infinitesimal generator of Brownian motion is $\\frac{1}{2}\\small\\triangle$. Unfortunately, I have no background in semigroup theory, and the expositions of semigroup theory I have ...\n2answers\n5k views\n\n### Transformation of state-space that preserves Markov property\n\nI am solving a problem in Mathematical Statistics by Jun Shao Let $\\{X_n \\}$ be a Markov chain. Show that if $g$ is a one-to-one Borel function, then $\\{g(X_n )\\}$ is also a Markov chain. Give an ...\n1answer\n526 views\n\n0answers\n131 views\n\n2answers\n11k views\n\n2answers\n148 views\n\n### Show that the carré du champ operator is nonnegative\n\nLet $(E,\\mathcal E)$ be a measurable space $\\mathcal M_b(E,\\mathcal E):=\\left\\{f:E\\to\\mathbb R\\mid f\\text{ is bounded and }\\mathcal E\\text{-measurable}\\right\\}$ $(\\kappa_t)_{t\\ge0}$ be a Markov ...\n1answer\n188 views\n\n### Building a hidden markov model with an absorbing state.\n\nI'm working on trying to implement a hidden markov model to model the affect of a specific protein that can cut an RNA when the ribosome is translating the RNA slowly. Some brief background: The ...\n1answer\n166 views\n\n### Markov property w.r.t. a countable state space\n\nBackground Let $\\left(X_t\\right)_{t \\in I}$ ($I\\subseteq\\mathbb R$) be an $E$-valued stochastic process ($E$ being a Polish space with the Borel $\\sigma$-algebra $\\mathcal{B}\\left(E\\right)$) equipped ...\n2answers\n970 views\n\n### Markov Chain and Forward and Backward Probabilities with Alice and Bob\n\nSystem Alice and Bob are moving independently from one city to another. There are $d$ cities, the probability of moving to another city (for each individual) is $m$ and each move is equiprobable (...\n0answers\n43 views\n+50\n\n2answers\n508 views\n\n### A Markov process which is not strong Markov process (follow up 2)\n\nIn https://mathoverflow.net/questions/43833/a-markov-process-which-is-not-a-strong-markov-process George Lowther's example: \"Consider the following continuous Markov process $X$, starting from ...\n2answers\n1k views\n\n2answers\n556 views\n\n### Markovian Gaussian stationary process with continuous paths\n\nCould you, please, help me figure out the following problem. We call a stationary Gaussian process $\\xi_t$ (with continuous paths) an Ornstein-Uhlenbeck process if its correlation function \\$\\mathbb{..."
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https://bayport-response.com/math-question-759 | [
"# Quadratic equation word problems solver\n\nIn this blog post, we will show you how to work with Quadratic equation word problems solver. Let's try the best math solver.\n\n## The Best Quadratic equation word problems solver\n\nQuadratic equation word problems solver is a mathematical instrument that assists to solve math equations. Algebra is one of the most difficult subjects for high school students. It can be very confusing, and it often involves memorizing a lot of formulas. The good news is that it doesn’t have to be this way! There are a lot of different ways you can solve algebra problems, and you can learn them all. When you learn another way to solve an algebra problem, you'll be able to see the math behind it. You’ll also understand why algebra works in the first place, which will make it easier to remember later on. By doing this, you'll be able to start solving algebra problems more easily and quickly. This will help you get better grades and make learning math less stressful. So, how do you solve algebra? First off, you want to practice by doing lots of practicing. Once you know how to solve an algebra problem, it will become easier for you to do so in the future. Second, you need to understand the concept behind it. If you don’t understand why something works in the first place, then it's going to be much harder for you to remember how to use that same method in the future. Finally, you need to identify your strengths and weaknesses when it comes to solving algebra problems. This will allow you to focus on what you're good at so that you can get better grades in the future! So,\n\nMake sure the app is appropriate for your child’s age and skill level. A 5-year-old who doesn’t know how to add two numbers together is probably not going to do well with an app designed for 8-year-olds who already know how to do this. 2. Check the quality of the content and features. Does the app offer clear, accurate explanations of each concept? Does it include interactive activities that reinforce what your child has learned? 3. Be sure to let your child try out different options, such as adding up sets or doing calculations in their head before moving on to the computer or paper. This will help them get used to manipulating numbers in their head and practicing strategies such as mental subtraction and counting by fives.\n\nOne of the most important things to look for in a math homework app is a reliable, easy-to-use interface. This will make it easier for you to stay organized and get things done efficiently. There are also plenty of other features that can come in handy, depending on your situation. For example, if you're having trouble with algebra or calculus, it's a good idea to check out an app that offers tutoring and practice problems.\n\nTo solve a right triangle, the Pythagorean theorem is used. This states that if a triangle has a hypotenuse of , the triangles legs are in the ratio . And so by substituting the values into this theorem, we can find out the lengths of the two legs of our triangle: The hypotenuse can also be found by using similar triangles. For example, if we know one leg is and want to find the hypotenuse, we would use the formula: Taking these steps shows that we have solved our right triangle. To see why this works, let’s use an example: We start with a right triangle with legs of and . We know that since the angle between them is 60°. We also know that since the angle between them is 60° and since they both bisect angles. This means that is equal to one-half times . Therefore, . Therefore, . Therefore, . Therefore, . Therefore, , which is what we wanted to find!\n\nThere are a lot of different algebra math solvers out there. Some are more complicated than others, but they all essentially do the same thing: they help you solve algebra problems. There are a lot of different features that these solvers can have, but the most important thing is that they are able to correctly solve the problems you give them. if you're struggling with algebra, then using one of these solvers can be a great way to get the help you need.\n\n## Math checker you can trust\n\nI've been using the app for a very long time and I'm really a fan of the design; I like it simple and fresh and the way it solves equation but, in the meantime, I'm using the version 7.10 because you guys make upper versions work in connection only to be honest that is not cool guys; why would you do that I hope if you could make the app work offline again in the coming updates, I really appreciate your efforts though",
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"Lilliana Bailey\nIt’s very good app helpful in exams, studies, homework’s, classwork’s, explanations, and much more. It also has books which can provide you more knowledge and it clears your concept very well. Just try this app once for your math problems it will surely solve it",
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"Quintessa Martinez"
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https://puzzling.stackexchange.com/questions/103976/matchsticks-fix-the-equation/103977 | [
"# Matchsticks: Fix the Equation\n\nHere's a simple puzzle I just created:\n\nMove just one matchstick to fix the next equation:",
null,
"First move a matchstick like so\n\nThen observe from the other direction\n\nGiving\n\nLI - L = I\na correct equation in Roman numerals.\n\n• Amazing! You were really fast! Just waiting to accept the answer! :) – Pspl Oct 19 at 9:13\n• Yes! Thank you, I've spent some time on it, but can't figure it out. Thank you! – Jingbo the dude Oct 19 at 18:09\n\nAlternatively you can\n\nmove the minus sign match and place it next to the equality sign, turning it into identity: 17 ≡ 17 (seventeen is identical to seventeen).\n\n• That's a good answer. The the spacing between the numbers on the picture makes it not a perfect one :) Good thinking though. – Pspl Oct 19 at 9:20\n• Another way to do the same thing is to use the minus matchstick to add a cross to a 7 so it just says 17=17 but with two different styles of 7 – Todd Wilcox Oct 20 at 5:30\n• @ToddWilcox You could also use that calculator-style 7 with the upper-left vertical line filled in (so it's like a serif). – Darrel Hoffman Oct 20 at 15:59\n\nRequired Equation will be like this:\n\nAttempt 1:\n\n$$1-1=0$$\n\nAttempt 2:\n\nIf 17 can not be 0 then I guess we just need to remove hyphen sign and put it aside, so it will become 17=17\n\nAttempt 3: (though answer is already given and accepted. I am trying another way)\n\nIf we move horizontal line of 7 and put it diagonal onto equal sign then the equation will be like: 1 - 1 ≠ 17\n\n• Good lateral thinking! You simply don't care about the spacing! Eh eh... However the \"perfect\" answer was already been discovered. – Pspl Oct 19 at 9:22\n\nMove the stick to form 17>=17",
null,
"We could also put one matchstick onto another matchstick so it covers it completely, making $$17=17$$."
] | [
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http://en.0086xc.com/news/72.html | [
"CN | EN\n\n#",
null,
"News Center\nHomepage\n/\nSelection and precautions of varistor\n\n# Selection and precautions of varistor\n\n• Categories:Industry trends\n• Author:\n• Origin:\n• Time of issue:2020-04-13 13:38\n• Views:\n\n(Summary description)Before choosing a varistor, you should first understand the following related technical parameters: Nominal voltage refers to the voltage value across the varistor under the specified temperature and DC current. Leakage current refers to the current value flowing in the varistor when the maximum continuous DC voltage is applied under the condition of 25°C. Grade voltage refers to the voltage peak value that appears at both ends of the varistor when it passes 8/20 grade current pulses . The flow rate is the peak current when the specified pulse current (8/20μs) waveform is applied. Surge environment parameters include the maximum surge current Ipm (or the maximum surge voltage Vpm and the surge source impedance Zo), the surge pulse width Tt, the minimum time interval Tm between two adjacent surges, and the preset value of the varistor During the working life, the total number of surge pulses N etc. Generally speaking, varistors are often used in parallel with the protected device or device. Under normal circumstances, the DC or AC voltage across the varistor should be lower than the nominal voltage, even when the power supply fluctuates the worst. It should not be higher than the maximum continuous working voltage selected in the rated value, and the nominal voltage value corresponding to the maximum continuous working voltage value is the selected value. For the application of overvoltage protection, the varistor voltage value should be greater than the voltage value of the actual circuit. Generally, the following formula should be used for selection: VmA=av/bc where: a is the circuit voltage fluctuation coefficient; v is the circuit DC working voltage (effective value when AC); b is the varistor voltage error; c is the aging coefficient of the component; the actual value of VmA calculated in this way is 1.5 times of the DC working voltage , The peak value should be considered in the AC state, so the calculation result should be enlarged by 1.414 times. In addition, you must pay attention to: (1) It must be ensured that the continuous working voltage will not exceed the maximum allowable value when the voltage fluctuation is maximum, otherwise the service life of the varistor will be shortened; (2) When a varistor is used between the power line and the ground, sometimes the voltage between the line and the ground rises due to poor grounding. Therefore, a varistor with a higher nominal voltage than the line-to-line use is usually used. The surge current absorbed by the varistor should be less than the maximum flow rate of the product.\n\n# Selection and precautions of varistor\n\n(Summary description)Before choosing a varistor, you should first understand the following related technical parameters: Nominal voltage refers to the voltage value across the varistor under the specified temperature and DC current. Leakage current refers to the current value flowing in the varistor when the maximum continuous DC voltage is applied under the condition of 25°C. Grade voltage refers to the voltage peak value that appears at both ends of the varistor when it passes 8/20 grade current pulses . The flow rate is the peak current when the specified pulse current (8/20μs) waveform is applied. Surge environment parameters include the maximum surge current Ipm (or the maximum surge voltage Vpm and the surge source impedance Zo), the surge pulse width Tt, the minimum time interval Tm between two adjacent surges, and the preset value of the varistor During the working life, the total number of surge pulses N etc.\n\nGenerally speaking, varistors are often used in parallel with the protected device or device. Under normal circumstances, the DC or AC voltage across the varistor should be lower than the nominal voltage, even when the power supply fluctuates the worst. It should not be higher than the maximum continuous working voltage selected in the rated value, and the nominal voltage value corresponding to the maximum continuous working voltage value is the selected value. For the application of overvoltage protection, the varistor voltage value should be greater than the voltage value of the actual circuit. Generally, the following formula should be used for selection: VmA=av/bc where:\n\na is the circuit voltage fluctuation coefficient; v is the circuit DC working voltage (effective value when AC); b is the varistor voltage error; c is the aging coefficient of the component; the actual value of VmA calculated in this way is 1.5 times of the DC working voltage , The peak value should be considered in the AC state, so the calculation result should be enlarged by 1.414 times.\n\nIn addition, you must pay attention to:\n\n(1) It must be ensured that the continuous working voltage will not exceed the maximum allowable value when the voltage fluctuation is maximum, otherwise the service life of the varistor will be shortened;\n\n(2) When a varistor is used between the power line and the ground, sometimes the voltage between the line and the ground rises due to poor grounding. Therefore, a varistor with a higher nominal voltage than the line-to-line use is usually used.\n\nThe surge current absorbed by the varistor should be less than the maximum flow rate of the product.\n\n• Categories:Industry trends\n• Author:\n• Origin:\n• Time of issue:2020-04-13 13:38\n• Views:\nInformation\nBefore choosing a varistor, you should first understand the following related technical parameters: Nominal voltage refers to the voltage value across the varistor under the specified temperature and DC current. Leakage current refers to the current value flowing in the varistor when the maximum continuous DC voltage is applied under the condition of 25°C. Grade voltage refers to the voltage peak value that appears at both ends of the varistor when it passes 8/20 grade current pulses . The flow rate is the peak current when the specified pulse current (8/20μs) waveform is applied. Surge environment parameters include the maximum surge current Ipm (or the maximum surge voltage Vpm and the surge source impedance Zo), the surge pulse width Tt, the minimum time interval Tm between two adjacent surges, and the preset value of the varistor During the working life, the total number of surge pulses N etc.\n\nGenerally speaking, varistors are often used in parallel with the protected device or device. Under normal circumstances, the DC or AC voltage across the varistor should be lower than the nominal voltage, even when the power supply fluctuates the worst. It should not be higher than the maximum continuous working voltage selected in the rated value, and the nominal voltage value corresponding to the maximum continuous working voltage value is the selected value. For the application of overvoltage protection, the varistor voltage value should be greater than the voltage value of the actual circuit. Generally, the following formula should be used for selection: VmA=av/bc where:\n\na is the circuit voltage fluctuation coefficient; v is the circuit DC working voltage (effective value when AC); b is the varistor voltage error; c is the aging coefficient of the component; the actual value of VmA calculated in this way is 1.5 times of the DC working voltage , The peak value should be considered in the AC state, so the calculation result should be enlarged by 1.414 times.\n\nIn addition, you must pay attention to:\n\n(1) It must be ensured that the continuous working voltage will not exceed the maximum allowable value when the voltage fluctuation is maximum, otherwise the service life of the varistor will be shortened;\n\n(2) When a varistor is used between the power line and the ground, sometimes the voltage between the line and the ground rises due to poor grounding. Therefore, a varistor with a higher nominal voltage than the line-to-line use is usually used.\n\nThe surge current absorbed by the varistor should be less than the maximum flow rate of the product.\n\nScan the QR code to read on your phone\n\nNext: None\nNext: None\nOnline Service\n-\n\nCustomer service hotline :\n\n86-754-88813426",
null,
"WeChat",
null,
"Mobile website"
] | [
null,
"http://en.0086xc.com/img/nclogo1.png",
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https://encyclopediaofmath.org/wiki/Equation | [
"# Equation\n\nAn analytical form of the problem of investigating which values of variables in functions give equal results. The arguments on which the functions depend are usually called the unknowns, and the values of the unknowns for which the values of the function are equal are called the solutions of the equation; for such values of the unknowns one also says that they satisfy the given equation. A solution $x_0$ of an equation of the form $f(x)=0$ is also called a root of $f$.\n\nThe set of solutions of a given equation depends on the domain $M$ of values admitted for the unknown. An equation need not have a solution in $M$; it is then said to be unsolvable in $M$. If an equation is solvable, it may have one or several, or even an infinite set of, solutions. For example, the equation $x^4-4=0$ is unsolvable in the domain of rational numbers, but it has two solutions: $x_1=\\sqrt2$, $x_2=-\\sqrt2$ in the domain of real numbers and four solutions $x_1=\\sqrt2$, $x_2=-\\sqrt2$, $x_3=i\\sqrt2$, $x_4=-i\\sqrt2$ in the domain of complex numbers. The equation $\\sin x=0$ has infinitely many solutions: $x_k=k\\pi$, $k=0,\\pm1,\\pm2,\\ldots,$ in the domain of real numbers.\n\nIf an equation has as solution all numbers of a domain $M$, then it is called an identity on $M$.\n\nA set of equations for which it is required to find values of the unknowns satisfying all equations simultaneously is called a system of equations; a set of values of the unknowns satisfying simultaneously all equations of the system is called a solution of the system. Two systems of equations (or two equations) are called equivalent if each solution of one system (equation) is a solution of the other system (equation), and conversely, where both systems (equations) are considered in one and the same domain.\n\nThe process of finding a solution of an equation usually consists in modifying it to an equivalent equation. In certain cases the given equation is reduced to another one for which the solution set is larger than that for the given equation. Therefore, if one performs this operation in solving an equation, it is possible for extra solutions to appear, and hence all solutions obtained from the transformed equation must be substituted in the original equation.\n\nThe most thoroughly studied are the algebraic equations (cf. Algebraic equation). Solving them was a principal task of algebra in the 16th century and 17th century. If $f(x)$ is a transcendental function, then the equation $f(x)=0$ is said to be transcendental; depending on the form of $f(x)$, it is called a trigonometric, logarithmic or exponential equation if these are the transcendental functions involved.\n\nFor the practical solution of equations one usually applies different approximate methods (cf., e.g., Linear algebra, numerical methods in).\n\nAmong systems of equations the simplest are systems of linear equations (cf. Linear equation). The solution of a system of equations (not necessarily linear) is reduced, generally speaking, to the solution of a single equation with the help of so-called elimination of the unknowns (cf. Elimination theory).\n\nIn number theory one considers equations over the integers, the study of the solutions of which forms the subject of the theory of Diophantine equations.\n\nIn the general case an equation is a list of problems the solution of which consists in finding elements $a$ from a certain set $A$ such that $F(a)=\\Phi(a)$, where $F$ and $\\Phi$ are given mappings of $A$ into another set $B$. If $A$ and $B$ are sets of numbers, there arise equations of the form considered above. If $A$ and $B$ are point sets in a higher-dimensional space, one obtains a system of equations. If $A$ and $B$ are sets of functions, then, depending on the character of the mapping, one obtains ordinary differential equations (cf. Differential equation, ordinary), partial differential equations (cf. Differential equation, partial), integral equations (cf. Integral equation), and other forms of equations.\n\nHow to Cite This Entry:\nEquation. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Equation&oldid=32613\nThis article was adapted from an original article by Material from the article \"Equation\" in BSE-3 (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article"
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https://www.projecteuclid.org/euclid.ijm/1255985222 | [
"Illinois Journal of Mathematics\n\nDense subsets of Banach $\\ast$-algebras\n\nBertram Yood\n\nAbstract\n\nSome subsets of a Banach ${}^{\\ast}$-algebra $A$ are shown to be dense. In the special case of the algebra of $L(H)$ of all bounded linear operators on a Hilbert space $H$, the set of all $T$ in $L(H)$ for which $T^{n}$ is quasi-normal for no positive integers $n$ is dense in $L(H)$.\n\nArticle information\n\nSource\nIllinois J. Math., Volume 43, Issue 2 (1999), 403-409.\n\nDates\nFirst available in Project Euclid: 19 October 2009\n\nPermanent link to this document\nhttps://projecteuclid.org/euclid.ijm/1255985222\n\nDigital Object Identifier\ndoi:10.1215/ijm/1255985222\n\nMathematical Reviews number (MathSciNet)\nMR1703195\n\nZentralblatt MATH identifier\n0940.46032\n\nSubjects\nPrimary: 46K05: General theory of topological algebras with involution\n\nCitation\n\nYood, Bertram. Dense subsets of Banach $\\ast$-algebras. Illinois J. Math. 43 (1999), no. 2, 403--409. doi:10.1215/ijm/1255985222. https://projecteuclid.org/euclid.ijm/1255985222"
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http://openvibe.inria.fr/forum/viewtopic.php?f=13&t=10048&p=16150&sid=7f7fd5daf9b49c40d64eb49715e7eec5 | [
"## Errors in Changing Matlab scripting file for tutorial 2 FFT\n\nConcerning processing components: filters, file load/save, visualizations, communication ...\nNagarjunV\nPosts: 9\nJoined: Wed Feb 06, 2019 11:53 am\n\n### Errors in Changing Matlab scripting file for tutorial 2 FFT\n\nHello,\n\nI am using openvibe tutorial 2 FFT for data acquisition and signal processing. I have done small modifications to the code \"tuto2_FFT_filter_Uninitialize.m\", such that I want to print amplitude of the peak at 8hz frequency from the mean plot (which occurs after stopping the acquisition scenario) into a text file. I am highlighting the colour as shown below which represents the changes I have made. When I play the scenario. I get an error that fprintf is an undefined variable. Can anyone please suggest me and help me if any odig=fication is required for this code to work?\n\nfunction box_out = tuto2_FFT_filter_Uninitialize(box_in)\ndisp('Uninitializing the box...')\nsyms j k\nFs = box_in.inputs{1}.header.sampling_rate; % Sampling frequency\nL = box_in.inputs{1}.header.nb_samples_per_buffer; % Length of signal\n\nNFFT = 2^nextpow2(L);\nf = Fs/2*linspace(0,1,NFFT/2+1);\n\nbox_in.user_data.mean_fft_matrix = box_in.user_data.mean_fft_matrix / box_in.user_data.nb_matrix_processed;\n\n%% we close the previous figure window and plot the mean FFT between 5 and 50Hz.\nclose(gcf);\nplot_range_fmin = box_in.settings(3).value;\nplot_range_fmax = box_in.settings(4).value;\nplot(f(plot_range_fmin*2:plot_range_fmax*2),2*abs(box_in.user_data.mean_fft_matrix(plot_range_fmin*2:plot_range_fmax*2)))\nj = f(plot_range_fmin*2:plot_range_fmax*2);\nk = 2*abs(box_in.user_data.mean_fft_matrix(plot_range_fmin*2:plot_range_fmax*2));\nk = [j;k];\nfileID = fopen('exptable.txt','w');\nfprintf(fileID,'%f %f\\n',k);\nfclose(fileID);\n\ntitle('MEAN Single-Sided Amplitude Spectrum of the corrupted signal (channel 1)')\nxlabel('Frequency (Hz)')\nylabel('Amplitude')\n\n% We pause the execution for 10 seconds (to be able to see the figure before the scenario is stopped)\npause(10);\n\nbox_out = box_in;\nend\n\nThibaut\nPosts: 226\nJoined: Wed Oct 31, 2018 9:14 am\n\n### Re: Errors in Changing Matlab scripting file for tutorial 2\n\nHi,\n\nCode: Select all\n\n``fprintf(fileID,'%f %f\\n',k);``\nyou have 2 %f and only one argument.\nThibaut\n\nNagarjunV\nPosts: 9\nJoined: Wed Feb 06, 2019 11:53 am\n\n### Re: Errors in Changing Matlab scripting file for tutorial 2\n\nthank you for your reply, but that will not be a problem.\nAccording to MATLAB, the 2 f% are for printing both J and K. For y=[g,h] if we give command f% f%, it prints both g and h side by side.\n\nThibaut\nPosts: 226\nJoined: Wed Oct 31, 2018 9:14 am\n\n### Re: Errors in Changing Matlab scripting file for tutorial 2\n\nMaybe try the function disp()"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.73962617,"math_prob":0.60973203,"size":2468,"snap":"2021-43-2021-49","text_gpt3_token_len":664,"char_repetition_ratio":0.1176948,"word_repetition_ratio":0.22123894,"special_character_ratio":0.26539707,"punctuation_ratio":0.1659919,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99239427,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-23T07:43:45Z\",\"WARC-Record-ID\":\"<urn:uuid:e8774b47-e106-41d1-ac15-023dbdea16a7>\",\"Content-Length\":\"32274\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1415feee-7794-4e85-9021-61d08d194f55>\",\"WARC-Concurrent-To\":\"<urn:uuid:06fda48d-4c45-42f1-813f-3361a9d6a8e5>\",\"WARC-IP-Address\":\"131.254.254.107\",\"WARC-Target-URI\":\"http://openvibe.inria.fr/forum/viewtopic.php?f=13&t=10048&p=16150&sid=7f7fd5daf9b49c40d64eb49715e7eec5\",\"WARC-Payload-Digest\":\"sha1:EZEIZUT2AMH6QC7RKWEANVSNN47RS2MU\",\"WARC-Block-Digest\":\"sha1:6N3IL6DEFBWE7NL43ZF5XOERZ2PBCQAR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585653.49_warc_CC-MAIN-20211023064718-20211023094718-00670.warc.gz\"}"} |
https://www.tutorialspoint.com/python-pandas-format-the-period-object-and-display-the-time-with-24-hour-format | [
"# Python Pandas - Format the Period object and display the Time with 24-Hour format\n\nTo format the Period object, use the period.strftime() method and to display the time with 24-Hour format, set the parameter as %H.\n\nAt first, import the required libraries −\n\nimport pandas as pd\n\nThe pandas.Period represents a period of time. Creating a Period object\n\nperiod = pd.Period(freq=\"S\", year = 2021, month = 9, day = 18, hour = 17, minute = 20, second = 45)\n\nDisplay the Period object\n\nprint(\"Period...\\n\", period)\n\n\nDisplay the result. Here, Time is displayed with 24-Hour format i.e. [00,23]\n\nprint(\"\\nString representation (24-Hour format)...\\n\", period.strftime('%H'))\n\n## Example\n\nFollowing is the code\n\nimport pandas as pd\n\n# The pandas.Period represents a period of time\n# Creating a Period object\nperiod = pd.Period(freq=\"S\", year = 2021, month = 9, day = 18, hour = 17, minute = 20, second = 45)\n\n# display the Period object\nprint(\"Period...\\n\", period)\n\n# display the result\n# Here, Time is displayed with 24-Hour format i.e. [00,23]\nprint(\"\\nString representation (24-Hour format)...\\n\", period.strftime('%H'))\n\n## Output\n\nThis will produce the following code\n\nPeriod...\n2021-09-18 17:20:45\n\nString representation (24-Hour format)...\n17"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5932946,"math_prob":0.90904033,"size":3125,"snap":"2023-14-2023-23","text_gpt3_token_len":787,"char_repetition_ratio":0.23454021,"word_repetition_ratio":0.36802974,"special_character_ratio":0.28192,"punctuation_ratio":0.106343284,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9923558,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-30T03:08:09Z\",\"WARC-Record-ID\":\"<urn:uuid:f0047427-23cf-4170-98cd-1fb50fef3d08>\",\"Content-Length\":\"38799\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1fb93fa7-8752-4c19-99ca-6b5d7dee4fed>\",\"WARC-Concurrent-To\":\"<urn:uuid:ca54da9e-1ada-457f-9e75-785bcad9221d>\",\"WARC-IP-Address\":\"192.229.210.176\",\"WARC-Target-URI\":\"https://www.tutorialspoint.com/python-pandas-format-the-period-object-and-display-the-time-with-24-hour-format\",\"WARC-Payload-Digest\":\"sha1:VX6R6CYJTEQTPIYDXEWFVOSNGOAYWMRG\",\"WARC-Block-Digest\":\"sha1:ZZKLHRP7K7CF7MY2OVZKTGLPLS36UZP5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949093.14_warc_CC-MAIN-20230330004340-20230330034340-00336.warc.gz\"}"} |
https://qqtim.club/2017/10/01/master-horse-training-standard/index.html | [
"ReZero's Utopia.\n\n# Master horse training standard\n\nWord count: 295Reading time: 1 min\n2017/10/01 Share",
null,
"## Persona\n\n1. Search: BFS, DFS, IDA*, Hash\n\n2. Thoughts design: Greedy, binary, Tree point, Merge sort, Ruler method\n\n3. Math: extended Euclidean， prime, Eulers function, Fast matrix power, Gauss elimination\n\n4. Data structure: STL(priority queue, map, multiset), Monotone structure(queue, stack, sliding window), disjoint set\n\n5. Dynamic planning： classic problem， Tree DP, status compress DP\n\n6. Math: Game(bash, nim, Fibonacci, Wythoff)， derangement， inverse element, combination of recursive\n\n7. Graphy theory: minimal spanning tree， maximum matching， shortest path(floyd, dijkstra, bellman-ford, SPFA)\n\n8. Computation geometry: line segment intersection, Cross product， Polygon area\n\n9. Data structure: segment tree, tree attay, KMP, Trie\n\n10. Graphy theory: network flow(max-flow, minimum cost flow), strongly connected components\n\n## Team Necessary\n\n1. Data structure: splay tree, AC automantic machine, suffix array, Heavy-light Decomposition， balanced tree(treap), chairman tree, block, RMQ, LCA\n\n2. Dynamic planing: digit DP, interval DP, data structure optimize DP(gradient optimize, monotonic queue, quadrilateral inequality, binary system)\n\n3. Computation geometry： convex closure, Half-plane Intersection, area of circles\n\n4. Math: SG function, matrix, Fermat’s little theorem， Chinese remainder theorem, pick theorem, probability and expectation(probability DP), FFT, Simpson, displace(polya theorem, Burnside lemma), combinatorial mathematics(interval counting, Catalan number， Stirling number, counting order)\n\n5. Graphy theory: connected component(cut point, bridge, Biconnect), 2SAT, Differential Constrain, minimal cut, Eulerian circuit\n\n6. Graphy theory: Network flow(upper and lower bound, max density subgraph, largest closed subgraph), most perfect matching\n\n## Team Stronger\n\n1. Sundry: simulated annealing, plug DP, DLX, divide and conquer tree\n\n2. Data sturcture: dynamic tree, kd tree, suffix automantic machine, Mo team, Persistent data structure\n\n3. Math: generation function, liner programming, Mobius inversion, fast number theoretic transform\n\n4. Graphy Theory: stable matching, the Kth short path, Second-best Minmum Spanning Tree, maximum clique, hamiltonian circuit, Minimum spanning tree\n\n5. Computation geometry: Rotating calipers, 3D convex hull, Minimum Circumscribed Circle, Affine transformation matrix， motion planning"
] | [
null,
"https://qqtim.club/assets/loading.svg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.59875435,"math_prob":0.9254656,"size":2266,"snap":"2020-45-2020-50","text_gpt3_token_len":513,"char_repetition_ratio":0.100353666,"word_repetition_ratio":0.0,"special_character_ratio":0.18667255,"punctuation_ratio":0.2550505,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9568089,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-27T11:47:18Z\",\"WARC-Record-ID\":\"<urn:uuid:588b9b15-70ae-46d4-812d-b228ccb6ef27>\",\"Content-Length\":\"83069\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:969fef9a-e865-4aec-97d3-2a82edc52eea>\",\"WARC-Concurrent-To\":\"<urn:uuid:97fc4358-73bb-42fd-a0fe-9c07d266b58d>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"https://qqtim.club/2017/10/01/master-horse-training-standard/index.html\",\"WARC-Payload-Digest\":\"sha1:H476LBVGQFYRBSX45ZKGV2TGREW6LTE5\",\"WARC-Block-Digest\":\"sha1:3MDJY2EHRPU6K5EG2CFSVYMRFUA5GPRS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107894175.55_warc_CC-MAIN-20201027111346-20201027141346-00411.warc.gz\"}"} |
https://encyclopediaofmath.org/wiki/Analytic_sheaf | [
"# Analytic sheaf\n\nA sheaf $F$ on an analytic space $X$ such that for any point $x \\in X$ the set $F _ {x}$ is a module over the ring ${\\mathcal O} _ {x}$ of germs of holomorphic functions at the point $x$, and such that the mapping $(f , \\alpha ) \\rightarrow f \\alpha$, defined on the set of pairs $( f, \\alpha )$ where $f \\in {\\mathcal O} _ {x}$, $\\alpha \\in F _ {x}$, is a continuous mapping of ${\\mathcal O} \\times F$ into $F$ for $x \\in X$."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.628081,"math_prob":1.0000087,"size":1164,"snap":"2022-40-2023-06","text_gpt3_token_len":354,"char_repetition_ratio":0.118965514,"word_repetition_ratio":0.03141361,"special_character_ratio":0.30756015,"punctuation_ratio":0.16744186,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000077,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-28T16:57:51Z\",\"WARC-Record-ID\":\"<urn:uuid:5cd20ac6-9f2f-4cf0-aa7d-101e2689a2a7>\",\"Content-Length\":\"14625\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:00e56916-a208-43c8-b815-d27d82505ffc>\",\"WARC-Concurrent-To\":\"<urn:uuid:114d43a0-fc22-4744-a049-732400d456eb>\",\"WARC-IP-Address\":\"34.96.94.55\",\"WARC-Target-URI\":\"https://encyclopediaofmath.org/wiki/Analytic_sheaf\",\"WARC-Payload-Digest\":\"sha1:RWCAX4LUTEMFAPZDC6LVFR4RUJU76D7M\",\"WARC-Block-Digest\":\"sha1:GMZTATDOHN3P67RZMPU4HEQEEF3QL32N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335257.60_warc_CC-MAIN-20220928145118-20220928175118-00504.warc.gz\"}"} |
https://www.nag.com/numeric/py/nagdoc_latest/naginterfaces.library.mv.rot_procrustes.html | [
"# naginterfaces.library.mv.rot_procrustes¶\n\nnaginterfaces.library.mv.rot_procrustes(stand, pscale, x, y)[source]\n\nrot_procrustes computes Procrustes rotations in which an orthogonal rotation is found so that a transformed matrix best matches a target matrix.\n\nFor full information please refer to the NAG Library document for g03bc\n\nhttps://www.nag.com/numeric/nl/nagdoc_29/flhtml/g03/g03bcf.html\n\nParameters\nstandstr, length 1\n\nIndicates if translation/normalization is required.\n\nThere is no translation or normalization.\n\nThere is translation to the origin (i.e., to zero).\n\nThere is translation to origin and then to the centroid after rotation.\n\nThere is unit normalization.\n\nThere is translation and normalization (i.e., there is standardization).\n\nThere is translation and normalization to scale, then translation to the centroid after rotation (i.e., they are matched).\n\npscalestr, length 1\n\nIndicates if least squares scaling is to be applied after rotation.\n\nScaling is applied.\n\nNo scaling is applied.\n\nxfloat, array-like, shape\n\n, the matrix to be rotated.\n\nyfloat, array-like, shape\n\nThe target matrix, .\n\nReturns\nxfloat, ndarray, shape\n\nIf , will be unchanged.\n\nIf , , or , will be translated to have zero column means.\n\nIf or , will be scaled to have unit sum of squares.\n\nIf , will be scaled to have the same sum of squares as .\n\nyfloat, ndarray, shape\n\nIf , will be unchanged.\n\nIf or , will be translated to have zero column means.\n\nIf or , will be scaled to have unit sum of squares.\n\nIf or , will be translated and then after rotation translated back.\n\nThe output should be the same as the input except for rounding errors.\n\nyhatfloat, ndarray, shape\n\nThe fitted matrix, .\n\nrfloat, ndarray, shape\n\nThe matrix of rotations, , see Further Comments.\n\nalphafloat\n\nIf the scaling factor, ; otherwise is not set.\n\nThe residual sum of squares.\n\nresfloat, ndarray, shape\n\nThe residuals, , for .\n\nRaises\nNagValueError\n(errno )\n\nOn entry, .\n\nConstraint: or .\n\n(errno )\n\nOn entry, .\n\nConstraint: , , , , or .\n\n(errno )\n\nOn entry, .\n\nConstraint: .\n\n(errno )\n\nOn entry, and .\n\nConstraint: .\n\n(errno )\n\nOnly one distinct point (centred at zero) in array.\n\n(errno )\n\nOnly one distinct point (centred at zero) in array.\n\n(errno )\n\ncontains only zero-points when least squares scaling is applied.\n\n(errno )\n\nThe singular value decomposition has failed to converge.\n\nNotes\n\nLet and be matrices. They can be considered as representing sets of points in an -dimensional space. The matrix may be a matrix of loadings from say factor or canonical variate analysis, and the matrix may be a postulated pattern matrix or the loadings from a different sample. The problem is to relate the two sets of points without disturbing the relationships between the points in each set. This can be achieved by translating, rotating and scaling the sets of points. The matrix is considered as the target matrix and the matrix is rotated to match that matrix.\n\nFirst the two sets of points are translated so that their centroids are at the origin to give and , i.e., the matrices will have zero column means. Then the rotation of the translated matrix which minimizes the sum of squared distances between corresponding points in the two sets is found. This is computed from the singular value decomposition of the matrix:\n\nwhere and are orthogonal matrices and is a diagonal matrix. The matrix of rotations, , is computed as:\n\nAfter rotation, a scaling or dilation factor, , may be estimated by least squares. Thus, the final set of points that best match is given by:\n\nBefore rotation, both sets of points may be normalized to have unit sums of squares or the matrix may be normalized to have the same sum of squares as the matrix. After rotation, the results may be translated to the original centroid.\n\nThe th residual, , is given by the distance between the point given in the th row of and the point given in the th row of . The residual sum of squares is also computed.\n\nReferences\n\nKrzanowski, W J, 1990, Principles of Multivariate Analysis, Oxford University Press\n\nLawley, D N and Maxwell, A E, 1971, Factor Analysis as a Statistical Method, (2nd Edition), Butterworths"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.84793913,"math_prob":0.9804832,"size":4314,"snap":"2023-14-2023-23","text_gpt3_token_len":987,"char_repetition_ratio":0.13573086,"word_repetition_ratio":0.15934066,"special_character_ratio":0.22693557,"punctuation_ratio":0.1826809,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9928441,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-09T02:01:31Z\",\"WARC-Record-ID\":\"<urn:uuid:44aa3487-5d67-46e3-922b-a004993e5ff4>\",\"Content-Length\":\"123007\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6573b36e-3b01-4780-8e4b-d345704d8854>\",\"WARC-Concurrent-To\":\"<urn:uuid:c2049c0f-c65c-4494-ad96-11c9013f9ba7>\",\"WARC-IP-Address\":\"78.129.168.4\",\"WARC-Target-URI\":\"https://www.nag.com/numeric/py/nagdoc_latest/naginterfaces.library.mv.rot_procrustes.html\",\"WARC-Payload-Digest\":\"sha1:QEGNY3J5VBZZ3RQAGQ7B4COF3CYQFDP2\",\"WARC-Block-Digest\":\"sha1:SR3TIUVIXRHXNSFOOCVUC6X433AG5XM3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224655244.74_warc_CC-MAIN-20230609000217-20230609030217-00665.warc.gz\"}"} |
https://www.pryor.com/blog/tag/excel-formula-syntax/ | [
"# Tag Archives: Excel Formula Syntax\n\n### Excel Formula Syntax – The Language of Formulas and Functions",
null,
"Excel makes it a simple task to perform mathematical operations. Using Excel formula syntax, you can calculate and analyze data in your worksheet. As a reminder: Formulas are equations that combine values and cell references with operators to calculate a result. Functions are prebuilt formulas that can be quickly fed values without the need to write the…"
] | [
null,
"https://pryormediacdn.azureedge.net/blog/generic-blog-post.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8387343,"math_prob":0.984683,"size":428,"snap":"2021-31-2021-39","text_gpt3_token_len":79,"char_repetition_ratio":0.12028302,"word_repetition_ratio":0.0,"special_character_ratio":0.17523365,"punctuation_ratio":0.06849315,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99605477,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-27T00:42:48Z\",\"WARC-Record-ID\":\"<urn:uuid:b3d9c29c-49d3-45c1-ba06-4c5c4e0d3741>\",\"Content-Length\":\"146392\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dbbd85a2-ae8e-47af-acf7-1a06d584a7ce>\",\"WARC-Concurrent-To\":\"<urn:uuid:498a910b-5566-44ae-a7e4-58c9cefae57d>\",\"WARC-IP-Address\":\"52.165.130.142\",\"WARC-Target-URI\":\"https://www.pryor.com/blog/tag/excel-formula-syntax/\",\"WARC-Payload-Digest\":\"sha1:LLBIM5KKRWBQH54BIYD4Q3NKI7JSGTQT\",\"WARC-Block-Digest\":\"sha1:VROWU4LH26SZSVTB4QTIUPKXOWQ7E5QC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780058222.43_warc_CC-MAIN-20210926235727-20210927025727-00454.warc.gz\"}"} |
https://direct.physicsclassroom.com/Teacher-Toolkits/Vibrational-Motion/Vibrational-Motion-Complete-ToolKit | [
"## Vibrational Motion - Complete Toolkit\n\n### Objectives\n\n1. To identify several examples of vibrating objects and to use terms such as equilibrium position, restoring force, fixed end, and damping to describe their motion.\n2. To describe an object undergoing periodic motion using terms such as sinusoidal, cycle, period, amplitude, and damping and to relate these terms to the position-time graph.\n3. To define period, frequency, and amplitude and to determine their values from verbal and graphical descriptions of a vibrating object's motion.\n4. To describe and explain the motion of a simple pendulum using such representations as a free-body diagrams, position-time graphs, velocity-time graphs, energy tables, equations for period, and terms such as position, velocity, and acceleration.\n5. To describe and explain the motion of a vibrating mass on a spring using such representations as a free-body diagrams, position-time graphs, velocity-time graphs, energy tables, energy bar charts, equations for period, and terms such as position, velocity, and acceleration.\n\n### Interactive Simulations\n\n1.",
null,
"Physics Interactive: Mass on a Spring\n\nThis simulation from The Physics Classroom's Physics Interactives section animates the vibrational motion of a mass on a spring. The position-time and velocity-time graph of the motion are represented in real-time. Bar charts showing the kinetic energy, gravitational potential energy, and elastic potential energy are also shown in real-time. Users can alter the mass that is hung on the spring, the stiffness of the spring, and the amount of damping. Two springs are included for conducting side-by-side comparative studies. This variable-rich environment allows students to explore a variety of relationships related to vibrating masses on springs. The Physics Classroom has prepred a ready-to-use exercise that focuses on the question of what variables affect the period and the frequency of the vibrating mass.\n\n2.",
null,
"PhET: Hooke’s Law Simulation\n\nAs 17th-century physicist Robert Hook determined, “As the extension, so the force.” This newer HTML5 simulation lets students stretch and compress springs to explore the relationships among force, spring constant, displacement, and potential energy. It will help them gain insight into the meaning of “restoring force” an area of documented student misconception. It also promotes understanding of the predictable mathematical relationships that underlie Hooke’s Law. Teachers: The simulation can be set up as a spring system in either series or parallel. In addition, click the “Energy” tab to explore how potential energy stored in the spring changes with spring constant (k) and displacement.\n\n3.",
null,
"PhET: Pendulum Lab\n\nThis simulation displays one or two pendulums to explore how the period of a simple pendulum depends on the length of the string, mass of the pendulum bob, and amplitude of the swing. Use the Photogate timer to easily measure the period! You can vary the friction or jump to Planet X to explore the effect of changing gravity on a pendulum. Teachers: You will want to check out the PhET “Gold Star” teacher-submitted activities for great lesson plans to accompany this simulation.\n\n4.",
null,
"Hooke’s Law Digital Lab\n\nThis two-hour digital lab for high school physics was created specifically to accompany the PhET simulation “Masses and Springs”. In the first lesson, students explore how displacement of a spring is mathematically related to the load applied to it. In the next day’s exploration, learners analyze the energy of a mass oscillating on a spring by observing distribution and transfer of kinetic, potential, and gravitational potential energy. Turn-Key Resource: includes explicit directions for use of the simulation, problem sets, rubric, and answer key.\n\n### Video and Animations\n\n1.",
null,
"Direct Measurement Videos: SHM with Motion Graphs\n\nOnce again, veteran HS physics teacher Peter Bohacek has created a great DVM (Direct Measurement Video) to support conceptual understanding of the kinematics of simple harmonic motion (SHM). This video shows a low-friction glider moving on an air track. Springs on each side apply forces to the glider that produce SHM. Graphs of position, velocity, and acceleration vs. time are simultaneously displayed.\n\n2.",
null,
"PhysClips: Simple Harmonic Motion\n\nThis animation-based tutorial was recently rewritten to HTML. It would be a good choice for students with disabilities or reading difficulties – it presents information in a non-textual way, but without sacrificing rigor. The animations are nicely constructed, blending video and diagrams, with narration by a physics instructor with a lively Australian accent! This module takes the learner through a 4-part learning cycle: 1) Overview of simple harmonic motion, 2) Angular velocity, amplitude, and phase, 3) displacement, velocity, and acceleration in SHM, and 4) Acceleration and vibration. You’ll also find links to content support for teachers, chladni patterns, and phasor addition.\n\n3.",
null,
"Direct Measurement Video: Spring Force\n\nHere you’ll find seven high-resolution videos that allow students to use digital rulers and frame-counters to analyze applied force on springs and make precise measurements of quantities such as position and time. This set of videos, produced by HS teacher Peter Bohacek, allows students to measure the force and elongation of seven steel springs. Their data can be used to find the spring force constant of each spring. The resource can also be used to demonstrate Hooke’s Law.\n\n4.",
null,
"Veritasium: When Is a Bungee Jumper’s Acceleration Max?\n\nAt what point in a bungee jump is acceleration the greatest? Physics education researcher and YouTube icon Derek Muller brings us another cool “think problem” that lets you integrate concepts of kinematics, gravitational acceleration, spring tension, and the restoring force. The 1-minute video offers students 5 choices – Acceleration is greatest at A) Immediately after leaping off the platform, B) When the bungee cord becomes taut, C) At the fastest point, D) At the very bottom of the jump, or E) On the rebound. Teachers: Most students erroneously pick “C”, confusing velocity with acceleration. Each choice links to a short video explaining why it was correct or incorrect. Link to full video of Dr. Muller’s jump: Full Bungee Jump Video\n\n5.",
null,
"Circus Physics: Pendulum Motion\n\nThis video-based resource examines factors that affect the amplitude and period of a pendulum. It provides a highly visual way to explore pendulum motion as a trapeze artist swings on a bar/rope system. Watch what happens to the pendulum period as her center of mass changes when she sits on the bar or moves to the rope below. The accompanying activity guide introduces the math associated with pendulum motion. (Includes a Teacher’s Guide with discussion questions and classroom activities.)\n\n### Labs and Investigations\n\n1. The Physics Classroom, The Laboratory, A Wiggle in Time\n\nStudents observe the motion of a mass on a spring using a motion detector and describe the motion with words, graphs, and numbers.\n\n2. The Physics Classroom, The Laboratory, Period of a Pendulum\n\nStudents investigate the variables that affect the period of a pendulum and determine a mathematical equation that relates period to the variables that affect it.\n\n### Problem-Based Learning Activity: Real-World Applications\n\n1.",
null,
"Problem-Based Learning: Bungee Jumping\n\nThis activity for introductory physics presents a rough design for a bungee jump from a 20-meter tower. Students will work cooperatively to figure out the parameters for a safe jump. While some information is given, learners are expected to research certain aspects, such as the medically-recommended maximum acceleration for an untrained jumper. In keeping with the PBL method, students sift through information to separate useful from irrelevant data, locate missing information on their own, and the apply physics in finding solutions (including determining the spring constant of the bungee cord). Note to Teachers: The Student Guide is freely accessible. Access to the Teacher’s Guide with answer key is available by contacting the authors.\n\n### Interactive Digital Homework Problems\n\n1.",
null,
"Mass on a Vertical Spring Problem\n\nIf you haven’t yet seen Gary Gladding’s Interactive Examples, get ready to be impressed. Each problem consists of a conceptual exploration, strategic analysis that encourages critical thinking, and explicit help with setting up the calculations. In this interactive problem, a mass hangs from a vertical spring. If the spring is stretched and then released, what is the speed of the block when it returns to its original position for the first time? Students will use the Conservation of Mechanical Energy method to solve the problem. The author anticipates conceptual roadblocks and helps students recognize when and how to use energy conservation principles to solve physics problems.\n\n2.",
null,
"Block and Spring SHM Problem\n\nThis resource presents a similar, but not identical system to the previous problem. A block is attached to a massless spring on a friction-free surface. Students are given the initial velocity and distance from equilibrium. At what time will the block next pass through the x = 0 point? This problem was designed to help learners make the connection between the oscillation of a mass on a spring and the sinusoidal nature of simple harmonic motion. It provides carefully sequenced “help” support that includes free-body diagram, Displacement vs. Time graphs depicting SHM, and explicit help in using the Work-Kinetic Energy Theorem to solve the problem.\n\n### Classroom Learning Module: Understanding Periodic Motion\n\n1. TeachEngineering: Into the Swing of Things\n\nAfter watching a 1940 clip of the Galloping Gertie bridge collapse and a teacher demo with a simple pendulum, student groups do Internet research to identify examples of harmonic or periodic motion. Their task is to answer guided questions about factors that affect pendulum motion; basic properties of periodic motion; and how engineers exploit periodic motion to gain a mechanical advantage. This lesson includes background information, content support for teachers, vocabulary definitions, and both pre-lesson and post-lesson assessment ideas.\n\n2.",
null,
"TeachEngineering: Android Pendulums\n\nHere you’ll find the hands-on activity that accompanies the first lesson. Students will use Android smartphones or tablets as the bobs of a simple pendulum system that hangs from the ceiling. Why Android devices? Because they can be loaded with the AccelDataCapture app that was developed by the author of this module. The lesson explains how to alter one variable while keeping all other parameters constant. Students will identify the independent/dependent variables, collect data, and use the simple pendulum equation. Detailed set-up instructions, data sheets, and assessments with answer key are provided.\n\n### Problem-Solving Exercises:\n\n1. The Calculator Pad, Wave Basics, Problems #1-4\n\n### Science Reasoning Activities:\n\n1. The Period of a Pendulum\n\n2. Mass on a Spring\n\n### Historical Context:\n\n1.",
null,
"The Story Behind the Science: Pendulum Motion\n\nBet you’ve never heard of this wonderful collection of 30 science stories developed to help students explore key science concepts through the eyes of the real people who were involved. This “Story” explores the humble pendulum and its significant role in the development of modern science. Inspired by Galileo’s classic experimentation, the pendulum provided the Western world’s first accurate means of timekeeping. But perhaps more importantly, the story of the pendulum brings to light the shift to the use of mathematics to understand the natural world – the methodological core of the Scientific Revolution. It also delves into the value of idealization in science.\n\n### Common Misconception:\n\n1. Confusion of Amplitude and Period\n\nA vibrating object tends to lose energy over the course of time. This is referred to as damping. Damping causes the amplitude of vibration to decrease from one cycle to the next cycle. Students observing this will often use the phrase \"slowing down\" to describe the decrease in amplitude. They may even suggest that the period is decreasing. Such statements need to be corrected for a variety of reasons. First, the vibrating object is constant increasing and decreasing its speed from cycle to cycle. The increases in speed occur as the object is approaching its equilibrium position and these increases defy any description of the object \"slowing down.\" Second, the tendency to confuse a decrease in amplitude with a decrease in period demonstrates a huge confusion on the student's part. The period of vibration remains constant from cycle to cycle; damping doesn't alter the period of vibration.\n\n### Standards:\n\nA. Next Generation Science Standards (NGSS)\n\nPerformance Expectations\nEnergy\n• HS-PS3-2 Develop and use models to illustrate that energy at the macroscopic scale can be accounted for as a combination of energy associated with the motions of particles (objects) and energy associated with the relative positions of particles (objects).\n\nWaves\n• HS-PS4-1 Use mathematical representations to support a claim regarding relationships among the frequency, wavelength, and speed of waves traveling in various media.\n\nDisciplinary Core Ideas\nEnergy Conservation\n• HS-PS3.B.3 Mathematical expressions, which quantify how the stored energy in a system depends on its configuration (e.g., relative positions of charged particles, compression of a spring) and how kinetic energy depends on mass and speed, allow the concept of conservation of energy to be used to predict and describe system behavior.\n\nWave Properties\n• HS-PS4.A.1 The wavelength and frequency of a wave are related to one another by the speed of travel of the wave, which depends on the type of wave and the medium through which it is passing.\n\nCrosscutting Concepts\n\nSystems and System Models\n• When investigating or describing a system, the boundaries and initial conditions of the system need to be defined and their inputs and outputs analyzed and described using models.\n• Models can be used to predict the behavior of a system, but these predictions have limited precision and reliability due to the assumptions and approximations inherent in models.\n\nEnergy and Matter\n• The total amount of energy and matter in closed systems is conserved.\n• The transfer of energy can be tracked as energy flows through a system.\n\nScience and Engineering Practices\nPractice #2: Developing and Using Models\n• Develop and use a model based on evidence to illustrate the relationships between systems or between components of a system.\n\nPractice #3: Planning and Carrying Out Investigations\n• Plan and conduct an investigation individually and collaboratively to produce data to serve as the basis for evidence, and in the design: decide on types, how much, and accuracy of data needed to produce reliable measurements and consider limitations on the precision of the data (e.g., number of trials, cost, risk, time), and refine the design accordingly.\n\nPractice #4: Analyzing and Interpreting Data\n• Analyze data using tools, technologies, and/or models (e.g. computational, mathematical) in order to make valid and reliable scientific claims.\n\nPractice #5: Using Mathematics and Computational Thinking\n• Create or revise a simulation of a phenomenon, designed device, process, or system.\n• Use mathematical models and/or computer simulations to predict the effects of a design solution on systems and/or the interactions between systems.\n\nPractice #6: Constructing Explanations\n• Construct an explanation based on valid and reliable evidence obtained from a variety of sources (including students' own investigations, models, theories, simulations, peer review) and the assumption that theories and laws that describe the natural world operate today as they did in the past and will continue to do so in the future.\n• Apply scientific reasoning to link evidence to the claims to assess the extent to which the reasoning and data support the explanation or conclusion.\n\nPractice #8: Obtaining, Evaluating, and Communicating Information: High School\n• Communicate scientific and technical information (e.g. about the process of development and the design and performance of a proposed process or system) in multiple formats (including orally, graphically, textually, and mathematically)."
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https://www.diplomaincs.in/2019/11/notes-for-computer-science-and_23.html | [
"## 11/24/19\n\n### Problems of Operating System Numerical\n\nPROBLEMS of Operating System Unit: 02\n\nBefore going to problems we will discuss some abbreviations and their meaning are as follows: -\n\n1. TAT Turn Around Time\n2. BT Burst Time\n3. AT Arrival Time\n4. WT Waiting Time\n5. FT/CT Finish Time / Completion Time\n\nSome basic formula that we use: -\n\n1. Turn Around Time(TAT) = Completion time (CT) – Arrival Time (AT)\n2. Waiting Time (WT) = Turn Around Time (TAT) – Burst Time (BT)\nProblem 1.1: -\nConsider the following snapshot of an OS with 5 processes:\n Process Arrival Time Burst time P0 0 2 P1 2 3 P2 4 6 P3 6 4 P4 8 5\n\nCompute and prepare a chart for Average Waiting time and Average Turnaround time of each process for algorithm: -\n1. FCFS\n2. SJF\n3. Round Robin (Quantum = 2)\n\nSOLUTION: -\n\n1. Using FCFS: -\nUsing FCFS first we have to prepare GANTT CHART which is as follow: -\nGantt chart: -\n\nNow we can say that: -\n1. 2 is the completion time of P0 (process 0)\n2. 5 is the completion time of P1 (processn1)\n3. 11 is the completion time of P2 (process 2)\n4. 15 is the completion time of P3 (process 3)\n5. 20 is the completion time of P4 (process 4)\n\n Process AT BT CT TAT=CT-AT WT=TAT-BT P0 0 2 2 2-0= 2 2-2= 0 P1 2 3 5 5-2= 3 3-3= 0 P2 4 6 11 11-4= 7 7-6= 1 P3 6 4 15 15-6= 9 9-4= 5 P4 8 5 20 20-8= 12 12-5= 7 Total number of process= 5 TAT= 33 WT= 13\n\nAverage Turnaround Time= Total TAT/ Number of process\n= (33/5)\n=6.6 Milliseconds\nAverage Turnaround Time= Total WT/ Number of process\n= (13/5)\n=2.6 Milliseconds\n\n2. Using SJN (preemptive type): -\n\nUsing SJN first we have to prepare GANTT CHART which is as follow: -\n\nGantt chart: -\n\nNOTE: - First add smallest AT and it’s BT, then selects the smallest BT from the table in ascending order and prepare GANTT CHART.\n\nNow we can say that: -\n1. 2 is the completion time of P0 (process 0)\n2. 5 is the completion time of P1 (processn1)\n3. 20 is the completion time of P2 (process 2)\n4. 9 is the completion time of P3 (process 3)\n5. 14 is the completion time of P4 (process 4)\n\n Process AT BT CT TAT=CT-AT WT=TAT-BT P0 0 2 2 2-0= 2 2-2= 0 P1 2 3 5 5-2= 3 3-3= 0 P2 4 6 20 20-4= 16 16-6= 10 P3 6 4 9 9-6= 3 3-4= 1 P4 8 5 14 14-8= 6 6-5= 1 Total number of process= 5 TAT= 30 WT= 12\n\nAverage Turnaround Time= Total TAT/ Number of process\n= (30/5)\n=6.0 Milliseconds\nAverage Turnaround Time= Total WT/ Number of process\n= (12/5)\n=2.4 Milliseconds\n\n3. Round Robin (Quantum time/time-slice= 2)\n\nNow we can say that: -\n1. 2 is the completion time of P0 (process 0)\n2. 11 is the completion time of P1 (processn1)\n3. 19 is the completion time of P2 (process 2)\n4. 15 is the completion time of P3 (process 3)\n5. 20 is the completion time of P4 (process 4)\n\n Process AT BT CT TAT=CT-AT WT=TAT-BT P0 0 2 2 2-0= 2 2-2= 0 P1 2 3 11 11-2= 9 9-3= 6 P2 4 6 19 19-4= 15 15-6= 9 P3 6 4 15 15-6= 9 9-4= 5 P4 8 5 20 20-8= 12 12-5= 7 Total number of process= 5 TAT= 47 WT= 27\n\nAverage Turnaround Time= Total TAT/ Number of process\n= (47/5)\n=9.4 Milliseconds\nAverage Turnaround Time= Total WT/ Number of process\n= (27/5)\n=5.4 Milliseconds\n\nProblem 1.2: -(Try yourself)\n\nConsider the following snapshot of an OS with 5 processes:\n Process Arrival Time CPU Burst time P0 0 4 P1 1 6 P2 3 3 P3 9 6 P4 12 6\n\nCompute and prepare a chart for Average Waiting time and Average Turnaround time of each process for algorithm: -\n1. FCFS\n2. SJF\n3. Round Robin (Quantum = 2)\n1. FCFS: -\n\nAverage Turnaround Time =9.2 Milliseconds\n\nAverage Turnaround Time= 4.2 Milliseconds\n\n2. SJN: -\n\nAverage Turnaround Time = 8.6 Milliseconds\n\nAverage Turnaround Time= 3.6 Milliseconds\n\n3. Round Robin: -\n\nAverage Turnaround Time = Milliseconds\n\nAverage Turnaround Time Milliseconds\n\nProblem 1.3: -(Try yourself)\n\nConsider the following snapshot of an OS with 5 processes:\n Process Arrival Time CPU Burst time P0 0 3 P1 1 5 P2 3 2 P3 9 5 P4 12 5\n\nCompute and prepare a chart for Average Waiting time and Average Turnaround time of each process for algorithm: -\nUse Non-preemptive and Preemptive both type scheduling\n1. FCFS\n2. SJF\n3. Round Robin (Quantum = 2)\nNOTE:- Both preemptive and non-preemptive have same result. Average waiting time for SJN is small as compare to FCFS so SJN scheduling is best.\n\nAverage Turnaround Time =6.2 Milliseconds\n\nAverage Turnaround Time= 2.2 Milliseconds\n\n2. SJN: -\n\nAverage Turnaround Time = 5.6 Milliseconds\n\nAverage Turnaround Time= 1.6 Milliseconds\n\n3. Round Robin: -\n\nAverage Turnaround Time = Milliseconds\n\nAverage Turnaround Time Milliseconds\n\nProblem 1.4: -(Try yourself)\n\nConsider the following snapshot of an OS with 5 processes:\n Process Arrival Time CPU Burst time P0 0 1 P1 1 9 P2 2 1 P3 3 9\nCompute and prepare a chart for Average Waiting time and Average Turnaround time of each process for algorithm: -\n1. FCFS\n2. SJF\n\n1. FCFS: -\n\nAverage Turnaround Time =9.0 Milliseconds\n\nAverage Turnaround Time= 4.0 Milliseconds\n\n2. SJN: -\n\nAverage Turnaround Time = 7.25 Milliseconds\n\nAverage Turnaround Time= 2.5 Milliseconds\n\nProblem 1.5: -\n\nConsider the following snapshot of an OS with 5 processes:\n Process Arrival Time CPU Burst time P1 0 6 P2 0 8 P3 0 7 P4 0 3\n\nCompute and prepare a chart for Average Waiting time and Average Turnaround time of each process for algorithm: -\n1. SJF (Use Non-preemptive Scheduling)\n\nSolution: -\nUsing SJF scheduling, we would schedule these processes according to the following GANT CHART: -\nGantt chart: -\n P4 P1 P3 P2\n0 3 9 16 24\n\nThe waiting time for the above Gantt Chart is as follows: -\n1. 3 milliseconds for process P1,\n2. 16 milliseconds for process P2,\n3. 9 milliseconds for process P3,\n4. 0 milliseconds for process P4.\nAverage waiting time = (3+16+9+0)/4\n= 28/4\n= 7 milliseconds\nSimilarly, we can calculate Average TAT, And FCFS AWT & ATAT.\n\nProblem 1.6: - (Try yourself)\nConsider the following snapshot of an OS with 4 processes:\n Process Arrival Time CPU Burst time P1 0 8 P2 1 4 P3 2 9 P4 3 35\n\nCompute and prepare a chart for Average Waiting time and Average Turnaround time of each process for algorithm: -\n1. SJF (Use Non-preemptive Scheduling)\n\nProblem 1.7: -\nConsider the following snapshot of an OS with 5 processes:\n Process Burst Time Priority P1 10 3 P2 1 1 P3 2 3 P4 1 4 P5 5 2\n\nThis process are assumed to have arrived on the order P1,P2,P3,P4 and P5, all at time 0.(means arrival time of each process is 0)\nCompute and prepare a chart for Average Waiting time and Average Turnaround time of each process for algorithm: -\n1. FCFS\n2. SJF\n3. Priority scheduling\n4. Round Robin (Quantum = 2)\n\nSOLUTION: -\nIn the problem priority of processes are given but in the FCFS, SJF, and RR there is no need of priority. We are additionally solving this problem for priority scheduling.\nUsing FCFS: -\nUsing FCFS first we have to prepare GANTT CHART which is as follow: -\nGantt chart: -\n\nNow we can say that: -\n1. 10 is the completion time of P1 (process 1)\n2. 11 is the completion time of P2 (processn2)\n3. 13 is the completion time of P3 (process 3)\n4. 14 is the completion time of P4 (process 4)\n5. 19 is the completion time of P5 (process 5)\n\n Process AT BT CT TAT=CT-AT WT=TAT-BT P1 0 10 10 10-0=10 10-10= 0 P2 0 1 11 11-0=11 11-1= 10 P3 0 2 13 13-0=13 13-2= 11 P4 0 1 14 14-0=14 14-1= 13 P5 0 5 19 19-0=19 19-5= 14 Total number of process= 5 TAT= 67 WT= 48\n\nAverage Turnaround Time= Total TAT/ Number of process\n= (67/5)\n=13.4 Milliseconds\nAverage Turnaround Time= Total WT/ Number of process\n= (48/5)\n=9.6 Milliseconds\nUsing SJN: -\n\nUsing SJN first we have to prepare GANTT CHART which is as follow: -\nGantt chart: -\n\nNOTE: - First add smallest AT and it’s BT, then selects the smallest BT from the table in ascending order and prepare GANTT CHART.\nNow we can say that: -\n1. 19 is the completion time of P1 (process 1)\n2. 1 is the completion time of P2 (processn2)\n3. 4 is the completion time of P3 (process 3)\n4. 2 is the completion time of P4 (process 4)\n5. 9 is the completion time of P5 (process 5)\n Process AT BT CT TAT=CT-AT WT=TAT-BT P1 0 10 19 19-0=19 19-10= 9 P2 0 1 1 1-0=1 1-1= 0 P3 0 2 4 4-0=4 4-2= 2 P4 0 1 2 2-0=2 2-1= 1 P5 0 5 9 9-0=9 9-5= 4 Total number of process= 5 TAT= 35 WT= 16\n\nAverage Turnaround Time= Total TAT/ Number of process\n= (35/5)\n=7.0 Milliseconds\nAverage Turnaround Time= Total WT/ Number of process\n= (16/5)\n=3.2 Milliseconds\n\nPriority Scheduling: -\n\nUsing the priority scheduling first we have to prepare GANTT CHART, according to the priority given in the table which is as follow: -\nGantt chart: -\n\nNOTE: - First add smallest AT and it’s BT, then selects the smallest BT from the table in ascending order and prepare GANTT CHART.\nNow we can say that: -\n1. 16 is the completion time of P1 (process 1)\n2. 1 is the completion time of P2 (processn2)\n3. 18 is the completion time of P3 (process 3)\n4. 19 is the completion time of P4 (process 4)\n5. 6 is the completion time of P5 (process 5)\n\n Process AT BT CT TAT=CT-AT WT=TAT-BT P1 0 10 16 16-0=16 16-10= 6 P2 0 1 1 1-0=1 1-1= 0 P3 0 2 18 18-0=18 18-2= 16 P4 0 1 19 19-0=19 19-1= 18 P5 0 5 6 6-0=6 6-5= 1 Total number of process= 5 TAT= 60 WT= 50\n\nAverage Turnaround Time= Total TAT/ Number of process\n= (60/5)\n=12.0 Milliseconds\nAverage Turnaround Time= Total WT/ Number of process\n= (50/5)\n=8.2 Milliseconds\n\n1. Round Robin (Quantum time/time-slice= 2)\nQuantum is allocated to each process for execution. (it is in milliseconds)\n\nNow we can say that: -\n1. 16 is the completion time of P1 (process 1)\n2. 1 is the completion time of P2 (processn2)\n3. 18 is the completion time of P3 (process 3)\n4. 19 is the completion time of P4 (process 4)\n5. 6 is the completion time of P5 (process 5)\n\n Process AT BT CT TAT=CT-AT WT=TAT-BT P1 0 10 19 19-0=19 19-10= 9 P2 0 1 3 3-0=3 3-1= 2 P3 0 2 5 5-0=5 5-2= 3 P4 0 1 6 6-0=6 6-1= 5 P5 0 5 15 15-0=15 15-5= 10 Total number of process= 5 TAT= 48 WT= 29\n\nAverage Turnaround Time= Total TAT/ Number of process\n= (48/5)\n=9.6 Milliseconds\nAverage Turnaround Time= Total WT/ Number of process\n= (29/5)\n=5.8 Milliseconds\n\nProblem 1.8: - (Try yourself)\nConsider the following snapshot of an OS with 5 processes:\n Process Burst Time Priority P1 10 3 P2 1 1 P3 2 4 P4 1 5 P5 5 2\n\nThis process are assumed to have arrived on the order P1,P2,P3,P4 and P5, all at time 0.(means arrival time of each process is 0)\nCompute and prepare a chart for Average Waiting time and Average Turnaround time of each process for algorithm: -\n1. FCFS\n2. SJF\n3. Priority scheduling\n4. Round Robin (Quantum = 1)\n\n1. FCFS: -\n\nAverage Turnaround Time =13.4 Milliseconds\n\nAverage Turnaround Time= 9.6 Milliseconds\n\n2. SJN: -\n\nAverage Turnaround Time = 7.0 Milliseconds\n\nAverage Turnaround Time= 3.2 Milliseconds\n\n3. Round Robin: -\n\nAverage Turnaround Time =9.2 Milliseconds\n\nAverage Turnaround Time=5.4 Milliseconds\n\nProblem 1.9: -\nConsider the following snapshot of an OS with 5 processes:\n Process Arrival Time CPU Burst time P1 0 24 P2 0 3 P3 0 3\nCompute and prepare a chart for Average Waiting time and Average Turnaround time of each process for algorithm RR scheduling: -\nSOLUTION: -\n1. If we use a time quantum of 4 milliseconds, then process P1 gets the first 4 milliseconds. Since it requires another 20 milliseconds, it is preemptive after the first time quantum, and the CPU is given to the process in the queue, process P2.\n2. Since process P2 does not need 4 milliseconds, it quite before its time quantum expires. The CPU is then given to the next process, process P3.\n3. Once each process has received 1 time quantum, the CPU is returned to the P1 process for an additional time quantum.\n4. The resulting RR schedule is: -\nRound Robin (Quantum time/time-slice= taken as 4)\n\nNow we can say that: -\n1. 30 is the completion time of P1 (process 1)\n2. 7 is the completion time of P2 (processn2)\n3. 10 is the completion time of P3 (process 3)\n\n Process AT BT CT TAT=CT-AT WT=TAT-BT P1 0 24 30 30-0=30 30-24= 6 P2 0 3 7 7-0=7 7-3= 4 P3 0 3 10 10-0=10 10-3= 7 Total number of process= 5 TAT= 47 WT= 17\n\nAverage Turnaround Time= Total TAT/ Number of process\n= (47/5)\n=9.4 Milliseconds\nAverage Turnaround Time= Total WT/ Number of process\n= (17/5)\n=3.4 Milliseconds\n\nProblem 1.10: -\nConsider the following snapshot of an OS with 5 processes:\n Process Arrival Time Burst time P1 0 10 P2 1 29 P3 3 3 P4 9 7 P5 12 12\n\nCompute and prepare a chart for Average Waiting time and Average Turnaround time of each process for algorithm: -\n1. FCFS\n2. SJF\n3. Round Robin (Quantum = 10 milliseconds)\nWhich algorithm would give minimum AWT and ATAT?\n\nProblem 1.11: -\nConsider the following snapshot of an OS with 5 processes:\n Process Arrival Time Burst time P1 0 1 P2 1 9 P3 2 1 P4 3 9 P5 3 9\n\nCompute and prepare a chart for Average Waiting time and Average Turnaround time of each process for algorithm: -\n1. FCFS\n2. SJF\n3. Round Robin (Quantum = 4 milliseconds)\n\nProblem 1.11: -\nConsider the following snapshot of an OS with 5 processes:\n Process Arrival Time Burst time Priority P1 0 4 6 P2 1 5 5 P3 2 2 3 P4 3 1 4 P5 4 3 2 P6 5 6 1\n\nCompute and prepare a chart for Average Waiting time and Average Turnaround time of each process for algorithm: -\n1. FCFS\n2. SJF\n3. Priority\n4. Round Robin (Quantum = 4 milliseconds)\n\nFor any problem you can comment we will definitely reply on your problems.\n\n(Note:- Update available soon, comment for any type of help)\n\nCOA : -\n\n1.",
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"Sir,try urself ques ans??\n\n2.",
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"Ok I will upload it, within some days\n\nPlease do not enter any spam link in the comment box and use English and Hindi language for comment.\n\n## Latest Update\n\n### Key Components of XML",
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"https://www.blogger.com/img/blogger_logo_round_35.png",
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"https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhghAQkNr1VUA-YGAJb6wFHVo3a08wyLZ2YdjGC9K47bnTgkq3znzT9cnPL58LiKmmEDNTdV_ECc-d0Sv4YqKwf8OB-217a_Fe2zfYgb_6wkkyWUwNGIC0iA9Ea9p4hVQ/s45-c/Mohit.jpg",
null,
"https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5dgckCAABSZJnLfnnhjXDCfMfEhHNbU_a-KINQBYwnAuSiRnMpHcEny3Eru9ujxLyB6TxfEA9UNxugMw0-vYuAiTzJO6eGmYxEUhlb9-j5c7gOC7THAeoXvHqFrY80lzyacXnI6BrUMcS18IWCfbtmH2VKcWDGP-GR87X1C9MnkwQZulAERF2-F7cqA/w367-h207/XML.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.714959,"math_prob":0.9714106,"size":14722,"snap":"2023-40-2023-50","text_gpt3_token_len":5444,"char_repetition_ratio":0.21205327,"word_repetition_ratio":0.58690095,"special_character_ratio":0.39430785,"punctuation_ratio":0.10552107,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.994507,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,9,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-28T14:04:46Z\",\"WARC-Record-ID\":\"<urn:uuid:8d6224ee-5f79-4eef-a947-9b2b43634c73>\",\"Content-Length\":\"787605\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a7b2d179-3339-4199-9f55-618164c8a3ad>\",\"WARC-Concurrent-To\":\"<urn:uuid:18c31b2c-8970-4de8-be85-b501a8512fdd>\",\"WARC-IP-Address\":\"142.251.16.121\",\"WARC-Target-URI\":\"https://www.diplomaincs.in/2019/11/notes-for-computer-science-and_23.html\",\"WARC-Payload-Digest\":\"sha1:66VDW5D2LNYITXALEH6LA55R6LSTZRCG\",\"WARC-Block-Digest\":\"sha1:GVEXIFE5YSWPGCFXDPG7FMW22X2UTHAS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679099514.72_warc_CC-MAIN-20231128115347-20231128145347-00071.warc.gz\"}"} |
https://vi.stackexchange.com/questions/30798/how-to-check-last-char-is-in-vimscript | [
"# How to check last char is { in vimscript?\n\nI want to create a `{}` completion myself, and I did:\n\n``````inoremap <CR> if_open_curly_on_left()? <CR>}<ESC>O : <CR>\nfunction! s:if_open_curly_on_left() abort\nlet col = col('.') - 1\nreturn !col || getline('.')[col - 1] =~# '{'\nendfunction\n``````\n\nwhich is modified by other's code, since I don't really understand the meaning inside the function.\n\nWhen I typed enter I got:",
null,
"How to fix this?\n\nThe logic is wrong here:\n\n``````getline('.')[col - 1] =~# '{'\n``````\n\nThis is saying anywhere before my cursor do I have a `{` character.\n\nProbably want something that says does the end have a `{` character\n\n``````getline('.')[col - 1] =~# '{\\$'\ngetline('.')[col - 1] == '{'\n``````\n\nAlthough in theory that could still be weird if you put your insert cursor between `{}` characters and hit `<cr>`. So I imagine this needs to be adjusted a bit to become more robust\n\nMaybe you the logic and be something like:\n\n``````col('.') == (col('\\$') - 1) && strpart(getline('.'), -1) == '{'\n``````\n\nThis will check that your in the final column and it ends with `{`\n\nAside:\n\nWhy `col('.')-2` to get the previous character?\n\nTL;DR: column position is 1-indexed and strings/arrays are 0-indexed\n\n`col('.')` give you the current cursor position with the first column being `1`\n\n`getline('.')` will return a string or an array of characters with the first index being `0`\n\nAssume the following line and you cursor on the `r`:\n\n``````bar\n``````\n\nSo if we do a `getline('.')` and split it e.g. `split(getline('.'), '\\zs')` we get:\n\n``````['b', 'a', 'r']\n``````\n\nSince we are on the `r` character then `col('.')` gives us `3` which is outside the bounds of our array by one. We compensate for the column starting on position one by subtracting one, `col('.') - 1`, to give the current position. To give the \"previous character\" then subtract by `2`\n\n• By your kind hint at last line(, after many try-and-error), I got: `return line[col] != '}' && line[col-1] == '}'`. Regarding the original one, why it's `col-1` in `getline('.')[col-1]` when I already did `let col = col('.')-1`? Mar 29 at 21:51\n\nAfter some search this work(, but might be simplified further):\n\n``````\" My try\ninoremap <silent><expr> <CR>\n\\ <SID>if_open_curly_on_left()? \"\\<CR>}\\<ESC>O\":\n\\ \"\\<CR>\"\nfunction! s:if_open_curly_on_left() abort\nlet col = col('.') - 1\nreturn getline('.')[col - 1] == '{'\nendfunction\n``````"
] | [
null,
"https://i.stack.imgur.com/2aAoK.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8305453,"math_prob":0.8361041,"size":1312,"snap":"2021-31-2021-39","text_gpt3_token_len":335,"char_repetition_ratio":0.13990825,"word_repetition_ratio":0.0,"special_character_ratio":0.30868903,"punctuation_ratio":0.108474575,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98792416,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-18T13:15:57Z\",\"WARC-Record-ID\":\"<urn:uuid:4b3b23cd-1390-41a5-82de-f361540f509d>\",\"Content-Length\":\"176206\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0bdf58c8-fe83-447e-9b0d-cf745a7d26d6>\",\"WARC-Concurrent-To\":\"<urn:uuid:16ed3508-4cde-48a8-be45-38499e190cb5>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://vi.stackexchange.com/questions/30798/how-to-check-last-char-is-in-vimscript\",\"WARC-Payload-Digest\":\"sha1:YCHMM4RIBFNGGZJPCEZGWLSBJKS7SDWB\",\"WARC-Block-Digest\":\"sha1:TWBEV5I7TA7FAZCZ57MLEM5BLDHCOWL5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056476.66_warc_CC-MAIN-20210918123546-20210918153546-00106.warc.gz\"}"} |
https://cs.appstate.edu/~sjg/class/3610/hf08.html | [
"Dr. Sarah's Math 3610 Class Highlights\n\n### Dr. Sarah's Math 3610 Class Highlights The following is NOT HOMEWORK unless you miss part or all of the class. See the Main Class Web Page for ALL homework and due dates.\n\n• Tues Dec 9 Go over test 2. Abstract presentations. Course evaluations.\n• Tues Dec 2 Finish Applications Go over response to the assignment: Choose a short (new to our class) Euclidean Sketchpad exploration, web applet, or computer exploration related to Euclidean geometry, and be prepared to present it. Be sure to put it in context by discussing why it is interesting or important.\nAustin: http://www.cut-the-knot.org/Curriculum/Geometry/TangentTriangleToEllipse.shtml\nBrett: http://nlvm.usu.edu/en/nav/category_g_4_t_3.html\nCasey: http://www.cut-the-knot.org/Curriculum/Geometry/TangentTriangleToEllipse.shtml\nCayce: http://www.analyzemath.com/Geometry/properties_triangles.html\nDarrell: \"Two Trees.gsp\" under Investigations\nDewey: http://www.members.shaw.ca/ron.blond/SimilarTriangles.APPLET/index.html\nEmily: http://aleph0.clarku.edu/~djoyce/java/elements/usingApplet.html\nEdgar: Soccer Ball Application\nKaty: http://www.cut-the-knot.org/Curriculum/Geometry/HingedPythagoras2.shtml\nKimberly: http://www.frontiernet.net/~imaging/pythagorean.html\nLianna: http://faculty.evansville.edu/ck6/GIAJSP/EulerLine.html and http://aleph0.clarku.edu/~djoyce/java/Geometry/eulerline.html\nMandi: http://www.saltire.com/applets/simtri1/simtri1.htm\nToni: http://www.cut-the-knot.org/Curriculum/Geometry/HingedPythagoras2.shtml\n\nTake questions on Test 2. If time remains, then search for references for the final project.\n\n• Thur Dec 4 Test 2\n• Tues Nov 25 Finish the Euclidean proof presentations. Discuss parallel projects. If time remains, discuss applications of hyperbolic geometry. How to Sew a 2-Holed Cloth Torus.\n• Tues Nov 18 Lab directions\n\n• Thur Nov 20 Review lab work via the pictures listed in the Lab directions and the Hyperbolic parallel axiom image and the Pythagorean theorem image. Begin the Euclidean proof presentations.\n• Tues Nov 11 Begin hyperbolic geometry via the Escher worksheet.\n• What are the shortest distance paths in hyperbolic geometry?\nImage of Shortest Distance Paths\n• Is parallel the same as equidistant in hyperbolic geometry?\nImage of Equidistant 1\nImage of Equidistant 2\nReview our Euclidean proof that parallel means equidistant and discuss what goes wrong in hyperbolic geometry.\n\n• Thur Nov 13 Prove Playfair's postulate in Euclidean geometry and examine the relationship with Euclid's 5th in spherical geometry and Euclidean geometry.\n• Tues Nov 4 Taxicab activities in Sketchpad.\n\n• Thur Nov 6 Discuss folding activities of the sum of the angles in a triangle is 180 degrees. Discuss a proof using Euclidean axioms. Discuss what goes wrong on the sphere. Begin parallels in Euclidean geometery and review Playfair's postulate as well as Euclid's 5th. Prove that parallel lines are equidistant.\n• Tues Oct 28 Meet in 205. Finish using the triangles to examine the area of regular polygons on the sphere and Colorado and Wyoming. Reservoir problems. Go over the proof that the perpendicular bisectors are concurrent. Begin taxicab geometry via moving in Tivo, and play a few games of taxicab treasure hunt. If time remains, then begin taxicab activities in Sketchpad\n\n• Thur Oct 30 Discuss metric perspectives and coordinate geometry and do the missing square activity. Review Minesweeper and create an inconsistent game. Fill in a partial game to show that consistency does not imply uniqueness. Discuss Godel's 1930 theorem. Review taxicab Discuss taxicab circles and the relationship to the strategy of the game. Highlight the possible number of intersections of taxicab circles for different examples. US law is Euclidean. SAS in taxicab geometry.\n• Tues Oct 21 Project 5 timeline presentation sessions and peer and self-evaluation.\n\n• Thur Oct 23 Finish presentations. Use the triangles to examine the area of regular polygons on the sphere. Discuss Colorado and Wyoming.\n• Tues Oct 14 Sphere activity 1. Sphere activity 2. Sphere activity 1 and examine consequences, including whether the difference between the angle sum and pi is detectable for a 1 mile square area triangle in Kansas. AAA on the sphere.\n• Thur Oct 7 Test 1. Work on Project 5.\n\n• Thur Oct 9 [Of the five Platonic solids] So their combinations with themselves and with each other give rise to endless complexities, which anyone who is to give a likely account of reality must survey. [Plato, The Timaeus] Euclidean angle defect. applet 1 and applet 2. Begin measurement. Quotations from Archimedes. Measurements with and without metric perspectives. How were circumference, area and volume formulas obtained via axiomatic perspectives and before coordinate geometry and calculus II? Orange activity. Orange Activity and Archimedes polygonal method. Worksheet on Archimedes and Cavalieri's Principle.\n• Tues Sep 30 Ask students to share their ideas about Wile - how did they ensure the chase would always begin? How did they ensure Wile would catch the RR when the RR runs faster? Burden of Proof. Begin Euler's formula and platonic solids. Show there are 5 convex regular polyhedra in Euclidean geometry, but additional polyhedra in spherical geometry (infinitely many).\n\n• Thur Oct 2 Take questions on test 1. Go over proof from project 4. Review the platonic solids - and how to remember the faces and vertices (and from there calculate the edges using Euler's formula). Continue platonic solids.\n• Tues Sep 23 Finish presentations. Discuss similarity postulates. Similarity of quadrilaterals. Look at a proof of SAS and discuss what goes wrong on the sphere for large triangles. Applications of similarity: Sibley The Geometric Viewpoint p. 55 number 6. Sliding a Ribbon Wrapped around a Rectangle and Sliding a Ribbon Wrapped around a Box. Read the proof of the trig identity and then fill in the details and reasons using similarity, trig and the Pythagorean theorem. Note that the Pythagorean theorem is a consequence of similarity as in Project 4.\n\n• Thur Sep 25 Introduction to geometric similarity and its application to geometric modeling via. Mathematics Methods and Modeling for Today's Mathematics Classroom 6.3. Go over p. 214 Project 1, and the example on p. 212. Work on models for p. 216 number 4 (Loggers).\n• Tues Sep 16 Take questions. Nova's \"The Proof\" video and notes.\n\n• Thur Sep 18 Andrew Wiles worksheet. A second example. Begin similarity. Introduction to \"same shape\". Fig 8.4 Fig 8.21 Fig 8.32 Use the Triangle_Similarity.gsp file (control click and save the file. Then open it from Sketchpad) to complete the Similar Triangles - SSS, SAS, SSA worksheet. Groups prepare short presentations on SSS, SAS, AA, SSA, AAS, ASA, HL (Hypotenuse and leg of a right triangle - ie SSA in a right triangle).\n• Tues Sep 9 Meet in 205. Discuss the homework readings. Review the paper folding argument for Proposition 11.\nGo over an application - a proof that the perpendicular bisectors are concurrent.\nBuild a right triangle in Sketchpad and investigate the Pythagorean Theorem.\nGo through Behold Pythagoras!, Puzzled Pythagoras, and then Shear Pythagoras. Click on Contents to get to the other Sketches.\nGo through Euclid's proof. Discuss Sibley Geometric Viewpoint p. 7 # 10 on Project 2. If time remains, then an introduction to extensions of the Pythagorean Theorem including a review of the Greenwaldian Theorem, as well as the Scarecrow's Theorem, Pappus on Sketchpad.\n\n• Thur Sep 11 Finish Pappus' Theorem. Continue extensions of the Pythagorean Theorem. Review the Greenwaldian Theorem, and examine the Scarecrow's Theorem. Discuss the homework readings. Go over images and quotations. Highlight that the Yale tablet is Sibley The Geometric Viewpoint 1.1 3 and The 'hsuan-thu' [Zhou Bi Suan Jing] is similar to Bhaskara's diagram in Sibley The Geometric Viewpoint 1.1 10, and the connection of Eratosthenes to Wallace and West Roads to Geometry 1.1 8. Fermat's Last Theorem.\n• Tues Sept 2 Project 1 Presentations. Peer review.\n\n• Thur Sept 4 Collect self-evaluation for project 1. Review ASULearn solutions. Review Euclid's Proposition 1 in Sketchpad. Hand out folding arguments. Use a paper folding argument for Proposition 11. Euclid's Book 1 Proposition 11. Go over Sketchpad's built in version of Proposition 11 as well as a ray versus a line in Sketchpad.\n• Tues Aug 26 Fill out information sheet. Form groups of 2 or 3 people and prepare to come up to the front of the room and present a partner's\n1) Name\n2) Something that will help us remember them\nNext discuss how can we tell the earth is round without technology?\nMention the related problem on Project 2 for Friday [Wallace and West Roads to Geometry 1.1 8].\nWhere is North? Also discuss 8/08 article Cows Tend To Face North-South\nBegin the Geometry of the Earth Project. Groups choose their top three problems and turn these in to Dr. Sarah.\nInduction versus deduction. An introduction to minesweeper games as an axiomatic system.\nAxiom 1) Each square is a number or a mine.\nAxiom 2) A numbered square represents the number of neighboring mines in the blocks immediately above, below, left, right, or diagonally touching.\nExamine game 1. History of Euclid's elements and the societal context of philosophy and debate within Greek society. Intro to Geometric Constructions. Begin Euclid's Proposition 1 by hand and by a proof.\n\n• Thur Aug 28 Take questions on the syllabus. Students are called on in random order to state and then prove something about a specific square in game 2 of minesweeper. Euclid's Proposition 1 in Sketchpad. At 4pm go to 205. Students complete Proposition 1 in Sketchpad and then work on project 1."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7797821,"math_prob":0.62124306,"size":10270,"snap":"2022-27-2022-33","text_gpt3_token_len":2617,"char_repetition_ratio":0.11825443,"word_repetition_ratio":0.030872483,"special_character_ratio":0.20574489,"punctuation_ratio":0.14112692,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95123,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-19T06:07:47Z\",\"WARC-Record-ID\":\"<urn:uuid:2dc26d13-91aa-41c7-ba55-90816e10aed3>\",\"Content-Length\":\"15948\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6df6cddc-c4cb-43b5-af61-c718a33b81e3>\",\"WARC-Concurrent-To\":\"<urn:uuid:b2b7e95c-e910-44a0-8a84-3ff99c0d7982>\",\"WARC-IP-Address\":\"152.10.10.40\",\"WARC-Target-URI\":\"https://cs.appstate.edu/~sjg/class/3610/hf08.html\",\"WARC-Payload-Digest\":\"sha1:OH3JOFOLFTJCVI5MYCFFXPROE62WK2XL\",\"WARC-Block-Digest\":\"sha1:LL2RX7CBOHDEXCFIBJQZD7E54ZXTO7DZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573623.4_warc_CC-MAIN-20220819035957-20220819065957-00369.warc.gz\"}"} |
https://clay6.com/qa/130714/two-functions-f-r-to-r-g-r-to-r-are-defined-as-follows-br-f-x-begin-0-quad- | [
"",
null,
"# Two functions $f : R \\to R, \\; g: R \\to R$ are defined as follows <br> $f(x) = \\begin{cases} 0 , & \\quad \\text{x } \\text{ is rational}\\\\ 1, & \\quad \\text{x} \\text{ is irrational} \\end{cases}$ <br> $g(x) = \\begin{cases} -1 , & \\quad \\text{x } \\text{ is rational}\\\\ 0, & \\quad \\text{x} \\text{ is irrational} \\end{cases}$ <br> Then $(fog) (\\pi) + (gof)(e)$ is equal to :\n( A ) $-1$\n( B ) $0$\n( C ) $2$\n( D ) $1$"
] | [
null,
"https://clay6.com/images/down_arrow_square.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.50257844,"math_prob":1.0000088,"size":894,"snap":"2021-21-2021-25","text_gpt3_token_len":312,"char_repetition_ratio":0.13707866,"word_repetition_ratio":0.12345679,"special_character_ratio":0.33557048,"punctuation_ratio":0.09756097,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000061,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-18T05:15:57Z\",\"WARC-Record-ID\":\"<urn:uuid:515714a5-f8ae-4d8d-b0cb-b3aa2a503524>\",\"Content-Length\":\"18191\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:45f7d038-3121-4326-99be-e2f96ae857d3>\",\"WARC-Concurrent-To\":\"<urn:uuid:b65f97fb-9991-4570-a9eb-e149e10df2d9>\",\"WARC-IP-Address\":\"139.162.17.55\",\"WARC-Target-URI\":\"https://clay6.com/qa/130714/two-functions-f-r-to-r-g-r-to-r-are-defined-as-follows-br-f-x-begin-0-quad-\",\"WARC-Payload-Digest\":\"sha1:NNEWJAEZCC3I62S2AXPWQL7CUDYQ2IYY\",\"WARC-Block-Digest\":\"sha1:4Z46PCVEN7OBQJAHOA2MVZRKOQZ3O53I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989820.78_warc_CC-MAIN-20210518033148-20210518063148-00294.warc.gz\"}"} |
https://kashi.savingadvice.com/2006/03/ | [
"User Real IP - 3.236.218.88\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => Array\n(\n => 95.142.117.29\n)\n\n => Array\n(\n => 150.129.60.107\n)\n\n => Array\n(\n => 94.205.243.22\n)\n\n => Array\n(\n => 115.42.71.143\n)\n\n => Array\n(\n => 117.217.195.59\n)\n\n => Array\n(\n => 182.77.112.56\n)\n\n => Array\n(\n => 182.77.112.108\n)\n\n => Array\n(\n => 41.80.69.10\n)\n\n => Array\n(\n => 117.5.222.121\n)\n\n => Array\n(\n => 103.11.0.38\n)\n\n => Array\n(\n => 202.173.127.140\n)\n\n => Array\n(\n => 49.249.249.50\n)\n\n => Array\n(\n => 116.72.198.211\n)\n\n => Array\n(\n => 223.230.54.53\n)\n\n => Array\n(\n => 102.69.228.74\n)\n\n => Array\n(\n => 39.37.251.89\n)\n\n => Array\n(\n => 39.53.246.141\n)\n\n => Array\n(\n => 39.57.182.72\n)\n\n => Array\n(\n => 209.58.130.210\n)\n\n => Array\n(\n => 104.131.75.86\n)\n\n => Array\n(\n => 106.212.131.255\n)\n\n => Array\n(\n => 106.212.132.127\n)\n\n => Array\n(\n => 223.190.4.60\n)\n\n => Array\n(\n => 103.252.116.252\n)\n\n => Array\n(\n => 103.76.55.182\n)\n\n => Array\n(\n => 45.118.166.70\n)\n\n => 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=> 172.247.129.90\n)\n\n => Array\n(\n => 116.90.74.214\n)\n\n => Array\n(\n => 180.178.142.242\n)\n\n => Array\n(\n => 111.119.183.59\n)\n\n => Array\n(\n => 117.5.103.189\n)\n\n => Array\n(\n => 203.110.93.146\n)\n\n => Array\n(\n => 188.163.97.86\n)\n\n => Array\n(\n => 124.253.90.47\n)\n\n => Array\n(\n => 139.167.249.160\n)\n\n => Array\n(\n => 103.226.206.55\n)\n\n => Array\n(\n => 154.28.188.191\n)\n\n => Array\n(\n => 182.190.197.205\n)\n\n => Array\n(\n => 111.119.183.33\n)\n\n => Array\n(\n => 14.253.254.64\n)\n\n => Array\n(\n => 117.237.197.246\n)\n\n => Array\n(\n => 172.105.53.82\n)\n\n => Array\n(\n => 124.253.207.164\n)\n\n => Array\n(\n => 103.255.4.33\n)\n\n => Array\n(\n => 27.63.131.206\n)\n\n => Array\n(\n => 103.118.170.99\n)\n\n => Array\n(\n => 111.119.183.55\n)\n\n => Array\n(\n => 14.182.101.109\n)\n\n => Array\n(\n => 175.107.223.199\n)\n\n => Array\n(\n => 39.57.168.94\n)\n\n => Array\n(\n => 122.182.213.139\n)\n\n => Array\n(\n => 112.79.214.237\n)\n\n => Array\n(\n => 27.6.252.22\n)\n\n => Array\n(\n => 89.163.212.83\n)\n\n => Array\n(\n => 182.189.23.1\n)\n\n => Array\n(\n => 49.15.222.253\n)\n\n => Array\n(\n => 125.63.97.110\n)\n\n => Array\n(\n => 223.233.65.159\n)\n\n => Array\n(\n => 139.99.159.18\n)\n\n => Array\n(\n => 45.118.165.137\n)\n\n => Array\n(\n => 39.52.2.167\n)\n\n => Array\n(\n => 39.57.141.24\n)\n\n => Array\n(\n => 27.5.32.145\n)\n\n => Array\n(\n => 49.36.212.33\n)\n\n => Array\n(\n => 157.33.218.32\n)\n\n => Array\n(\n => 116.71.4.122\n)\n\n => Array\n(\n => 110.93.244.176\n)\n\n => Array\n(\n => 154.73.203.156\n)\n\n => Array\n(\n => 136.158.30.235\n)\n\n => Array\n(\n => 122.161.53.72\n)\n\n => Array\n(\n => 106.203.203.156\n)\n\n => Array\n(\n => 45.133.7.22\n)\n\n => Array\n(\n => 27.255.180.69\n)\n\n => Array\n(\n => 94.46.244.3\n)\n\n => Array\n(\n => 43.242.178.157\n)\n\n => Array\n(\n => 171.79.189.215\n)\n\n => Array\n(\n => 37.117.141.89\n)\n\n => Array\n(\n => 196.92.32.64\n)\n\n => Array\n(\n => 154.73.203.157\n)\n\n => Array\n(\n => 183.83.176.14\n)\n\n => Array\n(\n => 106.215.84.145\n)\n\n => Array\n(\n => 95.142.120.12\n)\n\n => Array\n(\n => 190.232.110.94\n)\n\n => Array\n(\n => 179.6.194.47\n)\n\n => Array\n(\n => 103.62.155.172\n)\n\n => Array\n(\n => 39.34.156.177\n)\n\n => Array\n(\n => 122.161.49.120\n)\n\n => Array\n(\n => 103.58.155.253\n)\n\n => Array\n(\n => 175.107.226.20\n)\n\n => Array\n(\n => 206.81.28.165\n)\n\n => Array\n(\n => 49.36.216.36\n)\n\n => Array\n(\n => 104.223.95.178\n)\n\n => Array\n(\n => 122.177.69.35\n)\n\n => Array\n(\n => 39.57.163.107\n)\n\n => Array\n(\n => 122.161.53.35\n)\n\n => Array\n(\n => 182.190.102.13\n)\n\n => Array\n(\n => 122.161.68.95\n)\n\n => Array\n(\n => 154.73.203.147\n)\n\n => Array\n(\n => 122.173.125.2\n)\n\n => Array\n(\n => 117.96.140.189\n)\n\n => Array\n(\n => 106.200.244.10\n)\n\n => Array\n(\n => 110.36.202.5\n)\n\n => Array\n(\n => 124.253.51.144\n)\n\n => Array\n(\n => 176.100.1.145\n)\n\n => Array\n(\n => 156.146.59.20\n)\n\n => Array\n(\n => 122.176.100.151\n)\n\n => Array\n(\n => 185.217.117.237\n)\n\n => Array\n(\n => 49.37.223.97\n)\n\n => Array\n(\n => 101.50.108.80\n)\n\n => Array\n(\n => 124.253.155.88\n)\n\n => Array\n(\n => 39.40.208.96\n)\n\n => Array\n(\n => 122.167.151.154\n)\n\n => Array\n(\n => 172.98.89.13\n)\n\n => Array\n(\n => 103.91.52.6\n)\n\n => Array\n(\n => 106.203.84.5\n)\n\n => Array\n(\n => 117.216.221.34\n)\n\n => Array\n(\n => 154.73.203.131\n)\n\n => Array\n(\n => 223.182.210.117\n)\n\n => Array\n(\n => 49.36.185.208\n)\n\n => Array\n(\n => 111.119.183.30\n)\n\n => Array\n(\n => 39.42.107.13\n)\n\n => Array\n(\n => 39.40.15.174\n)\n\n => Array\n(\n => 1.38.244.65\n)\n\n => Array\n(\n => 49.156.75.252\n)\n\n => Array\n(\n => 122.161.51.99\n)\n\n => Array\n(\n => 27.73.78.57\n)\n\n => Array\n(\n => 49.48.228.70\n)\n\n => Array\n(\n => 111.119.183.18\n)\n\n => Array\n(\n => 116.204.252.218\n)\n\n => Array\n(\n => 73.173.40.248\n)\n\n => Array\n(\n => 223.130.28.81\n)\n\n => Array\n(\n => 202.83.58.81\n)\n\n => Array\n(\n => 45.116.233.31\n)\n\n => Array\n(\n => 111.119.183.1\n)\n\n => Array\n(\n => 45.133.7.66\n)\n\n => Array\n(\n => 39.48.204.174\n)\n\n => Array\n(\n => 37.19.213.30\n)\n\n => Array\n(\n => 111.119.183.22\n)\n\n => Array\n(\n => 122.177.74.19\n)\n\n => Array\n(\n => 124.253.80.59\n)\n\n => Array\n(\n => 111.119.183.60\n)\n\n => Array\n(\n => 157.39.106.191\n)\n\n => Array\n(\n => 157.47.86.121\n)\n\n => Array\n(\n => 47.31.159.100\n)\n\n => Array\n(\n => 106.214.85.144\n)\n\n => Array\n(\n => 182.189.22.197\n)\n\n => Array\n(\n => 111.119.183.51\n)\n\n => Array\n(\n => 202.47.35.57\n)\n\n => Array\n(\n => 42.108.33.220\n)\n\n => Array\n(\n => 180.178.146.158\n)\n\n => Array\n(\n => 124.253.184.239\n)\n\n => Array\n(\n => 103.165.20.8\n)\n\n => Array\n(\n => 94.178.239.156\n)\n\n => Array\n(\n => 72.255.41.142\n)\n\n => Array\n(\n => 116.90.107.102\n)\n\n => Array\n(\n => 39.36.164.250\n)\n\n => Array\n(\n => 124.253.195.172\n)\n\n => Array\n(\n => 203.142.218.149\n)\n\n => Array\n(\n => 157.43.165.180\n)\n\n => Array\n(\n => 39.40.242.57\n)\n\n => Array\n(\n => 103.92.43.150\n)\n\n => Array\n(\n => 39.42.133.202\n)\n\n => Array\n(\n => 119.160.66.11\n)\n\n => Array\n(\n => 138.68.3.7\n)\n\n => Array\n(\n => 210.56.125.226\n)\n\n => Array\n(\n => 157.50.4.249\n)\n\n => Array\n(\n => 124.253.81.162\n)\n\n => Array\n(\n => 103.240.235.141\n)\n\n => Array\n(\n => 132.154.128.20\n)\n\n => Array\n(\n => 49.156.115.37\n)\n\n => Array\n(\n => 45.133.7.48\n)\n\n => Array\n(\n => 122.161.49.137\n)\n\n => Array\n(\n => 202.47.46.31\n)\n\n => Array\n(\n => 192.140.145.148\n)\n\n => Array\n(\n => 202.14.123.10\n)\n\n => Array\n(\n => 122.161.53.98\n)\n\n => Array\n(\n => 124.253.114.113\n)\n\n => Array\n(\n => 103.227.70.34\n)\n\n => Array\n(\n => 223.228.175.227\n)\n\n => Array\n(\n => 157.39.119.110\n)\n\n => Array\n(\n => 180.188.224.231\n)\n\n => Array\n(\n => 132.154.188.85\n)\n\n => Array\n(\n => 197.210.227.207\n)\n\n => Array\n(\n => 103.217.123.177\n)\n\n => Array\n(\n => 124.253.85.31\n)\n\n => Array\n(\n => 123.201.105.97\n)\n\n => Array\n(\n => 39.57.190.37\n)\n\n => Array\n(\n => 202.63.205.248\n)\n\n => Array\n(\n => 122.161.51.100\n)\n\n => Array\n(\n => 39.37.163.97\n)\n\n => Array\n(\n => 43.231.57.173\n)\n\n => Array\n(\n => 223.225.135.169\n)\n\n => Array\n(\n => 119.160.71.136\n)\n\n => Array\n(\n => 122.165.114.93\n)\n\n => Array\n(\n => 47.11.77.102\n)\n\n => Array\n(\n => 49.149.107.198\n)\n\n => Array\n(\n => 192.111.134.206\n)\n\n => Array\n(\n => 182.64.102.43\n)\n\n => Array\n(\n => 124.253.184.111\n)\n\n => Array\n(\n => 171.237.97.228\n)\n\n => Array\n(\n => 117.237.237.101\n)\n\n => Array\n(\n => 49.36.33.19\n)\n\n => Array\n(\n => 103.31.101.241\n)\n\n => Array\n(\n => 129.0.207.203\n)\n\n => Array\n(\n => 157.39.122.155\n)\n\n => Array\n(\n => 197.210.85.120\n)\n\n => Array\n(\n => 124.253.219.201\n)\n\n => Array\n(\n => 152.57.75.92\n)\n\n => Array\n(\n => 169.149.195.121\n)\n\n => Array\n(\n => 198.16.76.27\n)\n\n => Array\n(\n => 157.43.192.188\n)\n\n => Array\n(\n => 119.155.244.221\n)\n\n => Array\n(\n => 39.51.242.216\n)\n\n => Array\n(\n => 39.57.180.158\n)\n\n => Array\n(\n => 134.202.32.5\n)\n\n => Array\n(\n => 122.176.139.205\n)\n\n => Array\n(\n => 151.243.50.9\n)\n\n => Array\n(\n => 39.52.99.161\n)\n\n => Array\n(\n => 136.144.33.95\n)\n\n => Array\n(\n => 157.37.205.216\n)\n\n => Array\n(\n => 217.138.220.134\n)\n\n => Array\n(\n => 41.140.106.65\n)\n\n => Array\n(\n => 39.37.253.126\n)\n\n => Array\n(\n => 103.243.44.240\n)\n\n => Array\n(\n => 157.46.169.29\n)\n\n => Array\n(\n => 92.119.177.122\n)\n\n => Array\n(\n => 196.240.60.21\n)\n\n => Array\n(\n => 122.161.6.246\n)\n\n => Array\n(\n => 117.202.162.46\n)\n\n => Array\n(\n => 205.164.137.120\n)\n\n => Array\n(\n => 171.237.79.241\n)\n\n => Array\n(\n => 198.16.76.28\n)\n\n => Array\n(\n => 103.100.4.151\n)\n\n => Array\n(\n => 178.239.162.236\n)\n\n => Array\n(\n => 106.197.31.240\n)\n\n => Array\n(\n => 122.168.179.251\n)\n\n => Array\n(\n => 39.37.167.126\n)\n\n => Array\n(\n => 171.48.8.115\n)\n\n => Array\n(\n => 157.44.152.14\n)\n\n => Array\n(\n => 103.77.43.219\n)\n\n => Array\n(\n => 122.161.49.38\n)\n\n => Array\n(\n => 122.161.52.83\n)\n\n => Array\n(\n => 122.173.108.210\n)\n\n => Array\n(\n => 60.254.109.92\n)\n\n => Array\n(\n => 103.57.85.75\n)\n\n => Array\n(\n => 106.0.58.36\n)\n\n => Array\n(\n => 122.161.49.212\n)\n\n => Array\n(\n => 27.255.182.159\n)\n\n => Array\n(\n => 116.75.230.159\n)\n\n => Array\n(\n => 122.173.152.133\n)\n\n => Array\n(\n => 129.0.79.247\n)\n\n => Array\n(\n => 223.228.163.44\n)\n\n => Array\n(\n => 103.168.78.82\n)\n\n => Array\n(\n => 39.59.67.124\n)\n\n => Array\n(\n => 182.69.19.120\n)\n\n => Array\n(\n => 196.202.236.195\n)\n\n => Array\n(\n => 137.59.225.206\n)\n\n => Array\n(\n => 143.110.209.194\n)\n\n => Array\n(\n => 117.201.233.91\n)\n\n => Array\n(\n => 37.120.150.107\n)\n\n => Array\n(\n => 58.65.222.10\n)\n\n => Array\n(\n => 202.47.43.86\n)\n\n => Array\n(\n => 106.206.223.234\n)\n\n => Array\n(\n => 5.195.153.158\n)\n\n => Array\n(\n => 223.227.127.243\n)\n\n => Array\n(\n => 103.165.12.222\n)\n\n => Array\n(\n => 49.36.185.189\n)\n\n => Array\n(\n => 59.96.92.57\n)\n\n => Array\n(\n => 203.194.104.235\n)\n\n => Array\n(\n => 122.177.72.33\n)\n\n => Array\n(\n => 106.213.126.40\n)\n\n => Array\n(\n => 45.127.232.69\n)\n\n => Array\n(\n => 156.146.59.39\n)\n\n => Array\n(\n => 103.21.184.11\n)\n\n => Array\n(\n => 106.212.47.59\n)\n\n => Array\n(\n => 182.179.137.235\n)\n\n => Array\n(\n => 49.36.178.154\n)\n\n => Array\n(\n => 171.48.7.128\n)\n\n => Array\n(\n => 119.160.57.96\n)\n\n => Array\n(\n => 197.210.79.92\n)\n\n => Array\n(\n => 36.255.45.87\n)\n\n => Array\n(\n => 47.31.219.47\n)\n\n => Array\n(\n => 122.161.51.160\n)\n\n => Array\n(\n => 103.217.123.129\n)\n\n => Array\n(\n => 59.153.16.12\n)\n\n => Array\n(\n => 103.92.43.226\n)\n\n => Array\n(\n => 47.31.139.139\n)\n\n => Array\n(\n => 210.2.140.18\n)\n\n => Array\n(\n => 106.210.33.219\n)\n\n => Array\n(\n => 175.107.203.34\n)\n\n => Array\n(\n => 146.196.32.144\n)\n\n => Array\n(\n => 103.12.133.121\n)\n\n => Array\n(\n => 103.59.208.182\n)\n\n => Array\n(\n => 157.37.190.232\n)\n\n => Array\n(\n => 106.195.35.201\n)\n\n => Array\n(\n => 27.122.14.83\n)\n\n => Array\n(\n => 194.193.44.5\n)\n\n => Array\n(\n => 5.62.43.245\n)\n\n => Array\n(\n => 103.53.80.50\n)\n\n => Array\n(\n => 47.29.142.233\n)\n\n => Array\n(\n => 154.6.20.63\n)\n\n => Array\n(\n => 173.245.203.128\n)\n\n => Array\n(\n => 103.77.43.231\n)\n\n => Array\n(\n => 5.107.166.235\n)\n\n => Array\n(\n => 106.212.44.123\n)\n\n => Array\n(\n => 157.41.60.93\n)\n\n => Array\n(\n => 27.58.179.79\n)\n\n => Array\n(\n => 157.37.167.144\n)\n\n => Array\n(\n => 119.160.57.115\n)\n\n => Array\n(\n => 122.161.53.224\n)\n\n => Array\n(\n => 49.36.233.51\n)\n\n => Array\n(\n => 101.0.32.8\n)\n\n => Array\n(\n => 119.160.103.158\n)\n\n => Array\n(\n => 122.177.79.115\n)\n\n => Array\n(\n => 107.181.166.27\n)\n\n => Array\n(\n => 183.6.0.125\n)\n\n => Array\n(\n => 49.36.186.0\n)\n\n => Array\n(\n => 202.181.5.4\n)\n\n => Array\n(\n => 45.118.165.144\n)\n\n => Array\n(\n => 171.96.157.133\n)\n\n => Array\n(\n => 222.252.51.163\n)\n\n => Array\n(\n => 103.81.215.162\n)\n\n => Array\n(\n => 110.225.93.208\n)\n\n => Array\n(\n => 122.161.48.200\n)\n\n => Array\n(\n => 119.63.138.173\n)\n\n => Array\n(\n => 202.83.58.208\n)\n\n => Array\n(\n => 122.161.53.101\n)\n\n => Array\n(\n => 137.97.95.21\n)\n\n => Array\n(\n => 112.204.167.123\n)\n\n => Array\n(\n => 122.180.21.151\n)\n\n => Array\n(\n => 103.120.44.108\n)\n\n => Array\n(\n => 49.37.220.174\n)\n\n => Array\n(\n => 1.55.255.124\n)\n\n => Array\n(\n => 23.227.140.173\n)\n\n => Array\n(\n => 43.248.153.110\n)\n\n => Array\n(\n => 106.214.93.101\n)\n\n)\n```\nArchive for March, 2006: kashi's journal\n Layout: Blue and Brown (Default) Author's Creation\n Home > Archive: March, 2006\n\n# Archive for March, 2006\n\n## March recap, April goals\n\nMarch 31st, 2006 at 05:28 pm\n\nI did very poorly on my March goals.\n\nMARCH GOALS\n~ Pay double on my smaller college loan: Yes, more than double\n~ Limit groceries/Target/eating out to \\$250 (was \\$327 in Feb): Not even close, way more than in February!\n~ Drive to work no more than 5 times: drove 6 times\n~ Put \\$300 into savings: only \\$100 due to car repair\n~ Donate \\$25 to one organization of choice: Yes\n\nAPRIL GOALS\n~ Pay double on my smaller college loan\n~ Limit groceries/Target/eating out to \\$300\n~ Drive to work no more than 5 times\n~ Put \\$600+ into savings (if tax return comes)\n\nAnd my next Big Goal, get \\$1000 back into my emergency fund by June. I had to drain it to get my car fixed.\n\n## no keyboard\n\nMarch 29th, 2006 at 07:42 pm\n\nSo, the eBay guy selling the keyboard decided after I had sent payment that he needed an extra \\$50 for shipping. Uh, no. If you have a UPS calculator on your eBay page, and that calculator determines that shipping is \\$25, don't come back to me after the auction (and after I paid) and say shipping is suddenly \\$75. Luckily, he is fully refunding my money, and so graciously said he will not leave me negative feedback. Totally sketchy. I think I will lay off the eBay scene for a while. The only plus is that I have an extra \\$200 now. A bummer, though.\n\nThis really is the crappiest week. sigh* I guess when it rains, it pours, huh? I'd like to put my head under the covers and not come out for a few weeks.\n\n## *sigh*\n\nMarch 28th, 2006 at 03:24 pm\n\nThanks for all your thoughts. I got the dreaded phone call this morning. I am going to either visit my sister this weekend or have her come visit again. That's all I can really do, I guess. I told my parents to consider coming to visit next month, too. It would be good for them to get away. I wish this was the end of it...but they have a 28 year old horse. *sigh* It is particularly difficult when all of the animals are up there in years.\n\nI'm glad there are people here who understand what it's like to lose a pet. My parents are getting no sympathy whatsoever from co-workers, and my sister's \"friend\" rolled her eyes at her! No compassion. Maybe we get too attached...but I'd venture to guess that those people are too afraid to let themselves GET attached in the first place.\n\nI went home early yesterday. A migraine started building around 9:30am and I knew I wouldn't make it through the day. I slept - a LOT - and I'm feeling a bit better today. This is the weirdest cold I've ever had. It hit me like a freight train but I'm recovering fairly quickly.\n\nLet's see...in other news...I won a keyboard on eBay this morning! I have been planning to return my hypnosis CDs (the woman's voice is way too annoying to even concentrate), and my massage has been canceled again this month, so it all evens out in the end. I got a great deal on a 76-key professional keyboard, stand, and gig bag.",
null,
"Looking forward to playing again.\n\nSO still hasn't done his taxes. I want to get some sand between my toes, but that Mexican island trip might have to wait until next year. Maybe we can land a cheap flight to San Diego for a long weekend instead.\n\nMarch 27th, 2006 at 03:39 pm\n\nWhat a crappy weekend.\n\nI got really sick on Friday night and it stuck all weekend. I am still feeling ill today and am contemplating going home early. We'll see if the meds kick in or not.\n\nMy sister's childhood pet is dying. My parents are going through the awful pain of watching it happen and my sister is devastated. Needless to say, we spent a lot of time crying and not much else. I feel helpless. I know exactly how she feels, since my cat passed away two years ago. I wish there was something I could do. My parents are in very rough shape.\n\nWe did go to the zoo yesterday as a distraction procedure for my sister. Here's a little lion yawn to represent Mondays across the globe:",
null,
"## Friday snippets\n\nMarch 24th, 2006 at 05:05 pm\n\nMy friend still hasn't had her baby yet. He's due tomorrow. I can't wait! I took some photos of her giant preggo self for posterity.",
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"I'm hopefully meeting her for lunch today - I bet it's the last time I'll see her before the baby comes!\n\nReading about Cercis' gossip session with her coworker makes me long for that kind of connection at work again. I miss having someone (or several someones) to talk to! It seems like the only person I talk to on any given day is SO. I'm certainly not complaining, because I'm lucky to have him in my life, but I miss having interactions with other humans. My cats do plenty of talking too, but it's more along the lines of \"FEED ME FEED ME FEED ME FEED ME!\" It's just bizarre, being surrounded by hundreds of people and not conversing with any of them. Lonely in a crowd, anyone?\n\nI've been ogling 76-key keyboards on eBay lately. Anyone own one? I'd love to have a piano but don't want to lug it up or down the stairs, and we don't have room for one anyway. I'd like to have a large keyboard instead but don't want to pay full price. There are a few nice ones online...I'm just waiting to see how much they go for. If I could land one for \\$150, I'd be thrilled.\n\nSO still hasn't done his taxes, so the Mexico trip is still up in the air. I refuse to talk to him about it until he does his taxes and finds out how much he has to pay in. Perhaps that will motivate him to do them this weekend!\n\nThe playing gig will bring in \\$125! I'm super excited. Now I'm contemplating how to find more gigs like this. I know there are churches out there that love woodwind groups. I just need to find other (sane) players who want to form one. I put an ad up on craigslist...we'll see what comes of it.\n\nVivo para el fin de semana! Have a good one!\n\n## sister sister\n\nMarch 23rd, 2006 at 04:17 pm\n\nI can't believe there are 8 more days in March. On one hand, the year is going by at lightening speed, but on the other hand, I need more money! I haven't budgeted well at all this month.\n\nMy sister is coming to visit for the weekend. I have nothing planned and I'm not sure I want to plan anything. I'm already overbudget and she doesn't need constant diversions. Maybe I will leaf through the Entertainment Book and see if there is anything worthwhile. Otherwise, we will just chill. She likes to cook, so she can have free reign in the kitchen as far as I am concerned!\n\nI may have a paid playing gig coming up in a few weeks. I don't know how much it is going to pay, but it's a church service, and I haven't done one of those in years. I am trying to be a little more assertive about getting paid for my talents and not just handing them out for free. The book I'm reading is definitely helping. You can't get ahead if you do everything for free.\n\nI put a post-it note up on my computer that says \"Pursue Prosperity!\" There is nothing wrong with making more money!\n\n## Neutrogena refund\n\nMarch 21st, 2006 at 05:31 pm\n\nI've decided that Neutrogena is a great company. I bought their new mineral powder foundation but wasn't happy with it at all. I wrote them a nice complaint email and told them I was disappointed in the product. They sent me a refund check immediately! I am impressed. I've loved their other products that I've tried, so I figured it was worth an email to them, and it was!\n\nAs predicted, the Target/groceries/eating out budget for this month was a bust. We went grocery shopping last night and bought a lot of things to build up the pantry and freezer again. I am happy though, because it will keep us stocked for a while, which means I am more apt to cook.\n\nOf course, that was negated by the Indian takeout we picked up for dinner, but that was wonderful as always, and I don't regret it. How could I anyway, when SO so sweetly insisted on paying?\n\nI'm reading Secrets of Six-Figure Women (thanks to the person who recommended it!). It's fascinating. I need to develop more self-confidence in my abilities, that's for sure. I also have to stop thinking that making that much money is an impossible goal! It's possible! I am an intelligent, educated, strong woman! (rinse, lather, and repeat)\n\n## Happy Spring!\n\nMarch 20th, 2006 at 05:58 pm\n\nFriday's evening spending consisted of all of \\$12 for dinner. Not bad. We ended up staying at my house all night. No spend days on Saturday and Sunday, yay! I have to go to Target though (we're out of kitty litter and several other things, like milk and eggs), so I am totally not going to meet my goal this month. If I spend as much or less than I did in February, I'll be ok with that.\n\nOn Saturday I made homemade donuts and french bread. I made dinner both Saturday and Sunday, too. I love cooking on the weekends! I need to build up my pantry a little more, to make it easier.\n\nWe did get another five inches of snow, but it is melting already. Hopefully it will be gone soon! I can't wait for summer (funny thing to say on the first day of spring, I suppose).\n\nI think I could do the 8 day/7 night Mexico trip for \\$1075. I'd need to dip into some savings to pay for it. Still debating whether or not I should go for it! I'd really like to! If I really scrimp in April, I think I could.\n\n## bills, babies, & buckets of snow\n\nMarch 16th, 2006 at 03:22 pm\n\nMy credit card bill is massive this month, but it's all because of my car repair. Luckily, I have the emergency fund to fall back on. It is quite nice to have that safety net - I've never had it in the past. Now comes the task of building it back up again. Speaking of car repair, I updated the totals in my previous post. It makes me sick to see how much they charged me vs. what they should have charged me.\n\nMore snow today! We're expecting 5-8 inches, and it's been falling all night. I hope we don't get that much. I made it to work this morning, though, and on time, so I'm pretty pleased with myself. The bus showed up this time (though 20 minutes late)!\n\nI'm doing great on my driving challenge - I've only driven to work twice so far.",
null,
"I will not be able to meet my savings goal because of the car repair, but I can live with that. I'm still putting away \\$75, which is better than nothing. I'm doing fairly well on my eating out/Target/groceries challenge this month. I have about \\$42 left to spend.\n\nWith two weeks left, though, I hope I can make it. I'm going out with friends tomorrow night to celebrate St. Patty's Day, but we're going to avoid the crowded Irish pubs. Hopefully we can have a good time rather inexpensively.\n\nA friend of mine is about to have a baby, and I'm so excited! I bought a gift a few months ago when baby stuff was on clearance. I am looking forward to the phone call!\n\nOtherwise, not much going on...this is a slow month, and I'm greatful for it.\n\n## snow photos\n\nMarch 14th, 2006 at 03:49 am",
null,
"",
null,
"## run down of car repairs\n\nMarch 14th, 2006 at 03:13 am\n\nThis is for Russell (and anyone else interested in car repairs):\n\n- Lube, oil & filter (\\$29)\n- Tire rotation (\\$24)\n- Replace exhaust flange (\\$70 labor, \\$32 parts)\n- Exhaust replace, converter back (\\$100 labor, \\$125 parts)\n- Replace LS inner tie rod (\\$199 labor, \\$74 parts)\n- 4 wheel alignment (\\$90)\n- Brake adjust & clean (\\$50)\n- Belts R&R - A/C belt (\\$50 labor, \\$70 parts)\n\n- Supplies \\$47\n- Disposal \\$2.50\n\nThat's the repair shop's list on the invoice. \\$900 total WITH a \\$92 discount. Ouch!\n\n## snow\n\nMarch 13th, 2006 at 07:09 pm\n\nUnfortunately I have no idea how to properly link photos, but I have a great one showing how much snow we've gotten overnight. I couldn't get my car out of the driveway, so I hiked to the bus stop, only to wait an enternity and never have a bus show up. I went back home and have been in bed ever since! Delightful.\n\nNot much to report on the savings/spending front. I've been lying low after shelling out the \\$900 for my car. It runs so much better now. I'm hoping it will last for at least another 20,000 miles. Pretty please?\n\nMarch 9th, 2006 at 04:10 pm\n\nYikes, thanks for that article, DivaJen! I've been without Excedrin since Monday. No super excruciating headaches to speak of since then. I will definitely start limiting myself more - is that why my massage friend is always telling me my stomach is bad??\n\nGood news:\nI had a job interview yesterday and it went quite well. I'm not sure if I want the job, though. It would be slightly more interesting than what I'm doing now, but I can't see myself staying there forever, either. It may give me more graphic arts and Macintosh experience, though, which appeals to publishers. We'll see. They are going to let me know in the next couple of weeks. If it's more money, I'll definitely take it.\n\nI brought my car in to the shop last night (finally). Russell will be pleased to know that I am no longer attempting to asphyxiate myself by driving the Deathmobile. However, the shop called this morning and it's going to cost \\$900 to fix (and that is WITH a AAA discount). Ouch. I was expecting it to be high, and I have an emergency fund to fall back on, but I am still bummed. I hate handing over my carefully saved money for something like this.\n\nI may decide to use my tax refund for a Mexico trip instead of using it to pay off part of one of my college loans. I am tired of being practical! and I want to go to Mexico!!! I guess we'll see how I feel once I have the money in my hot little hands.\n\n## potential\n\nMarch 6th, 2006 at 08:57 pm\n\nI have a potential opportunity coming up. I'll post more when I know more.\n\nWhat a relaxing weekend. Dinner with the in-laws was wonderful - they opened up more than usual. I should have stayed out of the stores yesterday, though. I bought a magazine, a book, and a cookbook - tsk tsk! Also picked up the dreaded Girl Scout cookies and some crafting things. I'm way overbudget, as always. I rarely plan for these \"others\" like books and beads and cookies. They just happen, and then I scramble to cover the deficit. It's a recurring theme.\n\nI was thinking about using my tax refund for my college loans, but you know what? I'd much rather go on vacation. I found a beautiful, secluded beach resort in Mexico that is just what the doctor ordered. We'll see.\n\nI was going to attempt to live without Excedrin for one week...and failed miserably on the second day. My head was just killing me yesterday. I haven't touched the bottle today, though, so here's me, trying again. I want to make sure they aren't just rebound headaches.\n\n## Hallelujah, Friday\n\nMarch 3rd, 2006 at 03:48 pm\n\nFor mjrube94 (my favorite devil's advocate):\n1) SO is all for moving. Especially somewhere warm and by the ocean. He misses it, too.\n2) There are other schools in other places, namely Canada. Brrrr. I found a few other schools in Los Angeles, and they are options, too.\n3) I'm still researching all the schools and options...but I will definitely do my research on placements.\n4) This particular school has two sessions - Tues/Thurs full day classes and four-nights-per-week classes. If I took the night classes, I could definitely work a day job, and that's what I'd do.\n\nHaving said all that, I'm not sure that I will definitely set off in this direction, but it's simmering in the back of my mind. Also simmering is the idea that I should move out east (since that's where the publishing jobs are) and give this publishing thing one more shot. I'm also considering throwing up my hands and staying home all day (not really, but it would be nice).\n\nYesterday I was completely despondent, thus no post here. \"In a funk,\" as another blogger put it. I need to sit down and sort through all the things going on in my head. Buy new car or fix my car? - Buy a house here or move elsewhere? - SO wants to get married, do I? - Publishing career or give it up already? - Mexico trip or save money? AAAARRRRGGGGGG!!!!!\n\nThank God for Fridays (and the weekend).\n\n## crazy dream\n\nMarch 1st, 2006 at 09:36 pm\n\nSince this publishing job thing isn't really working out too well (unless I move out east), I have been daydreaming a bit.\n\nToday I stumbled across a professional makeup artist school. Being a makeup artist is my second chosen profession. I used to do makeup for all the school plays and have done makeup for weddings, parties, etc. I love, love, love makeup.\n\nI'd have to move to Hollywood and shell out \\$12,000 for a 12-month program (four nights per week). After that, though, I could do theatre makeup, or movies, or fashion shows. It would be so awesome to do prosthetic makeup - I built my Halloween costume one year, and it rocked.\n\nI probably wouldn't make any more than I am right now, I'd have to move to LA, and who knows what kind of work I would find after I finished the program. Still...it's tempting...and hard to make that leap. SO is already all for it!",
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"## ahh March\n\nMarch 1st, 2006 at 05:33 pm"
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https://file.scirp.org/Html/4-1880485_64134.htm | [
" Application of Artificial Neural Network for Analysis of Triangular Plate with Hole Considering Different Geometrical and Loading Parameters\n\nOpen Journal of Civil Engineering\nVol.06 No.01(2016), Article ID:64134,11 pages\n10.4236/ojce.2016.61004\n\nApplication of Artificial Neural Network for Analysis of Triangular Plate with Hole Considering Different Geometrical and Loading Parameters\n\nSaket Rusia, Krishna Kant Pathak\n\nDepartment of Civil and Environmental Engineering, National Institute of Technical Teachers’ Training and Research, Bhopal, India",
null,
"",
null,
"",
null,
"",
null,
"Received 6 February 2016; accepted 26 February 2016; published 29 February 2016\n\nABSTRACT\n\nIn this study, Artificial Neural Network has been employed for analysis of triangular plate with different geometrical and loading parameters. Plates, having different sizes of concentric holes are analyzed. Finite element analysis for 81 cases is carried out using ANSYS Workbench 15.0 software. Using these data of FEM analysis an Artificial Neural Network has been trained. The successfully trained network is further used for analysis of four new cases which are also validated by using ANSYS Workbench 15.0 software.\n\nKeywords:\n\nArtificial Neural Networks, Finite Element Analysis, Triangular Plate, ANSYS",
null,
"1. Introduction\n\nRegardless of the powerful analysis software now available those allow us to find out the numerical solution of various problems, including problems of structural analysis, the development of methods of approximate solution which would provide solutions in the form of simple analytic expressions is very important. One of the methods is artificial neural network also known as ANN. These are a functional abstraction of the biologic neural structures of the central nervous system.\n\nScientists have long been inspired by the human brain. In 1943, Warren S. McCulloch, a neuroscientist, and Walter Pitts , a logician, developed the first conceptual model of an Artificial Neural Network. In their paper, “A logical calculus of the ideas imminent in nervous activity”, they described the concept of a neuron, a single cell living in a network of cells that receives inputs, processes those inputs, and generates an output. Their work, and the work of many scientists and researchers that followed, did not mean to accurately describe how the biological brain works. Rather, an Artificial Neural Network was designed as a computational model based on the brain to solve certain kinds of problems.\n\nANNs are powerful pattern recognizers and classifiers. Garrett has given an interesting engineering definition of the ANN as: “a computational mechanism able to acquire, represent, and compute mapping from one multivariate space of information to another, given a set of data representing that mapping”. Their computing abilities have been proven in the fields of prediction and estimation, pattern recognition, and optimization. They are suitable particularly for problems too complex to be modeled and solved by classical mathematics and traditional procedures. Neural networks can be hardware (neurons are represented by physical components) or software based (computer models), and can use a variety of topologies and learning algorithms. Neural networks have been used for various structural analysis like fully stressed design of trusses, buckling behavior of plates, stress concentration factor analysis for membranes etc.\n\nIn Figure 1, an Artificial Neural Network consisting of an input layer with three neurons, one hidden layer with four neurons, and an output layer with two neurons is shown. There would be a state function and transfer function like summation function, sigmoid squashing function respectively. Here, a training algorithm is needed that can be a back-propagation algorithm. Neurons are the processing elements of network. Neuron consists of a set of weighted input connections, a bias input, a state function, a nonlinear transfer function, and an output. Figure 2 shows the structure of a neuron.\n\nP. Emmanuel Nicholas et al. proposed a novel approach to study neural network based buckling strength prediction of laminated composite plate with central cut-out. The laminated composite plates with holes analyzed using finite element analysis by optimizing the parameters like thickness, orientation, material and the stacking sequence to obtain the desired characteristics for these structures. They showed that using finite element analysis makes the process a more tedious job and thus proposed to construct the Artificial Neural Network to predict the buckling behavior of the composite plate. Hojjat Adeli presented the first journal article on neural network application in civil/structural engineering in 1989.\n\nFigure 1. Neural network structure.\n\nFigure 2. Structure of a neuron.\n\nIn many previous research papers, membrane with holes or cutouts analysed using finite element software like ANSYS , also stress field around circular holes in plates with arbitrary thickness has been studied but most of the time loading considered is to be in plane loading, However, it seems to be difficult to locate a work that quantifies the use of ANNs for analysis of equivalent stress, strain and directional deformation in a triangular plate subjected to vertical surface pressure without performing finite element analysis. The Artificial Neural Network is used as an alternative analysis tool to analyze plates with hole since it can handle uncertainty through the probability method. In some of the following research papers finite element analysis has been performed for plates and membranes with cut-outs. Zuxing Pan, Yuansheng Cheng and Jun Liu , dealt with a complex variable method and proposed stress functions to obtain the solution for stress distribution around rectangular hole in finite plate subjected to uniaxial tension. They analyzed effect of hole sizes, hole orientation and plate’s aspect ratio on stress distribution. Jeom Kee Paik examined the ultimate strength of metallic plates with central circular cut-out under shear loading. The influence of boundary conditions on the buckling load for rectangular plates of various cut-out shape, length/thickness ratio, and ply orientation was examined by Buket Okutan Baba . Boundary conditions considered and their various combinations were clamped, and pinned. The plates were subjected to in-plane compression load. The results of experimentation were validated using numerical analysis by ANSYS. A.V Singh, U.K Paul , presented the results of their study which was based on generalized work-energy method for rectangular plates with circular cut-out. Optimum design of holes and notches by considering fatigue life was presented by Hwai Chung Wu, and Bin Mu . V.G. Ukadgaonker, and D.K.N. Rao gave a general solution for bending of symmetric laminates with holes considering any shape of hole in symmetric laminates subjected to remotely apply bending or twisting moments. Moments around circular, elliptical, triangular, square, rectangular and several irregular shaped holes in cross-ply and angle ply symmetric laminates are obtained. Hsuan-Teh Hu, and Bor-Horng Lin studied the buckling resistance of symmetrically laminated plates with a given material system subjected to uniaxial compression. The research was done with plates having different plate thicknesses, aspect ratios, central circular cut-outs and different end conditions.\n\nIn this study, Artificial Neural Network has been employed for analysis of maximum equivalent von Mises stress, strain and directional deformation in equilateral triangular plate with different geometrical and loading patterns. Plates, having different size concentric holes are analyzed. Finite element analysis for 81 cases is carried out using ANSYS Workbench 15.0 software. Using these data of FEM analysis an Artificial Neural Network has been trained. The successfully trained network is further used for analysis of five new cases which are also validated using ANSYS Workbench 15.0 software.\n\n2. Structural Modelling and Analysis\n\nModeling, meshing and analysis contours of plate are shown in Figure 3. Plate is a polygon of three sides having concentric hole. Size of edges of plate, thickness, size of hole diameter and loading pressure are varying parameters. In Table 1, isotropic elastic constants values are shown, those were used given to as elemental properties. Other, in use entities for finite element modeling is provided in Table 2. In total 81 cases are generated, and these are stated in Table 3.\n\nGeometry of the plates are created using ANSYS workbench 15.0 geometry tool, design modeler and then analysed using ANSYS mechanical or multi-physics too. In these models, fixed edge support condition is provided. Varying loading pressure is acting in -z direction where plates are lying in X-Y plane.\n\nIn Figure 3, A is showing the model view, B is the meshed structure, in C, Loading has been shown and in D, E, F, contour of variation in maximum equivalent von Mises stress, strain and directional deformation are shown respectively.\n\n3. Finite Element Analysis\n\nFinite element analysis has been performed using ANSYS Workbench 15.0 and results for following parameters are recorded,\n\n1) Maximum Equivalent von Mises Stress\n\n2) Maximum Equivalent von Mises Strain\n\n3) Directional Deformation in Z-Direction\n\nOutput results are tabulated in Table 4. These values have been used as training data for Artificial Neural Network, in the next section.\n\nFigure 3. Modeling, meshing & analyses of plate.\n\nTable 1. Isotropic elastic constants.\n\nTable 2. Entities for finite element modeling.\n\n4. Application of Neural Network\n\nThe input, output data given in Table 3 and Table 4 are used for training of the neural network. A 4-6-3 size back propagation neural has been trained. The input parameters are edge dimension, hole diameter, thickness of plate and pressure applied and output parameters are maximum equivalent von Mises stress, maximum equivalent von Mises strain and directional deformation in Z-direction. For ANN, an in-house developed software has been used. The error tolerance is kept 0.005. It took 1996082 epochs to converge to this tolerance. Thus trained network is used for fully analysis of four new cases of model plates, given in Table 5.\n\nTable 3. Input for finite element models.\n\nTable 4. Data recorded as output of finite element analyses.\n\nValidation of results has been performed in Table 6, also percentage (%) variation is carried out between the analysed parameters Artificial Neural Network and ANSYS Workbench. Graphical representation of analysed parameters for new cases model plates are shown in Figure 4, Figure 6, Figure 8 and percentage variation are shown in Figure 5, Figure 7, and Figure 9. It can be observed that averages of absolute positive errors are 3.85, 4.2, and 3.98 for maximum equivalent von Mises stress, maximum equivalent von Mises strain and directional deformation in -Z direction respectively which are small values. Thus it has been proved that the use of Artificial Neural Network can avoid the lengthy and tedious complex modeling and analysis using costly FEM software.\n\nFurthermore, correlation analysis of ANSYS and ANN results have been also carried out and shown in Figures 10-12.\n\nThese regression maps are between ANSYS as observed values and ANN as predicted values. The more variance that is accounted for by the regression model the closer the data points will fall to the fitted regression line. Theoretically, if a model could explain 100% of the variance, the fitted values would always equal the observed values and, therefore, all the data points would fall on the fitted regression line. It can be observed that the sum of squared residuals (R2) for all three output parameters are close to 1 that accounts for 100% of the variance. Hence it can be proved that neural network predictions are close to FEM results.\n\nTable 5. New model cases for validation.\n\nTable 6. Validation of results.\n\nFigure 4. Max. equivalent von Mises stress by ANN and ANSYS for model case 1.\n\nFigure 5. % Variation of max. equivalent stress for ANN and ANSYS for model case 1.\n\nFigure 6. Max. equivalent von Mises strain by ANN and ANSYS for model case 2.\n\nFigure 7. % Variation of max. equivalent strain for ANN and ANSYS for model case 2.\n\nFigure 8. Directional deformation in -Z direction by ANN and ANSYS for model case.\n\nFigure 9. % Variation of directional deformation in -Z direction of ANN and ANSYS for model case 3.\n\nFigure 10. Regression analyses for max. equivalent von Mises stress between ANN and ANSYS for new models.\n\nFigure 11. Regression analyses for Max. equivalent von Mises strain between ANN and ANSYS for new models.\n\nFigure 12. Regression analyses for directional deformation in -Z direction between ANN and ANSYS for new models.\n\n5. Conclusions\n\nFollowings are the salient conclusions of this study:\n\n1) Artificial Neural Network (ANN) is a very powerful tool for stress analysis of triangular plates with concentric cut-outs.\n\n2) Artificial Neural Network approach is easy and fast whereas traditional techniques are tedious and time consuming and require greater skills.\n\n3) The differences between the maximum equivalent von Mises stress, strain and directional deformation calculated by ANN and ANSYS Workbench 15.0 are low.\n\n4) Using ANN, dependency upon costly analysis and design packages can be avoided.\n\nCite this paper\n\nSaketRusia,Krishna KantPathak, (2016) Application of Artificial Neural Network for Analysis of Triangular Plate with Hole Considering Different Geometrical and Loading Parameters. Open Journal of Civil Engineering,06,31-41. doi: 10.4236/ojce.2016.61004\n\nReferences\n\n1. 1. McCulloch, W.S. and Pitts, W.H. (1943) A Logical Calculus of the Ideas Imminent in Nervous Activity. The Bulletin of Mathematical Biophysics, 5, 115-133.\nhttp://dx.doi.org/10.1007/BF02478259\n\n2. 2. Garrett, J.H. (1994) Where and Why Artificial Neural Networks Are Applicable in Civil Engineering. Journal of Computing in Civil Engineering, 8, 129-130.\nhttp://dx.doi.org/10.1061/(ASCE)0887-3801(1994)8:2(129)\n\n3. 3. Nicholas, P.E., Padmanaban, K.P., Vasudevan, D. and Selvaraj, I.J. (2014) Neural Network Based Buckling Strength Prediction of Laminated Composite Plate with Central Cutout. Applied Mechanics and Materials, 592-594, 560-564.\nhttp://dx.doi.org/10.4028/www.scientific.net/AMM.592-594.560\n\n4. 4. Adeli, H. (2001) Neural Networks in Civil Engineering: 1989-2000. Computer-Aided Civil and Infrastructure Engineering, 16, 126-142.\nhttp://dx.doi.org/10.1111/0885-9507.00219\n\n5. 5. ANSYS (2013) Workbench User’ s Guide. ANSYS Work. 15.0 15317, 724-746.\n\n6. 6. Pan, Z., Cheng, Y. and Liu, J. (2013) Stress Analysis of a Finite Plate with a Rectangular Hole Subjected to Uniaxial Tension Using Modified Stress Functions. International Journal of Mechanical Sciences, 75, 265-277.\nhttp://dx.doi.org/10.1016/j.ijmecsci.2013.06.014\n\n7. 7. Paik, J.K. (2008) Ultimate Strength of Perforated Steel Plates under Combined Biaxial Compression and Edge Shear Loads. Thin-Walled Structures, 46, 207-213.\nhttp://dx.doi.org/10.1016/j.tws.2007.07.010\n\n8. 8. Baba, B.O. (2007) Buckling Behavior of Laminated Composite Plates. Journal of Reinforced Plastics and Composites, 26, 1637-1655.\nhttp://dx.doi.org/10.1177/0731684407079515\n\n9. 9. Singh, A. and Paul, U. (2003) Finite Displacement Static Analysis of Thin Plate with an Opening––A Variational Approach. International Journal of Solids and Structures, 40, 4135-4151.\nhttp://dx.doi.org/10.1016/S0020-7683(03)00204-X\n\n10. 10. Wu, H.C. and Mu, B. (2003) On Stress Concentrations for Isotropic/Orthotropic Plates and Cylinders with a Circular Hole. Composites Part B: Engineering, 34, 127-134.\nhttp://dx.doi.org/10.1016/S1359-8368(02)00097-5\n\n11. 11. Ukadgaonker, V.G. and Rao, D.K.N. (2000) A General Solution for Moments around Holes in Symmetric Laminates. Composite Structures, 49, 41-54.\nhttp://dx.doi.org/10.1016/S0263-8223(99)00124-5\n\n12. 12. Hu, H.-T. and Lin, B.-H. (1995) Buckling Optimization of Symmetrically Laminated Plates with Various Geometries and End Conditions. Composites Science and Technology, 55, 277-285.\nhttp://dx.doi.org/10.1016/0266-3538(95)00105-0"
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https://somme2016.org/how-to/what-is-an-8-bit-unsigned-integer/ | [
"# What is an 8-bit unsigned integer?\n\n## What is an 8-bit unsigned integer?\n\nAn 8-bit unsigned integer has a range of 0 to 255, while an 8-bit signed integer has a range of -128 to 127 – both representing 256 distinct numbers. It is important to note that a computer memory location merely stores a binary pattern.\n\n## What is integer Labview?\n\nIntegers represent whole numbers and can be positive or negative. Refer to the Numeric Data Types Table for more information about numeric data type bits, digits, and range.\n\nWhat is unsigned 64 bit integer?\n\nThe maximum (unsigned) 64-bit integer is 18446744073709551615 . This is (2^64)-1, which is essentially the square of (2^32)-1, which is “about” 4 billion. In general, you can estimate that every 10 bits represents 3 decimal digits.\n\n### What is 8-bit unsigned binary?\n\nUnsigned binary numbers are, by definition, positive numbers and thus do not require an arithmetic sign. An m-bit unsigned number represents all numbers in the range 0 to 2m − 1. For example, the range of 8-bit unsigned binary numbers is from 0 to 25510 in decimal and from 00 to FF16 in hexadecimal.\n\n### What is an 8 byte integer?\n\n8 byte unsigned integer. uintptr_t. Unsigned integer of size equal to a pointer. These type aliases are equivalent to using the name of the corresponding base type in the previous table and are appropriately defined for each data model. For example, the type name uint8_t is an alias for the type unsigned char.\n\nHow does LabVIEW transmit data?\n\nLabVIEW follows a dataflow model for running VIs. When a node executes, it produces output data and passes the data to the next node in the dataflow path. The movement of data through the nodes determines the execution order of the VIs and functions on the block diagram.\n\n## What is the 8 bit integer limit?\n\n256\nWith 8 bits, the maximum number of values is 256 or 0 through 255.\n\n## What is an 8-bit number?\n\n8 bits, can represent positive numbers from 0 to 255. hexadecimal. A representation of 4 bits by a single digit 0..\n\nWhich of the following is 8 byte?\n\nDiscussion :: Datatypes – General Questions (Q. No. 2)\n\n2. Which of the following is an 8-byte Integer?\n[A]. Char [B]. Long [C]. Short [D]. Byte [E]. Integer Answer: Option B Explanation: No answer description available for this question. Workspace Report errors Name : Email: Workspace Report"
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https://msp.org/akt/2016/1-1/p03.xhtml | [
"#### Vol. 1, No. 1, 2016\n\n Download this article",
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"For screen For printing",
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"Recent Issues Volume 4, Issue 4 Volume 3, Issue 4 Volume 3, Issue 3 Volume 3, Issue 2 Volume 3, Issue 1 Volume 2, Issue 4 Volume 2, Issue 3 Volume 2, Issue 2 Volume 2, Issue 1",
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"The Journal About the Journal Editorial Board Subscriptions Ethics Statement Submission Guidelines Submission Form Editorial Login Ethics Statement ISSN: 2379-1691 (e-only) ISSN: 2379-1683 (print) Author Index To Appear Other MSP Journals\nOn some negative motivic homology groups\n\n### Tohru Kohrita\n\nVol. 1 (2016), No. 1, 19–41\nDOI: 10.2140/akt.2016.1.19\n##### Abstract\n\nFor an arbitrary separated scheme $X$ of finite type over a finite field ${\\mathbb{F}}_{q}$ and a negative integer $j$, we prove, under the assumption of resolution of singularities, that ${H}_{-1}\\left(X,ℤ\\left(j\\right)\\right)$ is canonically isomorphic to ${H}_{-1}\\left({\\pi }_{0}\\left(X\\right),ℤ\\left(j\\right)\\right)$ if $j=-1$ or $-2$, and ${H}_{i}\\left(X,ℤ\\left(j\\right)\\right)$ vanishes if $i\\le -2$ and $i-j\\le 1$. As the group ${H}_{-1}\\left({\\pi }_{0}\\left(X\\right),ℤ\\left(j\\right)\\right)$ is explicitly known, this gives a explicit calculation of motivic homology of degree $-1$ and weight $-1$ or $-2$ of an arbitrary scheme over a finite field.\n\n##### Keywords\nmotivic homology, schemes over finite fields\nPrimary: 14F42\nSecondary: 19E15\n##### Milestones",
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https://www.my-courses.net/2021/05/solution-exercise-34-python-algorithm-which-determines-the-list-of-words-containing-at-least-two-digits.html | [
"# Solution Exercise 34: python algorithm which determines the list of words containing at least two digits",
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"#### Exercise 34\n\nWrite a Python algorithm as a function which takes a string text T as a parameter and which returns the list of words containing at least two digits. Example: if T = ‘Python2.7 is replaced by Python3.X since 2018’, the function returns the list [‘Python2.7’, ‘2018’]\n\n#### Solution\n\n``# creation of a function which calculates the number of digits on a stringdef number_digit (s): # initialization of the number of digits nbrDigit = 0 for x in s: if x.isdigit(): nbrDigit = nbrDigit + 1 return nbrDigit# creation of the function which returns the list of words containing at least two digitsdef list2Digits(s): # initialization of the searched words list listWord = [] # convert string s to a list L = s.split () for word in L: if number_digit(word) >= 2: listWord.append (word) return listWord# ExampleT = \"Python2.7 has been replaced by Python3.X since 2018\"print (list2Digits(T)) # prints: ['Python2.7', '2018']``\nYounes Derfoufi\nmy-courses.net"
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https://www.blog.sindibad.tn/the-3-6-9-theory-by-nikola-tesla-explains-universe-secret-code/ | [
"# 369 Theory of Tesla – Universe Secret Code\n\nThe 369 Theory of Tesla also called as Tesla Code 369. The 3 6 9 theory by Nikola Tesla explains the secret code of the universe, which can be used to understand the mysteries of the universe. The video also explains how to find digital root of a number. (Digital root method).\nAfter watching the video, you will understand how our whole universe is controlled by numbers 3, 6 and 9 and how they are present everywhere in the universe.\nYou will also know how the numbers 3, 6, 9 represent a higher dimensional world while rest of the numbers belong to our three-dimensional world.\nI will also discuss, how number 9 is the controller of all the other numbers and how it can control even space and time?\nNikola Tesla, the greatest mind of all the time and also a man of mysteries said that- if you knew the magnificence of 3, 6 and 9, you would have a key to the universe !\nHad he found some code of the universe?\nHere I will provide many mathematical and physical evidences and examples to prove, how the number 9 is present everywhere and is the controller of all the other numbers and the universe.\nOur universe is so mathematical, almost everything in nature is highly symmetrical and follows geometrical patterns. We can see these patterns in the form of golden ratio and Fibonacci sequence present in many things in nature.\nEverything is made up of particles and if we go deeper at even more microscopic level, we reach into the quantum world, which is pure mathematics.\nAll the Living things in this world are made up of cells. Multicellular organisms are formed by the division of a single cell. A single cell divides into 2 cells, then after division of both the cells we get 4 cells, then 8, 16, 32, 64 and 128 cells are produced. In this way the doubling sequence continues and finally we have a full multicellular organism.\nAccording to some researchers, numbers- 1, 2, 4, 8, 7 and 5 represent our three dimensional world where we live in while numbers 3 6 9 belong to a higher dimensional world.\nWhether we double or half the numbers, if we start with number 3, we get numbers 3 and 6 only.\nEven if we double or half the numbers starting from number 6, we will get the same numbers 3 and 6 every time. But surprisingly even this time 9 is absent in the sequence.\nActually number 9 controls the numbers 3 and 6. And numbers 1, 2, 4 are controlled by number 3\nWhile numbers 8, 7 and 5 are controlled by number 6. In this way number 9 controls all the numbers.\nWhether we double or half the numbers, starting from the number 9, we always get the number 9 only. It Implies that number 9 represents itself! Number 9 is present everywhere in the universe!\nwhen we were dividing the circle in half repeatedly, we were reaching the singularity of space. Since every time we got number 9, it means number 9 follows the space up to its singularity.\nAnd when we were increasing the number of sides of the polygon, the polygon was looking like a circle. As the number of sides of the polygon approaches infinity it becomes a circle. As in all the polygons we got number 9, it means number 9 also follows space upto infinity.\nSo it is clear that number 9 is present from singularity of space to the Infinity of space.\nNow I will give you some more examples, that will help you in believing that number 9 exists at all the places in the universe.\nspeed of light 186282miles/sec=1+8+6+2+8+2=27=2+7=9\nDiameter of the earth 7920 miles= 7+9+2+0=18=1+8=9\nDiameter of the Sun 864000 miles= 8+6+4+0+0+0=18=1+8=9\nDiameter of the Moon 2160 miles= 2+1+6+0=9\nNumber of planets in our solar system= 9 (controversial)\nDoes number 9 has any relationship with time also? (They all reduce to 9)\nSeconds in one hour 3600\nseconds in a day 86400\nseconds in a week 604800\nseconds in a month 2592000\nSeconds in a year 31536000\nMinutes in a day 1440\nMinutes in a week 10080\nMinutes in a month 43200\nMinutes in a year 525600\nand so on…\nBonus example:\nTotal Duration of pregnancy – 9 months\nIt proves that number 9 is present everywhere in space and time and also controls it.\nIf we talk about Hindu religion, we will find number 9 at different places. like in Om frequency 432 hz and its shape. In 108 sacred number etc.\nNumber 9 represents everything and also nothingness.\nActually number 9 is same as Number 0 in some sense.\nOne more thing, As we get 0 on multiplying any number by 0. In the same way we get number 9 as the digital root on multiplying it with number 9.\n0=9\nIf we add all the numbers except number 9, that is from number 1 to number 8; the sum reduces to number 9\n1 + 2 + 3 + = 36 = 9\nHence we can say that, number 9 represents all the other numbers. It represents itself and also nothingness and everything!!! Number 9 is the king of all the numbers that controls all the other numbers and space time..\n\n## Like it? Share with your friends!",
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"Dislike\n1486\nDislike",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9126666,"math_prob":0.9773809,"size":4930,"snap":"2022-27-2022-33","text_gpt3_token_len":1241,"char_repetition_ratio":0.16280958,"word_repetition_ratio":0.03943045,"special_character_ratio":0.25841784,"punctuation_ratio":0.089060985,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9892451,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-20T06:42:12Z\",\"WARC-Record-ID\":\"<urn:uuid:03a53db1-a3fc-4f79-9860-de26ca6fa196>\",\"Content-Length\":\"483312\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3f306001-7e8a-439f-85ab-9df177c8c3f4>\",\"WARC-Concurrent-To\":\"<urn:uuid:32500473-cf3d-4074-94fd-363c7077fc44>\",\"WARC-IP-Address\":\"46.105.57.169\",\"WARC-Target-URI\":\"https://www.blog.sindibad.tn/the-3-6-9-theory-by-nikola-tesla-explains-universe-secret-code/\",\"WARC-Payload-Digest\":\"sha1:SXTZ2IEGKMTGJ5GSU5HYZKWHH462JHDV\",\"WARC-Block-Digest\":\"sha1:MOLN55KPQXKAS24AWATCLD7U5RUPGO2C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573908.30_warc_CC-MAIN-20220820043108-20220820073108-00680.warc.gz\"}"} |
https://www.studysquare.co.uk/test/Maths/OCR/A-level/Numerical-integration | [
"",
null,
"NEXT TOPIC →\n\n# Numerical integration for OCR A-level Maths",
null,
"1. Trapezium rule\n2. Underestimates and overestimates\n3. Calculating area under a curve\n4. Numerical integration problems in context\n\nThe Trapezium Rule can be used to approximate the value of a definite integral. This is done by dividing up the area into equally wide strips, each of which is considered to be a trapezium. The area under the curve, and thus the definite integral, is given by multiplying the width of the strips with the sum of the average of the first and last vertical and the sum of the rest of the verticals. The more strips used to calculate the area, the more accurate the approximation will be.",
null,
"The Trapezium Rule only gives an approximation of the area under the curve and not an exact number. If the function is concave up, the approximation will be an overestimate, whereas if the function is concave down, it will be an underestimate.",
null,
"The area bounded between two curves can be found by subtracting the lower function by the upper one and integrating between the limits.",
null,
"The Trapezium Rule can be used in modelling a variety of situations.",
null,
"# 1\n\nHow can using the Trapezium rule to evaluate an integral be made more accurate?\n\nThe more strips used in the Trapzeium rule, the more accurate the approximation will be.",
null,
"# 2\n\nCalculate the shaded area of the given graph.\n\nThe area is given by the integral ∫⁰₋₁x³ + x² + 5 − (x²/2 + 4)dx = ∫⁰₋₁x³ + x²/2 + 1dx = [x⁴/4 + x³/6 + x]⁰₋₁ = 0⁴/4 + 0³/6 + 0 − ((−1)⁴/4 + (−1)³/6 − 1) = 11/12.",
null,
"# 3\n\nA marble is rolling for 12 seconds. Its speed is recorded every 3 seconds and given in the table. Calculate the distance travelled by the marble.\n\nThe distance travelled by the marble can be approximated using the Trapezium rule.\nDistance = 3(1/2(0.3 + 1.6) + 0.7 + 0.9 + 1.2) = 11.25.",
null,
"# 4\n\nWhich of the two given uses of the Trapezium rule will give a more accurate approximation of the area under the curve?\n\nDiagram A uses more strips, therefore will give a more accurate approximation of the area.",
null,
"# 5\n\nA ball is rolling for 16 seconds. The speed of the ball is recorder ever 4 seconds. Calculate the distance travelled by the ball during its motion.\n\nThe distance travelled by the ball can be approximated using the Trapezium rule.\nDistance = 4(1/2(0.2 + 1.3) + 0.5 + 0.8 + 1) = 12.2 m.",
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"End of page"
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https://math.stackexchange.com/questions/2661492/corollary-to-markovs-inequality | [
"# Corollary to Markov's Inequality [closed]\n\nHere is a particular case of Markov's Inequality that I failed to prove:\n\nLet $X$ be a non-negative integer-valued random variable with $\\mathbb{E}(X)\\leq m$ then $$\\mathbb{P}(X=0)\\geq 1-m$$\n\nDoes anyone has an idea how to prove this? Thank you.\n\n## closed as off-topic by user21820, The Phenotype, zz20s, Ethan Bolker, NamasteFeb 22 '18 at 18:47\n\nThis question appears to be off-topic. The users who voted to close gave this specific reason:\n\n• \"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.\" – user21820, The Phenotype, zz20s, Ethan Bolker, Namaste\nIf this question can be reworded to fit the rules in the help center, please edit the question.\n\n$$m\\geq\\mathsf EX=\\sum_{n=0}^{\\infty}n\\mathsf P(X=n)\\geq\\sum_{n=1}^{\\infty}\\mathsf P(X=n)=\\mathsf P(X\\geq1)=1-\\mathsf P(X=0)$$\n• To find it as corollary of Markovs inequality subtitute $a=1$ in $aP(X\\geq a)\\leq EX$ and work out. – drhab Feb 22 '18 at 12:05"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7700315,"math_prob":0.95419896,"size":709,"snap":"2019-26-2019-30","text_gpt3_token_len":238,"char_repetition_ratio":0.11914894,"word_repetition_ratio":0.7234042,"special_character_ratio":0.2961918,"punctuation_ratio":0.05263158,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99585074,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-18T21:21:16Z\",\"WARC-Record-ID\":\"<urn:uuid:8fea4b2b-ccc9-4e0a-bb41-c3c269d56c31>\",\"Content-Length\":\"124202\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:38e1804e-fb8b-45ee-89fe-7b32920ba425>\",\"WARC-Concurrent-To\":\"<urn:uuid:a6a521c6-a3ec-489c-9b36-da75ee80b5ec>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/2661492/corollary-to-markovs-inequality\",\"WARC-Payload-Digest\":\"sha1:MDZ3RHPBAR6TJOJFXPZC5RON6F2DNKV3\",\"WARC-Block-Digest\":\"sha1:PVU2ID66FFDAL4BBIW7WBFBMJZY5OLIT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525829.33_warc_CC-MAIN-20190718211312-20190718233312-00519.warc.gz\"}"} |
http://www.mathematics21.org/view/volume-1/index-286.html | [
"",
null,
"19.3. POINTFREE FUNCOID AS CONTINUATION\n\n286\n\n19.3. Pointfree funcoid as continuation\n\nProposition\n\n1510\n\n.\n\nLet\n\nf\n\nbe a pointfree funcoid. Then for every\n\nx\n\nSrc\n\nf\n\n,\n\ny\n\nDst\n\nf\n\nwe have\n\n1\n\n. If (Src\n\nf,\n\nZ\n\n) is a filtrator with separable core then\n\nx\n\n[\n\nf\n\n]\n\ny\n\n⇔ ∀\n\nX\n\nup\n\nZ\n\nx\n\n:\n\nX\n\n[\n\nf\n\n]\n\ny\n\n.\n\n2\n\n. If (Dst\n\nf,\n\nZ\n\n) is a filtrator with separable core then\n\nx\n\n[\n\nf\n\n]\n\ny\n\n⇔ ∀\n\nY\n\nup\n\nZ\n\ny\n\n:\n\nx\n\n[\n\nf\n\n]\n\nY\n\n.\n\nProof.\n\nWe will prove only the second because the first is similar.\n\nx\n\n[\n\nf\n\n]\n\ny\n\ny\n\n6\n\nDst\n\nf\n\nh\n\nf\n\ni\n\nx\n\n⇔ ∀\n\nY\n\nup\n\nZ\n\ny\n\n:\n\nY\n\n6 h\n\nf\n\ni\n\nx\n\n⇔ ∀\n\nY\n\nup\n\nZ\n\ny\n\n:\n\nx\n\n[\n\nf\n\n]\n\nY.\n\nCorollary\n\n1511\n\n.\n\nLet\n\nf\n\nbe a pointfree funcoid and (Src\n\nf,\n\nZ\n\n0\n\n), (Dst\n\nf,\n\nZ\n\n1\n\n) be\n\nfiltrators with separable core. Then\n\nx\n\n[\n\nf\n\n]\n\ny\n\n⇔ ∀\n\nX\n\nup\n\nZ\n\n0\n\nx, Y\n\nup\n\nZ\n\n1\n\ny\n\n:\n\nX\n\n[\n\nf\n\n]\n\nY.\n\nProof.\n\nApply the proposition twice.\n\nTheorem\n\n1512\n\n.\n\nLet\n\nf\n\nbe a pointfree funcoid. Let (Src\n\nf,\n\nZ\n\n0\n\n) be a binarily\n\nmeet-closed filtrator with separable core which is a meet-semilattice and\n\nx\n\nSrc\n\nf\n\n: up\n\nZ\n\n0\n\nx\n\n6\n\n=\n\nand (Dst\n\nf,\n\nZ\n\n1\n\n) be a primary filtrator over a boolean lattice.\n\nh\n\nf\n\ni\n\nx\n\n=\n\nDst\n\nf\n\nl\n\nhh\n\nf\n\nii\n\nup\n\nZ\n\n0\n\nx.\n\nProof.\n\nBy the previous proposition for every\n\ny\n\nDst\n\nf\n\n:\n\ny\n\n6\n\nDst\n\nf\n\nh\n\nf\n\ni\n\nx\n\nx\n\n[\n\nf\n\n]\n\ny\n\n⇔ ∀\n\nX\n\nup\n\nZ\n\n0\n\nx\n\n:\n\nX\n\n[\n\nf\n\n]\n\ny\n\n⇔ ∀\n\nX\n\nup\n\nZ\n\n0\n\nx\n\n:\n\ny\n\n6\n\nDst\n\nf\n\nh\n\nf\n\ni\n\nX.\n\nLet’s denote\n\nW\n\n=\n\nn\n\ny\n\nu\n\nDst\n\nf\n\nh\n\nf\n\ni\n\nX\n\nX\n\nup\n\nZ\n\n0\n\nx\n\no\n\n. We will prove that\n\nW\n\nis a generalized filter\n\nbase over\n\nZ\n\n1\n\n. To prove this enough to show that\n\nV\n\n=\n\nn\n\nh\n\nf\n\ni\n\nX\n\nX\n\nup\n\nZ\n\n0\n\nx\n\no\n\nis a generalized\n\nfilter base.\n\nLet\n\nP\n\n,\n\nQ ∈\n\nV\n\n. Then\n\nP\n\n=\n\nh\n\nf\n\ni\n\nA\n\n,\n\nQ\n\n=\n\nh\n\nf\n\ni\n\nB\n\nwhere\n\nA, B\n\nup\n\nZ\n\n0\n\nx\n\n;\n\nA\n\nu\n\nZ\n\n0\n\nB\n\nup\n\nZ\n\n0\n\nx\n\n(used the fact that it is a binarily meet-closed and theorem\n\n535\n\nand\n\nR v P u\n\nDst\n\nf\n\nQ\n\nfor\n\nR\n\n=\n\nh\n\nf\n\ni\n\n(\n\nA\n\nu\n\nZ\n\n0\n\nB\n\n)\n\nV\n\nbecause Dst\n\nf\n\nis strongly separable by\n\nproposition\n\n579\n\nSo\n\nV\n\nis a generalized filter base and thus\n\nW\n\nis a generalized filter\n\nbase.\n\nDst\n\nf\n\n/\n\nW\n\n⇔ ⊥\n\nDst\n\nf\n\n/\n\nd\n\nDst\n\nf\n\nW\n\nby theorem\n\n572\n\nThat is\n\nX\n\nup\n\nZ\n\n0\n\nx\n\n:\n\ny\n\nu\n\nDst\n\nf\n\nh\n\nf\n\ni\n\nX\n\n6\n\n=\n\nDst\n\nf\n\ny\n\nu\n\nDst\n\nf\n\nDst\n\nf\n\nl\n\nhh\n\nf\n\nii\n\nup\n\nZ\n\n0\n\nx\n\n6\n\n=\n\nDst\n\nf\n\n.\n\nComparing with the above,\n\ny\n\nu\n\nDst\n\nf\n\nh\n\nf\n\ni\n\nx\n\n6\n\n=\n\nDst\n\nf\n\ny\n\nu\n\nDst\n\nf\n\nDst\n\nf\n\nl\n\nhh\n\nf\n\nii\n\nup\n\nZ\n\n0\n\nx\n\n6\n\n=\n\nDst\n\nf\n\n.\n\nSo\n\nh\n\nf\n\ni\n\nx\n\n=\n\nd\n\nDst\n\nf\n\nhh\n\nf\n\nii\n\nup\n\nZ\n\n0\n\nx\n\nbecause Dst\n\nf\n\nis separable (proposition\n\n579\n\nand the\n\nfact that\n\nZ\n\n1\n\nis a boolean lattice).\n\nTheorem\n\n1513\n\n.\n\nLet (\n\nA\n\n,\n\nZ\n\n0\n\n) and (\n\nB\n\n,\n\nZ\n\n1\n\n) be primary filtrators over boolean\n\nlattices.\n\n1\n\n. A function\n\nα\n\nB\n\nZ\n\n0\n\nconforming to the formulas (for every\n\nI, J\n\nZ\n\n0\n\n)\n\nα\n\nZ\n\n0\n\n=\n\nB\n\n,\n\nα\n\n(\n\nI\n\nt\n\nJ\n\n) =\n\nαI\n\nt\n\nαJ"
] | [
null,
"http://www.mathematics21.org/view/volume-1/index286.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.52261895,"math_prob":0.6925785,"size":2321,"snap":"2019-43-2019-47","text_gpt3_token_len":1295,"char_repetition_ratio":0.16400518,"word_repetition_ratio":0.32708934,"special_character_ratio":0.39551917,"punctuation_ratio":0.104166664,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994043,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-16T15:04:14Z\",\"WARC-Record-ID\":\"<urn:uuid:d8e4bd74-6c22-4942-b1d2-26e253e2adb6>\",\"Content-Length\":\"55243\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:00dac9d2-6b21-4033-90ef-c538627ee0ec>\",\"WARC-Concurrent-To\":\"<urn:uuid:c14b281e-c3e9-418f-bea8-f8accc90eaf3>\",\"WARC-IP-Address\":\"104.236.49.103\",\"WARC-Target-URI\":\"http://www.mathematics21.org/view/volume-1/index-286.html\",\"WARC-Payload-Digest\":\"sha1:XK3TPSCX3YLEEWUZGWBLLF6BQBQHD3T5\",\"WARC-Block-Digest\":\"sha1:FEAQKVSZ2O5FL6JPDDDMJ5NPBOKT4F6H\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986668994.39_warc_CC-MAIN-20191016135759-20191016163259-00551.warc.gz\"}"} |
https://unix.stackexchange.com/questions/27013/displaying-seconds-as-days-hours-mins-seconds/170299 | [
"Displaying seconds as days/hours/mins/seconds?\n\nIs it possible to easily format seconds as a human-readable time in bash?\n\nI don't want to format it as a date, but as the number of days/hours/minutes, etc...\n\n• Could you provide an example/several examples please? – gabe. Dec 16 '11 at 22:51\n• Are you saying you have an interval, or a date since the epoch? – Paul Tomblin Dec 16 '11 at 23:22\n\nYou can use something like this:\n\nfunction displaytime {\nlocal T=\\$1\nlocal D=\\$((T/60/60/24))\nlocal H=\\$((T/60/60%24))\nlocal M=\\$((T/60%60))\nlocal S=\\$((T%60))\n(( \\$D > 0 )) && printf '%d days ' \\$D\n(( \\$H > 0 )) && printf '%d hours ' \\$H\n(( \\$M > 0 )) && printf '%d minutes ' \\$M\n(( \\$D > 0 || \\$H > 0 || \\$M > 0 )) && printf 'and '\nprintf '%d seconds\\n' \\$S\n}\n\nExamples:\n\n\\$ displaytime 11617\n3 hours 13 minutes and 37 seconds\n\\$ displaytime 42\n42 seconds\n\\$ displaytime 666\n11 minutes and 6 seconds\n\nEasiest and cleanest way is this one liner (here assuming GNU date):\n\nIf the number of seconds is, say:\n\nseconds=123456789 # as in one of the answers above\n\neval \"echo \\$(date -ud \"@\\$seconds\" +'\\$((%s/3600/24)) days %H hours %M minutes %S seconds')\"\n\n--> output: 1428 days 21 hours 33 minutes 09 seconds\n\nCredit goes to Stéphane Gimenez but if someone would like to display seconds only if a period is less than a minute here is my modified version that I use (also with fixed pluralization):\n\nconverts()\n{\nlocal t=\\$1\n\nlocal d=\\$((t/60/60/24))\nlocal h=\\$((t/60/60%24))\nlocal m=\\$((t/60%60))\nlocal s=\\$((t%60))\n\nif [[ \\$d > 0 ]]; then\n[[ \\$d = 1 ]] && echo -n \"\\$d day \" || echo -n \"\\$d days \"\nfi\nif [[ \\$h > 0 ]]; then\n[[ \\$h = 1 ]] && echo -n \"\\$h hour \" || echo -n \"\\$h hours \"\nfi\nif [[ \\$m > 0 ]]; then\n[[ \\$m = 1 ]] && echo -n \"\\$m minute \" || echo -n \"\\$m minutes \"\nfi\nif [[ \\$d = 0 && \\$h = 0 && \\$m = 0 ]]; then\n[[ \\$s = 1 ]] && echo -n \"\\$s second\" || echo -n \"\\$s seconds\"\nfi\necho\n}\n\nAn alternative example in POSIX:\n\nconverts(){\nt=\\$1\n\nd=\\$((t/60/60/24))\nh=\\$((t/60/60%24))\nm=\\$((t/60%60))\ns=\\$((t%60))\n\nif [ \\$d -gt 0 ]; then\n[ \\$d = 1 ] && printf \"%d day \" \\$d || printf \"%d days \" \\$d\nfi\nif [ \\$h -gt 0 ]; then\n[ \\$h = 1 ] && printf \"%d hour \" \\$h || printf \"%d hours \" \\$h\nfi\nif [ \\$m -gt 0 ]; then\n[ \\$m = 1 ] && printf \"%d minute \" \\$m || printf \"%d minutes \" \\$m\nfi\nif [ \\$d = 0 ] && [ \\$h = 0 ] && [ \\$m = 0 ]; then\n[ \\$s = 1 ] && printf \"%d second\" \\$s || printf \"%d seconds\" \\$s\nfi\nprintf '\\n'\n}\n• This worked really well. I simply pasted the function into my script and ran convert \\$s to get pretty output. – flickerfly Oct 19 '16 at 19:25\n\nI'd do it like this:\n\n\\$ seconds=123456789; echo \\$((seconds/86400))\" days \"\\$(date -d \"1970-01-01 + \\$seconds seconds\" \"+%H hours %M minutes %S seconds\")\n1428 days 21 hours 33 minutes 09 seconds\n\\$\n\nHere's the one liner above, broken down so that it's easier to understand:\n\n\\$ seconds=123456789\n\\$ echo \\$((seconds/86400))\" days\"\\\n\\$(date -d \"1970-01-01 + \\$seconds seconds\" \"+%H hours %M minutes %S seconds\")\n\nIn the above I'm echoing out the output of another command that's run inside the \\$( ... ) sub-command. That sub-command is doing this, calculating the number of days (seconds/86400), then using the date command in another sub-command \\$(date -d ... ), to generate the hours, minutes, and seconds for a given number of seconds.\n\nI modified the displaytime function above... as follows:\n\nseconds2time ()\n{\nT=\\$1\nD=\\$((T/60/60/24))\nH=\\$((T/60/60%24))\nM=\\$((T/60%60))\nS=\\$((T%60))\n\nif [[ \\${D} != 0 ]]\nthen\nprintf '%d days %02d:%02d:%02d' \\$D \\$H \\$M \\$S\nelse\nprintf '%02d:%02d:%02d' \\$H \\$M \\$S\nfi\n}\n\nbecause I always want to see HH:MM:SS, even if they are zeros.\n\nI'm building on atti's answer which I liked as an idea.\n\nYou can do this with the bash builtin printf which will take the seconds since the epoch as an argument. No need to fork to run date.\n\nYou have to set the timezone to UTC for printf because it formats the time in your local timezone and you will get the wrong answer if you are not in UTC time.\n\n\\$ seconds=123456789\n\\$ TZ=UTC printf \"%d days %(%H hours %M minutes %S seconds)T\\n\" \\$((seconds/86400)) \\$seconds\n1428 days 21 hours 33 minutes 09 seconds\n\nIn my local time (which is currently NZDT - +1300) the answer is wrong if I do not set the timezone\n\n\\$ seconds=123456789\n\\$ printf \"%d days %(%H hours %M minutes %S seconds)T\\n\" \\$((seconds/86400)) \\$seconds\n1428 days 09 hours 33 minutes 09 seconds\n\nWith and without setting the timezone\n\n\\$ seconds=\\$(( 3600 * 25))\n\\$ printf \"%d days %(%H hours %M minutes %S seconds)T\\n\" \\$((seconds/86400)) \\$seconds\n1 days 13 hours 00 minutes 00 seconds\n\n\\$ TZ=UTC printf \"%d days %(%H hours %M minutes %S seconds)T\\n\" \\$((seconds/86400)) \\$seconds\n1 days 01 hours 00 minutes 00 seconds\n• Note that bash's printf introduced this %(datefmt)T notation beginning with bash-4.2-alpha. – bishop Mar 27 at 15:40\n\nHere one\n\nsecs=378444\necho \\$((\\$secs/86400))d \\$((\\$((\\$secs - \\$secs/86400*86400))/3600))h:\\$((\\$((\\$secs - \\$secs/86400*86400))%3600/60))m:\\$((\\$((\\$secs - \\$secs/86400*86400))%60))s\n\nOutput:\n\n4d 9h:7m:24s\n\ndate --date '@1005454800' gives you Sun Nov 11 00:00:00 EST 2001, which is 1005454800 seconds after the Unix epoch. You can format that with the date +FORMAT option.\n\n• That's a date, not a duration, which the question was asking about. – Gilles Dec 17 '11 at 23:02"
] | [
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https://ge-spark.com/power-station/frequent-question-what-is-the-electric-flux-through-a-sphere-of-radius.html | [
"# Frequent question: What is the electric flux through a sphere of radius?\n\nContents\n\nConsidering a Gaussian surface in the form of a sphere at radius r > R, the electric field has the same magnitude at every point of the surface and is directed outward. The electric flux is then just the electric field times the area of the spherical surface.\n\n## What is the electric flux through a sphere that has a radius?\n\nWhat is the electric flux through a sphere that has a radius of 1.00 m and carries a charge of +1.00 µC at its centre? Φ = EA E = 8.99 x 109 x 1 x 10-6/ 12 E = 8.99 x 103 N/C. The area that the electric field lines penetrate is the surface area of the sphere of radius 1.00 m.\n\n## Does electric flux depend on radius of sphere?\n\nGauss’s Law\n\nPoint Charge Inside a Spherical Surface: – The flux is independent of the radius R of the sphere.\n\n## What is the electric flux through the spherical Gaussian surface of radius of a?\n\nThe electric flux through a spherical Gaussian surface of radius R centered about an amount of charge Q is 1200 Nm2 /C.\n\n## What is flux in sphere?\n\nConsidering a Gaussian surface in the form of a sphere at radius r > R , the electric field has the same magnitude at every point of the surface and is directed outward. The electric flux is then just the electric field times the area of the spherical surface.\n\n## How do you calculate electric flux?\n\nKnow the formula for electric flux.\n\n1. The Electric Flux through a surface A is equal to the dot product of the electric field and area vectors E and A.\n2. The dot product of two vectors is equal to the product of their respective magnitudes multiplied by the cosine of the angle between them.\n\n## How do you calculate total electric flux?\n\nThe total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux through an area is defined as the electric field multiplied by the area of the surface projected in a plane perpendicular to the field.\n\n## Does electric flux depend on distance?\n\nThe magnitude of the electric field everywhere is the same, because the distance from the charge is the same at each point, so we can pull that out of the integral, and we’re left with EA. … The flux doesn’t depend on the distance r.\n\n## What happens to flux when radius is doubled?\n\nEven if the radius is doubled, then the outward electric flux will remain the same as [dfrac{q}{{{varepsilon _0}}}]. This is because the outward electric flux is independent of the distribution of the charges and the separation between them inside the closed surface. Hence, option D is the correct answer.\n\nTHIS IS UNIQUE: What happens to the energy transfer as they move from producer to the higher level of consumer?\n\n## What happens to the flux through the sphere and the magnitude?\n\nWhat happens to the flux through the sphere and the magnitude of the electric field at the surface of the sphere? The flux and field both increase. … The flux decreases and the field remains the same.\n\n## What is the total flux through the sphere?\n\nBeing a scalar quantity, the total flux through the sphere will be equal to the algebraic sum of all these flux i.e. This expression shows that the total flux through the sphere is 1/eO times the charge enclosed (q) in the sphere. The total flux through closed sphere is independent of the radius of sphere .\n\n## What is the electric flux through the Gaussian surface?\n\nAccording to Gauss’s law, the flux of the electric field →E through any closed surface, also called a Gaussian surface, is equal to the net charge enclosed (qenc) divided by the permittivity of free space (ϵ0): … 4: The electric flux through any closed surface surrounding a point charge q is given by Gauss’s law."
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https://www.mdpi.com/2073-4441/11/4/677 | [
"Next Article in Journal\nA New Conceptual Model for Slope-Infiltration\nNext Article in Special Issue\nA Quantitative Flood-Related Building Damage Evaluation Method Using Airborne LiDAR Data and 2-D Hydraulic Model\n\nFont Type:\nArial Georgia Verdana\nFont Size:\nAa Aa Aa\nLine Spacing:\nColumn Width:\nBackground:\nArticle\n\n# Development of a Large Flood Regionalisation Model Considering Spatial Dependence—Application to Ungauged Catchments in Australia\n\n1\nEnvironment and Infrastructure Division, Cumberland Council, Merrylands, NSW 2170, Australia\n2\nSchool of Computing, Engineering and Mathematics, Western Sydney University; Locked Bag 1797, Penrith NSW 2750, Australia\n*\nAuthor to whom correspondence should be addressed.\nWater 2019, 11(4), 677; https://doi.org/10.3390/w11040677\nReceived: 12 February 2019 / Revised: 19 March 2019 / Accepted: 29 March 2019 / Published: 1 April 2019\n(This article belongs to the Special Issue Flood Modelling: Regional Flood Estimation and GIS Based Techniques)\n\n## Abstract\n\n:\nEstimation of large floods is imperative in planning and designing large hydraulic structures. Due to the limited availability of observed flood data, estimating the frequencies of large floods requires significant extrapolation beyond the available data. This paper presents the development of a large flood regionalisation model (LFRM) based on observed flood data. The LFRM assumes that the maximum observed flood data over a large number of sites in a region can be pooled together by accounting for the at-site variations in the mean and coefficient of variation. The LFRM is enhanced by adding a spatial dependence model, which accounts for the net information available for regional analysis. It was found that the LFRM, which accounts for spatial dependence and that pools 1 or 3 maxima from a site, was able to estimate the 1 in 1000 annual exceedance probability flood quantile with consistency, showing a positive bias on average (5–7%) and modest median relative errors (30–33%).\n\n## 1. Introduction\n\nThe estimation of rare to very rare floods is needed for many engineering applications, such as planning and designing large hydraulic structures, dam spillways, and flood control levees. Due to the limited availability of observed flood and rainfall data, the estimation of rare to very rare flood frequencies remains a challenging task. All methods used to estimate rare to very rare floods involve significant extrapolation beyond recorded flood and rainfall data. The term ‘rare’ flood(s) refers to floods with annual exceedance probabilities (AEPs) of 1 in 50 to 1 in 100 . Floods in the AEP range from 1 in 100 to the ‘credible limit of extrapolation’ (AEP in the order of 1 in 2000) are referred to as ‘very rare’ floods, while floods from the credible limit of extrapolation to the probable maximum flood (PMF) are termed ‘extreme’ floods. Due to knowledge and data limitations and the uncertainty involved in extrapolating beyond available data, the errors in final estimates can be quite high. The average size of recorded flood data series for Australian small to medium sized catchments is about 33 years .\nTo make better use of the available flood data and to be able to transfer this information to ungauged catchments, regional estimation methods are used, such as the index flood method [3,4]. The basic idea is that if a region is relatively homogenous, then the estimation of large to rare flood quantiles at a given site may be improved by using the larger observations at other sites as well (i.e., a trade-off between space and time). Some studies both in the past and present, and on an international scale, have looked at the advantages and disadvantages of different regional models for rare, very rare, and extreme floods (e.g., [5,6,7,8,9,10,11]).\nA new probabilistic model (PM) was introduced by specifically for this sort of analysis. Majone and Tomirotti originally calibrated the PM for Italian rivers, and extended the method using 7300 historical series of annual maximum flows observed at gauging stations belonging to different geographical areas around the world. This model is based on the assumption that the standardised maximum values (Qmax) of the annual maximum flood series (AMFS) from a large number of individual sites in a region are independent and can be pooled. The PM concept is identical to the basic concept of station-year methods: observed data from an assumed homogenous region are pooled and a non-parametric flood frequency curve is fitted on a probability plot. The traditional approach in the station-year methods is to achieve an acceptable degree of homogeneity within the region by standardising by the at-site mean or median values. The novel aspect of the PM’s standardisation is to take into account not only the at-site mean, but also the at-site coefficient of variation (CV) values of the time series data. This unique form of standardisation allows the pooling of more data from many stations compared to the standard index methods. It is known that the station-year method suffers from problems associated with inter-site dependence. In the PM technique presented by Majone and Tomirotti , it was assumed that the individual values in the standardised and pooled data series are independent. This assumption may be valid if the data being pooled comes from stations that are spread over a very large region, as with Majone and Tomirotti .\nThe ‘large flood regionalisation model’ (LFRM) described by is a modified version of the PM technique that was applied to a large set of catchments in Australia. Detailed examination showed that the values in the pooled LFRM data series used in this study tended to cluster in some years, with very few events in other years. This appears to violate the assumption of independent distribution of the events in time and indicates that some of the events occurring in the same year might have resulted from the same hydro-meteorological events. The testing of the LFRM by has demonstrated that if the Australian LFRM data series is assumed to be independent, the LFRM tends to underestimate the at-site flood frequency estimates.\nOn the basis of these findings from the initial application, the LFRM was further developed by (i) coupling it with a spatial dependence model ([13,14]) that reflects the reduction in the net information due to spatial dependence (e.g., [15,16,17,18,19,20]); (ii) pooling more data by taking the top 3 maximum values in a region; and (iii) combining it with Bayesian generalised least squares (BGLSR) and the region of influence (ROI) approach to develop regional prediction equations, so that the LFRM can be applied to ungauged catchments. In the ROI approach, a separate region can be formed for each of the gauged catchments by drawing an appropriate number of nearby stations. The advantages of the ROI approach are discussed in more detail in and [13,22,23].\nPoints (i), (ii), and (iii) above are, in essence, the main innovations of the LFRM model being presented in this paper. An advantage of the LFRM proposed in this paper is that it offers an alternative to traditional approaches of rare flood estimation methods based on rainfall runoff models, where time and resource constraints may not permit the development of detailed rainfall-based methods. Moreover, there is no guarantee that rainfall-based methods provide the best possible estimates.\nThe remainder of the paper is organised as follows: Section 2 presents the data used in the development of the LFRM and the spatial dependence model. The methodology, concept, and results of the spatial dependence models are given in Section 3. In Section 4, the results of the LFRM and the results of the enhanced LFRM in the light of spatial dependence are also provided. Section 4 also details the results associated with the use of the enhanced LFRM for ungauged catchment estimation. Conclusions are drawn in Section 5.\n\n## 2. Data Description\n\nA total of 654 gauging stations in Australia with reasonable record lengths (19 to 96 years with a mean of 34 years) suitable for regional flood frequency analysis (RFFA) were assembled as a part of Australian Rainfall and Runoff (ARR) revision ‘Project 5 Regional Flood Methods’ . The selected 654 sites are shown in Figure 1. The streamflow data of these sites were prepared following stringent procedures, as described in . Streamflows at these sites are essentially unregulated and have not been affected by major land use changes. The catchment areas range between 0.1 and 7406 km2 (mean: 350 km2). One does expect that the useful information for RFFA increases with the increasing number of stations in the region; however, the net information does not increase proportionally with the increasing number of stations within a given region, due to spatial dependence between data at gauging stations. While the shorter record lengths in this study (<25 years) would introduce notable uncertainty in parameter estimation for at-site frequency analysis, they were included, as they still contain useful additional information for the pooled data set of large flood events. From the 654 stations, two datasets for the LFRM were established: (a) 626 stations with reasonable concurrent record lengths were selected for use in the development of the LFRM and constant spatial dependence model. (b) The remaining 28 stations (6 from each of the states of New South Wales (NSW), Victoria (VIC), Queensland (QLD), Western Australia (WA), and 4 from Tasmania (TAS)) located around Australia were put aside for testing and validation with the LFRM. All 626 catchments were used to develop the prediction equations for the model parameters (i.e., mean and CV) of the LFRM (for ungauged catchments) using the BGLSR and the ROI approach.\nWith AMFS data, large errors are often associated with the highest flows in the data set because of the nature of the rating curve extrapolation errors. As will be discussed in the methodology section, the LFRM uses the largest 1 to 3 observed maxima values from each station in the region. Therefore, any errors in these observations can introduce significant error into the LFRM final quantile estimates. As can be read in , a rating ratio check was introduced and used to cull stations with significant rating curve error.\n\n## 3. Methods\n\n#### 3.1. Identification of A Suitable Parent Distribution\n\nThe LFRM concept is primarily non-parametric, and therefore an assumption regarding a particular distribution is not required. However, a probability distribution is fitted to the AMFS in order to derive a generic relationship for the effective number of stations Ne, which is used to adjust the plotting position of the LFRM points. By comparing a range of methods and distributions, it was found that the generalised extreme value (GEV) distribution is quite appropriate to approximate the AMFS in Australia based on (i) L-moments ratio diagram, the L-moment goodness-of-fit test, and (ii) the Anderson Darling goodness-of-fit Monte Carlo test. Based on visual inspections of the standardised (as per index flood) flood frequency curves developed, the GEV appeared to be a good candidate to describe the AMFS data for different Australian states. The GEV distribution seemed to capture the higher flows much better than the other competing three-parameter distributions. Further information and discussion can be found in [25,26].\n\n#### 3.2. Estimating Inter-Site Dependence\n\nSpatial dependence can be accounted for through the use of a spatial dependence model, which defines the effective number of independent stations in a region (Ne) [13,18,20]. The value of Ne can be calculated from the relative position of two frequency curves, the regional maximum curve (this is formed by pooling the highest values of the standardised maxima from the stations in the streamflow gauging network considered each year), and the regional average curve (which is formed by the average curve for the streamflow network considered, i.e., the average of the standardised at-site curves) (or typical curve) . This can be measured on a Gumbel plot (YT) by ln(Ne), the horizontal separation between the two frequency curves, as seen Figure 2. As shown by Dales and Reed that the value of Ne is constant irrespective of quantile or average recurrence interval (ARI) if the annual maxima are GEV distributed and the shape parameters of the regional maximum and regional average distributions are equal. In this study, a relatively simple model of spatial dependence was obtained by ignoring the possible variation of Ne with ARI/annual exceedence probability (AEP).\nFor each regional network of stations, a fixed value of Ne is calculated from the first probability weighted moments ($β o$-mean) $β 0 r$ and $β 0 t$ of the regional maximum (superscript r) and typical point (superscript t) data:\n$N e = [ ( β 0 r − x bound ) ( β 0 t − x bound ) ] 1 − κ$\nwhere κ is the shape parameter of the GEV distribution, and xbound is the bound of the typical point GEV distribution .\nNetwork sizes of N = 2, 4, and 8 stations were used to determine a relationship between N and Ne. Based on concurrent record lengths between sites (average 18 years) a maximum number of 8 stations in a network were adopted for the experiments carried out. To establish an indication of the typical degree of dependence in a network of size N, different forms of constrained and random sampling were adopted, and then Ne was related to the average correlation coefficient (ρ) of concurrent annual maxima at pairs of stations for the particular network size. The different forms of constrained and random sampling were (i) using a ROI approach to pool stations closest to the station of interest using N = 2, 4, and 8, (ii) pooling the closest 20 stations and then randomly sampling N = 2, 4, and 8 sites, and (iii) totally random sampling N = 2, 4, and 8 sites from the region. The different sampling approaches provide different information about ρ. For example, the ROI and random ROI method was useful when investigating small networks which are highly correlated. The total random network samples were over a broader range of correlation values. Further details on this can be seen in [13,14]. This was carried out on the real dataset and on a simulated dataset, with the main purpose of establishing a suitable model to describe the spatial dependence in each region/state and then combining the results to form one relationship to use for all of Australia. For the simulated data, the multi-site maxima were generated according to using a GEV distribution. For the generated dataset, the constant correlation coefficient varied from 0 to 0.5 in steps of 0.1. The overall steps in the analysis carried out in this paper are summarised in a flowchart given in Figure 3.\n\n#### 3.3. Establishing and Generalising Ne\n\nThe Ne values were obtained for these different network sizes by fitting the mean as described in Section 3.2 for the real and simulated datasets for the different networks and regions. It was found that the total random network exhibits less spatial dependence than both the ROI and random ROI networks. Indeed, sites that are closer together are more likely to show more spatial dependence. Importantly, the same features as detailed above were seen with the simulated data; however, the simulated data showed less spatial dependence in the ‘total random network’ as compared to the real dataset.\nTo avoid regional variations in the derived spatial dependence model, a regional approach was used to combine all the experimental results together for all of Australia. Regression analysis was used to relate Ne to the average correlation coefficient (ρ) of concurrent AMFS at pairs of stations for the different networks and regions for each of the adopted experiments (this includes the real data and simulated data). To derive the regression equation, it was determined to be more appropriate to build a general model that relates the ratio LN(Ne)/LN(N) to the average correlation coefficient (ρ), similar to that of Dales and Reed .\nThe form of the constant Ne model is given by Equation (2) which was calibrated by combining all the models for each of the Australian states into one generic equation. The final form of Equation (2) was identified by investigating the real and simulated data sets:\n$LN ( N e ) LN ( N ) = a + b ρ ¯$\nOverall, the one variable model (see Equation (2)) provided a relatively good fit for the experimental data. The fitted parameters of the constant Ne model for Australia (overall) are given in Table 1 for the real and simulated datasets. The final parameter values for the general Australian spatial dependence model were found by combining the different network values of the ratio LN(Ne)/LN(N) and developing a regression equation of the form represented by Equation (2), then taking the average of the coefficient values of the developed regression equations.\nThe coefficient of determination (R2) values for the final models (see Table 1) fitted to the real and simulated data sets are quite high, suggesting that the use of the constant Ne model should result in improved Ne estimates compared to the values calculated directly from the AMFS data in each station network.\nThe comparison of the fitted Ne values for the real data computed using Equation (1) and those by the spatial dependence Equation (2) are shown in Figure 4. The figure below illustrates that the scatter in the spatial dependence model estimates increases with increasing N. The scatter may also be attributed to natural and sampling variability from site to site, given that the concurrent record length for analysis was very modest. Figure 4 and Table 1 show the overall satisfactory performance of Equation (2), as far as practical application is concerned.\n\n## 4. Results\n\n#### 4.1. Development and Calibration of LFRM for Australian Data\n\nThe selected Qmax (1 and 3) (i.e., the top 1 and 3 maximum data points from each station’s AMFS data, referred to as Qmax), are first standardised by the at-site average of the AMFS data (mean), and then plotted in the (CV, Qmax/mean) plane. Figure 5 shows such a plot for (1 and 3 max) the study data set, consisting of 626 data points (1 max), 1878 data points (3 max) from 626 sites, which suggests the following relationship:\n$Q max mean = c + α CV ψ$\nThe coefficients () of Equation (3) were estimated by the maximum likelihood method for each of the plots in Figure 5. The estimated coefficients, along with their R2 values, are provided in Table 2.\nThe R2 values in Table 2 suggest that the estimated coefficients provide a reasonably good fit to the experimental data; this is more evident, however, when pooling the top 1 AMFS. When pooling 3 top maxima, a greater scatter is noticed, as can be seen in Figure 5; this is also supported by the drop in R2 value. An important note is made here on whether the weaker relationship with CV is compensated for later on by having additional data points to define the lower end of the distribution. What can be observed from Table 2 is that the exponent $ψ$ is appreciably greater than unity (as would be the case for a Gumbel distribution for 1 maxima) and decreases slightly with the pooling of more data (i.e., 3 max).\nBased on Figure 5, and assuming that a large part of the scatter can be explained by variations in the average recurrence interval (ARI/AEP) of the AMFS data, the best way to model the scatter is to search for a LFRM function in the form of:\n$Q max mean = c + f ( ARI ) CV ψ$\nwhere it is assumed that $f ( ARI / AEP )$ is a function of the ARI/AEP only, and can be substituted for the coefficient $α$. From Equation (3), the calibration procedure is based on the introduction of a new standardised variable, which can be defined by:\n$Y max = ( Q max mean ) − c CV ψ$\nwhere are based on the coefficients according to the number of annual maxima pooled (e.g., 1 or 3 maxima). This form of standardisation (Equation (5)) takes into account not only differences in the mean values, but also of the CV, raised to the power appropriate for a specific regional data set. As expected, as a result of this new standardisation, $Y max$ was practically uncorrelated with the CV, as was confirmed by the very small R2 of 0.0037 referring to the same set of data points for using the top 3 annual maxima. The following plotting position formula (Equations (6)–(8)), proposed by Majone and Tomirotti , was applied to estimate the ARI or the empirical non-exceedance frequency (AEP) of each of the $Y max$ values in the pooled data sets (i.e., max of 1 and 3) from the N = 626 sites. In order to define the form of the distribution of the variable, the top 1 and 3 annual maxima values of each site’s data were used. Here, the major assumption made is that the ith value of the $Y max$ series is independent of the other values and that the $Y max$ values belong to the same population. It follows that the plotting position of the $Y max$ can be provided by the following equations (Majone and Tomoirotti ):\n$P ( Y max ≤ y max ) = P ( Y ≤ y max ) n a$\nwhere $Y$ is the at-site standardised annual maximum and na is the site sample size (taken as the average of the site samples sizes, which is 34 for this study). Now, sorting the pooled sample of standardised maxima consisting of N = 626 (and L = 626 or 1878) in decreasing order and define as the mth ranked value in the pooled sample. The ARI of $y max ( m )$ (expressed as T years) can be estimated using\n$m N = 1 − P ( Y max ≤ y max ( m ) ) = 1 − P ( Y ≤ y max ( m ) ) n a = 1 − ( 1 − 1 T ) n a$\n$T = 1 ( 1 − ( 1 − m N ) 1 n a )$\nFrom this definition, the estimated ARI/AEP values would ideally be assumed to be representative of actual return periods. However, this may not be the case for the Australian flood data set, as many of the gauging sites used here are very close together spatially (see Figure 1) and hence there would be significant inter-site dependence. The plot of $Y max$ vs. YT (where YT is the Gumbel reduced variate and is used as a surrogate for ARI or AEP), where YT = −ln[−ln(1−1/T)] is shown in Figure 6 for $Y max$ (L = 626 and 1838). The plots for L = 626 and L = 1838 sites in Figure 6 (bottom curves for all two plots) are in line with what would be expected from using the additional data points. Clearly, the impact of using a greater number of maxima, e.g., 3 maxima, seems to provide a very smooth empirical distribution that is fitted closely by the distribution function. The plots also reveal that the experimental data can be approximated by a second-degree polynomial function of YT as given by Equation (9), whose model coefficients and R2 values can be seen in Table 3 for the different pooling of the annual maxima (i.e., top 1 and 3 maxima):\n$Y max = C 1 ( Y T ) 2 + C 2 ( Y T ) + C 3$\nwhich, in terms of Qmax/mean, takes the following form\n$Q max mean = c + ( C 1 ( Y T ) 2 + C 2 ( Y T ) + C 3 ) CV ψ$\nEquations (9) and (10) yield the analytical expression of the LFRM model for the study data, set using the top 1 and 3 annual maxima. The appropriate values of the coefficients in Table 3 are substituted into Equations (9) and (10). However, this formulation does not allow for the effects of the inter-site dependence.\n\n#### 4.2. Revision of LFRM for Spatial Dependence\n\nThe LFRM for the study data in its current form (see Equations (9) and (10)) does not allow for the effect of inter-site dependence. In this section, spatial dependence is accounted for through the use of the spatial dependence model derived in the previous section (see Equation (2)). For this study, the use and calculation of Ne for application with the LFRM is illustrated. Firstly, the average correlation for each pair of sites was calculated for the region by computing the correlation coefficient from a regional relationship with distance for all of the Australian states. The average correlation coefficient was found to be $ρ ¯ =$ 0.26. Secondly, using Equation (2) along with the coefficients for the Australian spatial dependence model given in Table 1 (using the real and simulated data) and $ρ ¯ =$ 0.26, the Ne was estimated. The calculated Ne value, along with the effective record length, is given in Table 4. From Table 3, it can be seen that the results from the real data match reasonably well with the simulated data.\nUsing the calculated Ne value of 207 (from the real dataset) in Equations (7) and (8) instead of the total number of stations (N = 626), we can now estimate the new plotting position of the pooled data points for 1 and 3 maxima. The new interpolated curve for Equations (9) and (10) has new coefficient values. The revised coefficient values of the LFRM have now been corrected for the spatial dependence in the dataset. The appropriate values of the coefficients of Equations (9) and (10) are given in Table 3. Differences are clearly seen in the coefficients of the LFRM when comparing the results of the dataset using N and Ne sites; this is due to the reduction of the total useful information (i.e., the effective number of stations). The new interpolated frequency curves can be seen in Figure 6 (both panels, top curves).\nWhat is striking in Figure 6 is the shift upwards in the frequency curve of the pooled data. Taking the 1 max plot for example, if one compares the Ymax value of approximately 4, it can be seen that, if one ignores the spatial dependence, the flood magnitude risk may be notably underestimated (for N sites Ymax = 4, AEP = 1 in 87, for Ne sites Ymax = 4, AEP = 1 in 8.3). For the pooling of the 1 max and correcting for spatial dependence (see max of 1 plot in Figure 6) it was found that the range of Ymax values for which the fitted model (referred to as LFRM_Ne henceforth) might be considered reliable is approximately 2.2 to 7, which corresponds to AEPs of 1 in 10 to approximately 1 in 2000.\nFigure 7 shows the behavior of the dimensionless quantiles derived from Equations (9) and (10) for AEPs of 1 in 100, 1 in 500, and 1 in 1000 for all the pooled data, (i.e., 1 and 3 max), and for the estimated quantiles using N and Ne. The dimensionless quantiles for the world model (referred to as the PM (world), based on 7300 gauging stations around the world) developed by Majone and Timorotti are also superimposed for comparison. The comparison with the PM (world) curves in Figure 5 indicates that the LFRM_Ne can explain a good amount of the scatter in these plots, as the set of curves (1 in 100 and 1 in 500 AEP curves) for this extended AEP range (including the 1 in 1000-AEP) captures most of the upper part of the points in the pooled data set of the Qmax/mean values. The flatter slopes in Figure 7 for 3 max (bottom panel), are consistent with what was shown in Figure 5 and seems to reflect a weaker relationship of Qmax/mean with CV. Comparison of the curves for max of 1 and 3 for Ne and N seems to indicate that allowance for spatial dependence has a smaller influence on slope. Figure 7 also indicates that the extra data i.e., 3 max provides slightly better definition of the left-hand tail of the distribution (where the top few points in the right-hand tail are mostly common in all 2 data sets (1 and 3 maxima). Further investigation also revealed that the LFRM_Ne can provide reasonably accurate growth curve estimation for CV values in the ranges 0.60–1.60 (approximately 81% (505 out of 626) of the study catchments fall in this range). However, the LFRM_Ne can perform poorly for some catchments, with CV values greater than 1.70.\n\n#### 4.3. Application of the LFRM to Ungauged Catchments\n\nOur interest is the application of Equations (9) and (10) to ungauged catchments, which requires the estimation of the mean flood and CV for the ungauged catchment in question. The BGLSR and the ROI approach, as discussed in , were used to develop the prediction equations for the mean flood and CV of the AMFS data as a function of catchment and climatic characteristics (predictor variables). The prediction equation for the mean flood used a ROI of 30–40 stations, while 65–80 stations were used for the CV, based on the findings from past studies (e.g., [21,22,23,27]) and which state was being analysed.\nThe regression equations are presented in general form below:\nThe prediction equations developed above using the ROI approach, and Equations (9) and (10) (LFRM_Ne model), were applied to the 28 test catchments, which were not used in developing the prediction equations. To make the comparison more useful and to benchmark the LFRM_Ne model, the developed prediction equations were also used to estimate the mean flood and CV with the PM (world) model developed by Majone and Tomirotti . It must be pointed out however, that the PM (world) model does not contain any of the data used to develop the Australian LFRM. The validation analysis was undertaken for AEPs to 1 in 1000. AEPs in the range of 1 in 50 to 1 in 100 were compared with at-site flood frequency analysis (FFA) (obtained from the fitted log Pearson type 3) distribution using the FLIKE software . Validating beyond the AEP 1 in 100 with at-site FFA estimates was not viewed as reliable, given the very large extrapolation errors involved. Any validation results obtained beyond AEP 1 in 100 would be of little significance for most of the stations.\nFor the lower AEPs (1 in 500 and 1 in 1000), comparison was made against the results obtained from another regional method where the parameters of the LP3 distribution (i.e., mean, standard deviation, and skew) were regressed against catchment characteristics (known as the PRT—see [21,26] for more details) and flood quantiles were then derived for the 1 in 500 and 1 in 1000 AEPs. The extrapolation of these distributions to the low AEPs also involves a large degree of uncertainty. To assess how well the derived large flood estimates could approximate the observed flood estimates, two numerical measures were applied. Relative bias (BIASr) was used to assess whether the predicted rare flood quantiles by the LFRM_Ne or PM (world) models systematically under- or overestimated the at-site FFA or the PRT estimates on average, considering all the 28 test catchments.\nThe relative error values (REr), with respect to the at-site FFA or the regional parameter regression technique (PRT) estimate, were also obtained. This is by no means the true error of the LFRM_Ne or PM (world) models; the estimated errors represented here by both the BIASr and REr may be taken as a reasonable indication of consistency of the LFRM_Ne or PM (world) models as compared to FFA and PRT estimates. Here, both the FFA and PRT estimates are associated with a higher degree of uncertainty due to considerable extrapolation involved. It is worth noting here that in calculating the median relative error (REr), the sign of the relative errors was ignored.\nTable 5 summarises the various error statistics with the LFRM_N (i.e., no spatial dependence) and LFRM_Ne models (considering the pooling of 1 and 3 maxima) and the PM (world) model based on the 28 test catchments. If spatial dependence is ignored in the Australian dataset, it is observed that the estimation for the AEP of 1 in 1000 using the LFRM_N model suffers from major underestimation on average (e.g., BIASr of −27%) for the ungauged catchment case. Moreover, from Table 5, it can be seen for 1 max and when the pooling of more data is undertaken (i.e., 3 maxima), and spatial dependence (LFRM_Ne) is compensated for, the BIASr is well corrected. For example, from Table 5, for the 1 in 1000 AEP, the BIASr for 1 and 3 max and LFRM_Ne are a 5 and 7% overestimation on average, respectively.\nFocusing on the 3 max results, for the AEPs of 1 in 50 to 1 in 1000-, the BIASr values are positive on average for the LFRM_Ne, while for the PM (world) models, there are a couple cases of underestimation on average. When compared to the results of preliminary LFRM models (i.e., ), the results obtained here present a significant improvement. As found in Haddad et al. , the underestimation on average was up to 40%. By pooling more data and also accounting for the inter-site dependence in the LFRM model, the underestimation problem, to a large extent, has been rectified. The results as benchmarked against the PM (world) model are reassuring; this places a higher degree of confidence in the estimates given by the LFRM_Ne model developed here.\nThe REr values in Table 5 show acceptable results, which are comparable to similar regional models for the larger AEP ranges (Rahman et al. ). Focusing on the 3 max results, the REr values range from 30% to 60% (which are also very comparable to the PM (world) model). It should be noted that in the PM (world) data set most of the stations were so well separated that they were mostly independent of each other, and this was the reason why Majone and Timorotti did not need to work out an effective number of sites. The LFRM_Ne model in this study has refined the approach of the PM (world) model, as significant inter-site dependence exists between stations in the Australian data set.\nAn error bar plot of the BIASr values is given in Figure 8, which displays the central tendency and variability of the sample BIASr values over the 28 independent test catchments. Here, Figure 8 displays the mean value (circle symbol) with a 95% limit bar for flood quantiles AEP 1 in 100 to 1 in 1000. While the mean values appear to be different for the two methods (i.e., LFRM_Ne and PM (world) models), the difference is modest because the error bars overlap, suggesting the LFRM_Ne model to be very comparable and even better than the PM (world) model. Moreover, it proves that consistency is achieved for the 3 maxima pooling LFRM_Ne model as the mean values and the spread of BIASr values are very similar to the PM (world) model. What is noteworthy is the difference between LFRM_Ne and LFRM_N. The mean values were found to be statistically different, which suggests that the LFRM_Ne has corrected the negative bias quite well and justifies the use of the LFRM_Ne. It is envisaged that as a part of the future assessment of the LFRM_Ne, model comparisons will be made against design flood estimates obtained by alternative methods (e.g., spillway design and dam safety studies based on design rainfall-based approaches).\n\n## 5. Conclusions\n\nThe large flood regionalisation model (LFRM) proposed here can be viewed as an alternative to traditional approaches of large flood estimation, where time and resource constraints may not permit the development of rainfall-based methods. This paper presented the further development and application of a simplified LRFM that pools the top 3 annual maxima flood values from many sites in a region to define the regional curve growth combined with a spatial dependence model for annual maximum flow data. To apply the LFRM to ungauged catchments, Bayesian generalised least squares regression was used to estimate the mean flood and coefficient of variation of annual floods. The extended LFRM coupled with a spatial dependence model offers an alternative method of regional flood estimation for AEPs down to 1 in 1000 years. It has been demonstrated that there is positive bias when estimating the 1 in 1000 AEP flood quantiles. The results obtained in this study represent a step forward for rare to very rare flood estimation for Australian catchments in the absence of detailed flood studies. Further analysis, testing, and future work will also include enhancing the LFRM with the new data being collated for ARR Revision Project 5, comparing results from the LFRM at catchments where detailed flood studies have been undertaken extending to the range of extreme floods (e.g., spillway adequacy assessment), and deriving uncertainty limits to provide a practical and simple tool for rare to very rare flood estimation.\n\n## Author Contributions\n\nK.H. and A.R. developed the concept. K.H. carried out all the modelling and analysis tasks, while A.R. reviewed the results and also contributed to writing.\n\n## Funding\n\nThis research received no external funding.\n\n## Acknowledgments\n\nAuthors gratefully acknowledge Erwin Weinmann’s review of an earlier version of this paper and his helpful comments. The authors would like to acknowledge Australian Rainfall and Runoff (ARR) Project 5 team, in particular, Nanada Nandakumar and George Kuczera for their suggestions on the work presented in this paper. The authors would also like to thank three anonymous reviewers and assistant editor whose comments improved the manuscript notably.\n\n## Conflicts of Interest\n\nAuthors declare no conflict of interest.\n\n## References\n\n1. Nathan, R.; Weinmann, E. Estimation of Very Rare to Extreme Floods. In Australian Rainfall and Runoff—A Guide to Flood Estimation (Book 8); Commonwealth of Australia: New South Wales, Australia, 2016. [Google Scholar]\n2. Rahman, A.; Haddad, K.; Kuczera, G.; Weinmann, P.E. Regional flood methods for Australia: data preparation and exploratory analysis. Australian Rainfall and Runoff Revision Projects; Project 5 Regional Flood Methods; Stage I Report No. P5/S1/003, Nov 2009; Engineers Australia, Water Engineering: New South Wales, Australia, 2009; pp. 1–181. 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Geographical distribution of the selected 654 stations from all over Australia.\nFigure 1. Geographical distribution of the selected 654 stations from all over Australia.\nFigure 2. Example plot of regional maximum and typical growth curves and the effective number of independent stations on a Gumbel plot for a random network of 2 and 4 gauging sites in Tasmania.\nFigure 2. Example plot of regional maximum and typical growth curves and the effective number of independent stations on a Gumbel plot for a random network of 2 and 4 gauging sites in Tasmania.\nFigure 3. Flowchart with the different methods described in the methodology.\nFigure 3. Flowchart with the different methods described in the methodology.\nFigure 4. Comparison of directly computed Ne from the annual maximum flood series (AMFS) data and Ne by the constant Ne model. The black line represents a 1:1 line.\nFigure 4. Comparison of directly computed Ne from the annual maximum flood series (AMFS) data and Ne by the constant Ne model. The black line represents a 1:1 line.\nFigure 5. Scatter of Qmax/mean data in the (CV(Q), Qmax/mean) plane, and non-linear interpolation function, which is represented with a red line. (Top panel) Maxima 1; (bottom panel) Maxima 3.\nFigure 5. Scatter of Qmax/mean data in the (CV(Q), Qmax/mean) plane, and non-linear interpolation function, which is represented with a red line. (Top panel) Maxima 1; (bottom panel) Maxima 3.\nFigure 6. Frequency distribution of standardised Ymax values using N and Ne stations. (Top panel) Maxima 1; (bottom panel) Maxima 3.\nFigure 6. Frequency distribution of standardised Ymax values using N and Ne stations. (Top panel) Maxima 1; (bottom panel) Maxima 3.\nFigure 7. Various Qmax/mean quantiles derived from the LFRM_Ne model and PM (World) model. (Top panel) Maxima 1—no spatial dependence; (middle panel) Maxima 1—spatial dependence accounted for; (bottom panel) Maxima 3—spatial dependence accounted for.\nFigure 7. Various Qmax/mean quantiles derived from the LFRM_Ne model and PM (World) model. (Top panel) Maxima 1—no spatial dependence; (middle panel) Maxima 1—spatial dependence accounted for; (bottom panel) Maxima 3—spatial dependence accounted for.\nFigure 8. Error bar plot of BIASr values with the LFRM_Ne and PM (world) models for the 28 test catchments. (Top panel) Maxima 1—no spatial dependence; (middle panel) Maxima 1—spatial dependence accounted; (bottom panel) Maxima 3—spatial dependence accounted for.\nFigure 8. Error bar plot of BIASr values with the LFRM_Ne and PM (world) models for the 28 test catchments. (Top panel) Maxima 1—no spatial dependence; (middle panel) Maxima 1—spatial dependence accounted; (bottom panel) Maxima 3—spatial dependence accounted for.\nTable 1. Properties of the constant Ne spatial dependence model.\nTable 1. Properties of the constant Ne spatial dependence model.\nRegionReal DataSimulated Data\nabR2abR2\nAustralia1−0.66881−0.6399\nTable 2. Coefficients of non-linear interpolation from Figure 5.\nTable 2. Coefficients of non-linear interpolation from Figure 5.\nQmax-AMFSc$α$$ψ$R2 (%)\n113.251.3787\n312.351.2070\nTable 3. Coefficients and R2 values of Ymax polynomial interpolation from Figure 6 for N and Ne sites.\nTable 3. Coefficients and R2 values of Ymax polynomial interpolation from Figure 6 for N and Ne sites.\n Ne sites C1 C2 C3 R2 1 −0.0078 0.504 2.57 0.985 3 −0.010 0.954 0.861 0.995 N sites C1 C2 C3 R2 1 −0.0263 0.787 0.52 0.997 3 −0.053 1.13 −0.603 0.999\nTable 4. Total record length (L) and effective record length (Le) for all the Australian datasets.\nTable 4. Total record length (L) and effective record length (Le) for all the Australian datasets.\nRegionNLConstant Ne Model—Real DataConstant Ne Model—Simulated Data\nAustralia Ne*LeNe*Le\n6262104920769692287654\n(33%) (36%)\n* Ne values in parentheses are percentages of N.\nTable 5. Summary of error statistics obtained from independent testing associated with the large flood regionalisation model (LFRM) model.\nTable 5. Summary of error statistics obtained from independent testing associated with the large flood regionalisation model (LFRM) model.\n 1 Max LFRM_N AEP (1 in Y) BIASr (%) REr (%) Model LFRM_N World Model LFRM_N World Model 1 in 50 −2 12 61 56 1 in 100 −16 −2 66 55 1 in 200 −18 6 46 33 1 in 500 −20 5 47 33 1 in 1000 −27 −1 49 34 1 Max LFRM_Ne AEP (1 in Y) BIASr (%) REr (%) Model LFRM_N World Model LFRM_N World Model 1 in 50 40 12 66 56 1 in 100 18 −2 66 55 1 in 200 22 6 28 33 1 in 500 15 5 29 33 1 in 1000 5 −1 33 34 3 Max LFRM_Ne AEP (1 in Y) BIASr (%) REr (%) Model LFRM_N World Model LFRM_N World Model 1 in 50 31 12 58 56 1 in 100 14 −2 60 55 1 in 200 15 6 30 33 1 in 500 15 5 31 33 1 in 1000 7 −1 30 34\n\n## Share and Cite\n\nMDPI and ACS Style\n\nHaddad, K.; Rahman, A. Development of a Large Flood Regionalisation Model Considering Spatial Dependence—Application to Ungauged Catchments in Australia. Water 2019, 11, 677. https://doi.org/10.3390/w11040677\n\nAMA Style\n\nHaddad K, Rahman A. Development of a Large Flood Regionalisation Model Considering Spatial Dependence—Application to Ungauged Catchments in Australia. Water. 2019; 11(4):677. https://doi.org/10.3390/w11040677\n\nChicago/Turabian Style\n\nHaddad, Khaled, and Ataur Rahman. 2019. \"Development of a Large Flood Regionalisation Model Considering Spatial Dependence—Application to Ungauged Catchments in Australia\" Water 11, no. 4: 677. https://doi.org/10.3390/w11040677\n\nNote that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here."
] | [
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https://research.nu.edu.kz/en/publications/complex-digital-laguerre-filter-design-with-weighted-least-square | [
"# Complex digital Laguerre filter design with weighted least square error subject to magnitude and phase constraints\n\nResearch output: Contribution to journalArticle\n\n6 Citations (Scopus)\n\n### Abstract\n\nComplex coefficient optimal digital Laguerre filter design with arbitrary asymmetric frequency response is developed. The optimization scheme is based on the weighted least square error criterion with complex Chebyshev error used as its magnitude and phase constraints. Developed version of the active set method has been exploited to solve the resultant semi-infinite quadratic programming problem. A quick and practical approach is proposed to evaluate the suboptimal value of the Laguerre parameter close to its optimum value. The design technique can be applied to design of asymmetric FIR filters just by setting the Laguerre parameter equal to zero. The accuracy of the processes has been verified by illustrating some numerical examples.\n\nOriginal language English 796-810 15 Signal Processing 88 4 https://doi.org/10.1016/j.sigpro.2007.09.020 Published - Apr 2008 Yes\n\n### Fingerprint\n\nDigital filters\nFIR filters\nFrequency response\n\n### Keywords\n\n• Complex digital Laguerre filter\n• Optimum Laguerre parameter\n• Peak constraint weighted least square design\n\n### ASJC Scopus subject areas\n\n• Signal Processing\n• Electrical and Electronic Engineering\n\n### Cite this\n\nIn: Signal Processing, Vol. 88, No. 4, 04.2008, p. 796-810.\n\nResearch output: Contribution to journalArticle\n\n@article{6ec7f7bb8c554354a23361df5790551a,\ntitle = \"Complex digital Laguerre filter design with weighted least square error subject to magnitude and phase constraints\",\nabstract = \"Complex coefficient optimal digital Laguerre filter design with arbitrary asymmetric frequency response is developed. The optimization scheme is based on the weighted least square error criterion with complex Chebyshev error used as its magnitude and phase constraints. Developed version of the active set method has been exploited to solve the resultant semi-infinite quadratic programming problem. A quick and practical approach is proposed to evaluate the suboptimal value of the Laguerre parameter close to its optimum value. The design technique can be applied to design of asymmetric FIR filters just by setting the Laguerre parameter equal to zero. The accuracy of the processes has been verified by illustrating some numerical examples.\",\nkeywords = \"Complex digital Laguerre filter, Optimum Laguerre parameter, Peak constraint weighted least square design\",\nyear = \"2008\",\nmonth = \"4\",\ndoi = \"10.1016/j.sigpro.2007.09.020\",\nlanguage = \"English\",\nvolume = \"88\",\npages = \"796--810\",\njournal = \"Signal Processing\",\nissn = \"0165-1684\",\npublisher = \"Elsevier\",\nnumber = \"4\",\n\n}\n\nTY - JOUR\n\nT1 - Complex digital Laguerre filter design with weighted least square error subject to magnitude and phase constraints\n\nAU - Zollanvari, Amin\n\nPY - 2008/4\n\nY1 - 2008/4\n\nN2 - Complex coefficient optimal digital Laguerre filter design with arbitrary asymmetric frequency response is developed. The optimization scheme is based on the weighted least square error criterion with complex Chebyshev error used as its magnitude and phase constraints. Developed version of the active set method has been exploited to solve the resultant semi-infinite quadratic programming problem. A quick and practical approach is proposed to evaluate the suboptimal value of the Laguerre parameter close to its optimum value. The design technique can be applied to design of asymmetric FIR filters just by setting the Laguerre parameter equal to zero. The accuracy of the processes has been verified by illustrating some numerical examples.\n\nAB - Complex coefficient optimal digital Laguerre filter design with arbitrary asymmetric frequency response is developed. The optimization scheme is based on the weighted least square error criterion with complex Chebyshev error used as its magnitude and phase constraints. Developed version of the active set method has been exploited to solve the resultant semi-infinite quadratic programming problem. A quick and practical approach is proposed to evaluate the suboptimal value of the Laguerre parameter close to its optimum value. The design technique can be applied to design of asymmetric FIR filters just by setting the Laguerre parameter equal to zero. The accuracy of the processes has been verified by illustrating some numerical examples.\n\nKW - Complex digital Laguerre filter\n\nKW - Optimum Laguerre parameter\n\nKW - Peak constraint weighted least square design\n\nUR - http://www.scopus.com/inward/record.url?scp=37648999292&partnerID=8YFLogxK\n\nUR - http://www.scopus.com/inward/citedby.url?scp=37648999292&partnerID=8YFLogxK\n\nU2 - 10.1016/j.sigpro.2007.09.020\n\nDO - 10.1016/j.sigpro.2007.09.020\n\nM3 - Article\n\nAN - SCOPUS:37648999292\n\nVL - 88\n\nSP - 796\n\nEP - 810\n\nJO - Signal Processing\n\nJF - Signal Processing\n\nSN - 0165-1684\n\nIS - 4\n\nER -"
] | [
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https://studyhippo.com/chem-102-lab-exam-1-exp-1-3/ | [
"# Chem 102 Lab – Exam 1 (exp. 1-3) – Flashcards",
null,
"Jaxon Craft\n\n## Unlock all answers in this set\n\nquestion\nPurpose of evaporation lab\nlooking at how much energy to pull molecules apart\nquestion\nintermolecular forces\nhold molecules together - different types include london-disperson, dipole-dipole, hydrogen bonds- and ion-dipole\nquestion\nlondon dispersion forces (LDF)\n- type of intermolecular force - everything has these - depend on instantaneous dipole moments (only happen when an electron happens to be in one place at one moment in time) which makes them weak and easy to break - get stronger as the molecule gets stronger since there are more of them (increase in molecular weight = increase in LDF)\nquestion\ndipole-dipole\n- type of intermolecular force - stronger than LDF because it is a permanent partial negative charge on one spot and a permanent partial positive charge on the other\nquestion\nhydrogen bond\n- type of intermolecular force - special case of dipole-dipole that occurs between hydrogen & O, F, or N - happens due to hydrogen being REALLY partially positive and O, F, or N being REALLY partially negative - if water didn't have a bent structure that allowed it to have 2 lone pairs (allowing 2 hydrogen bonds), then it would be a gas at room temp - strongest intermolecular force\nquestion\nion-dipole\n- type of intermolecular force - occurs between an ion and a polar molecule - stronger than dipole-dipole because in dipole-dipole, each of the molecules are only partially negative or positive whereas in ion-dipole one is totally positive and the other is totally negative so they have a stronger bond to the partial negative and partial positives from the polar molecule\nquestion\norder of strengths of intermolecular forces from weakest to strongest\nLDF < dipole-dipole< ion-dipole < h-bond\nquestion\nalkane\n- contain C and H - only intermolecular force present is LDF - the more carbons and hydrogens, the stronger the LDF will be (aka the higher the boiling point will be)\nquestion\nalcohols\n- contain C, H, and an OH group - intermolecular forces presents are LDF and h-bond\nquestion\nthe stronger the IMF, the ________ the change in temperature will be because...\nsmaller, it would take more time/energy to evaporate a substance with stronger intermolecular forces\nquestion\nwhat causes there to be a difference in states\n- kinetic energy, intermolecular forces, etc - depends on whats holding one molecule to another\nquestion\nwhat state has the molecules attracted closely but still able to move\nliquids\nquestion\nwhat state has the molecules not attracted at all\ngas\nquestion\nwhat state has the molecules attracted closely so that there is basically no movement\nsolid\nquestion\nunit cell\n- smallest unit that, when stacked repeatedly, makes up the entire crystal - 14 known unit cells - 3 types of cubic unit cells -> simple cubic, body centered, and face centered\nquestion\ncrystal\n- anything with a repeating pattern\nquestion\nhow do we look at a crystal?\n- x ray crystallography - have wave lengths on the size order of an atom\nquestion\nsimple cubic (primitive) cell\n- formation if you have an atom at each of the 8 corners of a perfect cube - only occurs in polonium - corners shared with 8 cubers (each individual corner is 1/8 of an atom) - one net atom in each cube -> weight of unit cell is weight of one atom of polonium\nquestion\ncoordination number\nhow many atoms does any 1 atom touch\nquestion\nnet atoms in simple cubic cell\n1\nquestion\ncoordination number in simple cubic cell\n6\nquestion\nlength of an edge in simple cubic cell\n2r\nquestion\nvolume occupied by atoms in a simple cubic cell\n4/3πr³\nquestion\nvolume of whole cell in a simple cubic cell\n8r³\nquestion\nbody centered cell\n- same as simple cubic with an atom on each corner but also has one in the direct center\nquestion\nnet atoms in a body centered cell\n1/8 on each corner = 1 total 1 in the middle = 1 total = 2 net atoms\nquestion\ncoordination number in a body centered cell\n8\nquestion\nedge length in a body centered cell\n4r/√3\nquestion\nvolume occupied by atoms in a body centered cell\n8/3πr³\nquestion\nvolume of whole cell in a body centered cell\n64r³/3√3\nquestion\nface centered cell\n- any face of the cell has an atom in the center and at every corner - looks like the 5 side of the dice\nquestion\nnet atoms in a face centered cell\n1/8 on each corner = 1 atom 1/2 on each face = 3 atoms = 4 net atoms\nquestion\ncoordination number in a face centered cell\n12\nquestion\nedge length in a face centered cell\n4r/√2\nquestion\nvolume occupied by atoms in a face centered cell\n16/3πr³\nquestion\nvolume of whole cell in a face centered cell\n64r³/2√2\nquestion\nempirical formula\nsimplest formula (ratio)\nquestion\nNaCl (solid state)\n- Cl⁻ forms face centered cell (net 4 atoms) - Na⁺ fills in the holes on the edges and one in the dead center ( 3 ions from edges (1/4 on each edge) and 1 ion from the center = net 4 atoms) - ratio of 4Na⁺ to 4Cl⁻ = empirical formula of NaCl\nquestion\nsolid state structures for ions\n- one ion forms the base cubic unit cell and the other ion fills in the holes -> usually the larger ion\nquestion\noctahedral holes\nholes on the edges and center of a face centered cell\nquestion\ncoordination number for an atom in an octahedral hole\n6\nquestion\nCuCl (solid state)\n- Cl⁻ is face centered - Cu⁺ is in the corners but inside the cell\nquestion\ntetrahedral holes\n- whole atom in the holes -> on the inside corners of the cell\nquestion\ncoordination number for an atom in a tetrahedral hole\n4\nquestion\nallotropes\ndifferent forms of the same element -> has different properties - i.e. carbon has a bunch of different allotropes\nquestion\nmost stable allotrope of carbon\ngraphite\nquestion\ngraphite\n- 2d structure - sheets of fused 6 membrane rings\nquestion\ndiamond\n- every carbon is bonded to 4 carbons - 3 dimensional\nquestion\ncolligative properties\ndepend on the # of solute particles in a solution - freezing point depression - boiling point elevation - vapor pressure decrease - osmotic pressure\nquestion\nsolvent\npresent in larger quantity in a solution - tends to be water\nquestion\nsolute\npresent in smaller quantity in a solution\nquestion\nwhy do we put salt in water for pasta?\nraises the boiling point\nquestion\nwhy do we put salt on the road before snow?\nlowers the freezing point\nquestion\nmolarity\n(mole solute)/(L solution)\nquestion\nmolality\n(moles solute)/(Kg solvent)\nquestion\nelevation of boiling point\nadding a solute increases the boiling point of a solvent - this is because it lowers the amount of surface area that the water has to evaporate from because there are also solid particles that are on the surface but won't form a vapor so they ultimately lower the vapor pressure\nquestion\nequation for change in boiling point\n∆t= (Kbp)(m)(i) i = 1 if you aren't dissolving an ionic compound, if its an ionic compound then its the number of particles the compound breaks into. i.e.: NaCl is 2 because it breaks in Na⁺ and Cl⁻\nquestion\nwhy does have NaCl have a bigger impact on boiling point than C₁₂H₂₂O₁₁?\nbecause NaCl is an ionic compound so it dissociates in water which makes it have twice as many molecules floating around as the other which will stay as one molecule\nquestion\nfreezing point depression\nadding a solute decreases the freezing point of a solvent - this is because there are solid particles on the surface that aren't going to freeze because they are already solid but aren't allowing water particles to freeze so it ultimately takes more energy to get the solution froze\nquestion\n∆t = (Kfp)(m)(i)\nquestion\nwhat is the edge length of a simple cubic unit cell in terms of \"r\"?\n2r\nquestion\nif the volume of a sphere is 4/3πr³, what is the volume of a face centered unit cell that is occupied by atoms\n16/3πr³\nquestion\nthe \"holes\" on the edges of a face centered cubic unit cell are called: a. octahedral holes b. tetrahedral holes c. pentagonal holes d. hexagonal holes\na. octahedral holes\nquestion\nthe carbons in diamond have what type of hybridization? a. sp b. sp² c. sp³ d. sp^4\nc. sp³\nquestion\nin the NaCl unit cell, in what type of arrangement or holes are the Cl⁻ ions found? a. simple cubic b. face centered. c. body centered\nb. face centered\nquestion\nthe presence of a solute ______ the frezing point of solution compared to freezing point of the pure solvent a. raises b. lowers c. has no effecent on\nb. lowers\nquestion\nthe solvent in the colligative properties experiment was a. lauric acid b. an unknown c. water\na. lauric acid\nquestion\nwhat is the freezing point of a solution in which 90.0g of glucose, C₆H₁₂O₆, is dissolved in 500 g of water. Kf water = 1.86°C/m MM glucose - 180g/mole a. 1.86°C b. -1.86°C c. 3.72°C d. -3.72°C e. 0.93°C\nb. -1.86°C\nquestion\nwhat is the molar mass of a substance if 0.0200g is equal to 3.34 x 10⁻⁶ moles? a. 1.67 x 10⁻⁴ b. 6.68 x 10⁻⁸ c. 5900\nc. 5900\nquestion\ncolligative properties depend on a. the number of solute particles b. the molar mass of the substance\na. the number of solute particles\nquestion\nin experiment 1, n-pentane and 1-butanol had almost identical molecular weights but significantly different ∆t values. Explain the difference in the ∆t values of these substances based on their intermolecular forces\n- n-pentane has only LDF - 1-butanol had LDF and H-bonds - h-bonds are stronger than LDF so 1-butanol has a smaller ∆t value since it takes more energy to break its intermolecular forces\nquestion\nwhich of the alcohols studied in experiment 1 has the strongest intermolecular forces of attraction? the weakest intermolecular forces? explain\n- 1-butanol has the highest molecular weight out of the alcohols which means it has the strongest intermolecular forces - methanol has the lowest molecular weight out of the alcohols which means it has the weakest intermolecular forces - when molecules have identical IMF types you have to look at the molecular weight of the molecules (higher MW = stronger IMF)\nquestion\nWhich of the two alkanes studied in experiment 1 has the stronger intermolecular forces? the weaker? explain\n- n-hexane has the stronger IMF forces because it has a higher molecular weight, which gives it more LDF - by default this makes n-pentane have the weaker IMF since it has the lower molecular weight and less LDF\nquestion\nhow many carbon atoms are in the diamond unit cell?\n8\nquestion\nwhat is the coordination number of each carbon in diamond?\n4\nquestion\nwhat is the hybridization of each carbon in diamond?"
] | [
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"https://studyhippo.com/wp-content/uploads/authors/Jaxon Craft.jpg",
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https://learnfromblogs.com/the-effects-of-math-homework-on-student-study | [
"# The effects of Math homework on student study\n\nAre you among the people who hate Math? Without any hesitation, you will say yes. Mathematics is the most boring subject for every student in the world. It is even more boring to do math homework. Whenever there is loads of homework, all the parents chase hard to get it done properly from their kids. They also send their kids to the best of math classes for scoring good marks in exams. Today, we will discuss the importance of math practice and how it can help in scoring good marks in exams.\n\n### Importance of Math Practice for the students\n\n#### Reasoning\n\nOne of the biggest advantages of math practice is that students get the logical reasoning for every question. Today, there are many ways by which students can get math homework help. There is modern equipment to teach the concepts of mathematics such as cards.\n\n#### Helps in handling various situations\n\nBy learning math, students can handle even the most difficult situations in life. Math can also help the students in inventions and discoveries of science. It is also important in all scientific operations and theories. Today, math is applicable in every field whether it is science, technology, or medicine.\n\n#### Helps in framing economics\n\nMath helps economists in making public policies. It further helps in making some of the most important decisions in the economy of the country. Cash reserve ratio and other analyses involve math as a major study. Math also helps in better functioning of the economy.\n\n#### Makes lateral thinking better\n\nMath helps the students in developing their lateral thinking. Today, one should have vast knowledge about other things rather than studies. With math, the students can gain the ability to use the available resources smartly and get the maximum results.\n\n#### Gain better focus\n\nWhen you solve math questions with answers, your focus becomes better at work. The focus is essential for every student these days. Math will help in increasing the concentration levels in the work. Students will learn to be more organized with math. They will also gain a feeling of helpfulness by learning math.\n\n#### Time management\n\nUnlike before, there are various new techniques nowadays by which students can gain an interest in mathematics. They can take the help of a math solver online and other experts. This subject will make the students more punctual in doing all their regular activities. Further, they will not feel tired of doing boring work also.\n\n#### Helps in understanding the complex things\n\nBecoming an expert in math will help the kids to understand even complex things. They will not get afraid or lose their self-confidence even in the toughest times of life. Math will increase the kids’ confidence in all things.\n\nApart from this, math also helps students in pursuing a career in science, engineering, medical, pharmacy, or MBA.\n\n### How can math homework help students?\n\nHave you ever thought why teachers always give you a bundle of math homework? There are some of the main reasons why it is so. Read below to know how math homework helps the students.\n\n#### Increases enthusiasm\n\nAny work done with passion, zeal, and enthusiasm proves to be the best. Solving math questions with answers makes the students more enthusiastic than before. They will love everything that they do. This will automatically reflect in their work. In another way, math increases the interest in you for every work.\n\nThese days the projects and assignments in high schools and colleges involve a lot of research. One has to find the information in the nook and corner for a unique project. Math homework will help the students to explore more of websites and other web pages. They will know to utilize every resource in a fruitful manner.\n\n#### Independence\n\nThe Internet has made possible everything these days. Students can easily get math homework answers online. There are various math experts who will help the students in getting done the math homework. In this way, students will get more independent in doing everything. They will not need the help of parents or teachers in small things.\n\n#### Revision\n\nFor scoring good marks in exams, it is necessary to revise the chapters taught earlier. Revision and practice are important in the subject of math. The students will get a better idea of the concepts with math homework. Whatever they learned can be easily grasped by doing math homework regularly.\n\n#### Helps the parents too\n\nAs a parent, you are always concerned about what your kid learns in school. Maths homework will also help the parents to know in detail what their students learned in the school. They can pay more attention to their kids which improves their performance in school.\n\n#### Improves memory\n\nToday, there is a Math word problem solver for solving the difficulties of students in math subjects. A sharp memory is necessary to do any work efficiently. Math homework will sharpen the memory and the students can remember everything in a precise way. The students can develop more their intellectual thinking with math.\n\n#### Prepares for exams\n\nPractice makes the man perfect is the quote which matches with the subject of math completely. Students have to practice math questions with answers for scoring good marks in exams. Doing math homework will make your kids ready for the exams beforehand. They don’t have to rush at the last moment of the exam.\n\n#### Helps the teachers too\n\nJust like students, teachers also have benefits if they give math homework to their students. It will further help the teachers in finding simple ways to teach math concepts.\n\n### Tips to score well in math exams\n\nA night before the math exam is very important for all students. If you are a student or were the student before, you must be aware of the tension that a math exam brings in one’s life. But, there are some expert tips by which you can score good marks in maths exams without stress or tension.\n\n#### Regular homework\n\nYou can now use the Photomath tool for doing the math homework. It makes the work of every student much better and simpler than before. For scoring good marks in exams, it is necessary that you should practice the mathematical equations every day without a break. Regular practice in math is the key to success. You can also get math homework answers from the internet.\n\n#### Revise the notes every now and then\n\nAs a student, you should make a habit of reviewing the math notes on a regular basis. After learning the chapters in the classroom, you should revise them at night, and doing this every day will help you gain a better knowledge of the subject.\n\n#### Use other math sources\n\nThe students have a lot of study material on every subject. For scoring excellent marks in the exams, you should refer to other materials rather than just textbooks. You can take math homework help from online tutorials which solve your questions anytime and anywhere. There are CD-ROMs, math guides, and other study materials for excelling in math for students which prove to be of great help during exams.\n\n#### Learn old concepts to master the new ones\n\nThe subject of math has succession. Whatever you learned in the earlier class may come in the next class. The new concepts of math have a relation with the old ones. In order to learn new math chapters well, you should review the previous years’ chapters. This will help you when get stuck up in solving the problems. You can also take the help of a math solver online when you are practicing the subject.\n\n#### Utilize free time\n\nAbove we discussed how math practice can help the students for scoring good marks in exams. There is another golden tip for gaining mastery in math subjects. You should review the chapters which you learned in the previous class or which are coming in the next class. Knowing the math concepts beforehand will help to know how to solve each problem.\n\n#### Take as much help as you can\n\nMath subject sounds more interesting with a group of friends. You can make a group and discuss the concepts in a clear manner. There are many online math teachers available on different websites who will help in math homework answers. There is also a math problem solver which will help you in getting mastery in the subject.\n\n#### Think of various steps to solve the problems\n\nStudents should find out the different ways to solve math problems. Math has various ways by which a problem can be solved. You can memorize the steps which will solve the problem in an easy way. You should find easy steps to solve math problems.\n\n### Important tips for math homework\n\n#### Do homework on time\n\nAs a parent, you should instruct your kids to do math homework on time. It is also the responsibility of the students to do work on time. This will save much of their time and they can practice more sums in a day. You can pick noon or evening hours for practicing math sums after lunch or play.\n\n#### Change the place\n\nMath seems boring if you practice it daily in one place. As a teacher, you can choose the place where the students can learn the subject in a better way. As a parent, you can choose a park where your kids will love doing maths sums. You can get them a math word problem solver by which they will take more interest in the subject.\n\n#### Practice with a fresh mind\n\nToday, the students are loaded with assignments, internals, and projects. They have to handle various tasks in a limited time frame. So, if you are a student, you should practice math sums when you really want to. You should have a fresh mind before practicing math. As a parent, you should allow your kids to play for a while or do some interesting things to boost their moods.\n\n#### Never lose confidence\n\nMath often makes the students nervous and worried. They may get demotivated if there is not an answer to a question. As a student, you should have self-confidence. You should try to solve the problems in numerous ways. You can get help from a math solver online or other websites which will make you better in the subject. The parents should also praise their kids which will increase their motivation and confidence in every situation.\n\n#### Set time\n\nProcrastination is often dangerous, especially for students. The teachers should set a timer to get the work done by their students. The students also should solve math problems within a given time. This will help in time management and they can also perform better in exams.\n\nThis is one of the most important tips for parents. Whenever there is math homework, you should ask the teachers to give their feedback after checking it. You can discuss with teachers about the progress of your child. This will also help your kids in getting better in weak areas. You can call the teacher or send an email asking for their feedback on math homework.\n\n#### Plan it\n\nIf you do not understand a certain math sum, you should circle it or mark it with a pen or pencil for asking your teachers later. You can prepare a note of sums which you didn’t understand in the previous class.\n\n##### Conclusion\n\nThese Math homework help tips are beneficial for students, parents, and teachers. Apart from that, there are also online guides and other books that will help the students in solving the sums of algebra and geometry easily. Math is the subject of practice and hard work. You should keep practicing without a break for excelling in it.\n\n###### 0+ homework tips Posts\n\nMaximum 0/300 words\n\n### Related Opinion\n\nIn your student life, as it is, you get very less free time, and you do not want to spend all your time doing your homework. If you are involved in extracurricular activities, finding time for academics becomes even more challenging. Homework for students is a compulsory thing in most schools. No matter how much you dislike homework, teachers are going to assign it. Most of the students are engaged in other activities such as after-school jobs, taking hobby classes, or going to tuitions. Sometimes, their schedule does not permit them to complete their assignment due to which they face...\n\nView more\n\nThis question had been asked to me in my school years when I was a child, I would have said ‘No’ in a very loud voice. But now, when I am a parent of two wonderful kids, saying direct ‘No’ to homework is very difficult for me. With time, I realized that homework has some good parts to play in a child’s life. But children often fail to understand it, and it is exactly what I did when I was a kid.But when I grew up and became a teacher, some questions frequently haunted my mind: Is homework harmful or...\n\nView more\n\nDo you like the idea of homework? To me, sometimes I feel that homework is essential, but when I see a child glued to the desk for a long with a very strained look on the face, I am not too fond of the idea of homework. Childhood is meant for enjoyment and learning and not for stress. However, an actual fact is that this is the age to learn the value of dedication and hard work. So what to do? Is there any middle way? How much homework is right and how much homework is too much? These questions...\n\nView more\n\nMath is not so difficult subject, but it certainly makes the students disinterested. Geometry includes theorems, formulas, shapes, angles, and much more. Doing geometry homework is always a tricky and confusing thing for students. They have to first understand the concepts of geometry, theorems, and formulas, after which they can do homework in a better way. But, today, there are many tips and tricks to learn geometry lessons faster and better. They are as follows:i.) Save timeTime is an important concept in doing geometry problems. Saving time will help you to understand the subject in depth. For learning geometry lessons...\n\nView more"
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https://link.springer.com/article/10.1007/s11424-017-5309-3 | [
"Advertisement\n\nJournal of Systems Science and Complexity\n\n, Volume 30, Issue 5, pp 1160–1172\n\n# A relative error estimation approach for multiplicative single index model\n\nArticle\n\n## Abstract\n\nAs an alternative to absolute error methods, such as the least square and least absolute deviation estimations, a product relative error estimation is proposed for a multiplicative single index regression model. Regression coefficients in the model are estimated via a two-stage procedure and their statistical properties such as consistency and normality are studied. Numerical studies including simulation and a body fat example show that the proposed method performs well.\n\n## Keywords\n\nAsymptotic properties least product relative error relative errors single index model\n\n## Preview\n\nUnable to display preview. Download preview PDF.\n\n## References\n\n1. \nChen K N, Guo S J, Lin Y Y, et al., Least absolute relative error estimation, Journal of the American Statistical Association, 2010, 105(491): 1104–1112.\n2. \nChen K N, Lin Y Y, Wang Z F, et al., Least product relative error estimation, Journal of Multivariate Analysis, 2016, 144: 91–98.\n3. \nPark H S and Stefanski L A, Relative-error prediction, Statistics & Probability Letters, 1998, 40(3): 227–236.\n4. \nYe J M, Price models and the value relevance of accounting information, SSRN Electronic Journal, http://ssrn.com/abstract=1003067, 2007.Google Scholar\n5. \nZhang Q Z and Wang Q H, Local least absolute relative error estimating approach for partially linear multiplicative model, Statistica Sinica, 2013, 23(3): 1091–1116.\n6. \nWang Z F, Liu W X, and Lin Y Y, A change-point problem in relative error-based regression, TEST, 2015, 24(4): 835–856.\n7. \nHristache M, Juditsky A, and Spokoiny V, Direct estimation of the index coefficient in a singleindex model, Annals of Statistics, 2001, 29(3): 595–623.\n8. \nRuppert D, Wand M P, and Carroll R J, Semiparametric Regression, Cambridge University Press, London, 2003.\n9. \nStute W and Zhu L X, Nonparametric checks for single-index models, The Annals of Statistics, 2005, 33(3): 1048–1083.\n10. \nZhu L X and Xue L G, Empirical likelihood confidence regions in a partially linear single-index model, Journal of the Royal Statistical Society: Series B (Statistical Methodology), 2006, 68(3): 549–570.\n11. \nWang J L, Xue L G, Zhu L X, et al., Estimation for a partial linear single-index model, The Annals of Statistics, 2010, 38(1): 246–274.\n12. \nChang Z Q, Xue L G, and Zhu L X, On an asymptotically more efficient estimation of the single-index model, Journal of Multivariate Analysis, 2010, 101(8): 1898–1901.\n13. \nKong E and Xia Y C, A single-index quantile regression model and its estimation, Econometric Theory, 2012, 28(4): 730–768.\n14. \nHärdle W K, Müller M, Sperlich S, et al., Nonparametric and Semiparametric Models, Springer Science & Business Media, Berlin, 2012.\n15. \nFan J Q and Gijbels I, Local Polynomial Modelling and Its Applications: Monographs on Statistics and Applied Probability, CRC Press, London, 1996.\n16. \nCarroll R J, Fan J Q, Gijbels I, et al., Generalized partially linear single-index models, Journal of the American Statistical Association, 1997, 92(438): 477–489.\n17. \nYu Y and Ruppert D, Penalized spline estimation for partially linear single-index models, Journal of the American Statistical Association, 2002, 97(460): 1042–1054.\n18. \nLi K C, Sliced inverse regression for dimension reduction, Journal of the American Statistical Association, 1991, 86(414): 316–327.\n19. \nXia Y C, Tong H, Li W K, et al., An adaptive estimation of dimension reduction space, Journal of the Royal Statistical Society: Series B (Statistical Methodology), 2002, 64(3): 363–410.\n20. \nLi T T, Yang H, Wang J L, et al., Correction on estimation for a partial-linear single-index model, The Annals of Statistics, 2011, 39(6): 3441–3443.\n\n## Copyright information\n\n© Institute of Systems Science, Academy of Mathematics and Systems Science, CAS and Springer-Verlag GmbH Germany 2017\n\n## Authors and Affiliations\n\n1. 1.Department of Statistics and Finance, School of ManagementUniversity of Science and Technology of ChinaHefeiChina\n\n## Personalised recommendations\n\n### Citearticle",
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"https://link.springer.com/track/controlled/article/denied/10.1007/s11424-017-5309-3",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.57649803,"math_prob":0.61196756,"size":4592,"snap":"2019-35-2019-39","text_gpt3_token_len":1257,"char_repetition_ratio":0.18177855,"word_repetition_ratio":0.05592105,"special_character_ratio":0.26110628,"punctuation_ratio":0.19810201,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95064676,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-19T04:26:05Z\",\"WARC-Record-ID\":\"<urn:uuid:58a4b7b6-6f3c-4258-b60c-d74a56f7dfde>\",\"Content-Length\":\"92424\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:34d12fe5-f58f-49ea-8158-f915710192c8>\",\"WARC-Concurrent-To\":\"<urn:uuid:8cfc0292-f6da-48f6-9a63-e51829ce239b>\",\"WARC-IP-Address\":\"151.101.200.95\",\"WARC-Target-URI\":\"https://link.springer.com/article/10.1007/s11424-017-5309-3\",\"WARC-Payload-Digest\":\"sha1:QX2R4CC22GLLLQJMW5OIKHYITX2OPA6Y\",\"WARC-Block-Digest\":\"sha1:5DJKANCNNP37F73H7K7VPPBMEM5I7OWP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027314641.41_warc_CC-MAIN-20190819032136-20190819054136-00141.warc.gz\"}"} |
http://www.orocos.org/stable/documentation/rtt/v2.x/api/html/classRTT_1_1types_1_1carray.html | [
"Orocos Real-Time Toolkit 2.6.0\nRTT::types::carray< T > Class Template Reference\n\nWraps a C array such that we can return a C array from a DataSource. More...\n\n`#include <rtt/types/carray.hpp>`\n\nList of all members.\n\n## Public Types\n\ntypedef T value_type\n\n## Public Member Functions\n\ncarray (value_type *t, std::size_t s)\nCreate an C array wrapper.\ncarray ()\nCreates an empty carray.\ncarray (boost::serialization::array< T > const &orig)\nWe are constructible from boost::serialization::array<T> Makes a shallow copy in order to keep the reference to the original data.\ntemplate<std::size_t N>\ncarray (boost::array< T, N > &orig)\nWe are constructible from boost::array<T,N> Makes a shallow copy in order to keep the reference to the original data.\nvoid init (value_type *t, std::size_t s)\n(Re-)initialize this carray to a new address and size.\nvalue_type * address () const\nThe address of the first element of the array.\nstd::size_t count () const\nconst carray< T > & operator= (const carray< T > &orig)\nAssignment only copies max(this->count(), orig.count()) elements from orig to this object.\n\n## Detailed Description\n\n### template<class T> class RTT::types::carray< T >\n\nWraps a C array such that we can return a C array from a DataSource.\n\nInspired on boost::serialization::array.\n\nDefault copy-constructible (shallow copy).\n\nAn assignment (operator=) makes a 'deep copy', while copy construction makes a 'shallow copy', where the copy refers to the same C array. This was chosen in the spirit of this class, where it keeps track of the original data, but when assigned from another carray, is meant as copying the data itself.\n\nYou can also point to parts of arrays. An uninitialized carray object returns null for both count() and address().\n\nParameters:\n T The data type of the array.\n\nDefinition at line 69 of file carray.hpp.\n\n## Constructor & Destructor Documentation\n\ntemplate<class T>\n RTT::types::carray< T >::carray ( value_type * t, std::size_t s ) ` [inline]`\n\nCreate an C array wrapper.\n\nParameters:\n t Pointer to first element of array. s Length of array. If zero, the parameter t will be ignored.\n\nDefinition at line 80 of file carray.hpp.\n\ntemplate<class T>\n RTT::types::carray< T >::carray ( ) ` [inline]`\n\nCreates an empty carray.\n\nYou are not allowed to read or write (operator=) to this array until it has been initialized.\n\nSee also:\ninit() in order to initialize it later on.\n\nDefinition at line 91 of file carray.hpp.\n\ntemplate<class T>\n RTT::types::carray< T >::carray ( boost::serialization::array< T > const & orig ) ` [inline]`\n\nWe are constructible from boost::serialization::array<T> Makes a shallow copy in order to keep the reference to the original data.\n\nParameters:\n orig\n\nDefinition at line 99 of file carray.hpp.\n\ntemplate<class T>\ntemplate<std::size_t N>\n RTT::types::carray< T >::carray ( boost::array< T, N > & orig ) ` [inline]`\n\nWe are constructible from boost::array<T,N> Makes a shallow copy in order to keep the reference to the original data.\n\nParameters:\n orig\n\nDefinition at line 112 of file carray.hpp.\n\n## Member Function Documentation\n\ntemplate<class T>\n value_type* RTT::types::carray< T >::address ( ) const` [inline]`\n\nThe address of the first element of the array.\n\nReturns:\nnull if count() == 0, the address otherwise.\n\nDefinition at line 130 of file carray.hpp.\n\nReferenced by RTT::types::carray< T >::operator=().\n\ntemplate<class T>\n std::size_t RTT::types::carray< T >::count ( ) const` [inline]`\nReturns:\nThe number of elements in the array.\n\nDefinition at line 138 of file carray.hpp.\n\nReferenced by RTT::types::carray< T >::operator=().\n\ntemplate<class T>\n const carray& RTT::types::carray< T >::operator= ( const carray< T > & orig ) ` [inline]`\n\nAssignment only copies max(this->count(), orig.count()) elements from orig to this object.\n\nIf orig.count() is smaller than this->count() the contents of the remaining elements is left unmodified. If it's greater, the excess elements are ignored.\n\nDefinition at line 149 of file carray.hpp.\n\nThe documentation for this class was generated from the following file:"
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https://www.edinformatics.com/science_projects/labs/ideas3.htm | [
"",
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"Today is\nScience Project Resources:\n\nNews\n\n • Science (general) • Health • Science (specific) • Technology\n\nScience Project Ideas\n\nExpanding a School Laboratory Experiment into a Science Project\n\nHere are some examples of School Laboratory activities that can be expanded into a science project. The following experiments are all from our integrated Math and Science Activity Center. ll these experiments are centered on the central theme-- Applications to Mathematical Relationships\n\nLab 1: The Spring Constant:\n\nProblem: What is the relationship between how much a spring stretches and the force pulling on the spring?\n\nLab 2: The Pendulum:\n\nProblem: What is the relationship between the period of a pendulum and the length of the string of the pendulum?\n\n:\n\nProblem: What is the relationship between the mass of a ball and its volume assuming a constant density?\n\nLab 3: Density of a Liquid:\n\nProblem: What is the relationship between the volume of a liquid and its density.\n\nLab 4: Light Intensity:\n\nProblem: What is the relationship between the intensity of a beam of light and the distPance from a light source?\n\nLab 5: Acceleration:\n\nProblem: What is a the relationship between how the distance travels and the time in travel for an accelerating object?\n\nLab 6: Polarization:\n\nProblem: What is the relationship between how much light passes through a Polaroid filter and the angle the filter is rotated?\n\nLab 7: Ohms Law :\n\nProblem: What is the relationship between current, voltage when there is a constant resistance in an electric circuit.\n\nProblem: What is the relationship between the decay of radioactive material and the time allowed for the decay?\n\nLab 9: Water Pressure:\n\nProblem: What is the relationship between water pressure and depth of water?\n\nLab 10: Attractive and Repulsive Forces:\n\nProblem: What is the relationship between the distance between two magnets and the force between them?\n\nLab 11: Damping Motion:\n\nProblem: What is the relationship between the height a ball bounces and the number of times it has bounced?"
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https://groupprops.subwiki.org/w/index.php?title=Nonstandard_definitions_of_group&oldid=10509 | [
"# Nonstandard definitions of group\n\n(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)\nThis is a survey article related to:group\nView other survey articles about group\n\nImportant terms have multiple definitions, from different perspectives. Some of these definitions describe the objects as special cases of more complicated objects.\n\nBy a definition of group we'll loosely mean something that gives rise to groups, and yields the same notion of isomorphism.\n\n## As automorphisms of a structure\n\n### Subgroup of a symmetric group\n\nNaive definition: A group is a collection of permutations on a set, that contains the identity permutation, is closed under inversion, and is closed under composition. In other words, a group is a subgroup of a symmetric group.\n\nJustification for this definition: Cayley's theorem, which states that every group can be embedded as a subgroup of a symmetric group, via the regular action: the action on itself by left multiplication.\n\nProblem with this definition: The chief problem with this definition is that it doesn't recognize that the same group could arise in totally different ways as a subgroup of a symmetric group. For instance, in the symmetric group on four elements, the Klein-four group occurs in two distinct ways: one as a normal subgroup including the double transpositions, and the other, as a direct product of 2-element subgroups generated by disjoint transpositions (e.g.,",
null,
"$\\langle (1 2), (3 4) \\rangle$).\n\nModification to the definition: The definition can be rectified by including what it means for two groups to be isomorphic.\n\n### Subgroup of a general linear group\n\nNaive definition: A group is a collection of invertible linear transformation on a vector space, that contains the identity transformation, and is closed under composition and inversion of transformations. In other words, a group is a subgroup of a general linear group.\n\nJustification for this definition: Every group can be embedded in a symmetric group, and the symmetric group on a set embeds into the vector space with that set as basis (note: the set may be infinite, so the vector space may be infinite-dimensional).\n\nProblem with this definition: As in the case of the previous definition, the definition fails to capture the notion of isomorphism of groups.\n\nModification to the definition: The definition can be rectified by including what it means for two groups to be isomorphic.\n\n### Category with one object\n\nDefinition': A group is a category with one object, where every morphism is invertible. Two groups are isomorphic if the corresponding categories are equivalent as categories (or equivalently, are isomorphic as categories).\n\n## Group with (absent) additional structure\n\n### Discrete topological group\n\nDefinition: A group is a discrete topological group.\n\nJustification for this definition: We can embed the category of groups inside the category of topological groups by sending each group to the corresponding discrete topological group. In fact, this embedding is the left-adjoint functor to the forgetful functor from topological groups to groups, that sends a topological group to its underlying group. The notion of discrete topological group also resonates, for instance, with the general idea that covering spaces and fundamental groups are a special case of principal bundles for a topological group.\n\n### Group object in the category of sets\n\nDefinition: A group is a group object in the category of sets (here, the monoidal product is Cartesian product).\n\nJustification for the definition: The notion of group object was invented to generalize the idea of groups to structures that are sets with additional structure, or are structures without an underlying set. When we specialize to the category of sets, we recover the original notion of group.\n\n### Hopf algebra in the category of sets\n\nDefinition: A group is a Hopf algebra in the category of sets.\n\n## As strengthenings of its weakenings\n\nAlso see: Varying group\n\n1. A group is a monoid where every element has a two-sided inverse\n2. A group is an associative algebra loop, or equivalently, is an associative quasigroup"
] | [
null,
"https://groupprops.subwiki.org/w/images/math/a/b/2/ab27943d31357341fc14fc5bdcc1c0fb.png ",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9178628,"math_prob":0.87804025,"size":4118,"snap":"2021-04-2021-17","text_gpt3_token_len":824,"char_repetition_ratio":0.16820613,"word_repetition_ratio":0.13846155,"special_character_ratio":0.18406993,"punctuation_ratio":0.10273973,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97966254,"pos_list":[0,1,2],"im_url_duplicate_count":[null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-19T03:17:02Z\",\"WARC-Record-ID\":\"<urn:uuid:2038833a-5540-4b96-9a5d-7ede8cdb7347>\",\"Content-Length\":\"29682\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:57bb4371-881b-41b7-a4e5-b55815b299e1>\",\"WARC-Concurrent-To\":\"<urn:uuid:a85037b1-d6e4-4c44-9575-dec5962a2649>\",\"WARC-IP-Address\":\"96.126.114.7\",\"WARC-Target-URI\":\"https://groupprops.subwiki.org/w/index.php?title=Nonstandard_definitions_of_group&oldid=10509\",\"WARC-Payload-Digest\":\"sha1:MO5NYBR6H5OAJN7C7PZ7DSZXTHUFPERC\",\"WARC-Block-Digest\":\"sha1:2P7VFUOTCBUNVAMQDLJJSK5SRK25ZINE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703517559.41_warc_CC-MAIN-20210119011203-20210119041203-00159.warc.gz\"}"} |
https://www.stat.math.ethz.ch/pipermail/r-help/2008-October/177553.html | [
"# [R] Bootstrap\n\nDaniel Malter daniel at umd.edu\nWed Oct 22 02:31:26 CEST 2008\n\n```Two data frames, data.x and data.y with one random variable and one\nidentifier variable each. Two methods to merge them, one the \"brute force\"\napproach, the other using R's merge() function:\n\ndata.x=data.frame(rnorm(100,0,1),1:100)\n##data.x is already orderer by id\n\nnames(data.x)=c(\"x\",\"x.id\")\n\ndata.y=data.frame(rnorm(100,0,1),sample(1:100,replace=F))\n##data.y is not ordered by id\n\nnames(data.y)=c(\"y\",\"y.id\")\n\n##order y by id\ndata.y.new=data.y[order(data.y\\$y.id),]\n\n##after ordering y, just bind the datasets together\nmerged.data1=data.frame(data.x,data.y.new)\nmerged.data1\n##This works only if the identifier variable is unique\n\n##more elegant solution to merge datasets by merge()\nmerged.data2=merge(data.x,data.y,by.x=\"x.id\",by.y=\"y.id\")\nmerged.data2\n##This also works if identifier is not unique\n\nCheers,\nDaniel\n\n-------------------------\ncuncta stricte discussurus\n-------------------------\n\n-----Ursprüngliche Nachricht-----\nVon: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] Im\nAuftrag von Marcioestat\nGesendet: Tuesday, October 21, 2008 5:24 PM\nAn: r-help at r-project.org\nBetreff: [R] Bootstrap\n\nHi listers,\nI've been work on a bootstrap estimator and I don't know how to make a\ncertain manipulation at the data.\nAs an example I have the following data and samples... I need now to\nidentify the vector in wich I sampled the id, I mean that I have to merge\nthe files having the key as the id of the samples in order to find the\nvector that were selected.\nMárcio\n\nid<-seq(1:49)\nx<-runif(49)\nu<-runif(49)\ndata<-cbind(id,x,u)\nB<-10\nsampling<-lapply(1:B, function(i) sample(id, replace=T))\n\n--\nView this message in context:\nhttp://www.nabble.com/Bootstrap-tp20099684p20099684.html\nSent from the R help mailing list archive at Nabble.com.\n\n______________________________________________\nR-help at r-project.org mailing list\nhttps://stat.ethz.ch/mailman/listinfo/r-help"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5837285,"math_prob":0.6368859,"size":2174,"snap":"2022-05-2022-21","text_gpt3_token_len":602,"char_repetition_ratio":0.13686636,"word_repetition_ratio":0.0,"special_character_ratio":0.31048757,"punctuation_ratio":0.20445344,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95245296,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-19T16:47:03Z\",\"WARC-Record-ID\":\"<urn:uuid:5a58db68-e819-4dd0-935d-0e302fc393ad>\",\"Content-Length\":\"4968\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:518ba204-5365-4037-aaee-5036fe53a92c>\",\"WARC-Concurrent-To\":\"<urn:uuid:b0d8b414-df4d-4ec8-a017-8f25c87ea8bf>\",\"WARC-IP-Address\":\"129.132.119.195\",\"WARC-Target-URI\":\"https://www.stat.math.ethz.ch/pipermail/r-help/2008-October/177553.html\",\"WARC-Payload-Digest\":\"sha1:YGW6ZNXY6TP2XAHNJS5NIEHLHGFG3Y6P\",\"WARC-Block-Digest\":\"sha1:RJ57ZKOBKHOIZUZ4JVG7JQXSXQ7ERSFX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662529538.2_warc_CC-MAIN-20220519141152-20220519171152-00400.warc.gz\"}"} |
https://www.colorhexa.com/03f296 | [
"# #03f296 Color Information\n\nIn a RGB color space, hex #03f296 is composed of 1.2% red, 94.9% green and 58.8% blue. Whereas in a CMYK color space, it is composed of 98.8% cyan, 0% magenta, 38% yellow and 5.1% black. It has a hue angle of 156.9 degrees, a saturation of 97.6% and a lightness of 48%. #03f296 color hex could be obtained by blending #06ffff with #00e52d. Closest websafe color is: #00ff99.\n\n• R 1\n• G 95\n• B 59\nRGB color chart\n• C 99\n• M 0\n• Y 38\n• K 5\nCMYK color chart\n\n#03f296 color description : Vivid cyan - lime green.\n\n# #03f296 Color Conversion\n\nThe hexadecimal color #03f296 has RGB values of R:3, G:242, B:150 and CMYK values of C:0.99, M:0, Y:0.38, K:0.05. Its decimal value is 258710.\n\nHex triplet RGB Decimal 03f296 `#03f296` 3, 242, 150 `rgb(3,242,150)` 1.2, 94.9, 58.8 `rgb(1.2%,94.9%,58.8%)` 99, 0, 38, 5 156.9°, 97.6, 48 `hsl(156.9,97.6%,48%)` 156.9°, 98.8, 94.9 00ff99 `#00ff99`\nCIE-LAB 84.854, -68.677, 31.159 37.292, 65.721, 39.573 0.262, 0.461, 65.721 84.854, 75.414, 155.596 84.854, -74.13, 54.808 81.069, -59.76, 27.807 00000011, 11110010, 10010110\n\n# Color Schemes with #03f296\n\n• #03f296\n``#03f296` `rgb(3,242,150)``\n• #f2035f\n``#f2035f` `rgb(242,3,95)``\nComplementary Color\n• #03f21e\n``#03f21e` `rgb(3,242,30)``\n• #03f296\n``#03f296` `rgb(3,242,150)``\n• #03d7f2\n``#03d7f2` `rgb(3,215,242)``\nAnalogous Color\n• #f21e03\n``#f21e03` `rgb(242,30,3)``\n• #03f296\n``#03f296` `rgb(3,242,150)``\n• #f203d7\n``#f203d7` `rgb(242,3,215)``\nSplit Complementary Color\n• #f29603\n``#f29603` `rgb(242,150,3)``\n• #03f296\n``#03f296` `rgb(3,242,150)``\n• #9603f2\n``#9603f2` `rgb(150,3,242)``\nTriadic Color\n• #5ff203\n``#5ff203` `rgb(95,242,3)``\n• #03f296\n``#03f296` `rgb(3,242,150)``\n• #9603f2\n``#9603f2` `rgb(150,3,242)``\n• #f2035f\n``#f2035f` `rgb(242,3,95)``\nTetradic Color\n• #02a667\n``#02a667` `rgb(2,166,103)``\n• #02c077\n``#02c077` `rgb(2,192,119)``\n• #03d986\n``#03d986` `rgb(3,217,134)``\n• #03f296\n``#03f296` `rgb(3,242,150)``\n• #12fca2\n``#12fca2` `rgb(18,252,162)``\n• #2cfcac\n``#2cfcac` `rgb(44,252,172)``\n• #45fdb6\n``#45fdb6` `rgb(69,253,182)``\nMonochromatic Color\n\n# Alternatives to #03f296\n\nBelow, you can see some colors close to #03f296. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #03f25a\n``#03f25a` `rgb(3,242,90)``\n• #03f26e\n``#03f26e` `rgb(3,242,110)``\n• #03f282\n``#03f282` `rgb(3,242,130)``\n• #03f296\n``#03f296` `rgb(3,242,150)``\n• #03f2aa\n``#03f2aa` `rgb(3,242,170)``\n• #03f2be\n``#03f2be` `rgb(3,242,190)``\n• #03f2d2\n``#03f2d2` `rgb(3,242,210)``\nSimilar Colors\n\n# #03f296 Preview\n\nText with hexadecimal color #03f296\n\nThis text has a font color of #03f296.\n\n``<span style=\"color:#03f296;\">Text here</span>``\n#03f296 background color\n\nThis paragraph has a background color of #03f296.\n\n``<p style=\"background-color:#03f296;\">Content here</p>``\n#03f296 border color\n\nThis element has a border color of #03f296.\n\n``<div style=\"border:1px solid #03f296;\">Content here</div>``\nCSS codes\n``.text {color:#03f296;}``\n``.background {background-color:#03f296;}``\n``.border {border:1px solid #03f296;}``\n\n# Shades and Tints of #03f296\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000906 is the darkest color, while #f5fffb is the lightest one.\n\n• #000906\n``#000906` `rgb(0,9,6)``\n• #001d12\n``#001d12` `rgb(0,29,18)``\n• #01301e\n``#01301e` `rgb(1,48,30)``\n• #01442a\n``#01442a` `rgb(1,68,42)``\n• #015736\n``#015736` `rgb(1,87,54)``\n• #016a42\n``#016a42` `rgb(1,106,66)``\n• #027e4e\n``#027e4e` `rgb(2,126,78)``\n• #02915a\n``#02915a` `rgb(2,145,90)``\n• #02a466\n``#02a466` `rgb(2,164,102)``\n• #02b872\n``#02b872` `rgb(2,184,114)``\n• #03cb7e\n``#03cb7e` `rgb(3,203,126)``\n• #03df8a\n``#03df8a` `rgb(3,223,138)``\n• #03f296\n``#03f296` `rgb(3,242,150)``\nShade Color Variation\n• #0dfca0\n``#0dfca0` `rgb(13,252,160)``\n• #20fca7\n``#20fca7` `rgb(32,252,167)``\n• #33fcaf\n``#33fcaf` `rgb(51,252,175)``\n• #47fdb7\n``#47fdb7` `rgb(71,253,183)``\n• #5afdbe\n``#5afdbe` `rgb(90,253,190)``\n• #6dfdc6\n``#6dfdc6` `rgb(109,253,198)``\n• #81fdcd\n``#81fdcd` `rgb(129,253,205)``\n• #94fed5\n``#94fed5` `rgb(148,254,213)``\n• #a8fedd\n``#a8fedd` `rgb(168,254,221)``\n• #bbfee4\n``#bbfee4` `rgb(187,254,228)``\n• #cefeec\n``#cefeec` `rgb(206,254,236)``\n• #e2fff4\n``#e2fff4` `rgb(226,255,244)``\n• #f5fffb\n``#f5fffb` `rgb(245,255,251)``\nTint Color Variation\n\n# Tones of #03f296\n\nA tone is produced by adding gray to any pure hue. In this case, #74817c is the less saturated color, while #03f296 is the most saturated one.\n\n• #74817c\n``#74817c` `rgb(116,129,124)``\n• #6b8a7e\n``#6b8a7e` `rgb(107,138,126)``\n• #619480\n``#619480` `rgb(97,148,128)``\n• #589d82\n``#589d82` `rgb(88,157,130)``\n• #4ea785\n``#4ea785` `rgb(78,167,133)``\n• #45b087\n``#45b087` `rgb(69,176,135)``\n• #3cb989\n``#3cb989` `rgb(60,185,137)``\n• #32c38b\n``#32c38b` `rgb(50,195,139)``\n• #29cc8d\n``#29cc8d` `rgb(41,204,141)``\n• #1fd68f\n``#1fd68f` `rgb(31,214,143)``\n• #16df92\n``#16df92` `rgb(22,223,146)``\n• #0ce994\n``#0ce994` `rgb(12,233,148)``\n• #03f296\n``#03f296` `rgb(3,242,150)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #03f296 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5278222,"math_prob":0.7780102,"size":3695,"snap":"2019-35-2019-39","text_gpt3_token_len":1613,"char_repetition_ratio":0.12652397,"word_repetition_ratio":0.011049724,"special_character_ratio":0.5564276,"punctuation_ratio":0.23522854,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9857262,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-18T13:58:23Z\",\"WARC-Record-ID\":\"<urn:uuid:61640841-0248-4b44-bcdf-f2c47af34c30>\",\"Content-Length\":\"36267\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3704a0b0-372e-4b05-9716-8848e74a08ff>\",\"WARC-Concurrent-To\":\"<urn:uuid:020c87c3-68ee-42b1-a871-c05045ec2f75>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/03f296\",\"WARC-Payload-Digest\":\"sha1:UBKYJ55VP55ZLWQQEU6V4I3U47W4LMPD\",\"WARC-Block-Digest\":\"sha1:ED3ROJEIOD66YFNY2OCGEQYQ4ZCH5553\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027313889.29_warc_CC-MAIN-20190818124516-20190818150516-00486.warc.gz\"}"} |
https://www.dreamincode.net/forums/topic/414012-ieee-754-and-machine-numbers/ | [
"# IEEE-754 and machine numbers\n\nPosted 08 December 2018 - 01:10 PM\n\nI've been trying to wrap my head around machine numbers like the unit roundoff (u) and epsilon (e) in combination with the IEEE 754 standard. My textbook states some things that don't really make sense to me.\n\nUnit roundoff according to my textbook is:\n\n• for single precision (mantissa is 23 bit): u = 6e-8\n• for double precision (mantissa is 52 bit): u = 2e-16\n\nI've been trying to derive a formula for these results with two relations:\n\n• my textbook states: \"In binary arithmetic with rounding we usually have e = 2*u\"\n• e = 2^-n, n being the amount of mantissa bits\n\nThese combined results would then give: u = 2^-(n+1), again with n being the amount of mantissa bits. Checking this formule with the given results of u for different precisions:\n\nfor single: u = 2^-(23+1) = 5.96e-8, this result checks out.\nfor double: u = 2^-(52+1) = 1.11e-16, this result doesn't check out.\n\nCould someone please help me derive a correct formule for the unit roundoff, or point me to some mistakes I have been making? All help is appreciated.\n\nIs This A Good Question/Topic? 0\n\nPage 1 of 1\n\n .related ul { list-style-type: circle; font-size: 12px; font-weight: bold; } .related li { margin-bottom: 5px; background-position: left 7px !important; margin-left: -35px; } .related h2 { font-size: 18px; font-weight: bold; } .related a { color: blue; }"
] | [
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https://astarmathsandphysics.com/ib-maths-notes/polynomials/1016-finding-a-coefficient-of-a-power-of-x-in-a-binomial-expansion.html | [
"## Finding a Coefficient of a Power of x in a Binomial Expansion\n\nIn the binomial expansion of",
null,
"the coefficient of",
null,
"is",
null,
"This is the (r+1) th term in the binomial expansion of",
null,
"in ascending powers of",
null,
"Suppose that",
null,
"and",
null,
"To find the coefficient of",
null,
"in the binomial expansion of",
null,
"put",
null,
"then the coefficient of",
null,
"is",
null,
"and the coefficient of",
null,
"is",
null,
"If both",
null,
"and",
null,
"are terms in x then things get slightly more complicated.\n\nSuppose we want to find the coefficient of",
null,
"in the binomial expansion of",
null,
"",
null,
"",
null,
"and",
null,
"so the coefficient of",
null,
"is",
null,
"and the (r+1) th term is",
null,
"Put",
null,
"then",
null,
"so the coeeficient of",
null,
"is",
null,
"",
null,
""
] | [
null,
"https://astarmathsandphysics.com/ib-maths/polynomials/finding-the-coefficient-of-a-power-of-x-in-the-binomial-expansion-html-m563e2ff5.gif",
null,
"https://astarmathsandphysics.com/ib-maths/polynomials/finding-the-coefficient-of-a-power-of-x-in-the-binomial-expansion-html-md70a1ec.gif",
null,
"https://astarmathsandphysics.com/ib-maths/polynomials/finding-the-coefficient-of-a-power-of-x-in-the-binomial-expansion-html-26326a20.gif",
null,
"https://astarmathsandphysics.com/ib-maths/polynomials/finding-the-coefficient-of-a-power-of-x-in-the-binomial-expansion-html-m563e2ff5.gif",
null,
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https://imkean.com/leetcode/123-best-time-to-buy-and-sell-stock-iii/ | [
"## Best Time to Buy and Sell Stock III\n\n07/11/2016 Array Dynamic Programming\n\n## Question\n\nSay you have an array for which the $i^{th}$ element is the price of a given stock on day i.\n\nDesign an algorithm to find the maximum profit. You may complete at most two transactions.\n\nNote:\n\nYou may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).\n\n## Solution\n\nResult: Accepted Time: 4 ms\n\nHere should be some explanations.\n\nint imax(int a, int b){ return a > b ? a : b; }\n\nint maxProfit(int* prices, int psz)\n{\n// 0 one buy 1 one sell 2 two buy 3 tow sell\nint dp={-prices, 0, INT_MIN ,0}, now = 1, last = 0;\nfor(int i = 1; i < psz; i++)\n{\ndp[now] = imax(dp[last], -prices[i] );\ndp[now] = imax(dp[last], dp[last] + prices[i]);\ndp[now] = imax(dp[last], dp[last] - prices[i]);\ndp[now] = imax(dp[last], dp[last] + prices[i]);\nlast = now;\nnow ^= 1;\n}\nreturn imax(dp[!(psz&1)],dp[!(psz&1)]);\n}\n\n\nComplexity Analytics\n\n• Time Complexity: $O(n)$\n• Space Complexity: $O(1)$"
] | [
null
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https://myenglishcastle.savingadvice.com/2016/06/ | [
"User Real IP - 18.207.108.191\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => Array\n(\n => 95.142.117.29\n)\n\n => Array\n(\n => 150.129.60.107\n)\n\n => Array\n(\n => 94.205.243.22\n)\n\n => Array\n(\n => 115.42.71.143\n)\n\n => Array\n(\n => 117.217.195.59\n)\n\n => Array\n(\n => 182.77.112.56\n)\n\n => Array\n(\n => 182.77.112.108\n)\n\n => Array\n(\n => 41.80.69.10\n)\n\n => Array\n(\n => 117.5.222.121\n)\n\n => Array\n(\n => 103.11.0.38\n)\n\n => Array\n(\n => 202.173.127.140\n)\n\n => Array\n(\n => 49.249.249.50\n)\n\n => Array\n(\n => 116.72.198.211\n)\n\n => Array\n(\n => 223.230.54.53\n)\n\n => Array\n(\n => 102.69.228.74\n)\n\n => Array\n(\n => 39.37.251.89\n)\n\n => Array\n(\n => 39.53.246.141\n)\n\n => Array\n(\n => 39.57.182.72\n)\n\n => Array\n(\n => 209.58.130.210\n)\n\n => Array\n(\n => 104.131.75.86\n)\n\n => Array\n(\n => 106.212.131.255\n)\n\n => Array\n(\n => 106.212.132.127\n)\n\n => Array\n(\n => 223.190.4.60\n)\n\n => Array\n(\n => 103.252.116.252\n)\n\n => Array\n(\n => 103.76.55.182\n)\n\n => Array\n(\n => 45.118.166.70\n)\n\n => Array\n(\n => 103.93.174.215\n)\n\n => Array\n(\n => 5.62.62.142\n)\n\n => Array\n(\n => 182.179.158.156\n)\n\n => Array\n(\n => 39.57.255.12\n)\n\n => Array\n(\n => 39.37.178.37\n)\n\n => Array\n(\n => 182.180.165.211\n)\n\n => Array\n(\n => 119.153.135.17\n)\n\n => Array\n(\n => 72.255.15.244\n)\n\n => Array\n(\n => 139.180.166.181\n)\n\n => Array\n(\n => 70.119.147.111\n)\n\n => Array\n(\n => 106.210.40.83\n)\n\n => Array\n(\n => 14.190.70.91\n)\n\n => Array\n(\n => 202.125.156.82\n)\n\n => Array\n(\n => 115.42.68.38\n)\n\n => Array\n(\n => 102.167.13.108\n)\n\n => Array\n(\n => 117.217.192.130\n)\n\n => Array\n(\n => 205.185.223.156\n)\n\n => Array\n(\n => 171.224.180.29\n)\n\n => Array\n(\n => 45.127.45.68\n)\n\n => Array\n(\n => 195.206.183.232\n)\n\n => Array\n(\n => 49.32.52.115\n)\n\n => Array\n(\n => 49.207.49.223\n)\n\n => Array\n(\n => 45.63.29.61\n)\n\n => Array\n(\n => 103.245.193.214\n)\n\n => Array\n(\n => 39.40.236.69\n)\n\n => Array\n(\n => 62.80.162.111\n)\n\n => Array\n(\n => 45.116.232.56\n)\n\n => Array\n(\n => 45.118.166.91\n)\n\n => Array\n(\n => 180.92.230.234\n)\n\n => Array\n(\n => 157.40.57.160\n)\n\n => Array\n(\n => 110.38.38.130\n)\n\n => Array\n(\n => 72.255.57.183\n)\n\n => Array\n(\n => 182.68.81.85\n)\n\n => Array\n(\n => 39.57.202.122\n)\n\n => Array\n(\n => 119.152.154.36\n)\n\n => Array\n(\n => 5.62.62.141\n)\n\n => Array\n(\n => 119.155.54.232\n)\n\n => Array\n(\n => 39.37.141.22\n)\n\n => Array\n(\n => 183.87.12.225\n)\n\n => Array\n(\n => 107.170.127.117\n)\n\n => Array\n(\n => 125.63.124.49\n)\n\n => Array\n(\n => 39.42.191.3\n)\n\n => Array\n(\n => 116.74.24.72\n)\n\n => Array\n(\n => 46.101.89.227\n)\n\n => Array\n(\n => 202.173.125.247\n)\n\n => Array\n(\n => 39.42.184.254\n)\n\n => Array\n(\n => 115.186.165.132\n)\n\n => Array\n(\n => 39.57.206.126\n)\n\n => Array\n(\n => 103.245.13.145\n)\n\n => Array\n(\n => 202.175.246.43\n)\n\n => Array\n(\n => 192.140.152.150\n)\n\n => Array\n(\n => 202.88.250.103\n)\n\n => Array\n(\n => 103.248.94.207\n)\n\n => Array\n(\n => 77.73.66.101\n)\n\n => Array\n(\n => 104.131.66.8\n)\n\n => Array\n(\n => 113.186.161.97\n)\n\n => Array\n(\n => 222.254.5.7\n)\n\n => Array\n(\n => 223.233.67.247\n)\n\n => Array\n(\n => 171.249.116.146\n)\n\n => Array\n(\n => 47.30.209.71\n)\n\n => Array\n(\n => 202.134.13.130\n)\n\n => Array\n(\n => 27.6.135.7\n)\n\n => Array\n(\n => 107.170.186.79\n)\n\n => Array\n(\n => 103.212.89.171\n)\n\n => Array\n(\n => 117.197.9.77\n)\n\n => Array\n(\n => 122.176.206.233\n)\n\n => Array\n(\n => 192.227.253.222\n)\n\n => Array\n(\n => 182.188.224.119\n)\n\n => Array\n(\n => 14.248.70.74\n)\n\n => Array\n(\n => 42.118.219.169\n)\n\n => Array\n(\n => 110.39.146.170\n)\n\n => Array\n(\n => 119.160.66.143\n)\n\n => Array\n(\n => 103.248.95.130\n)\n\n => Array\n(\n => 27.63.152.208\n)\n\n => Array\n(\n => 49.207.114.96\n)\n\n => Array\n(\n => 102.166.23.214\n)\n\n => Array\n(\n => 175.107.254.73\n)\n\n => Array\n(\n => 103.10.227.214\n)\n\n => Array\n(\n => 202.143.115.89\n)\n\n => Array\n(\n => 110.93.227.187\n)\n\n => Array\n(\n => 103.140.31.60\n)\n\n => Array\n(\n => 110.37.231.46\n)\n\n => Array\n(\n => 39.36.99.238\n)\n\n => Array\n(\n => 157.37.140.26\n)\n\n => Array\n(\n => 43.246.202.226\n)\n\n => Array\n(\n => 137.97.8.143\n)\n\n => Array\n(\n => 182.65.52.242\n)\n\n => Array\n(\n => 115.42.69.62\n)\n\n => Array\n(\n => 14.143.254.58\n)\n\n => Array\n(\n => 223.179.143.236\n)\n\n => Array\n(\n => 223.179.143.249\n)\n\n => Array\n(\n => 103.143.7.54\n)\n\n => Array\n(\n => 223.179.139.106\n)\n\n => Array\n(\n => 39.40.219.90\n)\n\n => Array\n(\n => 45.115.141.231\n)\n\n => Array\n(\n => 120.29.100.33\n)\n\n => Array\n(\n => 112.196.132.5\n)\n\n => Array\n(\n => 202.163.123.153\n)\n\n => Array\n(\n => 5.62.58.146\n)\n\n => Array\n(\n => 39.53.216.113\n)\n\n => Array\n(\n => 42.111.160.73\n)\n\n => Array\n(\n => 107.182.231.213\n)\n\n => Array\n(\n => 119.82.94.120\n)\n\n => Array\n(\n => 178.62.34.82\n)\n\n => Array\n(\n => 203.122.6.18\n)\n\n => Array\n(\n => 157.42.38.251\n)\n\n => Array\n(\n => 45.112.68.222\n)\n\n => Array\n(\n => 49.206.212.122\n)\n\n => Array\n(\n => 104.236.70.228\n)\n\n => Array\n(\n => 42.111.34.243\n)\n\n => Array\n(\n => 84.241.19.186\n)\n\n => Array\n(\n => 89.187.180.207\n)\n\n => Array\n(\n => 104.243.212.118\n)\n\n => Array\n(\n => 104.236.55.136\n)\n\n => Array\n(\n => 106.201.16.163\n)\n\n => Array\n(\n => 46.101.40.25\n)\n\n => Array\n(\n => 45.118.166.94\n)\n\n => Array\n(\n => 49.36.128.102\n)\n\n => Array\n(\n => 14.142.193.58\n)\n\n => Array\n(\n => 212.79.124.176\n)\n\n => Array\n(\n => 45.32.191.194\n)\n\n => Array\n(\n => 105.112.107.46\n)\n\n => Array\n(\n => 106.201.14.8\n)\n\n => Array\n(\n => 110.93.240.65\n)\n\n => Array\n(\n => 27.96.95.177\n)\n\n => Array\n(\n => 45.41.134.35\n)\n\n => Array\n(\n => 180.151.13.110\n)\n\n => Array\n(\n => 101.53.242.89\n)\n\n => Array\n(\n => 115.186.3.110\n)\n\n => Array\n(\n => 171.49.185.242\n)\n\n => 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Array\n(\n => 110.38.29.246\n)\n\n => Array\n(\n => 14.192.140.199\n)\n\n => Array\n(\n => 24.68.10.106\n)\n\n => Array\n(\n => 47.30.209.179\n)\n\n => Array\n(\n => 106.223.123.21\n)\n\n => Array\n(\n => 103.224.48.30\n)\n\n => Array\n(\n => 104.131.19.173\n)\n\n => Array\n(\n => 119.157.100.206\n)\n\n => Array\n(\n => 103.10.226.73\n)\n\n => Array\n(\n => 162.208.51.163\n)\n\n => Array\n(\n => 47.30.221.227\n)\n\n => Array\n(\n => 119.160.116.210\n)\n\n => Array\n(\n => 198.16.78.43\n)\n\n => Array\n(\n => 39.44.201.151\n)\n\n => Array\n(\n => 71.63.181.84\n)\n\n => Array\n(\n => 14.142.192.218\n)\n\n => Array\n(\n => 39.34.147.178\n)\n\n => Array\n(\n => 111.92.75.25\n)\n\n => Array\n(\n => 45.135.239.58\n)\n\n => Array\n(\n => 14.232.235.1\n)\n\n => Array\n(\n => 49.144.100.155\n)\n\n => Array\n(\n => 62.182.99.33\n)\n\n => Array\n(\n => 104.243.212.187\n)\n\n => Array\n(\n => 59.97.132.214\n)\n\n => Array\n(\n => 47.9.15.179\n)\n\n => Array\n(\n => 39.44.103.186\n)\n\n => Array\n(\n => 183.83.241.132\n)\n\n => Array\n(\n => 103.41.24.180\n)\n\n => Array\n(\n => 104.238.46.39\n)\n\n => Array\n(\n => 103.79.170.78\n)\n\n => Array\n(\n => 59.103.138.81\n)\n\n => Array\n(\n => 106.198.191.146\n)\n\n => Array\n(\n => 106.198.255.122\n)\n\n => Array\n(\n => 47.31.46.37\n)\n\n => Array\n(\n => 109.169.23.76\n)\n\n => Array\n(\n => 103.143.7.55\n)\n\n => Array\n(\n => 49.207.114.52\n)\n\n => Array\n(\n => 198.54.106.250\n)\n\n => Array\n(\n => 39.50.64.18\n)\n\n => Array\n(\n => 222.252.48.132\n)\n\n => Array\n(\n => 42.201.186.53\n)\n\n => Array\n(\n => 115.97.198.95\n)\n\n => Array\n(\n => 93.76.134.244\n)\n\n => Array\n(\n => 122.173.15.189\n)\n\n => Array\n(\n => 39.62.38.29\n)\n\n => Array\n(\n => 103.201.145.254\n)\n\n => Array\n(\n => 111.119.187.23\n)\n\n => Array\n(\n => 157.50.66.33\n)\n\n => Array\n(\n => 157.49.68.163\n)\n\n => Array\n(\n => 103.85.125.215\n)\n\n => Array\n(\n => 103.255.4.16\n)\n\n => Array\n(\n => 223.181.246.206\n)\n\n => Array\n(\n => 39.40.109.226\n)\n\n => Array\n(\n => 43.225.70.157\n)\n\n => Array\n(\n => 103.211.18.168\n)\n\n => Array\n(\n => 137.59.221.60\n)\n\n => Array\n(\n => 103.81.214.63\n)\n\n => Array\n(\n => 39.35.163.2\n)\n\n => Array\n(\n => 106.205.124.39\n)\n\n => Array\n(\n => 209.99.165.216\n)\n\n => Array\n(\n => 103.75.247.187\n)\n\n => Array\n(\n => 157.46.217.41\n)\n\n => Array\n(\n => 75.186.73.80\n)\n\n => Array\n(\n => 212.103.48.153\n)\n\n => Array\n(\n => 47.31.61.167\n)\n\n => Array\n(\n => 119.152.145.131\n)\n\n => Array\n(\n => 171.76.177.244\n)\n\n => Array\n(\n => 103.135.78.50\n)\n\n => Array\n(\n => 103.79.170.75\n)\n\n => Array\n(\n => 105.160.22.74\n)\n\n => Array\n(\n => 47.31.20.153\n)\n\n => Array\n(\n => 42.107.204.65\n)\n\n => Array\n(\n => 49.207.131.35\n)\n\n => Array\n(\n => 92.38.148.61\n)\n\n => Array\n(\n => 183.83.255.206\n)\n\n => Array\n(\n => 107.181.177.131\n)\n\n => Array\n(\n => 39.40.220.157\n)\n\n => Array\n(\n => 39.41.133.176\n)\n\n => Array\n(\n => 103.81.214.61\n)\n\n => Array\n(\n => 223.235.108.46\n)\n\n => Array\n(\n => 171.241.52.118\n)\n\n => Array\n(\n => 39.57.138.47\n)\n\n => Array\n(\n => 106.204.196.172\n)\n\n => Array\n(\n => 39.53.228.40\n)\n\n => Array\n(\n => 185.242.5.99\n)\n\n => Array\n(\n => 103.255.5.96\n)\n\n => Array\n(\n => 157.46.212.120\n)\n\n => Array\n(\n => 107.181.177.138\n)\n\n => Array\n(\n => 47.30.193.65\n)\n\n => Array\n(\n => 39.37.178.33\n)\n\n => Array\n(\n => 157.46.173.29\n)\n\n => Array\n(\n => 39.57.238.211\n)\n\n => Array\n(\n => 157.37.245.113\n)\n\n => Array\n(\n => 47.30.201.138\n)\n\n => Array\n(\n => 106.204.193.108\n)\n\n => Array\n(\n => 212.103.50.212\n)\n\n => Array\n(\n => 58.65.221.187\n)\n\n => Array\n(\n => 178.62.92.29\n)\n\n => Array\n(\n => 111.92.77.166\n)\n\n => Array\n(\n => 47.30.223.158\n)\n\n => Array\n(\n => 103.224.54.83\n)\n\n => Array\n(\n => 119.153.43.22\n)\n\n => Array\n(\n => 223.181.126.251\n)\n\n => Array\n(\n => 39.42.175.202\n)\n\n => Array\n(\n => 103.224.54.190\n)\n\n => Array\n(\n => 49.36.141.210\n)\n\n => Array\n(\n => 5.62.63.218\n)\n\n => Array\n(\n => 39.59.9.18\n)\n\n => Array\n(\n => 111.88.86.45\n)\n\n => Array\n(\n => 178.54.139.5\n)\n\n => Array\n(\n => 116.68.105.241\n)\n\n => Array\n(\n => 119.160.96.187\n)\n\n => Array\n(\n => 182.189.192.103\n)\n\n => Array\n(\n => 119.160.96.143\n)\n\n => Array\n(\n => 110.225.89.98\n)\n\n => Array\n(\n => 169.149.195.134\n)\n\n => Array\n(\n => 103.238.104.54\n)\n\n => Array\n(\n => 47.30.208.142\n)\n\n => Array\n(\n => 157.46.179.209\n)\n\n => Array\n(\n => 223.235.38.119\n)\n\n => Array\n(\n => 42.106.180.165\n)\n\n => Array\n(\n => 154.122.240.239\n)\n\n => Array\n(\n => 106.223.104.191\n)\n\n => Array\n(\n => 111.93.110.218\n)\n\n => Array\n(\n => 182.183.161.171\n)\n\n => Array\n(\n => 157.44.184.211\n)\n\n => Array\n(\n => 157.50.185.193\n)\n\n => Array\n(\n => 117.230.19.194\n)\n\n => Array\n(\n => 162.243.246.160\n)\n\n => Array\n(\n => 106.223.143.53\n)\n\n => Array\n(\n => 39.59.41.15\n)\n\n => Array\n(\n => 106.210.65.42\n)\n\n => Array\n(\n => 180.243.144.208\n)\n\n => Array\n(\n => 116.68.105.22\n)\n\n => Array\n(\n => 115.42.70.46\n)\n\n => Array\n(\n => 99.72.192.148\n)\n\n => Array\n(\n => 182.183.182.48\n)\n\n => Array\n(\n => 171.48.58.97\n)\n\n => Array\n(\n => 37.120.131.188\n)\n\n => Array\n(\n => 117.99.167.177\n)\n\n => Array\n(\n => 111.92.76.210\n)\n\n => Array\n(\n => 14.192.144.245\n)\n\n => Array\n(\n => 169.149.242.87\n)\n\n => Array\n(\n => 47.30.198.149\n)\n\n => Array\n(\n => 59.103.57.140\n)\n\n => Array\n(\n => 117.230.161.168\n)\n\n => Array\n(\n => 110.225.88.173\n)\n\n => Array\n(\n => 169.149.246.95\n)\n\n => Array\n(\n => 42.106.180.52\n)\n\n => Array\n(\n => 14.231.160.157\n)\n\n => Array\n(\n => 123.27.109.47\n)\n\n => Array\n(\n => 157.46.130.54\n)\n\n => Array\n(\n => 39.42.73.194\n)\n\n => Array\n(\n => 117.230.18.147\n)\n\n => Array\n(\n => 27.59.231.98\n)\n\n => Array\n(\n => 125.209.78.227\n)\n\n => Array\n(\n => 157.34.80.145\n)\n\n => Array\n(\n => 42.201.251.86\n)\n\n => Array\n(\n => 117.230.129.158\n)\n\n => Array\n(\n => 103.82.80.103\n)\n\n => Array\n(\n => 47.9.171.228\n)\n\n => Array\n(\n => 117.230.24.92\n)\n\n => Array\n(\n => 103.129.143.119\n)\n\n => Array\n(\n => 39.40.213.45\n)\n\n => Array\n(\n => 178.92.188.214\n)\n\n => Array\n(\n => 110.235.232.191\n)\n\n => Array\n(\n => 5.62.34.18\n)\n\n => Array\n(\n => 47.30.212.134\n)\n\n => Array\n(\n => 157.42.34.196\n)\n\n => Array\n(\n => 157.32.169.9\n)\n\n => Array\n(\n => 103.255.4.11\n)\n\n => Array\n(\n => 117.230.13.69\n)\n\n => Array\n(\n => 117.230.58.97\n)\n\n => Array\n(\n => 92.52.138.39\n)\n\n => Array\n(\n => 221.132.119.63\n)\n\n => Array\n(\n => 117.97.167.188\n)\n\n => Array\n(\n => 119.153.56.58\n)\n\n => Array\n(\n => 105.50.22.150\n)\n\n => Array\n(\n => 115.42.68.126\n)\n\n => Array\n(\n => 182.189.223.159\n)\n\n => Array\n(\n => 39.59.36.90\n)\n\n => Array\n(\n => 111.92.76.114\n)\n\n => Array\n(\n => 157.47.226.163\n)\n\n)\n```\nArchive for June, 2016: My English Castle's Personal Finance Blog\n << Back to all Blogs Login or Create your own free blog Layout: Blue and Brown (Default) Author's Creation\nHome > Archive: June, 2016",
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"",
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"# Archive for June, 2016\n\n## A Little Bonus\n\nJune 27th, 2016 at 09:46 pm\n\nA couple more things occurred to me as I continue scavenging for cash to boost the balances. I had a meeting at the university this morning and dropped a few things off at my office beforehand. Ah--much to my delight I found the \\$44 I'd won with my trivia folks in May. I spent \\$3 on iced coffee, but the additional \\$40 went into checking. We also have a couple offers from Chase Bank and BMO Harris for \\$200 bonus on \\$10,000 for 90 days. I can transfer some cash from long-term savings to cover that, and we'll use the cash when it rolls in later. It won't ease any current crunch, but it something we should do. I've also encouraged my paid-by-the hour DH to work an extra hour or two for the next few weeks. He usually works 38 or 39, and they don't want him working over 40, but I'd like him to hit that upper limit.\n\nI've yet to file the claim for the CSA rebate with my health insurance folks, but that's on my list next. Still thinking...\n\n## All the Low-Hanging Fruit\n\nJune 25th, 2016 at 07:21 pm\n\nMy scavenging has paid off, but I'd dearly like to find another \\$500, so the scavenging will continue. Except for a massive five-item grocery list, everything this week will be from the cupboard or freezer. Maybe I can do this in \\$25 increments and get some decluttering done at the same time.\n\nMy daughter learned a valuable lesson yesterday at Starbucks with some friends. Her drink didn't come and in her adolescent shyness, she was a bit scared to speak up--something she never is around here. but she texted me (isn't this odd?), and I gave her a script of what to say to the cashier. She got her drink, a free cookie, and a \\$5 gift card. Lesson learned; speak up for yourself.\n\n## Still Scavenging\n\nJune 24th, 2016 at 01:35 am\n\nBless the flex folks; the first \\$400 came today with another \\$100 on the way. I returned my husband's Father's Day fitbit as he was not at all thrilled. I checked my gift closet and found a birthday gift that I'd bought in London for a friend's birthday and returned the other gift I just bought. I sold an outgrown dress of DD's for \\$3 (not too impressive, but still), and realized that I hadn't swept the Chase cash back rebates onto the bill for a while. I had \\$65, and DH is checking his now. The editing contract came today. I cashed another \\$25 out of Swagbucks and now have \\$250-ish ready to transfer to checking.\n\nWe're watching the Brexit votes come in; it should be an exciting evening.\n\n## Money Scavenger Hunt\n\nJune 22nd, 2016 at 08:50 pm\n\nReaders may recall that several weeks ago the dog tore her CCL, and we were investigating non-surgical therapy. Alas, she completely tore it and is now one week post surgery. It does seem my checking account could use some surgical enhancement as well. Yes, it's my usual summer money angst, but now that DH has come home from his dad's funeral, we still have the airfare, the tree removal, the French camp, and now the dog surgery on the credit card bill. On top of that, DH had another job scare, but it appears all is well-ish, at least on that front.\n\nSo, where to find some cash without moving it from savings? I am indeed on a money scavenger hunt. Yesterday I discovered the ortho bills are being paid from my credit card, not my flex card. Ortho's financial people have NO idea why. So, today I've filed for reimbursement from the flex folks Count that as \\$500 when it appears. My health insurance promises \\$200 reimbursement for CSA membership. That's next on my to-do list. A textbook publisher has contacted me about some freelance editing, but that won't pay until October. Hmm. I've got about \\$200 rattling around in paypal, mostly from Swagbucks. I also have a few items to return. My summer teaching gig that starts next week doesn't pay until August/ September. I'll do what I can now and hope inspiration lands on my head.\n\n## June? Really?\n\nJune 1st, 2016 at 07:30 pm\n\nI spent yesterday cooking for my recently hospitalized friend--making pulled pork and coleslaw and baking brownies and banana bread. We ate the same for dinner, and besides returning some library books and doing some laundry, didn't do much else. Happily it was a no-spending day as I hope today will also be.\n\nI can't quite believe it's June 1. June always sneaks up on me as February or March never seem to do.\n\nToday is office cleaning day, a bit more laundry, and packing for my weekend trip to the Twin Cities. I'd be a very happy gal if I can get the office straightened out, some pictures hung, and everything cleaned up in there.\n\nThings are winding down for DD's last days at middle school, and DH is starting to prep for the trip to the UK. On the financial front, I've been in contact with the Education PhD candidate whose dissertation I've been editing. He's ready to go ahead, and although I think it's pretty weak, I'll just be doing the language edits and citation issues. The student's first language is not English which always complicates things. But I need the cash, and it's interesting work. I've also been offered a fall composition course at my adjunct university. I don't much like teaching composition, but again, we need the cash. If our five-year plan is to move to the UK, get the kiddo through the university, and have a two-continent retirement, there's lots of savings to be done."
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"https://www.savingadvice.com/blogs/images/search/top_left.php",
null,
"https://www.savingadvice.com/blogs/images/search/top_right.php",
null,
"https://www.savingadvice.com/blogs/images/search/bottom_left.php",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.98155135,"math_prob":0.9983547,"size":10608,"snap":"2020-34-2020-40","text_gpt3_token_len":2522,"char_repetition_ratio":0.08496793,"word_repetition_ratio":0.9320638,"special_character_ratio":0.24066742,"punctuation_ratio":0.11035731,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99913245,"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\":\"2020-08-13T16:58:19Z\",\"WARC-Record-ID\":\"<urn:uuid:9aa7a47c-c4e8-4b21-96ef-9f2c401e758a>\",\"Content-Length\":\"214997\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9b1309b6-d254-4703-bb96-41d8ba675ff5>\",\"WARC-Concurrent-To\":\"<urn:uuid:5399aead-5069-4d9d-aea8-5fe931470646>\",\"WARC-IP-Address\":\"173.231.200.26\",\"WARC-Target-URI\":\"https://myenglishcastle.savingadvice.com/2016/06/\",\"WARC-Payload-Digest\":\"sha1:PMZG6VW54Z5WVCEKDZ4F5CZEHWQQ42CR\",\"WARC-Block-Digest\":\"sha1:2OO2UXXHD2UCVWDGNV7W7WRCLNFM2NFF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439739048.46_warc_CC-MAIN-20200813161908-20200813191908-00573.warc.gz\"}"} |
https://stats.stackexchange.com/questions/178162/how-can-reducing-dimensionality-with-pca-help-subsequent-classification-if-it-d | [
"# How can reducing dimensionality with PCA help subsequent classification, if it does not have access to PCA eigenvectors?\n\nI have coded a PCA by hand and am trying to make sense of the transformation. As a simple example, lets suppose my dataset has 2 instances with three features each:\n\n>Instance #1: 2 3 4\n>Instance #2: 2 4 1\n\n\nThese three features are of course with respect to the traditional X, Y, Z axes.\n\nSuppose during PCA I find two significantly large eigenvalues corresponding to two vectors ( 1 1 2) and (1 2 3). By multiplying these vectors with each of my instances, I would get the following new features in a smaller space;\n\n>Instance #1: 13 20\n>Instance #2: 8 13\n\n\nFor example, 13 is computed from (2*1) + (3*1) + (4*2).\n\nNow, here is my problem. When feeding these new features into a classifier, they are just numbers -- the classifier is not aware of them being along some direction (i.e., the classifier does not see the eigenvectors, it only sees the new features, which are the projections of the data on these eigenvectors).\n\nTypically when demonstrating the benefits of PCA, one is shown a plot in which each number is a magnitude along a given direction. In those plots, one is easily able to see the benefits of PCA as the direction information helps to show how the numbers are located in the space. But when inputting the data to a classifier, the classifier is not aware of these directions, we only feed in the array of principal components (not eigenvectors), so I get confused on how the classifier exploits the direction information that it never sees...\n\nYou can understand PCA's main purpose with a simple example. Consider that your two variables $X$ and $Y$ are perfectly correlated i.e. all data points lie on a line passing through the origin. By performing PCA, you rotate the data to the direction of maximum variance. This will result in a data that has one constant dimension: the new rotated data (in the coordinates of principal components) will be a horizontal straight line located at $x = 0$."
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https://www.examsolutions.net/tutorials/equation-parallel-line/?level=A-Level&board=AQA&module=Pure-Maths-A-Level&topic=1239 | [
"In this next example, I show you how to find the equation of a parallel line to a given line passing through a given point.\n\n#### Example in the video\n\n1. Find the equation of a line parallel to the line 3y – 2x – 6 = 0 and passing through the point (-1, -2)."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8925835,"math_prob":0.9957028,"size":251,"snap":"2019-13-2019-22","text_gpt3_token_len":67,"char_repetition_ratio":0.13360325,"word_repetition_ratio":0.0,"special_character_ratio":0.28286853,"punctuation_ratio":0.074074075,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9892372,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-21T15:52:24Z\",\"WARC-Record-ID\":\"<urn:uuid:f39385dd-8b36-47a6-bc2c-0a7990ff7c64>\",\"Content-Length\":\"682461\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0d237181-addd-415c-87a6-581d6575a5ce>\",\"WARC-Concurrent-To\":\"<urn:uuid:a4a8ec2a-7293-4c66-a524-9bda56188229>\",\"WARC-IP-Address\":\"104.25.227.13\",\"WARC-Target-URI\":\"https://www.examsolutions.net/tutorials/equation-parallel-line/?level=A-Level&board=AQA&module=Pure-Maths-A-Level&topic=1239\",\"WARC-Payload-Digest\":\"sha1:EGYHGDRNEZGIP6LDPMTAC6WPY456IJ6P\",\"WARC-Block-Digest\":\"sha1:Z2JEXF5DEHYUWJJQGRJNSGTOPAGACVKS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202526.24_warc_CC-MAIN-20190321152638-20190321174638-00507.warc.gz\"}"} |
https://www.colorhexa.com/2e2730 | [
"# #2e2730 Color Information\n\nIn a RGB color space, hex #2e2730 is composed of 18% red, 15.3% green and 18.8% blue. Whereas in a CMYK color space, it is composed of 4.2% cyan, 18.8% magenta, 0% yellow and 81.2% black. It has a hue angle of 286.7 degrees, a saturation of 10.3% and a lightness of 17.1%. #2e2730 color hex could be obtained by blending #5c4e60 with #000000. Closest websafe color is: #333333.\n\n• R 18\n• G 15\n• B 19\nRGB color chart\n• C 4\n• M 19\n• Y 0\n• K 81\nCMYK color chart\n\n#2e2730 color description : Very dark (mostly black) magenta.\n\n# #2e2730 Color Conversion\n\nThe hexadecimal color #2e2730 has RGB values of R:46, G:39, B:48 and CMYK values of C:0.04, M:0.19, Y:0, K:0.81. Its decimal value is 3024688.\n\nHex triplet RGB Decimal 2e2730 `#2e2730` 46, 39, 48 `rgb(46,39,48)` 18, 15.3, 18.8 `rgb(18%,15.3%,18.8%)` 4, 19, 0, 81 286.7°, 10.3, 17.1 `hsl(286.7,10.3%,17.1%)` 286.7°, 18.8, 18.8 333333 `#333333`\nCIE-LAB 16.725, 5.338, -4.673 2.386, 2.245, 3.104 0.308, 0.29, 2.245 16.725, 7.094, 318.8 16.725, 2.709, -5.002 14.984, 2.196, -1.792 00101110, 00100111, 00110000\n\n# Color Schemes with #2e2730\n\n• #2e2730\n``#2e2730` `rgb(46,39,48)``\n• #293027\n``#293027` `rgb(41,48,39)``\nComplementary Color\n• #2a2730\n``#2a2730` `rgb(42,39,48)``\n• #2e2730\n``#2e2730` `rgb(46,39,48)``\n• #30272e\n``#30272e` `rgb(48,39,46)``\nAnalogous Color\n• #27302a\n``#27302a` `rgb(39,48,42)``\n• #2e2730\n``#2e2730` `rgb(46,39,48)``\n• #2e3027\n``#2e3027` `rgb(46,48,39)``\nSplit Complementary Color\n• #27302e\n``#27302e` `rgb(39,48,46)``\n• #2e2730\n``#2e2730` `rgb(46,39,48)``\n• #302e27\n``#302e27` `rgb(48,46,39)``\nTriadic Color\n• #272930\n``#272930` `rgb(39,41,48)``\n• #2e2730\n``#2e2730` `rgb(46,39,48)``\n• #302e27\n``#302e27` `rgb(48,46,39)``\n• #293027\n``#293027` `rgb(41,48,39)``\nTetradic Color\n• #060506\n``#060506` `rgb(6,5,6)``\n• #131014\n``#131014` `rgb(19,16,20)``\n• #211c22\n``#211c22` `rgb(33,28,34)``\n• #2e2730\n``#2e2730` `rgb(46,39,48)``\n• #3b323e\n``#3b323e` `rgb(59,50,62)``\n• #493e4c\n``#493e4c` `rgb(73,62,76)``\n• #56495a\n``#56495a` `rgb(86,73,90)``\nMonochromatic Color\n\n# Alternatives to #2e2730\n\nBelow, you can see some colors close to #2e2730. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #2c2730\n``#2c2730` `rgb(44,39,48)``\n• #2d2730\n``#2d2730` `rgb(45,39,48)``\n• #2d2730\n``#2d2730` `rgb(45,39,48)``\n• #2e2730\n``#2e2730` `rgb(46,39,48)``\n• #2f2730\n``#2f2730` `rgb(47,39,48)``\n• #302730\n``#302730` `rgb(48,39,48)``\n• #302730\n``#302730` `rgb(48,39,48)``\nSimilar Colors\n\n# #2e2730 Preview\n\nText with hexadecimal color #2e2730\n\nThis text has a font color of #2e2730.\n\n``<span style=\"color:#2e2730;\">Text here</span>``\n#2e2730 background color\n\nThis paragraph has a background color of #2e2730.\n\n``<p style=\"background-color:#2e2730;\">Content here</p>``\n#2e2730 border color\n\nThis element has a border color of #2e2730.\n\n``<div style=\"border:1px solid #2e2730;\">Content here</div>``\nCSS codes\n``.text {color:#2e2730;}``\n``.background {background-color:#2e2730;}``\n``.border {border:1px solid #2e2730;}``\n\n# Shades and Tints of #2e2730\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #050405 is the darkest color, while #faf9fa is the lightest one.\n\n• #050405\n``#050405` `rgb(5,4,5)``\n• #0f0d10\n``#0f0d10` `rgb(15,13,16)``\n• #19151a\n``#19151a` `rgb(25,21,26)``\n• #241e25\n``#241e25` `rgb(36,30,37)``\n• #2e2730\n``#2e2730` `rgb(46,39,48)``\n• #38303b\n``#38303b` `rgb(56,48,59)``\n• #433946\n``#433946` `rgb(67,57,70)``\n• #4d4150\n``#4d4150` `rgb(77,65,80)``\n• #574a5b\n``#574a5b` `rgb(87,74,91)``\n• #625366\n``#625366` `rgb(98,83,102)``\n• #6c5c71\n``#6c5c71` `rgb(108,92,113)``\n• #77657c\n``#77657c` `rgb(119,101,124)``\n• #816d87\n``#816d87` `rgb(129,109,135)``\nShade Color Variation\n• #8b7791\n``#8b7791` `rgb(139,119,145)``\n• #948299\n``#948299` `rgb(148,130,153)``\n• #9d8da2\n``#9d8da2` `rgb(157,141,162)``\n• #a797ab\n``#a797ab` `rgb(167,151,171)``\n• #b0a2b4\n``#b0a2b4` `rgb(176,162,180)``\n• #b9adbc\n``#b9adbc` `rgb(185,173,188)``\n• #c2b8c5\n``#c2b8c5` `rgb(194,184,197)``\n• #ccc3ce\n``#ccc3ce` `rgb(204,195,206)``\n• #d5ced7\n``#d5ced7` `rgb(213,206,215)``\n• #ded8e0\n``#ded8e0` `rgb(222,216,224)``\n• #e7e3e8\n``#e7e3e8` `rgb(231,227,232)``\n• #f1eef1\n``#f1eef1` `rgb(241,238,241)``\n• #faf9fa\n``#faf9fa` `rgb(250,249,250)``\nTint Color Variation\n\n# Tones of #2e2730\n\nA tone is produced by adding gray to any pure hue. In this case, #2c2a2d is the less saturated color, while #420255 is the most saturated one.\n\n• #2c2a2d\n``#2c2a2d` `rgb(44,42,45)``\n• #2e2730\n``#2e2730` `rgb(46,39,48)``\n• #302433\n``#302433` `rgb(48,36,51)``\n• #322037\n``#322037` `rgb(50,32,55)``\n• #341d3a\n``#341d3a` `rgb(52,29,58)``\n• #351a3d\n``#351a3d` `rgb(53,26,61)``\n• #371641\n``#371641` `rgb(55,22,65)``\n• #391344\n``#391344` `rgb(57,19,68)``\n• #3b1047\n``#3b1047` `rgb(59,16,71)``\n• #3d0c4b\n``#3d0c4b` `rgb(61,12,75)``\n• #3f094e\n``#3f094e` `rgb(63,9,78)``\n• #410651\n``#410651` `rgb(65,6,81)``\n• #420255\n``#420255` `rgb(66,2,85)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #2e2730 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
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https://studysoup.com/tsg/13733/introductory-chemistry-5-edition-chapter-8-problem-46p | [
"×\n×\n\n# Determine the theoretical yield of C when each of the",
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"ISBN: 9780321910295 34\n\n## Solution for problem 46P Chapter 8\n\nIntroductory Chemistry | 5th Edition\n\n• Textbook Solutions\n• 2901 Step-by-step solutions solved by professors and subject experts\n• Get 24/7 help from StudySoup virtual teaching assistants",
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"Introductory Chemistry | 5th Edition\n\n4 5 1 387 Reviews\n20\n2\nProblem 46P\n\nPROBLEM 46P\n\nDetermine the theoretical yield of C when each of the initial quantities of A and B is allowed to react in the generic reaction:\n\n2 A + 3 B → 2 C\n\n(a) 2 mol A; 4 mol B\n\n(b) 3 mol A; 3 mol B\n\n(c) 5 mol A; 6 mol B\n\n(d) 4 mol A; 5 mol B\n\nStep-by-Step Solution:\n\nSolution 46P\n\nStep 1:\n\n(a)2 mol A; 4 mol B\n\nFor these you need to set up proportions to find the theoretical yield. The general reaction\n\n2A + 3B",
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"2C\n\nMeans that 2 moles of A and 3 moles of B react to make 2 moles of C. The proportion you need to set up is this:\n\nmoles of reactant in general reaction = moles of reactant in problem\n\nmoles of product in general reaction = moles of product being asked for\n\nSo we have for problem A:\n\n2 mol A = 2 mol A\n\n2 mol C = x\n\nx = 2 mol C\n\nand:\n\n3 mol B = 4 mol B\n\n2 mol C x\n\nx = 8/3 mol C (2.667 mol)\n\nAs you can see, I set up proportions for both Reactant A and Reactant B as given in the problem. The theoretical yield is the proportion which gives the least amount of moles of product, which in this case is the 2 moles of Reactant A, which gave 2 moles of Product C, whereas the 4 moles of Reactant B would have given 2.667 moles of C (thus, A is called the limiting reactant, since, in the reaction given by problem A, all of Reactant A would have reacted with most of B to yield 2 moles of C). So the theoretical yield is 2 moles of C.\n\nStep 2:\n\n(b) 3 mol A; 3 mol B\n\nHere,\n\n2 mol A = 3 mol A\n\n2 mol C x\n\nx = 3 mol C\n\nand:\n\n3 mol B = 4 mol B\n\n2 mol C x\n\nx = 8/3 = 2.6 mol C\n\nHere the theoretical yield is 2.6 moles\n\nStep 3 of 4\n\nStep 4 of 4\n\n##### ISBN: 9780321910295\n\nUnlock Textbook Solution"
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"https://studysoup.com/cdn/24cover_2610068",
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"https://studysoup.com/cdn/24cover_2610068",
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"https://chart.googleapis.com/chart",
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https://www.colorhexa.com/01fa72 | [
"# #01fa72 Color Information\n\nIn a RGB color space, hex #01fa72 is composed of 0.4% red, 98% green and 44.7% blue. Whereas in a CMYK color space, it is composed of 99.6% cyan, 0% magenta, 54.4% yellow and 2% black. It has a hue angle of 147.2 degrees, a saturation of 99.2% and a lightness of 49.2%. #01fa72 color hex could be obtained by blending #02ffe4 with #00f500. Closest websafe color is: #00ff66.\n\n• R 0\n• G 98\n• B 45\nRGB color chart\n• C 100\n• M 0\n• Y 54\n• K 2\nCMYK color chart\n\n#01fa72 color description : Vivid cyan - lime green.\n\n# #01fa72 Color Conversion\n\nThe hexadecimal color #01fa72 has RGB values of R:1, G:250, B:114 and CMYK values of C:1, M:0, Y:0.54, K:0.02. Its decimal value is 129650.\n\nHex triplet RGB Decimal 01fa72 `#01fa72` 1, 250, 114 `rgb(1,250,114)` 0.4, 98, 44.7 `rgb(0.4%,98%,44.7%)` 100, 0, 54, 2 147.2°, 99.2, 49.2 `hsl(147.2,99.2%,49.2%)` 147.2°, 99.6, 98 00ff66 `#00ff66`\nCIE-LAB 86.795, -77.233, 50.982 37.233, 69.588, 27.388 0.277, 0.519, 69.588 86.795, 92.543, 146.571 86.795, -78.765, 79.071 83.42, -66.314, 38.928 00000001, 11111010, 01110010\n\n# Color Schemes with #01fa72\n\n• #01fa72\n``#01fa72` `rgb(1,250,114)``\n• #fa0189\n``#fa0189` `rgb(250,1,137)``\nComplementary Color\n• #0cfa01\n``#0cfa01` `rgb(12,250,1)``\n• #01fa72\n``#01fa72` `rgb(1,250,114)``\n• #01faef\n``#01faef` `rgb(1,250,239)``\nAnalogous Color\n• #fa010c\n``#fa010c` `rgb(250,1,12)``\n• #01fa72\n``#01fa72` `rgb(1,250,114)``\n• #ef01fa\n``#ef01fa` `rgb(239,1,250)``\nSplit Complementary Color\n• #fa7201\n``#fa7201` `rgb(250,114,1)``\n• #01fa72\n``#01fa72` `rgb(1,250,114)``\n• #7201fa\n``#7201fa` `rgb(114,1,250)``\n• #89fa01\n``#89fa01` `rgb(137,250,1)``\n• #01fa72\n``#01fa72` `rgb(1,250,114)``\n• #7201fa\n``#7201fa` `rgb(114,1,250)``\n• #fa0189\n``#fa0189` `rgb(250,1,137)``\n• #01ae4f\n``#01ae4f` `rgb(1,174,79)``\n• #01c75b\n``#01c75b` `rgb(1,199,91)``\n• #01e166\n``#01e166` `rgb(1,225,102)``\n• #01fa72\n``#01fa72` `rgb(1,250,114)``\n• #16fe80\n``#16fe80` `rgb(22,254,128)``\n• #30fe8d\n``#30fe8d` `rgb(48,254,141)``\n• #49fe9b\n``#49fe9b` `rgb(73,254,155)``\nMonochromatic Color\n\n# Alternatives to #01fa72\n\nBelow, you can see some colors close to #01fa72. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #01fa34\n``#01fa34` `rgb(1,250,52)``\n• #01fa49\n``#01fa49` `rgb(1,250,73)``\n• #01fa5d\n``#01fa5d` `rgb(1,250,93)``\n• #01fa72\n``#01fa72` `rgb(1,250,114)``\n• #01fa87\n``#01fa87` `rgb(1,250,135)``\n• #01fa9c\n``#01fa9c` `rgb(1,250,156)``\n• #01fab0\n``#01fab0` `rgb(1,250,176)``\nSimilar Colors\n\n# #01fa72 Preview\n\nThis text has a font color of #01fa72.\n\n``<span style=\"color:#01fa72;\">Text here</span>``\n#01fa72 background color\n\nThis paragraph has a background color of #01fa72.\n\n``<p style=\"background-color:#01fa72;\">Content here</p>``\n#01fa72 border color\n\nThis element has a border color of #01fa72.\n\n``<div style=\"border:1px solid #01fa72;\">Content here</div>``\nCSS codes\n``.text {color:#01fa72;}``\n``.background {background-color:#01fa72;}``\n``.border {border:1px solid #01fa72;}``\n\n# Shades and Tints of #01fa72\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #001007 is the darkest color, while #fbfffd is the lightest one.\n\n• #001007\n``#001007` `rgb(0,16,7)``\n• #002310\n``#002310` `rgb(0,35,16)``\n• #003719\n``#003719` `rgb(0,55,25)``\n• #004a22\n``#004a22` `rgb(0,74,34)``\n• #005e2b\n``#005e2b` `rgb(0,94,43)``\n• #007134\n``#007134` `rgb(0,113,52)``\n• #01853d\n``#01853d` `rgb(1,133,61)``\n• #019845\n``#019845` `rgb(1,152,69)``\n• #01ac4e\n``#01ac4e` `rgb(1,172,78)``\n• #01bf57\n``#01bf57` `rgb(1,191,87)``\n• #01d360\n``#01d360` `rgb(1,211,96)``\n• #01e669\n``#01e669` `rgb(1,230,105)``\n• #01fa72\n``#01fa72` `rgb(1,250,114)``\n• #11fe7c\n``#11fe7c` `rgb(17,254,124)``\n• #24fe87\n``#24fe87` `rgb(36,254,135)``\n• #38fe92\n``#38fe92` `rgb(56,254,146)``\n• #4bfe9c\n``#4bfe9c` `rgb(75,254,156)``\n• #5ffea7\n``#5ffea7` `rgb(95,254,167)``\n• #72feb2\n``#72feb2` `rgb(114,254,178)``\n• #86ffbd\n``#86ffbd` `rgb(134,255,189)``\n• #99ffc7\n``#99ffc7` `rgb(153,255,199)``\n``#adffd2` `rgb(173,255,210)``\n• #c0ffdd\n``#c0ffdd` `rgb(192,255,221)``\n• #d4ffe7\n``#d4ffe7` `rgb(212,255,231)``\n• #e7fff2\n``#e7fff2` `rgb(231,255,242)``\n• #fbfffd\n``#fbfffd` `rgb(251,255,253)``\nTint Color Variation\n\n# Tones of #01fa72\n\nA tone is produced by adding gray to any pure hue. In this case, #75867d is the less saturated color, while #01fa72 is the most saturated one.\n\n• #75867d\n``#75867d` `rgb(117,134,125)``\n• #6b907c\n``#6b907c` `rgb(107,144,124)``\n• #62997b\n``#62997b` `rgb(98,153,123)``\n• #58a37a\n``#58a37a` `rgb(88,163,122)``\n``#4ead79` `rgb(78,173,121)``\n• #45b678\n``#45b678` `rgb(69,182,120)``\n• #3bc077\n``#3bc077` `rgb(59,192,119)``\n• #31ca76\n``#31ca76` `rgb(49,202,118)``\n• #28d376\n``#28d376` `rgb(40,211,118)``\n• #1edd75\n``#1edd75` `rgb(30,221,117)``\n• #14e774\n``#14e774` `rgb(20,231,116)``\n• #0bf073\n``#0bf073` `rgb(11,240,115)``\n• #01fa72\n``#01fa72` `rgb(1,250,114)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #01fa72 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5379196,"math_prob":0.68816495,"size":3691,"snap":"2019-51-2020-05","text_gpt3_token_len":1594,"char_repetition_ratio":0.13100082,"word_repetition_ratio":0.011049724,"special_character_ratio":0.55432135,"punctuation_ratio":0.23404256,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98332465,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-08T00:50:42Z\",\"WARC-Record-ID\":\"<urn:uuid:6b81c43b-6582-4b1b-9e80-404c7e1cf2f8>\",\"Content-Length\":\"36263\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef2dea9d-8a54-4b1e-9112-50dc69e64440>\",\"WARC-Concurrent-To\":\"<urn:uuid:703d704d-8de6-4506-858c-e9b3239eb21a>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/01fa72\",\"WARC-Payload-Digest\":\"sha1:CX5VDWW2WSFOBNPTLHUSDTVG2GAVRGEH\",\"WARC-Block-Digest\":\"sha1:XFW3ILQ2LFE26IYJHFSJ6T3OGKT36ED7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540503656.42_warc_CC-MAIN-20191207233943-20191208021943-00065.warc.gz\"}"} |
https://mathematica.stackexchange.com/questions/201148/find-and-replace-particular-multiplication-of-factors-in-big-multiplication | [
"# Find and replace particular multiplication of factors in big multiplication\n\nI have a sum of terms where each is a multiplication of several factors. For example my expression in FullForm is\n\nexpr = Plus[Times[Derivative[a][x], Derivative[0, 1][z][t, x]],\nTimes[R, z[t, x], Derivative[a][x], Derivative[0, 1][\\[Sigma]][t, x]],\nTimes[2, R, a[x], Derivative[0, 1][z][t, x], Derivative[0, 1][\\[Sigma]][t, x]],\nTimes[Power[R, 2], a[x], z[t, x], Power[Derivative[0, 1][\\[Sigma]][t, x], 2]],\nTimes[a[x], Derivative[0, 2][z][t, x]], Times[R, a[x], z[t, x], Derivative[0, 2][\\[Sigma]][t, x]], Derivative[1, 0][z][t, x],\nTimes[R, z[t, x], Derivative[1, 0][\\[Sigma]][t, x]]]\n\n\nI want to know if in any of the terms I have a multiplication of the form\n\nTimes[z[t, x],Derivative[a][x]]\n\n\nand replace it with another expression in that case.\n\nI am stuck even trying to find my particular multiplication. I have tried with\n\nCases[expr, Times[z[t, x],Derivative[a][x]],\\[Infinite]]\n\n\nBut since the exact expression I am looking for does not appear in the full form I only get\n\nout = {}\n\n\nAs a start I would be happy if I could only find if my desired factor appears in any term. Then once I can do that it would be even better to replace it, but let us go one step at a time.\n\nexpr /. Times[p___, Derivative[a][x], z[t, x], q___] :> Times[p, newexpr, q]\n\n\nReplace newexpr as desired.\n\nNote the triple underscore on patterns p___ and q___, which means something like: \"This is a pattern which may contain any number of elements in sequence, if it even appears at all.\" This allows us to match forms like Times[a'[x], z[t,x]], Times[..., a'[x], z[t,x]], and Times[a'[x], z[t,x], ...] simultaneously. This is largely just required because Times tags agglomerate together if they end up next to each other, so it's difficult to predict how large they are going to be. Because Mathematica does various things to put them into a canonical ordering and considers Times commutative, however, it shouldn't be an issue with what order you specify the replacement in.\n\nYou can also use Replace with simpler pattern:\n\nReplace[expr, s_. Derivative[a][x] z[t, x] :> s newexpr, All]\n\n\n(a^′)[x] (z^(0,1))[t,x]+newexpr R (σ^(0,1))[t,x]+2 R a[x] (z^(0,1))[t,x] (σ^(0,1))[t,x]+R^2 a[x] z[t,x] (σ^(0,1))[t,x]^2+a[x] (z^(0,2))[t,x]+R a[x] z[t,x] (σ^(0,2))[t,x]+(z^(1,0))[t,x]+R z[t,x] (σ^(1,0))[t,x]\n\n TeXForm @ %\n\n\n$$a'(x) z^{(0,1)}(t,x)+R^2 a(x) \\sigma ^{(0,1)}(t,x)^2 z(t,x)+2 R a(x) \\sigma ^{(0,1)}(t,x) z^{(0,1)}(t,x)+R a(x) \\sigma ^{(0,2)}(t,x) z(t,x)+a(x) z^{(0,2)}(t,x)+\\text{newexpr} R \\sigma ^{(0,1)}(t,x)+R \\sigma ^{(1,0)}(t,x) z(t,x)+z^{(1,0)}(t,x)$$\n\n• Can I just ask what the s_. stands for?\n– I.C.\nJun 28 '19 at 21:03\n• @I.C. s_. is the pattern that stands for the Default argument ( for Times it is 1). So s_. foo matches both 5 foo (where a multiplicative factor (5) is there) and foo (where the multiplicative factor is the default argument of Times, i.e. 1). Hope this helps.\n– kglr\nJun 29 '19 at 0:01"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8206362,"math_prob":0.99892235,"size":1182,"snap":"2021-43-2021-49","text_gpt3_token_len":393,"char_repetition_ratio":0.2113752,"word_repetition_ratio":0.0,"special_character_ratio":0.34856176,"punctuation_ratio":0.19736843,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99988925,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-04T10:46:19Z\",\"WARC-Record-ID\":\"<urn:uuid:194a49e1-d25f-45f2-b3e3-946dcf1a4022>\",\"Content-Length\":\"145996\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0b34151c-d3fe-4184-8a67-53be53f84023>\",\"WARC-Concurrent-To\":\"<urn:uuid:33e072ce-b94e-4fd8-8e31-04b192a91517>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://mathematica.stackexchange.com/questions/201148/find-and-replace-particular-multiplication-of-factors-in-big-multiplication\",\"WARC-Payload-Digest\":\"sha1:P4J4ZRVUUSLMG2LZM2YTL2ZZIV7K2QOV\",\"WARC-Block-Digest\":\"sha1:KFBONUFHWNZFHVHGLKEG7RMNE52JCQAJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362969.51_warc_CC-MAIN-20211204094103-20211204124103-00206.warc.gz\"}"} |
http://www.informit.com/articles/article.aspx?p=2923213&seqNum=3 | [
"",
null,
"• Print\nThis chapter is from the book\n\n3.3 Modules (C++20)\n\nWe are finally about to get a better way of expressing physical modules in C++. The language feature, called modules is not yet ISO C++, but it is an ISO Technical Specification [ModulesTS]. Implementations are in use, so I risk recommending it here even though details are likely to change and it may be years before everybody can use it in production code. Old code, in this case code using #include, can “live” for a very long time because it can be costly and time consuming to update.\n\nConsider how to express the Vector and use() example from §3.2 using modules:\n\n// file Vector.cpp:\n\nmodule; // this compilation will define a module\n\n// ... here we put stuff that Vector might need for its implementation ...\n\nexport module Vector; // defining the module called \"Vector\"\n\nexport class Vector {\npublic:\nVector(int s);\ndouble& operator[](int i);\nint size();\nprivate:\ndouble* elem; // elem points to an array of sz doubles\nint sz;\n};\n\nVector::Vector(int s)\n:elem{new double[s]}, sz{s} // initialize members\n{\n}\n\ndouble& Vector::operator[](int i)\n{\nreturn elem[i];\n}\n\nint Vector::size()\n{\nreturn sz;\n}\n\nexport int size(const Vector& v) { return v.size(); }\n\nThis defines a module called Vector, which exports the class Vector, all its member functions, and the non-member function size().\n\nThe way we use this module is to import it where we need it. For example:\n\n// file user.cpp:\n\nimport Vector; // get Vector's interface\n#include <cmath> // get the standard-library math function interface including sqrt()\n\ndouble sqrt_sum(Vector& v)\n{\ndouble sum = 0;\nfor (int i=0; i!=v.size(); ++i)\nsum+=std::sqrt(v[i]); // sum of square roots\nreturn sum;\n}\n\nI could have imported the standard library mathematical functions also, but I used the old-fashioned #include just to show that you can mix old and new. Such mixing is essential for gradually upgrading older code from using #include to import.\n\nThe differences between headers and modules are not just syntactic.\n\n• A module is compiled once only (rather than in each translation unit in which it is used).\n\n• Two modules can be imported in either order without changing their meaning.\n\n• If you import something into a module, users of your module do not implicitly gain access to (and are not bothered by) what you imported: import is not transitive.\n\nThe effects on maintainability and compile-time performance can be spectacular."
] | [
null,
"https://www.facebook.com/tr",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.79845047,"math_prob":0.86354876,"size":2995,"snap":"2019-43-2019-47","text_gpt3_token_len":669,"char_repetition_ratio":0.12470745,"word_repetition_ratio":0.0,"special_character_ratio":0.24641068,"punctuation_ratio":0.15041323,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97310406,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-15T12:31:58Z\",\"WARC-Record-ID\":\"<urn:uuid:4cd4087d-33a5-40aa-b60f-d7082c2dec17>\",\"Content-Length\":\"43245\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:408298eb-ea09-44c2-a70b-4001a190a6d4>\",\"WARC-Concurrent-To\":\"<urn:uuid:ca2eba1f-35d9-413c-9da3-9dba29d6e965>\",\"WARC-IP-Address\":\"159.182.165.92\",\"WARC-Target-URI\":\"http://www.informit.com/articles/article.aspx?p=2923213&seqNum=3\",\"WARC-Payload-Digest\":\"sha1:II54MNSN5OGCJODCPRFMSZOMVLTZMREG\",\"WARC-Block-Digest\":\"sha1:FCAYMBIVUCT6KN6ZV3O5VXZJUG7B733U\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986658566.9_warc_CC-MAIN-20191015104838-20191015132338-00415.warc.gz\"}"} |
https://physics.stackexchange.com/questions/160945/when-is-it-appropriate-to-drop-pressure-terms-when-applying-conservation-of-mome | [
"# When is it appropriate to drop pressure terms when applying conservation of momentum to a fluid?\n\nI'm trying to wrap my head around pressure forces in incompressible, irrotational, invicid flow. Applying conservation of momentum to a control volume gives me\n\n\\begin{equation} \\frac{\\partial}{\\partial t} \\int_{c.v.} \\rho \\vec{v} \\mathrm{d}V + \\int_{c.s} \\rho \\vec{v} (\\vec{v} - \\vec{v}_{c.v.})\\cdot \\hat n \\mathrm{d}A = -\\int_{c.s.} P \\hat n \\mathrm{d}A + \\int_{c.s.} \\overline{\\overline{\\tau}}^{\\prime} \\cdot \\hat n \\mathrm{d}A + \\int_{c.v.} \\rho \\vec{g} \\mathrm{d}V\\,. \\end{equation}\n\nWhen is it it acceptable to let pressure force terms on the right hand side of the equation equal zero? I know that if there is a uniform pressure on a closed surface, the force is zero, but that is the only case I really understand. Does it matter if the surface on which I'm evaluating the force is collinear with a solid body or perhaps a streamline?\n\nFor the kind of flow you mention conservation of mass gives you the equation\n\n\\begin{equation} \\nabla\\cdot\\boldsymbol{v}=0 \\end{equation}\n\nwhere $\\boldsymbol{v}$ is the velocity field, in integral form it would be\n\n\\begin{equation} \\oint{\\boldsymbol{v}\\cdot\\boldsymbol{n}dA}=0 \\end{equation}\n\nand for \"conservation\" of linear momentum you have\n\n\\begin{equation} \\frac{D\\boldsymbol{v}}{Dt}=-\\nabla\\frac{P}{\\rho} \\end{equation}\n\nif you have an homogeneous pressure field, there would not be a gradient, so in that case you will have the equation\n\n\\begin{equation} \\frac{D\\boldsymbol{v}}{Dt}=0 \\end{equation}\n\nwhich means that velocity is constant along stream lines"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.78322446,"math_prob":0.9993786,"size":840,"snap":"2020-24-2020-29","text_gpt3_token_len":265,"char_repetition_ratio":0.13397129,"word_repetition_ratio":0.0,"special_character_ratio":0.2904762,"punctuation_ratio":0.11666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999924,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-03T14:14:34Z\",\"WARC-Record-ID\":\"<urn:uuid:11c458ed-d680-425d-8815-10bf2f00046d>\",\"Content-Length\":\"142572\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b218afc5-8020-4677-a349-3d7b2112329b>\",\"WARC-Concurrent-To\":\"<urn:uuid:7309f5e2-f061-4221-a10a-77121e66b6fb>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/160945/when-is-it-appropriate-to-drop-pressure-terms-when-applying-conservation-of-mome\",\"WARC-Payload-Digest\":\"sha1:YPXRRWTKXNUENT3QNPVXEHRJPXM2OCXK\",\"WARC-Block-Digest\":\"sha1:YK7ZGV5U6DN5AWKRVLTVQEP672MKE5J7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655882051.19_warc_CC-MAIN-20200703122347-20200703152347-00464.warc.gz\"}"} |
https://stats.stackexchange.com/questions/223214/why-are-all-clt-problems-using-a-single-random-sample-when-the-clt-requires-rep | [
"# Why are all CLT problems using a single random sample when the CLT requires “repeatedly drawing random samples”?\n\nIn our course notes, it says: \"The CLT states that if random samples of size $n$ are repeatedly drawn from any population with mean $\\mu$ and variance $\\sigma^2$, then when $n$ is large the distribution of the sample means will be approximately normal.\"\n\nI bolded \"repeatedly\" because that's the part I'm confused about. In the examples and the homework problems, it seems that even when we have $n$ supposedly large enough ($>30$), we're still only talking about a single sample of size $n$.\n\nFor example, one of the problems states: \"Suppose a population has mean $\\mu=5$ and standard deviation $\\sigma = 2$, then supposed a random sample of size 38 is selected. What is the probability that the sample mean is between $4$ and $6$?\"\n\nThen we go on to use the CLT to solve the problem, supposedly because $n > 30$ allows us to use the CLT. But from that previous paragraph in italic I quoted, it seems that the CLT only works when many many samples of size n are drawn from a population, even when $n > 30$.\n\nHere we're only randomly selecting a single sample, yet it seems to be good enough to use the CLT. Why is that?\n\nEDIT: When I say \"all CLT problems\" I mean all the ones I ran into, including the famous swan problem, in which we randomly select $n>30$ swans once.\n\n• You ask about the meaning of probability (explained as a long-run phenomenon) in a context where just a single result (such as the sample mean) is available. Here is a perfect analogy. When we say a coin is \"fair\" we mean that in a sufficiently long run of repeated independent flips, the proportion of heads will be close to $1/2$. If I ask you about the chance that one flip of this fair coin will be heads, if you are rational you will answer \"1/2\". But that's only a statement about one flip, not a long run of flips! Why was your answer of \"1/2\" good enough? – whuber Jul 11 '16 at 18:19\n• Possible duplicate of Why does the central limit theorem work with a single sample? – Greenparker Jul 11 '16 at 18:20\n• @whuber, you have a knack for understanding the source of people's confusions :) and you're right, that's exactly the same thing; for some reason I couldn't see it. I think stuff is thrown at us so quickly (summer course) that I'm getting confused with all the terminology. – jeremy radcliff Jul 11 '16 at 18:56\n• @Greenparker, you're right, it's a very similar question. Thanks for the link. – jeremy radcliff Jul 11 '16 at 18:58\n• You're confused about the meaning of the word \"sample.\" The trials $X_1,\\ldots,X_n$ contitute one sample of size $n.$ $\\qquad$ – Michael Hardy May 27 '19 at 17:41\n\nThe CLT is a statement about fluctuations of averages. Specifically, a single sample of size $n$ will give you an expression:\n$$\\mu_n:=\\frac{X_1+\\cdots+X_n}{n},$$\nwhich gives you an estimate of the true mean (average). Here $\\mu_n$ is a random variable, with some complicated distribution. The CLT states that as $n$ gets larger, $\\mu_n$ will be closer and closer to the true mean $\\mu$, with fluctuations that look like a normal distribution, centered on the mean with variance $\\sigma^2/n$. Said another way, $\\mu_n$ is approximately distributed like $N(\\mu,\\sigma^2/n)$. In other words, if you were to repeat your sampling a bunch of times, and plot a histogram of the $\\mu_n$'s that you drew each time, you'd get something that looks like a normal distribution with the above parameters.\nThus it makes perfect sense to ask questions like: what is the probability that $\\mu_n>2$? In this case we'd use the CLT to approximate the distribution of $\\mu_n$ as a normal distribution. This means that you could numerically verify the CLT by repeatedly drawing samples of size $n$ and computing $\\mu_n$, thereby plotting a histogram and calculating probabilities from it.\n• @AlexR. I think that makes sense, thank you. Does it matter whether we know that the population is normally distributed if $n$ is large enough? In the \"swan problem\", we are told \"the weight of an swan is normally distributed\", then told we get a random sample of $30+$ swans, and we go on to use the CLT. but if our sample of size $n$ is already $>30$, does it even matter what the original population's distribution looks like? For ex if we weren't told the swan weights are N distributed, would it matter? – jeremy radcliff Jul 11 '16 at 18:53"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.96395624,"math_prob":0.9929958,"size":1267,"snap":"2020-10-2020-16","text_gpt3_token_len":315,"char_repetition_ratio":0.122723676,"word_repetition_ratio":0.0,"special_character_ratio":0.2549329,"punctuation_ratio":0.09765625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99916196,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-17T23:28:26Z\",\"WARC-Record-ID\":\"<urn:uuid:69365cdf-29f8-4109-9e56-4db273f83a42>\",\"Content-Length\":\"149056\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aeb3b862-87e1-4f91-8118-a7da1eb11990>\",\"WARC-Concurrent-To\":\"<urn:uuid:77e2bc32-f5f3-42e4-a43e-2abf5b7b56a4>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/223214/why-are-all-clt-problems-using-a-single-random-sample-when-the-clt-requires-rep\",\"WARC-Payload-Digest\":\"sha1:UNJXGLOH5G3JOKY7EO4MZQY5V426RESA\",\"WARC-Block-Digest\":\"sha1:2TQ5EMW77NNH6HXG2T7WLISTQRF4LKIS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875143373.18_warc_CC-MAIN-20200217205657-20200217235657-00312.warc.gz\"}"} |
https://optimization-online.org/tag/copositive-matrix/ | [
"## Exactness of Parrilo’s conic approximations for copositive matrices and associated low order bounds for the stability number of a graph\n\nDe Klerk and Pasechnik (2002) introduced the bounds $\\vartheta^{(r)}(G)$ ($r\\in \\mathbb{N}$) for the stability number $\\alpha(G)$ of a graph $G$ and conjectured exactness at order $\\alpha(G)-1$: $\\vartheta^{(\\alpha(G)-1)}(G)=\\alpha(G)$. These bounds rely on the conic approximations $\\mathcal{K}_n^{(r)}$ by Parrilo (2000) for the copositive cone $\\text{COP}_n$. A difficulty in the convergence analysis of $\\vartheta^{(r)}$ is the bad behaviour … Read more"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77128947,"math_prob":0.9997027,"size":593,"snap":"2023-14-2023-23","text_gpt3_token_len":174,"char_repetition_ratio":0.1188455,"word_repetition_ratio":0.0,"special_character_ratio":0.2866779,"punctuation_ratio":0.03,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999503,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-21T17:55:27Z\",\"WARC-Record-ID\":\"<urn:uuid:95cdf5da-4bbf-4e4b-bb42-d1281345bc7f>\",\"Content-Length\":\"81451\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:46b4373c-3f05-4383-b92a-6590f4d4644a>\",\"WARC-Concurrent-To\":\"<urn:uuid:9d4ac38c-d23f-4433-928c-0ed2644843a7>\",\"WARC-IP-Address\":\"128.104.153.102\",\"WARC-Target-URI\":\"https://optimization-online.org/tag/copositive-matrix/\",\"WARC-Payload-Digest\":\"sha1:SX4M2JXGRPEMLI2GKG6TPXQGV6E4TI4M\",\"WARC-Block-Digest\":\"sha1:7MR5G6P5V3E4ZNYLGOEXSYV2GJ7M4FZF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943704.21_warc_CC-MAIN-20230321162614-20230321192614-00337.warc.gz\"}"} |
https://www.technetiumcr225.xyz/wiki/Lego_Star_Wars_II:_The_Original_Trilogy | [
"# Polydivisible number\n\nIn mathematics a polydivisible number (or magic number) is a number in a given number base with digits abcde... that has the following properties:\n\n1. Its first digit a is not 0.\n2. The number formed by its first two digits ab is a multiple of 2.\n3. The number formed by its first three digits abc is a multiple of 3.\n4. The number formed by its first four digits abcd is a multiple of 4.\n5. etc.\n\n## Definition\n\nLet $n$",
null,
"be a natural number, and let $k=\\lfloor \\log _{b}{n}\\rfloor +1$",
null,
"be the number of digits in the number in base $b$",
null,
". $n$",
null,
"is a polydivisible number if for all $0\\leq i",
null,
",\n\n${\\frac {n-(n{\\bmod {b}}^{k-i-1})}{b^{k-i-1}}}\\equiv 0{\\bmod {i}}$",
null,
".\n\nFor example, 10801 is a seven-digit polydivisible number in base 4, as\n\n${\\frac {10801-(10801{\\bmod {4}}^{7-0-1})}{4^{7-0-1}}}={\\frac {10801-(10801{\\bmod {4}}^{6})}{4^{6}}}={\\frac {10801-(10801{\\bmod {4}}096)}{4096}}={\\frac {10801-2609}{4096}}={\\frac {8192}{4096}}=2\\equiv 0{\\bmod {1}}$",
null,
"${\\frac {10801-(10801{\\bmod {4}}^{7-1-1})}{4^{7-1-1}}}={\\frac {10801-(10801{\\bmod {4}}^{5})}{4^{5}}}={\\frac {10801-(10801{\\bmod {1}}024)}{1024}}={\\frac {10801-561}{1024}}={\\frac {10240}{1024}}=10\\equiv 0{\\bmod {2}}$",
null,
"${\\frac {10801-(10801{\\bmod {4}}^{7-2-1})}{4^{7-2-1}}}={\\frac {10801-(10801{\\bmod {4}}^{4})}{4^{4}}}={\\frac {10801-(10801{\\bmod {2}}56)}{256}}={\\frac {10801-49}{256}}={\\frac {10752}{256}}=42\\equiv 0{\\bmod {3}}$",
null,
"${\\frac {10801-(10801{\\bmod {4}}^{7-3-1})}{4^{7-3-1}}}={\\frac {10801-(10801{\\bmod {4}}^{3})}{4^{3}}}={\\frac {10801-(10801{\\bmod {6}}4)}{64}}={\\frac {10801-49}{64}}={\\frac {10752}{64}}=168\\equiv 0{\\bmod {4}}$",
null,
"${\\frac {10801-(10801{\\bmod {4}}^{7-4-1})}{4^{7-4-1}}}={\\frac {10801-(10801{\\bmod {4}}^{2})}{4^{2}}}={\\frac {10801-(10801{\\bmod {1}}6)}{16}}={\\frac {10801-1}{16}}={\\frac {10800}{16}}=675\\equiv 0{\\bmod {5}}$",
null,
"${\\frac {10801-(10801{\\bmod {4}}^{7-5-1})}{4^{7-5-1}}}={\\frac {10801-(10801{\\bmod {4}}^{1})}{4^{1}}}={\\frac {10801-(10801{\\bmod {4}})}{4}}={\\frac {10801-1}{4}}={\\frac {10800}{4}}=2700\\equiv 0{\\bmod {6}}$",
null,
"${\\frac {10801-(10801{\\bmod {4}}^{7-6-1})}{4^{7-6-1}}}={\\frac {10801-(10801{\\bmod {4}}^{0})}{4^{0}}}={\\frac {10801-(10801{\\bmod {1}})}{1}}={\\frac {10801-0}{1}}={\\frac {10801}{1}}=10801\\equiv 0{\\bmod {7}}$",
null,
"## Enumeration\n\nFor any given base $b$",
null,
", there are only a finite number of polydivisible numbers.\n\n### Maximum polydivisible number\n\nThe following table lists maximum polydivisible numbers for some bases b, where A−Z represent digit values 10 to 35.\n\nBase $b$",
null,
"Maximum polydivisible number () Number of base-b digits ()\n2 102 2\n3 20 02203 6\n4 222 03014 7\n5 40220 422005 10\n10 36085 28850 36840 07860 36725 25\n12 6068 903468 50BA68 00B036 20646412 28\n\n### Estimate for $F_{b}(n)$",
null,
"and $\\Sigma (b)$",
null,
"",
null,
"Graph of number of $n$",
null,
"-digit polydivisible numbers in base 10 $F_{10}(n)$",
null,
"vs estimate of $F_{10}(n)$",
null,
"Let $n$",
null,
"be the number of digits. The function $F_{b}(n)$",
null,
"determines the number of polydivisible numbers that has $n$",
null,
"digits in base $b$",
null,
", and the function $\\Sigma (b)$",
null,
"is the total number of polydivisible numbers in base $b$",
null,
".\n\nIf $k$",
null,
"is a polydivisible number in base $b$",
null,
"with $n-1$",
null,
"digits, then it can be extended to create a polydivisible number with $n$",
null,
"digits if there is a number between $bk$",
null,
"and $b(k+1)-1$",
null,
"that is divisible by $n$",
null,
". If $n$",
null,
"is less or equal to $b$",
null,
", then it is always possible to extend an $n-1$",
null,
"digit polydivisible number to an $n$",
null,
"-digit polydivisible number in this way, and indeed there may be more than one possible extension. If $n$",
null,
"is greater than $b$",
null,
", it is not always possible to extend a polydivisible number in this way, and as $n$",
null,
"becomes larger, the chances of being able to extend a given polydivisible number become smaller. On average, each polydivisible number with $n-1$",
null,
"digits can be extended to a polydivisible number with $n$",
null,
"digits in ${\\frac {b}{n}}$",
null,
"different ways. This leads to the following estimate for $F_{b}(n)$",
null,
":\n\n$F_{b}(n)\\approx (b-1){\\frac {b^{n-1}}{n!}}.$",
null,
"Summing over all values of n, this estimate suggests that the total number of polydivisible numbers will be approximately\n\n$\\Sigma (b)\\approx {\\frac {b-1}{b}}(e^{b}-1)$",
null,
"Base $b$",
null,
"$\\Sigma (b)$",
null,
"Est. of $\\Sigma (b)$",
null,
"Percent Error\n2 2 ${\\frac {1}{2}}(e^{2}-1)\\approx 3.1945$",
null,
"59.7%\n3 15 ${\\frac {2}{3}}(e^{3}-1)\\approx 12.725$",
null,
"-15.1%\n4 37 ${\\frac {3}{4}}(e^{4}-1)\\approx 40.199$",
null,
"8.64%\n5 127 ${\\frac {4}{5}}(e^{5}-1)\\approx 117.93$",
null,
"−7.14%\n10 20456 ${\\frac {9}{10}}(e^{10}-1)\\approx 19823$",
null,
"-3.09%\n\n## Specific bases\n\nAll numbers are represented in base $b$",
null,
", using A−Z to represent digit values 10 to 35.\n\n### Base 2\n\nLength n F2(n) Est. of F2(n) Polydivisible numbers\n1 1 1 1\n2 1 1 10\n\n### Base 3\n\nLength n F3(n) Est. of F3(n) Polydivisible numbers\n1 2 2 1, 2\n2 3 3 11, 20, 22\n3 3 3 110, 200, 220\n4 3 2 1100, 2002, 2200\n5 2 1 11002, 20022\n6 2 1 110020, 200220\n7 0 0 $\\varnothing$",
null,
"### Base 4\n\nLength n F4(n) Est. of F4(n) Polydivisible numbers\n1 3 3 1, 2, 3\n2 6 6 10, 12, 20, 22, 30, 32\n3 8 8 102, 120, 123, 201, 222, 300, 303, 321\n4 8 8 1020, 1200, 1230, 2010, 2220, 3000, 3030, 3210\n5 7 6 10202, 12001, 12303, 20102, 22203, 30002, 32103\n6 4 4 120012, 123030, 222030, 321030\n7 1 2 2220301\n8 0 1 $\\varnothing$",
null,
"### Base 5\n\nThe polydivisible numbers in base 5 are\n\n1, 2, 3, 4, 11, 13, 20, 22, 24, 31, 33, 40, 42, 44, 110, 113, 132, 201, 204, 220, 223, 242, 311, 314, 330, 333, 402, 421, 424, 440, 443, 1102, 1133, 1322, 2011, 2042, 2200, 2204, 2231, 2420, 2424, 3113, 3140, 3144, 3302, 3333, 4022, 4211, 4242, 4400, 4404, 4431, 11020, 11330, 13220, 20110, 20420, 22000, 22040, 22310, 24200, 24240, 31130, 31400, 31440, 33020, 33330, 40220, 42110, 42420, 44000, 44040, 44310, 110204, 113300, 132204, 201102, 204204, 220000, 220402, 223102, 242000, 242402, 311300, 314000, 314402, 330204, 333300, 402204, 421102, 424204, 440000, 440402, 443102, 1133000, 1322043, 2011021, 2042040, 2204020, 2420003, 2424024, 3113002, 3140000, 3144021, 4022042, 4211020, 4431024, 11330000, 13220431, 20110211, 20420404, 24200031, 31400004, 31440211, 40220422, 42110202, 44310242, 132204314, 201102110, 242000311, 314000044, 402204220, 443102421, 1322043140, 2011021100, 3140000440, 4022042200\n\nThe smallest base 5 polydivisible numbers with n digits are\n\n1, 11, 110, 1102, 11020, 110204, 1133000, 11330000, 132204314, 1322043140, none...\n\nThe largest base 5 polydivisible numbers with n digits are\n\n4, 44, 443, 4431, 44310, 443102, 4431024, 44310242, 443102421, 4022042200, none...\n\nThe number of base 5 polydivisible numbers with n digits are\n\n4, 10, 17, 21, 21, 21, 13, 10, 6, 4, 0, 0, 0...\nLength n F5(n) Est. of F5(n)\n1 4 4\n2 10 10\n3 17 17\n4 21 21\n5 21 21\n6 21 17\n7 13 12\n8 10 8\n9 6 4\n10 4 2\n\n### Base 10\n\nThe polydivisible numbers in base 10 are\n\n1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 102, 105, 108, 120, 123, 126, 129, 141, 144, 147, 162, 165, 168, 180, 183, 186, 189, ... (sequence A144688 in the OEIS)\n\nThe smallest base 10 polydivisible numbers with n digits are\n\n1, 10, 102, 1020, 10200, 102000, 1020005, 10200056, 102000564, 1020005640, 10200056405, 102006162060, 1020061620604, 10200616206046, 102006162060465, 1020061620604656, 10200616206046568, 108054801036000018, 1080548010360000180, 10805480103600001800, ... (sequence A214437 in the OEIS)\n\nThe largest base 10 polydivisible numbers with n digits are\n\n9, 98, 987, 9876, 98765, 987654, 9876545, 98765456, 987654564, 9876545640, 98765456405, 987606963096, 9876069630960, 98760696309604, 987606963096045, 9876062430364208, 98485872309636009, 984450645096105672, 9812523240364656789, 96685896604836004260, ... (sequence A225608 in the OEIS)\n\nThe number of base 10 polydivisible numbers with n digits are\n\n9, 45, 150, 375, 750, 1200, 1713, 2227, 2492, 2492, 2225, 2041, 1575, 1132, 770, 571, 335, 180, 90, 44, 18, 12, 6, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ... (sequence A143671 in the OEIS)\nLength n F10(n) Est. of F10(n)\n1 9 9\n2 45 45\n3 150 150\n4 375 375\n5 750 750\n6 1200 1250\n7 1713 1786\n8 2227 2232\n9 2492 2480\n10 2492 2480\n11 2225 2255\n12 2041 1879\n13 1575 1445\n14 1132 1032\n15 770 688\n16 571 430\n17 335 253\n18 180 141\n19 90 74\n20 44 37\n21 18 17\n22 12 8\n23 6 3\n24 3 1\n25 1 1\n\n## Programming example\n\nThe example below searches for polydivisible numbers in Python.\n\ndef find_polydivisible(base: int) -> List[int]:\n\"\"\"Find polydivisible number.\"\"\"\nnumbers = []\nprevious = []\nfor i in range(1, base):\nprevious.append(i)\nnew = []\ndigits = 2\nwhile not previous == []:\nnumbers.append(previous)\nfor i in range(0, len(previous)):\nfor j in range(0, base):\nnumber = previous[i] * base + j\nif number % digits == 0:\nnew.append(number)\nprevious = new\nnew = []\ndigits = digits + 1\nreturn numbers\n\n\n## Related problems\n\nPolydivisible numbers represent a generalization of the following well-known problem in recreational mathematics:\n\nArrange the digits 1 to 9 in order so that the first two digits form a multiple of 2, the first three digits form a multiple of 3, the first four digits form a multiple of 4 etc. and finally the entire number is a multiple of 9.\n\nThe solution to the problem is a nine-digit polydivisible number with the additional condition that it contains the digits 1 to 9 exactly once each. There are 2,492 nine-digit polydivisible numbers, but the only one that satisfies the additional condition is\n\n381 654 729\n\nOther problems involving polydivisible numbers include:\n\n• Finding polydivisible numbers with additional restrictions on the digits - for example, the longest polydivisible number that only uses even digits is\n480 006 882 084 660 840 40\n• Finding palindromic polydivisible numbers - for example, the longest palindromic polydivisible number is\n300 006 000 03\n• A common, trivial extension of the aforementioned example is to arrange the digits 0 to 9 to make a 10 digit number in the same way, the result is 3816547290. This is a pandigital polydivisible number."
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https://blog.softhubsolution.com/write-a-java-program-to-calculate-the-difference-between-two-dates-in-days/ | [
"# Write a Java program to calculate the difference between two dates in days\n\n## Write a Java program to calculate the difference between two dates in days?\n\nThis Java program calculates the number of days between two dates.\n\n• First, two Calendar objects are created using the Calendar.getInstance() method, which returns a Calendar object representing the current date and time in the default time zone. The set() method is used to set the year, month, and day of each Calendar object to the desired values.\n• Next, the difference between the two Calendar objects is calculated by subtracting the milliseconds of one Calendar object from the other and taking the absolute value. The Math.abs() method is used to ensure a positive value.\n• The TimeUnit.DAYS.convert() method is then used to convert the milliseconds difference to days. The resulting value is stored in a long variable called Days.\n• Finally, the number of days between the two dates is printed to the console using System.out.println().\n\nNote that this approach calculates the difference between two dates as the number of 24-hour periods, and does not consider daylight saving time or other factors that can affect the length of a day. For more precise date and time calculations, it is recommended to use the java.time package, which provides a more accurate and flexible API for working with dates and times.\n\n```import java.time.*;\nimport java.time.temporal.ChronoUnit;\nimport java.util.Calendar;\nimport java.util.concurrent.TimeUnit;\n\nclass Calculate_Days\n{\n\npublic static void main(String[] args)\n{\nCalendar cal1 = Calendar.getInstance();\ncal1.set(2022, 10, 1);\nCalendar cal2 = Calendar.getInstance();\ncal2.set(2023, 10, 1);\n\nlong Ms = Math.abs(cal1.getTimeInMillis() - cal2.getTimeInMillis());\nlong Days = Math.abs(TimeUnit.DAYS.convert(Ms, TimeUnit.MILLISECONDS));\n\nSystem.out.println(\"Difference in days is: \" + Days);\n}\n}```\n\n### Output\n\nDifference in days is: 365\n\nThis program calculates the difference in days between two dates. The program uses the Java time library to access the current date and time, and to perform the date calculations.\n\nThe program first creates two Calendar objects, representing the two dates whose difference is to be calculated. The set() method sets the year, month, and day of the two calendars.\n\nNext, the program calculates the difference between the two dates in milliseconds using the getTimeInMillis() method of the Calendar class. This difference is then converted to days using the TimeUnit class, which provides a convenient way to convert between different units of time.\n\nFinally, the program prints the difference in days using the System.out.println() method.",
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"http://blog.softhubsolution.com/wp-content/uploads/2022/12/IMG_20220407_092219_249.webp",
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http://www.myrkraverk.com/blog/2017/12/xemacs-sorting-key-value-lines-by-value/ | [
"",
null,
"# XEmacs: Sorting Key-Value Lines by Value\n\nXEmacs 21.4.24 [direct ftp download] and the latest stable release (2015) is the version I’m personally using. The directions here may well apply to GNU Emacs as well; I don’t know.\n\nMost Emacs users are familiar with the command `M-x sort-lines` which alphabetically sorts the lines highlighted in the current buffer.\n\nHowever, I had the wish to sort `key: values` as follows, by the value.\n\n```foo: 12\nbaz: 7\nbar: 2\n```\n\nAs you can see, in this instance, the value is numeric and lexicographical sorting of numbers results in\n\n```foo: 12\nbar: 2\nbaz: 7\n```\n\nthat is, alphabetical and not numeric.\n\nIn order to “fix this” we have to delve into the sorting internals of XEmacs.\n\nFirst, let’s look at the function sort-lines, which is fairly straight forward.1\n\n```(defun sort-lines (reverse beg end)\n;; [documentation string elided]\n(interactive \"P\\nr\")\n(save-excursion\n(save-restriction\n(narrow-to-region beg end)\n(goto-char (point-min))\n(sort-subr reverse 'forward-line 'end-of-line))))\n```\n\nThe most important thing to note here is the use of the function `sort-subr`. The rest of the code is simply boiler plate to limit the sort to the highlighted region. The `(interactive \"P\\nr\")` is to read the start and end of the region into the argument values `beg` and `end`. For our purposes, we can simply treat this as “magic”; the effect of the newline embedded in the string is to separate the `reverse` (read with `\"P\"`) from the region (read with `\"r\"`) and that means people can sort in reverse alphabetical order with `C-u M-x sort-lines`. The code we’ll develop in here does not have this feature; adding it can be considered an exercise for the reader.\n\nAlright, now we know how `sort-lines` works, and we know that we can use `sort-subr` to sort by values, the question is how do we do it?\n\nFirst, and obviously, we read the documentation string with `C-h f sort-subr`. This tells us that there are two variables we can make use of,1\n\nSTARTKEYFUN moves from the start of the record to the start of the key. It may return either a non-nil value to be used as the key, or else the key is the substring between the values of point after STARTKEYFUN and ENDKEYFUN are called. If STARTKEYFUN is nil, the key starts at the beginning of the record.\n\nand\n\nCOMPAREFUN compares the two keys. It is called with two strings and should return true if the first is “less” than the second, just as for `sort’. If nil or omitted, the default function accepts keys that are numbers (compared numerically) or strings (compared lexicographically).\n\nThe first thing we note here is the `startkeyfun`. It’ll allow us to limit the sort comparison to the value part of the lines. The trick here is to just move the point past the `:` (colon). We can do that with `search-forward`. Since in my case, and the example here, all the lines do have a colon, we’ll not consider the case where it might be missing in the line, hence we don’t impose any limit on the search (`nil`) nor do we care about errors (we allow them with `nil`); however, we limit the count to exactly one.\n\nThat leaves us with a call that looks like\n\n``` (search-forward \":\" nil nil 1)\n```\n\nand to apply that to `sort-subr` we define an entirely new function, `my-sort-key-value-lines`. We wrap `search-forward` in a lambda for simplicity. Notice that we return `nil` explicitly from the lambda, because otherwise `sort-subr` will use its return value (the location of point in the buffer after the colon) and sort from that.2\n\n```(defun my-sort-key-value-lines (beg end)\n(interactive \"r\")\n(save-excursion\n(save-restriction\n(narrow-to-region beg end)\n(goto-char (point-min))\n(sort-subr nil 'forward-line 'end-of-line\n(lambda ()\n(search-forward \":\" nil nil 1)\n;; returns point, so we explicitly return\nnil)))))\n```\n\nAnd this is the code that results in lexicographical sorting of the values, but we want numeric sorting. There are at least two ways to fix that.\n\nFirst, we can use the comparison function, to compare both arguments as numbers. We just have to convert the arguments to integers, and then compare them.2\n\n```(defun my-sort-key-value-lines (beg end)\n(interactive \"r\")\n(save-excursion\n(save-restriction\n(narrow-to-region beg end)\n(goto-char (point-min))\n(sort-subr nil 'forward-line 'end-of-line\n(lambda ()\n(search-forward \":\" nil nil 1)\nnil)\nnil\n(lambda (a b)\n(< (string-to-number a)\n(string-to-number b)))))))\n```\n\nIn the code above we do that in the second lambda. The `nil` between them is the `end-of-key` function, which we don’t need to define because it’s the same as the `end-of-record` (represented by `'end-of-line` in the above code).\n\nThe simpler method, is to do the conversion in the previous lambda, and use the default comparison function. Which results in the third revision.2\n\n```(defun my-sort-key-value-lines (beg end)\n(interactive \"r\")\n(save-excursion\n(save-restriction\n(narrow-to-region beg end)\n(goto-char (point-min))\n(sort-subr nil 'forward-line 'end-of-line\n(lambda ()\n(search-forward \":\" nil nil 1)\n(string-to-number (buffer-substring (point) (point-at-eol))))))))\n```\n\nYou can now just drop this into your `~/.xemacs/init.el` and use the command `M-x my-sort-key-value-lines` to sort `key: value` lines, whenever you have the need. And this leaves us with the desired numerical sort order.\n\n```bar: 2\nbaz: 7\nfoo: 12\n```\n\n1 This code is GPL.\n\n2 This code can be considered WTFPL 2.0; at least the parts inside the lambdas and the rest is just boilerplate.\n\n## One thought on “XEmacs: Sorting Key-Value Lines by Value”\n\n1.",
null,
"myrkraverk says:\n\nWhen I first wrote this code, I was not aware of `M-x sort-numeric-fields`, described on Xah Lee’s page and mentioned in the on-line manual; however, I had spaces in my real world keys, and as of 21.4, `M-x sort-numerc-fields` is hard coded to use white space as field separators. I therefore had to code my own solution anyway.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
] | [
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null,
"http://0.gravatar.com/avatar/09392d9b44953476b9ee42f907803a33",
null
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https://freebooksummary.com/nteract-ons-of-charged-part-cles-w-th-matter-biology-essay-essay | [
"This material is available only on Freebooksummary\n\n# Nteract Ons Of Charged Part Cles W Th Matter Biology Essay",
null,
"The whole doc is available only for registered users OPEN DOC\n\nBook: The Natural\n\nTopics: Essay\n\nPages: 39 Words: 9553 Views: 284\nAccess Full Document\n\nPlease Sign Up\nto get full document.\n\nAccess Full Document\n\nPlease Sign Up\nto get full document.\n\nCharged atoms are cardinal to the medical usage of radiation. Even if the primary radiation is a photon beam, it is the charged atoms, known as secondary radiation in such instances, that cause the biological consequence, whether it is cell killing or other alterations that may finally bring on malignant neoplastic disease. In fact, charged atoms are frequently termed ionizing radiation, and photons (and neutrons) termed non-ionising or indirectly ionizing.\n\nThe coevals of X raies, i.e. bremsstrahlung, is a charged-particle interaction. Radiotherapy is sometimes delivered by primary charged atom beams, normally megavoltage negatrons, where negatron interactions with affair are evidently important (Mayles et al. 2007).\n\n## Collision Losingss\n\nCoulomb interactions with the edge atomic negatrons are the chief manner that charged atoms (negatrons, protons, etc.) lose energy in the stuffs and energies of involvement in radiation therapy. The atom creates a trail of ionizations and excitements along its way. Occasionally, the energy transportation to the atomic negatron is sufficient to make a alleged delta beam (or I?-ray) which is a (secondary) negatron with an appreciable scope of its ain. This is schematically illustrated in Figure 2.1 (Evans 1955, ICRU 1970).\n\nThe physical theoretical account of the Coulomb interaction between the fast charged atom and a edge negatron in the medium is shown in Figure 2.2. The negatron is assumed to be free, and its binding energy assumed to be negligible compared to the energy it receives. The entrance negatron is traveling at a velocity V in a way antonym to axis x. The primary atom imparts a net urge to the edge negatron in a way perpendicular to its way (Nahum 1985).\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.1.png\n\nFigure 2.1. Diagrammatic representation of the path of a charged atom in affair\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.2.png\n\nFigure 2.2. Interaction between a fast primary charged atom and a edge negatron\n\nUsing classical, non-relativistic hit theory, from Newton ‘s 2nd jurisprudence, (i.e. the alteration in impulse is equal to the impulse [ the clip built-in of the force ]) and from the Coulomb jurisprudence for the force between charged atoms, it can be shown that the energy transportation Q is given by:\n\n* MERGEFORMAT (.)\n\nwhere B, the distance of closest attack, is known as the impact parametric quantity; m is the mass of the negatron; omega is the charge on the primary atom; V is the speed of the primary atom; and the changeless, K is changeless looking in the Coulomb-force look (=8.9875A-109 Nm2C-2). It should be noted that the mass of the primary atom does non come in into Equation 3.1, which every bit applies to negatrons, protons (that both have z=1), and other heavier charged atoms. Equation 3.1 leads to the undermentioned classical look for the cross-section per negatron, derived function in the energy transportation Q:\n\n* MERGEFORMAT (.)\n\nThe full relativistic, quantum-mechanical cross-section for Coulomb interactions between free negatrons, due to Moller (1932), is:\n\n* MERGEFORMAT (.)\n\nwhere T is the negatron kinetic energy; Iµ=Q/T is the energy transportation in units of the negatron kinetic energy; I„=T/mec2 is the kinetic energy in units of the negatron remainder mass; V is the negatron speed.\n\nThe Moller look (Equation 2.3) is valid provided that the negatron energy is much greater than the adhering energies of the atoms in the medium. The binding energies besides set a lower bound to the energy transportation possible.\n\n## Collision Stoping Power\n\nThe happening of really frequent, little energy losingss along the way of any charged atom in affair leads of course to the construct of halting power, defined as the mean energy loss, dE, per unit distance, Ds, along the path of the atom. This is normally expressed as the mass hit halting power, written (1/I?) (dE/ds) gap or Scol/I?, which is calculated from\n\n* MERGEFORMAT (.)\n\nwhere NA is Avogadro ‘s figure, and Z and A are atomic and aggregate figure, severally. Qmax for an negatron with kinetic energy E0 is equal to E0/2. The rating of the minimal energy transportation Qmin represents a major trouble.\n\nThe full quantum-mechanical look for the negatron mass hit halting power (Berger and Seltzer 1964; ICRU 1984a, 1984b) is given by\n\n* MERGEFORMAT (.)\n\nwhere\n\n* MERGEFORMAT (.)\n\nand the excess measures non defined so far are\n\nmec2, rest mass energy of the negatron\n\nI?=V/c\n\nrhenium, negatron radius (=e2/mec2=2.818A-10-15 m)\n\nI, average excitement energy\n\nI?, density-effect rectification\n\nThe average excitement energy or possible, I, is an norm of the passage energies Ei weighted by their oscillator strengths fi harmonizing to the followers:\n\n* MERGEFORMAT (.)\n\nIt is efficaciously the geometric mean of all the ionization and excitement potencies of the atoms in the engrossing medium; it is, of class, the more exact opposite number of Bohr ‘s average characteristic frequence that was discussed supra. In general, I can non be derived theoretically except in the simplest instances such as monatomic gases. Alternatively, it must be derived from measurings of halting power or scope. The most recent values of I, based mostly on experimental informations, are given in ICRU (1984b). For illustration the best current estimation of the I-value for H2O is 75.0 electron volt. By and large, the I-value additions as Z additions (Table 2.1).\n\nTable 2.1. Average excitement energies, I, and other measures relevant to the rating of the hit halting power of selected human tissues and other stuffs of\n\ndosimetric involvement\n\nMaterial\n\nI (electron volt)\n\nDensity (g cm-3)\n\nAdipose tissue (ICRP)\n\n63.2\n\n0.558468\n\n0.920\n\nAir (prohibitionist)\n\n85.7\n\n0.499190\n\n1.205A-10-3\n\nBone, compact (ICRU)\n\n91.9\n\n0.530103\n\n1.850\n\nBone, cortical (ICRP)\n\n106.4\n\n0.521299\n\n1.850\n\nFerrous-sulphate dosemeter solution\n\n76.3\n\n0.553282\n\n1.024\n\nLithium fluoride\n\n94.0\n\n0.462617\n\n2.635\n\nMuscle, skeletal (ICRP)\n\n75.3\n\n0.549378\n\n1.040\n\nMuscle, striated (ICRU)\n\n74.7\n\n0.550051\n\n1.040\n\nPhotographic emulsion\n\n331.0\n\n0.454532\n\n3.815\n\nPMMA (lucite, Lucite)\n\n74.0\n\n0.539369\n\n1.190\n\nPolystyrene\n\n68.7\n\n0.537680\n\n1.060\n\nWater (liquid)\n\n75.0\n\n0.555087\n\n1.000\n\nThe indispensable characteristics of the mass hit halting power are retained in the undermentioned simplified look:\n\n* MERGEFORMAT (.)\n\nComparing this look with Equation 2.5, the addition at diminishing subrelativistic energies due to the (1/V2) factor can be identified. This is merely explained by the fact that slow negatrons spend more clip traveling past an atom than make fast 1s, and accordingly the urge is greater and therefore more energy is lost. At relativistic energies, there is a more gradual addition in the fillet power which is known as the relativistic rise (Figure 2.3 Nahum (1985)).\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.3.png\n\nFigure 2.3. Collision energy loss as a map of negatron kinetic energy\n\n## Density Effect\n\nThe denseness or polarisation consequence (Fermi 1940; Sternheimer 1961; ICRU 1984a) reduces the value of Scol at relativistic energies in condensed media via the term I? in Equation 2.5 and Equation 2.8. It is connected to the relativistic rise in the halting power. If the fillet medium has a high denseness, (i.e. condensed media as opposed to gases) so the electric field seen by the atoms distant from the fast atom path is reduced due to the polarisation of the intervening atoms (as illustrated in Figure 2.4 (Nahum 1985)). Consequently, the part of these distant hits to the halting power will be reduced. This decrease in hit halting power is known as the polarisation or denseness consequence.\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.4.png\n\nFigure 2.4. Conventional account of the mechanism of the denseness consequence\n\nFigure 2.5 shows the fluctuation of Scol/I? with energy for air and H2O, two substances with similar atomic composings and similar I-values. The relativistic rise in the hit halting power in the condensed medium is much less marked compared to that in the gas because of the denseness consequence. Consequently, the ratio of mass halting powers, H2O to air, is strongly energy dependent above around 0.5 MeV (Nahum 1983).\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.5.png\n\nFigure 2.5. The fluctuation of mass hit halting power with negatron kinetic energy for air and for H2O\n\n## Electron Stopping-Power Data for Substances of Medical Interest\n\nTable 2.1 lists the relevant parametric quantities for assorted human tissues and some other substances of dosimetric involvement taken from ICRU (1984a) every bit good as for H2O as a comparing. It can be seen that the I-values all autumn between 73 electron volts and 75 electron volt with the exclusion of adipose tissue (high H content) and bone with its high Ca content. In fact, the I-value is about relative to the average atomic figure. Given the similarity of the values of I,, and the (mass) denseness (the latter being involved in the density-effect rectification I?) in the tabular array, the values of (Scol/I?) must besides be really similar. This is really convenient, as it means that the electron energy loss over a given distance in the organic structure can be derived from that in H2O by merely multiplying by the denseness, presuming that radiation losingss are besides really similar.\n\nThe values of the (mass) stopping-power ratio, smed, air, for assorted substances of involvement in medical dosimetry, as a map of negatron kinetic energy in the megavoltage part are shown in Figure 2.6. The ratio is virtually independent of energy except for that of air; this is really convenient for dosemeter response rating and intervention planning intents. These medium-to-water stopping-power ratios are likely to happen direct application in the transition of Monte-Carlo-derived dose distributions in patients to water-equivalent doses (Siebers et al. 2000).\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.6.png\n\nFigure 2.6. Ratios of mass hit halting powers, medium to H2O, for assorted substances of medical and dosimetric involvement\n\nPostscript: Polystyrene, PMMA: Polymethyl methacrylate (Lucite).\n\n## Restricted Stoping Power\n\nIn restricted halting power, merely energy transportations below a certain value I” are included and it is calculated by puting Qmax equal to I” in Equation 2.4. The full look is once more given by Equation 2.5 and Equation 2.6, but with the F (I„) term modified to:\n\n* MERGEFORMAT (.)\n\nThe look for the restricted fillet power still includes the density-effect rectification factor I? .\n\n## Radiative Losses (Bremsstrahlung)\n\nThe acceleration of the negatrons in the strong electric field of a nucleus leads to the production of bremsstrahlung. The acceleration is relative to the atomic charge, Z, divided by the mass, m, of the traveling atom. The strength of radiation produced is so relative to (Z/m) 2. This is a comparatively unimportant energy loss mechanism below approximately 10 MeV in low-Z stuffs, and it is wholly negligible for heavy charged atoms. The cross-section, I?rad, for this wholly non-classical procedure is highly complicated. One important characteristic is that, really about:\n\n* MERGEFORMAT (.)\n\nTherefore, on norm, the losingss will be appreciably larger than for hits. This means that considerable energy-loss straggling due to radiation losingss can be expected.\n\n## Radiation Stoping Power\n\nIn an precisely correspondent manner to that for hit losingss in the old subdivision,\n\na radiative fillet power, (dE/ds) rad or Srad, and besides a mass radiation halting power\n\n(Srad/I?) can be defined. The general signifier of the mass radiative halting power for high energies (complete showing: I„ & gt; & gt; 1/I± Z1/3) is given by:\n\n* MERGEFORMAT (.)\n\nwhere a is the all right construction invariable (I± a‰? 1/137). From an review of Equation 2.11, it can be seen that the radiative halting power additions about linearly with kinetic energy in the MeV part, in contrast to the weak logarithmic energy dependance of the hit halting power. Approximately, it can be written:\n\n* MERGEFORMAT (.)\n\nwhere B is a really easy changing map of E and Z. The factor, Z2/A, causes an addition in Srad/I? for higher Z (Figure 2.7).\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.7.png\n\nFigure 2.7. A comparing of the mass radiative and mass hit halting powers, Srad/I? and Scol/I? severally, for C, Cu, and lead\n\n## Radiaton Yield\n\nA utile measure is the fraction of the initial negatron energy, E0, that is lost to bremsstrahlung in decelerating down to rest. This fraction is known as the Radiation Yield, I’ (E0), and is given by:\n\n* MERGEFORMAT (.)\n\nThe dependance of I’ (E0) on E0 and on Z is about additive, which closely corresponds to the relation between Srad/I?, E, and Z.\n\nThe radiation output is involved in ciphering the dosimetric measure, g, which is the fraction of energy transferred (by photons) to a medium in the signifier of negatron kinetic energy that is later re-radiated as bremsstrahlung.\n\n## Angular Distribution of Bremsstrahlung Photons\n\nThe angular distribution of the emitted photons is really strongly forward-peaked at relativistic negatron energies with a average value I? a‰? mc2/E where Tocopherol is the entire energy of the negatrons. This forward-peaking is the ground for the flattening filter in a additive gas pedal intervention caput.\n\n## Entire Stopping Power\n\nThe hit and radiative fillet powers are often summed to give the sum halting power, written (dE/ds) tot or Stot:\n\n* MERGEFORMAT (.)\n\nFigure 2.8 shows the entire mass halting power (labelled “ Entire Loss ”), aggregate hit halting power, and several restricted mass hit halting powers (I” = 10 keV, 1 keV and 100 electron volt) for H2O against electron kinetic energy E for values between 10-5 MeV and 104 MeV. It can be seen that Stot/I? varies easy with E over the energy scope of primary involvement in radiation therapy (from 1.937 MeV cm2 g-1 at 4 MeV to merely 2.459 MeV cm2 g-1 at 25 MeV).\n\nSeveral characteristics should be noted:\n\nRadiation losingss merely become of import above around 10 MeV in H2O\n\nThe relativistic rise in the hit losingss is little because of the denseness consequence\n\nCollision losingss restricted to I” & lt; 10 keV merely consequence in a modest decrease in halting power compared to the unrestricted Scol which emphasizes the predomination of really little losingss\n\nThe approximative value for the electronic halting power in H2O in the MeV part is about 2 MeV cm-1; the value in tissue is really similar.\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.8.png\n\nFigure 2.8. Mass halting power in H2O for negatrons\n\n## Energy-Loss Sidetracking\n\nIt is of import to recognize that halting power is an mean value for the energy loss per unit distance. Fluctuations will happen about this average value in any existent state of affairs. This gives rise to what is known as energy-loss straggling (Figure 2.9) (Berger and Wang 1988).\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.9.png\n\nFigure 2.9. Energy widening because of energy-loss straggling after the transition of a monoenergetic negatron beam (energy E0) through a thin absorber\n\n## Continuous-Slowing-Down-Approximation (CSDA) Scope\n\nCharged atoms lose energy in a quasi-continuous manner along their paths in affair, finally coming to rest. This means that, unlike photons with their exponential fading, charged atoms do hold a finite, moderately chiseled scope. Mathematically it has been found convenient to specify the alleged continuous-slowing-down-approximation (CSDA) scope, r0 in the undermentioned manner:\n\n* MERGEFORMAT (.)\n\nThis represents the mean pathlength travelled in coming to rest by a charged atom holding kinetic energy, E0. Note that for negatrons, as opposed to heavy charged atoms, this will ever be well greater than the mean incursion deepness because of the pronounced angular warps that negatrons suffer in decelerating down. The CSDA scope r0 is about relative to E0 in the curative energy scope because of the comparatively slow fluctuation of Stot in this energy scope.\n\n## Elastic Nuclear Dispersing\n\nWhen a charged atom base on ballss near to the atomic karyon, at a distance much smaller than the atomic radius, the Coulomb interaction will now be between the fast atom and the atomic charge instead than with one of the edge negatrons. In the instance of negatrons, this causes appreciable alterations of way, but about ne’er (with the exclusion of the bremsstrahlung procedure) any alteration in energy. The sprinkling is fundamentally elastic, the energy lost being the negligible sum required to fulfill impulse preservation between the really light negatron and the positively charged karyon.\n\nThis interaction procedure is basically Rutherford dispersing with differential cross-section:\n\n* MERGEFORMAT (.)\n\n## Application to an Electron Depth-Dose Curve\n\nFigure 2.10 illustrates the natural philosophies of negatron interactions as they apply to electron beams used in radiation therapy; it shows three different depth-dose curves obtained through Monte-Carlo simulation, matching to different estimates about negatron conveyance natural philosophies for a 30 MeV wide, monoenergetic, and parallel negatron beam in H2O (Nahum and Brahme 1985, Seltzer 1978).\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.10.png\n\nFigure 2.10. The consequence of assorted estimates on the negatron depth-dose curve for a wide, 30 MeV negatron beam in H2O (CSDA range r0 = 13.1cm)\n\nThe curve labelled “ CSDA straight in front ” corresponds to straight paths and shows the Bragg Peak, usually associated with heavy charged atoms; this highly simple estimate illustrates really clearly the behavior of the entire halting power Stot as the negatron energy bit by bit decreases with deepness. The gradual lessening in dosage with depth mirrors the lessening in entire halting power with falling negatron energy. At an energy stopping point to that of the negatron remainder mass (0.511 MeV), nevertheless, the hit halting power goes through a lower limit and so rises quickly (chiefly because of the 1/I? 2 term in Equation 2.5.\n\nThe “ CSDA multiple dispersing ” curve involves directional alterations through multiple sprinkling (Equation 2.16), but does non affect any secondary atom conveyance or any simulation of energy-loss straggling. The addition in dose off from the surface is wholly due to the increasing mean asynclitism of the negatron paths with deepness and the fact that the beam is wide (i.e. there is sidelong dispersing equilibrium); this is sometimes known as spread buildup). At around z/r0=0.7, the planar fluence starts to diminish as negatron paths begin to make their terminal. The maximal occurs as a consequence of a competition between the spread build-up and the lessening in the planar fluence because of negatrons making the terminal of their scope.\n\nThe “ No knock-on conveyance ” curve does non include the coevals and conveyance of knock-on negatrons or delta beams, but it does include radiative losingss (i.e. bremsstrahlung) and one sees the alleged bremsstrahlung tail beyond the practical scope. Besides, now the incline of the dose slump is much reduced; this is chiefly due to the incorporation of energy-loss straggling.\n\nFinally, the unlabeled full curve corresponds to a simulation including the full negatron conveyance natural philosophies (of most relevancy in this energy part). The consequence of imitating I?-ray conveyance is clearly seen in the build-up stopping point to the surface; this is correspondent to the much more marked one in megavoltage photon beams where the scopes of the mainly Compton negatrons are significantly greater than those of the preponderantly low-energy I?-rays.\n\n## INTERACTIONS OF PHOTONS WITH MATTER\n\nPhoton interactions are stochastic (i.e. random) by nature. Unlike negatrons, they may undergo a few, one, or no interactions as they pass through affair. In each interaction, secondary ionising atoms are created. These may be charged atoms (normally negatrons) or uncharged atoms (normally photons). The charged atoms deposit their energies near to the interaction site and contribute to the local energy deposition, whereas, secondary photons may be transported some distance before interacting.\n\nSecondary photons are of import because they contribute to the photon fluence indoors and around an irradiated organic structure and to dose when they interact and produce secondary negatrons. The comparative importance of secondary photons depends on the energies of the primary photons. In external beam therapy utilizing megavoltage beams, the dominant part to the captive doses within the patient is due to primary photons.\n\n## Photon Interaction Cross-Sections\n\nPhotons interact with assorted mark entities such as atomic negatrons, nuclei, atoms or molecules. The chance of interaction with a mark entity is normally expressed in footings of the cross-section I?. The type of mark for the interaction is marked, when necessary, by adding an index to I?. Therefore, eI? and aI? designate the cross-section per negatron and per atom, severally. The relation between them is given by as=ZA-es where Z is the atomic figure of the atom.\n\nPhoton interactions can be characterized as soaking up or sprinkling procedures. In a full soaking up procedure, the incoming photon loses all its energy and the energy is transferred to the mark entity. Secondary atoms are emitted during or later to the interaction. In a full sprinkling procedure, an incoming photon interacts with a mark entity and its way of gesture, energy and impulse may be changed because of this interaction. The photon, nevertheless, is non absorbed, and alterations of energy and impulse are governed by the Torahs of relativistic kinematics. The chief soaking up procedures are photoelectric (pe) soaking up, brace (brace), and three (trip) production. The chief sprinkling procedures are consistent (coh) and incoherent (incoh) sprinkling. Nuclear photo-effect (phn) is an soaking up procedure that is largely neglected but needs to be considered in some instances. The entire interaction crosssection, independent of which procedure occurs, is the amount of the cross-sections for the single procedures:\n\n* MERGEFORMAT (.)\n\nThe unit of cross-section is m2. Although it does non belong to the International System of Units, the barn (1 barn = 10-24 cm2 = 10-28 M2) is still often used.\n\nIn a sprinkling procedure, the distribution of scattered photons may non be isotropous, but may alternatively be anisotropic in some manner related to the way of the incoming photon and its polarization. In order to quantify such effects, the cross-section is regarded as a map of the solid angle I© in the way of the scattered photon and the construct of the differential crosssection dI?/dI© is introduced. The differential cross-section is defined in a manner correspondent to the entire cross-section with (dI?/dI©) dI© related to the chance that the photon spreads into solid angle dI©. It follows that:\n\n* MERGEFORMAT (.)\n\nwhere I? is the sprinkling (polar) angle, and I• is an azimuthal angle. In many state of affairss, the sprinkling will, on norm, have no azimuthal dependance, and the equation can so be written\n\n* MERGEFORMAT (.)\n\nEquation 2.19 may besides be written\n\n* MERGEFORMAT (.)\n\nwhere\n\n* MERGEFORMAT (.)\n\nThe measure dI?/dI? is besides referred to as a differential cross-section.\n\n## Photoelectric Absorption\n\nPhotoelectric soaking up is illustrated in Figure 2.11. In this procedure, an incoming photon interacts with an atom and is absorbed. An atomic negatron is ejected with kinetic energy T from one of the atomic shells. Its kinetic energy is given by:\n\n* MERGEFORMAT (.)\n\nHere, hI? is the energy of the interacting photon, and EB is the adhering energy of the atomic negatron. The procedure can non happen with a free negatron. The atom is needed in order to conserve impulse. Because of the heavy mass of the karyon, the energy transferred to the atom is negligible.\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.11.png\n\nFigure 2.11. Photoelectric soaking up\n\nIn general, the cross-section I?pe for photoelectric soaking up additions strongly with diminishing photon energy. Figure 2.12 shows this cross-section for lead. The cross-section displays a series of discontinuities at energies matching to the adhering energies of the negatrons in the atomic shells. These discontinuities are known as soaking up borders. Below the soaking up border, the photon does non hold sufficient energy to emancipate an negatron from the shell. At energies merely above the border, the photon has sufficient energy to emancipate the negatron. Therefore, the cross-section suddenly increases because the figure of negatrons that can take portion in the soaking up procedure additions. The soaking up border is most marked at the K shell in a high atomic figure stuff. The L-shell has three sub-shells and, correspondingly, three soaking up borders are seen in Figure 2.12 at the energies 13.04 keV, 15.20 keV, and 15.86 keV of the L sub-shells in lead. At energies above the K soaking up border, approximately 80 % of the interactions take topographic point in the K shell.\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.12.png\n\nFigure 2.12. The entire photoelectric soaking up cross-section for lead as a map of photon energy\n\nThe cross-section for photoelectric soaking up depends strongly on atomic figure. Above the K soaking up border, the cross-section per atom as a map of photon energy and atomic figure is about given by\n\n* MERGEFORMAT (.)\n\nThe cross-section additions as the 4th power of the atomic figure and is reciprocally relative to the 3rd power of photon energy. This points to the strong impact of this procedure at low photon energies, peculiarly, at high atomic Numberss.\n\nThe angular distribution of the photoelectrons is peaked at angles of Iˆ/2 to the forward way at low photon energies, but it becomes progressively frontward directed as the photon energy additions.\n\nAfter photoelectric soaking up, a vacancy is left in the atomic shell. This vacancy is later filled with an negatron from an outer shell. The energy released is equal to the difference in the binding energies of an negatron in the two shells (e.g. Ek-EL in a passage from the L to the K shell). The energy released is carried off either by the emanation of a photon or an negatron. The photon is known as a characteristic X ray because of its fixed energy determined by the atomic figure of the atom and the shells involved. Characteristic X raies are isotropically emitted. At energies instantly above an soaking up border, they may transport a significant fraction of the incident photon energy. Electrons emitted after electronic rearrangement are known as Auger negatrons. They are besides isotropically emitted. The kinetic energy of an Auger negatron is equal to the energy released in the passage minus its binding energy.\n\n## Compton Interaction and Scattering Procedures\n\nIn a sprinkling procedure, the photon changes its way of gesture. If its energy is reduced, the sprinkling is called incoherent. The sprinkling may besides happen without energy loss and is so referred to as coherent sprinkling. The footings elastic and Rayleigh dispersing have besides been used for this procedure.\n\nFor photon energies that well exceed the binding energies of the atomic negatrons, the kinematics of the sprinkling procedure is normally described by sing the mark negatron to be free and at remainder at the minute of hit. In this instance, the sprinkling is incoherent because the photon will lose energy upon being scattered. At lower photon energies, the adhering energies of the atomic negatrons can non be neglected. The photon can so disperse from single edge negatrons (incoherent dispersing) or from all the edge electrons together, dispersing in stage (consistent sprinkling). In the latter instance, the whole atom takes portion in the sprinkling procedure to conserve impulse.\n\n## Incoherent Dispersing\n\nIn incoherent sprinkling, the photon transportations portion of its energy to an atomic negatron that is ejected from the atomic shell. The procedure was foremost described by Compton who assumed the negatron to be free and at remainder at the minute of hit. In this estimate, the procedure is besides known as Compton dispersing. The kinematics of Compton sprinkling is illustrated in Figure 2.13.\n\n* MERGEFORMAT (.)\n\nwhere I? = hI?/ (m0c2) and m0 is the remainder mass of the negatron.\n\nCalciferol: \\_Ph.DTEZFiguresFig2.13.png\n\nFigure 2.13. Dispersing angles and energies for Compton spread\n\nThe cross-section for Compton sprinkling is named after Klein and Nishina who foremost derived an look for its value. The differential Klein-Nishina cross-section per negatron is given by\n\n* MERGEFORMAT (.)\n\nAt low energies (hI?a†’0), this reduces to\n\n* MERGEFORMAT (.)\n\nThis cross-section is known as the classical Thomson differential cross-section. The entire Klein-Nishina cross-section per negatron may be obtained by incorporating Equation 2.25, replacing for hI? ‘ utilizing Equation 2.24. The consequence is:\n\n* MERGEFORMAT (.)\n\nThe differential Klein-Nishina cross-section is shown in Figure 2.14.\n\nCalciferol: \\_Ph.DTEZFiguresFig2.14.png\n\nFigure 2.14. Cross-sections for Compton dispersing from free negatrons\n\nThe influence of negatron binding on the incoherent dispersing cross-section is normally quantified by the incoherent dispersing map S (x, Z). The differential sprinkling cross-section for incoherent dispersing per atom is so given by\n\n* MERGEFORMAT (.)\n\nThe incoherent dispersing map is by and large assumed to be a map of the impulse transportation and the atomic figure, Z. It is tabulated in footings of the impulse transportation related measure x given by\n\n* MERGEFORMAT (.)\n\nwhere I» is the wavelength of the primary photon.\n\n## Coherent Dispersing\n\nIn coherent sprinkling, the photon is jointly scattered by the atomic negatrons. Basically, no energy is lost by the photon as it transfers momentum ten to the atom while being scattered through the angle I?. The sprinkling from the different negatrons is in stage, and the attendant angular warp is determined by an intervention pattern feature of the atomic figure of the atom. The differential cross-section for coherent sprinkling is obtained as the merchandise of the differential Thomson dispersing cross-section and the atomic signifier factor F squared\n\n* MERGEFORMAT (.)\n\nThe atomic signifier factor is, like the incoherent dispersing map, a cosmopolitan map of ten. It takes its maximal value in the forward way (I?=0) where F (0, Z) =Z. It decreases to zero as ten additions; with increasing impulse transportation ten, it gets progressively hard for all negatrons to disperse in stage without absorbing energy.\n\n## Pair and Triplet Production\n\nPair production is illustrated in Figure 2.15. In brace production, the photon is absorbed in the electric field of the karyon. An negatron (electron) -positron brace is created and emitted with the amount of their kinetic energies, T-+T+, being determined by the demand for preservation of energy\n\n* MERGEFORMAT (.)\n\nCalciferol: \\_Ph.DTEZFiguresFig2.15.png\n\nFigure 2.15. Pair production\n\nFrom the above Equation 2.31, it is clear that the procedure has a threshold value of 2m0c2 (1.02 MeV), the minimal energy required to make two negatrons. On norm, the electron-positron brace about every bit portions the kinetic energy available.\n\nThe procedure of brace production may besides happen in the electric field of an atomic negatron. The atomic negatron will flinch with sufficient energy to be ejected from the atomic shell. Three negatrons appear as a consequence of the interaction and, consequently, the procedure is called three production. Triplet production has an energy threshold at 4m0c2 (2.04 MeV).\n\nThe cross-section for brace production in the atomic field is zero below threshold. It so quickly increases with increasing energy and, good above threshold, varies about as the square of the atomic charge Z. The cross-section for three production, at energies above threshold, about varies as Z.\n\n## Nuclear Photoeffect\n\nWhen the photon energy exceeds that of the adhering energy of a nucleon, it can be absorbed in a atomic reaction. As a consequence of the reaction, one or more nucleons (neutrons and/or protons) are ejected. The cross-section for the atomic photoeffect depends on both the atomic figure, Z, and the atomic mass, A, and therefore on the isotopic copiousness in a sample of a given component. The cross-section has an energy threshold, and it is shaped as a elephantine resonance extremum. The peak occurs between 5 and 40 MeV, depending on the component, and it can lend between 2 % (high-Z component) and 6 % (low-Z component) to the entire cross-section.\n\n## The Entire Atomic Cross-Section\n\nThe entire atomic cross-section and its partial cross-sections are given in Figure 2.16 for the elements C (Z=6) and lead (Z=82).\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.16.png\n\nFigure 2.16. The entire and partial cross-sections for C (a) and lead (B) for photon energies from 10 keV to 100 MeV\n\n## Macroscopic Behaviour\n\nPhotons incident on an absorber will either interact in it (bring forthing secondary negatrons and/or scattered photons) or else base on balls through it without interacting. The figure of photons transmitted undisturbed through an absorber of thickness T of a given component and denseness can, for mono-energetic photons, be derived in the undermentioned manner (see Figure 2.17). The figure of primary photons, dI¦, interacting in a thin bed dx at deepness ten is relative to the thickness of the bed and the figure of photons incident on the bed so that\n\n* MERGEFORMAT (.)\n\nThe additive fading coefficient m is a belongings of the stuff and depends on photon energy. The subtraction mark indicates that photons are removed from the beam. Integrating the equation from x=0 to x=t gives the figure, I¦ (T), of primary photons that are transmitted through the absorber. This figure decreases exponentially with increasing thickness T harmonizing to\n\n* MERGEFORMAT (.)\n\nwith I¦0=I¦ (0) the figure of incident photons.\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.17.png\n\nFigure 2.17. Calculation of photon transmittal through a slab of affair\n\nThe additive fading coefficient is the chance per unit length for interaction and is related to the entire atomic cross-section, I?tot, through the relation\n\n* MERGEFORMAT (.)\n\nwhere N is the figure of mark entities per unit volume. It is given by N= (NA/A) I? .\n\nThe mass fading coefficient, I?/I?, obtained by spliting I? with I?, is independent of the existent denseness of the absorber and makes this measure attractive for usage in digests.\n\nThe incursion power of a photon beam is normally expressed by agencies of the average free way. This is defined as the mean distance,, travelled by the photon before it interacts. For mono-energetic photons it is given by\n\n* MERGEFORMAT (.)\n\n## Principles and Basic Concepts in Radiation Dosimetry\n\nThe accurate finding of captive dosage is important to the success of radiation therapy because of the comparatively steep sigmoidal dose-response curves for both tumors control and normal-tissue harm. There are many different stairss involved in the finding of the absorbed dose distribution in the patient. One of the most of import of these involves measurings with a sensor (frequently termed a dosemeter) in a apparition (frequently H2O, sometimes water-like plastic) placed in the radiation field.\n\n## Absorbed Dose\n\nICRU (1980, 1998) defines absorbed dose as the quotient of by, where is the average energy imparted by ionizing radiation to affair of mass:\n\n(2.1)\n\nThe unit of captive dosage is the “ grey ” which is 1 Joule per kg (J kg-1); the old unit is the “ rad ” which is 10-2 grey (sometimes referred as a centigray).\n\nThe energy imparted, Iµ, by ionizing radiation to the affair in a volume is defined by ICRU (1980, 1998) as:\n\n(2.2)\n\nwhere\n\nis the amount of the energies (excepting remainder mass energies) of all those charged and uncharged ionising atoms that enter the volume (known as the beaming energy)\n\nis the amount of the energies (excepting remainder mass energies) of all those charged and uncharged ionising atoms that leave the volume, and\n\nis the amount of all alterations (lessenings: positive mark, additions: negative mark) of the remainder mass energy of karyon and simple atoms in any atomic transmutations that occur in the volume.\n\nFigure 6.2 illustrates the construct of energy imparted. In the left portion of the figure which represents a Compton interaction within the volume V, the energy imparted is given by\n\n(2.3)\n\nwhere is the kinetic energy of the charged particle-of initial kinetic energy T-upon go forthing the volume V. Note that the photon does non look as this is non emitted within the volume V. The term is non involved here.\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.1.png\n\nFigure 2.1. An illustration of the construct of the energy imparted to an simple volume by radiation\n\nThe volume on the right in the figure involves I?-ray emanation () from a radioactive atom, brace production (kinetic energies T1 and T2), and obliteration radiation as the antielectron comes to rest. The energy imparted in this instance is given by\n\n(2.4)\n\nwhere\n\n(2.5)\n\nIn Equation 2.4, the nothing is the Rin term; ‘1.022 MeV ‘ is the Rout term consisting the two obliteration I?-rays. The I?-ray energy arises from a lessening in the remainder mass of the karyon; the term is due to the creative activity of an electron-positron brace, i.e. an addition in rest mass energy. Finally the term is due to the obliteration of an negatron and a antielectron.\n\n## Kerma and Exposure\n\nThe measure kerma, can be thought of as a measure towards absorbed dose. It is conceptually really near to exposure, the first radiation measure to be officially defined (Greening 1981). All practising infirmary physicists will come across kerma (in air in a cobalt-60 I?-ray beam) in the context of graduating ionization Chamberss at a National Standards Laboratory. The formal definition (ICRU 1980, 1998) follows:\n\nThe kerma, K, is the quotient dEtr by diabetes mellitus, where dEtr is the amount of the initial kinetic energies of all the charged ionising atoms liberated by uncharged ionising atoms in a stuff of mass diabetes mellitus:\n\n(2.6)\n\nThe units of kerma are the same as for absorbed dosage, i.e. J kg-1 or grey (Gy). Kerma applies merely to indirectly ionizing atoms which, for our intents, about ever average photons, although neutrons besides fall into this class.\n\nExposure is conceptually closely related to air kerma. Exposure, normally denoted by X, is the quotient of dQ by diabetes mellitus where dQ is the absolute value of the entire charge of the ions of one mark produced in air when all the negatrons (electrons and antielectrons) liberated by photons in air of mass diabetes mellitus are wholly stopped in air. Until the late seventiess, all ionization Chamberss were calibrated in footings of exposure; later this was replaced by air kerma.\n\nFigure 2.2 illustrates the construct of kerma (and exposure). It is the initial kinetic energies that are involved; the eventual destiny of the charged atoms (i.e. if they do or make non go forth the simple volume), does non impact kerma. In the volume in the figure, the initial kinetic energies of the two negatrons labelled e1 contribute to the kerma, as both were generated in the volume. The fact that one of these negatrons leaves the volume with a residuary kinetic energy T is irrelevant. None of the kinetic energy of the negatron come ining the volume with kinetic energy T contributes to kerma as this negatron was generated outside the volume.\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.2.png\n\nFigure 2.2. Illustration of the constructs of kerma and exposure\n\nIt is of import to gain that kerma includes the energy that the charged atoms will finally re-radiate in the signifier of bremsstrahlung photons. Kerma can be partitioned as follows (Attix 1979):\n\n(2.7)\n\nwhere degree Celsius refers to hit losingss and R to radiation losingss. The hit kerma Kc is related to the (entire) kerma by\n\n(2.8)\n\nThe measure g is the fraction of the initial kinetic energy of the negatrons that is re-radiated as bremsstrahlung (in the peculiar medium of involvement).\n\nExposure and air kerma can be related to each other as follows. Multiplying the charge dQ by the average energy required to bring forth one ion brace, divided by the negatron charge, i.e. W/e, yields the hit portion of the energy transferred, i.e. and hence\n\n(2.9)\n\nor\n\n(2.10)\n\n## Atom Fluence\n\nTo cipher absorbed dose we require measures that describe the radiation field; these are known as ‘field measures ‘ (Greening 1981). Particle fluence is a really of import basic measure, affecting the figure of atoms per unit country. The construct is illustrated in Figure 2.3.\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.3.png\n\nFigure 2.3. Word picture of the radiation field at a point P in footings of the radiation tracking a sphere centred at P\n\nLet N be the outlook value of the figure of atoms striking a finite sphere environing point P (during a finite clip interval). If the domain is reduced to an minute one at P with a cross sectional country of district attorney, so the fluence I¦ is given by\n\n(2.11)\n\nwhich is normally expressed in units of m-2 or cm-2 (ICRU 1980, 1998). Fluence is a scalar quantity-the way of the radiation is non taken into history.\n\nWe will be run intoing fluence, derived function in energy, frequently written as I¦E:\n\n(2.12)\n\nin which instance the (entire) fluence is given by:\n\n(2.13)\n\nIt should be noted that fluence can besides be expressed as the quotient of the amount of the path lengths I”s of the atoms traversing the simple domain and the volume of the domain (Chilton 1978):\n\n(2.14)\n\nThis signifier is highly utile when sing alleged pit integrals which involve measuring the fluence averaged over a volume.\n\n## Energy Fluence\n\nEnergy fluence is merely the merchandise of fluence and atom energy. Let R be the outlook value of the entire energy (excepting remainder mass energy) carried by all the N atoms in Figure 6.4. Then the energy fluence I? is given by (ICRU 1980, 1998)\n\n(2.15)\n\nIf merely a individual energy E of atoms is present, so and.\n\nIn a similar manner, one can specify the above measures per unit of clip, i.e. fluence rate and energy fluence rate.\n\n## Planar Fluence\n\nPlanar fluence is the figure of atoms traversing a fixed plane in either way (i.e., summed by scalar add-on) per unit country of the plane. One can besides specify a vector measure matching to net flow but this is of small usage in dosimetry because it requires scalar, non vector, add-on of the effects of single atoms.\n\nPlanar fluence is a peculiarly utile construct when covering with beams of charged atoms. In certain state of affairss, e.g. at little deepnesss in a parallel negatron beam, one can state that the planar fluence remains changeless as the deepness increases; whereas, fluence by and large increases due to the alteration in way of the negatron paths. The difference between the two measures is illustrated in Figure 2.4.\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.4.png\n\nFigure 2.4. Conventional 2D illustration of the construct of two-dimensional fluence. The figure of atoms traversing the two horizontal lines (of equal length) is the same. This shows that the planar fluence in the original beam way remains changeless, whereas the fluence, illustrated by the entire path length in the two circles, is much greater on the downstream side of the dispersing medium.\n\n## Relation between Fluence and Kerma\n\nSee the conventional Figure 2.5 demoing N photons, each of energy E, traversing sheer a thin bed (of stuff med) of thickness deciliter and country district attorney. To pull out energy from atom paths and reassign it to the medium, we require an interaction coefficient. We can utilize the ICRU (1980, 1998) definition of the mass energy-transfer coefficient I?tr/I?:\n\n(2.16)\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.5.png\n\nFigure 2.5. Illustration of N radiation paths (photons) of energy E traversing a thin bed of stuff med, thickness deciliter, country district attorney, mass diabetes mellitus, denseness I?\n\nIdentifying the fraction of incident radiant energy dRtr/R as dEtr/ (NE), and doing a simple rearrangement, we have\n\n(2.17)\n\nThen, spliting both sides by the mass of the bed diabetes mellitus and rearranging once more:\n\n(2.18)\n\nReplacing diabetes mellitus by I?dV we can compose:\n\n(2.19)\n\nwhere we recognise the left manus side as kerma (in medium med) and the look in the square brackets as the amount of the path lengths divided by the volume, ensuing in\n\n(2.20)\n\nor in footings of energy fluence I?:\n\n(2.21)\n\nNote that the perpendicular incidence in Figure 2.5 was assumed for simpleness; Equation 2.20 and Equation 2.21 are valid for arbitrary angles of incidence.\n\nUp to this point, we have confined ourselves to atoms traversing the thin bed with a individual energy. In the more practical instance of a spectrum of energies, described by the fluence derived function in energy, I¦E, we evaluate the Kmed from:\n\n(2.22)\n\nwhere the energy dependance of (I?tr/I?) med has been shown explicitly. Therefore, we have now arrived at a relationship linking kerma and fluence for photons.\n\nIt will besides be necessary to cipher the hit kerma, Kc from photon fluence. To make this, one replaces the mass energy transportation coefficient I?tr/I? by the mass energy soaking up coefficient I?en/I?, where energy absorbed is defined to except that portion of the initial kinetic energy of charged atoms converted to bremsstrahlung photons. The two coefficients are related by:\n\n(2.23)\n\nwhich is of course the same factor that relates Kc and K. It hence follows that:\n\n(2.24)\n\nand, likewise, for the built-in over I¦E in the instance of a spectrum of incident photons:\n\n(2.25)\n\n## Relation between Kerma and Absorbed Dose\n\nHaving established a relationship between kerma and fluence in the old subdivision, if absorbed dosage can be related to kerma so a relationship will eventually be established between absorbed dosage and fluence for photons. However, the absorbed dose Dmed in medium med concerns the (mean) value of energy imparted to an simple volume, whereas kerma concerns energy transferred as the charged atoms can go forth the simple volume (or thin bed), taking a fraction of the initial kinetic energy with them. This is illustrated in Figure 2.6. Note that the measure denoted in the figure by is the net energy transferred and excludes that portion of the initial kinetic energy converted into bremsstrahlung photons. It is equal to as we have seen above.\n\nIn Figure 6.7, allow the energy imparted to the bed be denoted by 3, the (cyberspace) kinetic energy go forthing the bed be denoted by and the (cyberspace) kinetic energy come ining the bed on charged atoms be denoted by. Then, from Equation 6.2, we have:\n\n(2.26)\n\nIf now the negatron path that leaves the bed is replaced by an indistinguishable path that enters the bed we can compose:\n\n(2.27)\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.6.png\n\nFigure 2.6. Conventional illustration of how secondary negatrons created by photons can reassign (kinetic) energy to a thin bed, leave that bed, and besides enter the bed from outside. If the energy go forthing the bed () is precisely replaced by energy entrance (), so charged atom equilibrium is said to be and absorbed dosage can be equated to collision kerma. This is schematically represented on the figure, where the full length of the path of a secondary negatron is precisely equal to the partial paths of negatrons (in bold) within the bed bounds.\n\nIt so follows that:\n\n(2.28)\n\nThe equality between energy go forthing and energy come ining on charged atoms is known as charged atom equilibrium (CPE). This can be realised under certain fortunes.\n\nDividing both sides of Equation 6.27 by the mass of the bed or volume component, and\n\naltering from stochastic to mean measures, we can compose:\n\n(2.29)\n\nwhich is a really of import consequence: under the particular status of charged atom equilibrium, the absorbed dosage is equal to the hit kerma.\n\nConsequently, replacing kerma by absorbed dose in Equation 2.24 and Equation 2.25, for monoenergetic photons it follows that\n\n(2.30)\n\nand, likewise, for the built-in over I¦E in the instance of a spectrum of incident photons:\n\n(2.31)\n\nEquation 6.30 and Equation 6.31 are really of import relationships in radiation dosimetry.\n\n## Charged Particle Equilibrium\n\nCharged atom equilibrium (CPE), besides known as electronic equilibrium, is said to be in a volume V in an irradiated medium if each charged atom of a given type and energy go forthing V is replaced by an indistinguishable atom of the same energy come ining V.\n\nThere are, nevertheless, many state of affairss where there is a deficiency of CPE. Strictly talking, it is impossible to split the deficiency of CPE into separate constituents (i.e. longitudinal or sidelong disequilibrium). However, this differentiation may be utile for a better apprehension, particularly for high energy photons where the secondary negatrons are chiefly peaked frontward.\n\nFigure 2.7 illustrates schematically how charged atom equilibrium can really be achieved in a photon beam. Naturally, the figure is a immense over-simplication as in pattern there will be a whole spectrum of secondary negatron energies and waies. However, the statements are non basically altered by demoing merely one negatron get downing in each voxel, (labelled A to G in the figure) and going in a consecutive line. In each voxel, one negatron is generated and hence, the kerma will be changeless as it is assumed here that there is no or negligible photon fading. Merely a fraction of the negatron path deposits energy in voxel A; hence, the dosage is low and there is clearly no replacing for the portion of the path that leaves the volume. In voxel B a new negatron starts but here there is besides portion of the negatron path which started upstream in voxel A. Hence the dosage is higher than in voxel A; in voxel C the dosage is higher still. However, in voxel D, where the negatron which started in voxel A comes to rest, all the subdivisions of an negatron path are present. This means that the amount of the kinetic energies go forthing this volume must be precisely balanced by the amount of the kinetic energies come ining and staying in the volume, i.e. CPE is first attained in D. Subsequent voxels (E, F, G, etc.) will incorporate forms of negatron paths indistinguishable to those in D and, hence, CPE must besides use in these volumes. At the deepness of voxel D, the absorbed dosage is now equal to kerma (purely this ought to be collision kerma) and this will besides be the instance in the subsequent voxels E, F, G, etc., in the absence of photon fading.\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.7.png\n\nFigure 2.7. Diagram demoing the physique up to bear down atom equilibrium for the idealized instance of no fading of the photon beam and one heterosexual negatron path generated in each slab labelled A to G\n\nIn the state of affairs of a photon beam true CPE is purely impossible to accomplish in pattern. Attenuation means that the photon fluence does non stay changeless and, hence, the figure of secondary atoms (negatrons) get downing at different deepnesss besides can non be changeless. Table 2.1 illustrates the grade of photon fading in H2O thicknesses guaranting transeunt electronic equilibrium for photon beams of different energies. The grade of CPE failure increases as the photon energy additions. Consequently, experimental finding of exposure (now air kerma) is non attempted at maximal photon energies above about 3 MeV. Even below this energy, little corrections have to be made for the consequence of photon fading.\n\nTable 2.1. Approximate thickness of H2O required to set up transient charged atom equilibrium. For bremsstrahlung beams of different maximal energies; the concluding column gives about the fading of photons in that thickness of H2O.\n\nMaximum Energy of Photons (MeV)\n\nApproximate Thickness of Water for Equilibrium (millimeter)\n\nApproximate Photon Attenuation (%)\n\n0.3\n\n0.1\n\n0.03\n\n0.6\n\n0.4\n\n0.1\n\n1\n\n0.8\n\n0.3\n\n2\n\n2.5\n\n0.8\n\n3\n\n8\n\n2\n\n6\n\n15\n\n4\n\n8\n\n25\n\n6\n\n10\n\n30\n\n7\n\n15\n\n50\n\n9\n\n20\n\n60\n\n11\n\n30\n\n80\n\n13\n\nEven though rigorous CPE may non be, in many state of affairss it is really good approximated, such as at depths beyond the dose upper limit in media irradiated by photons below around 1 MeV in energy. At higher energies, the peers mark in Equation 2.29 can be replaced by a proportionality mark. This is so termed transient charged atom equilibrium (TCPE):\n\n(2.32)\n\nThe measures K, Kc and dose D are plotted as a map of deepness in Figure 6.8. The build-up part and the part of TCPE, where the D and Kc curves become parallel to each other are illustrated.\n\nIf radiative interactions and scattered photons are ignored it can be shown (Greening\n\n1981) that:\n\n(2.33)\n\nwhere I? is the common incline of the D, K and Kc curves and is the average distance the secondary charged atoms carry their energy in the way of the primary beams while lodging it as dosage (Attix 1986). This invariable of proportionality between dosage and hit kerma is normally denoted by I?, i.e.\n\n(2.34)\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.8.png\n\nFigure 2.8. Variation of kerma, K, hit kerma, Kc, and dose, D, with deepness in a beam of indirectly ionizing radiation such as a photon beam\n\nAnother state of affairs where the dosage is different from the hit kerma because CPE is non achieved is for beams with really little cross subdivisions. Figure 2.9 and Figure 2.10 show the consequences from a Monte-Carlo simulation (utilizing the EGSnrc codification) of depth-dose fluctuation for a cylindrical cobalt-60 I?-ray beam (energies 1.17 and 1.33 MeV, every bit weighted) incident sheer on the terminal of a cylinder of H2O with radius 20 centimeter and length 50 centimeter. The dosage has been scored along the cardinal axis in cylindrical columns of increasing rad, R, with 1 millimeters depth intervals for the first centimeter so 0.5 centimeters down to\n\n10 centimeter deepness (the consequences at greater deepnesss are non shown).\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.9.png\n\nFigure 2.9. Cobalt-60 I?-ray beam in H2O incident on a H2O cylinder (radius 20 centimeter, length 50 centimeter): depth-dose curves obtained from Monte-Carlo simulation (108 photon histories for each instance) when the tantamount beam radius, R, is 0.05 centimeter. The upper curve, with no conveyance of secondary negatrons, corresponds to the kerma. The lower curve includes full negatron conveyance and shows clearly the deficiency of CPE.\n\nIn Figure 2.9 the radius of the marking volume is really little, 0.05 centimeter. The upper curve corresponds to no negatron conveyance (which is achieved by puting the negatron conveyance cutoff to 0.5 MeV); therefore the measure scored corresponds precisely to (H2O) kerma. Upon exchanging negatron conveyance back on (by merely cut downing the negatron conveyance cutoff to a appropriately low value: 50 keV was chosen) the buildup of dosage in the first 3 millimeter, due to electron conveyance, can now be observed. At the same clip, nevertheless, the absolute value of the dosage has fallen by more than a factor 2. The radius of merely half a millimeter is in fact equal to merely a fraction of the scopes of the highest energy secondary negatrons created (virtually 100 % of these will be due to Compton interaction) which is insufficient for the constitution of CPE or, set another manner, the negatrons will preponderantly go forth the marking parts without being balanced by an equal figure come ining them, as the beam is so narrow. The captive dosage is accordingly much lower than the kerma in this really particular geometry.\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.10.png\n\nFigure 2.10. Depth-dose curves obtained for cobalt-60 I?-ray beams from Monte-Carlo simulation (108 photon histories) with tantamount rad, R, of 0.5 and 10.0 centimeter incident on a H2O cylinder (radius 20 centimeter, length 50 centimeter) with full negatron conveyance (ECUT-KE=50 keV). The increased dosage and decreased effectual fading for the larger radius is due to photon (Compton) spread.\n\nIn Figure 2.10, the tantamount beam rad, R, are 0.5 centimeters and 10.0 centimeter, severally. The depth-dose curve for the larger radius lies well above that for the smaller 1. The latter was chosen to be big plenty for the constitution of (partial) CPE. The decreased effectual fading at the larger beam radius is explained by an increasing build-up of Compton-scattered photons with deepness. The dependance of depth-dose curves on field size is a well-known and of import phenomenon in external-beam radiation therapy.\n\nIt must be stressed that for big collimated beams where CPE is achieved on the beam axis, it is still partly lost at the border, which contributes to enlarging the penumbra part.\n\n## Stoping Power and Cema\n\nMentioning back to Figure 2.5, consider that we now have N negatron paths incident sheer on the thin bed of medium med and thickness deciliter, etc. Alternatively of the mass energy-transfer coefficient for photons, for charged atoms the measure of relevancy is halting power, the energy lost per unit path length and we denote this energy by dEl to separate it from dEtr, which is used for indirectly ionizing radiation. We are interested in energy locally deposited in the thin bed so it is clearly appropriate to use the hit halting power, Scol, instead than the entire halting power as the latter would include the energy lost in the signifier of bremsstrahlung that would get away the thin bed (this is correspondent to the difference between kerma and hit kerma). Thus we can compose\n\n(2.35)\n\nNote that, unlike the correspondent Equation 2.17 for photons, we do non necessitate the energy of the atoms. Dividing both sides by the mass of the bed diabetes mellitus and showing this as I?dV on the right manus side, we obtain\n\n(2.36)\n\nwhich can be rearranged to give:\n\n(2.37)\n\nwhere, as in the instance of indirectly ionizing radiation, the measure in the square brackets is the fluence I¦ and therefore:\n\n(2.38)\n\nUntil late, there was no equivalent of kerma for the instance of charged atoms. However, the measure cema, converted energy per unit mass, was proposed by Kellerer et Al. (1992). ICRU (1998) defines cema as the energy lost by charged atoms, excepting secondary negatrons, in electronic hits in a mass diabetes mellitus of a stuff. ‘Secondary negatrons ‘ refer to the delta beams generated by the incident primary negatrons, and their kinetic energies have already been included in dEl. Therefore, cema is equal to dEl/dm and, accordingly, to the merchandise of electron fluence and mass hit halting power.\n\nCema is non needfully equal to absorbed dose, as some of the delta beams can go forth the thin bed, merely as secondary negatrons can make in the instance of the primary radiation being photons (Figure 2.6). To affect absorbed dose, it must follow that any charged atom kinetic energy go forthing the thin bed or simple volume is replaced by an precisely equal sum come ining the bed and being deposited in it or imparted to it. Consequently, it must be assumed that there is delta-ray equilibrium in order to be able to compare cema with absorbed dosage and therefore we can compose, for a medium m:\n\n(2.39)\n\nor, in the instance of polyenergetic negatron radiation:\n\n(2.40)\n\nwhere I¦E is the fluence, derived function in energy.\n\n## Delta-Ray Equilibrium\n\nNaturally, delta-ray equilibrium must ever be if charged atom equilibrium exists. However, where the primary radiation consists of charged atoms CPE can ne’er be achieved except in the instead particular instance of uniformly distributed B beginnings in a big medium. For the beams of high-energy negatrons used in radiation therapy, the energy of the primary negatrons decreases continuously with deepness and therefore there can non be equilibrium. However, the scopes of the delta beams are preponderantly highly short and about all the energy transferred through hit losingss, i.e. the cema, is deposited locally. One demand merely look at how close the ratio LI”/Scol is to integrity for really little values of I” to be convinced of the above (Figure 3.8), i.e. most of the hits result in really little energy losingss and negatrons with these low energies have highly short scopes. Therefore, deltaray equilibrium is by and large fulfilled to a high grade in media irradiated by negatron beams.\n\nOne state of affairs where delta-ray equilibrium is decidedly non a good estimate, nevertheless, is really near to the apparition surface in an negatron beam. The appreciable scope in the forward way of the most energetic delta beams consequences in a little but discernable delta-ray physique up (see Figure 3.16).\n\n## Cavity Theory\n\nWhen a measuring is made with a sensor, the sensor stuff will, in general, differ from that of the medium into which it is introduced. The signal from a radiation sensor will by and large be relative to the energy absorbed in its sensitive stuff and therefore to the absorbed dosage in this stuff, Ddet.\n\nThe sensor can be thought of as a pit introduced into the unvarying medium of involvement; this name stems from the fact that gas-filled ionization Chamberss dominated the development of the topic (Greening 1981) and the associated theory, which relates Ddet to Dmed, is known as pit theory. In its most general signifier, the purpose of pit theory is to find the factor fQ given by\n\n(2.41)\n\nfor an arbitrary sensor ‘det ‘, in an arbitrary medium ‘med ‘, and in an arbitrary radiation quality Q (photons or negatrons). Figure 2.11 illustrates the state of affairs schematically.\n\nDegree centigrades: UsersTurkayDesktopFiguresFig2.11.png\n\nFigure 2.11. The general state of",
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"# shinyPsychometric: simulating how experimental choices affect results uncertainty\n\nIn the R community, Shiny is trending a lot. Shiny is an R package for developing interactive web applications. I wanted to give it a try, and ended up with an app to simulate the impact of data collection choices on the outcome reliability of a curve fitting process.\n\nI wrote the code behind it for a Master class on research methods in Psychology. During this year-long class, students learn about research by practicing it under the supervision of a staff member. This was the first time they had to design and run a real experiment. I wrote the code to help them make informed choices. Indeed, subtle design choices (e.g. the number of items presented to a participant) may dramatically change the reliability of the results. The app is there to let them experiment before collecting their experimental data.\n\nExperimental data may be far less rock solid that they appear, and this is not obvious to most people. Even more importantly from an educational perspective, the app illustrates the need to quantify/display results uncertainty (see e.g. a paper by Vivien Marx in this year Nature Methods1).\n\n# The app\n\nThe app uses the psychometric function as an example of curve fitting and parameter estimation.",
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"Gustav Theodor Fechner (1801–1887), founder of the psychophysics\n\nThe psychometric function relates the strength of an external stimulus and the observer’s responses. Measuring it is a classical method in psychophysics since the mid 19th century and Gustav Fechner.\n\nA standard way to measure one’s psychometric function is to collect an observer’s responses to a number of predetermined stimulus intensities (aka the method of constant stimuli). As an example, a small light dot can be displayed at various light intensities on a steady dark background, and the observer is asked to report the dot presence. The various light intensities are chosen in order to go from ‘not visible at all’ to ‘detectable most of the time’. The order of presentation is randomized, and each level is repeated many times. The observer’s responses (in our example, the probability of detecting the light) are plotted against the stimulus strengths (the light intensity, or contrast), and a sigmoid curve is fitted. This curve is the observer’s psychometric function for light/contrast detection. For historical reasons, a threshold is often estimated: it is the stimulus strength corresponding to a given level of performance (e.g. the light contrast at which the dot is detected 75% of the time).\n\n## Experimental choices",
null,
"The app left panel includes most of the standard choices an experimenter faces when planning to measure a psychometric function with the method of constant stimuli.\n\nAs in any experiment, one faces choices when planning to measure a psychometric function: the number of tested stimulus intensities, their particular values, the number of times each intensity is presented, etc. Those choices may greatly affect the reliability of the curve parameters and threshold estimates. There is an abundance of methodological articles to guide the researcher (see e.g. the special issue of Perception & Psychophysics on ‘Threshold Estimation’). This app offers a way to directly experiment with design choices by simulation.\n\nMost standard choices are presented in the app left panel. For each parameter, the default values provided are representative of what can be found in the literature. The ‘performance range’ may deserve some explanation. It is the range (in theoretical probability of the participant’s answer, i.e. along the y axis) in which data will be sampled. The program computes the corresponding values on the stimulus intensity x axis, and places n simulated intensities at equal distances within this interval.\n\n## The simulation (or the gist of it)\n\nThe app runs a simulation. In this section, I’ll give you the gist of it. How it is coded is shown below.\n\nEverything starts with a predefined psychometric function. This is the ‘real’ psychometric function of our simulated participant. It is shown in red on the plot. When running an experiment, this function is obviously not known, and the experimenter aims at estimating it. However, we know it and randomly draw data from it. This way, we can evaluate how close are the values estimated on the simulated data to the ‘real’ ones from the original function.\n\nFor this example, I chose a cumulative Gaussian, with mean arbitrarily set to 0 and standard deviation to 1.\n\nSo here is the flow:\n\n1. the user modifies an experimental parameter, or hits the ‘Generate new data sample’ button. This sets the particular experimental choices for the simulation;\n2. At each chosen stimulus intensity, the observer’s answers are simulated by randomly sampling from a binomial distribution. It is a discrete probability function of the number of successes in a sequence of n independent trials. The probability of a correct response at each stimulus intensity is given by the ‘real’ known psychometric function;\n3. A curve is fitted to the data by maximum likelihood. The curve is a cumulative Gaussian with at least two parameters: its mean and standard deviation. A more general description of the psychometric function is presented below, and adds parameters. The estimated curve is displayed in grey on the upper part of the figure. You can compare it to the known function from which data has been sampled ;\n4. A threshold is estimated. It is simply the stimulus intensity corresponding to a selected performance level, the threshold criterion. The estimated threshold is shown in the lower part of the figure, along with its real value (in red). You can see how close/far they are from each other as you change the experimental parameters or generate a new sample.\n\n## Showing variability\n\nOne way to have a feeling of how variable are the results is simply to run the simulation multiple times. After having set your virtual experiment, you can click on the ‘Generate new sample’ button a few times, and watch how the fitted curve changes from one sample to another. It is surprising how variable they are, even with a fair amount of items. However, this give you only a hint of the outcome variability.\n\nA better way is to display the distribution of the results of multiple simulation runs. You can do this by increasing the number of generated data sets from 1 up to 500. Note that this may take some time to compute. Now all estimated curves are displayed on the top panel, with the alpha level adjusted to avoid over-plotting. The bottom panel shows a histogram and a kernel density of the resulting distribution of threshold estimates. Remember that all the data sets have been generated under the same conditions, and represent various possible outcomes of the same experiment. You can assess the impact of different data collection scenarios directly from your browser, just as it has been done more quantitatively in various publications.\n\nA classical and highly cited one (Wichmann and Hill, 20012) discusses the importance to estimate a third parameter of the psychometric function: the lapsing rate. In this more general formalization (see below) of the psychometric function, the function is allowed to asymptote at the maximal performance of",
null,
"$1-\\lambda$ instead of 1.",
null,
"$\\lambda$ captures the rate at which observers lapse, responding incorrectly irrespective of the stimulus intensity. The app includes the possibility to estimate",
null,
"$\\lambda$ by checking the box “Is the lapsing rate a free parameter?”. Whichmann et Hill concluded that failing to estimate it results in seriously biased psychometric function’s parameters. You can try it yourself by checking/unchecking the box.\n\n# Running the app\n\nThe app is available in two versions:\n\n• one that you can directly access in your browser (link here). It runs on a remote R server hosted and powered by Rstudio (It is currently under beta testing);\n• one that you run locally on your computer. It requires to have R and some additional packages installed. Installation instructions are listed at the app github repos.\n\n# Having a look at the code\n\nin this section, I will highlight some general aspects of the code. You can have a look at the whole code hosted on github, which is fully commented.\n\nA general form of the psychometric function is",
null,
"$\\Psi(x;\\alpha,\\beta,\\gamma,\\lambda) = \\gamma + (1 - \\gamma - \\lambda)F(x;\\alpha,\\beta)$\n\nwhere\n\n•",
null,
"$F$ is a sigmoid function mapping the stimulus intensity",
null,
"$x$ to the range [0;1], typically a cumulative distribution function of a distributional family. In this app, we use a cumulative Gaussian ;\n•",
null,
"$\\alpha$ and",
null,
"$\\beta$ are the location parameters of",
null,
"$F$. In the case of a cumulative Gaussian, they correspond to its mean (",
null,
"$\\mu$) and standard deviation (",
null,
"$\\sigma$); and\n•",
null,
"$\\gamma$ and",
null,
"$\\lambda$ adjust the range of",
null,
"$F$.",
null,
"$\\gamma$ is the base rate of performance in the absence of signal. In the case when the participant is forced to choose between alternatives,",
null,
"$\\gamma$ is the guessing rate. The psychometric function asymptotes at",
null,
"$1-\\lambda$.",
null,
"$\\lambda$ captures the rate at which observers lapse, responding incorrectly irrespective of the stimulus intensity.\n\nIn our R implementation, we use",
null,
"$exp(\\beta)$ instead of",
null,
"$\\beta$, to make sure that it is always positive.\n\n# parameters of the psychometric function used in this simulation\n# Rem: in the functions that define the psychometric function, we use exp(beta), instead of beta, to force beta to be non negative in the fitting procedure. so here we define beta as log(beta)\np <- c(0, log(1), 0.01, 0)\nnames(p) <- c('alpha', 'logbeta', 'lambda', 'gamma')\n\n# psychometric function\n# x are the stimulus intensities\n# p is a vector of 4 paramaters: alpha, log(beta), lambda and gamma\n# REM Note that we use exp(beta) in the function, to make sure that it is always positive.\n# Here we use a cumulative Gaussian (pnorm) as sigmoid function.\n# the output is a vector of accurate response probabilities\nppsy <- function(x, p) p + (1- p - p) * pnorm(x, p, exp(p))\n\nSimilar to the way the classical distribution functions are defined in R (e.g. pnorm(), qnorm() and rnorm()), we declared a quantile function qpsy() and a random generation rpsy() for the psychometric function.\n\n# function to convert probabilities from the psychometric function\n# to the underlying cummulative distribution.\n# Useful to compute a threshold\n# prob is the response probability from $\\Psi$ (in the range [gamma; 1-lambda])\n# the output is a vector of corresponding probabilities on $F$ (in the range [0;1])\nprobaTrans <- function(prob, lambda, gamma) (prob-gamma) / (1-gamma-lambda)\n\n# Quantile function (inverse of the psychometric function)\n# prob is a vector of probabilities\n# p is a vector of 4 paramaters: alpha, log(beta), gamma and lambda\n# Here we use a cumulative Gaussian (qnorm) as sigmoid function.\n# the output is a vector of stimulus intensities\nqpsy <- function(prob, p) qnorm(probaTrans(prob, p, p), p, exp(p))\n\n# random generation from the psychometric function\n# We sample random values from a binomial distribution with probabilities given by the ---known--- underlying psychometric function.\n# x are the stimulus intensities\n# p is a vector of 4 paramaters: alpha, log(beta), gamma and lambda\n# nObs is the number of observations for each stimulus level x\nrpsy <- function(x, p, nObs)\n{\nprob <- ppsy(x, p)\nrbinom(prob, nObs, prob)\n}\n\nThere are multiple ways to fit a psychometric function in R. We opted for direct (constrained) log likelihood maximization.\n\n# define the likelihood function\n# p is the parameters vector, in the order (alpha, log(beta), lambda, gamma)\n# df is a 3-column matrix with the stimulus intensities (df[, 1]), and the number of correct and incorrect responses (df[, 2:3])\n# opts is a list of options (set by the user of the shiny app)\nlikelihood <- function(p, df, opts) {\n\n# do we estimate lambda?\nif (opts$estimateLambda) { # if yes, we set gamma (the last parameter) to 0 psi <- ppsy(df[, 1], c(p, 0)) } else { # we set both gamma and lambda to 0 psi <- ppsy(df[, 1], c(p, 0, 0)) } # bayesian flat prior # implements the constraints put on lambda, according to Wichmann & Hill if (opts$estimateLambda && p > 0.06) {\n-log(0)\n} else {\n# compute the log likelihood\n-sum(df[,2]*log(psi) + df[,3]*log(1-psi))\n}\n}\n\n# A function to fit the psychometric function\n# There are multiple ways to do it. Here we directly maximize the log likelihood. But see e.g. Knoblauch and Maloney (2012) for some R code examples on how to use glm with special link functions.\n# df is a 3-column matrix with the stimulus intensities (df[, 1]), and the number of correct and incorrect responses (df[, 2:3])\n# opts is a list of options (set by the user of the shiny app)\nfitPsy <- function(df, opts)\n{\nif(opts$estimateLambda){ # estimating$\\alpha$,$\\beta$and$\\lambda$optim(c(1, log(3), 0.025), likelihood, df=df, opts=opts, control=list(parscale=c(1, 1, 0.001))) } else { # estimating$\\alpha$,$\\beta\\$\noptim(c(1, log(3)), likelihood, df=df, opts=opts)\n}\n}\n\nOther ways include\n\n• using glm() with the binomial error family and standard link (e.g. logit for a cumulative Gaussian);\n• using glm() with custom links from the psyphy package;\n• a Bayesian inference procedure provided in package PsychoFun;\n• etc.\n\nand are described in e.g.\n\n• Knoblauch, K., & Maloney, L. T. (2012). Modeling Psychophysical Data in R. New York: Springer-Verlag. doi:10.1007/978-1-4614-4475-6\n• Yssaad-Fesselier, R., & Knoblauch, K. (2006). Modeling psychometric functions in R. Behavior research methods, 38(1), 28–41. doi:10.3758/BF03192747\n\n1. Marx, V. (2013). Data visualization: ambiguity as a fellow traveler. Nature methods, 10(7), 613–615. doi:10.1038/nmeth.2530.\n2. Wichmann, F. A., & Hill, N. J. (2001). The psychometric function: I. Fitting, sampling, and goodness of fit. Perception & Psychophysics, 63(8), 1293–1313. doi:10.3758/BF03194544.\nAdvertisements"
] | [
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https://www.hepdata.net/search/?q=&collaboration=CMS&page=1&phrases=Proton-Proton+Scattering | [
"Showing 10 of 27 results\n\n#### Searches for additional Higgs bosons and for vector leptoquarks in $\\tau\\tau$ final states in proton-proton collisions at $\\sqrt{s}$ = 13 TeV\n\nThe collaboration\nCMS-HIG-21-001, 2022.\nInspire Record 2132368\n\nThree searches are presented for signatures of physics beyond the standard model (SM) in $\\tau\\tau$ final states in proton-proton collisions at the LHC, using a data sample collected with the CMS detector at $\\sqrt{s}$ = 13 TeV, corresponding to an integrated luminosity of 138 fb$^{-1}$. Upper limits at 95% confidence level (CL) are set on the products of the branching fraction for the decay into $\\tau$ leptons and the cross sections for the production of a new boson $\\phi$, in addition to the H(125) boson, via gluon fusion (gg$\\phi$) or in association with b quarks, ranging from $\\mathcal{O}$(10 pb) for a mass of 60 GeV to 0.3 fb for a mass of 3.5 TeV each. The data reveal two excesses for gg$\\phi$ production with local $p$-values equivalent to about three standard deviations at $m_\\phi$ = 0.1 and 1.2 TeV. In a search for $t$-channel exchange of a vector leptoquark U$_1$, 95% CL upper limits are set on the dimensionless U$_1$ leptoquark coupling to quarks and $\\tau$ leptons ranging from 1 for a mass of 1 TeV to 6 for a mass of 5 TeV, depending on the scenario. In the interpretations of the $M_\\mathrm{h}^{125}$ and $M_\\mathrm{h, EFT}^{125}$ minimal supersymmetric SM benchmark scenarios, additional Higgs bosons with masses below 350 GeV are excluded at 95% CL.\n\n313 data tables\n\nExpected and observed $95\\%\\text{ CL}$ upper limits on the product of the cross sections and branching fraction for the decay into $\\tau$ leptons for $gg\\phi$ production in a mass range of $60\\leq m_\\phi\\leq 3500\\text{ GeV}$, in addition to $\\text{H}(125)$. The central $68$ and $95\\%$ intervals are given in addition to the expected median value. In this case, $bb\\phi$ production rate has been profiled. The peak in the expected $gg\\phi$ limit is tribute to a loss of sensitivity around $90\\text{ GeV}$ due to the background from $Z/\\gamma^\\ast\\rightarrow\\tau\\tau$ events. Numerical values provided in this table correspond to Figure 10a of the publication.\n\nExpected and observed $95\\%\\text{ CL}$ upper limits on the product of the cross sections and branching fraction for the decay into $\\tau$ leptons for $bb\\phi$ production in a mass range of $60\\leq m_\\phi\\leq 3500\\text{ GeV}$, in addition to $\\text{H}(125)$. The central $68$ and $95\\%$ intervals are given in addition to the expected median value. In this case, $gg\\phi$ production rate has been profiled. Numerical values provided in this table correspond to Figure 10b of the publication.\n\nExpected and observed $95\\%\\text{ CL}$ upper limits on the product of the cross sections and branching fraction for the decay into $\\tau$ leptons for $gg\\phi$ production in a mass range of $60\\leq m_\\phi\\leq 3500\\text{ GeV}$, in addition to $\\text{H}(125)$. The central $68$ and $95\\%$ intervals are given in addition to the expected median value. In this case, $bb\\phi$ production rate has been fixed to zero. Numerical values provided in this table correspond to Figure 37 of the auxilliary material of the publication.\n\nMore…\n\n#### Measurement of the top quark pole mass using $\\mathrm{t\\bar{t}}$+jet events in the dilepton final state in proton-proton collisions at $\\sqrt{s}$ = 13 TeV\n\nThe collaboration\nCMS-TOP-21-008, 2022.\nInspire Record 2106483\n\nA measurement of the top quark pole mass $m_\\mathrm{t}^\\text{pole}$ in events where a top quark-antiquark pair ($\\mathrm{t\\bar{t}}$) is produced in association with at least one additional jet ($\\mathrm{t\\bar{t}}$+jet) is presented. This analysis is performed using proton-proton collision data at $\\sqrt{s}$ = 13 TeV collected by the CMS experiment at the CERN LHC, corresponding to a total integrated luminosity of 36.3 fb$^{-1}$. Events with two opposite-sign leptons in the final state (ee, $\\mu\\mu$, e$\\mu$) are analyzed. The reconstruction of the main observable and the event classification are optimized using multivariate analysis techniques based on machine learning. The production cross section is measured as a function of the inverse of the invariant mass of the $\\mathrm{t\\bar{t}}$+jet system at the parton level using a maximum likelihood unfolding. Given a reference parton distribution function (PDF), the top quark pole mass is extracted using the theoretical predictions at next-to-leading order. For the ABMP16NLO PDF, this results in $m_\\mathrm{t}^\\text{pole}$ = 172.94 $\\pm$ 1.37 GeV.\n\n10 data tables\n\nAbsolute differential cross section as a function of the rho observable at parton level.\n\nCovariance matrix for the total uncertainty for the measurement of the absolute differential cross section as a function of the rho observable at parton level.\n\nCovariance matrix for the statistical uncertainty for the measurement of the absolute differential cross section as a function of the rho observable at parton level.\n\nMore…\n\n#### Measurement of differential cross sections for the production of a Z boson in association with jets in proton-proton collisions at $\\sqrt{s}$ = 13 TeV\n\nThe collaboration\nCMS-SMP-19-009, 2022.\nInspire Record 2078067\n\nA measurement is presented of the production of Z bosons that decay into two electrons or muons in association with jets, in proton-proton collisions at a centre-of-mass energy of 13 TeV. The data were recorded by the CMS Collaboration at the LHC with an integrated luminosity of 35.9 fb$^{-1}$. The differential cross sections are measured as a function of the transverse momentum ($p_\\mathrm{T}$) of the Z boson and the transverse momentum and rapidities of the five jets with largest $p_\\mathrm{T}$. The jet multiplicity distribution is measured for up to eight jets. The hadronic activity in the events is estimated using the scalar sum of the $p_\\mathrm{T}$ of all the jets. All measurements are unfolded to the stable particle-level and compared with predictions from various Monte Carlo event generators, as well as with expectations at leading and next-to-leading orders in perturbative quantum chromodynamics.\n\n70 data tables\n\nMeasured cross section as a function of exclusive jet multiplicity, $N_{\\text{jets}}$, and breakdown of the relative uncertainty.\n\nBin-to-bin correlation in the measured cross section as a function of exclusive jet multiplicity, $N_{\\text{jets}}$.\n\nMeasured cross section as a function of the rapidity absolute value of the first jet, $|y(\\text{j}_1)|$, and breakdown of the relative uncertainty.\n\nMore…\n\n#### Search for resonances decaying to three W bosons in proton-proton collisions at $\\sqrt{s}$ = 13 TeV\n\nThe collaboration Tumasyan, Armen ; Adam, Wolfgang ; Andrejkovic, Janik Walter ; et al.\nPhys.Rev.Lett. 129 (2022) 021802, 2022.\nInspire Record 2015402\n\nA search for resonances decaying into a W boson and a radion, where the radion decays into two W bosons, is presented. The data analyzed correspond to an integrated luminosity of 138 fb$^{-1}$ recorded in proton-proton collisions with the CMS detector at $\\sqrt{s} =$ 13 TeV. One isolated charged lepton is required, together with missing transverse momentum and one or two massive large-radius jets, containing the decay products of either two or one W bosons, respectively. No excess over the background estimation is observed. The results are combined with those from a complementary channel with an all-hadronic final state, described in an accompanying paper. Limits are set on parameters of an extended warped extra-dimensional model. These searches are the first of their kind at the LHC.\n\n11 data tables\n\nPost-fit distributions of the reconstructed $\\ell\\nu$+jets system ($m_{\\mathrm{j}\\ell\\nu}$, $m_{\\mathrm{jj}\\ell\\nu}$) in data and simulation for SR4.\n\nObserved upper limits at 95\\% \\CL on the signal cross section $\\times$ branching fraction as functions of the $m_{\\mathrm{W}_{\\mathrm{KK}}}$ and $m_{\\mathrm{R}}$ resonance masses after combinign with an analysis of the all-hadronic final state.\n\nExpected median lower limit contour on the $m_{\\mathrm{W}_{\\mathrm{KK}}}$ and $m_{\\mathrm{R}}$ plane after combinign with an analysis of the all-hadronic final state.\n\nMore…\n\n#### Version 3 Measurement of the inclusive and differential $\\mathrm{t\\bar{t}}\\gamma$ cross sections in the dilepton channel and effective field theory interpretation in proton-proton collisions at $\\sqrt{s}$ =13 TeV\n\nThe collaboration Tumasyan, Armen ; Adam, Wolfgang ; Andrejkovic, Janik Walter ; et al.\nJHEP 05 (2022) 091, 2022.\nInspire Record 2013377\n\nThe production cross section of a top quark pair in association with a photon is measured in proton-proton collisions in the decay channel with two oppositely charged leptons (e$^\\pm\\mu^\\mp$, e$^+$e$^-$, or $\\mu^+\\mu^-$). The measurement is performed using 138 fb$^{-1}$ of proton-proton collision data recorded by the CMS experiment at $\\sqrt{s} =$ 13 TeV during the 2016-2018 data-taking period of the CERN LHC. A fiducial phase space is defined such that photons radiated by initial-state particles, top quarks, or any of their decay products are included. An inclusive cross section of 175.2 $\\pm$ 2.5 (stat) $\\pm$ 6.3 (syst) fb is measured in a signal region with at least one jet coming from the hadronization of a bottom quark and exactly one photon with transverse momentum above 20 GeV. Differential cross sections are measured as functions of several kinematic observables of the photon, leptons, and jets, and compared to standard model predictions. The measurements are also interpreted in the standard model effective field theory framework, and limits are found on the relevant Wilson coefficients from these results alone and in combination with a previous CMS measurement of the $\\mathrm{t\\bar{t}}\\gamma$ production process using the lepton+jets final state.\n\n64 data tables\n\nObserved and predicted event yields as a function of $p_{T}(\\gamma)$ in the $e\\mu$ channel, after the fit to the data.\n\nObserved and predicted event yields as a function of $p_{T}(\\gamma)$ in the $ee$ channel, after the fit to the data.\n\nObserved and predicted event yields as a function of $p_{T}(\\gamma)$ in the $\\mu\\mu$ channel, after the fit to the data."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.78207487,"math_prob":0.99241406,"size":369579,"snap":"2022-40-2023-06","text_gpt3_token_len":84251,"char_repetition_ratio":0.19107354,"word_repetition_ratio":0.96920055,"special_character_ratio":0.22288604,"punctuation_ratio":0.088811725,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9964924,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-03T07:31:58Z\",\"WARC-Record-ID\":\"<urn:uuid:5f67e2fe-6647-41b1-8f43-61e313f9d75f>\",\"Content-Length\":\"1049186\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1707c0f6-746f-4b0c-8dc2-2ccb89a1b40c>\",\"WARC-Concurrent-To\":\"<urn:uuid:1884a7e5-39e2-450d-8a5e-d898a11febc5>\",\"WARC-IP-Address\":\"188.185.30.5\",\"WARC-Target-URI\":\"https://www.hepdata.net/search/?q=&collaboration=CMS&page=1&phrases=Proton-Proton+Scattering\",\"WARC-Payload-Digest\":\"sha1:ZCNKM2MRZU7H65HIHEOD54MDNGQP5Y3P\",\"WARC-Block-Digest\":\"sha1:AKXB4HMBBJRUZEUB7MCUNDDUQYUDXHF4\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500044.16_warc_CC-MAIN-20230203055519-20230203085519-00443.warc.gz\"}"} |
https://metanumbers.com/3714 | [
"## 3714\n\n3,714 (three thousand seven hundred fourteen) is an even four-digits composite number following 3713 and preceding 3715. In scientific notation, it is written as 3.714 × 103. The sum of its digits is 15. It has a total of 3 prime factors and 8 positive divisors. There are 1,236 positive integers (up to 3714) that are relatively prime to 3714.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 4\n• Sum of Digits 15\n• Digital Root 6\n\n## Name\n\nShort name 3 thousand 714 three thousand seven hundred fourteen\n\n## Notation\n\nScientific notation 3.714 × 103 3.714 × 103\n\n## Prime Factorization of 3714\n\nPrime Factorization 2 × 3 × 619\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 3714 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 3,714 is 2 × 3 × 619. Since it has a total of 3 prime factors, 3,714 is a composite number.\n\n## Divisors of 3714\n\n1, 2, 3, 6, 619, 1238, 1857, 3714\n\n8 divisors\n\n Even divisors 4 4 2 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 7440 Sum of all the positive divisors of n s(n) 3726 Sum of the proper positive divisors of n A(n) 930 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 60.9426 Returns the nth root of the product of n divisors H(n) 3.99355 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 3,714 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 3,714) is 7,440, the average is 930.\n\n## Other Arithmetic Functions (n = 3714)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 1236 Total number of positive integers not greater than n that are coprime to n λ(n) 618 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 520 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 1,236 positive integers (less than 3,714) that are coprime with 3,714. And there are approximately 520 prime numbers less than or equal to 3,714.\n\n## Divisibility of 3714\n\n m n mod m 2 3 4 5 6 7 8 9 0 0 2 4 0 4 2 6\n\nThe number 3,714 is divisible by 2, 3 and 6.\n\n## Classification of 3714\n\n• Arithmetic\n• Abundant\n\n### Expressible via specific sums\n\n• Polite\n• Non-hypotenuse\n\n• Square Free\n\n• Sphenic\n\n## Base conversion (3714)\n\nBase System Value\n2 Binary 111010000010\n3 Ternary 12002120\n4 Quaternary 322002\n5 Quinary 104324\n6 Senary 25110\n8 Octal 7202\n10 Decimal 3714\n12 Duodecimal 2196\n20 Vigesimal 95e\n36 Base36 2v6\n\n## Basic calculations (n = 3714)\n\n### Multiplication\n\nn×i\n n×2 7428 11142 14856 18570\n\n### Division\n\nni\n n⁄2 1857 1238 928.5 742.8\n\n### Exponentiation\n\nni\n n2 13793796 51230158344 190268808089616 706658353244833824\n\n### Nth Root\n\ni√n\n 2√n 60.9426 15.4863 7.80657 5.17569\n\n## 3714 as geometric shapes\n\n### Circle\n\n Diameter 7428 23335.8 4.33345e+07\n\n### Sphere\n\n Volume 2.14592e+11 1.73338e+08 23335.8\n\n### Square\n\nLength = n\n Perimeter 14856 1.37938e+07 5252.39\n\n### Cube\n\nLength = n\n Surface area 8.27628e+07 5.12302e+10 6432.84\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 11142 5.97289e+06 3216.42\n\n### Triangular Pyramid\n\nLength = n\n Surface area 2.38916e+07 6.03753e+09 3032.47"
] | [
null
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https://hero.handmade.network/forums/code-discussion/t/984-day_229__what_about_radix_sort#5617 | [
"Mārtiņš Možeiko\n2415 posts / 2 projects\nEdited by Mārtiņš Možeiko on\nIn episode 229 Casey said that we'll probably implement better sorting method than BubbleSort. Something that is O(n*log(n)) not O(n*n). So that would be something like MergeSort, HeapSort or QuickSort. Which is \"lowest you can possibly do\".\n\nBut what about Radix sort? It is usually lower than O(n*log(n)). It is practically O(n) sort (to be accurate it is O(k*n) where k is digit count used in RadixSort). Linear from element count! It is very simple to implement, and is very branch-prediction friendly. Usual sort algorithms are very unfriendly to branch-prediction, because they compare individual elements and do different stuff depending on branch. For non-trivial amount of elements (thousands of elements) RadixSort is faster than any QuickSort implementation.\n\nOnly disadvantage is that it is not in-place sort. It needs temporary memory where to store objects while sorting. But with out fast memory management it is very cheap operation. Typical implementation uses bytes as \"digits\" for RadixSort. Here's the sample implementation:\n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 // creates u32 value from r32 so that sorted u32 values give same order // as original r32 values when sorted // if float is negative, flips all bits // if float is positive, flips only sign bit inline u32 FloatToU32Key(r32 Key) { u32 KeyBits = *((u32*)&Key); u32 Mask = -s32(KeyBits >> 31) | 0x80000000; KeyBits ^= Mask; return KeyBits; } inline u32 GetByteN(u32 Value, u32 ByteIndex) { Assert(ByteIndex < 4); return (Value >> (8*ByteIndex)) & 0xFF; } internal void SortEntries(render_group *RenderGroup) { u32 Count = RenderGroup->PushBufferElementCount; tile_sort_entry *Entries = (tile_sort_entry *)(RenderGroup->PushBufferBase + RenderGroup->SortEntryAt); // allocate memory, so don't call this in a loop, but we know this method is // called once per frame (at least for now) // or another option would be to reserve memory in PushBuffer in render_group // structure next to where original entries are allocated tile_sort_entry *Temp = PushArray(TempArena, Count, tile_sort_entry); tile_sort_entry *Source = Entries; for (u32 DigitIndex=0; DigitIndex<4; DigitIndex++) { u32 Counts = {}; for (u32 Index=0; IndexSortKey); u32 Digit = GetByteN(Key, DigitIndex); Temp[Counts[Digit]++] = *Entry; } tile_sort_entry *Swap = Source; Source = Temp; Temp = Swap; } } \n\nNo comparisons at all. Except loop conditions which are very branch-predictable.\n\nTricky part about FloatToU32Key is explained here: http://stereopsis.com/radix.html\nIt can be avoided with a bit more code for creating Count array: http://www.codercorner.com/RadixSortRevisited.htm\n\nAnd if the key is not float but integer, then whole FloatToU32Key function can be avoided and Entry->SortKey can be used directly. Code becomes even simpler.\nGinger Bill\n222 posts / 3 projects\nI am ginger thus have no soul.\nThis is the first time that I have seen radix sort and I am now very interested.\n\nHow does perform in practice? I notice it uses temporary memory but a lot of sorting algorithms do anyway.\nMārtiņš Možeiko\n2415 posts / 2 projects\nEdited by Mārtiņš Možeiko on\nYeah, radix sort is my favorite sort. If you have integer or float keys, it is very simple to implement and it is very fast in practice. You can do it also for a strings, but implementation gets a bit tricky.\n\nHere's a small benchmark for this radix sort implementation and std::sort (QuickSort with optimizations for small element count): https://gist.github.com/mmozeiko/0bd42648536b164f5dc8\n\nWith GCC 5.3.0 it gives following times (in milliseconds):\n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Count Quick Radix 64 0.00 0.00 128 0.00 0.00 256 0.01 0.00 512 0.02 0.01 1024 0.04 0.01 2048 0.09 0.02 4096 0.19 0.06 8192 0.45 0.09 16384 0.83 0.19 32768 1.75 0.42 65536 3.88 0.85 131072 7.99 1.84 262144 17.24 4.10 524288 36.52 7.15 1048576 76.17 18.17 \n\nFaster than quick sort :)\nGinger Bill\n222 posts / 3 projects\nI am ginger thus have no soul.\nI think this is going to be my next favourite sort. However, if space is a problem, I think I will need to use something else.\n\nThank you for bringing this up.\nRoderic Bos\n70 posts\nDon't know if everybody knows this, but this seems a helpfull page for choosing/understanding a sorting algorithm:\n\nhttp://www.sorting-algorithms.com/\nAndrew Bromage\n183 posts / 1 project\nResearch engineer, resident maths nerd (Erdős number 3).\nmmozeiko\nIn episode 229 Casey said that we'll probably implement better sorting method than BubbleSort. Something that is O(n*log(n)) not O(n*n). So that would be something like MergeSort, HeapSort or QuickSort. Which is \"lowest you can possibly do\".\n\nThis was covered on day 232. One of the things that Casey mentioned during the Q&A on day 232 is that it's been proven that you can't do better than O(n log n) for a comparison-based sort.\n\nThe proof is surprisingly simple to understand if you understand just a little bit of information theory. So I'm going to sketch it out here.\n\nInformation theory is basically the theory of communicating over a communication channel. For analysing algorithms, this is handy, because you can think of your algorithm as communicating with a query operation, and with a little information theory, you can work out how much information it needs to get for your algorithm needs to do its job.\n\nThe main fact that you need to know is this: To communicate a number between 1 and N, you need to transfer at least log N bits of information. If it helps, for \"bits\" read \"digits\". To communicate a number between 1 and 1,000,000, you need to transmit six decimal digits. The base-10 logarithm of 1,000,000 is 6. But we're computer scientists, so all logarithms are base 2 unless otherwise specified.\n\nIn the discussion that follows, I'm going to assume that all of the sort keys are distinct, that is, that there are no two keys in the collection which test equal. Later, I'll mention (but not explain, much less prove) how it works in that case.\n\nFirst, let's think about binary search. You have a collection of N numbers, and you need to find one of them. That means that you need to discover a number between 1 and N (or 0 and N-1). To discover such a number, you need log N bits of information.\n\nNow suppose you have a query operation which returns a single bit of information. Like, oh, the \"less than or equal to\" operation, which takes two keys and returns a boolean. That means you need to call that query operator at least log N times. Each time you call it, you get one bit of information, and you need log N bits, so you need to call it log N times.\n\nSo the best search algorithm, under those circumstances, must take O(log N) time. Binary search takes O(log N) time. Binary search is therefore (asymptotically) optimal.\n\nNow let's look at sorting. If you have a list of N elements, and the keys are distinct, then there N! possible orderings. That's the factorial of N. Sorting those elements is equivalent to discovering which permutation the elements are in. That means that you need to discover a number between 1 and N!, which means you need to learn log(N!) bits.\n\nBy Stirling's approximation, log(N!) = N log N + O(lower order crap). It follows that any sort algorithm must discover N log N bits of information. If you have a query operator which returns one bit of information (e.g. a comparison), then you must call it O(N log N) times. It follows that any comparison-based sort requires O(N log N) comparisons.\n\nHeap sort and merge sort perform O(N log N) comparisons in the worst case, therefore they are (asymptotically) optimal.\n\nQED\n\nThe reason why radix sort could do better is that its query operator returns more than one bit of information.\n\nIf there are duplicate keys, then things are a little more complex. In the limiting case, if all of the keys in the list are equal, then it should only take O(N) time to discover this and then say \"we're done\". If you want to take this into account, then it turns out that comparison-based sorting takes O(NH) time, where H is the entropy of the key distribution. You are not expected to understand this.\n\nYou can analyse lots of algorithms this way. For example, the \"top k\" problem which Casey also mentioned on day 232 can be done in O(N log k) comparisons. An algorithm which does this is left as an exercise.\n\nmmozeiko\nUsual sort algorithms are very unfriendly to branch-prediction, because they compare individual elements and do different stuff depending on branch.\n\nMore to the point, those branches are impossible to predict, because the result of the comparison operation is the very information that you are trying to discover!\n\nHaving said that, many sort algorithms are better about this than others. Insertion sort, for example, almost always runs faster than bubble sort on real hardware because its basic operation is not \"compare and swap if necessary\". The basic structure of bubble sort is:\n\n 1 2 3 4 5 for (i = 0; i < n; ++i) { for (j = 0; j + 1 < n; ++j) { compare-and-swap-if-necessary(A[j], A[J+1]); } } \n\nThe \"swap if necessary\" operation is a mostly unpredictable branch which is executed O(n^2) times in the worst case. The basic structure of insertion sort, however, is:\n\n 1 2 3 4 5 6 7 8 9 for (i = 1; i < n; ++i) { tmp = A[i]; j = i - 1; while (j >= 0 && A[j] > tmp) { A[j+1] = A[j]; --j; } A[j+1] = tmp; } \n\nThat branch is still mostly unpredictable, but it's the termination condition for the inner loop, so it only runs O(n) times. Remember what Casey said about constant factors? A mispredicted branch is expensive, so an algorithm which mispredicts branches O(n) times will typically beat one that mispredicts branches O(n^2) times.\n\nYeah, I never use bubble sort.\n\nmmozeiko\nOnly disadvantage is that it is not in-place sort.\n\nThat is not true. The usual in-place variant of radix sort is sometimes called American flag sort.\n\nActually, you don't even need extra storage to store the digit counts! This is a very counter-intuitive result which was only settled a couple of years ago using one of the most brilliant pieces of bit hackery I've seen in recent times. Thanks to information theory, you need fewer bits to store an ascending sequence of integers than an unsorted sequence, and this extra space that you gain as the algorithm progresses is precisely equal to the extra space that you need to do radix sort!\n\nThey also give impractical algorithms for when you can't modify sort keys, but essentially this settles the problem from a theoretical point of view: Fixed sized integer sorting takes O(n) time with O(1) extra space in the worst case.\n\nmmozeiko\n 1 2 3 4 5 6 7 8 9 10 11 // creates u32 value from r32 so that sorted u32 values give same order // as original r32 values when sorted // if float is negative, flips all bits // if float is positive, flips only sign bit inline u32 FloatToU32Key(r32 Key) { u32 KeyBits = *((u32*)&Key); u32 Mask = -s32(KeyBits >> 31) | 0x80000000; KeyBits ^= Mask; return KeyBits; } \n\nI'm pretty sure this doesn't handle Inf or subnormal numbers correctly. Maybe that's not an issue...\nMārtiņš Možeiko\n2415 posts / 2 projects\nEdited by Mārtiņš Možeiko on\nYeah, I'm aware of algorithm to do radix sort inplace, but it does a lot more branching so I felt it will be slower than the trivial radix sort implementation with additional memory. Also afaik it is unstable sort, right?\n\nCool, the idea with not using any memory at all looks pretty cool. But I feel the implementation complexity will make it slower than simple 256 element array on stack.\n\nAs for infinity and denormals, I seriously doubt they would be present in our render buffer for sort keys. But if you don't like them, it should be pretty straight forward to replace them with normal integer keys - because our layers are discrete then the render command placement on layers can be assigned to simple integer values.\n\nHere's the new benchmark with unstable inplace radix sort: https://gist.github.com/mmozeiko/0bd42648536b164f5dc8\n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Count Quick Radix RadixInplace RadixInplOpt 64 0.00 0.00 0.00 0.00 128 0.00 0.00 0.01 0.00 256 0.01 0.00 0.02 0.01 512 0.02 0.01 0.05 0.01 1024 0.03 0.01 0.13 0.02 2048 0.07 0.02 0.26 0.04 4096 0.17 0.04 0.45 0.09 8192 0.35 0.07 0.77 0.19 16384 0.77 0.15 1.55 0.42 32768 1.61 0.29 3.33 0.89 65536 3.51 0.59 6.49 1.83 131072 7.32 1.22 10.74 3.28 262144 14.79 2.49 14.99 6.53 524288 29.43 4.89 23.31 12.66 1048576 58.25 9.77 38.81 25.92 \n\nWhile it's still faster than QuickSort for larger element count, it is slower than simpler radix sort with additional memory. RadixInplOpt is almost the same as RadixInplace, but with small optimization - when element count is relatively small in one bucket (for example, less than 64) then insertion sort is used. It improves performance. Pretty much the same optimization is used in std::sort algorithm.\nAndrew Bromage\n183 posts / 1 project\nResearch engineer, resident maths nerd (Erdős number 3).\nmmozeiko\nYeah, I'm aware of algorithm to do radix sort inplace, but it does a lot more branching so I felt it will be slower than the trivial radix sort implementation with additional memory. Also afaik it is unstable sort, right?\n\nThe third algorithm in the paper is in-place, no extra memory, and stable. It's also over three pages full of pseudocode and nobody should ever try to implement it.\n\nmmozeiko\nCool, the idea with not using any memory at all looks pretty cool. But I feel the implementation complexity will make it slower than simple 256 element array on stack.\n\nOf course. Ultimately, that paper is a CS theory paper.\n\nWhen CS theory papers give algorithms, they are generally existence proofs / galactic algorithms (sometimes both). A galactic algorithm (a term coined by Ken Regan, I believe) is an asymptotically \"better\" algorithm whose speedup will not be realised unless the size of the input is larger than the number of fundamental particles in this galaxy. An existence proof is merely to show that something is possible.\n\nIn this case, it is a proof that it is possible to stable-sort integers in O(n) time with constant extra memory. Since it's literally impossible to do better, the long-standing theoretical question is resolved.\n\nIncidentally, since we're talking about P vs NP, I'd like to talk for a moment about my favourite existence proof algorithm: Tarski's algorithm for quantifier elimination in real closed fields. The existence of this algorithm shows that the theory of the real numbers with field operations is decidable, which is awesome. However, is it a practical algorithm?\n\nWell, the complexity class called EXP or EXPTIME is, basically, O(2^n). DEXPTIME, or 2-EXPTIME, is O(2^(2^n)). You can similarly define 3-EXPTIME, and in general, k-EXPTIME. The union of all of the k-EXPTIMEs for all integers is the complexity class ELEMENTARY.\n\nTarski's algorithm is in the complexity class NONELEMENTARY. Let that sink in.\nMārtiņš Možeiko\n2415 posts / 2 projects\nEdited by Mārtiņš Možeiko on\nPseudonym73\nThe third algorithm in the paper is in-place, no extra memory, and stable.\n\nAre you sure its stable? I tried to implement it, but all I get is unstable sort. Well I'm not yet 100% sure it's not some bug in my code so...\n\nIt seems FreeBSD also implements radix sort from same paper. Here's the source https://svnweb.freebsd.org/base/s...?revision=256281&view=markup#l132\nIt explicitly has \"Unstable, in-place sort\" comment before r_sort_a function.\n\nAs far as I understand paper it is unfair to say algorithm C doesn't use memory. It uses additional O(logN) memory on stack.\nAndrew Bromage\n183 posts / 1 project\nResearch engineer, resident maths nerd (Erdős number 3).\nmmozeiko\nAre you sure its stable? I tried to implement it, but all I get is unstable sort. Well I'm not yet 100% sure it's not some bug in my code so...\n\nPossible miscommunication. I was referring to the Patrascu paper, Radix Sorting With No Extra Space.\nMārtiņš Možeiko\n2415 posts / 2 projects\nOK, yeah I was talking about American Flag Sort.\n3 posts\nEdited by fod669 on\nI had a question in the stream today about combining the count steps into one. I was curious so I tried it myself using mmozeiko's testing code.\n\nHere's the modified radix sort code. I just called it RadixSort5n because it makes 5 passes over the list.\n\n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 static void RadixSort5n(r32 *Entries, r32* Temp, u32 Count) { r32 *Source = Entries; u32 Counts = {}; for (u32 Index = 0; Index < Count; Index++) { u32 Key = FloatToU32Key(Source[Index]); Counts[GetByteN(Key, 0)]++; Counts[GetByteN(Key, 1)]++; Counts[GetByteN(Key, 2)]++; Counts[GetByteN(Key, 3)]++; } for (u32 DigitIndex = 0; DigitIndex < 4; DigitIndex++) { u32 TotalCount = 0; for (u32 Index = 0; Index < 256; Index++) { u32 CurrentCount = Counts[DigitIndex][Index]; Counts[DigitIndex][Index] = TotalCount; TotalCount += CurrentCount; } } for (u32 DigitIndex = 0; DigitIndex < 4; DigitIndex++) { for (u32 Index = 0; Index < Count; Index++) { u32 Key = FloatToU32Key(Source[Index]); u32 Digit = GetByteN(Key, DigitIndex); Temp[Counts[DigitIndex][Digit]++] = Source[Index]; } r32 *Swap = Source; Source = Temp; Temp = Swap; } } \n\nAnd here are the results:\n\n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Count Quick Radix Radix5n RadixInplace RadixInplOpt 64 0.00 0.00 0.00 0.00 0.00 128 0.00 0.00 0.00 0.01 0.00 256 0.01 0.00 0.00 0.03 0.01 512 0.02 0.01 0.01 0.07 0.01 1024 0.04 0.01 0.01 0.17 0.03 2048 0.09 0.03 0.03 0.35 0.06 4096 0.20 0.07 0.05 0.59 0.11 8192 0.43 0.12 0.10 1.05 0.22 16384 0.91 0.24 0.20 2.16 0.49 32768 1.95 0.50 0.37 4.74 1.20 65536 4.08 0.94 0.80 10.20 2.22 131072 8.95 2.04 1.62 15.59 4.08 262144 18.26 4.14 3.30 21.46 7.95 524288 36.02 7.52 6.73 31.38 15.23 1048576 69.79 15.94 13.10 51.89 29.97 \n\nSo overall it is a slight improvement, especially as the list gets larger. Which I suppose is to be expected as the algorithm is linear anyway, and the counting overhead gets less in comparison to the size of the list.\n\nThe only downside really is the cost of the extra 3 arrays.\nMārtiņš Možeiko\n2415 posts / 2 projects\nEdited by Mārtiņš Možeiko on\nNice!\nIf list gets larger you can improve it even more - do only 4 passes. Basically process 11 bits at a time, that will change element count in Counts array from 256 to 2048 and you'll need only 3 such Counts arrays.\n\nIt will use a bit more stack memory, but if you are sorting million (or whatever) elements, then additional temporary 2048 elems x3 on stack won't make anything worse.\n\nBtw, Casey didn't mention this on stream - but to easier understand Radix sort, you can start looking only at its inner loop. That code implements Counting sort: https://en.wikipedia.org/wiki/Counting_sort which is very easy to understand. Radix sort simply applies Counting sort multiple times for every digit in key. And because counting sort is stable sort, the result will still be sorted.\nGianluca Alloisio\n33 posts\nI've written a devlog about sorting for libLepre, and mentioned this thread. Thanks guys :) http://liblepre.tumblr.com/post/138434159025/devlog-33-sorting\nDumitru Frunza\n24 posts\nApprentice github.com/dfrunza/hoc\n\nOne of the things that Casey mentioned during the Q&A on day 232 is that it's been proven that you can't do better than O(n log n) for a comparison-based sort.\n\nThere's another proof of this in the book \"Data Structures and Algorithms\" by Alfred Aho et al., Section 8.6 \"A Lower Bound for Sorting by Comparisons\",\nwhich I find to be simple and accessible.\n\nHere's my interpretation of it.\n\nConsider the set of all possible initial orderings (also called 'permutations') of the elements of the input array.\nThe sorting algorithm can be thought of as a process by which this set is progressively \"narrowed\" down\nuntil only one ordering remains - the sorted ordering.\n\nAs an illustration, take the array A that contains 3 elements a,b,c in some order, and which are comparable by the '<' (less than) relation. Start with\nthe set of all possible orderings {abc,acb,bac,bca,cab,cba}, then ask a question about the relation between two elements of the array - say A<A?\nSplit the set in two - one containing all orderings which are consistent with the answer 'yes' and the other consistent with the answer 'no'.\nE.g. let A:=abc; if A<A then we know that a<b, and so pick the orderings (from the initial set) where a precedes b and make a new set {abc,acb,cab};\nif A>=A then b<a so pick the orderings where b precedes a and make the set {bac,bca,cba}. Proceed in this manner recursively,\nby asking a question and splitting the sets until the resulting sets contain only one ordering.\n\nThis process creates a binary tree called a 'decision tree', and the questions asked are specific to the sorting algorithm employed.\nThe running time of the algorithm, in the worst case, is proportional to the longest path in the decision tree. A balanced tree\nis optimal in this sense, because the longest path from root to a leaf is minimal among other possible tree configurations.\nThus the best possible algorithm will produce a balanced decision tree.\n\nIn a balanced binary tree, the length of a path from root to its leaves is log(x), where x is the number of leaves.\nThat's easy to see\n 1 2 3 4 5 6 * ------ 1 node, len(path)=log(1)=0 / \\ *----*----- 2 nodes, len(path)=log(2)=1 / \\ / \\ *---*-*---*-- 4 nodes, len(path)=log(4)=2 ........... x nodes, len(path)=log(x) \n\nLet n be the number of elements in the input array, then the total number of leaves in the decision tree is n! (n factorial),\nbecause that's the number of all possible initial orderings of the array, and each leaf in the decision tree contains only one (unique) ordering.\nThen length of the longest path in the decision tree is log(n!).\nNow n! can be (very roughly) approximated as n^n (n to the power of n), n! < n^n, so\nlog(n!) < log(n^n) = n*log(n)\n\nq.e.d."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.92940366,"math_prob":0.9327845,"size":6609,"snap":"2022-40-2023-06","text_gpt3_token_len":1557,"char_repetition_ratio":0.11960636,"word_repetition_ratio":0.026632302,"special_character_ratio":0.24164018,"punctuation_ratio":0.114179105,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96882224,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-07T12:35:11Z\",\"WARC-Record-ID\":\"<urn:uuid:893a663b-e224-47fb-850a-f7042c907ee1>\",\"Content-Length\":\"87584\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f42abc1d-850f-492f-a7c7-0d3ad9195c25>\",\"WARC-Concurrent-To\":\"<urn:uuid:f5d419fa-5f87-4b69-b020-25a01253407f>\",\"WARC-IP-Address\":\"104.21.89.60\",\"WARC-Target-URI\":\"https://hero.handmade.network/forums/code-discussion/t/984-day_229__what_about_radix_sort#5617\",\"WARC-Payload-Digest\":\"sha1:4CHO2SDRCA2MRAORZ2G3CHQHPQQSZHCB\",\"WARC-Block-Digest\":\"sha1:VTQPKFITTGZWKOSDDWUDYZIBZ6Z6CTHM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030338073.68_warc_CC-MAIN-20221007112411-20221007142411-00394.warc.gz\"}"} |
http://modelai.gettysburg.edu/2012/pig/ | [
"",
null,
"Pig is a folk jeopardy dice game described by John Scarne in 1945, and was an ancestor of the modern game Pass the Pigs® (originally called PigMania®).\n\nThe rules are simple: Two players race to reach 100 points. Each turn, a player repeatedly rolls a die until either a 1 is rolled or the player holds and scores the sum of the rolls (i.e. the turn total). At any time during a player's turn, the player is faced with two decisions:\n\n• roll - If the player rolls a\n• 1: the player scores nothing and it becomes the opponent's turn.\n• 2 - 6: the number is added to the player's turn total and the player's turn continues.\n• hold - The turn total is added to the player's score and it becomes the opponent's turn.\n\nTo familiarize yourself with play, you can play an optimal Pig opponent online. The key decision facing a player is how large a turn total should be risked to possibly get an even larger total. To learn more about the game of Pig, visit The Game of Pig web page.\n\n Summary Solving the Dice Game Pig - Pig, a popular folk jeopardy dice game, serves as a fun focus problem for this introduction to dynamic programming, value iteration, and basic concepts of reinforcement learning. Topics Dynamic programming, states, actions, rewards, discount factor, value iteration, reinforcement learning. Audience Introductory Artificial Intellegence students Difficulty Students selecting two exercises (1 from each set of 3 at the end of sections 2 and 3) can complete the exercises in about 1 week. For example, Pig Solitaire and Pig can be completed in less than 200 lines of Java code. Strengths Students acquire a deeper experiential knowledge of dynamic programming and value iteration through a set of fun challenge exercises that allow student choice of pursuit. Advanced exercises and recommended further lines of study are also provided. Weaknesses The presentation follows Knuth's literate programming style, and thus requires knowledge of the Java programming language. Dependencies Students must be mathematically literate, understanding summation notation, basic probability theory, and recursive definitions, as well as being comfortable with created notation. Students must also have intermediate programming skills and be familiar with Java syntax and semantics. Variants Advanced projects include analysis of the board game Risk, Yahtzee, Hog, and 10,000 (a.k.a. Farkle). Basic and advanced projects may be easily adapted to teach Monte Carlo and Temporal Difference Learning techniques.\n\n# Solving the Dice Game Pig: an introduction to dynamic programming and value iteration\n\nIndex:",
null,
"## Overview\n\nThe jeopardy dice game Pig is very simple to describe, yet the optimal policy for play is far from trivial. Using the computation of the optimal solution as a central challenge problem, we introduce dynamic programming and value iteration methods, applying them to similar problems using the Java language.\n\n## Objectives\n\nThe object of this project is to give the student a deep, experiential understanding of dynamic programming and value iteration through explanation, implementation examples, and implementation exercises\n\n• Dynamic Programming:\n• Demonstrate the need for dynamic programming through Fibonacci number computation.\n• Guide the student through the details of a Java solution to a simple Pig variant with an acyclic state space.\n• Provide problem solving exercises that will ground the student's understanding of dynamic programming in experience.\n• Value Iteration:\n• Introduce the method of value iteration.\n• Demonstrate its application to a simple coin variant of Pig called Piglet.\n• Guide the student through the details of a Java solution to Piglet.\n• Provide problem solving exercises that will ground the student's understanding of value iteration in experience.\n\n## Prerequisites\n\nThe student should understand basic probability and algebraic systems of equations (although knowledge of linear algebraic solution techniques are not required). The student should also understand the syntax of Java. Students who program in C++ should be able to follow most of the Java examples.\n\nBefore starting the project, the student will want to understand the definition of a Markov decision process, perhaps by covering the recommended background reading below.\n\nTo understand the terminology of Markov decision processes and general application of value iteration, we recommend that students read sections 17.1 (Sequential Decision Problems) and 17.2 (Value Iteration) of:\n\nStuart Russell and Peter Norvig. Artificial Intelligence: a modern approach, 3rd ed. Prentice Hall, Upper Saddle River, NJ, USA, 2010.\n\nStudents are encouraged to play the game of Pig in order to have a better understanding of the problem. An applet with an optimal computer player is available at the Game of Pig page.\n\n## Description",
null,
"The project is described in the file pig.pdf. You will need the free Adobe Acrobat Reader to view this file.",
null,
"Example code described in the project is available here:\n\nThis code was generated from the same source file that generated pig.pdf using the literate programming tool noweb.\n\nWe recommend having students do at least one dynamic programming exercise and one value iteration exercise.\n\nStudents seeking more extended challenges will also find descriptions of related advanced projects for dynamic programming and value iteration.\n\n## Syllabus\n\nHere is a sample syllabus used in CS 371 - Introduction to AI at Gettysburg College:\n\nFirst class: Tuesday 9/28\nPreparatory Reading: Sections 1-2 of pig.pdf.\nDynamic programming\nHomework due Thursday 9/30: One complete exercise from section 2.4\n\nSecond class: Thursday 9/30\nPreparatory Reading: Remaining sections of pig.pdf, Russell & Norvig sections 17.1, 17.2\nValue Iteration\nHomework due Thursday 10/7: One complete exercise from section 3.5\n(There was a two-day break 10/4-10/5. Without the break, Tuesday 10/5 rather than 10/7 would have been appropriate.)\n\nThe following additional coverage is not included in this project, but may provide the instructor additional ideas:\n\nThird class: Thursday 10/7:\nMountain Car Problem of Example 8.2 of Richard S. Sutton and Andrew G. Barto. Reinforcement Learning: an introduction. MIT Press, Cambridge, Massachusetts, 1998, pp. 214-215. (Online at http://www.cs.ualberta.ca/~sutton/book/8/node9.html See also: http://www-anw.cs.umass.edu/rlr/domains.html, http://www.cs.ualberta.ca/~sutton/MountainCar/MountainCar.html\n\nOur approach to this problem is to simply apply value iteration, discretizing the state space into a 400-by-400 grid of cells, and approximating the dynamics of each time step as a mapping from one cell to another. In other words, overlay the 2D state space (position, velocity) with a 400-by-400 grid. Simulate one time-step from each grid intersection and note which is the nearest intersection to the resulting state. This defines a mapping to discretely approximate the system dynamics. Then apply value iteration with a reward of -1 for each step that the system is not at the rightmost position."
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"http://modelai.gettysburg.edu/2012/pig/pigSoln.jpg",
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https://invernessgangshow.net/how-many-inches-is-7-5-cm/ | [
"## 7.5 Centimeter (cm) amounts to 2.9527575 customs (in)\n\nThe 7.5 centimeter to inches converter is a size converter native one unit come another. One centimeter is approximately 0.3937 inches.\n\nYou are watching: How many inches is 7.5 cm\n\nThe units of length must be convert from centimeters come inches. The 7.5 centimeter to inches is the most simple unit conversion friend will discover in primary school school. This is just one of the most usual operations in a wide selection of mathematics applications.\n\nThis article explains how to convert 7.5 centimeter to inches and also use the device for convert one unit native another, and the relationship between centimeters and also inches with in-depth explanations.",
null,
"## Why readjust the length from 7.5 centimeter to inches come inches?\n\nA centimeter (or centimeter) is a unit the length. It is one hundredth the a meter. However, the joined States supplies a common unit of length. Royal units are provided in the same means in good Britain.\n\nThe typical Imperial or us unit of measure up for size (or distance) is inches. If you have information about length in centimeters; and you need the same number in tantamount inch units, you have the right to use this converter.\n\n### The relationship in between inches and also cm\n\nTo transform 7.5 centimeters to inches or inches to centimeters, the relationship in between inches and also centimeters is that one customs in the metric mechanism is specifically 2.57.5 centimeters.\n\n1 inch = 2.57.5 cm\n\nTherefore,\n\n1 centimeter = 1 / 2.57.5 inch\n\nTo transform centimeters to inches, we must divide the value in centimeters through 2.57.5.\n\nIf the unit size is 1 cm, the equivalent length in inch is 1 centimeter = 0.393701 inches\n\n## How countless inches is 7.5cm\n\n### Convert 7.5 centimeter (centimeters) to inches (in)\n\nWith this length converter we can easily transform cm come inches favor 10 centimeter to inches, 16 cm to inches, 7.5 centimeter to inches, 7.5cm in inches etc.\n\nSince we know that a centimeter is roughly 0.393701 inches, the conversion native one centimeter to inches is easy. To convert centimeters to inches, main point the centimeter value provided by 0.393701.\n\nFor example, to convert 10 centimeters come inches, multiply 10 centimeters through 0.393701 to acquire the worth per inch.\n\n(i.e.) 10 x 0.393701 = 3.93701 inches.\n\nTherefore, 10 centimeters is equal to 3.93701 inches.\n\nNow consider an additional example: 7.5cm in inches is converted together follows:\n\n## How carry out I convert 7.5 cm to inches?\n\nTo convert 7.5 centimeter to in, just take the really measurement in cm and multiply this number by 2. 7.557.5. So you can transform how numerous inches is 7.5 cm manually.\n\nYou can likewise easily transform centimeters come inches using the following centimeters come inches conversion:\n\n### How many inches is 7.5 cm\n\nAs us know, 1 cm = 0.393701 inches\n\n## What is 7.5 centimeter in inches\n\nIn this way, 7.5 centimeters have the right to be convert to inch by multiplying 7.5 through 0.393701 inches.\n\n(i.e.) 7.5 centimeter to one inch = 7.5 x 0.393701 inches\n\n7.5 cm = customs = 2.9527575 inches\n\n### 7.5 centimeter is how numerous inches\n\nTherefore, 7.5 cm is how plenty of inches 7.5 cm is same to 11,811 inches.\n\n## Example of convert centimeters to inches\n\nThe following instances will assist you understand just how to transform centimeters to inches.\n\n### Convert 7.5 centimeter to inches\n\nReply:We know that 1 cm = 0.393701 inches.\n\nSee more: What Does Post Mean Before Or After ? What Does Post Mean Before Or After\n\nTo convert 7.5 centimeters come inches, main point 7.5 centimeters through 0.393701 inches.\n\n= 7.5 x 0.393701 inches\n\n= 2.9527575 inches\n\n7.5 cm is same to how plenty of inchesHow countless inches is 7.5 cm equal to7.5 come 7.5 cm is how many inchesWhat is 7.5 centimeter equal come in inches?Convert 7.5 centimeter to inches7.5 cm convert to inches"
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https://wpcalc.com/en/ideal-gas-law/ | [
"# Ideal Gas Law\n\nThe ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behavior of many gases under many conditions, although it has several limitations. It was first stated by Émile Clapeyron in 1834 as a combination of Boyle’s law, Charles’s law and Avogadro’s Law\n\n### Calculator of Ideal Gas Law\n\n I want to calculate Volume(V) Temperature(T) Pressure(P) Moles of Gas(n) Temperature K Pressure kPa Moles of Gas moles Volume L\n\n### Equation of ideal gas law\n\nThe state of an amount of gas is determined by its pressure, volume, and temperature. The modern form of the equation relates these simply in two main forms. The temperature used in the equation of state is an absolute temperature: in the SI system of units, Kelvin.\n\nGas Equation: PV = nRT\n\nwhere,\n\n• P = pressure,\n• V = volume,\n• n = moles of gas,\n• T = temperature,\n• R = 8.314 J K-1 mol-1, ideal gas constant.\n\n### Ideal Gas Law Example:\n\nCase 1: Find the volume from the 0.250 moles gas at 200kpa and 300K temperature.P = 200 kPa, n = 0.250 mol, T = 300K, R = 8.314 J K-1 mol-1\n\nStep 1:\n\nSubstitute the values in the below volume equation: Volume(V) = nRT / P = (0.250 x 8.314 x 300) / 200 = 623.55 / 200 Volume(V) = 3.12 L This example will guide you to calculate the volume manually.\n\nCase 2: Find the temperature from the 250ml cylinder contaning 0.50 moles gas at 153kpa.V = 250ml -> 250 / 1000 = 0.250 L, n = 0.50 mol, P = 153 kPa, R = 8.314 J K-1 mol-1\n\nStep 1:\n\nSubstitute the values in the below temperature equation: Temperature(T) = PV / nR = (153 x 0.250) / (0.50 x 8.314) = 38.25 / 4.16 Temperature(T) = 9.2 K This example will guide you to calculate the temperature manually.",
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] | [
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https://journalskuwait.org/kjs/index.php/KJS/article/view/243 | [
"### Extension of Mazhar’s theorem on summability factors\n\nMEHMET ALI SARIGOL\n\n#### Abstract\n\nBy (X; Y ) we denote the set of all sequences such thatPa nnis summable X wheneverPa n is summable Y , where X and Y are sum-mability methods. In this paper we characterize the set (jC; jk ;N; p n) fork > 1; > 1 and arbitrary positive sequence (pn) using functional analytictechniques, and so extend the known results of Mazhar ; Mehdi and theauthor [10; 12].\n\n#### Keywords\n\nAbsolute Cesàro summability; absolute Riesz summability; bounded linear operators; matrix transformations; summability factors.\n\nPDF\n\n#### References\n\nBor, H. & B. Kuttner, B. 1989. On the necessary condition for absolute weighted arithmetic mean\n\nsummability factors. Acta Mathematica Hungarica 54: 57-61.\n\nBosanquet, L. S. & Das, G. 1979. Absolutely summability factors for Nörlund means. Proceedings\n\nLondon Mathematical Society 838: 1-52.\n\nChow, H. C. 1954. Note on convergence and summability factors. Journal of the London Mathematical\n\nSociety 29: 459-476.\n\nFlett, T. M. 1957. On an extension of absolute summability and some theorems of Littlewood and Paley.\n\nProceedings London Mathematical Society 7: 113-141.\n\nJakimovsky, A. & Russel, D. C. 1972. Matrix mapping between BK-spaces, Bulletin London\n\nMathematical Society 4: 345-353.\n\nMazhar, S. M. 1971. On the Absolute summability factors of infinite series. Tohoku Mathematical Journal\n\n: 433-451.\n\nMehdi, M. R. 1960. Summability factors for generalized absolute summability I. Proceedings London\n\nMathematical Society. 10: 180-200.\n\nPeyerimhoff, A. 1954. Summierbarkeitsfactoren für absolut Cesàro summierbareReihen. Mathematische\n\nZeitschrift 59: 417-424.\n\nRhoades, B. E. & Savas, E. 2004. A characterization of absolute summability factors. Taiwanese Journal\n\nof Mathematics 8: 453-465.\n\nSarigol, M. A. 1993a. On two absolute Riesz summability factors of infinite series. Proceedings of the\n\nAmerican Mathematical Society 118: 485-488.\n\nSarigol, M. A. 1993b. A note on summability. Studia Scientiarum Mathematicarum Hungarica 28; 395-\n\nSarigol, M. A. & Bor, H. 1995. Characterization of absolute summability factors. Journal of Mathematical\n\nAnalysis and Applications 195: 537-545.\n\nSarigol, M. A. 2011a. Matrix transformations on fields of absolute weighted mean summability. Studia\n\nScientiarum Mathematicarum Hungarica 48; 331-341.\n\nSarigol, M. A. 2011b. Characterization of general summability factors and applications, Computers and\n\nMathematics with Applications 62: 2665-2670.\n\nStieglitz, M. & Tietz, H. 1977. Matrixtransformationen von Folgenraumen Eine Ergebnisübersicht.\n\nMathematische Zeitschrift 154: 1-16.\n\nSulaiman, W. T. 1992. On summability factors of infinite series. Proceedings of the American Mathematical\n\nSociety 115: 313-317."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.51716244,"math_prob":0.610243,"size":2227,"snap":"2019-13-2019-22","text_gpt3_token_len":676,"char_repetition_ratio":0.19163293,"word_repetition_ratio":0.05704698,"special_character_ratio":0.2833408,"punctuation_ratio":0.28275862,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9891767,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-26T12:06:29Z\",\"WARC-Record-ID\":\"<urn:uuid:44f86794-3d81-4f9b-aab7-250a3dc4b6f1>\",\"Content-Length\":\"19893\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec0033fa-6915-48b3-959d-0b0faef46052>\",\"WARC-Concurrent-To\":\"<urn:uuid:96cc7a52-475d-47ce-913d-bacc4e53351a>\",\"WARC-IP-Address\":\"166.62.76.66\",\"WARC-Target-URI\":\"https://journalskuwait.org/kjs/index.php/KJS/article/view/243\",\"WARC-Payload-Digest\":\"sha1:JCCBSVPC2X265VU7SI6K7PWV2A2YFCKL\",\"WARC-Block-Digest\":\"sha1:A65J47E53GM4LVB5HTYZOQRDPYG4JH6Q\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232259126.83_warc_CC-MAIN-20190526105248-20190526131248-00087.warc.gz\"}"} |
https://projecteuclid.org/euclid.aoap/1034801251 | [
"The Annals of Applied Probability\n\nCounterexamples in importance sampling for large deviations probabilities\n\nAbstract\n\nA guiding principle in the efficient estimation of rare-event probabilities by Monte Carlo is that importance sampling based on the change of measure suggested by a large deviations analysis can reduce variance by many orders of magnitude. In a variety of settings, this approach has led to estimators that are optimal in an asymptotic sense. We give examples, however, in which importance sampling estimators based on a large deviations change of measure have provably poor performance. The estimators can have variance that decreases at a slower rate than a naive estimator, variance that increases with the rarity of the event, and even infinite variance. For each example, we provide an alternative estimator with provably efficient performance. A common feature of our examples is that they allow more than one way for a rare event to occur; our alternative estimators give explicit weight to lower probability paths neglected by leading-term asymptotics.\n\nArticle information\n\nSource\nAnn. Appl. Probab., Volume 7, Number 3 (1997), 731-746.\n\nDates\nFirst available in Project Euclid: 16 October 2002\n\nPermanent link to this document\nhttps://projecteuclid.org/euclid.aoap/1034801251\n\nDigital Object Identifier\ndoi:10.1214/aoap/1034801251\n\nMathematical Reviews number (MathSciNet)\nMR1459268\n\nZentralblatt MATH identifier\n0892.60043\n\nSubjects\nPrimary: 60F10: Large deviations 60J15 65C05: Monte Carlo methods\n\nCitation\n\nGlasserman, Paul; Wang, Yashan. Counterexamples in importance sampling for large deviations probabilities. Ann. Appl. Probab. 7 (1997), no. 3, 731--746. doi:10.1214/aoap/1034801251. https://projecteuclid.org/euclid.aoap/1034801251"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.78583217,"math_prob":0.67450714,"size":1951,"snap":"2019-43-2019-47","text_gpt3_token_len":461,"char_repetition_ratio":0.10888547,"word_repetition_ratio":0.03816794,"special_character_ratio":0.24038954,"punctuation_ratio":0.15789473,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9526179,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-17T11:03:21Z\",\"WARC-Record-ID\":\"<urn:uuid:a78db2d5-36c1-4a62-80c6-e7d6b0b48ccc>\",\"Content-Length\":\"42424\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b3cb679a-bc68-40e3-b23d-1e82d6c22c05>\",\"WARC-Concurrent-To\":\"<urn:uuid:36e52a69-df49-4d41-92ee-98c26df9635f>\",\"WARC-IP-Address\":\"132.236.27.47\",\"WARC-Target-URI\":\"https://projecteuclid.org/euclid.aoap/1034801251\",\"WARC-Payload-Digest\":\"sha1:HO2SVQSIYMJTJXAL3ZUJYDW2G2TFJLHY\",\"WARC-Block-Digest\":\"sha1:NZEFCW3GXLK66NPZX2XO22QYPYV2A7FD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986673538.21_warc_CC-MAIN-20191017095726-20191017123226-00441.warc.gz\"}"} |
https://www.schoolsolver.com/question/71644-step-by-step-with-explanation/ | [
"### Step by step with explanation\n\nQuestion\n\nf(x) = x ^ 2 + 1, g(x) = 2x - 2h(x) = - x - 1Find the correspondencebetween the composite functions and theirformulas\n\nDetails"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.67895025,"math_prob":0.99953437,"size":322,"snap":"2023-40-2023-50","text_gpt3_token_len":106,"char_repetition_ratio":0.13836478,"word_repetition_ratio":0.71428573,"special_character_ratio":0.3167702,"punctuation_ratio":0.03846154,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95304626,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-02T11:42:07Z\",\"WARC-Record-ID\":\"<urn:uuid:33123b86-415e-4dc8-b02f-504d2a7c3e84>\",\"Content-Length\":\"48127\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8ecc98f7-2c8e-4f3c-8b20-19975e01ff77>\",\"WARC-Concurrent-To\":\"<urn:uuid:cefae988-6ef3-4a29-843e-96649114ecc2>\",\"WARC-IP-Address\":\"23.22.5.68\",\"WARC-Target-URI\":\"https://www.schoolsolver.com/question/71644-step-by-step-with-explanation/\",\"WARC-Payload-Digest\":\"sha1:KZMH5WXAI3WNTGXG3W6KTB3P4X53EYIH\",\"WARC-Block-Digest\":\"sha1:3LU74U7JYH2PCIQ7RZ54B4OZVFOHRNF2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100399.81_warc_CC-MAIN-20231202105028-20231202135028-00661.warc.gz\"}"} |
http://fb-list-archive.s3-website-eu-west-1.amazonaws.com/firebird-support/2015/6/127093.html | [
"Subject Re: [firebird-support] Soc Sec No comparison using Firebird Softtech Support 2015-06-26T11:28:18Z\n\nGreetings Bogdan,\n\nYes, SUBSTR() does exists in v1.5 so I converted it successfully, I've alwasy used SUBSTRING() before. However, when I tried to run a few simple comparisons with 1 character off in the strings it always returned OK for the result. So I declared a few new variables (sOldSSNChar, sNewSSNChar) to see what is going on while debugging and to my surprise these varaibles would hold the first characted for both the old and new SSN, but then the rest (2nd thru 11th position) would all be blank (empty string), thus thinking they matched. Any idea why this may be happening?.\n\nThanks,\nMike\n\nHere is the slightly modified SP:\n\nSET TERM ^^ ;\nCREATE PROCEDURE SPS_SOCIAL_NUMBER_COMPARISON (\nI1 VarChar(20),\nI2 VarChar(20))\nreturns (\nRESULT VarChar(100))\nAS\ndeclare variable x1 char(20);\n\ndeclare variable x2 char(20);\n\ndeclare variable e smallint;\n\ndeclare variable e1 smallint;\n\ndeclare variable e2 smallint;\n\ndeclare variable i smallint;\n\nDECLARE VARIABLE sOldSSNChar Char(1);\nDECLARE VARIABLE sNewSSNChar Char(1);\nbegin\ni = 0;\nx1 = i1;\nx2 = i2; e = 0;\n\nwhile (i < 20 and e < 3) do\n\nbegin\n\ni = i + 1;\n\nsOldSSNChar = substr(x1, i, 1);\nsNewSSNChar = substr(x2, i, 1);\n\nif (substr(x1, i, 1) <> substr(x2, i, 1)) then\n\nbegin\n\ne = e + 1;\n\nif (e = 1) then e1 = i;\n\nelse if (e = 2) then e2 = i;\n\nend\n\nend\n\nif (e = 0) then result = 'OK';\n\nelse if (e = 3) then result = 'Not equal';\n\nelse if (e = 1) then result = e1 || '. character ' || substr(i1, i, 1) || ' has changed';\n\nelse\n\nbegin\n\nif (substr(x1, e1,1) = substr(x2, e2, 1) and\n\nsubstr(x1, e2, 1) = substr(x2, e1, 1)) then\n\nresult = e1 || ' and ' || e2 || ' were swapped';\n\nend\n\nsuspend;\n\nend\n^^\nSET TERM ; ^^",
null,
"This email has been checked for viruses by Avast antivirus software. www.avast.com"
] | [
null,
"https://ec.yimg.com/ec",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6680742,"math_prob":0.97995645,"size":2105,"snap":"2023-14-2023-23","text_gpt3_token_len":701,"char_repetition_ratio":0.12851024,"word_repetition_ratio":0.17021276,"special_character_ratio":0.34204274,"punctuation_ratio":0.17281106,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9592042,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-30T07:58:39Z\",\"WARC-Record-ID\":\"<urn:uuid:b7232646-ce94-4f7d-9cbd-d6cd6778e4f0>\",\"Content-Length\":\"12228\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bebc8239-19e2-4e2c-94c4-83efa3d38a8d>\",\"WARC-Concurrent-To\":\"<urn:uuid:755427bf-4292-42fb-917f-debd296e28ae>\",\"WARC-IP-Address\":\"52.92.35.92\",\"WARC-Target-URI\":\"http://fb-list-archive.s3-website-eu-west-1.amazonaws.com/firebird-support/2015/6/127093.html\",\"WARC-Payload-Digest\":\"sha1:A4T4FPV2K5ZKBSKA6EIMZHRZSL5BN5YN\",\"WARC-Block-Digest\":\"sha1:TQ3SH6L5JN32HLK7OHVEXOD5S5HDSEX3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949107.48_warc_CC-MAIN-20230330070451-20230330100451-00536.warc.gz\"}"} |
https://turningtooneanother.net/2020/06/02/how-do-you-find-the-probability-of-individual-events/ | [
"# How do you find the probability of individual events?\n\n## How do you find the probability of individual events?\n\nDivide the number of events by the number of possible outcomes.\n\n1. Determine a single event with a single outcome.\n2. Identify the total number of outcomes that can occur.\n3. Divide the number of events by the number of possible outcomes.\n4. Determine each event you will calculate.\n5. Calculate the probability of each event.\n\nHow do you find the probability of multiple independent events?\n\nProbability of Two Events Occurring Together: Independent Just multiply the probability of the first event by the second. For example, if the probability of event A is 2/9 and the probability of event B is 3/9 then the probability of both events happening at the same time is (2/9)*(3/9) = 6/81 = 2/27.\n\nHow can you find the probability of one or two inclusive events?\n\nInclusive events are events that can happen at the same time. To find the probability of an inclusive event we first add the probabilities of the individual events and then subtract the probability of the two events happening at the same time.\n\n### How do you do multiple events in probability?\n\nHow To: Given a set of events, compute the probability of the union of mutually exclusive events.\n\n1. Determine the total number of outcomes for the first event.\n2. Find the probability of the first event.\n3. Determine the total number of outcomes for the second event.\n4. Find the probability of the second event.\n\nWhen the probability is 0.75 then an event is?\n\nAs the probability goes up from 0.5 to 1.0, the odds increase from 1.0 to approach infinity. For example, if the probability is 0.75, then the odds are 75:25, three to one, or 3.0. If the odds are high (million to one), the probability is almost 1.00.\n\nWhat is the formula for probability?\n\nP(A) is the probability of an event “A” n(A) is the number of favourable outcomes. n(S) is the total number of events in the sample space….Basic Probability Formulas.\n\nAll Probability Formulas List in Maths\nConditional Probability P(A | B) = P(A∩B) / P(B)\nBayes Formula P(A | B) = P(B | A) ⋅ P(A) / P(B)\n\n## When two events are independent the probability of both occurring is?\n\nWhen two events are independent, the probability of both occurring is the product of the probabilities of the individual events. where P(A and B) is the probability of events A and B both occurring, P(A) is the probability of event A occurring, and P(B) is the probability of event B occurring.\n\nWhat is the probability of at least 2 of the 3 events would occur?\n\n1/4\nIf P(A) = 1/3, P(B) = 1/2 and P(C) = 1/4, then what is the probability of exactly 2 events occurring out of 3 events. Hence, probability that exactly 2 events occur out of 3 is 1/4.\n\nWhat is the probability that only one of the two events occur?\n\nSo, the probability that event 1 occurs and event 2 does not is p1⋅(1−p2). Similarly, the probability of event 2 happening and event 1 not happening is p2⋅(1−p1). Thus, the sum of the two probabilities, which is the probability of exactly one of the events happening, is p1⋅(1−p2)+p2(1−p1)=p1+p2−2p1p2."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9268447,"math_prob":0.995126,"size":3035,"snap":"2022-40-2023-06","text_gpt3_token_len":765,"char_repetition_ratio":0.28901353,"word_repetition_ratio":0.1163227,"special_character_ratio":0.25469524,"punctuation_ratio":0.10459588,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999893,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-29T11:36:50Z\",\"WARC-Record-ID\":\"<urn:uuid:3856018a-e2fa-471b-b305-52ac4ab088c7>\",\"Content-Length\":\"35681\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f3e5cec6-4d93-49e1-b57c-249754787551>\",\"WARC-Concurrent-To\":\"<urn:uuid:9915ced0-9cea-47c7-b506-292a734f4103>\",\"WARC-IP-Address\":\"104.21.25.202\",\"WARC-Target-URI\":\"https://turningtooneanother.net/2020/06/02/how-do-you-find-the-probability-of-individual-events/\",\"WARC-Payload-Digest\":\"sha1:UQYY7L4U5ZG7POSFHMW2ZCMEQTIHL5OP\",\"WARC-Block-Digest\":\"sha1:SGMZAJUGETZN7ELYNACDIZNBFCELGEFY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335350.36_warc_CC-MAIN-20220929100506-20220929130506-00639.warc.gz\"}"} |
https://zbmath.org/?q=an:0644.12006 | [
"# zbMATH — the first resource for mathematics\n\nQuelques résultats recents concernant les fonctions d’Igusa. (Some recent results concerning the Igusa functions). (French) Zbl 0644.12006\nSémin. Théor. Nombres, Univ. Bordeaux I 1986-1987, Exp. No. 25, 10 p. (1987).\nLet K be a local field of characteristic 0, let $$F\\in K[x_ 1,...,x_ n]$$ and let $$| dx|$$ be a Haar measure on $$K^ n$$. When $$\\Phi$$ is a locally constant function with compact support, the Igusa function associated to F is defined by $$Z_{\\Phi,K}(s)=\\iint_{K^ n}\\Phi | F|^ s| dx|$$. - In the archimedean case the author had proven that the poles of $$Z_{\\Phi,K}$$ are in the form s-n with s a root of the Bernstein polynomial b associated to F.\nNow in the non archimedean case, such a link between the poles of $$Z_{\\Phi,K}$$ and the roots of b does not exist. However, when $$n=2$$, the author shows that a pole of $$Z_{\\Phi,K}$$ has a real part which is a root of b. He then considers possible generalizations of this theorem when $$n>2$$, and works on technical results about it.\nReviewer: A.Escassut\n\n##### MSC:\n 11S80 Other analytic theory (analogues of beta and gamma functions, $$p$$-adic integration, etc.) 12J25 Non-Archimedean valued fields 11S40 Zeta functions and $$L$$-functions"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77278847,"math_prob":0.9986504,"size":1331,"snap":"2021-43-2021-49","text_gpt3_token_len":399,"char_repetition_ratio":0.113790505,"word_repetition_ratio":0.0,"special_character_ratio":0.30277985,"punctuation_ratio":0.16014235,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9993415,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-27T06:41:31Z\",\"WARC-Record-ID\":\"<urn:uuid:ea8895af-2ac2-4e74-a9aa-c27722a76da9>\",\"Content-Length\":\"46673\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bb0e32d9-9ac9-4aee-bc1c-ae526a85994c>\",\"WARC-Concurrent-To\":\"<urn:uuid:a3d27fab-9b8d-438d-8398-b0c1e1c0ca9d>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=an:0644.12006\",\"WARC-Payload-Digest\":\"sha1:3T7LMZNVUS5CXMEWCTNCXRYPQ37NM3HE\",\"WARC-Block-Digest\":\"sha1:XOP7ML2BDKXUFUNHWVHXY2BTHISKT6AW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323588102.27_warc_CC-MAIN-20211027053727-20211027083727-00376.warc.gz\"}"} |
https://socratic.org/questions/electronic-configration-of-alminum-in-excited-stated | [
"# Electronic configuration of Aluminum in excited state?\n\nThere could be three different configuration.\n\n#### Explanation:\n\nAluminium can release three electrons (don't be confused; electrons are released one by one) at most, so we call it trivalent.\n\nWhenever Aluminium releases a electron, It needs a minimum energy supply, which is called ionisation potential.\n\nWhen it releases its first electron from its valence shell, the ionistation potential or ionisation enthalpy needed is called first ionisation potential.\n\nLike this, we can define second and third ionisation potentials.\n\nSo, every time it releases elctrons, it absorbs more energy and becomes excited.\n\nSo, Electronic Configuration after becoming a univalent ion ($A {l}^{+}$):\n$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2}$ [$3 p$ shell is empty now.]\n\nElectronic Configuration after becoming a bivalent ion ($A {l}^{2 +}$):\n$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{1}$\n\nElectronic Configuration after becoming a trivalent ion ($A {l}^{3 +}$):\n\n$1 {s}^{2} 2 {s}^{2} 2 {p}^{6}$ [Now $3 s$ is empty]\n\nThe Most Excited State here is the third one.\n\nJun 3, 2017\n\n$\\left[N e\\right] 3 {s}^{2} \\textcolor{red}{3 {p}^{0}} 4 {s}^{1}$\n\nfor the first excited state, $3 p \\to 4 s$ transition, changing the valence electron configuration from $3 {s}^{2} 3 {p}^{1}$ to $3 {s}^{2} 4 {s}^{1}$. This is represented as (ignoring the $3 s$):\n\n$\\underline{\\textcolor{w h i t e}{\\uparrow \\downarrow}} \\text{ \"ul(color(white)(uarr darr))\" \"ul(color(white)(uarr darr))\" \"ul(color(white)(uarr darr))\" } \\underline{\\textcolor{w h i t e}{\\uparrow \\downarrow}}$\n$\\underbrace{\\text{ \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" }}$\n$\\text{ \"\" \"\" \"\" \"\" \"\" } 3 d$\n\n$\\underline{\\uparrow \\textcolor{w h i t e}{\\downarrow}}$\n$4 s$\n$\\text{ }$\n\n$\\underline{\\cancel{\\uparrow} \\textcolor{w h i t e}{\\downarrow}} \\text{ \"ul(color(white)(uarr darr))\" } \\underline{\\textcolor{w h i t e}{\\uparrow \\downarrow}}$\n$\\underbrace{\\text{ \"\" \"\" \"\" \"\" \"\" \"\" \"\" }}$\n$\\text{ \"\" \"\" } 3 p$\n\nThe normal electron configuration of aluminum is $\\left[N e\\right] 3 {s}^{2} 3 {p}^{1}$. Atomic excitations follow the following selection rules:\n\n• The change in total angular momentum, $\\Delta L$, must be $\\boldsymbol{\\pm 1}$, where $L = | {\\sum}_{k} {m}_{l , k} |$, $k$ indicates the $k$th electron, and ${m}_{l}$ is the magnetic quantum number of the orbital.\n• There must be no change in total spin, $S = | {\\sum}_{i} {m}_{s , i} |$, where $k$ indicates the $k$th electron and ${m}_{s}$ is the spin quantum number of the electron.\n\nFor the total spin $S$ to not change, i.e. $\\Delta S = 0$, our only option for aluminum is to transition from a half-filled orbital to an empty orbital.\n\nInitially, since we have only one $3 p$ electron, there is no sum and we have:\n\n${S}_{i} = | {m}_{s , 1} |$\n\n$= \\frac{1}{2}$\n\nfor:\n\n$\\underline{\\textcolor{w h i t e}{\\uparrow \\downarrow}} \\text{ \"ul(color(white)(uarr darr))\" \"ul(color(white)(uarr darr))\" \"ul(color(white)(uarr darr))\" } \\underline{\\textcolor{w h i t e}{\\uparrow \\downarrow}}$\n$\\underbrace{\\text{ \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" }}$\n$\\text{ \"\" \"\" \"\" \"\" \"\" } 3 d$\n\n$\\underline{\\textcolor{w h i t e}{\\uparrow \\downarrow}}$\n$4 s$\n$\\text{ }$\n\n$\\underline{\\uparrow \\textcolor{w h i t e}{\\downarrow}} \\text{ \"ul(color(white)(uarr darr))\" } \\underline{\\textcolor{w h i t e}{\\uparrow \\downarrow}}$\n$\\underbrace{\\text{ \"\" \"\" \"\" \"\" \"\" \"\" \"\" }}$\n$\\text{ \"\" \"\" } 3 p$\n\nNow, since for a transition upwards, only $\\Delta L = \\pm 1$ is allowed, we should consider that right now, for the one $3 p$ electron,\n\n${L}_{i} = {m}_{l , 1} = | - 1 | = 1$.\n\nThe nearby orbitals in energy are $3 d$ and $4 s$. For aluminum, the $4 s$ orbital is closer in energy to the $3 p$ than the $3 d$ is.\n\nThat means the $3 p$ electron can transition upwards into the $4 s$ for the first excited state (which is what we want). This gives:\n\n$\\Delta L = {L}_{f} - {L}_{i} = 0 - 1 = - 1$,\n\nwhich is allowed. Therefore, the first excited state electron configuration is:\n\n$\\textcolor{b l u e}{\\left[N e\\right] 3 {s}^{2} \\textcolor{red}{3 {p}^{0}} 4 {s}^{1}}$\n\ndue to a $3 p \\to 4 s$ transition, with $\\Delta L = 0 - 1 = - 1$, and $\\Delta S = \\frac{1}{2} - \\frac{1}{2} = 0$. We have ended up with:\n\n$\\underline{\\textcolor{w h i t e}{\\uparrow \\downarrow}} \\text{ \"ul(color(white)(uarr darr))\" \"ul(color(white)(uarr darr))\" \"ul(color(white)(uarr darr))\" } \\underline{\\textcolor{w h i t e}{\\uparrow \\downarrow}}$\n$\\underbrace{\\text{ \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" }}$\n$\\text{ \"\" \"\" \"\" \"\" \"\" } 3 d$\n\n$\\underline{\\uparrow \\textcolor{w h i t e}{\\downarrow}}$\n$4 s$\n$\\text{ }$\n\n$\\underline{\\textcolor{w h i t e}{\\uparrow \\downarrow}} \\text{ \"ul(color(white)(uarr darr))\" } \\underline{\\textcolor{w h i t e}{\\uparrow \\downarrow}}$\n$\\underbrace{\\text{ \"\" \"\" \"\" \"\" \"\" \"\" \"\" }}$\n$\\text{ \"\" \"\" } 3 p$"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77629167,"math_prob":0.99913675,"size":3013,"snap":"2020-45-2020-50","text_gpt3_token_len":954,"char_repetition_ratio":0.32668662,"word_repetition_ratio":0.3243243,"special_character_ratio":0.3601062,"punctuation_ratio":0.08506224,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995999,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-25T07:59:22Z\",\"WARC-Record-ID\":\"<urn:uuid:79a5f10c-76ef-430c-8f28-b22fcc54a9f5>\",\"Content-Length\":\"41064\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:85cbb177-43d8-47fa-ad28-fbebda8c339d>\",\"WARC-Concurrent-To\":\"<urn:uuid:c5441897-7493-4c28-88a2-3fc8e79d5435>\",\"WARC-IP-Address\":\"216.239.36.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/electronic-configration-of-alminum-in-excited-stated\",\"WARC-Payload-Digest\":\"sha1:NIYJXR2PN3NKSAL6XAYAA75BRJLFD5RU\",\"WARC-Block-Digest\":\"sha1:FV3VPHSVNEXBPFNLLDHG5F5RQPQFZEO5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107888402.81_warc_CC-MAIN-20201025070924-20201025100924-00196.warc.gz\"}"} |
http://www.netlib.org/lapack/lapack-3.1.1/html/dgeqrf.f.html | [
"``` SUBROUTINE DGEQRF( M, N, A, LDA, TAU, WORK, LWORK, INFO )\n*\n* -- LAPACK routine (version 3.1) --\n* Univ. of Tennessee, Univ. of California Berkeley and NAG Ltd..\n* November 2006\n*\n* .. Scalar Arguments ..\nINTEGER INFO, LDA, LWORK, M, N\n* ..\n* .. Array Arguments ..\nDOUBLE PRECISION A( LDA, * ), TAU( * ), WORK( * )\n* ..\n*\n* Purpose\n* =======\n*\n* DGEQRF computes a QR factorization of a real M-by-N matrix A:\n* A = Q * R.\n*\n* Arguments\n* =========\n*\n* M (input) INTEGER\n* The number of rows of the matrix A. M >= 0.\n*\n* N (input) INTEGER\n* The number of columns of the matrix A. N >= 0.\n*\n* A (input/output) DOUBLE PRECISION array, dimension (LDA,N)\n* On entry, the M-by-N matrix A.\n* On exit, the elements on and above the diagonal of the array\n* contain the min(M,N)-by-N upper trapezoidal matrix R (R is\n* upper triangular if m >= n); the elements below the diagonal,\n* with the array TAU, represent the orthogonal matrix Q as a\n* product of min(m,n) elementary reflectors (see Further\n* Details).\n*\n* LDA (input) INTEGER\n* The leading dimension of the array A. LDA >= max(1,M).\n*\n* TAU (output) DOUBLE PRECISION array, dimension (min(M,N))\n* The scalar factors of the elementary reflectors (see Further\n* Details).\n*\n* WORK (workspace/output) DOUBLE PRECISION array, dimension (MAX(1,LWORK))\n* On exit, if INFO = 0, WORK(1) returns the optimal LWORK.\n*\n* LWORK (input) INTEGER\n* The dimension of the array WORK. LWORK >= max(1,N).\n* For optimum performance LWORK >= N*NB, where NB is\n* the optimal blocksize.\n*\n* If LWORK = -1, then a workspace query is assumed; the routine\n* only calculates the optimal size of the WORK array, returns\n* this value as the first entry of the WORK array, and no error\n* message related to LWORK is issued by XERBLA.\n*\n* INFO (output) INTEGER\n* = 0: successful exit\n* < 0: if INFO = -i, the i-th argument had an illegal value\n*\n* Further Details\n* ===============\n*\n* The matrix Q is represented as a product of elementary reflectors\n*\n* Q = H(1) H(2) . . . H(k), where k = min(m,n).\n*\n* Each H(i) has the form\n*\n* H(i) = I - tau * v * v'\n*\n* where tau is a real scalar, and v is a real vector with\n* v(1:i-1) = 0 and v(i) = 1; v(i+1:m) is stored on exit in A(i+1:m,i),\n* and tau in TAU(i).\n*\n* =====================================================================\n*\n* .. Local Scalars ..\nLOGICAL LQUERY\nINTEGER I, IB, IINFO, IWS, K, LDWORK, LWKOPT, NB,\n\\$ NBMIN, NX\n* ..\n* .. External Subroutines ..\nEXTERNAL DGEQR2, DLARFB, DLARFT, XERBLA\n* ..\n* .. Intrinsic Functions ..\nINTRINSIC MAX, MIN\n* ..\n* .. External Functions ..\nINTEGER ILAENV\nEXTERNAL ILAENV\n* ..\n* .. Executable Statements ..\n*\n* Test the input arguments\n*\nINFO = 0\nNB = ILAENV( 1, 'DGEQRF', ' ', M, N, -1, -1 )\nLWKOPT = N*NB\nWORK( 1 ) = LWKOPT\nLQUERY = ( LWORK.EQ.-1 )\nIF( M.LT.0 ) THEN\nINFO = -1\nELSE IF( N.LT.0 ) THEN\nINFO = -2\nELSE IF( LDA.LT.MAX( 1, M ) ) THEN\nINFO = -4\nELSE IF( LWORK.LT.MAX( 1, N ) .AND. .NOT.LQUERY ) THEN\nINFO = -7\nEND IF\nIF( INFO.NE.0 ) THEN\nCALL XERBLA( 'DGEQRF', -INFO )\nRETURN\nELSE IF( LQUERY ) THEN\nRETURN\nEND IF\n*\n* Quick return if possible\n*\nK = MIN( M, N )\nIF( K.EQ.0 ) THEN\nWORK( 1 ) = 1\nRETURN\nEND IF\n*\nNBMIN = 2\nNX = 0\nIWS = N\nIF( NB.GT.1 .AND. NB.LT.K ) THEN\n*\n* Determine when to cross over from blocked to unblocked code.\n*\nNX = MAX( 0, ILAENV( 3, 'DGEQRF', ' ', M, N, -1, -1 ) )\nIF( NX.LT.K ) THEN\n*\n* Determine if workspace is large enough for blocked code.\n*\nLDWORK = N\nIWS = LDWORK*NB\nIF( LWORK.LT.IWS ) THEN\n*\n* Not enough workspace to use optimal NB: reduce NB and\n* determine the minimum value of NB.\n*\nNB = LWORK / LDWORK\nNBMIN = MAX( 2, ILAENV( 2, 'DGEQRF', ' ', M, N, -1,\n\\$ -1 ) )\nEND IF\nEND IF\nEND IF\n*\nIF( NB.GE.NBMIN .AND. NB.LT.K .AND. NX.LT.K ) THEN\n*\n* Use blocked code initially\n*\nDO 10 I = 1, K - NX, NB\nIB = MIN( K-I+1, NB )\n*\n* Compute the QR factorization of the current block\n* A(i:m,i:i+ib-1)\n*\nCALL DGEQR2( M-I+1, IB, A( I, I ), LDA, TAU( I ), WORK,\n\\$ IINFO )\nIF( I+IB.LE.N ) THEN\n*\n* Form the triangular factor of the block reflector\n* H = H(i) H(i+1) . . . H(i+ib-1)\n*\nCALL DLARFT( 'Forward', 'Columnwise', M-I+1, IB,\n\\$ A( I, I ), LDA, TAU( I ), WORK, LDWORK )\n*\n* Apply H' to A(i:m,i+ib:n) from the left\n*\nCALL DLARFB( 'Left', 'Transpose', 'Forward',\n\\$ 'Columnwise', M-I+1, N-I-IB+1, IB,\n\\$ A( I, I ), LDA, WORK, LDWORK, A( I, I+IB ),\n\\$ LDA, WORK( IB+1 ), LDWORK )\nEND IF\n10 CONTINUE\nELSE\nI = 1\nEND IF\n*\n* Use unblocked code to factor the last or only block.\n*\nIF( I.LE.K )\n\\$ CALL DGEQR2( M-I+1, N-I+1, A( I, I ), LDA, TAU( I ), WORK,\n\\$ IINFO )\n*\nWORK( 1 ) = IWS\nRETURN\n*\n* End of DGEQRF\n*\nEND\n\n```"
] | [
null
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https://es.mathworks.com/matlabcentral/answers/16477-how-do-i-plot-lines-with-different-line-widths | [
"How do I plot lines with different line widths?\n\n9.483 views (last 30 days)\nLeor Greenberger on 22 Sep 2011\nAnswered: SHAILENDRA PANDEY on 11 Oct 2020\nHi,\nI want to do:\nplot(x1,y1,x2,y2,'LineWidth',8)\nbut the linewidth propery ends up applying to both lines. Do I have to use two plot functions with a hold on command to have line1 a different width than line2? Thanks.\nRamesh M on 28 Jul 2016\nit works thnx\n\nFangjun Jiang on 22 Sep 2011\nEdited: MathWorks Support Team on 8 Nov 2018\nTo plot two lines with different line widths, you can use either of these approaches.\n1. Return the two “Line” objects as an output argument from the “plot” function and then set the “LineWidth” property for each.\np = plot(x1,y1,x2,y2)\np(1).LineWidth = 5;\np(2).LineWidth = 10;\n2. Use the “hold on” command to plot the two lines separately. Specify the line width by setting the “LineWidth” property a name-value pair.\nplot(x1,y1,'LineWidth',5)\nhold on\nplot(x2,y2,'LineWidth',10)\nhold off\nMike Garrity on 8 Mar 2016\nJust FYI, there is an \"official\" syntax for setting a property to different values on different objects. However, it's really ugly, and doesn't work everywhere. For example, I don't think that the plot function accepts this form.\nIt looks like this:\nh = plot(x1,y1,x2,y2);\nset(h,{'LineWidth'},{5;10})\nThe property name and property value need to each be a cell array, and the shape of the value cell array has to match the shape of the handle cell array.\nThat said, you're really better off with 2 calls to set in this case.\n\nWayne King on 22 Sep 2011\nHi: You can use handles.\nh = plot(x1,y1,x2,y2);\nset(h(1),'linewidth',1);\nset(h(2),'linewidth',2);\n\nSHAILENDRA PANDEY on 11 Oct 2020\nx = 1:.01:10;\ny1 = sin(x);\ny2 = cos(x);\np = plot(x,y1,x,y2)\nset(p,{'LineWidth'},{5;10})\nLine width, specified as a positive value in points, where 1 point = 1/72 of an inch. If the line has markers, then the line width also affects the marker edges.\nThe line width cannot be thinner than the width of a pixel. If you set the line width to a value that is less than the width of a pixel on your system, the line displays as one pixel wide.\n\nHari Desanur on 15 Nov 2016\nEdited: Hari Desanur on 15 Nov 2016\nThe line width for a particular line can be set using line object handles. For example -\nl = plot(x1,y1,x2,y2);\nl(1).LineWidth = 3; % set line width of 3 for the first line (x1,y1)\nl(2).LineWidth = 6;\n\nMASTER WHOS on 14 Feb 2019\n1-What might be the problem? 2-What does linewidth property mean? 3-What to do?\n[I am using matlab R2016b]\nWhen I want to plot line with color and certain width, I recceive an error:\nTHE COMMAND\nplot(SNR_db,Pd_NSP_cat_mean(:,2),'b','LineWidth',2.5)\nTHE ERROR:\nError using plot\nThere is no LineWidth property on the Line class.\nError in PROJECTION_BASED_SHARING (line 76)\nplot(SNR_db ,Pd_NSP_cat_mean(:,1),'g','LineWidth ',2.5)"
] | [
null
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https://www.litscape.com/word_analysis/calculatedly | [
"# calculatedly in Scrabble®\n\nThe word calculatedly is playable in Scrabble®, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.\n\nCALCULATEDLY\n(243)\n\nCALCULATEDLY\n(243)\nCALCULATEDLY\n(216)\nCALCULATEDLY\n(162)\nCALCULATEDLY\n(144)\nCALCULATEDLY\n(100)\nCALCULATEDLY\n(96)\nCALCULATEDLY\n(88)\nCALCULATEDLY\n(88)\nCALCULATEDLY\n(84)\nCALCULATEDLY\n(84)\nCALCULATEDLY\n(80)\nCALCULATEDLY\n(80)\nCALCULATEDLY\n(80)\nCALCULATEDLY\n(80)\nCALCULATEDLY\n(69)\nCALCULATEDLY\n(66)\nCALCULATEDLY\n(56)\nCALCULATEDLY\n(52)\nCALCULATEDLY\n(48)\nCALCULATEDLY\n(48)\nCALCULATEDLY\n(48)\nCALCULATEDLY\n(46)\nCALCULATEDLY\n(44)\nCALCULATEDLY\n(44)\nCALCULATEDLY\n(36)\nCALCULATEDLY\n(30)\nCALCULATEDLY\n(28)\nCALCULATEDLY\n(27)\nCALCULATEDLY\n(26)\nCALCULATEDLY\n(26)\nCALCULATEDLY\n(26)\nCALCULATEDLY\n(23)\n\nCALCULATEDLY\n(243)\nCALCULATEDLY\n(216)\nCALCULATEDLY\n(162)\nCALCULATEDLY\n(144)\nCECALLY\n(106 = 56 + 50)\nCECALLY\n(104 = 54 + 50)\nCECALLY\n(101 = 51 + 50)\nCECALLY\n(101 = 51 + 50)\nCALCULATEDLY\n(100)\nACYLATE\n(98 = 48 + 50)\nACYLATE\n(98 = 48 + 50)\nACUTELY\n(98 = 48 + 50)\nACUTELY\n(98 = 48 + 50)\nCALCULATEDLY\n(96)\n(95 = 45 + 50)\nACYLATE\n(95 = 45 + 50)\nCECALLY\n(95 = 45 + 50)\nCECALLY\n(95 = 45 + 50)\nACUTELY\n(95 = 45 + 50)\nCECALLY\n(95 = 45 + 50)\nCECALLY\n(95 = 45 + 50)\nCECALLY\n(95 = 45 + 50)\nALLAYED\n(95 = 45 + 50)\n(94 = 44 + 50)\nALLAYED\n(94 = 44 + 50)\nCATCALL\n(94 = 44 + 50)\nCECALLY\n(92 = 42 + 50)\nCATCALL\n(92 = 42 + 50)\nCATCALL\n(92 = 42 + 50)\nCATCALL\n(92 = 42 + 50)\nCECALLY\n(90 = 40 + 50)\nACYLATE\n(90 = 40 + 50)\nCECALLY\n(90 = 40 + 50)\nACUATED\n(90 = 40 + 50)\nCAUDATE\n(90 = 40 + 50)\nACYLATE\n(89 = 39 + 50)\nACYLATE\n(89 = 39 + 50)\nACYLATE\n(89 = 39 + 50)\nACUTELY\n(89 = 39 + 50)\nACUTELY\n(89 = 39 + 50)\nACUTELY\n(89 = 39 + 50)\nACUTELY\n(89 = 39 + 50)\nACUTELY\n(89 = 39 + 50)\nACUTELY\n(89 = 39 + 50)\nALLAYED\n(89 = 39 + 50)\nACYLATE\n(89 = 39 + 50)\nACYLATE\n(89 = 39 + 50)\n(89 = 39 + 50)\nCAUDATE\n(89 = 39 + 50)\nACUATED\n(89 = 39 + 50)\nACYLATE\n(89 = 39 + 50)\nCECALLY\n(88 = 38 + 50)\nCALCULATEDLY\n(88)\nALLAYED\n(88 = 38 + 50)\nCALCULATEDLY\n(88)\nALLAYED\n(86 = 36 + 50)\nALLAYED\n(86 = 36 + 50)\nCECALLY\n(86 = 36 + 50)\nALLAYED\n(86 = 36 + 50)\nCATCALL\n(86 = 36 + 50)\nCATCALL\n(86 = 36 + 50)\nALLAYED\n(86 = 36 + 50)\nALLAYED\n(86 = 36 + 50)\n(86 = 36 + 50)\nALLAYED\n(86 = 36 + 50)\nCATCALL\n(86 = 36 + 50)\n(86 = 36 + 50)\nCAUDATE\n(86 = 36 + 50)\nCATCALL\n(86 = 36 + 50)\nCAUDATE\n(86 = 36 + 50)\nACUTELY\n(86 = 36 + 50)\n(86 = 36 + 50)\n(86 = 36 + 50)\n(86 = 36 + 50)\nLACTEAL\n(86 = 36 + 50)\nACUTELY\n(86 = 36 + 50)\nACYLATE\n(86 = 36 + 50)\n(86 = 36 + 50)\nCECALLY\n(86 = 36 + 50)\nACYLATE\n(86 = 36 + 50)\nACUATED\n(86 = 36 + 50)\nLACTEAL\n(86 = 36 + 50)\nCATCALL\n(86 = 36 + 50)\nACYLATE\n(84 = 34 + 50)\nALLAYED\n(84 = 34 + 50)\nCALCULATEDLY\n(84)\nCECALLY\n(84 = 34 + 50)\nCALCULATEDLY\n(84)\nACUTELY\n(84 = 34 + 50)\nCECALLY\n(84 = 34 + 50)\nCECALLY\n(84 = 34 + 50)\nCAUDATE\n(83 = 33 + 50)\nCAUDATE\n(83 = 33 + 50)\nCAUDATE\n(83 = 33 + 50)\nCAUDATE\n(83 = 33 + 50)\nACUATED\n(83 = 33 + 50)\nCAUDATE\n(83 = 33 + 50)\nACUATED\n(83 = 33 + 50)\n(83 = 33 + 50)\nACUATED\n(83 = 33 + 50)\nACUATED\n(83 = 33 + 50)\nALLAYED\n(83 = 33 + 50)\nACUATED\n(83 = 33 + 50)\nACUATED\n(83 = 33 + 50)\nCATCALL\n(83 = 33 + 50)\nACUTELY\n(82 = 32 + 50)\n(82 = 32 + 50)\n(82 = 32 + 50)\nACYLATE\n(82 = 32 + 50)\nACUTELY\n(82 = 32 + 50)\nCECALLY\n(82 = 32 + 50)\nCECALLY\n(82 = 32 + 50)\nCECALLY\n(82 = 32 + 50)\nACUATED\n(82 = 32 + 50)\nLACTEAL\n(80 = 30 + 50)\nLACTEAL\n(80 = 30 + 50)\nALLAYED\n(80 = 30 + 50)\nLACTEAL\n(80 = 30 + 50)\n(80 = 30 + 50)\nLACTEAL\n(80 = 30 + 50)\nLACTEAL\n(80 = 30 + 50)\n(80 = 30 + 50)\nLACTEAL\n(80 = 30 + 50)\n(80 = 30 + 50)\nLACTEAL\n(80 = 30 + 50)\nLACTEAL\n(80 = 30 + 50)\n(80 = 30 + 50)\nACUATED\n(80 = 30 + 50)\nACYLATE\n(80 = 30 + 50)\nCAUDATE\n(80 = 30 + 50)\nCECALLY\n(80 = 30 + 50)\nCECALLY\n(80 = 30 + 50)\nACUTELY\n(80 = 30 + 50)\nCALCULATEDLY\n(80)\nCECALLY\n(80 = 30 + 50)\nCALCULATEDLY\n(80)\nCATCALL\n(80 = 30 + 50)\nCALCULATEDLY\n(80)\nCALCULATEDLY\n(80)\nACUTELY\n(78 = 28 + 50)\nACUTELY\n(78 = 28 + 50)\nACUTELY\n(78 = 28 + 50)\nACUTELY\n(78 = 28 + 50)\nACYLATE\n(78 = 28 + 50)\nCATCALL\n(78 = 28 + 50)\nCAUDATE\n(78 = 28 + 50)\nCECALLY\n(78 = 28 + 50)\nACYLATE\n(78 = 28 + 50)\nCECALLY\n(78 = 28 + 50)\nCECALLY\n(78 = 28 + 50)\nCECALLY\n(78 = 28 + 50)\nCECALLY\n(78 = 28 + 50)\nCECALLY\n(78 = 28 + 50)\nCATCALL\n(78 = 28 + 50)\nACYLATE\n(78 = 28 + 50)\n(77 = 27 + 50)\nLACTEAL\n(77 = 27 + 50)\n(77 = 27 + 50)\n(77 = 27 + 50)\n(77 = 27 + 50)\n(77 = 27 + 50)\n(77 = 27 + 50)\n(77 = 27 + 50)\nALLAYED\n(76 = 26 + 50)\nALLAYED\n(76 = 26 + 50)\nCATCALL\n(76 = 26 + 50)\nACYLATE\n(76 = 26 + 50)\nACUTELY\n(76 = 26 + 50)\nACUTELY\n(76 = 26 + 50)\nCAUDATE\n(76 = 26 + 50)\n(76 = 26 + 50)\nALLAYED\n(76 = 26 + 50)\nACUATED\n(76 = 26 + 50)\nALLAYED\n(76 = 26 + 50)\nACYLATE\n(76 = 26 + 50)\nCATCALL\n(76 = 26 + 50)\nACYLATE\n(76 = 26 + 50)\nCATCALL\n(76 = 26 + 50)\nCATCALL\n(76 = 26 + 50)\n(76 = 26 + 50)\n(76 = 26 + 50)\nACYLATE\n(76 = 26 + 50)\nACYLATE\n(76 = 26 + 50)\nACUTELY\n(76 = 26 + 50)\nCAUDATE\n(76 = 26 + 50)\nALLAYED\n(76 = 26 + 50)\nACUTELY\n(76 = 26 + 50)\nCATCALL\n(76 = 26 + 50)\nACUATED\n(76 = 26 + 50)\n(76 = 26 + 50)\nALLAYED\n(76 = 26 + 50)\nLACTEAL\n(76 = 26 + 50)\nACUTELY\n(76 = 26 + 50)\n(76 = 26 + 50)\nACYLATE\n(76 = 26 + 50)\nCATCALL\n(74 = 24 + 50)\nACUTELY\n(74 = 24 + 50)\nCAUDATE\n(74 = 24 + 50)\nALLAYED\n(74 = 24 + 50)\nCATCALL\n(74 = 24 + 50)\n\n# calculatedly in Words With Friends™\n\nThe word calculatedly is playable in Words With Friends™, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.\n\nCALCULATEDLY\n(333)\n\nCALCULATEDLY\n(333)\nCALCULATEDLY\n(279)\nCALCULATEDLY\n(279)\nCALCULATEDLY\n(261)\nCALCULATEDLY\n(234)\nCALCULATEDLY\n(210)\nCALCULATEDLY\n(124)\nCALCULATEDLY\n(116)\nCALCULATEDLY\n(108)\nCALCULATEDLY\n(108)\nCALCULATEDLY\n(100)\nCALCULATEDLY\n(100)\nCALCULATEDLY\n(100)\nCALCULATEDLY\n(100)\nCALCULATEDLY\n(70)\nCALCULATEDLY\n(66)\nCALCULATEDLY\n(66)\nCALCULATEDLY\n(62)\nCALCULATEDLY\n(62)\nCALCULATEDLY\n(62)\nCALCULATEDLY\n(37)\nCALCULATEDLY\n(35)\nCALCULATEDLY\n(34)\nCALCULATEDLY\n(33)\nCALCULATEDLY\n(33)\nCALCULATEDLY\n(32)\nCALCULATEDLY\n(31)\nCALCULATEDLY\n(31)\nCALCULATEDLY\n(30)\nCALCULATEDLY\n(29)\nCALCULATEDLY\n(29)\nCALCULATEDLY\n(28)\n\nCALCULATEDLY\n(333)\nCALCULATEDLY\n(279)\nCALCULATEDLY\n(279)\nCALCULATEDLY\n(261)\nCALCULATEDLY\n(234)\nCALCULATEDLY\n(210)\nCECALLY\n(134 = 99 + 35)\nCALCULATEDLY\n(124)\nCECALLY\n(116 = 81 + 35)\nCATCALL\n(116 = 81 + 35)\nCALCULATEDLY\n(116)\nCATCALL\n(110 = 75 + 35)\nCECALLY\n(110 = 75 + 35)\nACYLATE\n(110 = 75 + 35)\nCATCALL\n(110 = 75 + 35)\nCALCULATEDLY\n(108)\nCALCULATEDLY\n(108)\nACUTELY\n(107 = 72 + 35)\nCAUDATE\n(107 = 72 + 35)\nLACTEAL\n(107 = 72 + 35)\nCATCALL\n(104 = 69 + 35)\nCECALLY\n(104 = 69 + 35)\nCECALLY\n(104 = 69 + 35)\nCATCALL\n(104 = 69 + 35)\nCATCALL\n(104 = 69 + 35)\nCECALLY\n(103 = 68 + 35)\nCECALLY\n(103 = 68 + 35)\nCECALLY\n(103 = 68 + 35)\nACUTELY\n(101 = 66 + 35)\nACUATED\n(101 = 66 + 35)\nACUTELY\n(101 = 66 + 35)\nALLAYED\n(101 = 66 + 35)\nCALCULATEDLY\n(100)\nCALCULATEDLY\n(100)\nCALCULATEDLY\n(100)\nCALCULATEDLY\n(100)\nACYLATE\n(98 = 63 + 35)\nCECALLY\n(98 = 63 + 35)\nACYLATE\n(98 = 63 + 35)\n(98 = 63 + 35)\nCATCALL\n(98 = 63 + 35)\n(98 = 63 + 35)\nCECALLY\n(98 = 63 + 35)\n(98 = 63 + 35)\nCYCLED\n(96)\nCAUDATE\n(95 = 60 + 35)\nACUTELY\n(95 = 60 + 35)\nCATCALL\n(95 = 60 + 35)\nCATCALL\n(95 = 60 + 35)\nACUTELY\n(95 = 60 + 35)\nCATCALL\n(95 = 60 + 35)\nACUATED\n(95 = 60 + 35)\nACUTELY\n(95 = 60 + 35)\nCECALLY\n(92 = 57 + 35)\nCECALLY\n(92 = 57 + 35)\nACYLATE\n(92 = 57 + 35)\nCATCALL\n(92 = 57 + 35)\nCATCALL\n(92 = 57 + 35)\nCECALLY\n(92 = 57 + 35)\n(92 = 57 + 35)\n(92 = 57 + 35)\nACUTELY\n(91 = 56 + 35)\nACUTELY\n(91 = 56 + 35)\nACUTELY\n(91 = 56 + 35)\n(89 = 54 + 35)\nALLAYED\n(89 = 54 + 35)\nCAUDATE\n(89 = 54 + 35)\nCAUDATE\n(89 = 54 + 35)\nALLAYED\n(89 = 54 + 35)\nLACTEAL\n(89 = 54 + 35)\nACUATED\n(89 = 54 + 35)\nACUATED\n(89 = 54 + 35)\nACUTELY\n(89 = 54 + 35)\n(87 = 52 + 35)\nACYLATE\n(87 = 52 + 35)\nACYLATE\n(87 = 52 + 35)\nACYLATE\n(87 = 52 + 35)\n(87 = 52 + 35)\n(87 = 52 + 35)\n(86 = 51 + 35)\n(86 = 51 + 35)\nCATCALL\n(86 = 51 + 35)\nACYLATE\n(86 = 51 + 35)\nACYLATE\n(86 = 51 + 35)\n(86 = 51 + 35)\nACYLATE\n(86 = 51 + 35)\n(86 = 51 + 35)\nCECALLY\n(85 = 50 + 35)\nCECALLY\n(85 = 50 + 35)\nACUATED\n(83 = 48 + 35)\nLACTEAL\n(83 = 48 + 35)\nCAUDATE\n(83 = 48 + 35)\nACUTELY\n(83 = 48 + 35)\nLACTEAL\n(83 = 48 + 35)\nLACTEAL\n(83 = 48 + 35)\nCAUDATE\n(83 = 48 + 35)\nCAUDATE\n(83 = 48 + 35)\nLACTEAL\n(83 = 48 + 35)\nCAUDATE\n(83 = 48 + 35)\nCAUDATE\n(83 = 48 + 35)\nLACTEAL\n(83 = 48 + 35)\nCAUDATE\n(83 = 48 + 35)\nACUTELY\n(83 = 48 + 35)\nLACTEAL\n(83 = 48 + 35)\nLACTEAL\n(83 = 48 + 35)\nACUTELY\n(83 = 48 + 35)\nACUATED\n(83 = 48 + 35)\nALLAYED\n(83 = 48 + 35)\n(83 = 48 + 35)\nACUATED\n(83 = 48 + 35)\nALLAYED\n(83 = 48 + 35)\n(83 = 48 + 35)\nALLAYED\n(83 = 48 + 35)\nALLAYED\n(83 = 48 + 35)\nACUATED\n(83 = 48 + 35)\nACUATED\n(83 = 48 + 35)\nALLAYED\n(83 = 48 + 35)\nALLAYED\n(83 = 48 + 35)\nCECALLY\n(81 = 46 + 35)\nCATCALL\n(81 = 46 + 35)\n(80 = 45 + 35)\nACYLATE\n(80 = 45 + 35)\nACYLATE\n(80 = 45 + 35)\nACYLATE\n(80 = 45 + 35)\nACUTELY\n(79 = 44 + 35)\nCECALLY\n(77 = 42 + 35)\nLACTEAL\n(77 = 42 + 35)\nALLAYED\n(77 = 42 + 35)\nLACTEAL\n(77 = 42 + 35)\nALLAYED\n(77 = 42 + 35)\n(77 = 42 + 35)\nCAUDATE\n(77 = 42 + 35)\nALLAYED\n(77 = 42 + 35)\nACUATED\n(77 = 42 + 35)\nCECALLY\n(77 = 42 + 35)\nCAUDATE\n(77 = 42 + 35)\nLACTEAL\n(77 = 42 + 35)\nACYLATE\n(77 = 42 + 35)\nACUATED\n(77 = 42 + 35)\nACUATED\n(77 = 42 + 35)\n(77 = 42 + 35)\nLACTEAL\n(77 = 42 + 35)\nALLAYED\n(77 = 42 + 35)\n(77 = 42 + 35)\nCECALLY\n(77 = 42 + 35)\n(77 = 42 + 35)\nCECALLY\n(77 = 42 + 35)\nCAUDATE\n(77 = 42 + 35)\nACUATED\n(77 = 42 + 35)\nDACTYL\n(75)\nCAECAL\n(75)\nCULLED\n(75)\nCECALLY\n(75 = 40 + 35)\nACUTELY\n(75 = 40 + 35)\n(75 = 40 + 35)\n(75 = 40 + 35)\nCAUDATE\n(75 = 40 + 35)\nLACTEAL\n(75 = 40 + 35)\nACUATED\n(75 = 40 + 35)\n(75 = 40 + 35)\nCATCALL\n(73 = 38 + 35)\nCECALLY\n(73 = 38 + 35)\n(73 = 38 + 35)\nCECALLY\n(73 = 38 + 35)\nCECALLY\n(73 = 38 + 35)\nACYLATE\n(73 = 38 + 35)\nCATCALL\n(73 = 38 + 35)\nCATCALL\n(73 = 38 + 35)\nCYCLED\n(72)\nCAUDAL\n(72)\nCALLED\n(72)\nCYCLED\n(72)\nDUCTAL\n(72)\nCYCLED\n(72)\n(71 = 36 + 35)\nACUTELY\n(71 = 36 + 35)\nACUTELY\n(71 = 36 + 35)\nCECALLY\n(71 = 36 + 35)\n(71 = 36 + 35)\n(71 = 36 + 35)\nACUTELY\n(71 = 36 + 35)\nALLAYED\n(71 = 36 + 35)\nCALCULATEDLY\n(70)\nCAECAL\n(69)\nCUTELY\n(69)\nCATCALL\n(69 = 34 + 35)\nCECALLY\n(69 = 34 + 35)\n(69 = 34 + 35)\nCATCALL\n(69 = 34 + 35)\nCECALLY\n(69 = 34 + 35)\nCECALLY\n(69 = 34 + 35)\nCATCALL\n(69 = 34 + 35)\nCECALLY\n(69 = 34 + 35)\nCECALLY\n(69 = 34 + 35)\nCATCALL\n(69 = 34 + 35)\n(69 = 34 + 35)\nCATCALL\n(69 = 34 + 35)\nCECALLY\n(69 = 34 + 35)\nACYLATE\n(69 = 34 + 35)\n(69 = 34 + 35)\n\n# Word Growth involving calculatedly\n\n## Shorter words in calculatedly\n\nat ate late calculate calculated\n\nla late calculate calculated\n\n## Longer words containing calculatedly\n\n(No longer words found)"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9398552,"math_prob":1.00001,"size":901,"snap":"2020-34-2020-40","text_gpt3_token_len":249,"char_repetition_ratio":0.26867336,"word_repetition_ratio":0.6,"special_character_ratio":0.1864595,"punctuation_ratio":0.18181819,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000079,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-01T03:48:52Z\",\"WARC-Record-ID\":\"<urn:uuid:c0ac3fe6-75c1-4528-8044-465f36362410>\",\"Content-Length\":\"156740\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1b6df89e-4809-4a0b-bff1-ad8154e866a8>\",\"WARC-Concurrent-To\":\"<urn:uuid:68681a2a-8b90-4156-a10b-c69206d0c825>\",\"WARC-IP-Address\":\"172.67.184.45\",\"WARC-Target-URI\":\"https://www.litscape.com/word_analysis/calculatedly\",\"WARC-Payload-Digest\":\"sha1:OVTGGH64XCPUEY2PZKRRE7Z2OLKXDWSM\",\"WARC-Block-Digest\":\"sha1:ORKQHCDC2YVUPY3542RXPVNKTDR33TA3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600402130615.94_warc_CC-MAIN-20201001030529-20201001060529-00015.warc.gz\"}"} |
https://math.stackexchange.com/users/77764/jenq?tab=topactivity | [
"### Questions (2)\n\n 4 Integrating $\\int \\left( \\frac{\\cos\\theta}{1+\\sin^2\\theta} \\right)^n \\, d\\theta$, for $n=2$ and $n=3$ 2 Why do we seek real-valued solutions to second-order linear homogeneous ODEs when the characteristic equation has complex roots?\n\n### Reputation (61)\n\nThis user has no recent positive reputation changes\n\nThis user has not answered any questions\n\n### Tags (4)\n\n 0 integration 0 calculus 0 trigonometry 0 ordinary-differential-equations\n\n### Accounts (8)\n\n Mathematics 61 rep 44 bronze badges Earth Science 21 rep 11 bronze badge MathOverflow 3 rep 33 bronze badges Stack Overflow 1 rep Mathematics Educators 1 rep"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7656809,"math_prob":0.95457214,"size":367,"snap":"2020-10-2020-16","text_gpt3_token_len":97,"char_repetition_ratio":0.14600551,"word_repetition_ratio":0.0,"special_character_ratio":0.27792916,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9867818,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-18T00:48:41Z\",\"WARC-Record-ID\":\"<urn:uuid:0e2ec8b3-b167-44bd-bd40-e304aa2fa1d1>\",\"Content-Length\":\"107433\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fa263d01-ce74-41f0-8e9e-48f6be54732c>\",\"WARC-Concurrent-To\":\"<urn:uuid:3e275261-5d5b-4738-b844-b57b84e844cb>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/users/77764/jenq?tab=topactivity\",\"WARC-Payload-Digest\":\"sha1:2IUFNTNT4RAC3J56J7UHUL6RJBH7QV6R\",\"WARC-Block-Digest\":\"sha1:6QCLC2QFISQTZY5T6TEJZQRYMRGUT22D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875143455.25_warc_CC-MAIN-20200217235417-20200218025417-00051.warc.gz\"}"} |
https://www.rdocumentation.org/packages/spatstat/versions/1.55-1/topics/update.ppm | [
"update.ppm\n\n0th\n\nPercentile\n\nUpdate a Fitted Point Process Model\n\nupdate method for class \"ppm\".\n\nKeywords\nmodels, methods, spatial\nUsage\n# S3 method for ppm\nupdate(object, …, fixdummy=TRUE, use.internal=NULL,\nenvir=environment(terms(object)))\nArguments\nobject\n\nAn existing fitted point process model, typically produced by ppm.\n\nArguments to be updated in the new call to ppm.\n\nfixdummy\n\nLogical flag indicating whether the quadrature scheme for the call to ppm should use the same set of dummy points as that in the original call.\n\nuse.internal\n\nOptional. Logical flag indicating whether the model should be refitted using the internally saved data (use.internal=TRUE) or by re-evaluating these data in the current frame (use.internal=FALSE).\n\nenvir\n\nEnvironment in which to re-evaluate the call to ppm.\n\nDetails\n\nThis is a method for the generic function update for the class \"ppm\". An object of class \"ppm\" describes a fitted point process model. See ppm.object) for details of this class.\n\nupdate.ppm will modify the point process model specified by object according to the new arguments given, then re-fit it. The actual re-fitting is performed by the model-fitting function ppm.\n\nIf you are comparing several model fits to the same data, or fits of the same model to different data, it is strongly advisable to use update.ppm rather than trying to fit them by hand. This is because update.ppm re-fits the model in a way which is comparable to the original fit.\n\nThe arguments ... are matched to the formal arguments of ppm as follows.\n\nFirst, all the named arguments in ... are matched with the formal arguments of ppm. Use name=NULL to remove the argument name from the call.\n\nSecond, any unnamed arguments in ... are matched with formal arguments of ppm if the matching is obvious from the class of the object. Thus ... may contain\n\n• exactly one argument of class \"ppp\" or \"quad\", which will be interpreted as the named argument Q;\n\n• exactly one argument of class \"formula\", which will be interpreted as the named argument trend (or as specifying a change to the trend formula);\n\n• exactly one argument of class \"interact\", which will be interpreted as the named argument interaction;\n\n• exactly one argument of class \"data.frame\", which will be interpreted as the named argument covariates.\n\nThe trend argument can be a formula that specifies a change to the current trend formula. For example, the formula ~ . + Z specifies that the additional covariate Z will be added to the right hand side of the trend formula in the existing object.\n\nThe argument fixdummy=TRUE ensures comparability of the objects before and after updating. When fixdummy=FALSE, calling update.ppm is exactly the same as calling ppm with the updated arguments. However, the original and updated models are not strictly comparable (for example, their pseudolikelihoods are not strictly comparable) unless they used the same set of dummy points for the quadrature scheme. Setting fixdummy=TRUE ensures that the re-fitting will be performed using the same set of dummy points. This is highly recommended.\n\nThe value of use.internal determines where to find data to re-evaluate the model (data for the arguments mentioned in the original call to ppm that are not overwritten by arguments to update.ppm).\n\nIf use.internal=FALSE, then arguments to ppm are re-evaluated in the frame where you call update.ppm. This is like the behaviour of the other methods for update. This means that if you have changed any of the objects referred to in the call, these changes will be taken into account. Also if the original call to ppm included any calls to random number generators, these calls will be recomputed, so that you will get a different outcome of the random numbers.\n\nIf use.internal=TRUE, then arguments to ppm are extracted from internal data stored inside the current fitted model object. This is useful if you don't want to re-evaluate anything. It is also necessary if if object has been restored from a dump file using load or source. In such cases, we have lost the environment in which object was fitted, and data cannot be re-evaluated.\n\nBy default, if use.internal is missing, update.ppm will re-evaluate the arguments if this is possible, and use internal data if not.\n\nValue\n\nAnother fitted point process model (object of class \"ppm\").\n\n• update.ppm\nExamples\n# NOT RUN {\ndata(nztrees)\ndata(cells)\n\n# fit the stationary Poisson process\nfit <- ppm(nztrees, ~ 1)\n\n# fit a nonstationary Poisson process\nfitP <- update(fit, trend=~x)\nfitP <- update(fit, ~x)\n\n# change the trend formula: add another term to the trend\nfitPxy <- update(fitP, ~ . + y)\n# change the trend formula: remove the x variable\nfitPy <- update(fitPxy, ~ . - x)\n\n# fit a stationary Strauss process\nfitS <- update(fit, interaction=Strauss(13))\nfitS <- update(fit, Strauss(13))\n\n# refit using a different edge correction\nfitS <- update(fitS, correction=\"isotropic\")\n\n# re-fit the model to a subset\n# of the original point pattern\nnzw <- owin(c(0,148),c(0,95))\nnzsub <- nztrees[,nzw]\nfut <- update(fitS, Q=nzsub)\nfut <- update(fitS, nzsub)\n\n# WARNING: the point pattern argument is called 'Q'\n\nranfit <- ppm(rpoispp(42), ~1, Poisson())\nranfit\n# different random data!\nupdate(ranfit)\n# the original data\nupdate(ranfit, use.internal=TRUE)\n\n# }\nDocumentation reproduced from package spatstat, version 1.55-1, License: GPL (>= 2)\n\nCommunity examples\n\nLooks like there are no examples yet."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.67965275,"math_prob":0.74183184,"size":5207,"snap":"2019-43-2019-47","text_gpt3_token_len":1304,"char_repetition_ratio":0.13665193,"word_repetition_ratio":0.04588235,"special_character_ratio":0.22431342,"punctuation_ratio":0.12462006,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96643853,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-15T22:25:27Z\",\"WARC-Record-ID\":\"<urn:uuid:0a54f2a8-78ec-4d73-b061-cc00956c2815>\",\"Content-Length\":\"21300\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7cdc4654-8a4c-4301-a397-1996b5ec4c26>\",\"WARC-Concurrent-To\":\"<urn:uuid:61be5e1e-369e-4261-bd0f-b48f480860bc>\",\"WARC-IP-Address\":\"35.174.49.12\",\"WARC-Target-URI\":\"https://www.rdocumentation.org/packages/spatstat/versions/1.55-1/topics/update.ppm\",\"WARC-Payload-Digest\":\"sha1:JAHL632TA5HPHFA5CBQ3TIMSUTPI6W7V\",\"WARC-Block-Digest\":\"sha1:WMFAS7GPZOY3LLZYEV63NFXEYVULUO3Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986660323.32_warc_CC-MAIN-20191015205352-20191015232852-00054.warc.gz\"}"} |
http://puzzleszone.blogspot.com/2008/06/heavy-marble-puzzle-what-is-minimum.html | [
"### Heavy Marble Puzzle- What is the minimum number of weighings needed to be certain of identifying the heavy marble?\n\n You have eight marbles and a two-pan balance. All the marbles weigh the same, except for one, which is heavier than all the others. The marbles are otherwise indistinguishable. You may make no assumptions about how much heavier the heavy marble is. What is the minimum number of weighings needed to be certain of identifying the heavy marble?\n\nThe first step in solving this problem is to realize that you can put more than one marble in each pan of the balance. If you have equal numbers of marbles in each pan, then the heavy marble must be in the group on the heavy side of the balance. This saves you from having to weigh each marble individually, and it enables you to eliminate many marbles in a single weighing.\n\nOnce you realize this, you are likely to devise a binary search-based strategy for finding the heavy marble. In this method, you begin by putting half of the marbles on each side of the balance. Then you eliminate the marbles from the light side of the balance and divide the marbles from the heavy side of the balance between the two pans. As shown in Figure 1, you continue this process until each pan holds only one marble, at which point the heavy marble is the only marble on the heavy side of the balance. Using this process you can always identify the heavy marble in three weighings.",
null,
"This may seem to be the correct answer. The solution wasn’t completely obvious, and it’s an improvement over weighing the marbles one by one. If you’re telling yourself that this seemed too easy, you’re right. The method described so far is a good start, but it’s not the best you can do.\n\nHow can you find the heavy marble in fewer than three weighings? Obviously, you’ll have to eliminate more than half the marbles at each weighing, but how can you do that?\n\nTry looking at this problem from an information flow perspective. Information about the marbles comes from the balance, and you use this information to identify the heavy marble. The more information you derive from each weighing, the more efficient your search for the marble will be. Think about how you get information from the balance: You place marbles on it and then look at the result. What are all the possible results? The left pan side could be heavier, the right side could be heavier, or both sides could weigh exactly the same. So there are three possible results, but so far you’ve been using only two of them. In effect, you’re only using 2/3 of the information that each weighing provides. Perhaps if you alter your method so that you use all of the information from each weighing you will be able to find the heavy marble in fewer weighings.\n\nUsing the binary search strategy, the heavy marble is always in one of the two pans, so there will always be a heavy side of the balance. In other words, you can’t take advantage of all the information the balance can provide if the heavy marble is always on the balance. What if you divided the marbles into three equal-sized groups, and weighed two of the groups on the balance? Just as before, if either side of the balance is heavier, you know that the heavy marble is in the group on that side. But now it’s also possible that the two groups of marbles on the balance will weigh the same - in this case, the heavy marble must be in the third group that’s not on the balance. Because you divided the marbles into three groups, keeping just the group with the heavy marble eliminates 2/3 of the marbles instead of half of them. This seems promising.\n\nThere’s still a minor wrinkle to work out before you can apply this process to the problem at hand. Eight isn’t evenly divisible by three, so you can’t divide the eight marbles into three equal groups. Why do you need the same number of marbles in each group? You need the same number of marbles so that when you put the groups on the balance the result doesn’t have anything to do with differing numbers of marbles on each side. Really, you need only two of the groups to be the same size. You’ll still want all three groups to be approximately the same size so you can eliminate approximately of the marbles after each weighing no matter which pile has the heavy marble.\n\nNow you can apply the three-group technique to the problem you were given. Begin by dividing the marbles into two groups of three, which you put on the balance, and one group of two, which you leave off. If the two sides weigh the same, the heavy marble is in the group of two, and you can find it with one more weighing, for a total of two weighings. On the other hand, if either side of the balance is heavier, the heavy marble must be in that group of three. You can eliminate all the other marbles, and place one marble from this group on either side of the balance, leaving the third marble aside. If one side is heavier, it contains the heavy marble; if neither side is heavier, the heavy marble is the one you didn’t place on the balance. This is also a total of two weighings, so you can always find the heavy marble in a group of eight using two weighings. An example of this process is illustrated in Figure 2",
null,
"Important Generalize your solution. What is the minimum number of weighings to find a heavy marble among n marbles?\n\nThis is the part where the interviewer determines whether you hit on the preceding solution by luck or because you really understand it. Think about what happens after each weighing. You eliminate 2/3 of the marbles and keep 1/3. After each weighing you have 1/3 as many marbles as you did before. When you get down to one marble, you’ve found the heavy marble.\n\nBased on this reasoning, you can reformulate the question as, “How many times do you have to divide the number of marbles by 3 before you end up with 1?” If you start with 3 marbles, you divide by 3 once to get 1, so it takes one weighing. If you start with 9 marbles you divide by 3 twice, so it takes two weighings. Similarly, 27 marbles require three weighings. What mathematical operation can you use to represent this “How many times do you divide by 3 to get to 1” process?\n\nBecause multiplication and division are inverse operations, the number of times you have to divide the number of marbles by 3 before you end up with 1 is the same as the number of times you have to multiply by 3 (starting at 1) before you get to the number of marbles. Repeated multiplication is expressed using exponents. If you want to express multiplying by 3 twice, you can write 32, which is equal to 9. When you multiply twice by 3 you get 9 - it takes two weighings to find the heavy marble among 9 marbles. In more general terms, it takes i weighings to find the heavy marble from among n marbles, where 3i = n. You know the value of n and want to calculate i, so you need to solve this for i. You can solve for i using logarithms, the inverse operation of exponentiation. If you take log3 of both sides of the preceding equation you get i = log3 n.\n\nThis works fine as long as n is a power of 3. However, if n isn’t a power of 3, then this equation calculates a noninteger value for i, which doesn’t make much sense, given that it’s extremely difficult to perform a fractional weighing. For example, if n = 8, as in the previous part of the problem, log3 8 is some number between 1 and 2 (1.893 to be a little more precise). From your previous experience, you know it actually takes two weighings when you have eight marbles. This seems to indicate that if you calculate a fractional number of weighings you should round it up to the nearest integer.\n\nDoes this make sense? Try applying it to n = 10 and see whether you can justify always rounding up. log3 9 is 2, so log3 10 will be a little more than two, or three if you round up to the nearest integer. Is that the correct number of weighings for 10 marbles? For 10 marbles, you would start out with two groups of 3 and one group of 4. If the heavy marble were in either of the groups of 3, you could find it with just one more weighing, but if it turned out to be in the group of 4 you might need as many as two more weighings for a total of 3, just as you calculated. In this case the fractional weighing seems to represent a weighing that you might need to make under some circumstances (if the heavy marble happens to be in the larger group) but not others. Because you’re trying to calculate the number of weighings needed to guarantee you’ll find the heavy marble, you have to count that fractional weighing as a full weighing even though you won’t always perform it, so it makes sense to always round up to the nearest integer. In programming, the function that rounds up to the nearest integer is often called ceiling, so you might express the minimum number of weighings needed to guarantee you’ll find the heavy marble among n marbles as ceiling(log3(n)).\n\n Tip For the group of 4 (out of the total of 10 marbles), you would divide the 4 marbles into two groups of 1 and one group of 2. If the heavy marble happened to be in the group of 2, you would need one more weighing (the third weighing) to determine which was the heavy marble. A fractional weighing may also represent a weighing that will always be performed but won’t eliminate a full 2 2/3 of the remaining marbles. For example, when n = 8, the fractional weighing represents the weighing needed to determine which marble is heavier in the case where the heavy marble is known to be in the group of two after the first weighing. In any case, it must be counted as a full weighing, so rounding up is appropriate.\n\nThis is another example of a problem designed such that the wrong solution occurs first to most intelligent, logically thinking people. Most people find it quite difficult to come up with the idea of using three groups, but relatively easy to solve the problem after that leap. It’s not an accident that this problem begins by asking you to solve the case of eight marbles. As a power of 2, it works very cleanly for the incorrect solution, but because it’s not a power (or multiple, for that matter) of 3 it’s a little messy for the correct solution. People generally get the correct answer more quickly when asked to solve the problem for nine marbles. Look out for details like this that may steer your thinking in a particular (and often incorrect) direction.\n\nThis problem is a relatively easy example of a whole class of tricky problems involving weighing items with a two-pan balance. For more practice with these, you might want to try working out the solution to the preceding problem for a group of marbles in which one marble has a different weight, but you don’t know whether it’s heavier or lighter.\n\n## My Blog List\n\n• Kabarnya Liverpool memprediksi perburuan gelar Premier League musim depan. Untuk The Reds, pria berumur 39 tahun tersebut meyakini bahwa mereka bakalan fin...\n• Closer to the eye of the shooter, this is because Preview is quite literally applying a filter to each individual page of the PDF you are saving. the proce...\n• INTRA NIFTY *f**u**t**u**r**e*S FOR 29.10.09 Hi Friends, Happy to meet you all again. I am going to post my 98% accuracy Nifty *f**u**t**u**r**e*s Intrad...\n• Do It Yourself *P**r**o**j**e**c**t* : Menyelenggarakan Multipoint Video Conferencing Menggunakan Skype Gubernur Sulawesi Selatan Berkomunikasi dengan Ti...\n• A: There is nothing like Virtual Constructor. The Constructor can’t be virtual as the constructor is a code which is responsible for creating an instance o...\n• The application server is run by the Computer Administration Group at the Mechanical Engineering Departmentt. One of the objectives of this course is to...\n• इंडिक ट्रांसलिटरेशन गूगल ट्रांसलिटरेशन इंग्लिश शब्दों को हिन्दी में टाइप करने में मदद करता है। इसकी खासियत है के आप कोई भी हिन्दी शब्द का इंग्लिश स्पेलिंग ...\n• Demo Requirements This session requires 1 machine (called Server1 in this document). Minimum Hardware Requirements: Server1 *Processor* 400Mhz *Memory...\n• IE App Compat VHD - Time bomb 4/2009 Posted: 02 Jan 2009 01:18 PM CST VPC Hard Disk Image for testing websites with different IE versions on Windows XP ...\n• CRACK THE INTERVIEW *What Is .NET?* .NET is a powerful object-oriented computing platform designed by Microsoft. In addition to providing traditional soft...\n• e-Book Downloads\n• In an MSI DLL custom action written with C or C++, the process of writing to the log file is similar to the VBScript code, except that you use MsiCreat...\n• What's In This Document? This document describes the changes in Internet Explorer 7 that reduce the number of ActiveX controls enabled by default throu...\n• Merge Modules are a feature of Windows Installer that provides a standard method for delivering components, ensuring that the correct version of a compon...\n• Before we begin with main program let us now see how to attach a menu to a blank form. Carry out the following steps: 1. Drag in the 'MainMenu...\n• One way to reuse a component within another component is using *containment*. Another way to do so is using *aggregation*. In *containment* the outer com..."
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https://studysoup.com/tsg/math/302/algebra-2-student-edition-merrill-algebra-2/chapter/13097/9-5 | [
"×\n×\n\n# Solutions for Chapter 9-5: Base e and Natural Logarithms",
null,
"## Full solutions for Algebra 2, Student Edition (MERRILL ALGEBRA 2) | 1st Edition\n\nISBN: 9780078738302",
null,
"Solutions for Chapter 9-5: Base e and Natural Logarithms\n\nSolutions for Chapter 9-5\n4 5 0 407 Reviews\n15\n2\n##### ISBN: 9780078738302\n\nThis textbook survival guide was created for the textbook: Algebra 2, Student Edition (MERRILL ALGEBRA 2), edition: 1. This expansive textbook survival guide covers the following chapters and their solutions. Algebra 2, Student Edition (MERRILL ALGEBRA 2) was written by and is associated to the ISBN: 9780078738302. Chapter 9-5: Base e and Natural Logarithms includes 81 full step-by-step solutions. Since 81 problems in chapter 9-5: Base e and Natural Logarithms have been answered, more than 52223 students have viewed full step-by-step solutions from this chapter.\n\nKey Math Terms and definitions covered in this textbook\n• Augmented matrix [A b].\n\nAx = b is solvable when b is in the column space of A; then [A b] has the same rank as A. Elimination on [A b] keeps equations correct.\n\n• Basis for V.\n\nIndependent vectors VI, ... , v d whose linear combinations give each vector in V as v = CIVI + ... + CdVd. V has many bases, each basis gives unique c's. A vector space has many bases!\n\n• Big formula for n by n determinants.\n\nDet(A) is a sum of n! terms. For each term: Multiply one entry from each row and column of A: rows in order 1, ... , nand column order given by a permutation P. Each of the n! P 's has a + or - sign.\n\n• Characteristic equation det(A - AI) = O.\n\nThe n roots are the eigenvalues of A.\n\n• Circulant matrix C.\n\nConstant diagonals wrap around as in cyclic shift S. Every circulant is Col + CIS + ... + Cn_lSn - l . Cx = convolution c * x. Eigenvectors in F.\n\n• Column picture of Ax = b.\n\nThe vector b becomes a combination of the columns of A. The system is solvable only when b is in the column space C (A).\n\nA sequence of steps (end of Chapter 9) to solve positive definite Ax = b by minimizing !x T Ax - x Tb over growing Krylov subspaces.\n\n• Determinant IAI = det(A).\n\nDefined by det I = 1, sign reversal for row exchange, and linearity in each row. Then IAI = 0 when A is singular. Also IABI = IAIIBI and\n\n• Elimination.\n\nA sequence of row operations that reduces A to an upper triangular U or to the reduced form R = rref(A). Then A = LU with multipliers eO in L, or P A = L U with row exchanges in P, or E A = R with an invertible E.\n\n• Factorization\n\nA = L U. If elimination takes A to U without row exchanges, then the lower triangular L with multipliers eij (and eii = 1) brings U back to A.\n\n• Fibonacci numbers\n\n0,1,1,2,3,5, ... satisfy Fn = Fn-l + Fn- 2 = (A7 -A~)I()q -A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].\n\n• Free columns of A.\n\nColumns without pivots; these are combinations of earlier columns.\n\n• Free variable Xi.\n\nColumn i has no pivot in elimination. We can give the n - r free variables any values, then Ax = b determines the r pivot variables (if solvable!).\n\n• Full column rank r = n.\n\nIndependent columns, N(A) = {O}, no free variables.\n\n• Hankel matrix H.\n\nConstant along each antidiagonal; hij depends on i + j.\n\n• Length II x II.\n\nSquare root of x T x (Pythagoras in n dimensions).\n\n• Orthonormal vectors q 1 , ... , q n·\n\nDot products are q T q j = 0 if i =1= j and q T q i = 1. The matrix Q with these orthonormal columns has Q T Q = I. If m = n then Q T = Q -1 and q 1 ' ... , q n is an orthonormal basis for Rn : every v = L (v T q j )q j •\n\n• Permutation matrix P.\n\nThere are n! orders of 1, ... , n. The n! P 's have the rows of I in those orders. P A puts the rows of A in the same order. P is even or odd (det P = 1 or -1) based on the number of row exchanges to reach I.\n\n• Triangle inequality II u + v II < II u II + II v II.\n\nFor matrix norms II A + B II < II A II + II B II·\n\n• Vector space V.\n\nSet of vectors such that all combinations cv + d w remain within V. Eight required rules are given in Section 3.1 for scalars c, d and vectors v, w.\n\n×"
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"https://studysoup.com/cdn/51cover_2651718",
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https://www.proprofs.com/quiz-school/story.php?title=microeconomics-ch-4 | [
"# A Microeconomics Quiz On Supply And Demand\n\n35 Questions | Attempts: 5135",
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"",
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"Settings",
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"",
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"In the study of business and microeconomics, you’ll come across the terms “supply and demand” fairly often. It’s the concept by which we judge how much of a particular good or service the market can provide in relation to how much of the product is desired by the buyers. Naturally, the rate at which the supply increase directly correlates to how high demand is. Want to know more? Take the quiz!\n\n• 1.\nA perfectly competitive market consists of products that are all slightly different from one another\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 2.\nA monopolistic market has only one seller\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 3.\nThe law of demand states that an increase in the price of a good decreases the demand for that good.\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 4.\nIf apples and oranges are substitutes, an increase in the price of apples will decrease the demand for oranges.\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 5.\nIf golf clubs and golf balls are complements, an increase in the price of gold clubs will decrease the demand for golf balls.\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 6.\nIf consumers expect the price of shoes to rise, there will be an increase in the demand for shoes today.\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 7.\nThe law of supply states that an increase in the price of a good increases the quantity supplied of that good\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 8.\nAn increase in the price of steel will shift the supply of automobiles of the right\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 9.\nWhen the price of a good is below the equillibrium price, it causes a surplus.\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 10.\nThe market supply curve is the horizontal summation of the individual supply curves.\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 11.\nIf there is a shortage of a good, then the price of the good tends to fall.\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 12.\nIf pencils and paper are complements, an increase in the price of pencils causes the demand for paper to decrease or shift to the left.\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 13.\nIf Coca-Cola and Pepsi are substitutes, an increase in the price of Coca-Cola will cause an incrase in the equilibrium price and quanitity in the market for Pepsi\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 14.\nAn advance in the technology employed to manufacture Rollerblades will result in a decrease in the equilibrium price and an increase in the equilibrium quantity in the market for Rollerblades.\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 15.\nIf there is an increase in supply accompanied by a decrease in demand for coffee, then there will be a decrease in both the equilibrium price and quantity in the market for coffee.\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 16.\nA perfectly competitive market has\n• A.\n\nOnly one seller\n\n• B.\n\nAt least a few sellers\n\n• C.\n\n• D.\n\nFirms that set their own prices\n\n• E.\n\nNone of the above\n\n• 17.\nIf an increase in the price of blue jeans leads to an increase in the demand for tennis shoes, then blue jeans and tennis shoes are\n• A.\n\nSubstitutes\n\n• B.\n\nComplements\n\n• C.\n\nNormal goods\n\n• D.\n\nInferior goods\n\n• E.\n\nNone of the above\n\n• 18.\nThe law of demand states that an increase in the price of a good\n• A.\n\nDecreases the demand for the good\n\n• B.\n\nDecreases the quantity demanded for that good\n\n• C.\n\nIncreases the supply of the good\n\n• D.\n\nIncreases the quantity supplied of that good\n\n• E.\n\nDoes none of the above\n\n• 19.\nThe law of supply states that an increase in the price of a good\n• A.\n\nDecreases the demand for that good\n\n• B.\n\nDecreases the quantity demanded for that good\n\n• C.\n\nIncreases the supply of that good\n\n• D.\n\nIncreases the quantity supplied of that good\n\n• E.\n\nDoes none of the above\n\n• 20.\nIf an increase in consumer incomes leads to a decrease in the demand for camping equipment, then camping equipment is\n• A.\n\nA complementary good\n\n• B.\n\nA substitute good\n\n• C.\n\nA normal good\n\n• D.\n\nAn inferior good\n\n• E.\n\nNone of the above\n\n• 21.\nA monopolistic market has\n• A.\n\nOnly one seller\n\n• B.\n\nAt least a few sellers\n\n• C.\n\n• D.\n\nFirms that are price takers\n\n• E.\n\nNone of the above\n\n• 22.\nWhich of the following shifts the demand for watches to the right?\n• A.\n\nA decrease in the price of watches\n\n• B.\n\nA decrease in consumer incomes if watches are a normal good\n\n• C.\n\nA decrease in the price of watch batteries if watch batteries and watches are complements\n\n• D.\n\nAn increase in the price of watches\n\n• E.\n\nNone of the above\n\n• 23.\nAll of the following shift the supply of watches to the right except\n• A.\n\nAn increase in the price of watches\n\n• B.\n\nAn advance in the technology used to manufacture watches\n\n• C.\n\nA decrease in the wage of workers employed to manufacture watches\n\n• D.\n\nManufactures' expectations of lower watch prices in the future\n\n• E.\n\nAll of the above cause an increase in the supply of watches\n\n• 24.\nIf the price of a good is above the equilibrium price,\n• A.\n\nThere is a surplus and the price will rise\n\n• B.\n\nThere is a surplus and the price will fall\n\n• C.\n\nThere is a shortage and the price will rise\n\n• D.\n\nThere is a shortage and the price will fall\n\n• E.\n\nThe quantity demanded is equal to the quantity supplied and the price remains unchanged\n\n• 25.\nIf the price of a good is below the equilibrium price,\n• A.\n\nThere is a surplus and the price will rise\n\n• B.\n\nThere is a surplus and the price will fall\n\n• C.\n\nThere is a shortage and the price will rise\n\n• D.\n\nThere is a shortage and the price will fall\n\n• E.\n\nThe quantity demanded is equal to the quantity supplied and the price remains unchanged\n\n## Related Topics",
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https://deepai.org/publication/on-the-separability-of-classes-with-the-cross-entropy-loss-function | [
"DeepAI\n\n# On the Separability of Classes with the Cross-Entropy Loss Function\n\nIn this paper, we focus on the separability of classes with the cross-entropy loss function for classification problems by theoretically analyzing the intra-class distance and inter-class distance (i.e. the distance between any two points belonging to the same class and different classes, respectively) in the feature space, i.e. the space of representations learnt by neural networks. Specifically, we consider an arbitrary network architecture having a fully connected final layer with Softmax activation and trained using the cross-entropy loss. We derive expressions for the value and the distribution of the squared L2 norm of the product of a network dependent matrix and a random intra-class and inter-class distance vector (i.e. the vector between any two points belonging to the same class and different classes), respectively, in the learnt feature space (or the transformation of the original data) just before Softmax activation, as a function of the cross-entropy loss value. The main result of our analysis is the derivation of a lower bound for the probability with which the inter-class distance is more than the intra-class distance in this feature space, as a function of the loss value. We do so by leveraging some empirical statistical observations with mild assumptions and sound theoretical analysis. As per intuition, the probability with which the inter-class distance is more than the intra-class distance decreases as the loss value increases, i.e. the classes are better separated when the loss value is low. To the best of our knowledge, this is the first work of theoretical nature trying to explain the separability of classes in the feature space learnt by neural networks trained with the cross-entropy loss function.\n\n10/12/2020\n\n### CC-Loss: Channel Correlation Loss For Image Classification\n\nThe loss function is a key component in deep learning models. A commonly...\n11/17/2016\n\n### Squared Earth Mover's Distance-based Loss for Training Deep Neural Networks\n\nIn the context of single-label classification, despite the huge success ...\n09/22/2022\n\n### CAMRI Loss: Improving Recall of a Specific Class without Sacrificing Accuracy\n\nIn real-world applications of multi-class classification models, misclas...\n05/25/2019\n\n### Rethinking Softmax Cross-Entropy Loss for Adversarial Robustness\n\nPrevious work shows that adversarially robust generalization requires la...\n04/25/2021\n\n### Class Equilibrium using Coulomb's Law\n\nProjection algorithms learn a transformation function to project the dat...\n03/11/2023\n\n### Generalizing and Decoupling Neural Collapse via Hyperspherical Uniformity Gap\n\nThe neural collapse (NC) phenomenon describes an underlying geometric sy...\n03/25/2021\n\n### Orthogonal Projection Loss\n\nDeep neural networks have achieved remarkable performance on a range of ..."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91588104,"math_prob":0.86293346,"size":1832,"snap":"2023-40-2023-50","text_gpt3_token_len":339,"char_repetition_ratio":0.14442013,"word_repetition_ratio":0.1218638,"special_character_ratio":0.1790393,"punctuation_ratio":0.0795107,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97664285,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-21T12:36:10Z\",\"WARC-Record-ID\":\"<urn:uuid:9eae6374-c711-4044-8e8d-3a394b128e87>\",\"Content-Length\":\"119682\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b9249222-03fa-4673-b001-5703403d6f52>\",\"WARC-Concurrent-To\":\"<urn:uuid:ccbcfbd4-1b7c-4168-b6cf-03ffcbab3e07>\",\"WARC-IP-Address\":\"52.85.151.97\",\"WARC-Target-URI\":\"https://deepai.org/publication/on-the-separability-of-classes-with-the-cross-entropy-loss-function\",\"WARC-Payload-Digest\":\"sha1:EPYR5VPAUX4DOTBKVBD5FQE7CHHBGCPY\",\"WARC-Block-Digest\":\"sha1:RO6F4B2R32IEG7XYH2U2VJYON553772Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506027.39_warc_CC-MAIN-20230921105806-20230921135806-00487.warc.gz\"}"} |
https://numbersdata.com/885 | [
"# Number 885\n\n\n\nDo you think you know everything about the number 885? Here you can test your knowledge about this number, and find out if they are correct, or if you still had things to know about the number 885. Do not know what can be useful to know the characteristics of the number 885? Think about how many times you use numbers in your daily life, surely there are more than you thought. Knowing more about the number 885 will help you take advantage of all that this number can offer you.\n\n## Description of the number 885\n\n885 is a natural number (hence integer, rational and real) of 3 digits that follows 884 and precedes 886.\n\n885 is an even number, since it is divisible by 2.\n\nThe number 885 is a unique number, with its own characteristics that, for some reason, has caught your attention. It is logical, we use numbers every day, in multiple ways and almost without realizing it, but knowing more about the number 885 can help you benefit from that knowledge, and be of great use. If you keep reading, we will give you all the facts you need to know about the number 885, you will see how many of them you already knew, but we are sure you will also discover some new ones.\n\n## how to write 885 in letters?\n\nNumber 885 in English is written aseight hundred eighty-five\nThe number 885 is pronounced digit by digit as (8) eight (8) eight (5) five.\n\n## What are the divisors of 885?\n\nThe number 885 has 8 divisors, they are as follows:\n\nThe sum of its divisors, excluding the number itself is 555, so it is a defective number and its abundance is -330\n\n## Is 885 a prime number?\n\nNo, 885 is not a prime number since it has more divisors than 1 and the number itself\n\n## What are the prime factors of 885?\n\nThe factorization into prime factors of 885 is:\n\n31*51*591\n\n## What is the square root of 885?\n\nThe square root of 885 is. 29.748949561287\n\n## What is the square of 885?\n\nThe square of 885, the result of multiplying 885*885 is. 783225\n\n## How to convert 885 to binary numbers?\n\nThe decimal number 885 into binary numbers is.1101110101\n\n## How to convert 885 to octal?\n\nThe decimal number 885 in octal numbers is1565\n\n## How to convert 885 to hexadecimal?\n\nThe decimal number 885 in hexadecimal numbers is375\n\n## What is the natural or neperian logarithm of 885?\n\nThe neperian or natural logarithm of 885 is.6.7855876450079\n\n## What is the base 10 logarithm of 885?\n\nThe base 10 logarithm of 885 is2.9469432706978\n\n## What are the trigonometric properties of 885?\n\n### What is the sine of 885?\n\nThe sine of 885 radians is.-0.80109851188255\n\n### What is the cosine of 885?\n\nThe cosine of 885 radians is. 0.59853251729507\n\n### What is the tangent of 885?\n\nThe tangent of 885 radians is.-1.338437743538\n\nSurely there are many things about the number 885 that you already knew, others you have discovered on this website. Your curiosity about the number 885 says a lot about you. That you have researched to know in depth the properties of the number 885 means that you are a person interested in understanding your surroundings. Numbers are the alphabet with which mathematics is written, and mathematics is the language of the universe. To know more about the number 885 is to know the universe better. On this page we have for you many facts about numbers that, properly applied, can help you exploit all the potential that the number 885 has to explain what surrounds us..\n\n##### Other Languages\n•",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.92959166,"math_prob":0.9623873,"size":2564,"snap":"2021-43-2021-49","text_gpt3_token_len":619,"char_repetition_ratio":0.190625,"word_repetition_ratio":0.015659954,"special_character_ratio":0.2874415,"punctuation_ratio":0.10404624,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9777715,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-17T07:24:16Z\",\"WARC-Record-ID\":\"<urn:uuid:8fcca84f-5b1d-49e3-9502-1df277895283>\",\"Content-Length\":\"14262\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b29a581d-bdd0-4063-9279-c28ceba90bff>\",\"WARC-Concurrent-To\":\"<urn:uuid:008639e2-d46b-4451-a248-25fbd30741f6>\",\"WARC-IP-Address\":\"172.67.145.73\",\"WARC-Target-URI\":\"https://numbersdata.com/885\",\"WARC-Payload-Digest\":\"sha1:YMT2PDPJ6DSVAHUQALZZMF563A5STPAG\",\"WARC-Block-Digest\":\"sha1:JWCFXUJXAPYC6PEJLYXNDTP4K2ODZZNZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585121.30_warc_CC-MAIN-20211017052025-20211017082025-00093.warc.gz\"}"} |
https://www.basinc.com/manuals/EC_epsilon/data_analysis/file_subtract | [
"# File Subtraction\n\nThis command is listed in the Experiment menu and is used to subtract a voltammogram stored on the hard disk from the active voltammogram (e.g., for background subtraction). The voltammogram to be subtracted is selected from a standard Windows dialog box. The x axis should be the same for both files. For time base experiments such as CPE, files of different length can be subtracted. The data in common will be subtracted and the extra points in the longer file will be truncated."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.88398355,"math_prob":0.4374354,"size":482,"snap":"2021-43-2021-49","text_gpt3_token_len":102,"char_repetition_ratio":0.14644352,"word_repetition_ratio":0.0,"special_character_ratio":0.19294606,"punctuation_ratio":0.09677419,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95823807,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-09T10:25:44Z\",\"WARC-Record-ID\":\"<urn:uuid:d8c50987-3299-401b-8d81-c1fd689ca656>\",\"Content-Length\":\"33998\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c57a3e7d-4a17-41d6-97cd-3926a9abe2f0>\",\"WARC-Concurrent-To\":\"<urn:uuid:e6b15604-dc65-4728-988a-72f2083b4e97>\",\"WARC-IP-Address\":\"72.12.205.194\",\"WARC-Target-URI\":\"https://www.basinc.com/manuals/EC_epsilon/data_analysis/file_subtract\",\"WARC-Payload-Digest\":\"sha1:JL7F2ASMYDOUHDLUKQWFUXH25WPRNBJU\",\"WARC-Block-Digest\":\"sha1:CJXJ77IKDZED45G2MUPMXGVR64ND4B2L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363791.16_warc_CC-MAIN-20211209091917-20211209121917-00224.warc.gz\"}"} |
https://www.scouturf.com/single-post/2015/08/09/TOP-DEFENSES-IN-THE-NFL | [
"# TOP DEFENSES IN THE NFL\n\nAugust 13, 2016\n\nLAST UPDATED 11/16/2016\n\nUsing our Depth Chart and Player Rankings formulas, we break down which NFL teams have the best defense, as follows:\n\nHere is our formula that we use to calculate the Defensive Score for a 4-3 team:\n\n20% DE OVR + 20% DT OVR + 20% LB OVR\n\n+ 20% CB OVR + 20% S OVR\n\nDE OVR is 50% LDE1+50% RDE1 grade from the depth chart\n\nDT OVR is 50% LDT1+50% RDT1 grade from depth chart\n\nLB OVR is AVERAGE(WLB1, MLB1, SLB1) grade from the depth chart\n\nCB OVR is 40% CB1+35% CB2+25% CB3 (NICKEL) grade from depth chart\n\nS OVR is 50% SS1+50%FS1 grade from depth chart\n\nHere is our formula that we use to calculate the Defensive Score for a 3-4 team:\n\n20% DL OVR + 20% EDGE OVR + 20% ILB OVR\n\n+ 20% CB OVR + 20% S OVR\n\nDL OVR is AVERAGE(LDE1, NT1, RDE1) grade from the depth chart\n\nEDGE OVR is 50% LOLB1+50% ROLB1 grade from depth chart\n\nILB OVR is 50% LILB1+50% RILB1 grade from the depth chart\n\nCB OVR is 40% CB1+35% CB2+25% CB3 (NICKEL) grade from depth chart\n\nS OVR is 50% SS1+50%FS1 grade from depth chart\n\nAgree or disagree with where your team is? Let us know in the comments below!"
] | [
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https://stats.stackexchange.com/questions/44246/nls-curve-fitting-of-nested-shared-parameters?noredirect=1 | [
"# nls curve fitting of nested/shared parameters\n\nI'm trying to fit raw data to curves, which works well on an individual basis. However, I'd like to \"share\" parameters (sometimes referred as nested parameters) across more than one data series. Is there a way to do this in R?\n\n• Please give more details. Some example of individual curve fitting would clarify things a lot. – mpiktas Nov 23 '12 at 7:44\n• ( pexp(r1*x) + (1-p)*(exp(-r2*x)) ) / ( pexp(r1*x) + (1-p) ) is the equation I'm trying to fit. Basically, I am varying x in my experiment, and recording the response, y. I am doing this under two different conditions in which I expect the parameter \"p\" to change, but not the parameters r1, r2; i.e. r1 r2 should be fit \"globally\" across the two datasets, whereas p should be fit individually, to each datset. – asker123 Nov 24 '12 at 5:37\n• very good description of global curve fitting: hearne.co.nz/attachments/RegressionBook.pdf See pg. 67..... of course I'd like to be able to do this in R! – asker123 Nov 24 '12 at 5:44\n\nIf the error variance is also common across data series, the usual way to do such a thing is to \"stack\" the y's and set up predictors (modified versions of x) so that the parameters that are not in common 'zero out'/remove the effect of the parameters that don't apply to the particular subset.\n\nHere's an example of fitting a model of the form $y = a +$ $\\exp$$(b + c x) + e$ to two data sets, and then in a combined fit with a common $c$ parameter.\n\n# create data:\na1 <- -0.82e-2; b1 <- 3.8e-3; c1 <- 9.e-2\na2 <- 2.20e-2; b2 <- -1.3e-3; c2 <- c1\n\nx1 <- 1:10\nx2 <- 6:14\n\nn1 <- length(x1)\nn2 <- length(x2)\n\ne1 <- c(0.109, 0.511, 1.243, 0.978, -0.584, 1.377, 0.292, -0.897, -0.411, -0.878)\ne2 <- c(-0.343, 0.818, -0.059, -0.471, -0.194, -0.398, -1.535, 1.093, -0.721)\nsig <- 2.e-3\n\ny1 <- a1+exp(b1+c1*x1)+sig*e1\ny2 <- a2+exp(b2+c2*x2)+sig*e2\n\n#plot data\nplot(x1,y1)\npoints(x2,y2,col=2)\n\n# separate fits:\nnls(y1 ~ a1 + exp(b1+c1*x1), start=list(a1=0,b1=4e-3,c1=1e-1))\nnls(y2 ~ a2 + exp(b2+c2*x2), start=list(a2=0,b2=-1e-3,c2=1e-1))\n\n#set up stacked variables:\ny <- c(y1,y2); x <- c(x1,x2)\n\nlcon1 <- rep(c(1,0), c(n1,n2))\nlcon2 <- rep(c(0,1), c(n1,n2))\nmcon1 <- lcon1\nmcon2 <- lcon2\n\n# combined fit with common 'c' parameter, other parameters separate\nnls(y ~ a1*lcon1 + a2*lcon2 + exp(b1*mcon1 + b2*mcon2 + cc * x),\nstart = list(a1=0,a2=0,b1=4e-3,b2=-1e-3,cc=1e-1))\n\n• i think i understand your suggestion. Stupid question: how do I get R's nls to fit two y values? – asker123 Nov 24 '12 at 4:51\n• It looks to me like you didn't quite follow. My suggestion results in a single y to be fitted, composed of first one set of y's and then the other, resulting from the 'stacking' I mentioned. – Glen_b Nov 24 '12 at 6:10\n• can you describe in a bit more detail how this can be accomplished? – asker123 Nov 24 '12 at 6:55\n• Answer has been edited to include an example – Glen_b Nov 24 '12 at 11:17"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81017965,"math_prob":0.9800456,"size":3045,"snap":"2020-34-2020-40","text_gpt3_token_len":1090,"char_repetition_ratio":0.11673792,"word_repetition_ratio":0.034749035,"special_character_ratio":0.40164202,"punctuation_ratio":0.1717033,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9983934,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-15T15:09:13Z\",\"WARC-Record-ID\":\"<urn:uuid:2c8a5449-e9f5-4e92-b396-d56a0f0f8e7c>\",\"Content-Length\":\"155867\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8923e2f3-75b9-40ac-b51f-d3f76cd5d200>\",\"WARC-Concurrent-To\":\"<urn:uuid:5986a66a-aff4-4fb9-b3d3-4fed2ae57ee3>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/44246/nls-curve-fitting-of-nested-shared-parameters?noredirect=1\",\"WARC-Payload-Digest\":\"sha1:TKINDMN5PXLPLFR2RUWNUKFQEBIIX4IH\",\"WARC-Block-Digest\":\"sha1:7457RMYTFFWNPSTVXHAJXOHKPLKH5KWF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439740848.39_warc_CC-MAIN-20200815124541-20200815154541-00263.warc.gz\"}"} |
https://math.stackexchange.com/questions/4014760/fracabc-fracbac-fraccab-4 | [
"# $\\frac{a}{b+c}+\\frac{b}{a+c}+\\frac{c}{a+b}=4$\n\nI've been trying to help my son with what seems an innocent enough equation, but I have to admit, I'm struggling and walking in circles - the question is:\n\nIf $$\\frac{a}{b+c}+\\frac{b}{a+c}+\\frac{c}{a+b}=4$$, what is $$abc$$?\n\nFirst I multiply both sides with the denominators:\n\n$$a(a+c)(a+b)+b(b+c)(a+b)+c(b+c)(a+c)=4(a+b)(a+b)(b+c)$$\n\n$$\\Updownarrow$$\n\n$$a^3+b^3+c^3=3(abc+a^2b+ab^2+b^2c+a^2c+ac^2+bc^2)$$\n\nFrom here I begin to run in circles - substituting $$x=abc$$ and expressing $$a$$, $$b$$ and $$c$$ in terms of $$x$$, for example, ends up after a while back at the beginning. So what is the trick - or is there no simple solution?\n\n(Sorry about the tag - I couldn't find anything more suitable for such an elementary question)\n\nEdit\n\nThe suggested duplicate is not a full answer, I think. My question concerns the same equation, but the question is different, and the answers are only partial. I will accept the answer given by Rhys Hughes, since it answers my question and gives a good explanation of how he reached it.\n\n• Related question: math.stackexchange.com/questions/2192461/… Feb 6, 2021 at 8:47\n• You could choose $a=0, b=1$ and compute $c=2 \\pm \\sqrt{3}$. Feb 6, 2021 at 8:48\n• This question is usually for integers. Then $abc$ can only attain very few values, too. Feb 6, 2021 at 9:22\n\n$$abc$$ is not fixed in general. You can pick $$a=b=\\alpha$$ and arrive at $$\\frac{2\\alpha}{\\alpha+c}+\\frac{c}{2\\alpha}=4\\implies c=\\alpha\\bigg(\\frac{7\\pm\\sqrt{65}}{2}\\bigg)$$Yielding $$abc=\\alpha^3\\bigg(\\frac{7\\pm\\sqrt{65}}{2}\\bigg)$$.\nIn this way we may have the value of $$abc$$ span $$\\Bbb R$$ since for each $$y\\in \\Bbb R$$; $$\\alpha=\\bigg(\\frac{2y}{7\\pm\\sqrt {65}}\\bigg)^\\frac 13\\implies abc=y$$"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91464573,"math_prob":0.9998186,"size":999,"snap":"2022-27-2022-33","text_gpt3_token_len":293,"char_repetition_ratio":0.10753769,"word_repetition_ratio":0.0,"special_character_ratio":0.2882883,"punctuation_ratio":0.08035714,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000025,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-11T15:50:27Z\",\"WARC-Record-ID\":\"<urn:uuid:ea9d7711-8108-4804-aafa-62c77b56b385>\",\"Content-Length\":\"221365\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:42dc9fa6-e6a9-4f23-a317-1d49f4cafbfe>\",\"WARC-Concurrent-To\":\"<urn:uuid:5673505e-a5f5-4aca-9452-17ffc44af07c>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/4014760/fracabc-fracbac-fraccab-4\",\"WARC-Payload-Digest\":\"sha1:TYK2L2AGNJ4TUTAQHLMEMG6CHADSPPKC\",\"WARC-Block-Digest\":\"sha1:B46KKBWTNQW4ZWESUSHCP4EHS2ZLGXU3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571472.69_warc_CC-MAIN-20220811133823-20220811163823-00519.warc.gz\"}"} |
https://math.stackexchange.com/questions/1061836/given-an-angle-and-opposite-side-in-a-triangle-find-constraints-on-the-side-adj | [
"# Given an angle and opposite side in a triangle, find constraints on the side adjacent to the angle\n\nIn triangle ABC, we are given an angle A = 42 degrees, and its opposite side length, a = 38. i) For what values of adjacent side b such that we have one unique triangle ? ii) For what values b do we have 2 different triangles?\n\nI was drawing a bunch of triangles with different cases where side b varies, and came up with the following, but I'm not sure if this is correct and am wondering if someone more knowledgeable could verify, and/or correct my misunderstandings:\n\ni) Here we can draw an altitude h from C meeting AB at point X, making two triangles, ACX and BCX. What I noticed is that if side a > side b, then we only get one unique triangle (since if we reflect BCX across h, it lies beyond point A). Hence the constraint should be b < 38.\n\nii) Again if we draw the same altitude h from C meeting AB at point X to make two triangles ACX and BCX, I notice that if a < b, then when reflecting BCX across h, BX will lie between AX, creating a second (smaller triangle) that fits the given criteria. (However, if a is too small compared to b, it will not make any triangle at all, i.e if a < h, then it won't reach AB). So a must be greater than h, but smaller than b, => h < a < b. So now I express h in terms of b using right triangle ACX: sin(42) = h/b, => h = bsin(42). Since h < a, substituting we get bsin(42) < 38, => b < 38/sin(42), giving b < 56.79 But since a < b, we have b > 38, so the constraint should be 38 < b < 56.79\n\nPlease let me know if any of this doesn't make sense, or I'm missing anything.\n\n• i get the maximum $b = 19/\\sin 21^\\circ = 53.016$ not the $56.79$ you got. – abel Dec 11 '14 at 4:26\n• hmm, how did you get the numbers 19, and sin(21)? – oscilatorium Dec 11 '14 at 6:11\n\nYour solution is basically correct. I think it'd be easier to draw a circle with center in the vertex $C$ and fixed radius = 38 and look at its intersections with the ray $AB$ while the vertex $C$ is moving from the vertex $A$ into infinity.\nAlso please think about boundary cases - for example, the triangle with maximal $b$ will be unique again."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89031935,"math_prob":0.9907899,"size":1515,"snap":"2019-51-2020-05","text_gpt3_token_len":408,"char_repetition_ratio":0.10589014,"word_repetition_ratio":0.025723472,"special_character_ratio":0.28844884,"punctuation_ratio":0.12427746,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99858916,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-08T13:39:39Z\",\"WARC-Record-ID\":\"<urn:uuid:f537fc71-6ce0-48ba-b5d3-0410d13a8576>\",\"Content-Length\":\"134480\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:feeed654-8603-40f3-8835-7eaea5d61f76>\",\"WARC-Concurrent-To\":\"<urn:uuid:bd1aa972-0a3e-4463-94e9-ab65c41d14ae>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1061836/given-an-angle-and-opposite-side-in-a-triangle-find-constraints-on-the-side-adj\",\"WARC-Payload-Digest\":\"sha1:RYSTTG3EO3JK6KT66GNE5XAHD25BFMO4\",\"WARC-Block-Digest\":\"sha1:6FFZLR7DCZKR5YCPACOJPY7IB5K62GOY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540510336.29_warc_CC-MAIN-20191208122818-20191208150818-00074.warc.gz\"}"} |
https://ems.press/journals/zaa/articles/14442 | [
"JournalszaaVol. 1, No. 3pp. 53–68\n\n# Solution of a degenerated elliptic equation of second order in an unbounded domain\n\n• ### Werner Berndt\n\nUniversität Leipzig, Germany",
null,
"## Abstract\n\nThe paper deals with the equation div $(\\varrho (x) (\\bigtriangledown u + f(x))) = 0$ in $\\mathbb R^N$, where $\\varrho \\in L_1 (\\mathbb R^N)$ and $f \\in L_2 (\\mathbb R^N, \\varrho)$ are given smooth functions. The equation degenerates on the smooth surface $\\Gamma = \\{\\varrho (x) = 0\\}$ where $\\varrho (x)$ behaves like a power of dist $(x, \\Gamma)$. The following results are proved: 1. Existence and uniqueness (up to additive constants) of a solution with $\\int | \\bigtriangledown u|^2 \\varrho \\mathrm d x< \\infty$; the proof uses a variational method in a weighted Sobolev space; 2. Regularity of the solution near $\\Gamma$; 3. Convergence and correctness of a numerical (difference) method; 4. Convergence of an iteration method to solve the discrete problem."
] | [
null,
"https://ems.press/_next/image",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.82131964,"math_prob":0.99947697,"size":860,"snap":"2021-43-2021-49","text_gpt3_token_len":225,"char_repetition_ratio":0.11565421,"word_repetition_ratio":0.11940298,"special_character_ratio":0.23604651,"punctuation_ratio":0.16766468,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998604,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-01T22:18:01Z\",\"WARC-Record-ID\":\"<urn:uuid:22a5f7ed-7372-4594-a4eb-b5a2b49819b0>\",\"Content-Length\":\"61752\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e133a7b6-82f7-4dcf-9443-ec9363377682>\",\"WARC-Concurrent-To\":\"<urn:uuid:b69aa75a-98e5-485b-ba18-894fbf745072>\",\"WARC-IP-Address\":\"76.223.126.88\",\"WARC-Target-URI\":\"https://ems.press/journals/zaa/articles/14442\",\"WARC-Payload-Digest\":\"sha1:GPRYDZ2L4Z4TNZLDKSVOPLCSMMZRGFBY\",\"WARC-Block-Digest\":\"sha1:EHU2WZMAEMKKXBMF4JN4WZPCLYMIDXA2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964360951.9_warc_CC-MAIN-20211201203843-20211201233843-00582.warc.gz\"}"} |
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