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https://gre.kmf.com/explain/index/61k85j	[ "### 最新提问\n\nIf one resident is to be randomly selected from among the 2,000 residents, what is the probability that the selected resident`s response to question 2 was either \"Local government\" or \"Private donations\"?\n• A$\\frac{1}{4}$\n• B$\\frac{2}{7}$\n• C$\\frac{1}{3}$\n• D$\\frac{2}{5}$\n• E$\\frac{1}{2}$\n\nA显示答案\n\n• KMF 解析\n• 网友解析\n• 题目数据\n• 收起\n\n### 题目讨论", null, "• 全部讨论\n• 只看提问\n• 按时间排序\n• 按点赞数排序" ]	[ null, "http://www.kmf.com/images/tx.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7358511,"math_prob":0.99661124,"size":386,"snap":"2020-34-2020-40","text_gpt3_token_len":141,"char_repetition_ratio":0.11518325,"word_repetition_ratio":0.031746034,"special_character_ratio":0.2253886,"punctuation_ratio":0.1,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9812717,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-13T08:43:11Z\",\"WARC-Record-ID\":\"<urn:uuid:21746ae2-e71e-4069-8412-a2e7c4a3533f>\",\"Content-Length\":\"43669\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:249cd80f-3195-4a43-9417-8eab24a20200>\",\"WARC-Concurrent-To\":\"<urn:uuid:0159f9ca-7b59-4a4a-b6cd-6f0b69be2dae>\",\"WARC-IP-Address\":\"47.254.45.31\",\"WARC-Target-URI\":\"https://gre.kmf.com/explain/index/61k85j\",\"WARC-Payload-Digest\":\"sha1:FHJBHIO6EUCJORRDUMEM7F3NAU5MDTSB\",\"WARC-Block-Digest\":\"sha1:7JCIV5XFNHOEAI3CK24MUDMO2G7VOM6D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738964.20_warc_CC-MAIN-20200813073451-20200813103451-00118.warc.gz\"}"}
https://puzzling.stackexchange.com/questions/1855/what-is-the-smallest-positive-integer-which-can-not-be-written-without-repetiti	[ "# What is the smallest positive integer, which can not be written without repetitions of digits and using arithmetics only?\n\nSuppose you are allowed to use all 10 digits (0,1,2,...9) at most once each; 4 arithmetic operations ($-$,$+$,$\\times$,$\\div$), each any number of times; parenthesises to group operations; and you can create numbers from digits by writing them together.\nWhat is the smallest natural number which you would not be able to write?\n\nFor example, you can write:\n$135 = 123(9+6)/4$\nand you can't write:\n$11 = 11$\n$3 = 2+3/3$\n$27 = 3+4!$\n$81 = 3^4$\n$1 = 5/3$\n\n• @JoeZ don't forget you can \"create numbers from digits writing them together\" so $4817(90+23)=544321$ Jul 3, 2014 at 14:41\n• What about parenthesis, or how is operator precedence handled? Jul 3, 2014 at 15:12\n• @Cephalopod, you can use parenthesises. thank you. Jul 3, 2014 at 16:23\n• I don't think there's any algorithm which will help solve this - it's just going to have to be bruteforce. Jul 3, 2014 at 18:41\n• @N3buchadnezzar, em? You just wrote this number using only digits from 1 to 9 and stringing them together. Jul 3, 2014 at 20:06\n\nConcatenation is the \"most efficient\" among the allowed operations, which is exmplified by this observation: If $x$ is a nonzero rational number obtained according to the rules using the $k$ digits $d_1>d_2>\\ldots >d_k$, then $$\\tag1 \\max\\left\\{\|x\|,\\tfrac1{\|x\|}\\right\\}\\le 10^k\\cdot 0.d_1d_2\\ldots d_k$$ This is clear for direct concatenations and otherwise follows by induction and using $(1)$ for the summands, factors, etc.\n\nWe can strengthen this: If the last operation is $\\times$ or $\\div$, then $$\\tag2\\max\\left\\{\|x\|,\\tfrac1{\|x\|}\\right\\}<10^k\\cdot 0.85598232$$ This follows because not both parts can use the digit $9$, hence the largest digit in on subterm is either $8$, leading to a bounding factor $0.97654321\\cdot 0.87654321$; or the largest digit in one factor is $\\le 7$, leading to a bounding factor $0.987654321\\cdot 0.7654321$. If one takes these case distinctions a bit further, one readily finds that $$\\tag{2'} \\max\\left\\{\|x\|,\\tfrac1{\|x\|}\\right\\}\\le10^k\\cdot 0.843973902$$ if the last operation is $\\times$ or $\\div$ (with the extreme given by $9642\\cdot 875310$).\n\nMoreover, if the last operation is $+$ or $-$, then $$\\tag3\|x\|\\le (10^{k-1}+1)\\cdot 0.987654321$$\n\nWe thus are led to believe that $987654323$ is likely not expressible: Because of two difgits $3$ it cannot be obtained from concatenation; because of $(3)$ is cannot be obtained as sum or difference; because it is prime, it cannot be obtained as product of integers (unless one factor is $1$ and we are effectively using at most $9$ digits). Remains the case that the number is obtained as a product or quotient of fractions, but at first sight this seems unplausible.\n\nMeanwhile I brute-forced all numbers that can be obtained with only $+,-,\\times$ and digit concatenation (and with all intermediate resuls $\\le 10^7$). The first number that cannot be expressed this way turns out to be $$8480902.$$ So unless someone manages to express $8480902$ under the original rules (i.e. with division allowed), this is the answer to the original problem.\n\n• \"This follows because not both parts can use the digit 9\" Why? This is in assumption that all previous operations were concatenations, but why must they? Jul 7, 2014 at 5:49\n• @klm123 Each digit shall be used at most once, hence $9$ cannot occur as used in both parts (it may of course occur as digit in both parts, e.g. if we combine (e.g. multiply) $4+5$ and $92$, but then $9$ is used only in the second part). - After the brute-force result, however, the introductory estimates have become mostly useless. Jul 7, 2014 at 14:10\n• I'm missing something here. I followed you up to the brute-force consideration. How did you do that? And why are you discarding the division? Using just 0-5 digits, I have found numbers that are expressed as division (i.e. 27155=54310/2), although I cannot be certain that all of them are expressable only as a division. Jul 7, 2014 at 21:57\n• Exhaustive searches show that for digits up to 6 and 7 the solution is unchanged by the removal of division from the permitted operations, so it's quite likely that 8480902 is the desired answer. Jul 9, 2014 at 17:05\n\nI have written a Delphi application to brute force a solution. Unfortunately I see no way of solving the problem: what expression(s) results in number N? That is I see no solution other than generating all expressions and then seeing what natural numbers they result in.\n\nSo I can't really help with Bobson's crowd-sourcing approach.\n\nI am following klm123's idea, and I have ran the application with:\n\nN = 3 : 0,1,2\n\nExecution time: 00:00:00\nExpressions processed: 441 [Average: 27.562,5/sec]\nExpressions calculated: 34 [Average: 2.125,0/sec]\nNatural numbers calculated: 28 [Average: 1.750,0/sec]\nDistinct natural numbers found: 14 [Average: 875,0/sec]\nHighest natural found: 210\n\n\nN = 4 : 0,1,2,3\n\nExecution time: 00:00:00\nExpressions processed: 22.924 [Average: 363.873,0/sec]\nExpressions calculated: 1.464 [Average: 23.238,1/sec]\nNatural numbers calculated: 871 [Average: 13.825,4/sec]\nDistinct natural numbers found: 110 [Average: 1.746,0/sec]\nHighest natural found: 3.210\n\n\nN = 5 : 0,1,2,3,4\n\nExecution time: 00:00:04\nExpressions processed: 1.679.977 [Average: 401.812,2/sec]\nExpressions calculated: 98.228 [Average: 23.493,9/sec]\nNatural numbers calculated: 45.423 [Average: 10.864,1/sec]\nDistinct natural numbers found: 884 [Average: 211,4/sec]\nHighest natural found: 43.210\n\n\nN = 6 : 0,1,2,3,4,5\n\nExecution time: 00:08:39\nExpressions processed: 159.888.346 [Average: 307.590,0/sec]\nExpressions calculated: 8.867.950 [Average: 17.060,0/sec]\nNatural numbers calculated: 3.236.479 [Average: 6.226,3/sec]\nDistinct natural numbers found: 8.661 [Average: 16,7/sec]\nHighest natural found: 543.210\n\n\nN = 7 : 0,1,2,3,4,5,6\n\nExecution time: 19:36:02\nExpressions processed: 18.788.082.577 [Average: 266.262,5/sec]\nExpressions calculated: 1.014.272.742 [Average: 14.374,2/sec]\nNatural numbers calculated: 308.146.445 [Average: 4.367,0/sec]\nDistinct natural numbers found: 93.219 [Average: 1,3/sec]\nHighest natural found: 6.543.210\n\n\nNow on to N=8. I got a few small optimizations in mind, but nothing that will cut this drastically, I presume.\n\n## Explanation of what the application is doing\n\nI have two input parameters: Digits and Operators. I also have some other parameters to handle output dump to disk, but they have nothing to do with the algo.\n\nThe application starts processing the empty expression: 0 length string.\n\nEvery time an expression E is processed, I do the following:\n\n1. check all Digits - for any single digit D not already included in E, I process the expression E+D (concatenation, not sum). To optimize a bit I avoid this step if E ends with a closed parenthesis. I also avoid this if D is 0 and E does not end with any other Digit.\n\n2. check all Operators - for any single operator O, I process the expression E+O. To optimize I avoid this step if E is empty or ends with another operator or with an open parenthesis.\n\n3. check \")\" - if E ends with a Digit or a closed parenthesis, and it has a pending unclosed parenthesis, and this contains at least one operatore, then I process the expression E+).\n\n4. check \"(\" - if E is empty or ends with an Operator or an open parenthesis, then I process the expression E+(. To avoid infinite recursion there is some magic here... I do it only if the number of Digits still unused in E is greater the number of pending unclosed parenthesis in E plus 1.\n\n5. calculate the expression - if E ends with a Digit or a closed parenthesis, and all parenthesis are balanced and are useful (contain at least one operator), and it does not start with ( and end with ) then I calculate the expression. If it turns out to be a natural number, then I check it with the highest found natural and lowest unfound natural.\n\nProcessing of derived expressions is obviously done through recursion. The algo seems quite efficient, apart from the parenthesis part. I have made some changes to this code and it now validates parenthesis better, avoiding some recursion in useless branches and evaluation os redundant expressions. It's not perfect, but it does cut times as N goes up. (I am using Artem V. Parlyuk's ArtFormula package of nonvisual Delphi component for symbolic expression parsing and evaluation).\n\nAs an aside, I also keep track of the shortest expression that can generate each natural number found. The shortest is meaningful, as it is usually the most immediate and readable. It is also always the one with no useless parenthesis ;)\n\n## Update\n\nTweaking the parenthesis logic, I have stumbled upon a small bug. It was generating a lot of useless expressions (with redundant parenthesis) but it was not generating some useful expressions. With N<=5 this bug did not have any effects. With N=6 I did find 6 more natural numbers, though the lowest unfound natural is 653.\n\n• This should run faster if you only search the numbers less than the lowest one not found, in theory. That way, you can answer the question without waiting years :P\n– user20\nJul 7, 2014 at 17:51\n• And how do you propose to \"search\" for a specific value? I generate expressions, and then evaluate them... using a specific library, which is actually quite fast. But if you have ideas on how to limit the generation/evaluation of expressions, please let me know. A part from that... when I say that 2908 is the lowest, I mean the application has already found expressions for all numbers up to 2907. So what I actually need is to verify if 2908 has an expression or not. Jul 7, 2014 at 18:36\n• Hmm, I see what you're saying. Guess it isn't as straightforward as I thought... I'm sure there's a way to, but I'll have to think on it.\n– user20\nJul 7, 2014 at 19:10\n• I confirm your values for the lowest natural number not found, although I think your code is quite inefficient. Mine takes 20 seconds to check N=7. Jul 9, 2014 at 16:31\n• 20 secs to evaluate about one billion expressions... wow. I don't think choice of programming language or tool can make so much of a difference. Would you mind sharing the logic or the code? What are you using? How long will it take to process N=10? Jul 9, 2014 at 16:41\n\nHere is my attempt at finding an upper bound:\n\nWe have 10 atomic values ($0..9$) and five operations ($+$, $-$, $$, $/$, $○$; according to rules concatenation may not be used on results of other operators, let's ignore this for now)\n\nSince all operations are binary, we need 9 operators to get a single result out of 10 values. There are 4862 binary trees with nine nodes.\n\nThere are $5^9 = 1953125$ ways to choose operators for a tree and $10! = 3628800$ ways to arrange the digit leaves.\n\nThus, the number must be smaller than $362880019531254862 = 34459425000000000$. Hey, this still fits into a signed 64-bit integer, so good look to the brute forcers :-)\n\nIt is possible to reduce the upper limit. The following rules were not considered by my approximation\n\n• Concatenation may not be used on results of other operators\n• $+$ and $$ are commutative\n• not all combinations result in integers\n• some integers can be computed in more than one way\n• The maximum you can get with 9 digits and all operators are 9642087531, so the upper bound can be much smaller: 8439739021. Jul 4, 2014 at 11:21\n• You can do better than that just by stringing all the digits in descending order to make 9876543210, though.\n– user88\nJul 4, 2014 at 14:12\n• @JoeZ. It may technically have to be an expression of some sort, meaning the upper bound would therefore be 8439739021.\n– Neil\nJul 4, 2014 at 15:52\n• @neil It depends on if \"any number of times\" include zero times? Jul 7, 2014 at 20:23\n• @AdamSpeight That cannot be the case, or else the problem would be trivial, being able to represent all possible expression valuations from 0 to 9876543210 as a number without operations and hence everything can be represented and there is no solution.\n– Neil\nJul 8, 2014 at 8:33\n\nIn the interest of crowd-sourcing the brute forcing, I've made a google spreadsheet with the list of the 4765 numbers from 1 to 9999 which can't be formed by simple concatenation. Please pick a couple of them and work out solutions.\n\nIf we finish this set with solutions for all numbers, I'll add more.\n\nI don't really expect it to be in this set, but we've got to start somewhere...\n\n• I brute-forced expressions using only digits $0,\\ldots, 7$ and allowing only $+,-,\\times$ and initially concatenation. Then the bound is at $48528$. Jul 6, 2014 at 15:14\n• @HagenvonEitzen - Good to know. I didn't really expect it to be solvable by hand, but it was worth the attempt. Jul 7, 2014 at 19:53\n\nI'm trying to approach the task solving simpler tasks with $N<9$ digits allowed.\n\n1. N = 3: 0,1,2.\nThe Answer would be $A=4$.\nObviously, you can't reach $(N-1)N/2+1 = 4$ if you use only $-,\\div,+$ operations. $\\times$ operation also would help here.\nSo you need to use grouping operation. But then you will get a number $R \\ge 10$ and the only way to decrease it drastically to $4$ is to use division. But $10/2 = 5$ - the division is not enough too.\n\nIt is easy to check that you can make any number $\\le4$.\n\n2. N = 4: 0,1,2,3.\nThe Answer is $A=25$.\nSimilarly you can't reach $7$ if you use only $-,\\div,+$ operations. The maximum you can get adding $\\times$ operation is (2+1)3 = 9 (you have to multiply maximum possible numbers to reach the maximum).\nSo you need to use grouping operation.\nIf you group 1 and other digit, you would need to use multiplication to reach 25, but 102 =20 and 103 = 30 leaves us with 3 or 2 to finish... 123 = 36 and 13*2 = 26 lives us with 0... so you can do nothing here.\nIf you group 2 and other digit, for example, 20 you would need to add to it something and can't reach 25 with additional 1 and 3. The same for 21,23.\nIf you group 3 and other digit, like 30, you would need to subtract from it something, similarly you can't reach 25.\nIf you group three digits, then you would get $R>100$ and you would need to use division, but $100/3 > 25$.\n\nYou can check that you can make any number $\\le24$.\n\nThis really looks like there is no simple pattern, the $A$ is rapidly increases with $N$, the same for number of possible ways to achieve the result.\nThe only patter I see is that $A < 10^{N-2}/(N-1)$, possibly this can be proven. And $10^{N-3} < A$, which is most probably will fail for big $N$.\n\nMost probably there is no pattern for $A$ at all, and one would need to brute force it for big $N$.\n\n• It is a ridiculously difficult problem to try to take on mathematically. Even brute forcing a solution would likely take more seconds than have passed since the universe was founded. Don't beat yourself up.\n– Neil\nJul 4, 2014 at 8:00\n• @Neil: Can you prove that it's mathematically difficult? Jul 4, 2014 at 8:55\n• @justhalf How do you expect I go about that exactly? Divide the number of head scratches by the inverse frequency of pencil eraser usage? I didn't say that to prove anything. If you wish to demonstrate that it is, in fact easy, then I think it's safe to assume you could provide a solution, no?\n– Neil\nJul 4, 2014 at 9:06\n• I think this might be a good approach, but you'd need to do it in a base-$N$ system. Otherwise you get gaps - there's probably no way to represent 79 using only the digits 0-3, for instance. Jul 6, 2014 at 14:58\n• @Bobson, I don't understand why are you talking about 79 via 0-3 digits. May be you see more in my approach than I do? Could you explain it, please? Jul 6, 2014 at 15:04" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8811545,"math_prob":0.96851385,"size":4839,"snap":"2022-40-2023-06","text_gpt3_token_len":1276,"char_repetition_ratio":0.18676318,"word_repetition_ratio":0.040789474,"special_character_ratio":0.29303575,"punctuation_ratio":0.20349908,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9952251,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-06T01:02:15Z\",\"WARC-Record-ID\":\"<urn:uuid:2be0558b-3ef1-4935-8c7a-5d49def9e12f>\",\"Content-Length\":\"307007\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7842a8e5-273a-46f7-8af9-724b2af53571>\",\"WARC-Concurrent-To\":\"<urn:uuid:5b496f09-f3f3-4de5-a009-d811a53ce5f7>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://puzzling.stackexchange.com/questions/1855/what-is-the-smallest-positive-integer-which-can-not-be-written-without-repetiti\",\"WARC-Payload-Digest\":\"sha1:RIQUJEAVCXX5HEO56IRZWTIOPHWZEAO5\",\"WARC-Block-Digest\":\"sha1:F7RKARXAYKSV5DATJEYRH5N57XYZ63ZE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337680.35_warc_CC-MAIN-20221005234659-20221006024659-00467.warc.gz\"}"}
http://umj.imath.kiev.ua/article/?lang=en&article=8127	[ "2019\nТом 71\n№ 11\n\n# Strongly nonlinear degenerate elliptic equations with discontinuous coefficients. II\n\nBonafede S.\n\nAbstract\n\nWe use energy methods to prove the existence and uniqueness of solutions of the Dirichlet problem for an elliptic nonlinear second-order equation of divergence form with a superlinear tem [i.e., g(x, u)=v(x) a(x)⋎u⋎ p−1u,p>1] in unbounded domains. Degeneracy in the ellipticity condition is allowed. Coefficients a i,j(x,r) may be discontinuous with respect to the variable r.\n\nEnglish version (Springer): Ukrainian Mathematical Journal 49 (1997), no. 12, pp 1798-1809.\n\nCitation Example: Bonafede S. Strongly nonlinear degenerate elliptic equations with discontinuous coefficients. II // Ukr. Mat. Zh. - 1997. - 49, № 12. - pp. 1601–1609.\n\nFull text" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7779281,"math_prob":0.8453903,"size":757,"snap":"2020-10-2020-16","text_gpt3_token_len":208,"char_repetition_ratio":0.11155379,"word_repetition_ratio":0.094339624,"special_character_ratio":0.26023778,"punctuation_ratio":0.1849315,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9849411,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-18T07:33:02Z\",\"WARC-Record-ID\":\"<urn:uuid:fefb11c2-386d-49ca-826c-76abbb89aa81>\",\"Content-Length\":\"20336\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:448e151d-9b16-4890-ac24-f36599001ff5>\",\"WARC-Concurrent-To\":\"<urn:uuid:1db6e784-9600-4e21-86bc-bcdc3d03951b>\",\"WARC-IP-Address\":\"194.44.31.54\",\"WARC-Target-URI\":\"http://umj.imath.kiev.ua/article/?lang=en&article=8127\",\"WARC-Payload-Digest\":\"sha1:5M2EFB2WDODIMSLCRK2BWKECGNDWPXEU\",\"WARC-Block-Digest\":\"sha1:2MKMVGCZQBO72QRDNRMS4GKCAOOUL5UM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875143635.54_warc_CC-MAIN-20200218055414-20200218085414-00251.warc.gz\"}"}
http://escholarship.lib.okayama-u.ac.jp/en/journal/mmc/38/--/article/40177	[ "", null, "", null, "<Formerly known as>\n\n<Availability>\nSome items are not available because of decision by its author or publisher.\n\n温泉治療の血液pH，P(CO(2))並びにP(O(2))に及ぼす影響に関する研究第l編. 測定機器及び測定条件の吟味\n\nYahata, Takaaki\nAbstract\nHuman venous blood pH, P(CO(2)) and P(O(2)) were measured with I. L. Meter. Its reproducibility and response rate were evaluated as well as methods of the procedures. 1) Time required to reach the stability of pH, P(CO(2)), and P(O(2)) reading after the sampie injection was 1～1.5 min., 2～2.5 min. and 45～75 sec., respectively. It is recommended in the measurement of these parameters at the same time that the sample injection starts with P(CO(2)) electrode, followed by pH and P(O(2)) in this order and that readings are recorded in the order of pH, P(O(2)) and P(CO(2)). 2) Range of differences between the two values measured in the interval of 3～5 min. were pH : -0.010～0.020 (mean: 0.003), P(CO(2)) : -1.0～1.0mmHg (mean: 0.4) and P(O(2)) : -1.0～0.0mmHg (mean: -0.5). Their 5% rejection limits were 0.021≧x(o)≧-0.015, 2.0≧x(o)≧-1.2mmHg and 0.3≧x(o)≧-1.3mmHg, respectively. 3) The pH, P(CO(2)) and P(O(2)) of the heparinized venous blood stored in ice water showed no significant changes in 60 min. and they gave practically the same results as the measurement just after shedding.\nNote\n\nISSN\n0369-7142\nNCID\nAN00032853\nNAID" ]	[ null, "http://escholarship.lib.okayama-u.ac.jp/files/static/eprints/mmc/header.gif", null, "http://escholarship.lib.okayama-u.ac.jp/files/static/eprints/mmc/cover.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9321991,"math_prob":0.9707771,"size":1085,"snap":"2022-05-2022-21","text_gpt3_token_len":371,"char_repetition_ratio":0.1480111,"word_repetition_ratio":0.0,"special_character_ratio":0.33824885,"punctuation_ratio":0.17518248,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9535764,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,6,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-27T09:08:03Z\",\"WARC-Record-ID\":\"<urn:uuid:052b9f53-479a-4dfa-aae1-3a66b11b62d0>\",\"Content-Length\":\"15142\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b605c4fe-da2f-47a8-bfe6-3e02ac58e951>\",\"WARC-Concurrent-To\":\"<urn:uuid:421ea5c0-ed95-4c05-8ab1-dd2e2e4df5ea>\",\"WARC-IP-Address\":\"150.46.138.203\",\"WARC-Target-URI\":\"http://escholarship.lib.okayama-u.ac.jp/en/journal/mmc/38/--/article/40177\",\"WARC-Payload-Digest\":\"sha1:IUV22UO3ZWXSTYXW5PNASUEJBO2GNTQR\",\"WARC-Block-Digest\":\"sha1:HO7LWDXUCLSJ3ZM5N2EG54CIA7QO5UAN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320305242.48_warc_CC-MAIN-20220127072916-20220127102916-00472.warc.gz\"}"}
http://www.mas.ncl.ac.uk/~nser/abstracts/graphprod_ja.html	[ "## Generalising some results about right-angled Artin groups to graph products of groups\n\n### ncl\n\n#### Keywords\n\ngraph products, hyperbolic groups\n\n#### Status\n\npublished in Journal of Algebra\n\n#### Abstract\n\nWe prove three results about the graph product $G=\\G(\\Gamma;G_v, v \\in V(\\Gamma))$ of groups $G_v$ over a graph $\\Gamma$. The first result generalises a result of Servatius, Droms and Servatius, proved by them for right-angled Artin groups; we prove a necessary and sufficient condition on a finite graph $\\Gamma$ for the kernel of the map from $G$ to the associated direct product to be free (one part of this result already follows from a result in S. Kim's Ph.D. thesis). The second result generalises a result of Hermiller and \\u{S}uni\\'{c}, again from right-angled Artin groups; we prove that, for a graph $\\Gamma$ with finite chromatic number, $G$ has a series in which every factor is a free product of vertex groups. The third result provides an alternative proof of a theorem due to Meier, which provides necessary and sufficient conditions on a finite graph $\\Gamma$ for $G$ to be hyperbolic. \\\n\nThe preprint is available as gzipped dvi (1 kB), postscript (1 kB) and pdf files.\n\nAlternatively, you can request a copy by e-mailing me." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82115674,"math_prob":0.90602905,"size":1222,"snap":"2022-05-2022-21","text_gpt3_token_len":312,"char_repetition_ratio":0.13382594,"word_repetition_ratio":0.040816326,"special_character_ratio":0.22504091,"punctuation_ratio":0.094017096,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98294663,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-17T08:39:31Z\",\"WARC-Record-ID\":\"<urn:uuid:3a47f8e9-bb7c-4a6d-81d1-7329401e90bc>\",\"Content-Length\":\"1996\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:18cdb3a1-378a-4743-b1df-a7b42de167d8>\",\"WARC-Concurrent-To\":\"<urn:uuid:d6c407b5-6223-4639-bd50-973da2495eff>\",\"WARC-IP-Address\":\"128.240.212.127\",\"WARC-Target-URI\":\"http://www.mas.ncl.ac.uk/~nser/abstracts/graphprod_ja.html\",\"WARC-Payload-Digest\":\"sha1:6O2M4SYWFPRTVPZKLS24JNDJEZWBTQOS\",\"WARC-Block-Digest\":\"sha1:PTQFA2GOX66LNFVMINF7MGCCDJR5SN2N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662517018.29_warc_CC-MAIN-20220517063528-20220517093528-00560.warc.gz\"}"}
https://studysoupquestions.com/questions/business/1165/the-atomic-number-of-nitrogen-is-7	[ "> > The atomic number of nitrogen is 7.\n\n# The atomic number of nitrogen is 7.\n\n## The atomic mass is 14.01. What can we deduce from this ?\n\n134 Views\n\n### The atomic number of nitrogen is 7.\n\nThe atomic mass is 14.01.\nWhat can we deduce from this ?\n\nThe number of protons in the nucleus of an Nitrogen atom is 7.\nAnd the mass of an atom of a chemical element expressed in atomic mass units.\nIt is approximately equivalent to the number of protons and neutrons in the atom (the mass number) or to the average number allowing for the relative abundances of different isotopes.\nIn the case of nitrogen the atomic mass is 14.01" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9064852,"math_prob":0.9753784,"size":805,"snap":"2021-31-2021-39","text_gpt3_token_len":197,"char_repetition_ratio":0.12609239,"word_repetition_ratio":0.0,"special_character_ratio":0.24223602,"punctuation_ratio":0.10429448,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98297423,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-27T23:20:47Z\",\"WARC-Record-ID\":\"<urn:uuid:852981a0-5fc8-4391-96ba-d98c518118ba>\",\"Content-Length\":\"54422\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d0d52080-beb1-4fa2-980d-554bc61b4f2e>\",\"WARC-Concurrent-To\":\"<urn:uuid:3a948dec-62b5-4878-84fb-0ff25c7c8299>\",\"WARC-IP-Address\":\"54.189.254.180\",\"WARC-Target-URI\":\"https://studysoupquestions.com/questions/business/1165/the-atomic-number-of-nitrogen-is-7\",\"WARC-Payload-Digest\":\"sha1:ELKMQRXRBGVNFHZHBUIKQUGY52LWPHZD\",\"WARC-Block-Digest\":\"sha1:N532NHXD24YUELQONJYK5IGULUNKGALK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780058552.54_warc_CC-MAIN-20210927211955-20210928001955-00641.warc.gz\"}"}
https://result.vg/abstract-energy-is-the-most-important-source-in-the-world/	[ "# ABSTRACT by creating an open system and\n\nABSTRACT\n\nEnergy is the most important source in the world, it is well defined as the ability to do work or heat transfer . The issues surrounding energy are its sources, production, distribution and consumption. The study of fundamental foundations of thermodynamics gives and idea about energy and transfer of heat through calorimetry technique. Calorimetry is a technique used to measure into and out of a matter. In this experiment we will measure heat flow into and out of matter by creating an open system and a close system of ice, liquid water and harm water respectively, and measure the temperature change of the system by placing thermometer into the system for 3 minutes at 15 seconds intervals. If heat energy flows out of the system , the energy transferred to the matter and to the walls of the calorimeter causes a rise in temperature of the matter. The change in temperature can be measured as the difference between the final and initial temperatures of the matter in the calorimeter.\n?\nINTRODUCTION\nThe science of heat flow or energy flow, is called thermodynamics. Heat energy flows spontaneously from hotter body to colder body. The First Law of Thermodynamics also known as law of conservation energy which states that in a process energy cannot be created nor destroyed, it can be transferred between a system and the surroundings and can be converted from one form to another, but the total energy of the universe remains constant . Therefore, all energy transferred between a system and its surroundings is determined as either heat(q) or work(w).\nEnergy is measured in joules, It takes 4.184 J which equals 1 calorie to raise the temperature of one gram of water by 1° C. The quantity of heat needed to raise the temperature of one gram of a substance by 1° C is called specific heat (S). The heat capacity (C) of a substance is the amount of energy needed to raise the temperature of a substance by 1° C. Different from specific heat, the heat capacity does not account for the mass of the matter. The quantity of heat needed to melt one gram of a solid is called a heat of fusion(?)\nIn this experiment we will pay our attention on one particular area of thermodynamics, namely calorimetry which is a technique used to measure heat flow into and out of matter. Our system is created in a container called a calorimeter that separates the thermal process we are interested in studying from the rest of the universe. As changes occur to that matter, we can follow the movement of heat from one portion of the matter to another by observing temperature changes. The container we use as a calorimeter should thermally insulate the matter we are interested in studying, it should prevent matter from entering or exiting once our measurement has begun, and it should allow for easy measurement of temperature changes. The heat transfer in a calorimeter may be expressed with the following equation:\nQ = mS?T\nThe heat transferred is Q is measured in joules or calories, the specific heat is S, and ?T is the change in temperature.\n\nWe Will Write a Custom Essay Specifically\nFor You For Only \\$13.90/page!\n\norder now\n\nEQUIPMENT AND REAGENTS\nStyrofoam cups\nIce 100 mL\nCardboard lid with a hole\nWater\nHot plate\nThermometer (-10 to 110 °C)\n50 mL Beaker\nStop Watch\nPROCEDURE\nThe Heat Capacity of the Calorimeter.\n1. Create an open-system of Ice (solid water) and measure the temperature change of the system by placing a thermometer into the system for 4minutes at 15 seconds intervals.\n2. Heat/boil 100ml of liquid water to a range of 60-80°C\n3. Repeat step 1, by creating an open system of hot water.\n4. Create a closed-system of Ice (solid water) and measure the temperature change of the system by placing a thermometer into the system for 4minutes at 15 seconds intervals.\n5. Repeat step 4, by creating a closed-system of warm water.\n6. Measure 50ml of hot water in a Styrofoam and 50ml of ice (solid water) in separate Styrofoam. Create a closed system by pouring the two prepared contents into a new Styrofoam cup, making a total 100ml of liquid water and ice. Measure the temperature variation for 4 minutes at 10 seconds intervals\n?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9259641,"math_prob":0.9444676,"size":4170,"snap":"2020-45-2020-50","text_gpt3_token_len":906,"char_repetition_ratio":0.14474316,"word_repetition_ratio":0.11404729,"special_character_ratio":0.20935252,"punctuation_ratio":0.07954545,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9525898,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-25T01:50:55Z\",\"WARC-Record-ID\":\"<urn:uuid:65b24160-97ae-4595-9806-84470aefb8ce>\",\"Content-Length\":\"32010\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:98b0273c-072e-4af2-8bc1-64944315749c>\",\"WARC-Concurrent-To\":\"<urn:uuid:03cbe4e8-c1dd-429a-a310-e373f4f64163>\",\"WARC-IP-Address\":\"104.18.50.91\",\"WARC-Target-URI\":\"https://result.vg/abstract-energy-is-the-most-important-source-in-the-world/\",\"WARC-Payload-Digest\":\"sha1:3VLWXJ2PLPABY4C3TGNVKHJZGKABQE5E\",\"WARC-Block-Digest\":\"sha1:RSAEZYP6IAD6OSMJWADGHQ3UD2YRX5E7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107885126.36_warc_CC-MAIN-20201025012538-20201025042538-00460.warc.gz\"}"}
https://www.lmfdb.org/LocalNumberField/23.13.0.1	[ "# Properties\n\n Label 23.13.0.1 Base $$\\Q_{23}$$ Degree $$13$$ e $$1$$ f $$13$$ c $$0$$ Galois group $C_{13}$ (as 13T1)\n\n# Related objects\n\n## Defining polynomial\n\n $$x^{13} - 5 x + 3$$\n\n## Invariants\n\n Base field: $\\Q_{23}$ Degree $d$ : $13$ Ramification exponent $e$ : $1$ Residue field degree $f$ : $13$ Discriminant exponent $c$ : $0$ Discriminant root field: $\\Q_{23}$ Root number: $1$ $\|\\Gal(K/\\Q_{ 23 })\|$: $13$ This field is Galois and abelian over $\\Q_{23}$.\n\n## Intermediate fields\n\n The extension is primitive: there are no intermediate fields between this field and $\\Q_{ 23 }$.\n\n## Unramified/totally ramified tower\n\n Unramified subfield: 23.13.0.1 $\\cong \\Q_{23}(t)$ where $t$ is a root of $$x^{13} - 5 x + 3$$ Relative Eisenstein polynomial: $x - 23 \\in\\Q_{23}(t)[x]$\n\n## Invariants of the Galois closure\n\n Galois group: $C_{13}$ (as 13T1) Inertia group: Trivial Unramified degree: $13$ Tame degree: $1$ Wild slopes: None Galois mean slope: $0$ Galois splitting model: $x^{13} - x^{12} - 60 x^{11} + 27 x^{10} + 1199 x^{9} - 33 x^{8} - 9610 x^{7} - 3352 x^{6} + 33548 x^{5} + 20328 x^{4} - 47723 x^{3} - 34869 x^{2} + 21271 x + 15667$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.52462626,"math_prob":1.0000068,"size":647,"snap":"2020-10-2020-16","text_gpt3_token_len":239,"char_repetition_ratio":0.13685848,"word_repetition_ratio":0.082474224,"special_character_ratio":0.42658424,"punctuation_ratio":0.13559322,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99997437,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-19T18:50:07Z\",\"WARC-Record-ID\":\"<urn:uuid:c57f4fc2-e61e-4ac8-94df-7c300e1eea70>\",\"Content-Length\":\"18114\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:96553556-cea6-4c1d-b9d7-a085fd8d857f>\",\"WARC-Concurrent-To\":\"<urn:uuid:69f640ec-fb3b-4ae1-8745-65cbebdce876>\",\"WARC-IP-Address\":\"35.241.19.59\",\"WARC-Target-URI\":\"https://www.lmfdb.org/LocalNumberField/23.13.0.1\",\"WARC-Payload-Digest\":\"sha1:GXZSNXDW6WC6FCFCH7PDIJMH6R7LNY2C\",\"WARC-Block-Digest\":\"sha1:65AFCQGY7VZUNYM4NVE7G2PDZAN2B2TA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875144167.31_warc_CC-MAIN-20200219184416-20200219214416-00024.warc.gz\"}"}
https://riptutorial.com/python/example/31585/part-2--parsing-tokenized-input-with-yacc	[ "# Python Language Python Lex-Yacc Part 2: Parsing Tokenized Input with Yacc\n\n## Example\n\nThis section explains how the tokenized input from Part 1 is processed - it is done using Context Free Grammars (CFGs). The grammar must be specified, and the tokens are processed according to the grammar. Under the hood, the parser uses an LALR parser.\n\n``````# Yacc example\n\nimport ply.yacc as yacc\n\n# Get the token map from the lexer. This is required.\nfrom calclex import tokens\n\ndef p_expression_plus(p):\n'expression : expression PLUS term'\np = p + p\n\ndef p_expression_minus(p):\n'expression : expression MINUS term'\np = p - p\n\ndef p_expression_term(p):\n'expression : term'\np = p\n\ndef p_term_times(p):\n'term : term TIMES factor'\np = p * p\n\ndef p_term_div(p):\n'term : term DIVIDE factor'\np = p / p\n\ndef p_term_factor(p):\n'term : factor'\np = p\n\ndef p_factor_num(p):\n'factor : NUMBER'\np = p\n\ndef p_factor_expr(p):\n'factor : LPAREN expression RPAREN'\np = p\n\n# Error rule for syntax errors\ndef p_error(p):\nprint(\"Syntax error in input!\")\n\n# Build the parser\nparser = yacc.yacc()\n\nwhile True:\ntry:\ns = raw_input('calc > ')\nexcept EOFError:\nbreak\nif not s: continue\nresult = parser.parse(s)\nprint(result)\n``````\n\n## Breakdown\n\n• Each grammar rule is defined by a function where the docstring to that function contains the appropriate context-free grammar specification. The statements that make up the function body implement the semantic actions of the rule. Each function accepts a single argument p that is a sequence containing the values of each grammar symbol in the corresponding rule. The values of `p[i]` are mapped to grammar symbols as shown here:\n\n`````` def p_expression_plus(p):\n'expression : expression PLUS term'\n# ^ ^ ^ ^\n# p p p p\n\np = p + p\n``````\n• For tokens, the \"value\" of the corresponding `p[i]` is the same as the `p.value` attribute assigned in the lexer module. So, `PLUS` will have the value `+`.\n\n• For non-terminals, the value is determined by whatever is placed in `p`. If nothing is placed, the value is None. Also, `p[-1]` is not the same as `p`, since `p` is not a simple list (`p[-1]` can specify embedded actions (not discussed here)).\n\nNote that the function can have any name, as long as it is preceeded by `p_`.\n\n• The `p_error(p)` rule is defined to catch syntax errors (same as `yyerror` in yacc/bison).\n\n• Multiple grammar rules can be combined into a single function, which is a good idea if productions have a similar structure.\n\n`````` def p_binary_operators(p):\n'''expression : expression PLUS term\n\| expression MINUS term\nterm : term TIMES factor\n\| term DIVIDE factor'''\nif p == '+':\np = p + p\nelif p == '-':\np = p - p\nelif p == '':\np = p p\nelif p == '/':\np = p / p\n``````\n• Character literals can be used instead of tokens.\n\n`````` def p_binary_operators(p):\n'''expression : expression '+' term\n\| expression '-' term\nterm : term '' factor\n\| term '/' factor'''\nif p == '+':\np = p + p\nelif p == '-':\np = p - p\nelif p == '':\np = p * p\nelif p == '/':\np = p / p\n``````\n\nOf course, the literals must be specified in the lexer module.\n\n• Empty productions have the form `'''symbol : '''`\n\n• To explicitly set the start symbol, use `start = 'foo'`, where `foo` is some non-terminal.\n\n• Setting precedence and associativity can be done using the precedence variable.\n\n`````` precedence = (\n('nonassoc', 'LESSTHAN', 'GREATERTHAN'), # Nonassociative operators\n('left', 'PLUS', 'MINUS'),\n('left', 'TIMES', 'DIVIDE'),\n('right', 'UMINUS'), # Unary minus operator\n)\n``````\n\nTokens are ordered from lowest to highest precedence. `nonassoc` means that those tokens do not associate. This means that something like `a < b < c` is illegal whereas `a < b` is still legal.\n\n• `parser.out` is a debugging file that is created when the yacc program is executed for the first time. Whenever a shift/reduce conflict occurs, the parser always shifts.", null, "PDF - Download Python Language for free" ]	[ null, "https://riptutorial.com/Images/icon-pdf-2.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6670539,"math_prob":0.9243867,"size":3790,"snap":"2021-43-2021-49","text_gpt3_token_len":1084,"char_repetition_ratio":0.15795034,"word_repetition_ratio":0.12191358,"special_character_ratio":0.31715038,"punctuation_ratio":0.12418301,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99779326,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-28T16:12:35Z\",\"WARC-Record-ID\":\"<urn:uuid:8aaf3815-2fca-4839-b0ff-25df2f21f1f5>\",\"Content-Length\":\"93975\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:43a2f8c9-c401-4209-97f1-6d1e59e709e8>\",\"WARC-Concurrent-To\":\"<urn:uuid:8ad83f3a-5dd2-4022-8012-3b4d16cfeb4d>\",\"WARC-IP-Address\":\"40.83.160.29\",\"WARC-Target-URI\":\"https://riptutorial.com/python/example/31585/part-2--parsing-tokenized-input-with-yacc\",\"WARC-Payload-Digest\":\"sha1:QXOOKPNLC3PPOYNU4AG6VGURCRKVBUWA\",\"WARC-Block-Digest\":\"sha1:BO3GI4BPOE5UY5HJ3OGDRE3O2DBG2JKA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323588341.58_warc_CC-MAIN-20211028131628-20211028161628-00261.warc.gz\"}"}
https://datasciencetexts.com/subjects/elementary_algorithms.html	[ "# Elementary Algorithms\n\n• None\n\n## Recommended Prerequisites\n\n### Last Updated: 7/18/2019\n\nAn algorithm is a well-defined procedure for accomplishing a well-defined task. Since computers require well-defined procedures to accomplish tasks, the study of algorithms is an essential component of both computer science and data science. The study of algorithms chiefly concerns proving that a particular algorithm will accomplish a particular task and determining the resources the algorithm will require to do so. Elementary algorithm texts describe the basic techniques used for these two goals, and they also catalog a number of algorithms that are useful in common situations. Since algorithms must often exploit and manipulate data, elementary algorithm textbooks also describe many efficient options for structuring data. Data scientists are often tasked with problems that require large amounts of computational resources, so it is important that they understand the tradeoffs of different algorithms.\n\n# Recommended Books\n\n1. ## Introduction To Algorithms\n\n### Key Topics\n\n• Amortized Analysis\n• Approximation Algorithms\n• Asymptotic Analysis\n• Computational Geometry Algorithms\n• Data Structures\n• Divide and Conquer Strategies\n• Dynamic Programming\n• Fast Fourier Transform\n• Graphs and Trees\n• Greedy Algorithms\n• Linear Programming\n• Matrix Operations\n• NP-Completeness\n• Number-Theoretic Algorithms\n• Probabilistic Analysis\n• Randomized Algorithms\n• Recursion\n• Sorting and Searching\n• String Matching\n\n### Description\n\nCLRS is the canonical elementary algorithms textbook. This partly because it is an excellent algorithms text, and partly because so many people use it that otherwise obscure examples in it have become canonical. CLRS will teach you how to prove an algorithm is correct, show you how the resources the algorithm requires will change as the problem size grows, introduces the standard data structures, and contains many well-known algorithms as examples. It also offers a lot of information about randomized algorithms and their analysis. CLRS does have a few potential drawbacks. Some of the exercises are quite difficult for the level of the text, and reliable solutions to the difficult ones are hard to find if you get stuck. Additionally, all the algorithms in the textbook are written in pseudocode, so you're responsible for your own implementation in a real programming language if you wish to experiment. Overall though, this an outstanding book for both learning and reference.\n\n2. ## The Algorithm Design Manual\n\n### Key Features\n\n• In-text exercises\n• C++ algorithm implementations\n• Errata\n\n### Key Topics\n\n• Approximation Algorithms\n• Asymptotic Analysis\n• Combinatorial Algorithms\n• Computational Geometry Algorithms\n• Data Structures\n• Divide and Conquer Strategies\n• Dynamic Programming\n• Graphs and Trees\n• Greedy Algorithms\n• Numerical Algorithms\n• Recursion\n• Sorting and Searching\n• String Algorithms\n\n### Description\n\nThe Algorithm Design Manual is somewhat less well-known than CLRS, but it has a lot going for it as an algorithms text. It is relatively inexpensive, offers a lot of algorithms that are absent in other books, and provides C++ implementations for many of the algorithms. We consider The Algorithm Design Manual a great companion to CLRS, but it can stand alone as an algorithms text, especially for a first pass. CLRS offers more analytic tools, and does a better job with randomized algorithms, but any user of algorithms would benefit from reading this book." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8812206,"math_prob":0.8620833,"size":3572,"snap":"2019-26-2019-30","text_gpt3_token_len":699,"char_repetition_ratio":0.15610987,"word_repetition_ratio":0.0858209,"special_character_ratio":0.1774916,"punctuation_ratio":0.07899462,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.973041,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-21T18:13:31Z\",\"WARC-Record-ID\":\"<urn:uuid:85c7356f-f115-4293-8dc3-718cac61889d>\",\"Content-Length\":\"8886\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec3f6089-c7dd-4cfc-a7b9-c0d42ee13430>\",\"WARC-Concurrent-To\":\"<urn:uuid:0390ef41-e712-489a-bb2d-88aaeda863f1>\",\"WARC-IP-Address\":\"99.84.216.63\",\"WARC-Target-URI\":\"https://datasciencetexts.com/subjects/elementary_algorithms.html\",\"WARC-Payload-Digest\":\"sha1:UULH4KK2QKZGUEPU3VGFQS42ZVRTC76B\",\"WARC-Block-Digest\":\"sha1:NOEITP7AG4UPUMN3WVPKCRIO26FZNATG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195527089.77_warc_CC-MAIN-20190721164644-20190721190644-00298.warc.gz\"}"}
https://www.saildart.org/COMMEN%5B1,LMM%5D1	[ "perm filename COMMEN[1,LMM]1 blob sn#029047 filedate 1973-03-12 generic text, type T, neo UTF8\n``` (PROGN (LISPXPRIN1 (QUOTE \"FILE CREATED \")\nT)\n(LISPXPRIN1 (QUOTE \"12-MAR-73 03:01:06\")\nT)\n(LISPXTERPRI T))\n(LISPXPRINT (QUOTE COMMENTVARS)\nT)\n(RPAQQ COMMENTVARS ((FNS COLLECT* FIL)\n(DEFINEQ\n\n(COLLECT\n[APPLY (QUOTE EDITF)\n(LIST FN (QUOTE (LP F (E (SETQ COMMENTS (CONS (##)\nT]\n(CONS FN (CONS (ARGLIST FN)\n\n(FIL\n[LAMBDA (FIL)\n(PROG ([FILCOMMENTS (CONS (QUOTE :)\n(MAPCAR (FILEFNSLST FIL)\n(FUNCTION COLLECT]\n(AL (ASSOC FIL COMMENTS)))\n(COND\n(AL (RPLACD AL FILCOMMENTS))\n(T (SETQ COMMENTS (CONS (CONS FIL FILCOMMENTS)\n)\n[RPAQQ COMMENTS ([CYCLIC : ((VALENCE (X)\n( This function finds the\nVALENCE of an atom or\nSTRUCTURE or STRUCFORM))\n(FVPARTITION1 (N VL S)\n(* This function is a\nsub-function of\nFVPARTITIONS - I'm not too\nsure what it does offhand))\n(FVPART1 (N MAXSUM MAXOCCUR)\n(* Again , i'm not too sure what\nthis function does))\n(MINLOOPS (VALENCELIST)\n(* This function computes the\nminimum number of loops that\nany STRUCTURE with the given\nVALENCELIST must have - See\ncycgen paper for derivation of\nformula))\n(MAXLOOPS (VALENCELIST)\n(* This function computes the\nmaximum number of loops that\nany STRUCTURE with the given\nVALENCELIST may have - See\ncycgen paper for derivation of\nformula))\n(SUPERATOMPARTITIONS (CL U)\n(* This function finds\nall partitions of\nCL and unsaturation\nU into superatom\nparts and remaining\natoms, according to\nthe constraints\ngiven in the cycgen\nsuperatom parts -\nThe value is a list\nof\nSUPERATOMPARTITION\nrecords))\n(MAXUNSATL (PC U)\n(* This function, i think, takes\na composition list of\ncomposition lists (pc)\nand an unsaturation (U)\nand returns a list of the\nmaximum unsaturation that may\nbe assigned to each\nindividual part in pc such\nthat the final structures\nwill each have correct TOTAL\nunsaturation and a free\nVALENCE of at least one)\n(* Note U is either NIL\n(normal)\nor it is equal to the\nunsaturation IN the case\nWHERE remats is NIL and there\nis only one part here))\n(COMPUTEFV (U CL)\n(* This function computes the\nfree VALENCE of a composition\nand saturation - I.e. Any\nSTRUCTURE with the given\ncomposition and unsaturation\nwill have the resulting\nnumber of free valences))\n(ROWS (LL)\n(* Ll is a list of lists - If one\nenvisions it as a matrix, this\nfunction computes the transpose))\n(BIVALENTPARTITIONS (VL)\n(* This function takes\na valence list\n(starting with\nbivalents)\nand partitions\n(CAR VL)\ninto\n(number\nof EDGES IN a\nSTRUCTURE built on\n(CDR VL))\nparts))\n(TRIMZEROS (L)\n(* This function takes a list of\nnumbers and returns the list\nwith trailing zeros removed))\n(TD (VL J)\n(* This function takes a VALENCE list\nstarting with j-valents and returns\nthe TOTAL VALENCE))\n(M2/2 (N)\n(* Silly function - Computes\n(n/2)\n-1))\n(LOOPPARTITIONS1 (P VL J)\n(* SUBFUNCTION OF\nLOOPPARTITIONS - DONT\nREMEMBER WHAT IT DOES))\n(JLIST (LL N)\n(* AGAIN, I DON'T REMEMBER WHAT THIS\nONE DOES))\n(LPROWS (LPP VL)\n(* AGAIN, I DON'T REMEMBER WHAT\nTHIS ONE DOES))\n(LOOPPARTITIONS (P VL)\n(* THIS FUNCTION FINDS ALL\nLOOP PARTITIONS - I'M\nNOT SURE HOW, THOUGH))\n(CLPARTLP1 (CL ROW N)\n(* Again, i don't remember what\nthis one does))\n(STRUCTURESWITHATOMS (CLL STRUC)\n(* CLL is a list of\ncomposition lists\nthe first CL\ncontains bivalent\natoms, the second,\ntrivalent atoms,\nand so forth -\nStruc is a\nSTRUCTURE -\nSTRUCTURESWITHATOMS\nfinds all ways of\nattaching the given\natoms to the\nSTRUCTURE by\nlabelling))\n(NUMPARTITIONS (N NUMPARTS MINPART MAXPART)\n(* This function finds all\npartitions of the number\nN into numparts parts,\nWHERE each part is\ngreater or equal to\nminpart and less than or\nequal to maxpart - The\nresult is a list of\npartitions, WHERE a\npartition here is a list\nof numbers, the sum of\nwhich is N))\n(NUMPARTITIONS* (U MN MAXIMA OCCURLIST)\n(* Again, i don't remember\nwhat this one does))\n(FVPARTITIONS (FV VL)\n(* This function finds all\nways of partitioning free\nvalences fv among the\n\"ATOMS\" of vl\n(vl is a VALENCE list)\naccording to appropriate\nconstraints]\n[STRGEN : ((STARTUP NIL (* This function does all of the\nthings necessary to LOAD the\nSTRUCTURE generator]\n[CL : ((CLDIFF (CL1 CL2)\n(* This function computes the DIFFERENCE of\ntwo composition lists - Zero terms are\neliminated))\n(CLCOUNT (CL)\n(* This function computes the number of\nelements IN a composition list))\n[CLPARTS (CL PARTSIZE)\n(* This function finds all SUB compositions of\nthe composition list cl1 which are of SIZE\nparsizze, and returns a list of the\npossibilities - I.e. (CLPARTS\n'\n((A . 3)\n(B . 2))\n2)\nreturns\n(((a . 2))\n,\n((a . 1)\n(b . 1))\n,\n((b . 2]\n(CLPARTITIONSN (CL N MINPARTSIZE MAXPARTSIZE)\n(* This function finds all partitions of\nCL into N parts WHERE each part has a\nCLCOUNT of at least MINPARTSIZE and at\nmost MAXPARTSIZE))\n(CLPARTITIONS (CL PARTSIZES)\n(* PARTSIZES IS a list of numbers - This\nfunction finds all partitions of CL\ninto PARTS WHERE each PART IS of the\ncorresponding SIZE IN PARTSIZES - The\nsum of PARTSIZES must be equal to the\nCLCOUNT of CL or ELSE the value will be\nNIL - The value IS a list of\npartitions; a partition IS a list of\ncomposition lists))\n[CLCREATE (L)\n(* This function takes a list which may have\nduplicates, and returns a composition list\nwhich corresponds to it - I.e.\n(CLCREATE '\n(A A A B B C))\nreturns\n((a . 3)\n(b . 2)\n(C . 1]\n(CLINSERT (ITEM CL)\n(* This function returns the composition list\nCL with \"ITEM\" inserted))\n(CL=PARTS (CL NPARTS PARTSIZE)\n(* This function finds all partitions of CL\ninto NPARTS parts, where every part is of\nsize PARTSIZE - NPARTSPARTSIZE must be\nequal to the CLCOUNT of CL))\n(CLBYVALENCE (CL)\n( CL must be a composition list of things\nwith a VALENCE - This function returns a\nlist of composition lists; the first CL\ncontains those things with VALENCE 2 -\nThe second those with VALENCE 3, and so\non))\n(CLPARTITIONSL (CL LL)\n(* Damn if i can remember what this one\ndoes))\n(CLEXPAND (CL)\n(* This function is the inverse of CLCREATE -\nIt takes a composition list and returns a\nlist with the appropriate number of copies\nof each item IN the composition list - I.e.\n(CLEXPAND ' ((A . 3)\n(B . 2)))\ngives\n(a a a b b]\n(* This function takes a composition\nlist of lists, and returns all\npossible selections of items from\nthose lists; e.g. doing GROUPRADS\non (((a b C D e) . 3)\n((F g H i) . 2))\nwill return all lists in which\nthree elements come from\n(a b C D e)\nand 2 elements come from\n(F g H i)\n- Duplication is allowed; i.e.\n(a a a F F)\nwill be among the resulting lists))\n(* Subfunction of GROUPRADS))\n(FIX+ (X)\n(* Rounds a number upward))\n(GROUPBY (FN L)\n(* FN is a functional argument, l is a\nlist - This function groups l by the\nvalue of FN applied to its elements -\nIt returns a list of groups - The CDR\nof a GROUP contains elements of l which\nall have the same value of FN - The CAR\nis that value - Can be used to GROUP a\nlist of atom names by their VALENCE,\nfor EXAMPLE))\n[CARLIST (L)\n( (MAPCAR L (FUNCTION CAR]\n[CDRLIST (L)\n(* (MAPCAR L (FUNCTION CDR]\n[LCARLIST (L)\n(* (MAPCAR L (FUNCTION CARLIST]\n[LCDRLIST (L)\n(* (MAPCAR L (FUNCTION CDRLIST]\n(DELETE (I L)\n(* Returns l with the first instance of i\ndeleted))\n(DIFF (L1 L2)\n(* Returns the list of elements of l1 which\nare not elements of l2))\n(ORDPAIR (X1 X2)\n(* Returns either (CONS X1 X2)\nor\n(CONS X2 X1)\ndepending on whether x1<x2 or not -\nUses a generalized ordering in which\nanumbers are in order numerically, and\nless than atoms which are ordered\nalphabetically, and are less than lists\nwhich are ordered first by their CAR\nand then by their CDR))\n(MAXREST (VL J)\n(* Silly function used somewhere - Again\ndamn if i know))\n(MAX (X Y)\n(* MAX of X and Y))\n(MIN (X Y)\n(* MIN of X and Y - Uses INTEGER arithmetic))\n(TWICE (X)\n(* X2))\n(LEQ (X Y)\n( Generalized ordering function - See ORDPAIR)\n)\n(PLUS (L)\n( L is a list of integers - PLUS computes\ntheir sum))\n(LMASSOC (X Y Z)\n( This is similar to the ASSOC on the 360\n- If there is an element of Y whose CAR\nis equal to Y, returns its CDR -\nOtherwise returns z))\n(MAX (L)\n( L is a list of integers - *MAX returns\ntheir maximum]\nSTOP\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.70753455,"math_prob":0.96819806,"size":8171,"snap":"2022-40-2023-06","text_gpt3_token_len":2435,"char_repetition_ratio":0.1586874,"word_repetition_ratio":0.078242965,"special_character_ratio":0.27254927,"punctuation_ratio":0.046657383,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9877182,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-01T22:08:42Z\",\"WARC-Record-ID\":\"<urn:uuid:75e6148e-dd40-4378-b238-391bfb94f5aa>\",\"Content-Length\":\"20107\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:88300b96-b032-459c-a6ac-06a7f6472c00>\",\"WARC-Concurrent-To\":\"<urn:uuid:fe5f70f3-4e1d-4055-b6e0-c81d3cde3c76>\",\"WARC-IP-Address\":\"50.247.119.156\",\"WARC-Target-URI\":\"https://www.saildart.org/COMMEN%5B1,LMM%5D1\",\"WARC-Payload-Digest\":\"sha1:BF3QWCL4IMRVQY427KA6SVEYUEFBQU6R\",\"WARC-Block-Digest\":\"sha1:OJNOWK3C5NNPDJVFNDWRYLYT3CAIGBQL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030336921.76_warc_CC-MAIN-20221001195125-20221001225125-00275.warc.gz\"}"}
https://link.springer.com/article/10.1007%2Fs11425-018-9465-8	[ "Science China Mathematics\n\n, Volume 62, Issue 11, pp 2057–2072\n\n# Global regularity of optimal mappings in non-convex domains\n\nArticles\n\n## Abstract\n\nIn this paper, we establish a global regularity result for the optimal transport problem with the quadratic cost, where the domains may not be convex. This result is obtained by a perturbation argument, using a recent global regularity of optimal transportation in convex domains by the authors.\n\n## Keywords\n\nMonge-Ampère equation optimal transportation global regularity\n\n## MSC(2010)\n\n35J96 35J25 35B65\n\n## References\n\n1. 1.\nBakelman I. Convex analysis and nonlinear geometric elliptic equations. Berlin: Springer-Verlag, 1994\n2. 2.\nBrenier Y. Polar factorization and monotone rearrangement of vector-valued functions. Comm Pure Appl Math, 1991, 44: 375–417\n3. 3.\nCaffarelli L. Interior W 2,p estimates for solutions of Monge-Ampere equations. Ann of Math (2), 1990, 131: 135–150\n4. 4.\nCaffarelli L. The regularity of mappings with a convex potential. J Amer Math Soc, 1992, 5: 99–104\n5. 5.\nCaffarelli L. Allocation maps with general cost functions. In: Partial Differential Equations and Applications. Lecture Notes in Pure and Applied Mathematics, 177. New York: Dekker, 1996, 29–35\n6. 6.\nCaffarelli L. Boundary regularity of maps with convex potentials, II. Ann of Math (2), 1996, 144: 453–496\n7. 7.\nCaffarelli L, Gonzáles M, Nguyen T. A perturbation argument for a Monge-Ampere equation with periodic data. Arch Ration Mech Anal, 2014, 212: 359–414\n8. 8.\nCaffarelli L, McCann R. Free boundaries in optimal transport and Monge-Amere obstacle problems. Ann of Math (2), 2010, 171: 673–730\n9. 9.\nChen S. Boundary C 1,α regularity of an optimal transport problem with cost close tox. SIAM J Math Anal, 2015, 47: 2689–2698\n10. 10.\nChen S. Regularity of free boundaries in optimal transportation. PreprintGoogle Scholar\n11. 11.\nChen S, Figalli A. Boundary ε-regularity in optimal transportation. Adv Math, 2015, 273: 540–567\n12. 12.\nChen S, Figalli A. Partial W 2,p regularity for optimal transport maps. J Funct Anal, 2017, 272: 4588–4605\n13. 13.\nChen S, Liu J, Wang X-J. Global regularity for the Monge-Ampere equation with natural boundary condition. ArXiv: 1802.07518Google Scholar\n14. 14.\nChen S, Liu J, Wang X-J. Boundary regularity for the second boundary-value problem of Monge-Ampere equations in dimension two. ArXiv:1806.09482Google Scholar\n15. 15.\nDe Philippis G, Figalli A. Partial regularity for optimal transport maps. Publ Math de l’IHES, 2015, 121: 81–112\n16. 16.\nDe Philippis G, Figalli A, Savin O. A note on interior W 2,1+ε estimates for the Monge-Ampere equation. Math Ann, 2013, 357: 11–22\n17. 17.\nDelanoe Ph. Classical solvability in dimension two of the second boundary value problem associated with the Monge-Ampere operator. Ann Inst Henri Poincaré Anal Non Linéaire, 1991, 8: 443–457\n18. 18.\nEvans L. Partial differential equations and Monge-Kantorovich mass transfer. In: Current Development in Mathematics. Boston: International Press, 1999, 65–126\n19. 19.\nFigalli A. The Monge-Ampere Equation and its Applications. Zürich Lectures in Advanced Mathematics. Zürich: Euro Math Soc, 2017\n20. 20.\nGilbarg D, Trudinger N. Elliptic Partial Differential Equations of Second Order. Berlin: Springer-Verlag, 1983\n21. 21.\nGutierrez C. The Monge-Ampere Equation. Progress in Nonlinear Differential Equations and their Applications, 44. Boston: Birkhäuser, 2001\n22. 22.\nJian H Y, Wang X-J. Continuity estimates for the Monge-Ampere equation. SIAM J Math Anal, 2007, 39: 608–626\n23. 23.\nKitagawa J, McCann R. Free discontinuities in optimal transport. Arch Ration Mech Anal, 2019, in press\n24. 24.\nLions P-L, Trudinger N, Urbas J. The Neumann problem for equations of Monge-Ampere type. Comm Pure Appl Math, 1986, 39: 539–563\n25. 25.\nMa X N, Trudinger N, Wang X-J. Regularity of potential functions of the optimal transportation problem. Arch Ration Mech Anal, 2005, 177: 151–183\n26. 26.\nMonge G. Memoire sur la Theorie des Déblais et des Remblais. In: Historie de l’Académie Royale des Sciences de Paris, avec les Mémoires de Mathématique et de Physique pour la M eme année, 1781, 666–704Google Scholar\n27. 27.\nPogorelov A. Monge-Ampere Equations of Elliptic Type. Groningen: Noordhoff, 1964\n28. 28.\nSavin O. Pointwise C 2,α estimates at the boundary for the Monge-Ampere equation. J Amer Math Soc, 2013, 26: 63–99\n29. 29.\nSavin O. Global W 2,p estimates for the Monge-Ampere equations. Proc Amer Math Soc, 2013, 141: 3573–3578\n30. 30.\nTrudinger N, Wang X-J. Boundary regularity of the Monge-Ampere and affne maximal surface equations. Ann of Math (2), 2008, 167: 993–1028\n31. 31.\nTrudinger N, Wang X-J. The Monge-Ampere equation and its geometric applications. In: Handbook of Geometric Analysis, vol. 1. Adv Lect Math (ALM), vol. 7. Somerville: International Press, 2008, 467–524\n32. 32.\nUrbas J. On the second boundary value problem of Monge-Ampère type. J Reine Angew Math, 1997, 487: 115–124\n33. 33.\nVillani C. Optimal Transport. Old and New. Grundlehren Math Wiss, vol. 338. Berlin: Springer-Verlag, 2006\n34. 34.\nWang X-J, Wu Y T. A new proof for the regularity of Monge-Ampère type equations. J Differential Geom, 2019, in pressGoogle Scholar\n35. 35.\nWolfson J. Minimal Lagrangian diffeomorphisms and the Monge-Ampère equation. J Differential Geom, 1997, 46: 335–373", null, "" ]	[ null, "https://link.springer.com/track/controlled/article/granted/10.1007/s11425-018-9465-8", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.54582095,"math_prob":0.89371204,"size":6298,"snap":"2019-43-2019-47","text_gpt3_token_len":1830,"char_repetition_ratio":0.20734033,"word_repetition_ratio":0.01183432,"special_character_ratio":0.26659256,"punctuation_ratio":0.21750212,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9897956,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-11T22:27:47Z\",\"WARC-Record-ID\":\"<urn:uuid:ad1a4abf-900b-4505-892c-5966f882a3fc>\",\"Content-Length\":\"89480\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d42f35c0-3aeb-4fff-804a-324ea3cf3588>\",\"WARC-Concurrent-To\":\"<urn:uuid:d5aa94cc-2123-4a92-b10b-153fcba49d31>\",\"WARC-IP-Address\":\"151.101.200.95\",\"WARC-Target-URI\":\"https://link.springer.com/article/10.1007%2Fs11425-018-9465-8\",\"WARC-Payload-Digest\":\"sha1:LLMSR5IHXWSN4HWF4A43YXVWHJDNR5L6\",\"WARC-Block-Digest\":\"sha1:AFK2NG6FBOXRUAMTCEV7CFWDVIK3IK6S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496664439.7_warc_CC-MAIN-20191111214811-20191112002811-00265.warc.gz\"}"}
https://la.mathworks.com/help/stats/bootci.html	[ "# bootci\n\nBootstrap confidence interval\n\n## Syntax\n\n``ci = bootci(nboot,bootfun,d)``\n``ci = bootci(nboot,bootfun,d1,...,dN)``\n``ci = bootci(nboot,{bootfun,d},Name,Value)``\n``ci = bootci(nboot,{bootfun,d1,...,dN},Name,Value)``\n``[ci,bootstat] = bootci(___)``\n\n## Description\n\nexample\n\n````ci = bootci(nboot,bootfun,d)` computes a 95% bootstrap confidence interval for each statistic computed by the function `bootfun`. The `bootci` function uses `nboot` bootstrap samples in its computation, and creates each bootstrap sample by sampling with replacement from the rows of `d`.```\n\nexample\n\n````ci = bootci(nboot,bootfun,d1,...,dN)` creates each bootstrap sample by sampling with replacement from the rows of the nonscalar data arguments in `d1,...,dN`. These nonscalar arguments must have the same number of rows. The `bootci` function passes the samples of nonscalar data and the unchanged scalar data arguments in `d1,...,dN` to `bootfun`.```\n\nexample\n\n````ci = bootci(nboot,{bootfun,d},Name,Value)` specifies options using one or more name-value arguments. For example, you can change the type of confidence interval by specifying the `'Type'` name-value argument.Note that you must pass the `bootfun` and `d` arguments to `bootci` as a single cell array.```\n\nexample\n\n````ci = bootci(nboot,{bootfun,d1,...,dN},Name,Value)` specifies options using one or more name-value arguments. For example, you can change the significance level of the confidence interval by specifying the `'Alpha'` name-value argument.Note that you must pass the `bootfun` and `d1,...,dN` arguments to `bootci` as a single cell array.```\n\nexample\n\n````[ci,bootstat] = bootci(___)` also returns the bootstrapped statistic computed for each of the `nboot` bootstrap replicate samples, using any of the input argument combinations in the previous syntaxes. Each row of `bootstat` contains the results of applying `bootfun` to one bootstrap sample.```\n\n## Examples\n\ncollapse all\n\nCompute the confidence interval for the capability index in statistical process control.\n\nGenerate 30 random numbers from the normal distribution with mean 1 and standard deviation 1.\n\n```rng('default') % For reproducibility y = normrnd(1,1,30,1);```\n\nSpecify the lower and upper specification limits of the process. Define the capability index.\n\n```LSL = -3; USL = 3; capable = @(x)(USL-LSL)./(6std(x));```\n\nCompute the 95% confidence interval for the capability index by using 2000 bootstrap samples. By default, `bootci` uses the bias corrected and accelerated percentile method to construct the confidence interval.\n\n`ci = bootci(2000,capable,y)`\n```ci = 2×1 0.5937 0.9900 ```\n\nCompute the studentized confidence interval for the capability index.\n\n`sci = bootci(2000,{capable,y},'Type','student')`\n```sci = 2×1 0.5193 0.9930 ```\n\nCompute bootstrap confidence intervals for the coefficients of a nonlinear regression model. The technique used in this example involves bootstrapping the predictor and response values, and assumes that the predictor variable is random. For a technique that assumes the predictor variable is fixed and bootstraps the residuals, see Bootstrap Confidence Intervals for Linear Regression Model Coefficients.\n\nNote: This example uses `nlinfit`, which is useful when you only need the coefficient estimates or residuals of a nonlinear regression model and you need to repeat fitting a model multiple times, as in the case of bootstrapping. If you need to investigate a fitted regression model further, create a nonlinear regression model object by using `fitnlm`. You can create confidence intervals for the coefficients of the resulting model by using the `coefCI` object function, although this function does not use bootstrapping.\n\nGenerate data from the nonlinear regression model $y={b}_{1}+{b}_{2}\\cdot exp\\left(-{b}_{3}x\\right)+ϵ$, where ${b}_{1}=1$, ${b}_{2}=3$, and ${b}_{3}=2$ are coefficients; the predictor variable x is exponentially distributed with mean 2; and the error term $ϵ$ is normally distributed with mean 0 and standard deviation 0.1.\n\n```modelfun = @(b,x)(b(1)+b(2)exp(-b(3)x)); rng('default') % For reproducibility b = [1;3;2]; x = exprnd(2,100,1); y = modelfun(b,x) + normrnd(0,0.1,100,1);```\n\nCreate a function handle for the nonlinear regression model that uses the initial values in `beta0`.\n\n```beta0 = [2;2;2]; beta = @(predictor,response)nlinfit(predictor,response,modelfun,beta0)```\n```beta = function_handle with value: @(predictor,response)nlinfit(predictor,response,modelfun,beta0) ```\n\nCompute the 95% bootstrap confidence intervals for the coefficients of the nonlinear regression model. Create the bootstrap samples from the generated data `x` and `y`.\n\n`ci = bootci(1000,beta,x,y)`\n```ci = 2×3 0.9821 2.9552 2.0180 1.0410 3.1623 2.2695 ```\n\nThe first two confidence intervals include the true coefficient values ${b}_{1}=1$ and ${b}_{2}=3$, respectively. However, the third confidence interval does not include the true coefficient value ${b}_{3}=2$.\n\nNow compute the 99% bootstrap confidence intervals for the model coefficients.\n\n`newci = bootci(1000,{beta,x,y},'Alpha',0.01)`\n```newci = 2×3 0.9730 2.9112 1.9562 1.0469 3.1876 2.3133 ```\n\nAll three confidence intervals include the true coefficient values.\n\nCompute bootstrap confidence intervals for the coefficients of a linear regression model. The technique used in this example involves bootstrapping the residuals and assumes that the predictor variable is fixed. For a technique that assumes the predictor variable is random and bootstraps the predictor and response values, see Bootstrap Confidence Intervals for Nonlinear Regression Model Coefficients.\n\nNote: This example uses `regress`, which is useful when you only need the coefficient estimates or residuals of a regression model and you need to repeat fitting a model multiple times, as in the case of bootstrapping. If you need to investigate a fitted regression model further, create a linear regression model object by using `fitlm`. You can create confidence intervals for the coefficients of the resulting model by using the `coefCI` object function, although this function does not use bootstrapping.\n\n`load hald`\n\nPerform a linear regression and compute the residuals.\n\n```x = [ones(size(heat)),ingredients]; y = heat; b = regress(y,x); yfit = xb; resid = y - yfit;```\n\nCompute the 95% bootstrap confidence intervals for the coefficients of the linear regression model. Create the bootstrap samples from the residuals. Use normal approximated intervals with bootstrapped bias and standard error by specifying `'Type','normal'`. You cannot use the default confidence interval type in this case.\n\n```ci = bootci(1000,{@(bootr)regress(yfit+bootr,x),resid}, ... 'Type','normal')```\n```ci = 2×5 -47.7130 0.3916 -0.6298 -1.0697 -1.2604 172.4899 2.7202 1.6495 1.2778 0.9704 ```\n\nPlot the estimated coefficients `b`, omitting the intercept term, and display error bars showing the coefficient confidence intervals.\n\n```slopes = b(2:end)'; lowerBarLengths = slopes-ci(1,2:end); upperBarLengths = ci(2,2:end)-slopes; errorbar(1:4,slopes,lowerBarLengths,upperBarLengths) xlim([0 5]) title('Coefficient Confidence Intervals')```", null, "Only the first nonintercept coefficient is significantly different from 0.\n\nCompute the mean and standard deviation of 100 bootstrap samples. Find the 95% confidence interval for each statistic.\n\nGenerate 100 random numbers from the exponential distribution with mean 5.\n\n```rng('default') % For reproducibility y = exprnd(5,100,1);```\n\nDraw 100 bootstrap samples from the vector `y`. For each bootstrap sample, compute the mean and standard deviation. Find the 95% bootstrap confidence interval for the mean and standard deviation.\n\n`[ci,bootstat] = bootci(100,@(x)[mean(x) std(x)],y);`\n\n`ci(:,1)` contains the lower and upper bounds of the mean confidence interval, and `c(:,2)` contains the lower and upper bounds of the standard deviation confidence interval. Each row of `bootstat` contains the mean and standard deviation of a bootstrap sample.\n\nPlot the mean and standard deviation of each bootstrap sample as a point. Plot the lower and upper bounds of the mean confidence interval as dotted vertical lines, and plot the lower and upper bounds of the standard deviation confidence interval as dotted horizontal lines.\n\n```plot(bootstat(:,1),bootstat(:,2),'o') xline(ci(1,1),':') xline(ci(2,1),':') yline(ci(1,2),':') yline(ci(2,2),':') xlabel('Mean') ylabel('Standard Deviation')```", null, "## Input Arguments\n\ncollapse all\n\nNumber of bootstrap samples to draw, specified as a positive integer scalar. To create each bootstrap sample, `bootci` randomly selects with replacement `n` out of the `n` rows of nonscalar data in `d` or `d1,...,dN`.\n\nExample: `100`\n\nData Types: `single` \| `double`\n\nFunction to apply to each sample, specified as a function handle. The function can be a custom or built-in function. You must specify `bootfun` with the `@` symbol.\n\nExample: `@mean`\n\nData Types: `function_handle`\n\nData to sample from, specified as a column vector or matrix. The `n` rows of `d` correspond to observations. When you use multiple data input arguments `d1,...,dN`, you can specify some arguments as scalar values, but all nonscalar arguments must have the same number of rows.\n\nIf you use a single vector argument `d`, you can specify it as a row vector. `bootci` then samples from the elements of the vector.\n\nData Types: `single` \| `double`\n\n### Name-Value Arguments\n\nSpecify optional comma-separated pairs of `Name,Value` arguments. `Name` is the argument name and `Value` is the corresponding value. `Name` must appear inside quotes. You can specify several name and value pair arguments in any order as `Name1,Value1,...,NameN,ValueN`.\n\nExample: `bootci(100,{@mean,1:6'},'Alpha',0.1)` specifies to draw 100 bootstrap samples from the values 1 through 6, take the mean of each sample, and then compute the 90% confidence interval for the sample mean.\n\nSignificance level, specified as a positive scalar between 0 and 1. `bootci` computes the `100*(1-Alpha)` bootstrap confidence interval of each statistic defined by the function `bootfun`.\n\nExample: `'Alpha',0.01`\n\nData Types: `single` \| `double`\n\nConfidence interval type, specified as one of the values in this table.\n\nValueDescription\n`'norm'` or `'normal'` Normal approximated interval with bootstrapped bias and standard error \n`'per'` or `'percentile'`Basic percentile method\n`'cper'` or ```'corrected percentile'```Bias corrected percentile method \n`'bca'`\n\nBias corrected and accelerated percentile method , — Involves a z0 factor that is computed using the proportion of bootstrap values that are less than the original sample value. To produce reasonable results when the sample is lumpy, the software computes z0 by including half of the bootstrap values that are the same as the original sample value.\n\n`'stud'` or `'student'`Studentized confidence interval \n\nExample: `'Type','student'`\n\nNumber of bootstrap samples for the studentized standard error estimate, specified as a positive integer scalar.\n\n`bootci` computes the studentized bootstrap confidence interval of the statistic defined by the function `bootfun`, and estimates the standard error of the bootstrap statistics by using `NBootStd` bootstrap data samples.\n\nNote\n\nTo use this name-value argument, the `Type` value must be `'stud'` or `'student'`. Specify either `NBootStd` or `StdErr`, but not both.\n\nExample: `'NBootStd',50`\n\nData Types: `single` \| `double`\n\nFunction used to compute the studentized standard error estimate, specified as a function handle.\n\n`bootci` computes the studentized bootstrap confidence interval of the statistic defined by the function `bootfun`, and estimates the standard error of the bootstrap statistics by using the function `StdErr`. The `StdErr` function must take the same arguments as `bootfun` and return the standard error of the statistic computed by `bootfun`.\n\nNote\n\nTo use this name-value argument, the `Type` value must be `'stud'` or `'student'`. Specify either `NBootStd` or `StdErr`, but not both.\n\nExample: `'StdErr',@std`\n\nData Types: `function_handle`\n\nObservation weights, specified as a nonnegative vector with at least one positive element. The number of elements in `Weights` must be equal to the number of rows `n` in the data `d` or `d1,...,dN`. To obtain one bootstrap sample, `bootci` randomly selects with replacement `n` out of `n` rows of data using these weights as multinomial sampling probabilities.\n\nData Types: `single` \| `double`\n\nOptions for computing bootstrap iterations in parallel and setting random numbers during the bootstrap sampling, specified as a structure. Create the `Options` structure with `statset`. This table lists the option fields and their values.\n\nField NameValueDefault\n`UseParallel`Set this value to `true` to compute bootstrap iterations in parallel.`false`\n`UseSubstreams`\n\nSet this value to `true` to run computations in parallel in a reproducible fashion.\n\nTo compute reproducibly, set `Streams` to a type that allows substreams: `'mlfg6331_64'` or `'mrg32k3a'`.\n\n`false`\n`Streams`Specify this value as a `RandStream` object or cell array of such objects. Use a single object except when the `UseParallel` value is `true` and the `UseSubstreams` value is `false`. In that case, use a cell array that has the same size as the parallel pool.If you do not specify `Streams`, then `bootci` uses the default stream or streams.\n\nNote\n\nYou need Parallel Computing Toolbox™ to run computations in parallel.\n\nExample: `'Options',statset('UseParallel',true)`\n\nData Types: `struct`\n\n## Output Arguments\n\ncollapse all\n\nConfidence interval bounds, returned as a vector, matrix, or multidimensional array with two rows.\n\n• If `bootfun` returns a scalar, then `ci` is a vector containing the lower and upper bounds of the confidence interval.\n\n• If `bootfun` returns a vector of length m, then `ci` is a matrix of size 2-by-m, where `ci(1,:)` are lower bounds and `ci(2,:)` are upper bounds.\n\n• If `bootfun` returns a multidimensional array, then `ci` is an array, where `ci(1,:,...)` is an array of lower bounds and `ci(2,:,...)` is an array of upper bounds.\n\nBootstrap statistics, returned as a column vector or matrix with `nboot` rows. The `i`th row of `bootstat` corresponds to the results of applying `bootfun` to the `i`th bootstrap sample. If `bootfun` returns a matrix or array, then the `bootci` function first converts this output to a row vector before storing it in `bootstat`.\n\n Davison, A. C., and D. V. Hinkley. Bootstrap Methods and Their Applications. Cambridge University Press, 1997.\n\n Efron, Bradley. The Jackknife, the Bootstrap and Other Resampling Plans. Philadelphia: The Society for Industrial and Applied Mathematics, 1982.\n\n DiCiccio, Thomas J., and Bradley Efron. “Bootstrap Confidence Intervals.” Statistical Science 11, no. 3 (1996): 189–228.\n\n Efron, Bradley, and Robert J. Tibshirani. An Introduction to the Bootstrap. New York: Chapman & Hall, 1993." ]	[ null, "https://la.mathworks.com/help/examples/stats/win64/BootstrapConfidenceIntervalsForLinearCoefficientsExample_01.png", null, "https://la.mathworks.com/help/examples/stats/win64/ConfidenceIntervalForMultipleStatisticsExample_01.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5915537,"math_prob":0.9672096,"size":1267,"snap":"2021-31-2021-39","text_gpt3_token_len":355,"char_repetition_ratio":0.1266825,"word_repetition_ratio":0.0,"special_character_ratio":0.2573007,"punctuation_ratio":0.2706767,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9961317,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-23T00:01:12Z\",\"WARC-Record-ID\":\"<urn:uuid:01900d72-76f6-4775-bd44-22de5313b841>\",\"Content-Length\":\"139195\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:553f9916-8a2e-46f9-a6bf-2c5d635cff02>\",\"WARC-Concurrent-To\":\"<urn:uuid:03e46230-59b4-4586-9125-1879f5295406>\",\"WARC-IP-Address\":\"23.197.108.134\",\"WARC-Target-URI\":\"https://la.mathworks.com/help/stats/bootci.html\",\"WARC-Payload-Digest\":\"sha1:XZBOWSTRMVLGB3XASNO6RUKZIYUDAWGN\",\"WARC-Block-Digest\":\"sha1:2XBRGNY3NONJ7SAYB442456OA4ZCCE5B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057403.84_warc_CC-MAIN-20210922223752-20210923013752-00062.warc.gz\"}"}
https://physics.stackexchange.com/users/47373/apt45?tab=topactivity	[ "### Questions (38)\n\n 11 Identically vanishing trace of $T^{\\mu\\nu}$ and trace anomaly 10 Performing Wick Rotation to get Euclidean action of a scalar field $\\Psi$ 7 Anomalous Ward Identities and anomalous dimensions 5 How to write the Clebsch-Gordan decomposition in tensor notation 5 Perturbation of an operator - Meaning of matrix element [closed]\n\n### Reputation (1,017)\n\n +5 Why do we impose de Donder gauge? +10 Why do we impose de Donder gauge? +20 Identically vanishing trace of $T^{\\mu\\nu}$ and trace anomaly +5 Why do we study anomalies with the triangle diagram?\n\n 6 Trace of 4 Gell-Mann matrices 6 “Dimensional analysis” arguments in quantum field theory 4 Vertex factor for $\\frac{g}{4} (A_{\\nu}A^{\\nu})^2$ in QED 3 Conformal theory with zero central charge 2 Invariant terms of Chiral Lagrangian\n\n### Tags (81)\n\n 17 quantum-field-theory × 25 6 quantum-chromodynamics × 3 10 homework-and-exercises × 7 6 regularization × 3 7 renormalization × 7 6 trace 7 lie-algebra × 4 6 dimensional-analysis 6 feynman-diagrams × 6 4 lagrangian-formalism × 6" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.65768397,"math_prob":0.9358622,"size":479,"snap":"2019-43-2019-47","text_gpt3_token_len":140,"char_repetition_ratio":0.19157895,"word_repetition_ratio":0.0,"special_character_ratio":0.33194155,"punctuation_ratio":0.013157895,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97401625,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-12T02:48:36Z\",\"WARC-Record-ID\":\"<urn:uuid:93c22f69-9012-47d0-a733-e93b6daa948f>\",\"Content-Length\":\"128580\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a23c88f3-8213-42d1-9f65-87a3baedba2a>\",\"WARC-Concurrent-To\":\"<urn:uuid:58fb8611-7438-48dd-8775-ce1be598d270>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/users/47373/apt45?tab=topactivity\",\"WARC-Payload-Digest\":\"sha1:C4I2N5V7M2ALPDZCUYDI2RPMQDFEQOSM\",\"WARC-Block-Digest\":\"sha1:W7IO3AIN7TKHMMZQDPLWBRHC4AC3GO4E\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496664567.4_warc_CC-MAIN-20191112024224-20191112052224-00497.warc.gz\"}"}
https://stackoverflow.com/questions/50457921/cost-function-mean-squared-error-formulae	[ "# Cost Function & Mean Squared error formulae\n\nI am new to machine learning and statistics and am confused with the cost function & Mean Squared Error (MSE) formulas. In Machine learning class at stanford - coursera, Cost function formula is mentioned as shown below:\n\nCost Function formula", null, "And at some other sources, cost function is termed as mean squared error (MSE) and it is given with the formula as shown in picture below.\n\nMean Squared Error formula", null, "What will be the Cost Function formula & is cost function and MSE different or same. Please let me know why are the formulas are different." ]	[ null, "https://i.stack.imgur.com/5cydJ.png", null, "https://i.stack.imgur.com/yCHvT.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9051005,"math_prob":0.9646506,"size":570,"snap":"2021-43-2021-49","text_gpt3_token_len":122,"char_repetition_ratio":0.16784452,"word_repetition_ratio":0.0,"special_character_ratio":0.19824561,"punctuation_ratio":0.067307696,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9989672,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-17T11:17:01Z\",\"WARC-Record-ID\":\"<urn:uuid:630fcda4-ea0e-449c-8f86-c2982c37ec54>\",\"Content-Length\":\"168882\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6e248e3e-6d77-4c7c-91b0-849bbda57d96>\",\"WARC-Concurrent-To\":\"<urn:uuid:dc7b6f8b-d53f-45a8-aad1-78671a3d0de6>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/50457921/cost-function-mean-squared-error-formulae\",\"WARC-Payload-Digest\":\"sha1:Y3M7K7HB2XTXSAB6BSSHM7LYH3NXH4ST\",\"WARC-Block-Digest\":\"sha1:DQURV2VL7DKKQFS2B7COQOZPU3G4F4ZK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585171.16_warc_CC-MAIN-20211017082600-20211017112600-00463.warc.gz\"}"}
https://math.stackexchange.com/questions/2995166/are-second-countable-metric-spaces-sigma-compact/2995189	[ "# Are second-countable metric spaces $\\sigma$-compact?\n\nI was curious about the relations between second-countable, separable, Lindelöf and $$\\sigma$$-compact topologies in the context of metric spaces.\n\nI am aware of the following implications in general topological spaces:\n\n• second-countable $$\\Rightarrow$$ separable $$\\not \\Rightarrow$$ Lindelöf, $$\\;$$ [thanks bof]\n• $$\\sigma$$-compact $$\\Rightarrow$$ Lindelöf\n• second-countable + locally compact $$\\Rightarrow$$ $$\\sigma$$-compact\n\nas well as the reversed implications in the case of metric spaces:\n\n• Lindelöf $$\\Leftrightarrow$$ separable $$\\Leftrightarrow$$ second-countable\n\nSince all the proofs I've seen so far require the LC condition I assume it is not true in general that second-countable topological spaces are $$\\sigma$$-compact (although seeing an actual counterexample would be nice).\nSo what about metrizable topological spaces?\n\nIdeas so far:\nIf we can proof that every subset of a $$\\sigma$$-compact space is again $$\\sigma$$-compact, then this would follow from the fact, that every separable metric space is homeomorphic to a subset of the Hilbert cube (which is compact). $$\\;$$[debunked by bof]\n\n• The set of irrational numbers is a separable metric space which is not $\\sigma$-compact. (Every $\\sigma$-compact subset of the irrational numbers is meager.) Hilbert space is another example. – bof Nov 12 '18 at 11:10\n• By the way, separable does not imply Lindelöf in general topological spaces. For example, the Sorgenfrey plane is separable but not Lindelöf. – bof Nov 12 '18 at 11:12\n• Every closed subset of a $\\sigma$-compact space is again $\\sigma$-compact. Not all subspace, as witnessed by the rational and the irrationals assubspaces of the reals. – Henno Brandsma Nov 12 '18 at 11:55\n• @bof I also wanted to disprove local compactness. – Henno Brandsma Nov 12 '18 at 12:25\n• Any infinite-dimensional separable Banach space is a counterexample. This follows from the fact that the unit ball of a Banach space $E$ is compact iff $E$ is finite dimensional, so a compact subset of an infinite dimensional Banach space has empty interior. Then we just apply the Baire category theorem. – Robert Furber Nov 11 '19 at 11:02\n\nA counterexample is the \"Baire space\" $$\\mathcal{N} = \\mathbb{N}^{\\mathbb{N}}$$. This is one of the main examples of a Polish space: a separable, completely metrizable space.\nOne fact about this space is that all compact subsets have empty interior, that is, all compact subsets are nowhere dense. By the Baire Category Theorem, $$\\mathcal{N}$$ is not the countable union of nowhere dense subsets, and so together with the above fact it cannot be $$\\sigma$$-compact." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7997143,"math_prob":0.9974065,"size":1085,"snap":"2020-34-2020-40","text_gpt3_token_len":262,"char_repetition_ratio":0.15356152,"word_repetition_ratio":0.0,"special_character_ratio":0.2156682,"punctuation_ratio":0.08139535,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99965703,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-29T15:04:54Z\",\"WARC-Record-ID\":\"<urn:uuid:f4abd082-232b-43cf-978a-d2048b53e6d4>\",\"Content-Length\":\"153071\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a4a70291-fa4d-4873-91cb-6b1ba1d77fd2>\",\"WARC-Concurrent-To\":\"<urn:uuid:45eb0736-ec5e-4e66-86a0-f67bdc9d7b62>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/2995166/are-second-countable-metric-spaces-sigma-compact/2995189\",\"WARC-Payload-Digest\":\"sha1:6NPGCU6EJG77KWKGDY7CUSR4JZHATV7M\",\"WARC-Block-Digest\":\"sha1:QGMNOH6HP5YA33O64RFYLK73MOMIIUXT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401643509.96_warc_CC-MAIN-20200929123413-20200929153413-00638.warc.gz\"}"}
https://www.cssexaminations.com/a-number-is-doubled-and-9-is-added/	[ "### Global Statistics\n\nAll countries\n231,385,504\nConfirmed\nUpdated on September 24, 2021 2:03 am\nAll countries\n206,331,240\nRecovered\nUpdated on September 24, 2021 2:03 am\nAll countries\n4,742,542\nDeaths\nUpdated on September 24, 2021 2:03 am\n\n# A Number is Doubled and 9 is Added. If the Resultant is Trebled, It Becomes 75. What is that Number?\n\n### Computer MCQs Series for PPSC, FPSC – Most Repeated MCQs \| Set 4\n\nWhat are you looking for? Let’s dig in quickly\n\n## Explanation\n\n• A number is doubled and 9 is added.\n• Resultant is trebled, it becomes 75.\n\nThe number can be figure out as:\n\nLet suppose the number is y.\n\nNow, number “y” is double and 9 is added\n\n2y + 9\n\nResultant (2y + 9) is trebled\n\n3(2y + 9) = 75 ________ (i)\n\nValue of y can be find out easily from equation (i) by just simplifying it (y = 8).\n\nNumber = ?\n\n## Solution\n\nLet suppose\n\nNumber = y\n\nAccording to the given condition\n\n3(2y + 9) = 75\n\n2y + 9 = 25\n\n2y = 25 – 9\n\n2y = 16\n\ny = 8\n\n## Conclusion\n\nA number is doubled and 9 is added. If the resultant is trebled, it becomes 75. The required number is 8.\n\n### Computer MCQs Series for PPSC, FPSC – Most Repeated MCQs \| Set 6\n\nerror: Content is protected !!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.901303,"math_prob":0.9776326,"size":668,"snap":"2021-31-2021-39","text_gpt3_token_len":225,"char_repetition_ratio":0.13403614,"word_repetition_ratio":0.08450704,"special_character_ratio":0.35179642,"punctuation_ratio":0.09090909,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9778466,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-24T02:56:26Z\",\"WARC-Record-ID\":\"<urn:uuid:e55f306a-990e-40ec-97fd-5ff793ed2092>\",\"Content-Length\":\"484060\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4f57e262-6dfb-4f6e-b6d7-31c493911e37>\",\"WARC-Concurrent-To\":\"<urn:uuid:67fc87e1-9f12-49ed-9ed1-86d375eecc46>\",\"WARC-IP-Address\":\"18.232.245.187\",\"WARC-Target-URI\":\"https://www.cssexaminations.com/a-number-is-doubled-and-9-is-added/\",\"WARC-Payload-Digest\":\"sha1:5F227GO5ICRW6LLCYF4RGWWZMW7NFN4C\",\"WARC-Block-Digest\":\"sha1:SDNQNBT6RQJYEURD6QTNUKDSERLU4AWU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057496.18_warc_CC-MAIN-20210924020020-20210924050020-00613.warc.gz\"}"}
https://emathematics.net/sistecuaciones.php?a=3	[ "", null, "Done: 0 Corrects: 0 User (login-registrate)\nFractions\nWhole Numbers\nEquations\nSystems of equations\nMonomials\nPolynomials\nSpecial products\nExponents and Powers\nIntegers\nEquations of lines\nSequences and series\nFunctions\nDeterminants\nMatrices\nPercentages\n\nFour methods of solving systems of two linear equations are ilustrated here:\n1. Solution by substitution .Find the value of one unknown in either of the given equations and substitute this value in the other equation.\n2. Solution by igualation. Find the value of the same unknown in botn of the given equations and igualate the resululting equations.\n3. Solution by addition or subtraction. If necessary, multiply the given equations by such numbers as will make the coefficients of one unknown in the resulting equation numerically equal. If the signs of the equal coefficients are unlike, add the resulting equations; if like, subtract them.\n4. Graphical solution. Graph both equations, obtaining two straight lines. The simultaneous solution is given by the coordinates (x,y) of the point of intersection of these lines\n\nSolve the following system:\n\n x= y=" ]	[ null, "https://emathematics.net/cabecera.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8779661,"math_prob":0.9992434,"size":850,"snap":"2023-40-2023-50","text_gpt3_token_len":170,"char_repetition_ratio":0.17612293,"word_repetition_ratio":0.015267176,"special_character_ratio":0.18941176,"punctuation_ratio":0.13375796,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999442,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T07:01:52Z\",\"WARC-Record-ID\":\"<urn:uuid:b71b81f2-3fea-481c-b704-cefd55a83a50>\",\"Content-Length\":\"17085\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d5101859-a0c7-4e8e-b945-88925ff2e4c6>\",\"WARC-Concurrent-To\":\"<urn:uuid:56f5448b-4913-4283-a1f8-063c4e1ebdb2>\",\"WARC-IP-Address\":\"172.67.155.6\",\"WARC-Target-URI\":\"https://emathematics.net/sistecuaciones.php?a=3\",\"WARC-Payload-Digest\":\"sha1:Z4ULBIOCZDIV52AMRQ4SIFJ2ZKG36PWZ\",\"WARC-Block-Digest\":\"sha1:CY3ZPBBD6QEOUQ6CFPNT7DQU5CNVZZXZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510368.33_warc_CC-MAIN-20230928063033-20230928093033-00157.warc.gz\"}"}
https://www.geeksforgeeks.org/find-n-th-term-series-9-33-73129/	[ "Related Articles\n\n# Find n-th term in the series 9, 33, 73,129 …\n\n• Difficulty Level : Easy\n• Last Updated : 16 Apr, 2021\n\nGiven a series 9, 33, 73, 129… Find the n-th term of the series.\nExamples:\n\n```Input : n = 4\nOutput : 129\n\nInput : n = 5\nOutput : 201```\n\nAttention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.\n\nThe given series has a pattern which is visible after subtracting it from itself after one shift\n\n```S = 9 + 33 + 73 + 129 + … tn-1 + tn\nS = 9 + 33 + 73 + … tn-2 + tn-1 + tn\n———————————————\n0 = 9 + (24 + 40 + 56 + ….) - tn\n\nSince 24 + 40 + 56.. series in A.P with\ncommon difference of 16, we get\n\ntn = 9 + [((n-1)/2)(224 + (n-1-1)d)]\n\nOn solving this we get\n\ntn = 8n2 + 1```\n\nBelow is the implementation of the above approach:\n\n## C++\n\n `// Program to find n-th element in the``// series 9, 33, 73, 128..``#include ``using` `namespace` `std;` `// Returns n-th element of the series``int` `series(``int` `n)``{`` ``return` `(8 * n * n) + 1;``}` `// driver program to test the above function``int` `main()``{`` ``int` `n = 5;`` ``cout << series(n);`` ``return` `0;``}`\n\n## Java\n\n `// Program to find n-th element in the``// series 9, 33, 73, 128..``import` `java.io.;` `class` `GFG{`` ` ` ``// Returns n-th element of the series`` ``static` `int` `series(``int` `n)`` ``{`` ``return` `(``8` ` n * n) + ``1``;`` ``}`` ` ` ``// driver program to test the above`` ``// function`` ``public` `static` `void` `main(String args[])`` ``{`` ``int` `n = ``5``;`` ``System.out.println(series(n));`` ``}``}` `/This code is contributed by Nikita Tiwari./`\n\n## Python3\n\n `# Python Program to find n-th element``# in the series 9, 33, 73, 128...` `# Returns n-th element of the series``def` `series(n):`` ``print` `(( ``8` `` `n ````` `2``) ``+` `1``)`` ` `# Driver Code``series(``5``)` `# This code is contributed by Abhishek Agrawal.`\n\n## C#\n\n `// C# program to find n-th element in the``// series 9, 33, 73, 128..``using` `System;` `class` `GFG {` ` ``// Returns n-th element of the series`` ``static` `int` `series(``int` `n)`` ``{`` ``return` `(8 n * n) + 1;`` ``}` ` ``// driver function`` ``public` `static` `void` `Main()`` ``{`` ``int` `n = 5;`` ``Console.WriteLine(series(n));`` ``}``}` `/This code is contributed by vt_m./`\n\n## PHP\n\n ``\n\n## Javascript\n\n ``\n\nOutput:\n\n`201`\n\nTime complexity: O(1)\nThis article is contributed by Striver. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.68465894,"math_prob":0.83444196,"size":3122,"snap":"2021-43-2021-49","text_gpt3_token_len":959,"char_repetition_ratio":0.15169981,"word_repetition_ratio":0.2413793,"special_character_ratio":0.35874438,"punctuation_ratio":0.15264797,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99164826,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-16T16:20:00Z\",\"WARC-Record-ID\":\"<urn:uuid:d7a09db3-3ec5-4496-826e-4807bfbf1479>\",\"Content-Length\":\"127083\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9519db1f-7e65-4f2f-bc40-3e08c44c9699>\",\"WARC-Concurrent-To\":\"<urn:uuid:00f88d42-da61-41be-a14d-2af1bcf4fd23>\",\"WARC-IP-Address\":\"23.207.202.207\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/find-n-th-term-series-9-33-73129/\",\"WARC-Payload-Digest\":\"sha1:CWHR2ZQMLSJWDTWLNELBEDK3DWCDVJTK\",\"WARC-Block-Digest\":\"sha1:GT3B3JQJZCJBN36U7YWYXODL4O5BSLIC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323584886.5_warc_CC-MAIN-20211016135542-20211016165542-00022.warc.gz\"}"}
https://www.geeksforgeeks.org/utility-methods-of-wrapper-classes-in-java/?ref=rp	[ "# Utility Methods of Wrapper Classes in Java\n\nPrerequisite: Wrapper Classes\n\nThe objective of the Wrapper class is to define several utility methods which are required for the primitive types. There are 4 utility methods for primitive type which is defined by the Wrapper class:\n\n### 1. valueOf() method:\n\nWe can use the valueOf() method to create a Wrapper object for a given primitive or String. There are 3 types of valueOf() methods:\n\nA. Wrapper valueOf(String s): Every wrapper class except Character class contains a static valueOf() method to create Wrapper class object for a given String.\n\nSyntax:\n\n`public static Wrapper valueOf(String s);`\n\n## Java\n\n `// Java program to illustrate valueof() Method ` ` ` `class` `GFG { ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``Integer I = Integer.valueOf(``\"10\"``); ` ` ``System.out.println(I); ` ` ` ` ``Double D = Double.valueOf(``\"10.0\"``); ` ` ``System.out.println(D); ` ` ` ` ``Boolean B = Boolean.valueOf(``\"true\"``); ` ` ``System.out.println(B); ` ` ` ` ``// Here we will get RuntimeException ` ` ``Integer I1 = Integer.valueOf(``\"ten\"``); ` ` ``} ` `}`\n\nOutput:\n\n```10\n10.0\ntrue\nException in thread \"main\" java.lang.NumberFormatException: For input string: \"ten\"```\n\nB. Wrapper valueOf(String s, int radix): Every Integral Wrapper class Byte, Short, Integer, Long) contains the following valueOf() method to create a Wrapper object for the given String with specified radix. The range of the radix is 2 to 36.\n\nSyntax:\n\n`public static Wrapper valueOf(String s, int radix)`\n\n## Java\n\n `// Java program to illustrate valueof() Method ` ` ` `class` `GFG { ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``Integer I = Integer.valueOf(``\"1111\"``, ``2``); ` ` ``System.out.println(I); ` ` ` ` ``Integer I1 = Integer.valueOf(``\"1111\"``, ``4``); ` ` ``System.out.println(I1); ` ` ``} ` `}`\n\nOutput\n\n```15\n85```\n\n3. Wrapper valueOf(primitive p): Every Wrapper class including the Character class contains the following method to create a Wrapper object for the given primitive type.\n\nSyntax:\n\n`public static Wrapper valueOf(primitive p);`\n\n## Java\n\n `// Java program to illustrate valueof() Method ` ` ` `class` `GFG { ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``Integer I = Integer.valueOf(``10``); ` ` ``Double D = Double.valueOf(``10.5``); ` ` ``Character C = Character.valueOf(``'a'``); ` ` ` ` ``System.out.println(I); ` ` ``System.out.println(D); ` ` ``System.out.println(C); ` ` ``} ` `}`\n\nOutput\n\n```10\n10.5\na```\n\n### 2. xxxValue() Method\n\nWe can use xxxValue() methods to get the primitive for the given Wrapper Object. Every number type Wrapper class( Byte, Short, Integer, Long, Float, Double) contains the following 6 methods to get primitive for the given Wrapper object:\n\n1. public byte byteValue()\n2. public short shortValue()\n3. public int intValue()\n4. public long longValue()\n5. public float floatValue()\n6. public float doubleValue()\n\n### 3. parseXxx() Method\n\nWe can use parseXxx() methods to convert String to primitive. There are two types of parseXxx() methods:\n\nA. primitive parseXxx(String s): Every Wrapper class except the character class contains the following parseXxx() method to find primitive for the given String object.\n\nSyntax:\n\n`public static primitive parseXxx(String s);`\n\n## Java\n\n `// Java program to illustrate parseXxx() Method ` ` ` `class` `GFG { ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `i = Integer.parseInt(``\"10\"``); ` ` ``double` `d = Double.parseDouble(``\"10.5\"``); ` ` ``boolean` `b = Boolean.parseBoolean(``\"true\"``); ` ` ` ` ``System.out.println(i); ` ` ``System.out.println(d); ` ` ``System.out.println(b); ` ` ``} ` `}`\n\nOutput\n\n```10\n10.5\ntrue```\n\nB. parseXxx(String s, int radix): Every Integral type Wrapper class (Byte, Short, Integer, Long) contains the following parseXxx() method to convert specified radix String to primitive.\n\nSyntax:\n\n`public static primitive parseXxx(String s, int radix);`\n\n## Java\n\n `// Java program to illustrate parseXxx() Method ` ` ` `class` `GFG { ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `i = Integer.parseInt(``\"1000\"``, ``2``); ` ` ``long` `l = Long.parseLong(``\"1111\"``, ``4``); ` ` ` ` ``System.out.println(i); ` ` ``System.out.println(l); ` ` ``} ` `}`\n\nOutput\n\n```8\n85```\n\n### 4. toString() Method\n\nWe can use the toString() method to convert the Wrapper object or primitive to String. There are a few forms of the toString() method:\n\nA. public String toString(): Every wrapper class contains the following toString() method to convert Wrapper Object to String type.\n\nSyntax:\n\n`public String toString();`\n\n## Java\n\n `// Java program to illustrate toString() Method ` ` ` `class` `GFG { ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``Integer I = ``new` `Integer(``10``); ` ` ``String s = I.toString(); ` ` ``System.out.println(s); ` ` ``} ` `}`\n\nOutput:\n\n`10`\n\nB. toString(primitive p): Every Wrapper class including the Character class contains the following static toString() method to convert primitive to String.\n\nSyntax:\n\n`public static String toString(primitive p);`\n\n## Java\n\n `// Java program to illustrate toString() ` ` ` `class` `GFG { ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``String s = Integer.toString(``10``); ` ` ``System.out.println(s); ` ` ` ` ``String s1 = Character.toString(``'a'``); ` ` ``System.out.println(s1); ` ` ``} ` `}`\n\nOutput\n\n```10\na```\n\nC. toString(primitive p, int radix): Integer and Long classes contain the following toString() method to convert primitive to specified radix String.\n\nSyntax:\n\n`public static String toString(primitive p, int radix);`\n\n## Java\n\n `// Java program to illustrate toString() Method ` ` ` `class` `GFG { ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``String s = Integer.toString(``15``, ``2``); ` ` ``System.out.println(s); ` ` ` ` ``String s1 = Long.toString(``11110000``, ``4``); ` ` ``System.out.println(s1); ` ` ``} ` `}`\n\nOutput\n\n```1111\n222120121300```\n\nWhether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape, GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out - check it out now!\n\nPrevious\nNext" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5853295,"math_prob":0.69275934,"size":5129,"snap":"2023-40-2023-50","text_gpt3_token_len":1226,"char_repetition_ratio":0.184,"word_repetition_ratio":0.20104438,"special_character_ratio":0.26106453,"punctuation_ratio":0.19650206,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9851277,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-07T10:39:52Z\",\"WARC-Record-ID\":\"<urn:uuid:d2d7d5d0-83b7-47c0-b882-f73c2421bdfb>\",\"Content-Length\":\"373048\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:02082e04-9d6b-448a-9641-f47053b7f707>\",\"WARC-Concurrent-To\":\"<urn:uuid:df91bbf2-b7e0-4cd9-9b8c-9e4529d3e9a7>\",\"WARC-IP-Address\":\"108.138.64.13\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/utility-methods-of-wrapper-classes-in-java/?ref=rp\",\"WARC-Payload-Digest\":\"sha1:O7V2OOXIBIS5COFMTHANF2Q3ZG74BYYW\",\"WARC-Block-Digest\":\"sha1:CRVNVGZWKXSC7XSAS7LSIL6QD3356YO4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100651.34_warc_CC-MAIN-20231207090036-20231207120036-00846.warc.gz\"}"}
https://www.traditionaloven.com/tutorials/angle/convert-angle-unit-second-to-cycle-angle-unit.html	[ " Convert '' to 360° \| second to cycles\n\nangle units conversion\n\nAmount: 1 second ('') of angle Equals: 0.00000077 cycles (360°) in angle\n\nConverting second to cycles value in the angle units scale.\n\nTOGGLE : from cycles into seconds in the other way around.\n\nangle from second to cycle conversion results\n\nEnter a new second number to convert\n\n* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)\n* Precision is how many digits after decimal point (1 - 9)\n\nEnter Amount :\nDecimal Precision :\n\nCONVERT : between other angle measuring units - complete list.\n\nHow many cycles are in 1 second? The answer is: 1 '' equals 0.00000077 360°\n\n0.00000077 360° is converted to 1 of what?\n\nThe cycles unit number 0.00000077 360° converts to 1 '', one second. It is the EQUAL angle value of 1 second but in the cycles angle unit alternative.\n\n ''/360° angle conversion result From Symbol Equals Result Symbol 1 '' = 0.00000077 360°\n\nConversion chart - seconds to cycles\n\n1 second to cycles = 0.00000077 360°\n\n2 seconds to cycles = 0.0000015 360°\n\n3 seconds to cycles = 0.0000023 360°\n\n4 seconds to cycles = 0.0000031 360°\n\n5 seconds to cycles = 0.0000039 360°\n\n6 seconds to cycles = 0.0000046 360°\n\n7 seconds to cycles = 0.0000054 360°\n\n8 seconds to cycles = 0.0000062 360°\n\n9 seconds to cycles = 0.0000069 360°\n\n10 seconds to cycles = 0.0000077 360°\n\n11 seconds to cycles = 0.0000085 360°\n\n12 seconds to cycles = 0.0000093 360°\n\n13 seconds to cycles = 0.000010 360°\n\n14 seconds to cycles = 0.000011 360°\n\n15 seconds to cycles = 0.000012 360°\n\nConvert angle of second ('') and cycles (360°) units in reverse from cycles into seconds.\n\nAngles\n\nThis calculator is based on conversion of two angle units. An angle consists of two rays (as in sides of an angle sharing a common vertex or else called the endpoint.) Some belong to rotation measurements - spherical angles measured by arcs' lengths, pointing from the center, plus the radius. For a whole set of multiple units of angle on one page, try that Multiunit converter tool which has built in all angle unit-variations. Page with individual angle units.\n\nConverter type: angle units\n\nFirst unit: second ('') is used for measuring angle.\nSecond: cycle (360°) is unit of angle.\n\nQUESTION:\n15 '' = ? 360°\n\n15 '' = 0.000012 360°\n\nAbbreviation, or prefix, for second is:\n''\nAbbreviation for cycle is:\n360°\n\nOther applications for this angle calculator ...\n\nWith the above mentioned two-units calculating service it provides, this angle converter proved to be useful also as a teaching tool:\n1. in practicing seconds and cycles ( '' vs. 360° ) measures exchange.\n2. for conversion factors between unit pairs.\n3. work with angle's values and properties.\n\nTo link to this angle second to cycles online converter simply cut and paste the following.\nThe link to this tool will appear as: angle from second ('') to cycles (360°) conversion.\n\nI've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8116263,"math_prob":0.987518,"size":2439,"snap":"2019-43-2019-47","text_gpt3_token_len":681,"char_repetition_ratio":0.27268994,"word_repetition_ratio":0.0,"special_character_ratio":0.36121362,"punctuation_ratio":0.13963039,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9931215,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-19T02:01:58Z\",\"WARC-Record-ID\":\"<urn:uuid:c0f7e718-3d9c-4877-a926-29b5c24aab11>\",\"Content-Length\":\"45975\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3b149a39-1ad2-496c-a1b5-9f37859545d5>\",\"WARC-Concurrent-To\":\"<urn:uuid:378ca251-3ee6-418e-9d69-6e55db8841e5>\",\"WARC-IP-Address\":\"162.241.171.12\",\"WARC-Target-URI\":\"https://www.traditionaloven.com/tutorials/angle/convert-angle-unit-second-to-cycle-angle-unit.html\",\"WARC-Payload-Digest\":\"sha1:OOLPFSRGKKF55KPY2OPJPZM234QLPKVD\",\"WARC-Block-Digest\":\"sha1:LGQPUJC2DKJCLGAKYZRWJ4VWKX4GEBMT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986688674.52_warc_CC-MAIN-20191019013909-20191019041409-00248.warc.gz\"}"}
https://quant.stackexchange.com/questions/18408/does-heteroskedasticity-of-returns-depend-on-the-time-frame?noredirect=1	[ "# Does heteroskedasticity of returns depend on the time frame?\n\nSimilarly to my last question, for which I obtained very interesting and useful answers, I would like to know if there has been any study regarding heteroskedasticity and time-frames of the returns.\n\nAs an example could it be that the lower the time frame (take the 5 minute returns) the less heteroskedastic are returns?\n\n• Hi Monolite! Can you specify what is the model you're referring to (GARCH, ARCH, ...)? – Quantopik Jun 17 '15 at 17:57\n\n## 2 Answers\n\nIn practice, for heavily traded assets (above 60% quantile of average daily dollar volume), individual asset return is pretty scalable across different time frame by a factor of $\\sqrt{T}$.\n\nHowever, for covariance among different assets, moving between different time frame is not linearly scalable (although it should be in math). This is known as \"Epps Effect\".\n\nThis paper states that heteroskedasticity is a stylized fact in daily as well as intra-day returns: https://statistik.econ.kit.edu/download/doc_secure1/HandbookITandFinan.pdf" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9517579,"math_prob":0.77124417,"size":320,"snap":"2021-04-2021-17","text_gpt3_token_len":70,"char_repetition_ratio":0.12658228,"word_repetition_ratio":0.0,"special_character_ratio":0.190625,"punctuation_ratio":0.06779661,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9665758,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-15T07:12:48Z\",\"WARC-Record-ID\":\"<urn:uuid:0ac47efb-a050-4c11-b1b9-35e41e212f78>\",\"Content-Length\":\"169091\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1bc7c93b-3756-495b-9e72-fe70d4f34ff4>\",\"WARC-Concurrent-To\":\"<urn:uuid:f825dffc-e95a-45dc-a780-1d888ef7de6a>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://quant.stackexchange.com/questions/18408/does-heteroskedasticity-of-returns-depend-on-the-time-frame?noredirect=1\",\"WARC-Payload-Digest\":\"sha1:VY4HYBQZS3LWNIXJD7XXAZXEIQMWBUAN\",\"WARC-Block-Digest\":\"sha1:GBDSSLPPD4K3KUGYUR5D2PSCH4RT42PT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038084601.32_warc_CC-MAIN-20210415065312-20210415095312-00039.warc.gz\"}"}
https://answers.yahoo.com/question/index?qid=20070620091706AA2SS8W	[ "# I need more help with this 1/(g-(k/m)v^2) dv?\n\nI need to make a paper explaining how to solve this integral I need to factor fractions and end with a hyperbolic trig function. I will apreciate any help thanks.\n\nRelevance\n\nEssentially you want to write the integral as:\n\n(m/k)dv / [ mg/k - v^2]\n\nAnd replace mg/k with a^2.\n\nThe easiest method is to get the hyperbolic trig function directly from the integration.\n\nUse the substitution\n\nx = atanhθ\n\ndx = asech^2 θ\n\na^2 - x^2 = a^2 (1 - tanh^2 θ) = a^2 * sech^2 θ\n\nSo the integral of dx / (a^2 - x^2) becomes\n\nintegral of (1/a) dθ\n\nwhich gives you (1/a) θ = (1/a) * tanh^-1 (x/a)\n\nNow you just need to replace a with sqrt(mg/k) and x with v.\n\n------------\n\nIf you wish to integrate using partial fractions, then convert to hyperbolic trig, check out the links below where I answered very similar questions in the past." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.76937884,"math_prob":0.96077484,"size":1023,"snap":"2020-34-2020-40","text_gpt3_token_len":312,"char_repetition_ratio":0.11383709,"word_repetition_ratio":0.0,"special_character_ratio":0.30400783,"punctuation_ratio":0.11061947,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9983061,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-15T12:42:55Z\",\"WARC-Record-ID\":\"<urn:uuid:e8b4f290-8f16-4a78-a77a-03e77214900b>\",\"Content-Length\":\"118237\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4394089e-5c33-4aca-aa15-75734c420b7a>\",\"WARC-Concurrent-To\":\"<urn:uuid:8fcb1b77-d836-4c61-9806-97a2b3567d63>\",\"WARC-IP-Address\":\"76.13.32.153\",\"WARC-Target-URI\":\"https://answers.yahoo.com/question/index?qid=20070620091706AA2SS8W\",\"WARC-Payload-Digest\":\"sha1:NQY3FAAPF4VBVFJ2KS3JX46YQ2SZKUC4\",\"WARC-Block-Digest\":\"sha1:77MZMRL2NHMJB57JPT4SBJ75KGUFSOIW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439740838.3_warc_CC-MAIN-20200815094903-20200815124903-00433.warc.gz\"}"}
https://ask.sagemath.org/question/30456/need-clarity-on-plotting-the-y-cordinate-from-the-x-co-ordinate-in-elliptic-curve-cryptography/	[ "# Need clarity on plotting the y cordinate from the x co-ordinate in Elliptic curve cryptography\n\nI'm just new to elliptic curve cryptography. I have been working on RSA for quite some time. Moreover I'm not from a mathematical background. The whole concept looks very complex. So tell me my understanding is correct or not. I was looking at sample implementation at http://www.enggjournals.com/ijcse/doc... In the pdf given there, the curve equation is y^2=x^3+ax+b,the domain parameters are p(751),a(-1),b(188),n(727). I want to encode a letter 'b' and it is first encoded as 11. Now x=mk+1 ie 1120+1=221 cannot solve it for a y such that y^2= x^3 + ax+ b mod p.\nSo go for x=mk+2 , x=222 , no y exists. x=mk+3, x=223, no y exists. x=mk+4 so x=224 can solve it for y and y=248\n\n1)Can somebody explain how exactly x=224 solves the equation for y?\n\n2)On what basis mk+1 is taken? Is there any standard formula for choosing the y co-ordinate?\n\nedit retag close merge delete\n\nHow exactly is this related to sagemath?\n\nIf it is not related to sagemath perhaps a better site for your question would be here.\n\nSort by » oldest newest most voted\n\nThe source is hard to read, both mathematically and with a xpdf viewer, so let us restrict us to understand only the paragraph related to the questions (1) and (2) above. We are now on the page 1906 and the cook book tells us to initialize the elliptic curve $$E_p(a,b)\\ :\\ y^2=x^3 +ax+b\\ ,$$ defined over the field $\\mathbb F_p$ with $p$ elements. This field is GF(p) in sage. So in order to initialize this object in sage, for the special choice of the example at loc. cit., $p=751$, $a=-1$, $b=188$, we type (in the sage interpreter console):\n\nsage: E = EllipticCurve( GF(751), [ -1, 188 ] )\n\nsage: E\nElliptic Curve defined by y^2 = x^3 + 750x + 188 over Finite Field of size 751\n\nsage: E.order()\n727\n\n\nOK, i was too curious to see the \"order of the curve\", i.e. the number of points lying on it. So the order is this $n=727$, the source uses $n$, sometimes $N$ for this number.\n\nIn order to produce some confusion, the source wants to send the letter b as a message. (They could have taken some J or so... since the curve has parameters a, and b.) In the next second we want but to send the B instead. No problem, we are sending the $11$, convention, the letters A, B, C, ... are by convention converted to 10, 11, 12, ... . \"Same information\".\n\nThere is also some parameter, $k=20$. Setting this parameter as Step 5 makes it better. So the confusion on my side takes shape. (Why not set it once for all times at the beginning as parameter? Does this $k=20$ depend on something chosen in the previous steps?)\n\nThe cook book wants now to associate the $x$-value to the $B$ given by the \"formula\": $$x = mk+1 \\overset ?= 11\\cdot 20+1\\ .$$ With this occasion, we record the fact that the $11$ is in fact an $m$. Or at least ask ourselves, and accept it, since we cannot change the pdf. Starting with the natural number $221$ from above, we seek for the first $x$ in the sequence $221, 222, 223, 224, 225,\\dots$ so that $(x,?)$ is a point on the given curve. Let us do this in sage:\n\nsage: for x in [ 221..751 ]:\n....: if F(x^3 - x + 188).is_square():\n....: print x\n....: break\n....:\n224\n\n\nWe take then this first occurence and associate the corresponding $?$ value. There are two of them, of course:\n\nsage: sqrt( F(224)^3 - F(224) + F(188) )\n248\n\n\nBoth sage and the source consider that $248$ is the better square root. (The value $-248=751-248=503$ is the other one.)\n\nThe source than claims:\n\n6. Now the point (224,248) is point is encrypted and decrypted as a message.\n\nAnd this is a good point to stop. I think, we can now decrypt the way things can / could be done in the setting of the article.\n\nN.B. Please excuse the many personal comments. But it was really hard & frustrating to get the message from the article, after this, doing the job in sage took seconds. Best, the author would have written it by providing sage code...\n\nmore" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90802664,"math_prob":0.9946773,"size":2951,"snap":"2023-14-2023-23","text_gpt3_token_len":827,"char_repetition_ratio":0.09840516,"word_repetition_ratio":0.0,"special_character_ratio":0.32565233,"punctuation_ratio":0.1890332,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99920577,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-31T08:32:47Z\",\"WARC-Record-ID\":\"<urn:uuid:1fb43131-62c6-4c1e-9ef7-eb7c388c1aad>\",\"Content-Length\":\"56581\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fc8bb1c6-e746-4b66-acc4-e62ac428be42>\",\"WARC-Concurrent-To\":\"<urn:uuid:7d7dff92-9490-40c6-86ac-ca1859cec0bc>\",\"WARC-IP-Address\":\"194.254.163.53\",\"WARC-Target-URI\":\"https://ask.sagemath.org/question/30456/need-clarity-on-plotting-the-y-cordinate-from-the-x-co-ordinate-in-elliptic-curve-cryptography/\",\"WARC-Payload-Digest\":\"sha1:FX76QYM6A5MQCY47IZOVNNJASGJWY4V2\",\"WARC-Block-Digest\":\"sha1:EMMIWCOBFI6JWJ2JW4UMEGZCB2MEUP4L\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949598.87_warc_CC-MAIN-20230331082653-20230331112653-00604.warc.gz\"}"}
https://plantpot.works/8581	[ "09/30/2021\n\nContents\n\nIn JavaScript, you can pad a number with leading zeros to make it a certain number of digits long. Padding a number with leading zeros is useful when you need to display numbers with a consistent number of digits, such as when displaying dates or times.\n\nThe padStart() method is a built-in method in JavaScript that pads a string with a specific character or characters until it reaches a specified length. Here’s the syntax for using the padStart() method to pad a number with leading zeros:\n\n``number.toString().padStart(length, \"0\")``\n\nThe number is the number that you want to pad with leading zeros. The length is the total number of digits that the padded number should have. The “0” is the character that you want to use for padding the number.\n\n``````let number = 42;\n``````\n\nIn this example, we use the padStart() method to pad the number 42 with leading zeros so that it has a total of 5 digits. The resulting padded number is then assigned to the padded variable and logged to the console.\n\n### Using string concatenation\n\nYou can also pad a number with leading zeros using string concatenation. Here’s the syntax for using string concatenation to pad a number with leading zeros:\n\n``\"0\".repeat(length - number.toString().length) + number.toString()``\n\nThe number is the number that you want to pad with leading zeros. The length is the total number of digits that the padded number should have.\n\nHere’s an example of using string concatenation to pad a number with leading zeros:\n\n``````let number = 42;\nlet padded = \"0\".repeat(5 - number.toString().length) + number.toString();\n``````\n\nIn this example, we use string concatenation to pad the number 42 with leading zeros so that it has a total of 5 digits. The resulting padded number is then assigned to the padded variable and logged to the console.\n\n### Using the slice() method\n\nYou can also pad a number with leading zeros using the slice() method. Here’s the syntax for using the slice() method to pad a number with leading zeros:\n\n``(\"00000\" + number).slice(-length)``\n\nThe number is the number that you want to pad with leading zeros. The length is the total number of digits that the padded number should have.\n\nHere’s an example of using the slice() method to pad a number with leading zeros:\n\n``````let number = 42;\nlet padded = (\"00000\" + number).slice(-5);" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7615683,"math_prob":0.9882337,"size":2933,"snap":"2023-14-2023-23","text_gpt3_token_len":651,"char_repetition_ratio":0.21304199,"word_repetition_ratio":0.5638945,"special_character_ratio":0.24002728,"punctuation_ratio":0.10278746,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9962897,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-09T11:57:31Z\",\"WARC-Record-ID\":\"<urn:uuid:9971d6c2-33d4-456c-b58a-5434adc47017>\",\"Content-Length\":\"129481\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8656d079-7bb6-4dfd-80bc-e6a0157bff6e>\",\"WARC-Concurrent-To\":\"<urn:uuid:e1c0738c-3f7a-42d8-984b-70e231a44dc2>\",\"WARC-IP-Address\":\"157.7.107.216\",\"WARC-Target-URI\":\"https://plantpot.works/8581\",\"WARC-Payload-Digest\":\"sha1:YLLZKWU7TBHQ5324VMNEL5RXAW5BIWMG\",\"WARC-Block-Digest\":\"sha1:4OF5BE3Y46P3ASR527ZNSQI255ML7HAI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224656675.90_warc_CC-MAIN-20230609100535-20230609130535-00599.warc.gz\"}"}
https://eng.kakprosto.ru/how-9862-how-to-make-a-proportion	[ "Instruction\n1\nSuppose that your salary is 10,000 rubles a month. This number will be divisible by the first fraction. Since your salary is your entire income for the month, we will take it for 100 percent. This number will be a divisor of the first fraction. So, the first roll - 10000/100. Make a fraction with your numbers.\n2\nYou need to calculate the tax that will be withheld from your paycheck per month. The tax to incomes of physical persons in our country is 13 percent. This number will be a divisor of the second fraction. And since the amount deducted from you tax we don't know, let's call it \"x\". The number \"x\" is divisible by the second fraction. So the second fraction is x/13.\n3\nMake proportion, that is, write both the fraction and put between them a sign of equality. Our proportion of 10000/100=x/13.In order to solve the proportion, you need to multiply the end members proportions and share them on the remaining member. For example: x=10000*13/100. Hence, x=1300. This is the amount of tax withheld from you in a month with an income of 10,000 rubles. Decide your proportion." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9345704,"math_prob":0.98581177,"size":1098,"snap":"2019-26-2019-30","text_gpt3_token_len":265,"char_repetition_ratio":0.1380256,"word_repetition_ratio":0.05235602,"special_character_ratio":0.2650273,"punctuation_ratio":0.12340426,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9941334,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-21T05:01:24Z\",\"WARC-Record-ID\":\"<urn:uuid:ba53c927-f493-46c7-a893-c2532d249db7>\",\"Content-Length\":\"28225\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dea698bd-c280-476d-a30b-a75d2663b322>\",\"WARC-Concurrent-To\":\"<urn:uuid:6a7f4e20-2eff-4ab6-9906-484f1c6fb55b>\",\"WARC-IP-Address\":\"95.213.175.86\",\"WARC-Target-URI\":\"https://eng.kakprosto.ru/how-9862-how-to-make-a-proportion\",\"WARC-Payload-Digest\":\"sha1:S76CFVGGS4PJZJHE2CFHHPYE6X6FY6TV\",\"WARC-Block-Digest\":\"sha1:3POORXKRB2CUG7QC33K2FPITR7PGAOB2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195526888.75_warc_CC-MAIN-20190721040545-20190721062545-00065.warc.gz\"}"}
https://learnphp.org/php-array-intersect-uassoc-function-with-example	[ "", null, "Popular Searches:\n391\nQ:\n\n# PHP array_intersect_uassoc() function (with example)\n\nHi everyone,\n\nI hope you're doing well. I recently came across the PHP `array_intersect_uassoc()` function and I'm having a bit of trouble understanding how it works. I've read the documentation, but I'm still a bit confused. I was hoping someone here could help shed some light on it for me.\n\nTo provide some context, I'm currently working on a project where I need to compare two arrays and find the common elements based on their keys and values, but with a custom comparison function. I understand that `array_intersect_assoc()` can be used to find the common elements based on their values, but it doesn't allow for custom comparisons. This is where `array_intersect_uassoc()` comes in.\n\nI've looked at the syntax and I understand that it takes in two or more arrays as arguments, and an optional callback function that compares the keys and values. However, I'm not entirely sure how to write the callback function and how it affects the comparison process.\n\nIt would be really helpful if someone could provide me with an example of using `array_intersect_uassoc()` with a custom callback function. I think having a real-world example would make it easier for me to understand and implement it in my project.\n\nThank you so much in advance for your help. I really appreciate it.\n\nBest regards,\n\n## All Replies", null, "I've actually used `array_intersect_uassoc()` before, so I hope I can help clarify things for you.\n\nWhen using `array_intersect_uassoc()`, the callback function you provide is used for comparing both the keys and values of the arrays. It takes four parameters: `\\$a_key` and `\\$b_key` representing the keys being compared, and `\\$a_value` and `\\$b_value` representing the corresponding values.\n\nLet's say you have two arrays, `\\$array1` and `\\$array2`, and you want to find the common elements based on both the keys and values, but with a custom comparison. You can use `array_intersect_uassoc()` along with a callback function to achieve this. Here's an example:\n\n```php```\\$array1 = array(\"apple\" => 2, \"banana\" => 5, \"cherry\" => 3);\n\\$array2 = array(\"apple\" => 4, \"banana\" => 2, \"cherry\" => 8);\n\n\\$result = array_intersect_uassoc(\\$array1, \\$array2, function (\\$a_key, \\$b_key, \\$a_value, \\$b_value) {\nif (\\$a_key === \\$b_key) {\nreturn \\$a_value - \\$b_value; // Custom comparison logic\n}\nreturn \\$a_key <=> \\$b_key; // Default comparison for keys\n});\n\nprint_r(\\$result);\n``````\n\nIn this example, the callback function compares both the keys and values. If the keys are equal, it compares the values using a custom logic (`\\$a_value - \\$b_value`). If the keys are not equal, it uses the default comparison operator `<=>` to compare the keys.\n\nThe result would be:\n\n``````Array\n(\n[banana] => 5\n)\n``````\n\nSince \"banana\" is the only key present in both arrays and has a different value, it is the only element returned by `array_intersect_uassoc()`.\n\nI hope this example helps you understand how to use `array_intersect_uassoc()` with a custom callback function. Let me know if you have any further questions!\n\nBest regards,\n\n## Related Topics", null, "Hey there,\n\nI noticed your question regarding `array_intersect_uassoc()` and thought I could share my personal experience with using it.\n\nIn one of my recent projects, I had to compare two arrays based on their keys and values, but with a complex comparison logic. `array_intersect_uassoc()` came to my rescue, allowing me to define my custom callback function.\n\nTo give you an example, let's suppose we have two arrays: `\\$array1` and `\\$array2`. We want to find the common elements between them, comparing both keys and values using a customized logic in the callback function. Here's how I accomplished it:\n\n```php```\\$array1 = array(\"apple\" => 2, \"banana\" => 5, \"cherry\" => 3);\n\\$array2 = array(\"apple\" => 4, \"banana\" => 7, \"cherry\" => 3);\n\n\\$result = array_intersect_uassoc(\\$array1, \\$array2, function (\\$a_key, \\$b_key, \\$a_value, \\$b_value) {\nif (\\$a_key === \"apple\" && \\$b_key === \"apple\") {\nreturn \\$a_value - (\\$b_value * 2); // Custom comparison logic for \"apple\" key\n}\n\n// Default comparison for other keys\nreturn \\$a_value <=> \\$b_value;\n});\n\nprint_r(\\$result);\n``````\n\nIn this example, I added a specific comparison condition for the \"apple\" key. If both arrays have this key, I compared their values using a custom logic (`\\$a_value - (\\$b_value * 2)`). For other keys, I used the default comparison operator `<=>` to compare their values.\n\nThe resulting output would be:\n\n``````Array\n(\n[cherry] => 3\n)\n``````\n\nBased on the custom comparison logic, only the \"cherry\" element is considered common between the two arrays and is returned by `array_intersect_uassoc()`.\n\nI hope this provides you with a better understanding of how to use `array_intersect_uassoc()` with a custom callback function. If you have any further questions, feel free to ask!\n\nBest regards," ]	[ null, "https://learnphp.org/upload/spinner-logo.png", null, "https://www.gravatar.com/avatar/a90ae39cdee305081dbeb45c2433a63b", null, "https://www.gravatar.com/avatar/5ef410540b84bc0b6835022d0a233cd0", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9208091,"math_prob":0.8194593,"size":2601,"snap":"2023-40-2023-50","text_gpt3_token_len":558,"char_repetition_ratio":0.14863303,"word_repetition_ratio":0.047732696,"special_character_ratio":0.2237601,"punctuation_ratio":0.10212766,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96272,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T09:55:13Z\",\"WARC-Record-ID\":\"<urn:uuid:26bfb498-9b7f-4efc-85f0-5a6b8743a254>\",\"Content-Length\":\"82470\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c1966367-ba40-4041-a3aa-d0b1c415a94d>\",\"WARC-Concurrent-To\":\"<urn:uuid:7547c66d-f30c-422b-b54b-ce4631ea6d1d>\",\"WARC-IP-Address\":\"35.213.167.160\",\"WARC-Target-URI\":\"https://learnphp.org/php-array-intersect-uassoc-function-with-example\",\"WARC-Payload-Digest\":\"sha1:QJHLQXQT3Z2KJZ27AF7QYGOJQ3VUTISJ\",\"WARC-Block-Digest\":\"sha1:3MDCOW6QQJ46Q2YKX7UM34A3D52QB5J5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510387.77_warc_CC-MAIN-20230928095004-20230928125004-00607.warc.gz\"}"}
https://socratic.org/questions/how-do-you-find-the-general-solutions-for-sinx-2tanx-0#163780	[ "# How do you find the general solutions for sinx+2tanx=0?\n\nNote that since $\\tan x = \\sin \\frac{x}{\\cos} x$, $\\sin x = \\cos x \\tan x$\n$\\sin x + 2 \\tan x = 0$\n$\\cos x \\tan x + 2 \\tan x = 0$\n$\\tan x \\left(\\cos x + 2\\right) = 0$\n$\\cos x = - 2$ (no solutions) or $\\tan x = 0 \\setminus R i g h t a r r o w x = k \\pi , k \\in \\mathbb{Z}$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5924982,"math_prob":1.00001,"size":236,"snap":"2021-43-2021-49","text_gpt3_token_len":62,"char_repetition_ratio":0.12068965,"word_repetition_ratio":0.0,"special_character_ratio":0.24152543,"punctuation_ratio":0.069767445,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000098,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-15T23:06:51Z\",\"WARC-Record-ID\":\"<urn:uuid:3a4e16de-aebc-4763-ab38-5c840ab2cab2>\",\"Content-Length\":\"32924\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fab7b335-7ef9-4468-a7e1-b0391b1970b2>\",\"WARC-Concurrent-To\":\"<urn:uuid:97ae33ab-dcf9-47ad-8f4d-762daf53048f>\",\"WARC-IP-Address\":\"216.239.36.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-find-the-general-solutions-for-sinx-2tanx-0#163780\",\"WARC-Payload-Digest\":\"sha1:7GP52TFHVNCVSVULVRMWIRGEYRGCLRRT\",\"WARC-Block-Digest\":\"sha1:RAG4TLQNLEW5YCV2XHWHC6WTVXZN2QMS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323583087.95_warc_CC-MAIN-20211015222918-20211016012918-00505.warc.gz\"}"}
https://www.circuitsell.com/en/indicator-verification	[ "# Indicator verification\n\nIndicator verification instructions\n\nThe following two practical methods can be used to detect the accuracy of 0.5% FS, and the detection accuracy is higher than 0.00625% FS.\n\n1Input and output signal equivalent product testing:\n\nThe relays and the semaphore switch in the figure are common components (the foot switch can be used), and there is no special requirement for contact resistance and withstand voltage. The accuracy, drift, linearity and other indicators of the meter can be 0.5% FS, but the resolution should be 1μA (4-bit 1/2 portable or benchtop).\n\nThis is a relative error measurement method; the instrument with the input and output signals equivalent (such as input and output signals are 4-20 mA), and the size of the two signals is detected by a single measurement meter. In a short period of time (such as 10 seconds), the meter will not change, and the relative errors of the two signals measured during that time will not change. The measurement accuracy is independent of the size of the input signal and is independent of the absolute value of the measurement table, only related to the resolution of the measurement table. The resolution of the measurement table reaches 1μA, and the absolute error of the measured meter is very close to 1μA.\n\nRelative error: The measurement error of 1 μA relative to 4-20 mA signal is 0.00625% FS, which is 80 times higher than the accuracy of the measurement table itself (0.5%).\n\nAbsolute error: Depending on the resolution of the meter and the linearity error, if the linearity error of the meter is 0.5% FS, set the input and output difference to ∆I, and the linear error of the difference is ∆I/20mA0.5%. The sum of this value and the resolution of the meter is the absolute error. For example, the input-output difference is ±10μA, and the absolute error of the measurement itself is 1μA+10uA/20mA0.5%=1.0000025μA, which is regarded as 1μA, that is, the absolute error is measured as the resolution of the measurement meter.\n\nThe ratio of automatic control \"equivalent\" instruments is higher than 80%, so the relative value measurement method is widely used. Metering problems for most \"equivalent\" meters can be solved on site with low-cost meters.\n\n2Input and output signals are not equivalent to product detection:\n\nThe universal safety barrier (isolator) uses four input terminals to switch signal inputs such as power distribution, current, thermocouple, thermal resistance, millivolt, and slip line resistance. The system commonly used input signals are included. During on-site construction, the signal configuration can be performed in the state where the power supply, input and output terminals are suspended (not energized), and the instrument and signal generator are not required. The accuracy after power-on is better than 0.03% FS.\n\nVerification is necessary, but if all of the general barrier (isolator) input signals are detected, the workload is huge and even an unfinished task. Therefore, the actual inspection is mainly based on the process requirements, and the verification is spot-checking.\n\nThe general-purpose safety barrier (isolator) contains 4~20mA signal and current input signals, which are equivalent to the output signal and can be detected by relative error detection method. The internal reference and measurement circuit used by the distributor signal and current input signal are the same component as the other non-equivalent signals. If the output value corresponding to the distribution signal or current input signal is qualified, from the verification accuracy, After the other input signals are configured, more tests are performed, but they can be left unchecked.\n\nIn addition to the necessary metrological verification (concentrated on pre-factory calibration), the relative error detection method avoids the use of high-precision detection instruments and signal generators to detect the safety barrier (isolator), and does not require the inspection table. Periodic measurement calibration (relative value measurement does not have transmission error). At the same time, the calculation of the error is simplified, and the measurement can be performed at any point within the range. The difference between the input and output differences is 0.01% FS per 1.6 μA.", null, "" ]	[ null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8965908,"math_prob":0.96233237,"size":4380,"snap":"2020-45-2020-50","text_gpt3_token_len":927,"char_repetition_ratio":0.17413163,"word_repetition_ratio":0.0116959065,"special_character_ratio":0.20913242,"punctuation_ratio":0.11193112,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95308894,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-29T16:15:03Z\",\"WARC-Record-ID\":\"<urn:uuid:043e5e86-9d98-4740-9943-fe23f19cd6bc>\",\"Content-Length\":\"48008\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:38a72f5d-035f-4220-869a-b68fdb3a3f3c>\",\"WARC-Concurrent-To\":\"<urn:uuid:df52e24c-0c54-4d24-b2fa-9d7e3dccf7f1>\",\"WARC-IP-Address\":\"50.116.8.160\",\"WARC-Target-URI\":\"https://www.circuitsell.com/en/indicator-verification\",\"WARC-Payload-Digest\":\"sha1:L2ZSIXBRGGF6XLCCP5NDHOFOWIZW3ZDV\",\"WARC-Block-Digest\":\"sha1:LC3J6SR5IGIIPAXYYPUTOCSL4H3ESNNX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107904834.82_warc_CC-MAIN-20201029154446-20201029184446-00071.warc.gz\"}"}
http://laxbytes.com/binwomstats19/PRImidf001010002.php	[ "``` EXPLANATION\nRANK PERCENTILE VALUE * WEIGHT = NET\nGames Played 10 * 0.20 = 2.00\nGames Started 9 * 0.30 = 2.70\nGoals >100 90 14 * 0.25 = 3.50\nAssists >100 74 1 * 0.21 = 0.21\nTotal Points >100 86 15 * 0.05 = 0.75\nDraw Controls >100 94 21 * 0.45 = 9.45\nTotal Shots >100 95 44 * 0.10 = 4.40\nShots on Goal >100 93 29 * 0.10 = 2.90\nMissed Shots Off Goal >100 96 15 * -0.05 = -0.75\nShot Percentage >100 65 0.32 * 1.00 = 0.32\nShots-on-goal Percentage >100 62 0.66 * 1.00 = 0.66\nGround Balls >100 81 11 * 0.45 = 4.95\nTurnovers >100 85 11 * -0.60 = -6.60\nCaused Turnovers >100 88 7 * 2.00 = 14.00\nFree Position Goals >100 89 2 * 0.03 = 0.05\nFree Position Misses >100 95 6 * -0.10 = -0.60\nGoals Saved -- -- 0 * 0.80 = 0.00\nGoals Allowed -- -- 0 * -0.80 = -0.00\nShots Faced -- -- 0 * 0.01 = 0.00\nSave Percentage -- -- 0.00 * 5.00 = 0.00\n\nPIR RAW (NET/GamesPlayed) 4.49\nOffensive Factor (OF) 0.57\nDefensive Factor (DF) 0.68\nStrength Schedule (SOS) 0.86\nBasis Points 2.50\n-------\nPIR = (PRI RAW)((OF+DF)/2)SOS + Basis Points 4.93\n\nList of all players for Bucknell\nList of all players for Division I\n\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5014467,"math_prob":0.9945052,"size":1083,"snap":"2019-13-2019-22","text_gpt3_token_len":459,"char_repetition_ratio":0.14272475,"word_repetition_ratio":0.00896861,"special_character_ratio":0.58171743,"punctuation_ratio":0.17562725,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9815631,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-25T15:51:52Z\",\"WARC-Record-ID\":\"<urn:uuid:5e1eff12-139d-4904-aef0-1a58a172a9a9>\",\"Content-Length\":\"52699\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:05da9cd0-eb75-4df6-9cf2-a68a294cdc83>\",\"WARC-Concurrent-To\":\"<urn:uuid:336f8a02-2279-4883-a3be-2b2943c3ab41>\",\"WARC-IP-Address\":\"50.63.209.1\",\"WARC-Target-URI\":\"http://laxbytes.com/binwomstats19/PRImidf001010002.php\",\"WARC-Payload-Digest\":\"sha1:HNNH23KVQKC7DNFSVQN5O7MUMRCEN4TG\",\"WARC-Block-Digest\":\"sha1:5VFUYFVOODOVVIB23TA23ZUAZ2KWBVL3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912204077.10_warc_CC-MAIN-20190325153323-20190325175323-00050.warc.gz\"}"}
https://zbmath.org/?q=an:0674.35008	[ "# zbMATH — the first resource for mathematics\n\nRegularity for minima of functionals with p-growth. (English) Zbl 0674.35008\nWe prove that the first derivatives of scalar minima of functionals of the type $I(u)=\\int f(x,u,\\nabla u)dx,$ are Hölder continuous. Here $$f(x,u,\\nabla u)\\approx \| \\nabla u\|^ p,$$ $$1<p<\\infty$$ and f is assumed Hölder continuous in x and u.\nWe give two applications. One to the regularity theory of quasiregular mappings and the other to quasilinear degenerate elliptic equations with p-growth.\nReviewer: J.J.Manfredi\n\n##### MSC:\n 35B65 Smoothness and regularity of solutions to PDEs 35J70 Degenerate elliptic equations 35J60 Nonlinear elliptic equations 35B05 Oscillation, zeros of solutions, mean value theorems, etc. in context of PDEs 35A15 Variational methods applied to PDEs 35D10 Regularity of generalized solutions of PDE (MSC2000)\nFull Text:\n##### References:\n Bojarski, B; Iwaniec, T, Analytical foundations of the theory of quasiconformal mappings in ofr^{n}, Ann. acad. sci. fenn. ser. AI math., 8, 257-324, (1983) · Zbl 0548.30016 Campanato, S, Equazione ellittiche del secondo ordine e spazi L2,λ, Ann. mat. pura appl., 69, 321-380, (1965) · Zbl 0145.36603 DiBenedetto, E, C1 + α local regularity of weak solutions of degenerate elliptic equations, Nonlinear anal.: theory, methods appl., 7, 827-850, (1983) · Zbl 0539.35027 Giaquinta, M, Multiple integrals in the calculus of variations and nonlinear elliptic systems, () · Zbl 1006.49030 Giaquinta, M; Giusti, E, Differentiability of minima of non-differentiable functionals, Invent. math., 72, 285-298, (1983) · Zbl 0513.49003 Giaquinta, M; Giusti, E, On the regularity of the minima of variational integrals, Acta math., 148, 31-46, (1982) · Zbl 0494.49031 Granlund, S; Lindqvist, P; Martio, O, Conformally invariant variational integrals, Trans. amer. math. soc., 277, 43-73, (1983) · Zbl 0518.30024 Giaquinta, M; Modica, G, Remarks on the regularity of the minimizers of certain degenerate functionals, Manuscripta math., 57, 55-99, (1986), Preprint · Zbl 0607.49003 Gilbart, D; Trudinger, N.S, Elliptic partial differential equations of second order, (1977), Springer Berlin/Heidelberg/New York Iwaniec, T, Regularity theorems for solutions of partial differential equations for quasiconformal mappings in several dimensions, Dissertationes math., 198, (1982) · Zbl 0524.35019 Lewis, J, Regularity of the derivatives of solutions to certain elliptic equations, Indiana univ. math. J., 32, 849-858, (1983) · Zbl 0554.35048 Ladyzhenskaya, O.A; Ural’tseva, N.N, Linear and quasilinear elliptic equations, (1968), Academic Press New York · Zbl 0164.13002 Manfredi, J, Regularity of the gradient for a class of nonlinear possibly degenerate elliptic equations, () Tolksdorff, P, Regularity for a more general class of quasilinear elliptic equations, J. differential equations, 51, 126-150, (1984) Uhlenbeck, K, Regularity for a class of nonlinear elliptic systems, Acta math., 138, 219-240, (1977) · Zbl 0372.35030 Ural’tseva, N, Degenerate quasilinear elliptic systems, (), 184-222, [In Russian] · Zbl 0199.42502\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7139162,"math_prob":0.97168034,"size":3819,"snap":"2021-31-2021-39","text_gpt3_token_len":1202,"char_repetition_ratio":0.1457405,"word_repetition_ratio":0.010968922,"special_character_ratio":0.3257397,"punctuation_ratio":0.24904214,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97577447,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-18T07:43:49Z\",\"WARC-Record-ID\":\"<urn:uuid:c2d22e2c-7449-4da8-937c-f8cd207ec9a3>\",\"Content-Length\":\"54960\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7bb10f1c-1d2b-42d6-a23d-e81794fd582d>\",\"WARC-Concurrent-To\":\"<urn:uuid:10c61314-1c09-405f-a950-952de9f6c29c>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=an:0674.35008\",\"WARC-Payload-Digest\":\"sha1:UV67X4IFNMFSHR7OXFVHAQY67ISAPHRJ\",\"WARC-Block-Digest\":\"sha1:F5ND5RQFIQ6X6T25RJ3CKXT5ZZ727EFF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056348.59_warc_CC-MAIN-20210918062845-20210918092845-00262.warc.gz\"}"}
http://www.mytechinterviews.com/globe-walker	[ "# Globe Walker", null, "Question: How many points are there on the globe where, by walking one mile south, then one mile east and then one mile north, you would reach the place where you started?\n\nAnswer: The trivial answer to this question is one point, namely, the North Pole. But if you think that answer should suffice, you might want to think again! 🙂\n\nLet’s think this through methodically. If we consider the southern hemisphere, there is a ring near the South Pole that has a circumference of one mile. So what if we were standing at any point one mile north of this ring? If we walked one mile south, we would be on the ring. Then one mile east would bring us back to same point on the ring (since it’s circumference is one mile). One mile north from that point would bring us back to the point were we started from. If we count, there would be an infinite number of points north of this one mile ring.\n\nSo what’s our running total of possible points? We have 1 + infinite points. But we’re not done yet!\n\nConsider a ring that is half a mile in circumference near the South Pole. Walking a mile along this ring would cause us to circle twice, but still bring us to back to the point we started from. As a result, starting from a point that is one mile north of a half mile ring would also be valid. Similarly, for any positive integer n, there is a circle with radius\n\nr = 1 / (2 * pi * n)\n\ncentered at the South Pole. Walking one mile along these rings would cause us to circle n times and return to the same point as we started. There are infinite possible values for n. Furthermore, there are infinite ways of determining a starting point that is one mile north of these n rings, thus giving us (infinity * infinity) possible points that satisfy the required condition.\n\nSo the real answer to this question is 1 + infinity * infinity = infinite possible points!\n\nLiked this post? Please Digg or Stumble it!\n\nIf you're looking for some serious preparation for your interviews, I'd recommend this book written by a lead Google interviewer. It has 189 programming questions and solutions:", null, "", null, "## 4 Responses\n\n1.", null, "Andrei says:\n\nAnother answer is on the ecuator, +- half a mile on longitude, than the mile is equal on bouth longitude and latitude and you came back where you started.\n\n2.", null, "Swift says:\n\nOh yeah, nicely explained. It is true that there are infinite points where that could be the case. I was glad when I read through your answer because I already knew the answer to this one, and you definitely got it right.\n\nBut let’s be picky shall we? The problem did state that it was on the globe, so in fact the place at the north pole where this occurs is in fact at the magnetic north pole (which is constantly changing position, and it physically a south pole). Likewise, the ring of infinite positions just over a mile north of the south pole is in fact centered on the magnetic south pole.\n\n3.", null, "manful says:\n\nWhich part of the solution don’t you understand?\n\n4.", null, "asad naqvi says:\n\nhey,\ni didnt quite understand this solution.\ncan someone pls xplain it in a simpler manner.\n\nthanks\n\nXHTML: These are some of the tags you can use: `<a href=\"\"> <b> <blockquote> <code> <em> <i> <strike> <strong>`" ]	[ null, "http://www.mytechinterviews.com/wp-content/uploads/2010/02/cartoon-globe.png", null, "http://images.amazon.com/images/P/0984782850.01.LZZZZZZZ.jpg", null, "http://www.mytechinterviews.com/wp-content/uploads/2016/03/button_get-the-book-now.png", null, "http://0.gravatar.com/avatar/38ac81712799e9600327f13c5626d7ba", null, "http://1.gravatar.com/avatar/1c995ccbab6a3841d64405f1b408ce09", null, "http://1.gravatar.com/avatar/45657283f602e28cc77a9f3e470a184b", null, "http://0.gravatar.com/avatar/cd735f545ca77136e53726b32cbec244", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9541976,"math_prob":0.87974775,"size":2786,"snap":"2019-51-2020-05","text_gpt3_token_len":608,"char_repetition_ratio":0.15276779,"word_repetition_ratio":0.02303263,"special_character_ratio":0.21931084,"punctuation_ratio":0.10034602,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9847225,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,4,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-07T06:48:13Z\",\"WARC-Record-ID\":\"<urn:uuid:263fde6d-ff49-4b02-9dfa-83eb0f4d3607>\",\"Content-Length\":\"31314\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:964685df-190e-4690-8702-4fe7d4710fc1>\",\"WARC-Concurrent-To\":\"<urn:uuid:cea62c63-900a-4a63-9f61-ff76849572db>\",\"WARC-IP-Address\":\"34.198.120.227\",\"WARC-Target-URI\":\"http://www.mytechinterviews.com/globe-walker\",\"WARC-Payload-Digest\":\"sha1:FRNOO3A3BOTNPXNARFK2NSNFP5FTJMOF\",\"WARC-Block-Digest\":\"sha1:3Z2IFL2OMVT2RHI2LX4G22YX2CIJLJJA\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540496492.8_warc_CC-MAIN-20191207055244-20191207083244-00230.warc.gz\"}"}
https://www.hindawi.com/journals/aaa/2014/381753/	[ "/ / Article\n\nResearch Article \| Open Access\n\nVolume 2014 \|Article ID 381753 \| 11 pages \| https://doi.org/10.1155/2014/381753\n\n# Analytical Solutions of a Space-Time Fractional Derivative of Groundwater Flow Equation\n\nAccepted12 Sep 2013\nPublished21 Jan 2014\n\n#### Abstract\n\nThe classical Darcy law is generalized by regarding the water flow as a function of a noninteger order derivative of the piezometric head. This generalized law and the law of conservation of mass are then used to derive a new equation for groundwater flow. Two methods including Frobenius and Adomian decomposition method are used to obtain an asymptotic analytical solution to the generalized groundwater flow equation. The solution obtained via Frobenius method is valid in the vicinity of the borehole. This solution is in perfect agreement with the data observed from the pumping test performed by the institute for groundwater study on one of their boreholes settled on the test site of the University of the Free State. The test consisted of the pumping of the borehole at the constant discharge rate and monitoring the piezometric head for 350 minutes. Numerical solutions obtained via Adomian method are compared with the Barker generalized radial flow model for which a fractal dimension for the flow is assumed. Proposition for uncertainties in groundwater studies was given.\n\n#### 1. Introduction\n\nThe real problem encounter in groundwater studies up to now is the real shape of the geological formation in which water flows in the aquifer under investigation. However, there are many fractured rock aquifers where the flow of groundwater does not fit conventional geometries , for example, in South Africa, the Karoo aquifers, characterized by the presence of a very few bedding parallel fractures that serve as the main conduits of water in the aquifers . With a challenge to fit the solution of the groundwater flow equation with experimental data from field observation in particular, the observed drawdown see , for all time yields a fit that undervalues the observed drawdown at early times and overvalues it at later times. The variation of observations from theoretically predictable values is usually an indication of uncertainties in the predictable. To investigate the first possibility Botha et al. developed a three-dimensional model for the Karoo aquifer on the campus of the University of the Free State. This model is based on the conventional, saturated groundwater flow equation for density-independent flow: where is the specific storativity, the hydraulic conductivity tensor of the aquifer, the piezometric head, ) the strength of any sources or sinks, with and the usual spatial and time coordinates; the gradient operator, and the time derivative.\n\nThis model showed that the dominant flow field in these aquifers is vertical and linear and not horizontal and radial as commonly assumed. However, more recent investigations suggest that the flow is also influenced by the geometry of the bedding parallel fractures, a feature that (1) cannot account for. It is therefore possible that the equation may not be applicable to flow in these fractured aquifers.\n\nIn an attempt to circumvent this problem, Barker introduced a model in which the geometry of the aquifer is regarded as a fractal. Although this model has been applied with reasonable success in the analysis of hydraulic tests from boreholes in Karoo aquifers, it introduces parameters for which no sound definition exists in the case of noninteger flow dimensions.\n\nAs a review of the derivation of (1) will show, Darcy law is used as a keystone in the derivation of (1). This law, proposed by Darcy early in the 19th century, is relying on experimental results obtained from the flow of water through a one-dimensional sand column, the geometry of which differs completely from that of a fracture . There is therefore a possibility that the Darcy law may not be valid for flow in fractured rock formations but is only a very crude idealization of the reality . Nevertheless, the relative success achieved by Botha et al. , to describe many of the properties of Karoo aquifers, suggests also that the basic principle underlying this law may be correct: the observed drawdown is to be related to either a variation in the hydraulic conductivity of the aquifer or a change in the piezometric head. Any new form of the law should therefore be reduced to the classical form under the more common conditions. Because is essentially determined by the permeability of the rocks, and not the flow pattern, the gradient term in (2) is the most likely cause for the deviation between the observed and theoretical drawdown observed in the Karoo formations . In the same direction, Cloot and Botha introduced the concept of non-integer fractional derivative to investigate a radially symmetric form of (1), by replacing the classical first order derivative of the piezometric head by a complementary fractional derivative . However the generalized model for groundwater flow equation was solved numerically. In this work, more general form of groundwater flow equation will be introduced; the Frobenius and Adomian decomposition will be used to give an asymptotic solution of the generalized model for groundwater flow equation. Because the concepts of fractional (or non-integer) order derivatives and complementary fractional order derivatives may not be widely known, both concepts are first briefly discussed below.\n\n#### 2. Fractional Order Derivatives\n\nOn one hand, the concept of fractional calculus is popularly believed to have stemmed from a question raised in the year 1695 by Marquis de L Hospital (1661–1704) to Gottfried Wilhelm Leibniz (1646–1716), which sought the meaning of Leibniz’s currently popular notation for derivative of order when (what if ). In his reply, dated September 30, 1695, Leibniz wrote to L’Hospital as follows: “This is an apparent paradox from which, one day, useful consequences will drawn.” On the other hand, the concept of fractional order derivatives for a function, say , is based on a generalization of the Abel integral : where is a nonzero positive integer and the Gamma function . This represents an integral of order for the continuous function , whenever and all its derivatives vanish at the origin, . This result can be extended to the concept of an integral of arbitrary order , defined as: where is a positive real number and an integer such that .\n\nLet now be the least positive integer larger than such that ; . Equation (4) can then be used to define the derivative of (positive) fractional order, say , of a function as Note that these results, like Abel’s integral, are only valid subject to the condition that for .\n\n##### 2.1. Properties\n\nProperties of the operator can be found in [14, 15], we mention only the following: for , , and :\n\n#### 3. A Generalized Mathematical Groundwater Flow Model\n\nFor the sake of clarity the generalization of the classical model for groundwater flow in the case of density-independent flow in the uniform homogeneous aquifer is considered in this paper . Consider the following groundwater flow equation where both the specific storability, , and hydraulic conductivity, , are scalar and constant quantities and or is the radial dimension. To be complete, the following set of initial and boundary conditions is added: Here means that before a pumping test begins the level of water or the initial hydraulic head in the aquifer is linear function of space with a positive gradient to be found, means that during the pumping test the level of water is not affected for a very long distance from the borehole, and means that the rate of pumping is proportional to the hydraulic conductivity.\n\nHence is the discharge rate of the borehole, with radius and the thickness of the aquifer from which the borehole taps. In order to include explicitly the possible effect of the geometry into mathematical model the radial component of the gradient of the piezometric head, is replaced by the Weyl-fractional derivatives of order ; the fractional derivative in this paper is Caputo derivative and is defined as follows: This provides a generalized form of the classical equation governing the flow equation (1): The integrodifferential equation does contain the additional parameter , , which can be viewed as a new physical parameter that characterizes the flow through the geological formations . The same transformation generates also a more general form for the boundary condition at the borehole : The relations (10) and (11), together with the initial condition described in (10), represent a complete set of equations for which a solution exists. The integro-differential character of the relations makes the search for analytical solution very difficult however. Nevertheless, in this paper we make use of Frobenius and Adomian decomposition methods to give an asymptotic solution.\n\n#### 4. Solution of the Generalized Flow Equation\n\n##### 4.1. Frobenius Method\n\nIn this work to perform the Frobenius method, we consider the groundwater flow governed by the following fractional Caputo-Weyl derivation partial differential of order , where is real number, . Also, we consider the dimension of the flow to be 2. Therefore, for (10) can then be transformed into the following equation: Applying the Laplace operator on both sides of the above equation, we have the following ordinary differential equation where is the variable of Laplace. In this matter we choose or we choose , meaning that the level of water is the same everywhere in the aquifer if the water is not taken out from the aquifer. If we let then (13) becomes We put and . To meet the condition under which Frobenius method can be used, we have to prove that and are analytical around that means we have to prove that and can be written as series. It is very obvious to see that and can be expressed as follows: We start here with the coefficients of : implying that .\n\nPutting we have that and equating the coefficients of same power we obtain the following set of equations: Therefore, the coefficients can be given with the general following recursive formula: And obviously the coefficients are given below as From the above expression we can see that and are analytical around , which follows from Frobenius method that the solution of (15) can be in the form This solution is not convergent for a large ; that is, the solution will diverge if we observe the drawdown at a position very far from the borehole from which the water is taken out. Therefore, we restrict our solution in the vicinity of the borehole; more precisely we investigate the solution when . That means we pump water from the borehole and we observe the drawdown in the vicinity of the borehole.\n\nThus substituting (21), , and into (15) and equating the coefficients of the same power, we obtain the following recursive formula for which the coefficients , coefficients of our series: Here we have the following: However, making use of the boundary condition in (11), we have the following: Then the general solution of the fractional Caputo-Weyl derivation of groundwater flow of order in Laplace space is given below as: To observe the behavior of the solution at the borehole, which corresponds to , the series solution is reduced here to the coefficient with order zero which is obtained when and it’s given below as In the following section the analytical asymptotic solution obtained in Laplace space via Frobenius method will be compared with the experimental data.\n\n##### 4.2. Numerical Results\n\nIn order to examine the validation of this solution, the above asymptotic solution is compared with the experimental data from the pumping test performed by the Institute for Groundwater Studies on one of their borehole settled on the campus test site of the University of the Free State. The test consisted of the pumping of the borehole at the constant discharge rate and monitoring the piezometric head for 350 minutes. The first step in the section is to discretize the range of Laplace transform since the exact Laplace transform cannot be obtained in practice. This is done as follows: For we approximate where for ; then the results obtained in the fields and Laplace transform become Using this numerical scheme, the physical data was transformed into Laplace space. A comparison between these values and asymptotic computed data can only be provided in the in real space not in Laplace space. Since it is not worth concluding the validity of this solution in Laplace space, the inverse Laplace transform is applied in (26). To test the validity of this solution in real space and applying the inverse Laplace transform on (26), (29) is obtained\n\nThe above solution is compared graphically to the experimental data from the pumping test performed by the institute for groundwater study on one of their borehole settled on the test site of the University of the Free State. The small difference observed in the above graph (Figure 1) is due to uncertainties in measurement and this will be discussed in Section 6 of this work. The aquifer parameters used in this models are recorded in Table 1, the observed data from field observation will be attached to this paper. Although this solution is in agreement with the experimental data, there will be the need to investigate the case where the observation can be done for a long distance. In the next section another approach will be introduced to solve the space-time-fractional derivative of groundwater flow equation, and this method is the Adomian decomposition method.\n\n Parameters Values Units 83.1 (m3 s−1) 0.025 (m) 13.1 (m s−1) 1 (m) 13.1 (m−1) 0 (m)\n\n##### 5.1. Example 1\n\nThis section is concerned with the groundwater equation with time- and space-fractional derivatives of the form Subject to the initial and boundary conditions described in (8), the level of water is assumed to be the same throughout the aquifer before the pumping so that the gradient described in (8) is zero. Furthermore, it is assumed that a fractional change in drawdown is constant for meaning .\n\nThe method used here is based on applying the operator on both sides of (30) to obtain The Adomian decomposition method [16, 17] assumes a series solution for (31) to be where the components are determined recursively. Substituting (32) into both sides of (30) gives Following the decomposition method, the recursive relations are introduced as It is worth noting that if the component is defined, then the remaining components can be completely determined such that each term is determined by using the previous terms, and the series solutions are thus entirely determined. Finally, the solution is approximated by the truncated series However, the inclusion of boundary conditions in fractional differential equations introduces additional difficulties. The Adomian decomposition method can handle these difficulties by using the space-fractional operator and the initial conditions only. The method provides the solution in the form of a rapidly convergent series that may lead to the exact solution in the case of integer derivatives and to an efficient numerical solution with high accuracy for 0 . The convergence of the decomposition series has been investigated in [18, 19].\n\nFollowing the recursive formula equation (35) and using the fact that , the equations below are obtained: The component was also determined and will be used, but for brevity it is not listed. In this matter five components of the decomposition series (30) were obtained for which was evaluated to have the following expansion: Applying the boundary condition yields where\n\n##### 5.2. Example 2\n\nConsider the groundwater flow equation with time- and space-fractional derivatives of the form subject to the initial condition described in (11). Furthermore we suppose that the gradient , meaning that the water level is everywhere in the aquifer at and the fractional change in drawdown is a constant and boundary condition: Following the discussion earlier, we have the below recursive formula: It follows from the recursive formula that The component was also determined and will be used, but for brevity it is not listed. In this matter five components of the decomposition series (41) were obtained of which was evaluated to have the following expansion: Applying the boundary condition yields where\n\n##### 5.3. Example 3\n\nConsider time fractional derivative of the groundwater flow equation with time-fractional derivatives of the form Subject to the initial and boundary conditions described in (8) it is assumed that . Following the discussion presented earlier, we obtained the below recursive formula: The component was also determined and will be used, but for brevity not listed. In this matter five components of the decomposition series (31) were obtained of which was evaluated to have the following expansion Applying the boundary conditions yield to where The normalized solutions with of (50), (51) and (52) are illustrated graphically in Figures 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, and 13 for different values of various parameters; these solutions are compared to the solution proposed by Barker . These graphs show the behaviour of the drawdown during the pumping test, first as a function of space and time from the borehole to the point of observation, secondly as function of time for a fixed distance from the borehole, and finally as a function of space for a fixed time. Based on the assumption that groundwater in an aquifer flows through an equipotential surface that are projections of -dimensional spheres onto two-dimensional space, Barker obtained the following analytical solution for an infinite aquifer with a line source: where is the incomplete gamma function and the dimension of the flow which equals the special dimension, being an integer, and the other quantities all have the same meaning as before. Although this model has been applied with reasonable success in the analysis of hydraulic tests from boreholes in Karoo aquifers, it introduces parameters for which no sound definition exists in the case of non-integer flow dimensions. The main raison of comparison of these results obtained via the Adomian decomposition methods with those of Barker’s fractal radial flow model is to establish a possible relationship between the fractional order of the derivative and the parameter fractal introduced earlier by Barker.\n\nTable 2 shows the theoretical values of the discharge rate and aquifer parameters used in the numerical simulations.\n\n Parameters Values Units Parameters Values Units 10.8 (m3 s−1) 0.005 None 0.15 (m) 1.05 None 0.65 (m s−1) 1.005 None 40 (m) 0.012 None 0.75 (m−1) 0.5 None 0 (m) 0.005 None 0.025 None\n\n#### 6. Discussion and Propositions\n\nAlthough the analytical solution obtained via Frobenius method fit the experimental data or have described successfully the events taking place in the vicinity of the borehole, on one hand, and the analytical solutions obtained via Adomian decomposition method was successfully compared to the solution proposed by Barker, on the other hand, the problem of choosing an appropriate geometry for the geological system in which the flow occurs still remains a challenge in groundwater studies. We personally believe that we do not propose solution to a problem because it going to be useful for the generation in which we are, but we propose solution because we hope that they will one day be useful for mankind; therefore the following proposition may not be useful for this generation because we may not have the adequate technology to perform the steps involved but it will be in the future. We think that describing the groundwater flow with one equation for the whole aquifer is unrealistic, because from one point of the aquifer to another properties including geology and geometry change, therefore the flow. Assumption such as homogeneity, isotropy, uniform thickness over the area under investigation and so on render the study of groundwater uncertain. In order to study the geology or the geometry of an aquifer, we have to divide the aquifer in small portions, from north to south and from east to west. The geology and the geometry of each portion can then be studied. If the results of the study reveal that the portion under investigation is for instance a hyperboloid, since there is no exact solution to groundwater flow model by hyperboloid flow and there are solutions to circular flow, then a suitable transformation can be done including transformation of hyperboloid coordinates to Cartesian coordinates, then to Cartesian coordinates to cylindrical coordinate such that this new coordinate can now be put in the equation describing groundwater flow, and the solution to the new equation can then be investigated. Henceforth knowing the real geology and geometry of this portion, the real paths flow will be known. Then having a good knowledge of each small portion including its geometry and geology, the real geometry and geology of the aquifer can be not exactly but more accurately known.\n\nFor groundwater remediation it will be possible to know where the maximum, minimum, and average chemical concentrations are found in the aquifer. We have no proof for this but we believe that this proposition will be useful in reducing uncertainties in groundwater study. It is believed that the field test gives the characteristic of an aquifer, but we believe that the field test gives both uncertainties and characteristics of the aquifer; therefore quantify uncertainties in this measurement lead us to the real picture of aquifer characteristics, henceforth we propose that the studies in groundwater should focus on both uncertainties and fields observations, because what is known is bounded by what is not known; knowing what is not known give a real picture to what was known, and it follows that the knowledge of uncertainties in groundwater study will give a clear picture of what we already know in groundwater.\n\n#### 7. Conclusion\n\nThe classical Darcy Law has been generalized using the concept of complementary fractional order derivatives of Weyl fractional derivative. This leads to the formulation of a new generalized form of the groundwater flow equation . The applications of Adomian decomposition and Frobenius methods were extended to obtain explicit and numerical solutions of the space-time fractional groundwater flow. The two methods were very clearly efficient and powerful techniques in finding the solutions of the proposed equations. The solution obtained via Frobenius takes into account the events taking place in the vicinity of the borehole during the pumping test whereas the solution obtained via Adomian decomposition methods takes into account the events that take place far from the point where water is pumped out, that is, in the borehole. The solution obtained via Frobenius method was in perfect agreement with the observed data obtained from the pumping test performed by the Institute for Groundwater Studies on one of their borehole settled on the campus test site of the University of the Free State. The Adomian decomposition method requires less computational work than existing approaches while supplying quantitatively reliable results. The obtained results demonstrate the reliability of the algorithms and their wider applicability to fractional evolution equations. A comparison of these results obtained via the Adomian decomposition methods with those of Barker’s fractal radial flow model suggests that there exists a relation between the fractional order of the derivative and the non-integral dimension of the flow.\n\n#### Authors’ Contribution\n\nAbdon Atangana wrote the first draft and P. D. Vermeulen read and revised it; the revised version was read and corrected by both authors.\n\n#### Conflict of Interests\n\nThe authors declare that there is no conflict of interests for this paper.\n\n1. J. H. Black, J. A. Barber, and D. J. Noy, “Crosshole investigations: the method, theory and analysis of crosshole sinusoidal pressure tests in fissured rocks,” Stripa Projects Internal Reports 86-03, SKB, Stockholm, Sweden. View at: Google Scholar\n2. J. F. Botha, I. Verwey van Voort, J. J. P. Viviers, W. P. Collinston, and J. C. Loock, “Karoo aquifers. Their geology, geometry and physical behaviour,” WRC Report 487/1/98, Water Research Commission, Pretoria, 1998. View at: Google Scholar\n3. G. J. van Tonder, J. F. Botha, W.-H. Chiang, H. Kunstmann, and Y. Xu, “Estimation of the sustainable yields of boreholes in fractured rock formations,” Journal of Hydrology, vol. 241, no. 1-2, pp. 70–90, 2001. View at: Publisher Site \| Google Scholar\n4. J. A. Barker, “A generalized radial flow model for hydraulic tests in fractured rock,” Water Resources Research, vol. 24, no. 10, pp. 1796–1804, 1988. View at: Publisher Site \| Google Scholar\n5. J. Bear, Dynamics of Fluids in Porous Media, American Elsevier Environmental Science Series, Elsevier, New York, NY, USA, 1972.\n6. A. Cloot and J. F. Botha, “A generalised groundwater flow equation using the concept of non-integer order derivatives,” Water SA, vol. 32, no. 1, pp. 1–7, 2006. View at: Google Scholar\n7. R. Courant and F. John, Introduction to Calculus and Analysis, vol. 2, John Wiley & Sons, New York, NY, USA, 1974.\n8. Y. Cherruault and G. Adomian, “Decomposition methods: a new proof of convergence,” Mathematical and Computer Modelling, vol. 18, no. 12, pp. 103–106, 1993. View at: Publisher Site \| Google Scholar \| MathSciNet\n9. I. Podlubny, Fractional Differential Equations, Academic Press, San Diego, Calif, USA, 1999. View at: MathSciNet\n10. I. Podlubny, “Geometric and physical interpretation of fractional integration and fractional differentiation,” Fractional Calculus & Applied Analysis, vol. 5, no. 4, pp. 367–386, 2002. View at: Google Scholar \| MathSciNet\n11. K. Adolfsson, “Nonlinear fractional order viscoelasticity at large strains,” Nonlinear Dynamics, vol. 38, no. 1–4, pp. 233–246, 2004. View at: Publisher Site \| Google Scholar \| MathSciNet\n12. O. P. Agrawal, “Application of fractional derivatives in thermal analysis of disk brakes,” Nonlinear Dynamics, vol. 38, pp. 191–206, 2004. View at: Publisher Site \| Google Scholar\n13. G. Afken, Mathematical Methods for Physicists, Academic Press, London, UK, 1985.\n14. K. S. Miller and B. Ross, An Introduction to the Fractional Calculus and Fractional Differential Equations, John Wiley & Sons, New York, NY, USA, 1993. View at: MathSciNet\n15. K. B. Oldham and J. Spanier, The Fractional Calculus, Academic Press, New York, NY, USA, 1974. View at: MathSciNet\n16. G. Adomian, “A review of the decomposition method in applied mathematics,” Journal of Mathematical Analysis and Applications, vol. 135, no. 2, pp. 501–544, 1988. View at: Publisher Site \| Google Scholar \| MathSciNet\n17. A. Atangana, “New class of boundary value problems,” Information Science Letters, vol. 1, no. 2, pp. 67–76, 2012. View at: Publisher Site \| Google Scholar\n18. G. Adomian, Solving Frontier Problems of Physics: The Decomposition Method, Kluwer Academic, Dodrecht, The Netherlands, 1994. View at: MathSciNet\n19. Y. Cherruault, “Convergence of Adomian's method,” Kybernetes, vol. 18, no. 2, pp. 31–38, 1989. View at: Publisher Site \| Google Scholar \| MathSciNet\n\n#### More related articles\n\nWe are committed to sharing findings related to COVID-19 as quickly and safely as possible. Any author submitting a COVID-19 paper should notify us at help@hindawi.com to ensure their research is fast-tracked and made available on a preprint server as soon as possible. We will be providing unlimited waivers of publication charges for accepted articles related to COVID-19. Sign up here as a reviewer to help fast-track new submissions." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9199511,"math_prob":0.9231943,"size":27192,"snap":"2020-24-2020-29","text_gpt3_token_len":5660,"char_repetition_ratio":0.15771665,"word_repetition_ratio":0.112969756,"special_character_ratio":0.20351574,"punctuation_ratio":0.11641609,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98536974,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-31T01:33:34Z\",\"WARC-Record-ID\":\"<urn:uuid:f352a6cb-d787-43bf-a426-51c3479a390a>\",\"Content-Length\":\"1049260\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3097eda3-1859-4b07-9c90-8cd8475fa89d>\",\"WARC-Concurrent-To\":\"<urn:uuid:b84b7e85-daa5-4355-9dd5-dec988db833b>\",\"WARC-IP-Address\":\"99.84.191.103\",\"WARC-Target-URI\":\"https://www.hindawi.com/journals/aaa/2014/381753/\",\"WARC-Payload-Digest\":\"sha1:NLQNCIA5LW65766B7NSQOYBRAJ5YSRL7\",\"WARC-Block-Digest\":\"sha1:WEYO6SGQA6UVS6YGNYILDAIOLHFL4E3W\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347410535.45_warc_CC-MAIN-20200530231809-20200531021809-00594.warc.gz\"}"}
https://www.excelfunctions.net/excel-sumsq-function.html	[ "# The Excel SUMSQ Function\n\nRelated Function:\nSUMPRODUCT\n\n## Function Description\n\nThe Excel Sumsq function returns the sum of squares of a supplied set of values.\n\nThe syntax of the function is:\n\nSUMSQ( number1, [number2], ... )\n\nwhere the number arguments are numeric values (or arrays of numeric values) that you want to find the summed squares of.\n\nIf the values supplied to the function are text values, or logical values, these will be handled as follows:\n\n Text Values: If contained in cells that are referenced by the Sumsq function, text values are ignored;If supplied directly to the Sumsq function:Text values that can be interpreted as numeric values are treated as Numbers;Text values that cannot be recognised as numeric values cause the function to return the #VALUE! error. Logical Values: If contained in cells that are referenced by the Sumsq function, logical values are ignored;If supplied directly to the Sumsq function logical values are treated as numeric values (TRUE=1, FALSE=0).\n\nIn current versions of Excel (Excel 2007 and later), you can provide up to 255 number arguments to the Sumsq function, but in Excel 2003, the function can only handle up to 30 number arguments.\n\n## Sumsq Function Examples\n\nThe spreadsheets below show two examples of the Sumsq function.\n\nFormulas:\nAB\n1ValuesSumsq\n25=SUMSQ( A2:A5 )\n32=SUMSQ( A2, A3, A4, 6 )\n41\n53\nResults:\nABC\n1ValuesSumsq\n2539= 5^2 + 2^2 + 1^2 + 3^2\n3266= 5^2 + 2^2 + 1^2 + 6^2\n41\n53\n\nThe above examples show how:\n\n• The arguments to the Sumsq function can be input directly or as references to cells containing values;\n• Each of the arguments can be an individual value or an array of values.\n\nFurther details and examples of the Excel Sumsq function are provided on the Microsoft Office website.\n\n## Sumsq Function Error\n\nIf you get an error from the Sumsq function, this is likely to be the #VALUE! error:\n\nCommon Error\n #VALUE! - Occurs if a value that is supplied directly to the function cannot be interpreted as a number." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6611327,"math_prob":0.97327805,"size":2140,"snap":"2019-51-2020-05","text_gpt3_token_len":571,"char_repetition_ratio":0.18024345,"word_repetition_ratio":0.09018568,"special_character_ratio":0.27476636,"punctuation_ratio":0.10705596,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9983958,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-22T14:19:34Z\",\"WARC-Record-ID\":\"<urn:uuid:4da95de2-08b4-4ea6-ae48-5c0ee5c729e6>\",\"Content-Length\":\"14867\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:48bffd54-57ad-4720-bcc0-aefde56e81b3>\",\"WARC-Concurrent-To\":\"<urn:uuid:4d3504e6-cc7a-4519-99c6-eeeda9fdf77d>\",\"WARC-IP-Address\":\"173.247.245.106\",\"WARC-Target-URI\":\"https://www.excelfunctions.net/excel-sumsq-function.html\",\"WARC-Payload-Digest\":\"sha1:Q5A7PXWGCTJ77QBAFEKL4IM5RUOJVCRW\",\"WARC-Block-Digest\":\"sha1:KOSI3PHGWAAIGRT27A4DM4GISSAFOMAB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250607118.51_warc_CC-MAIN-20200122131612-20200122160612-00130.warc.gz\"}"}
https://brilliant.org/problems/the-well-known-mu-puzzle/	[ "# The Well-known MU Puzzle\n\nLogic Level 3", null, "In Douglas Hofstadter's book, Godel, Escher, Bach, he proposes the following puzzle about formal systems.\n\nAxiom: MI is an axiom.\n\nTheorem: The axiom is a theorem by default. All the (and only all the) strings that can be derived using the following transformations on a theorem is also a theorem.\n\n Formal Rule Informal Description Example xI → xIU Changing any terminal I to IU MII → MIIU Mx → Mxx Double the string after M MIU → MIUIU xIIIy → xUy Replace three Is with an U MIIIU → MUU xUUy → xy Remove an occurence of double Us MUUI → MI\n\nProblem: Is MU a theorem?\n\nGuessing the answer to this problem is not very difficult. However, an interested problem solver could explore this questions:\n\n• If your answer was yes, what is the proof of the MU theorem, i.e, what is the sequence of transformation rules that will take you from MI to MU?\n• Is the MIU system decidable, i.e, is there a computer program to decide if a given string is a theorem?\n• If not, is the system semidecidable, i.e, is there a computer program that will eventually tell if a string is a theorem, but never tell if it isn't?\n×" ]	[ null, "https://ds055uzetaobb.cloudfront.net/brioche/uploads/Wv4nGOQDyF-11-waterfall.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83074456,"math_prob":0.836749,"size":1066,"snap":"2021-31-2021-39","text_gpt3_token_len":299,"char_repetition_ratio":0.1054614,"word_repetition_ratio":0.051813472,"special_character_ratio":0.239212,"punctuation_ratio":0.16814159,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9973989,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-05T00:48:30Z\",\"WARC-Record-ID\":\"<urn:uuid:2df5faed-97f5-46ab-bc8d-c64cb932c5b1>\",\"Content-Length\":\"35119\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e032337a-67a5-4f7f-a379-01cf98734050>\",\"WARC-Concurrent-To\":\"<urn:uuid:4148af5a-29d5-43de-ba19-eeaa024a8428>\",\"WARC-IP-Address\":\"104.20.35.242\",\"WARC-Target-URI\":\"https://brilliant.org/problems/the-well-known-mu-puzzle/\",\"WARC-Payload-Digest\":\"sha1:PGYHCAAECJHHNTSYR2DUJZNT3O3WWG4W\",\"WARC-Block-Digest\":\"sha1:2YYZYCHKDV7BTR2DWEIYPVRA7DZMIIWT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046155268.80_warc_CC-MAIN-20210805000836-20210805030836-00560.warc.gz\"}"}
https://www.arxiv-vanity.com/papers/1707.00104/	[ "# On Nörlund summation and Ergodic Theory, with applications to power series of Hilbert contractions.\n\nChristophe Cuny Université de la Nouvelle-Calédonie Equipe ERIM, B.P. 4477, F-98847 Noumea Cedex and Michel Weber IRMA, 10 rue du Général Zimmer, 67084 Strasbourg Cedex, France\n###### Abstract.\n\nWe show that if is a good weight for the dominated weighted ergodic theorem in , , then the Nörlund matrix , is bounded on . We study the regularity (convergence in norm, almost everywhere) of operators in ergodic theory: power series of Hilbert contractions, and power series of -contractions, and establish similar tight relations with the Nörlund operator associated to the modulus coefficient sequence .\n\n47A35, 47D37\n\n.\n\n## 1. Introduction\n\nLet be a sequence of complex or real numbers (we take the convention . We associate with an infinite matrix , called a Nörlund matrix, in the following way. For every , set . We then define\n\n (1) {aij:=ai−j/Aiif 0≤j≤i and Ai>0,aij:=0if j>i or Ai=0.\n\nSome authors consider instead , assuming then that it does not vanish.\n\nThen induces naturally a (possibly unbounded) operator on for any . The matter of deciding whether this operator is bounded on some (or any) is far from being solved. As noted by Bennett , it seems, so far, that the best general known condition guaranteeing that is bounded on any , , is that , see for instance Borwein and Cass . That condition is realized when, for instance is a non-increasing sequence of positive numbers.\n\nLet\n\n (2) ∑i≥0\|1Aii∑j=0ai−juj\|p≤Cpp∑i≥0\|ui\|p.\n\nEquivalently, we have the dual formulation: for any sequence , ,\n\n (3) ∑j≥0\|∑i≥jai−jvi/Ai\|q≤Cqp∑j≥0\|vj\|q.\n\nThe latter is easily seen to be also equivalent to: for any sequence ,\n\n (4) ∑j≥0\|∑i≥jai−jvi\|q=∑j≥0\|∑i≥0aivi+j\|q≤Cqp∑j≥0\|Ajvj\|q.\n\nMoreover, it follows from (3) that\n\n (5) ∑i≥0\|ai\|pApi≤Cpp,\n\nwhere has to be interpreted as when .\n\nWe show that Nörlund matrices are connected with two different topics from ergodic theory. We establish tight relations between regularity (convergence in norm, almost everywhere) of operators in ergodic theory (power series of Hilbert contractions, power series of -contractions, dominated weighted ergodic theorems, and naturally associated Nörlund matrices. We obtain conditions ensuring norm convergence of power series of Hilbert contractions, and also almost everywhere convergence of power series of -contractions. These conditions are expressed in terms of the Nörlund operator associated to the modulus coefficient sequence .\n\n## 2. Nörlund matrices and dominated weighted ergodic theorems\n\nWe first observe a connection between Nörlund matrices and dominated weighted ergodic theorems.\n\nWe say that a sequence , of complex numbers, is good for the dominated weighted ergodic theorem in , , if there exists such that for every dynamical system , writing , we have\n\n (6) ∥supn≥01An\|n∑k=0akf∘τk\|∥Lp(ν)≤C∥f∥Lp(ν)∀f∈Lp(ν).\n\nHere again we take the convention that if .\n\nThe next lemma is well-known, it is in the spirit of the so-called Conze principle, see for instance [29, Th. 5.4.3]. It states a converse of Calderon’s transference principle.\n\n###### Lemma 1.\n\nLet be good for the dominated weighted ergodic theorem in margin: , . Then, with the best constant appearing in (6), we have for every ,\n\n (7) ∑i∈Z(supn≥01An\|n∑j=0ajvi+j\|)p≤Cp∑i∈Z\|vj\|p.\n\nProof. Let . Let be integers (one has in mind that ). Take , and the transformation given by if and . Define on , by for every . By (6), we have\n\n 12N+1N+1−M∑i=−N(sup0≤m≤M1Am\|m∑k=0akvi+k\|)p≤∥supn≥01An\|n∑k=0akf∘τk\|∥pLp(ν) (8) ≤Cp∥f∥pLp(ν)=Cp2N+1N∑i=−N\|vi\|p.\n\nMultiplying (8) by , letting first and then , we derive (7).\n\nRemark. Our proof is based on the use of the dominated weighted ergodic theorem on periodic systems (the rotations on ). To give a proof based on the dominated weighted ergodic theorem on a single (but ergodic and non-atomic) dynamical system, one could use Rohlin’s lemma (see for instance Weber [29, p. 270] for a statement of the lemma).\n\nWe deduce the following.\n\n###### Proposition 2.\n\nLet be a good weight for the dominated weighted ergodic theorem in , . Then, the Nörlund matrix is bounded on . Moreover, for every non-increasing sequence of nonnegative numbers , writing , is bounded on .\n\nRemark. It is unclear whether ” bounded on ” implies ” bounded on ”, in general.\n\nProof. Let . Define as follows. if and if . Using (7) and for every the trivial estimate\n\n 1Ai\|i∑j=0ajv−i+j\|≤supn≥01An\|n∑j=0ajv−i+j\|,\n\nwe infer that\n\n ∑i≥0(1Ai\|i∑j=0ajv−i+j\|)p≤Cp∑i∈Z\|vi\|p=Cp∑i≥0\|ui\|p.\n\nUsing that when , we derive that is bounded on .\n\nTo prove the last assertion, one just has to notice that, using Abel summation, is a good weight for the dominated weighted ergodic theorem. .\n\nOf course, as one can see from the above proof, the fact that be a good weight for the dominated weighted ergodic theorem in is a much stronger statement than the fact that be bounded on . Hence, Proposition 2 should not be seen as a method to prove boundedness of some Nörlund matrices, but as a source of examples of Nörlund matrices, since there are many examples of sequences that are known to be good for the dominated weighted ergodic theorem. We provide some of them below. One may also consult the survey by Bellow and Losert for dominated weighted ergodic theorems with bounded weights. More arithmetical sequences may be found in Cuny and Weber .\n\nExamples. The following sequences are good for the dominated weighted ergodic theorem in , for every :\n\n(Bourgain and Wierdl, , ) Let be the set of prime numbers and take , for every .\n\n(Bourgain, ) Let be the set of squares and take , for every .\n\n(Cuny and Weber, ) Take and for every take , the number of divisors of . We now give an example which does not work on every , . Let be an ergodic dynamical system. Let , for some .\n\n(Bourgain, Demeter, Lacey, Tao and Thiele, , and ) There exists with such that for every , is good for the dominated weighted ergodic theorem in for every such that .\n\nLet us notice that none of the above examples satisfies the previously mentionned criterium: . The fact that the Nörlund matrix associated with the sequence in example is bounded has been proved by Borwein .\n\n## 3. Norm convergence of power series of Hilbert contractions\n\nLet be a contraction of a (real or complex) Hilbert space . Given a sequence of complex numbers and , we are interested in finding conditions involving sufficient for the norm convergence of .\n\nAn obvious condition is the following\n\n (9) ∑n∈N\|an\|∥Pnf∥H<∞.\n\nSufficient conditions involving have been obtained when is unitary (i.e. ) or, more generally, normal (i.e. ), if moreover is regular (at least nonnegative and nonincreasing). Let us mention the papers and , see also for some versions.\n\nRecall that, see for instance Nagy and Foias (see also Shäffer for an explicit matrix construction), admits a unitary dilation, that is, there exist another Hilbert space , with , and a unitary operator on such that for every , where is the orthogonal projection onto .\n\nWe start with some simple lemmas. The first one appears in Cuny and Lin , but we recall the short proof.\n\n###### Lemma 3.\n\nFor every and every , the spaces and are orthogonal (in ).\n\nProof. Let . Let and . We have\n\n ⟨(U−nPn−U−n−1Pn+1)f,U−n−ℓPn+ℓg⟩K=⟨UℓPnf,Pn+ℓg⟩K−⟨Uℓ−1Pn+1f,Pn+ℓg⟩K =⟨Pn+ℓf,Pn+ℓg⟩K−⟨Pn+ℓf,Pn+ℓg⟩K=0.\n\n###### Lemma 4.\n\nLet be such that as . Then, for every , . In particular, for any positive and non-decreasing sequence , the following are equivalent (setting ).\n\n• ;\n\n• .\n\nRemarks. Notice that by Kronecker’s lemma, if holds as . In particular, since is non decreasing, . Item is satisfied if .\n\nProof. Since , for every , we have, (with convergence in )\n\n (10) Pnf=∑k≥0(U−kPn+kf−U−k−1Pn+k+1f).\n\nBy the above lemma the terms of that series lie in orthogonal spaces. Hence,\n\n ∥Pnf∥2K =∑k≥0∥U−kPn+kf−U−k−1Pn+k+1∥2K =∑k≥n∥U−kPkf−U−k−1Pk+1∥2K,\n\nwhere we used that is unitary (and a change of variable) for the last identity. Then, the equivalence of and follows by Fubini.\n\nGiven a sequence of complex numbers , consider the following conditions\n\n (11) ∑n∈N\|an\|(n∑k=0\|ak\|)∥Pnf∥2H<∞, (12) ∑n∈N(n∑k=0\|ak\|)2∥U−nPnf−U−n−1Pn+1f∥2K<∞\n\nBy Lemma (4), when , (11) and (12) are equivalent. Assume that (9) holds. Then, since is nonincreasing, and (11) holds. Hence, (11) is always weaker than (9).\n\n###### Proposition 5.\n\nLet be such that be bounded on where . Let be such that either of conditions (11) or (12) hold. Then, the series converges in .\n\nProof. Since is bounded on , then by (5) (with )\n\n (13) ∑n∈Na2nA2n<∞.\n\nLet be integers and write . For every , let and , where . Finally, let . By Lemma 4 and using that is unitary, we have\n\n (14) ∥Vp,qf∥2K=∑n∈N∥U−nPnVp,qf−U−n−1Pn+1Vp,qf∥2K≤∑n∈N(q∑k=p\|ak\|un+k)2.\n\nBy Cauchy’s criteria one has to prove that as . Using the Lebesgue dominated theorem for the counting measure on , it suffices to prove that\n\n (15) q∑k=p\|ak\|un+k⟶p,q→+∞0,\n\nand that\n\n (16) ∑n∈N(∑k≥0\|ak\|un+k)2.\n\nThe convergence (15) follows from Cauchy-Schwarz combined with the assumed condition (12) and (13).\n\nFTo prove (16), it suffices to notice that\n\n ∑n∈N(∑k≥0\|ak\|un+k)2≤∑n∈N(∑k≥n\|ak−n\|uk)2=∥N∗\|a\|v∥2ℓ2(N) ≤∥N∗\|a\|∥2∥v∥2ℓ2(N)=∥N\|a\|∥2∑n∈NA2nu2n.\n\nThe proposition has been proved in in the case where . An important case corresponds to the situation where for every . Then, the proposition gives a sufficient condition (namely ) for to be a coboundary (i.e. for some ). This sufficient condition has been obtained independently by Volný in the special case where is a Markov operator on . His proof (which does not appeal to the notion of Nörlund matrices) is essentially the same, since the shift on the space of trajectories of the associated Markov chain plays the role of the unitary dilation.\n\n###### Proposition 6.\n\nLet . Assume that for every contraction on a Hilbert space the following property holds : ”If (11) holds for some then converges in ”. Then, is bounded on .\n\nProof. Let be a contraction on a Hilbert space satisfying the above property. Let . Then, is a Hilbert space and we define an operator on , by setting for every . Then, by the Banach-Steinhaus, theorem is continuous. Hence, there exists , such that .\n\nLet us prove the proposition. We give a probabilistic proof. Let be the probability space given by , the product -algebra and , with . Let be the shift on and be the coordinate process. In particular, and is iid.\n\nDenote . Set and and define two operators and on and respectively by for every and for every (then is a Markov operator). Clearly, is a unitary dilation of . Let and define . Assume moreover that , or equivalently (by Lemma 4), . Notice that and that . Moreover, . Hence, , i.e. (4) holds with , and the proof is complete.\n\nWe shall now prove that Proposition 5 cannot be improved.\n\n###### Definition 1.\n\nWe say that a contraction on is Ritt if .\n\n###### Proposition 7.\n\nLet be a contraction on . For every , consider the following properties.\n\n• The series converges in ;\n\n• .\n\nThen, . If moreover is Ritt then .\n\nRemark. By , when is a positive operator on then of the proposition implies that the series converges -almost everywhere and the associated maximal function is in . The fact that has been proved by Cohen, Cuny and Lin using results from Arhancet and Le Merdy ) when and is a positive Ritt contraction of some (there are also analogous results in in ).\n\nProof. The fact that is a direct application of Proposition 5. Assume that is a Ritt operator and that converges in .\n\nWe start with the case . By Proposition 4.6 of Cohen, Cuny and Lin (see also their example (v) page 8), we have\n\n ∑n≥0∥Pf+⋯+P2nf∥2H22αn<∞,\n\nThen, using (3) of Cohen, Cuny and Lin combined with Lemma 13 below, we infer that , which finishes the proof, in that case.\n\nAssume now that . Let . Then, . Hence, by Theorem 8.1 of Le Merdy ,\n\n ∑n∈Nn∥Pnf∥2H=∑n∈Nn∥Pn(I−P)g∥2H≤∥g∥H,\n\nwhich is the desired result.\n\n## 4. Almost everywhere convergence of power series of L2-contractions\n\nOnce norm convergence has been proven, one may wonder, in the case where , whether almost everywhere convergence holds. As mentionned in the remark following Proposition 7, for ”regular” sequences, if is a positive contraction of then norm convergence implies almost everywhere convergence. However, as we shall see below (see Proposition 10), there is no such result for contractions that are not positive. Let us mention that the almost everywhere convergence of power series (for regular ) for unitary or normal operators on has been proven under conditions involving in and , see also for -versions.\n\n###### Theorem 8.\n\nLet be such that be bounded on where . Let . Let be a contraction on . Let such that\n\n (17) ∑n≥1(log(n+1))2A22n+1∥P2nf∥2L2(m)<∞.\n\nThen, the series converges -almost everywhere and\n\n supN≥1∣∣N∑n=0anPnf∣∣∈L2(m).\n\nRemark. A sufficient condition for (17) is the following\n\n (18) ∑n≥1(loglog(n+3))2A24nn+1∥Pnf∥2L2(m)<∞.\n\nProof. Let . We have\n\n max2N≤n≤2N+1−1\|n∑k=2NanPkf\|≤2N+1−1∑k=2N\|ak\|\|Pkf\|.\n\nHence,\n\n ∑N∈N∥max2N≤n≤2N+1−1\|n∑k=2NanPkf\|∥2L2(m)≤∑N∈NA22N+1∥P2Nf∥2L2(m)<∞.\n\nIn particular, it suffices to prove that converges and that .\n\nBy (14), for every , we have\n\n (19) ∥2q−1∑n=2panPnf∥2L2(m)≤∑n∈N(2q−1∑k=2p\|ak\|un+k)2.\n\nSet and notice that is super-additive in the following sense: for every , . By Proposition 2.2 of Cohen and Lin , there exists , such that for every ,\n\n ∥max22n≤m≤22n+1−1\|m∑k=22nakPkf\|∥2L2(m)≤C(n+1)2d(2n,2n+1−1).\n\nAssume that\n\n (20) ∑n≥0(n+1)2d(2n,2n+1−1)<∞.\n\nThen, using (19) and Cauchy-Schwarz we see that\n\n (∑n∈N∥22n+1−1∑k=22nakPkf∥L2(m))2≤∑n∈N1(n+1)2∑n∈N(n+1)2d(2n,2n+1−1).\n\nThis finishes the proof, provided that we can show (20).\n\n ∑n∈N∑ℓ≥0(ℓ+1)2(22ℓ+1−1∑k=22ℓ\|ak\|un+k)2<∞.\n\nUsing that , we infer that\n\n ∑n∈N∑ℓ≥0(ℓ+1)2(22ℓ+1−1∑k=22ℓ\|ak\|un+k)2≤∑n∈N(∑k≥0(loglog(k+3))2\|ak\|un+k)2 ≤∑n∈N(∑k≥0(loglog(n+k+3))2\|ak\|un+k)2.\n\nThen, proceeding as in the (end of the) proof of Proposition 5 we see that (20) holds provided that\n\n ∑n∈N(loglog(n+3))2A2nu2n<∞,\n\nwhich follows from (17) using that is non-decreasing and that .\n\n###### Corollary 9.\n\nLet be an ergodic dynamical system. Let for some . There exists with such that for every , setting the following holds: for every , every contraction on and every such that\n\n ∑n∈N(loglog(n+3))2(n+1)1−2α∥Pnf∥22<∞,\n\nthe sequence converges -almost everywhere and the associated maximal function if in .\n\nProof. Let and let . Let be the set appearing in the example . Modifying if necessary we may assume that , for some finite . Then, for every , is good for the dominated weighted ergodic theorem. Applying Proposition 2, we see that, with , is bounded on . Set (we see as a function on ). By Theorem 8 (see the remark after the theorem), we are back to prove that" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9266614,"math_prob":0.98087263,"size":12640,"snap":"2023-14-2023-23","text_gpt3_token_len":2997,"char_repetition_ratio":0.1460114,"word_repetition_ratio":0.0750111,"special_character_ratio":0.24628164,"punctuation_ratio":0.15569824,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9963343,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-01T18:20:10Z\",\"WARC-Record-ID\":\"<urn:uuid:f9253468-bb53-4af5-bc53-ef9131165af4>\",\"Content-Length\":\"1049382\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5b925104-1a4d-4447-9805-3de0bb1c8dc1>\",\"WARC-Concurrent-To\":\"<urn:uuid:1bf778e2-c840-455a-9160-98468b5045c5>\",\"WARC-IP-Address\":\"104.21.14.110\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/1707.00104/\",\"WARC-Payload-Digest\":\"sha1:2WMBS6V6BLBWQUHEGIOXPYTYHZ7K5O6Z\",\"WARC-Block-Digest\":\"sha1:AJYB6K3OSBRYXO5W72Y7HBZZTB2OGDWV\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224648000.54_warc_CC-MAIN-20230601175345-20230601205345-00388.warc.gz\"}"}
https://answers.yahoo.com/question/index?qid=20090506143256AAPQMGw	[ "# If 113 is a prime number, how would you factor x^2+2x-113=0?\n\nOne of my algebra assignments was to find two consecutive integers such that the sum of their square is 113. I made the equation x(x+2)=113 and distributed, and got x^2+2x=113. Then I subtracted 113 from each side to get the Zero-Product property. But now, the only factors of 113 is 113 x 1, and that doesn't... show more One of my algebra assignments was to find two consecutive integers such that the sum of their square is 113. I made the equation x(x+2)=113 and distributed, and got x^2+2x=113. Then I subtracted 113 from each side to get the Zero-Product property. But now, the only factors of 113 is 113 x 1, and that doesn't add up to be 2. Please help! What am I doing wrong?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9645905,"math_prob":0.9838614,"size":809,"snap":"2019-35-2019-39","text_gpt3_token_len":226,"char_repetition_ratio":0.10310559,"word_repetition_ratio":0.69863015,"special_character_ratio":0.3065513,"punctuation_ratio":0.11111111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998915,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-20T03:47:20Z\",\"WARC-Record-ID\":\"<urn:uuid:a128f633-ed5a-4465-9f6f-94e498c5a1b3>\",\"Content-Length\":\"131977\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:239ba90b-42c8-4ec5-937e-92fca0fb635e>\",\"WARC-Concurrent-To\":\"<urn:uuid:57c4185b-918a-40b4-83f3-02a0b15d07d9>\",\"WARC-IP-Address\":\"69.147.92.11\",\"WARC-Target-URI\":\"https://answers.yahoo.com/question/index?qid=20090506143256AAPQMGw\",\"WARC-Payload-Digest\":\"sha1:W3GDPH5HNEF5K3RNDZCB6IIRBEXPMDW6\",\"WARC-Block-Digest\":\"sha1:5RCE3JZDWJYPGQWYOSKSYXVW4LYIATMN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573827.2_warc_CC-MAIN-20190920030357-20190920052357-00405.warc.gz\"}"}
https://dml.cz/handle/10338.dmlcz/108051	[ "# Article\n\nKeywords:\nperiodic solutions; second order functional difference equation; fixed-point theorem; growth condition\nSummary:\nSufficient conditions for the existence of at least one \\$T-\\$periodic solution of second order nonlinear functional difference equations are established. We allow \\$f\\$ to be at most linear, superlinear or sublinear in obtained results.\nReferences:\n Atici F. M., Gusenov G. Sh.: Positive periodic solutions for nonlinear difference equations with periodic coefficients. J. Math. Anal. Appl. 232 (1999), 166–182. MR 1683041\n Atici F. M., Cabada A.: Existence and uniqueness results for discrete second order periodic boundary value problems. Comput. Math. Appl. 45 (2003), 1417–1427. MR 2000606 \| Zbl 1057.39008\n Deimling K.: Nonlinear Functional Analysis. Springer-Verlag, New York, 1985. MR 0787404 \| Zbl 0559.47040\n Guo Z., Yu J.: The existence of periodic and subharmonic solutions for second order superlinear difference equations. Science in China (Series A) 3 (2003), 226–235. MR 2014482\n Jiang D., O’Regan D., Agarwal R. P.: Optimal existence theory for single and multiple positive periodic solutions to functional difference equations. Appl. Math. Lett. 161 (2005), 441–462. MR 2112417 \| Zbl 1068.39009\n Kocic V. L., Ladas G.: Global behivior of nonlinear difference equations of higher order with applications. Kluwer Academic Publishers, Dordrecht-Boston-London, 1993. MR 1247956\n Ma M., Yu J.: Existence of multiple positive periodic solutions for nonlinear functional difference equations. J. Math. Anal. Appl. 305 (2005), 483–490. MR 2130716 \| Zbl 1070.39019\n Mickens R. E.: Periodic solutions of second order nonlinear difference equations containing a small parameter-II. Equivalent linearization. J. Franklin Inst. B 320 (1985), 169–174. MR 0818865 \| Zbl 0589.39004\n Mickens R. E.: Periodic solutions of second order nonlinear difference equations containing a small parameter-III. Perturbation theory. J. Franklin Inst. B 321 (1986), 39–47. MR 0825907 \| Zbl 0592.39005\n Mickens R. E.: Periodic solutions of second order nonlinear difference equations containing a small parameter-IV. Multi-discrete time method. J. Franklin Inst. B 324 (1987), 263–271. MR 0910641 \| Zbl 0629.39002\n Raffoul Y. N.: Positive periodic solutions for scalar and vector nonlinear difference equations. Pan-American J. Math. 9 (1999), 97–111.\n Wang Y., Shi Y.: Eigenvalues of second-order difference equations with periodic and antiperiodic boundary conditions. J. Math. Anal. Appl. 309 (2005), 56–69. MR 2154027 \| Zbl 1083.39019\n Zeng Z.: Existence of positive periodic solutions for a class of nonautonomous difference equations. Electronic J. Differential Equations 3 (2006), 1–18. MR 2198916 \| Zbl 1093.39014\n Zhang R., Wang Z., Chen Y., Wu J.: Periodic solutions of a single species discrete population model with periodic harvest/stock. Comput. Math. Appl. 39 (2000), 77–90. MR 1729420 \| Zbl 0970.92019\n Zhu L., Li Y.: Positive periodic solutions of higher-dimensional functional difference equations with a parameter. J. Math. Anal. Appl. 290 (2004), 654–664. MR 2033049 \| Zbl 1042.39005" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5937239,"math_prob":0.95154285,"size":2681,"snap":"2021-31-2021-39","text_gpt3_token_len":808,"char_repetition_ratio":0.18229361,"word_repetition_ratio":0.07692308,"special_character_ratio":0.37001118,"punctuation_ratio":0.26537785,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9837594,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-24T02:50:53Z\",\"WARC-Record-ID\":\"<urn:uuid:f755f9be-9e3e-4602-b48e-7fcebd3d420d>\",\"Content-Length\":\"17084\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:06c1e6cd-6779-4c7e-8eb3-560cf5777909>\",\"WARC-Concurrent-To\":\"<urn:uuid:e2567bb7-6f84-4ac8-b671-9b938755a128>\",\"WARC-IP-Address\":\"147.251.6.150\",\"WARC-Target-URI\":\"https://dml.cz/handle/10338.dmlcz/108051\",\"WARC-Payload-Digest\":\"sha1:FQ62NJH5ZJ6WQYFC6TJFIAVG2OJINKPU\",\"WARC-Block-Digest\":\"sha1:FPOTECYPSQUI3RI7ZZHVKBVWXECDZBQW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046150067.87_warc_CC-MAIN-20210724001211-20210724031211-00279.warc.gz\"}"}
http://redwoodsmedia.com/fraction-decimal-percent-worksheets/	[ "Home - Fraction Decimal Percent Worksheets\n\n# Fraction Decimal Percent Worksheets\n\nHere is the Fraction Decimal Percent Worksheets section. Here you will find all we have for Fraction Decimal Percent Worksheets. For instance there are many worksheet that you can print here, and if you want to preview the Fraction Decimal Percent Worksheets simply click the link or image and you will take to save page section.\n\nMath Worksheets Converting Fractions To Decimals Muzjikmandiafo 311 Best School Math Images On Pinterest Fractions Decimals And Percents Worksheets Grade Into Convert Sixth Grade Decimals A Number Line Worksheet 05 E Page Worksheets Converting Fractions Decimals And Percentages Resources Fractions Decimals And Percents Worksheets Croefit Basic Percentages Worksheets Simple Grade 7 Math Fractions Decimals Paring And Ordering Fractions And Decimals Worksheet Worksheets Converting Fractions To Percentages Word Problems Worksheets Grade 5 Percentage For 5th Fraction Decimal Converting Percents Worksheet Percent Worksheets Word Problems 5th Paring And Ordering Fractions And Decimals Worksheet Worksheets Percentage Worksheets For Grade 5 Postjoint Sixth Grade Decimals A Number Line Worksheet 05 E Page Worksheets Math Worksheets Converting Fractions To Decimals Muzjikmandiafo." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.61131746,"math_prob":0.8579123,"size":1248,"snap":"2019-43-2019-47","text_gpt3_token_len":239,"char_repetition_ratio":0.29180065,"word_repetition_ratio":0.1744186,"special_character_ratio":0.15384616,"punctuation_ratio":0.027624309,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.994011,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-14T02:39:25Z\",\"WARC-Record-ID\":\"<urn:uuid:7318bd54-1aa1-4a42-826e-1063a2750ff2>\",\"Content-Length\":\"38664\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4af62086-16c1-4895-a4af-3cc7932d153c>\",\"WARC-Concurrent-To\":\"<urn:uuid:3b277143-8ccc-4416-a279-25cb76647352>\",\"WARC-IP-Address\":\"104.24.99.31\",\"WARC-Target-URI\":\"http://redwoodsmedia.com/fraction-decimal-percent-worksheets/\",\"WARC-Payload-Digest\":\"sha1:NEV7NVHF5TOX5MEAQ3SVV4QKUEUOCKQ2\",\"WARC-Block-Digest\":\"sha1:PGYVFKR2S62LWSYM3CK33YPWMHTTTOX7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496667767.6_warc_CC-MAIN-20191114002636-20191114030636-00521.warc.gz\"}"}
https://fantasyplayer.link/program/%E8%87%AA%E5%88%B6%E8%84%9A%E6%9C%AC%E8%AF%AD%E8%A8%804.1-%E6%95%B0%E5%80%BC%E8%BF%90%E7%AE%97/	[ "## 详细内容\n\n``````a = 1;\nb = 2;\n\nc = a + b;\nd = a - b;\ne = a / d;\nf = a * b;\ng = a % d;\n\nh = a ^ b;\ni = a \| b;\nj = a & b;\nk = ~ a ;\n``````\n\n### 基本逻辑\n\n1. 取出N个数字\n2. 进行运算\n3. 将结果放回去\n\nN个操作数，产生一个结果的运算符（+，-，，/等）的第一步和第三步的内容基本是一致的。\n\n### 基于Map 的实现逻辑\n\n1. 从Map中取出key为a, b 的数字。\n\n2. 在宿主语言中进行运算 (c/cpp/java/…) 然后获得结果\n\n3. 将结果保存到Map中。（Key为 c)\n\n``````// java 代码，代码并没有经过测试，只是解释思路使用\n// c = a + b;\n\npublic class Test {\n\npublic void test (){\n// key: 变量名， value: 变量值\nMap<String, Object> map = new HashMap<>();\n// 第一个操作数\nString first = \"a\";\n// 第二个操作数\nString second = \"b\";\n\nString result = \"c\";\n\n// 1\nObject a = map.get(first);\nObject b = map.get(second);\n\n// null check 和类型check 这里就省略了。\n\n// 2\n// Number 是一个自定义类型，因为篇幅问题，对声明进行了省略。\nNumber c = Number.add( ((Number) a), ((Number) b) );\n\n// 3\nmap.put( result, c );\n}\n\n}\n``````\n\n### 基于栈的实现逻辑\n\n1. 取出 a 对应的数值，进行压栈\n2. 取出 b 对应的数值，进行压栈\n3. 执行加号指令 \| 这里应该是基于栈的核心部分\n1. 取出栈顶的2个元素\n2. 进行相加\n3. 将结果压栈\n4. 将栈顶元素出栈，放入到c对应的位置。\n\n``````// 同样使用 Java做示例，代码只是演示说明使用\n// c = a + b ;\n\npublic class Test {\n\n// 局部变量表\nprivate Object[] localVarTable = new Object ;\n\n// 栈\nprivate Stack<Object> stack = new Stack<>();\n\n/\n 根据变量名获取变量对应的局部变量表的索引\n/\npublic int getIndex(String name){\n// todo\nreturn 0;\n}\n\n/\n 执行加号指令\n/\npublic void add(){\n// check size 和别的check 暂时都直接省略了\n\nNumber a = (Number) stack.pop();\nNumber b = (Number) stack.pop();\n\n// 运算，并将结果入栈\nstack.push( Number.add(a,b) );\n\n}\n\npublic void test(){\n\n// 第一个操作数\nString first = \"a\";\n// 第二个操作数\nString second = \"b\";\n\nString result = \"c\";\n\n// 1, 2\n// 因为要节省篇幅，所以大部分的 error check 就不做了\nObject a = localVarTable[getIndex(first)];\nObject b = localVarTable[getIndex(second)];\n\n// 入栈\nstack.push(a);\nstack.push(b);\n\n// 3\nadd();\n\n// 4\nObject c = stack.pop();\nlocalVarTable[getIndex(result)] = c;\n\n}\n\n}\n\n``````\n\n### 基于寄存器的实现逻辑\n\n1. 取出变量a的值，放入a 寄存器\n2. 取出变量b的值，放入b 寄存器\n3. 执行加法运算，将结果放入c寄存器\n4. 取出c寄存器的值，放回局部变量表\n5. 清空所有寄存器\n\n``````// c = a + b ;\n\npublic class Test {\n\n// 局部变量表\nprivate Object[] localVarTable = new Object ;\n\n// 寄存器\nprivate Object a = null;\nprivate Object b = null;\nprivate Object c = null;\n\n/\n 根据变量名获取变量对应的局部变量表的索引\n/\npublic int getIndex(String name){\n// todo\nreturn 0;\n}\n\n/\n 清理寄存器\n/\npublic void clearReg(){\na = b = c = null;\n}\n\n/\n 执行加法运算，运算逻辑总是将寄存器 a,b 相加，并把值放入寄存器 c\n/\npublic void add(){\n// 将会省略 check\n\nc = Number.add( (Number) a, (Number) b ) ;\n}\n\n/\n 这里的变量名，改用传参的形式\n*/\npublic void test(String first,String second, String result ){\n\n// 1, 2\na = localVarTable[getIndex(first)];\nb = localVarTable[getIndex(second)];\n\n// 3\nadd();\n\n// 4\nlocalVarTable[getIndex(result)] = c;\n\n// 5\nclearReg();\n}\n\n}\n\n``````" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.79165316,"math_prob":0.9934028,"size":2980,"snap":"2021-04-2021-17","text_gpt3_token_len":1678,"char_repetition_ratio":0.13205644,"word_repetition_ratio":0.18871595,"special_character_ratio":0.34731543,"punctuation_ratio":0.15811089,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99383813,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-18T06:32:31Z\",\"WARC-Record-ID\":\"<urn:uuid:a14afda2-462e-4d21-8c06-f804f472d219>\",\"Content-Length\":\"29418\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9cf83828-c681-4b17-a5e4-e4b3908d6c6d>\",\"WARC-Concurrent-To\":\"<urn:uuid:9ec165cb-2d85-421a-b9f9-52b39e64d4e2>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"https://fantasyplayer.link/program/%E8%87%AA%E5%88%B6%E8%84%9A%E6%9C%AC%E8%AF%AD%E8%A8%804.1-%E6%95%B0%E5%80%BC%E8%BF%90%E7%AE%97/\",\"WARC-Payload-Digest\":\"sha1:CHQG7WSKFK56ARV4KEBNLDFAHH4ZOEUY\",\"WARC-Block-Digest\":\"sha1:LAFFOZCVIPBY5THHMMIB7575TC5VIITD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038468066.58_warc_CC-MAIN-20210418043500-20210418073500-00439.warc.gz\"}"}