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https://answers.everydaycalculation.com/percent-of/4-425 | [
"# Answers\n\nSolutions by everydaycalculation.com\n\n## What is 4 percent of 425?\n\n4% of 425 is 17\n\n#### Working out 4% of 425\n\n1. Write 4% as 4/100\n2. Since, finding the fraction of a number is same as multiplying the fraction with the number, we have\n4/100 of 425 = 4/100 × 425\n3. Therefore, the answer is 17\n\nIf you are using a calculator, simply enter 4÷100×425 which will give you 17 as the answer.\n\nMathStep (Works offline)",
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"Download our mobile app and learn how to work with percentages in your own time:\nAndroid and iPhone/ iPad\n\n#### Find another\n\nWhat is % of\n\n© everydaycalculation.com"
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https://physics.stackexchange.com/questions/206210/otto-and-carnot-efficiency-comparison | [
"Otto and Carnot efficiency comparison\n\nLets say both engines operate between same temperature limits.\n\nCarnot eff: $$e_\\text{Carnot} = 1-\\frac{T_L}{T_H}$$ Otto eff: $$e_\\text{Otto} = 1-\\frac{T_4-T_1}{T_3-T_2}$$ assuming that for Otto cycle points are located as below on a P-V diagram: $$\\array{ 3 & 4 \\\\ 2 & 1 }$$ I assumed the following for the Temperature limits. \\begin{align} T_3&=T_H \\\\ T_1&=T_L \\end{align} Question is: is there any way that the Otto cycle can be more efficient than the Carnot? (If e.x. $T_4$ will be close to $T_1$ enough, so the numerator of the Otto efficiency will become zero. Then the Otto cycle is more efficient, but as far as I know the most efficient is Carnot)\n\nPlease also suggest how to use the $T_4$ and $T_2$. (or get rid of them to calculate eff)\n\n• Where did the formula for the Otto cycle efficiency come from? The derivation on Wikipedia looks different. – CuriousOne Sep 10 '15 at 15:30\n• Welcome to Physics.SE! I've prettied up your mathematics a bit; hit the 'edit' button to see how. – rob Sep 22 '15 at 3:11\n• The statement that the Carnot efficiency is maximal is known as Carnot's theorem. – ACuriousMind Jun 26 '16 at 21:59\n\nOtto cycle consists of two isochoric and two isentropic processes.",
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"If $T_4$ approaches to $T_1$, then $T_3$ will approach to $T_2$, because for Otto cycle, line $3\\to 4$ in $T-s$ diagram must always be vertically.\n\nThus, although the numerator of the Otto efficiency approaches to zero, but the fraction doesn't approach to zero, because the denominator approaches to zero too and we will have an indeterminate form.\n\nwell , the thermal efficiency of otto cycle can be reduced to\n\ne (otto) = 1- T1 / T2 when T1= TL\n\ne (carnot)= 1- T1 / T3\n\nT3 is greater than T2 , then carnot is still higher efficiency than a\n\nreversible otto cycle"
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https://www.tutorialspoint.com/finding-inverse-hyperbolic-cosine-of-specified-number-in-golang | [
"# Finding Inverse Hyperbolic Cosine of Specified Number in Golang\n\nGolang is a statically typed, compiled programming language that is popular among developers for its simplicity, concurrency support, and garbage collection. In this article, we will discuss how to find the inverse hyperbolic cosine of a specified number in Golang.\n\nThe inverse hyperbolic cosine function is used to find the angle whose hyperbolic cosine is equal to a given number. It is denoted as acosh(x), where x is the input value.\n\nGolang provides the math package that includes various mathematical functions, including the Inverse Hyperbolic Cosine function. We can use the math.Acosh() function to find the inverse hyperbolic cosine of a specified number in Golang.\n\n## Syntax\n\nThe syntax for using the math.Acosh() function is as follows −\n\nfunc Acosh(x float64) float64\n\n\nThe Acosh() function takes a float64 value as input and returns its inverse hyperbolic cosine as a float64 value.\n\n## Example 1\n\nLet's take an example to understand how to use the Acosh() function in Golang.\n\npackage main\n\nimport (\n\"fmt\"\n\"math\"\n)\n\nfunc main() {\nx := 1.5\n\n// Finding Inverse Hyperbolic Cosine of x\nresult := math.Acosh(x)\n\nfmt.Printf(\"Inverse Hyperbolic Cosine of %.2f is %.2f\", x, result)\n}\n\n\n## Output\n\nInverse Hyperbolic Cosine of 1.50 is 0.96\n\n\nIn the above example, we have used the math.Acosh() function to find the inverse hyperbolic cosine of 1.5. The output is 0.96\n\n## Example 2\n\npackage main\n\nimport (\n\"fmt\"\n\"math\"\n)\n\nfunc main() {\nx := 3.14\ny := 1 / math.Sqrt(x*x-1)\nresult := math.Log(x + y)\n\nfmt.Printf(\"The inverse hyperbolic cosine of %.2f is %.2f\\n\", x, result)\n}\n\n\n## Output\n\nThe inverse hyperbolic cosine of 3.14 is 1.25\n\n\nIn this example, we first define the value of x as 3.14. We then calculate the value of y using the formula 1 / sqrt(x^2 - 1). Finally, we calculate the inverse hyperbolic cosine of x using the formula ln(x + y) and store the result in the variable result. We then print out the result using the fmt.Printf function.\n\n## Conclusion\n\nIn this article, we have discussed how to find the inverse hyperbolic cosine of a specified number in Golang using the math.Acosh() function. We have also provided an example to illustrate how to use this function in Golang.\n\nUpdated on: 12-Apr-2023\n\n55 Views",
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https://excelbaby.com/learn/vba-len-function/ | [
"# Len function\n\nReturns a Long containing the number of characters in a string or the number of bytes required to store a variable.\n\n## Syntax\n\n`Len(string | varname)`\n\nThe Len function syntax has these parts:\n\nPart Description\nstring Any valid string expression. If string contains Null, Null is returned.\nvarname Any valid variable name. If varname contains Null, Null is returned. If varname is a Variant, Len treats it the same as a String and always returns the number of characters it contains.\n\n## Remarks\n\nOne (and only one) of the two possible arguments must be specified. With user-defined types, Len returns the size as it will be written to the file.\n\n## Example\n\nThe first example uses Len to return the number of characters in a string or the number of bytes required to store a variable. The Type...End Type block defining `CustomerRecord` must be preceded by the keyword Private if it appears in a class module. In a standard module, a Type statement can be Public.\n\n``````Type CustomerRecord ' Define user-defined type.\nID As Integer ' Place this definition in a\nName As String * 10 ' standard module.\nAddress As String * 30\nEnd Type\n\nDim Customer As CustomerRecord ' Declare variables.\nDim MyInt As Integer, MyCur As Currency\nDim MyString, MyLen\nMyString = \"Hello World\" ' Initialize variable.\nMyLen = Len(MyInt) ' Returns 2.\nMyLen = Len(Customer) ' Returns 42.\nMyLen = Len(MyString) ' Returns 11.\nMyLen = Len(MyCur) ' Returns 8.``````\n\nThe second example uses LenB and a user-defined function (LenMbcs) to return the number of byte characters in a string if ANSI is used to represent the string.\n\n``````Function LenMbcs (ByVal str as String)\nLenMbcs = LenB(StrConv(str, vbFromUnicode))\nEnd Function\n\nDim MyString, MyLen\nMyString = \"ABc\"\n' Where \"A\" and \"B\" are DBCS and \"c\" is SBCS.\nMyLen = Len(MyString)\n' Returns 3 - 3 characters in the string.\nMyLen = LenB(MyString)\n' Returns 6 - 6 bytes used for Unicode.\nMyLen = LenMbcs(MyString)\n' Returns 5 - 5 bytes used for ANSI.``````"
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https://school.stmarysparish.org/math-grade8 | [
"# Math\n\n• ## Algebra 1\n\nPresently, Saint Mary School is in the early stages of a curriculum review cycle. Saint Mary School has adopted the rigorous academic curriculum set forth by the Archdiocese of Hartford. In accordance with our mission, the Archdiocese of Hartford Curriculum reflects a continuum of learning from prekindergarten through high school graduation, and the standards are based on the teachings of the Catholic Church, infused with Catholic social teaching, rooted in basic skills, and immersed in technology and 21st century literacies. Furthermore, all curriculum documents are in alignment with national and state standards and exceed the Common Core State Standards.\n\nALGEBRA STANDARDS\n\nA1.Understand and describe patterns and functional relationships\n\nA2. Represent and analyze quantitative relationships in a variety of ways\n\nA3.Use operations, properties and algebraic symbols to determine equivalence and solve problems A4.Use properties and characteristics of two- and three-dimensional shapes and geometric theorems to describe relationships, communicate ideas, and solve problems\n\nA5.Use spatial reasoning, location, and geometric relationships to solve problems\n\nA6. Develop and apply units, systems, formulas and appropriate tools to estimate and measure\n\nExpressions Equations, Functions, and Inequalities\n\nStudents will:\n\n· interpret parts of an expression, such as terms, factors, and coefficients\n\n· make and justify predictions based on patterns\n\n· apply the appropriate properties of real numbers and the steps for order of operations to write, evaluate, simplify, add, subtract, multiply, and divide expressions\n\n· derive the formula for the sum of a finite geometric series and use to solve problems\n\n· understand that a function, y = f(x), is a rule that assigns to each input (domain) exactly one output (range) - the graph of a function is the set of ordered pairs consisting of an input and the corresponding\n\n· use function notation to evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context\n\n· solve one-variable linear equations and\n\n· analyze and solve linear equations, functions, and pairs of linear equations and functions\n\nProblem Solving\n\n· rational equations\n\nStatistics and Probability\n\n· create equations and inequalities in one or two variables and use them to solve problems\n\n· solve standard word problems using one or two variables including:\n\n· interpret the solution to identify the number of acceptable solutions (e.g., extraneous solutions)\n\n· evaluate solutions for reasonableness, accuracy, and completeness\n\n· Identify and interpret data with exponential behavior.\n\n· investigate patterns of association in two-variable data"
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https://gambling-promotion.codes/what-is-odd-betting/ | [
"# What is Odd in Betting?\n\nCommercial content | New Customers Only | 18+\n\nThe odds are used to calculate the profit associated with a bet. It is the inverse of the probability of an outcome. The lower the odds, the higher the probability of an outcome: Odds = 1 / Probability.\n\n## The different types of odds\n\nEuropean odds: These are odds in the form of a whole number or a decimal number (e.g. 1.5 or 3). To calculate the gain from the European odds, use the formula:\nEarnings\n= Bet amount x (Odds – 1)\n\nEnglish or fractional stake: This is a stake in the form of a fraction (e.g. 1/2 or 2/1). In order to calculate the profit associated with an English quota, use the following formula: Profit = Bet amount x Quota.\n\nAmerican stake: This is a stake that can be expressed as a positive or negative number (for example : -200 or 200). Negative odds indicate the stake required to obtain a profit of 100; positive odds indicate the profit corresponding to a stake of 100. To calculate the profit linked to an American stake, use the formulae: Gain = Amount wagered x 100 / (-share) for negative odds OR Gain = Amount wagered x 100 /share for positive odds"
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http://www.global-sci.org/intro/article_detail/getBib?article_id=179 | [
"@Article{AAMM-3-470, author = {Fayssal Benkhaldoun, Mohammed Seaïd and Slah Sahmim}, title = {Mathematical Development and Verification of a Finite Volume Model for Morphodynamic Flow Applications}, journal = {Advances in Applied Mathematics and Mechanics}, year = {2011}, volume = {3}, number = {4}, pages = {470--492}, abstract = {\n\nThe accuracy and efficiency of a class of finite volume methods are investigated for numerical solution of morphodynamic problems in one space dimension. The governing equations consist of two components, namely a hydraulic part described by the shallow water equations and a sediment part described by the Exner equation. Based on different formulations of the morphodynamic equations, we propose a family of three finite volume methods. The numerical fluxes are reconstructed using a modified Roe's scheme that incorporates, in its reconstruction, the sign of the Jacobian matrix in the morphodynamic system. A well-balanced discretization is used for the treatment of the source terms. The method is well-balanced, non-oscillatory and suitable for both slow and rapid interactions between hydraulic flow and sediment transport. The obtained results for several morphodynamic problems are considered to be representative, and might be helpful for a fair rating of finite volume solution schemes, particularly in long time computations.\n\n}, issn = {2075-1354}, doi = {https://doi.org/10.4208/aamm.10-m1056}, url = {http://global-sci.org/intro/article_detail/aamm/179.html} }"
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https://www.allkidsnetwork.com/best/top-4th-grade-multiplication-kids-activities | [
"## Multiplication Word Problems Worksheet\n\nUse multiplication to solve the 5 word problems in this worksheet.\n\n• Tuesday, April 25, 2017\n• All Kids Network\n• 1,141 Visits\n\n## Simple Multiplication Word Problems Worksheet\n\nUse simple multiplication to solve the 5 word problems.\n\n• Tuesday, April 25, 2017\n• All Kids Network\n• 1,523 Visits\n\n## Simple Multiplication Word Problems Worksheet\n\nUse simple multiplication to solve the 5 word problems.\n\n• Tuesday, April 25, 2017\n• All Kids Network\n• 2,147 Visits\n\n## Multiplication Word Problems\n\nPractice multiplication and problem solving with these multiplication word problems worksheets.\n\n• Tuesday, April 25, 2017\n• All Kids Network\n• 4,651 Visits\n\n## Multiple Digit Multiplication Worksheets\n\nThe problems in this collection of worksheets are a little more challenging. The multiplication problems in these range from three digit numbers times two digit numbers to four digit numbers multiplied by three digit numbers.\n\n• Tuesday, March 10, 2015\n• All Kids Network\n• 29,788 Visits"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8338849,"math_prob":0.7353201,"size":500,"snap":"2020-45-2020-50","text_gpt3_token_len":84,"char_repetition_ratio":0.21370968,"word_repetition_ratio":0.16901408,"special_character_ratio":0.166,"punctuation_ratio":0.074074075,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9706957,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-24T15:22:01Z\",\"WARC-Record-ID\":\"<urn:uuid:c7b48089-7d2c-4c35-9002-e19ad22b14fe>\",\"Content-Length\":\"87089\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:29ec1832-eba5-4c2c-a881-ac6cf2b1847c>\",\"WARC-Concurrent-To\":\"<urn:uuid:03a7e5de-7954-4e83-bdfb-5c9d4f5b1415>\",\"WARC-IP-Address\":\"18.210.183.196\",\"WARC-Target-URI\":\"https://www.allkidsnetwork.com/best/top-4th-grade-multiplication-kids-activities\",\"WARC-Payload-Digest\":\"sha1:VC7LOEIYVTE7VBJLYCKCSM42J7QEMENK\",\"WARC-Block-Digest\":\"sha1:Z55RVWTTXC23EETOGGQDPRHCSK6MHUMU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141176864.5_warc_CC-MAIN-20201124140942-20201124170942-00031.warc.gz\"}"} |
https://hackerranksolution.in/detectonlydupliucatesinalistamazon/ | [
"# Detect the Only Duplicate in a List - Amazon Top Interview Questions\n\n### Problem Statement :\n\n```You are given a list nums of length n + 1 picked from the range 1, 2, ..., n. By the pigeonhole principle, there must be a duplicate. Find and return it. There is guaranteed to be exactly one duplicate.\n\nBonus: Can you do this in O(n) time and O(1) space?\n\nConstraints\n\nn ≤ 10,000\n\nExample 1\n\nInput\nnums = [1, 2, 3, 1]\n\nOutput\n1\n\nExample 2\n\nInput\nnums = [4, 2, 1, 3, 2]\n\nOutput\n2```\n\n### Solution :\n\n``` ```Solution in C++ :\n\nint solve(vector<int>& nums) {\nint tortoise = nums, hare = nums;\n\ndo {\ntortoise = nums[tortoise];\nhare = nums[nums[hare]];\n} while (tortoise != hare);\n\ntortoise = nums;\n\nwhile (tortoise != hare) {\ntortoise = nums[tortoise];\nhare = nums[hare];\n}\n\nreturn hare;\n}```\n```\n\n``` ```Solution in Java :\n\nimport java.util.*;\n\nclass Solution {\npublic int solve(int[] nums) {\nint res = 0;\nfor (int i = 0; i < nums.length; i++) res ^= i ^ nums[i];\nreturn res;\n}\n}```\n```\n\n``` ```Solution in Python :\n\nclass Solution:\ndef solve(self, nums):\nq = sum(nums)\nn = len(nums)\nv = (n - 1) * (n) // 2\nreturn q - v```\n```\n\n## Queries with Fixed Length\n\nConsider an -integer sequence, . We perform a query on by using an integer, , to calculate the result of the following expression: In other words, if we let , then you need to calculate . Given and queries, return a list of answers to each query. Example The first query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is . The secon\n\n## QHEAP1\n\nThis question is designed to help you get a better understanding of basic heap operations. You will be given queries of types: \" 1 v \" - Add an element to the heap. \" 2 v \" - Delete the element from the heap. \"3\" - Print the minimum of all the elements in the heap. NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct element\n\nJesse loves cookies. He wants the sweetness of all his cookies to be greater than value K. To do this, Jesse repeatedly mixes two cookies with the least sweetness. He creates a special combined cookie with: sweetness Least sweet cookie 2nd least sweet cookie). He repeats this procedure until all the cookies in his collection have a sweetness > = K. You are given Jesse's cookies. Print t\n\n## Find the Running Median\n\nThe median of a set of integers is the midpoint value of the data set for which an equal number of integers are less than and greater than the value. To find the median, you must first sort your set of integers in non-decreasing order, then: If your set contains an odd number of elements, the median is the middle element of the sorted sample. In the sorted set { 1, 2, 3 } , 2 is the median.\n\n## Minimum Average Waiting Time\n\nTieu owns a pizza restaurant and he manages it in his own way. While in a normal restaurant, a customer is served by following the first-come, first-served rule, Tieu simply minimizes the average waiting time of his customers. So he gets to decide who is served first, regardless of how sooner or later a person comes. Different kinds of pizzas take different amounts of time to cook. Also, once h\n\n## Merging Communities\n\nPeople connect with each other in a social network. A connection between Person I and Person J is represented as . When two persons belonging to different communities connect, the net effect is the merger of both communities which I and J belongs to. At the beginning, there are N people representing N communities. Suppose person 1 and 2 connected and later 2 and 3 connected, then ,1 , 2 and 3 w"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8749174,"math_prob":0.9710435,"size":3645,"snap":"2023-40-2023-50","text_gpt3_token_len":908,"char_repetition_ratio":0.10958528,"word_repetition_ratio":0.005891016,"special_character_ratio":0.2617284,"punctuation_ratio":0.14496644,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95796144,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-25T18:47:13Z\",\"WARC-Record-ID\":\"<urn:uuid:dbe108b9-a396-4360-8504-6d2cd58313d5>\",\"Content-Length\":\"41018\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4d91de05-ee09-434b-8024-e81b79c5dc6d>\",\"WARC-Concurrent-To\":\"<urn:uuid:8c3a5e29-3309-4d31-a967-c4a7b960ce88>\",\"WARC-IP-Address\":\"18.213.98.197\",\"WARC-Target-URI\":\"https://hackerranksolution.in/detectonlydupliucatesinalistamazon/\",\"WARC-Payload-Digest\":\"sha1:CHNEOY3CR3FD7WLMN6SHJFEQVFJ3UDNP\",\"WARC-Block-Digest\":\"sha1:OIRD6ZSDZ2Q6G6HASFGQK2BQJUWIGXNT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510085.26_warc_CC-MAIN-20230925183615-20230925213615-00098.warc.gz\"}"} |
https://www.ques10.com/p/1804/mechanical-vibrations-question-paper-dec-2013-me-1/ | [
"Question Paper: Mechanical Vibrations : Question Paper Dec 2013 - Mechanical Engineering (Semester 6) | Mumbai University (MU)\n2\n\n## Mechanical Vibrations - Dec 2013\n\n### Mechanical Engineering (Semester 6)\n\nTOTAL MARKS: 100\nTOTAL TIME: 3 HOURS\n(1) Question 1 is compulsory.\n(2) Attempt any four from the remaining questions.\n(3) Assume data wherever required.\n(4) Figures to the right indicate full marks.\n1 (a) Fig. 1 shows a rectangular block of mass 'M' resting on top of a semi-cylindrical surface. If block is slightly tipped at same end, find frequency of oscillations.",
null,
"(8 marks)\n1 (b) A door along with 250cm high, 80cm wide and 4.5cm thick and weighing 40kg is fitted with an automatic door closer. The door opens against a spring with a modulus of 1kg-cm/radian. If the door is opened 90° and released, how long with the door to be within 1° of clearing (Return of door is critically damped).(8 marks) 1 (c) Write a note on critical speed of shaft.(4 marks) 2 (a) The clender bar in the system is rotated 5° from equilibrium and released. Determine time independent response of the system.",
null,
"m=2kg, L=80cm, r=10cm, K=20000N/m, C=300Ns/m.\n(12 marks)\n2 (b) An accelerometer has natural frequency 15kHz. Determine the highest frequency it can measure with 1% accuracy. Assume damping ratio 0.7.(8 marks) 3 (a) An aircraft instrument of mass 20kg is to be isolated from engine vibration. The engine runs at speed ranging from 1800rpm to 2500rpm. Natural rubber isolator with negligible damping is used. Determine the rubber stiffness for 90% isolation.(10 marks) 3 (b) Using Lagrange's equation, find the frequency of the system shown in fig. 3",
null,
"(10 marks)\n4 (a) Find the Eigen values and Eigen vectors of the system shown in fig. 4",
null,
"(12 marks)\n4 (b) Derive the equation of motion and find the steady state amplitude of mass 'm'.",
null,
"(8 marks)\n5 (a) The natural frequency of transverse vibration of beam in fig. 6 is 20rad/s. Find the natural frequency of vibration if another 30N is added at 40cm from the left support.\nHint:",
null,
"where Uij is the deflection of section i due to unit load at section j and Si is the distance of section i from left support and Zj is the distance of section j from right support.",
null,
"(15 marks)\n5 (b) Explain the 'motion transmissibility' with frequency response curve and formulae.(5 marks) 6 (a) A shaft carries five masses A, B, C, D and E which revolve at the same radius in planes which are equidistant from one another. The magnitude of masses in plane A, C and D are 50kg, 40kg and 80kg respectively. The angle between A and C is 90° and that between C and D is 135°. Determine the magnitude of masses in planes B and E and their positions to put the shaft in complete rotating balance.(10 marks) 6 (b) Explain follow jump phenomenon in cam.(6 marks) 6 (c) What do you mean by static and dynamic balancing?(4 marks) 7 (a) A five cylinder in-line engine running at 750rpm has successive cranks 144° apart, the distance between cylinder centre-lines being 375mm. The piston stroke is 225mm and the ratio of connecting rod to the crank is 4. Estimate the engine for balance of primary and secondary forces and couples. Find the maximum values of these and positions of central crank at which these maximum values occur. The reciprocating mass for each cylinder is 15kg.(10 marks) 7 (b) A helical spring has both ends securely fixed one vertically above the other and a mass is attached to the spring at some intermediate point. Show that the frequency at vibrations is minimum when the load point is at the centre of fixed ends(y=0.5l) where y=distance between fixed end and load point and l=distance between fixed ends.(6 marks) 7 (c) What is the basic principle used in Holzer's method.(4 marks)\n\n written 3.5 years ago by",
null,
"Team Ques10 ♦♦ 410"
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"https://i.imgur.com/zn4pUhg.png",
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"https://i.imgur.com/BrK9Bpq.png",
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"https://i.imgur.com/coLsl12.png",
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"https://i.imgur.com/hKqOm2Y.png",
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"https://i.imgur.com/7q70N6g.png",
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"https://i.imgur.com/5kQs5NV.png",
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"https://i.imgur.com/1nTR9oH.png",
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"https://secure.gravatar.com/avatar/7e863a59a8cfbcdf477c438aa01ad97a",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.88482803,"math_prob":0.9550269,"size":3610,"snap":"2019-35-2019-39","text_gpt3_token_len":924,"char_repetition_ratio":0.11758181,"word_repetition_ratio":0.009569378,"special_character_ratio":0.2628809,"punctuation_ratio":0.08815427,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9837366,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-22T07:18:14Z\",\"WARC-Record-ID\":\"<urn:uuid:644d9eae-05a6-45e9-a09b-b62df7f47e6a>\",\"Content-Length\":\"26419\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cdd59f65-118b-4531-9981-2de1af5f4a8c>\",\"WARC-Concurrent-To\":\"<urn:uuid:96bb470a-4750-48ed-a998-36c8dcd8e2b3>\",\"WARC-IP-Address\":\"104.31.92.239\",\"WARC-Target-URI\":\"https://www.ques10.com/p/1804/mechanical-vibrations-question-paper-dec-2013-me-1/\",\"WARC-Payload-Digest\":\"sha1:MUKUHOR2GX7AKIZCR4SGRDUETWHBUYOS\",\"WARC-Block-Digest\":\"sha1:4ABHRQYW6LUMYTPMYJ6YMVSAPM5KUHZK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514575168.82_warc_CC-MAIN-20190922053242-20190922075242-00426.warc.gz\"}"} |
http://chowkafat.net/Enumeration27.html | [
"# 點算的奧秘：非齊次遞歸關係的解\n\n《點算的奧秘：齊次遞歸關係的解》中，筆者介紹了求解「常系數一元 線性齊次遞歸關係」的方法，在本網頁筆者將介紹求解「常系數一元線性非齊次遞歸關係」的方法。在這裡首 先重溫「非齊次遞歸關係」的定義。當我們把一個「遞歸關係」寫成標準形式，即把含有an、 an−1 ...等的項統統移到等號左邊，並把其餘的項移到等號右邊後，如果在等號右邊的項不 等於0，則該「遞歸關係」是「非齊次」的。我們把這個在等號右邊的非零項稱為「非齊次遞歸關係」的「非齊 次部分」。\n\n「未定系數法」乃基於以下假設：「非齊次遞歸關係」的「特解」與該「遞歸關係」的「非齊次部分」具有相 同的形式，兩者僅在系數方面有差異。因此，假如給定「遞歸關係」的「非齊次部分」是「指數函數」3 × 2n，我們便假設pn也具有相同的形式，即pn = A × 2n；假如給定「遞歸關係」的「非齊次部分」是「3階多項式」n3 − 1，我們便 假設該「遞歸關係」的「特解」也具有相同的形式，即pn = An3 + Bn2 + Cn + D (註2)。假如給定「遞歸關係」的「非齊次部分」具有較複雜的形式(例如混合了「指數函數」和「多項 式」)，則pn亦要具有同類型的複雜形式，現把這些情況總結成下表：\n\nan = 2an−1 + n − 1\n\nan − 2an−1 = n − 1\n\n An + B = 2[A(n − 1) + B] + n − 1 = 2An − 2A + 2B + n − 1 = (2A + 1)n − 2A + 2B − 1\n\n A = 2A + 1 B = − 2A + 2B − 1\n\npn = −n − 1 □\n\nan = gn + pn\n\na1 = 0\n\nan − 2an−1 = 0\n\nx − 2 = 0\n\ngn = C × 2n\n\nan = C × 2n − n − 1\n\n a1 = C × 21 − 1 − 1 即，2C − 2 = 0\n\nan = 2n − n − 1\n\na4 = 24 − 4 − 1 = 11\n\n an = 2an−1 + 3n−1 − 2n−1 ，若n > 1 a1 = 0\n\nan − 2an−1 = 3n−1 − 2n−1\n\nx − 2 = 0\n\ngn = C × 2n\n\npn = A × 3n + B × 2n (註3)\n\npn = A × 3n + Bn × 2n\n\n A3n + Bn2n = 2[A3n−1 + B(n − 1)2n−1] + 3n−1 − 2n−1 3A(3n−1) + 2Bn(2n−1) = (2A + 1)3n−1 + (−2B − 1)2n−1 + 2Bn(2n−1) 3A(3n−1) = (2A + 1)3n−1 + (−2B − 1)2n−1\n\n 3A = 2A + 1 0 = −2B − 1\n\n pn = 3n − n / 2 × 2n = 3n − n × 2n−1\n\nan = C × 2n + 3n − n × 2n−1\n\n a1 = C × 20 + 30 − 0 × 20−1 即，C + 1 = 0\n\nan = −2n + 3n − n × 2n−1\n\na4 = −24 + 34 − 4 × 24−1 = 33\n\n an − 4an−1 + 5an−2 − 2an−3 = 2 + 2n，若n > 2 a0 = 6，a1 = 20，a2 = 52\n\nx3 − 4x2 + 5x − 2 = 0\n\n(x − 1)2(x − 2) = 0\n\n gn = C × 1n + Dn × 1n + E × 2n = C + Dn + E × 2n\n\npn = A + B × 2n\n\npn = An2 + Bn × 2n\n\n An2 + Bn2n − 4[A(n − 1)2 + B(n − 1)2n−1] + 5[A(n − 2)2 + B(n − 2)2n−2] − 2[A(n − 3)2 + B(n − 3)2n−3] = An2 + Bn2n − 4An2 + 8An − 4A − 4Bn2n−1 + 4B2n−1 + 5An2 − 20An + 20A + 5Bn2n−2 − 10B2n−2 − 2An2 + 12An − 18A − 2Bn2n−3 + 6B2n−3 = (1 − 4 + 5 − 2)An2 + (8 − 20 + 12)An + (−4 + 20 − 18)A + (8 − 16 + 10 − 2)Bn2n−3 + (16 − 20 + 6)B2n−3 = −2A + 2B2n−3\n\n−2A + 2B2n−3 = 2 + 8 × 2n−3\n\n pn = −n2 + 4n × 2n = −n2 + n × 2n+2\n\nan = C + Dn + E × 2n − n2 + n × 2n+2\n\n a0 = C + D × 0 + E × 20 − 02 + 0 × 20+2 即，C + E = 6\n\n a1 = C + D × 1 + E × 21 − 12 + 1 × 21+2 即，C + D + 2E + 7 = 20\n\n a2 = C + D × 2 + E × 22 − 22 + 2 × 22+2 即，C + 2D + 4E + 28 = 52\n\n an = 2 + 3n + 4 × 2n − n2 + n × 2n+2 = 2 + 3n − n2 + (1 + n) × 2n+2\n\na3 = 2 + 3 × 3 − 32 + (1 + 3) × 23+2 = 130\na4 = 2 + 3 × 4 − 42 + (1 + 4) × 24+2 = 318\n\na3 = 4a2 − 5a1 + 2a0 + 2 + 23 = 4 × 52 − 5 × 20 + 2 × 6 + 2 + 23 = 130\na4 = 4a3 − 5a2 + 2a1 + 2 + 24 = 4 × 130 − 5 × 52 + 2 × 20 + 2 + 24 = 318",
null,
"",
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"http://chowkafat.net/Picture/homepage.gif",
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| {"ft_lang_label":"__label__zh","ft_lang_prob":0.92916745,"math_prob":1.0000097,"size":4707,"snap":"2019-13-2019-22","text_gpt3_token_len":5118,"char_repetition_ratio":0.1146077,"word_repetition_ratio":0.0416,"special_character_ratio":0.4172509,"punctuation_ratio":0.0034324944,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998803,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-20T11:51:08Z\",\"WARC-Record-ID\":\"<urn:uuid:6e806cd6-4ad7-4fb9-b522-454d87c02db7>\",\"Content-Length\":\"17863\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0308ac0d-baa6-4676-bd86-cb5f68447b99>\",\"WARC-Concurrent-To\":\"<urn:uuid:25b3d182-f1c2-4163-82a4-7cf3a14fcd04>\",\"WARC-IP-Address\":\"98.137.244.36\",\"WARC-Target-URI\":\"http://chowkafat.net/Enumeration27.html\",\"WARC-Payload-Digest\":\"sha1:R5HV5SN3CRRSL6PPGOSXAOD6UCJHQ5OF\",\"WARC-Block-Digest\":\"sha1:XRQPYBZKMF5VPGK5TTZRI2I327JWTQ3R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202326.46_warc_CC-MAIN-20190320105319-20190320131319-00552.warc.gz\"}"} |
https://brilliant.org/problems/abcd-5432/ | [
"ABCD 5432\n\nGiven 4 positive integers $a, b, c$ and $d$ such that $a^5 =b^4, c^3 = d^2$ and $c-a = 19$, what is the value of $d-b$?\n\n×"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6240605,"math_prob":1.0000099,"size":271,"snap":"2019-43-2019-47","text_gpt3_token_len":106,"char_repetition_ratio":0.13857678,"word_repetition_ratio":0.0,"special_character_ratio":0.38745388,"punctuation_ratio":0.26136363,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000068,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-18T10:11:01Z\",\"WARC-Record-ID\":\"<urn:uuid:8a185292-b86f-4958-8245-ab5d41cb0206>\",\"Content-Length\":\"41533\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d5b242a8-0187-4e93-b7e1-b3b45d4ae773>\",\"WARC-Concurrent-To\":\"<urn:uuid:f27c2cfe-8c86-4ba2-ae40-36fde27b2341>\",\"WARC-IP-Address\":\"104.20.35.242\",\"WARC-Target-URI\":\"https://brilliant.org/problems/abcd-5432/\",\"WARC-Payload-Digest\":\"sha1:Y67WCIBBM4PBX6S4UDFXZ7ASITREQFY4\",\"WARC-Block-Digest\":\"sha1:XJUOSQ4JJWXKKCQMPANBIEO5NL5ROLFV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986679439.48_warc_CC-MAIN-20191018081630-20191018105130-00287.warc.gz\"}"} |
https://au.mathworks.com/help/simulink/slref/model_cmd.html | [
"# model\n\nExecute particular phase of simulation of model\n\n## Syntax\n\n```[sys,x0,str,ts] = model([],[],[],'sizes');\n[sys,x0,str,ts] = model([],[],[],'compile');\noutputs = model(t,x,u,'outputs');\nderivs = model(t,x,u,'derivs');\ndstates = model(t,x,u,'update');\nmodel([],[],[],'term');\n\n```\n\n## Description\n\nThe `model` command executes a specific phase of the simulation of a Simulink® model whose name is `model`. The command's last argument (`flag`) specifies the phase of the simulation to be executed. See Simulation Phases in Dynamic Systems for a description of the steps that Simulink software uses to simulate a model.\n\nThis command ignores the effects of state transitions and conditional execution. Therefore, it is not suitable for models which have such logic. Use this command for models which can be represented as simple dynamic systems. Such systems should meet these requirements.\n\n• All states in the model must be built-in non-bus data types. For a discussion on built-in data types, see About Data Types in Simulink.\n\n• If you are using vector format to specify the state, this command can access only non-complex states of `double` data type.\n\n• There is minimal amount of state logic (Stateflow, conditionally executed subsystems etc.)\n\n• The models are not mixed-domain models. That is, most blocks in the model are built-in Simulink blocks and do not include user-written S-functions or blocks from other Sim* products.\n\nFor models which do not comply with these requirements, using this command can cause Simulink to produce results which can only be interpreted by further analyzing and simplifying the model.\n\nNote\n\nThe state variable `x` can be represented in structure as well as vector formats. The variable follows the limitations of the format in which it is specified.\n\nThis command is also not intended to be used to run a model step-by-step, for example, to debug a model. Use the Simulink debugger if you need to examine intermediate results to debug a model.\n\n## Arguments\n\n `sys` Vector of model size data:`sys(1)` = number of continuous states`sys(2)` = number of discrete states`sys(3)` = number of outputs`sys(4)` = number of inputs`sys(5)` = reserved`sys(6)` = direct-feedthrough flag (1 = yes, 0 = no)`sys(7)` = number of Continuous, Discrete, Fixed in minor step, and Controllable sample times (= number of rows in `ts`) `x0` Vector containing the initial conditions of the system's states `str` Vector of names of the blocks associated with the model's states. The state names and initial conditions appear in the same order in `str` and `x0`, respectively. `ts` An `m`-by-`2` matrix containing the sample time (period, offset) information of the Continuous, Discrete, Fixed in minor step, and Controllable sample times in the model.For more information on the sample times in Simulink, see Types of Sample Time. outputs Outputs of the model at time step `t`. derivs Derivatives of the continuous states of the model at time `t`. dstates States of the model at time `t` returned as either a structure or an array. Simulink returns a structure when the model has states and `x` is either empty `([])` or in structure format. Otherwise, Simulink returns an array. If the return type is a vector or array, Simulink returns real double discrete states only.If the return type is a structure, Simulink returns a structure that contains both continuous and discrete states of built-in types only. Non-built-in types are omitted. `t` Time step, specified as real double in scalar format. `x` State vector, specified as real double in structure or vector format. `u` Inputs, specified as real double in vector format. `flag` Specification of the simulation phase to be executed:`'sizes'` executes the size computation phase of the simulation. This phase determines the sizes of the model's inputs, outputs, state vector, etc.`'compile'` executes the compilation phase of the simulation. The compilation phase propagates signal and sample time attributes.`'update'` computes the next values of the model's discrete states.`'outputs'` computes the outputs of the model's blocks at time `t`.`'derivs'`computes the derivatives of the model's continuous states at time step `t`.`'term'` causes Simulink software to terminate simulation of the model.NoteThe `output`, `update`, and `derivs` flags are valid only for single-tasking models. For more information on single-tasking and multi-tasking, see Tasking Modes (Simulink Coder).\n\n## Examples\n\nThe following command executes the compilation phase of the `vdp` model that comes with Simulink software.\n\n```vdp([], [], [], 'compile') ```\n\nThe following command terminates the simulation initiated in the previous example.\n\n```vdp([], [], [], 'term') ```\n\nNote\n\nSimulink does not let you close a model while it is compiling or simulating. For all phases except the `'sizes'` phase, before closing the model, you must invoke the model command with the `'term'` argument."
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https://www.proofwiki.org/wiki/Definition:Extended_Real_Number_Space | [
"# Definition:Topology on Extended Real Numbers\n\n## Definition\n\nLet $\\overline \\R$ denote the extended real numbers.\n\nThe (standard) topology on $\\overline \\R$ is the order topology $\\tau$ associated to the ordering on $\\overline \\R$.\n\n### Extended Real Number Space\n\nThe topological space $\\left({\\overline \\R, \\tau}\\right)$ may be referred to as the extended real number space.\n\n## Also see\n\n• Results about the extended real number space can be found here."
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https://www.pythonf.cn/read/14255 | [
"# 设计和历史常见问题-Python教程\n\n• 设计和历史常见问题\n\n• 为什么Python使用缩进来分组语句？\n\n• 为什么简单的算术运算得到奇怪的结果？\n\n• 为什么浮点计算不准确？\n\n• 为什么Python字符串是不可变的？\n\n• 为什么必须在方法定义和调用中显式使用“self”？\n\n• 为什么不能在表达式中赋值？\n\n• 为什么Python对某些功能（例如list.index()）使用方法来实现，而其他功能（例如len(List)）使用函数实现？\n\n• 为什么 join()是一个字符串方法而不是列表或元组方法？\n\n• 异常有多快？\n\n• 为什么Python中没有switch或case语句？\n\n• 难道不能在解释器中模拟线程，而非得依赖特定于操作系统的线程实现吗？\n\n• 为什么lambda表达式不能包含语句？\n\n• 可以将Python编译为机器代码，C或其他语言吗？\n\n• Python如何管理内存？\n\n• 为什么CPython不使用更传统的垃圾回收方案？\n\n• CPython退出时为什么不释放所有内存？\n\n• 为什么有单独的元组和列表数据类型？\n\n• 列表是如何在CPython中实现的？\n\n• 字典是如何在CPython中实现的？\n\n• 为什么字典key必须是不可变的？\n\n• 为什么 list.sort() 没有返回排序列表？\n\n• 如何在Python中指定和实施接口规范？\n\n• 为什么没有goto？\n\n• 为什么原始字符串（r-strings）不能以反斜杠结尾？\n\n• 为什么Python没有属性赋值的“with”语句？\n\n• 为什么 if/while/def/class语句需要冒号？\n\n• 为什么Python在列表和元组的末尾允许使用逗号？\n\n## 为什么Python使用缩进来分组语句？\n\nGuido van Rossum 认为使用缩进进行分组非常优雅，并且大大提高了普通Python程序的清晰度。大多数人在一段时间后就学会并喜欢上这个功能。\n\n```if (x <= y)\nx++;\ny--;\nz++;\n```\n\n## 为什么浮点计算不准确？\n\n>>>\n```>>> 1.2 - 1.0\n0.19999999999999996\n```\n\nCPython 中的 ``` float ``` 类型使用C语言的 ``` double ``` 类型进行存储。 ``` float ``` 对象的值是以固定的精度（通常为 53 位）存储的二进制浮点数，由于 Python 使用 C 操作，而后者依赖于处理器中的硬件实现来执行浮点运算。 这意味着就浮点运算而言，Python 的行为类似于许多流行的语言，包括 C 和 Java。\n\n>>>\n```>>> x = 1.2\n```\n\n``` x ``` 存储的值是与十进制的值 ``` 1.2 ``` (非常接近) 的近似值，但不完全等于它。 在典型的机器上，实际存储的值是：\n\n```1.0011001100110011001100110011001100110011001100110011 (binary)\n```\n\n```1.1999999999999999555910790149937383830547332763671875 (decimal)\n```\n\n## 为什么不能在表达式中赋值？\n\n```while chunk := fp.read(200):\nprint(chunk)\n```\n\n## 为什么Python对某些功能（例如list.index()）使用方法来实现，而其他功能（例如len(List)）使用函数实现？\n\n(a) 对于某些操作，前缀表示法比后缀更容易阅读 -- 前缀（和中缀！）运算在数学中有着悠久的传统，就像在视觉上帮助数学家思考问题的记法。比较一下我们将 x*(a+b) 这样的公式改写为 x*a+x*b 的容易程度，以及使用原始OO符号做相同事情的笨拙程度。\n\n(b) 当读到写有len(X)的代码时，就知道它要求的是某件东西的长度。这告诉我们两件事：结果是一个整数，参数是某种容器。相反，当阅读x.len()时，必须已经知道x是某种实现接口的容器，或者是从具有标准len()的类继承的容器。当没有实现映射的类有get()或key()方法，或者不是文件的类有write()方法时，我们偶尔会感到困惑。\n\nhttps://mail.python.org/pipermail/python-3000/2006-November/004643.html\n\n## 为什么 join()是一个字符串方法而不是列表或元组方法？\n\n```\", \".join(['1', '2', '4', '8', '16'])\n```\n\n```\"1, 2, 4, 8, 16\"\n```\n\n```\"1, 2, 4, 8, 16\".split(\", \")\n```\n\n``` join() ``` 是字符串方法，因为在使用该方法时，您告诉分隔符字符串去迭代一个字符串序列，并在相邻元素之间插入自身。此方法的参数可以是任何遵循序列规则的对象，包括您自己定义的任何新的类。对于字节和字节数组对象也有类似的方法。\n\n## 异常有多快？\n\n```try:\nvalue = mydict[key]\nexcept KeyError:\nmydict[key] = getvalue(key)\nvalue = mydict[key]\n```\n\n```if key in mydict:\nvalue = mydict[key]\nelse:\nvalue = mydict[key] = getvalue(key)\n```\n\n## 为什么Python中没有switch或case语句？\n\n```def function_1(...):\n...\n\nfunctions = {'a': function_1,\n'b': function_2,\n'c': self.method_1, ...}\n\nfunc = functions[value]\nfunc()\n```\n\n```def visit_a(self, ...):\n...\n...\n\ndef dispatch(self, value):\nmethod_name = 'visit_' + str(value)\nmethod = getattr(self, method_name)\nmethod()\n```\n\n## 为什么lambda表达式不能包含语句？\n\nPython的 lambda表达式不能包含语句，因为Python的语法框架不能处理嵌套在表达式内部的语句。然而，在Python中，这并不是一个严重的问题。与其他语言中添加功能的lambda表单不同，Python的 lambdas只是一种速记符号，如果您懒得定义函数的话。\n\n## 可以将Python编译为机器代码，C或其他语言吗？\n\nCython 将带有可选注释的Python修改版本编译到C扩展中。 Nuitka 是一个将Python编译成 C++ 代码的新兴编译器，旨在支持完整的Python语言。要编译成Java，可以考虑 VOC\n\n## Python如何管理内存？\n\nPython 内存管理的细节取决于实现。 Python 的标准实现 CPython 使用引用计数来检测不可访问的对象，并使用另一种机制来收集引用循环，定期执行循环检测算法来查找不可访问的循环并删除所涉及的对象。 ``` gc ``` 模块提供了执行垃圾回收、获取调试统计信息和优化收集器参数的函数。\n\n```for file in very_long_list_of_files:\nf = open(file)\nc = f.read(1)\n```\n\n```for file in very_long_list_of_files:\nwith open(file) as f:\nc = f.read(1)\n```\n\n## 列表是如何在CPython中实现的？\n\nCPython的列表实际上是可变长度的数组，而不是lisp风格的链表。该实现使用对其他对象的引用的连续数组，并在列表头结构中保留指向该数组和数组长度的指针。\n\n## 字典是如何在CPython中实现的？\n\nCPython的字典实现为可调整大小的哈希表。与B-树相比，这在大多数情况下为查找（目前最常见的操作）提供了更好的性能，并且实现更简单。\n\n## 为什么字典key必须是不可变的？\n\n• 哈希按其地址（对象ID）列出。这不起作用，因为如果你构造一个具有相同值的新列表，它将无法找到；例如:\n\n```mydict = {[1, 2]: '12'}\nprint(mydict[[1, 2]])\n```\n\n会引发一个 ``` KeyError ``` 异常，因为第二行中使用的 ``` [1, 2] ``` 的 id 与第一行中的 id 不同。换句话说，应该使用 ``` == ``` 来比较字典键，而不是使用 ``` is ```\n\n• 使用列表作为键时进行复制。这没有用的，因为作为可变对象的列表可以包含对自身的引用，然后复制代码将进入无限循环。\n\n• 允许列表作为键，但告诉用户不要修改它们。当你意外忘记或修改列表时，这将产生程序中的一类难以跟踪的错误。它还使一个重要的字典不变量无效： ``` d.keys() ``` 中的每个值都可用作字典的键。\n\n• 将列表用作字典键后，应标记为其只读。问题是，它不仅仅是可以改变其值的顶级对象；你可以使用包含列表作为键的元组。将任何内容作为键关联到字典中都需要将从那里可到达的所有对象标记为只读 —— 并且自引用对象可能会导致无限循环。\n\n```class ListWrapper:\ndef __init__(self, the_list):\nself.the_list = the_list\n\ndef __eq__(self, other):\nreturn self.the_list == other.the_list\n\ndef __hash__(self):\nl = self.the_list\nresult = 98767 - len(l)*555\nfor i, el in enumerate(l):\ntry:\nresult = result + (hash(el) % 9999999) * 1001 + i\nexcept Exception:\nresult = (result % 7777777) + i * 333\nreturn result\n```\n\n## 为什么 list.sort() 没有返回排序列表？\n\n```for key in sorted(mydict):\n... # do whatever with mydict[key]...\n```\n\n## 如何在Python中指定和实施接口规范？\n\nPython 2.6添加了一个 ``` abc ``` 模块，允许定义抽象基类 (ABCs)。然后可以使用 ``` isinstance() ``` ``` issubclass() ``` 来检查实例或类是否实现了特定的ABC。 ``` collections.abc ``` 模块定义了一组有用的ABCs 例如 ``` Iterable ``` ``` Container ``` , 和 ``` MutableMapping ```\n\n## 为什么没有goto？\n\n```class label(Exception): pass # declare a label\n\ntry:\n...\nif condition: raise label() # goto label\n...\nexcept label: # where to goto\npass\n...\n```\n\n## 为什么原始字符串（r-strings）不能以反斜杠结尾？\n\n```f = open(\"/mydir/file.txt\") # works fine!\n```\n\n```dir = r\"\\this\\is\\my\\dos\\dir\" \"\\\\\"\ndir = r\"\\this\\is\\my\\dos\\dir\\ \"[:-1]\ndir = \"\\\\this\\\\is\\\\my\\\\dos\\\\dir\\\\\"\n```\n\n## 为什么Python没有属性赋值的“with”语句？\n\nPython有一个 'with' 语句，它封装了块的执行，在块的入口和出口调用代码。有些语言的结构是这样的:\n\n```with obj:\na = 1 # equivalent to obj.a = 1\ntotal = total + 1 # obj.total = obj.total + 1\n```\n\nPython使用动态类型。事先不可能知道在运行时引用哪个属性。可以动态地在对象中添加或删除成员属性。这使得无法通过简单的阅读就知道引用的是什么属性：局部属性、全局属性还是成员属性？\n\n```def foo(a):\nwith a:\nprint(x)\n```\n\n```function(args).mydict[index][index].a = 21\nfunction(args).mydict[index][index].b = 42\nfunction(args).mydict[index][index].c = 63\n```\n\n```ref = function(args).mydict[index][index]\nref.a = 21\nref.b = 42\nref.c = 63\n```\n\n## 为什么 if/while/def/class语句需要冒号？\n\n```if a == b\nprint(a)\n```\n\n```if a == b:\nprint(a)\n```\n\n## 为什么Python在列表和元组的末尾允许使用逗号？\n\nPython 允许您在列表，元组和字典的末尾添加一个尾随逗号:\n\n```[1, 2, 3,]\n('a', 'b', 'c',)\nd = {\n\"A\": [1, 5],\n\"B\": [6, 7], # last trailing comma is optional but good style\n}\n```\n\n```x = [\n\"fee\",\n\"fie\"\n\"foo\",\n\"fum\"\n]\n```",
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http://mauricio.resende.info/abstracts/gmissub.html | [
"Algorithm 786: FORTRAN Subroutines for Approximate Solution of Maximum Independent Set Problems using GRASP\n\nM. G. C. Resende, T. A. Feo, and S. H. Smith\n\nACM Transactions on Mathematical Software, vol. 24, pp. 386-394, 1998\n\nABSTRACT\n\nLet G = (V, E) be an undirected graph, where V and E are the sets of vertices and edges of G, respectively. A subset S of the vertices in V is independent if all of its members are pairwise nonadjacent, i.e. have no edge between them. A solution to the NP-hard maximum independent set problem is an independent set of maximum cardinality. This paper describes a set of FORTRAN subroutines to find an approximate solution of a maximum independent set problem. A greedy randomized adaptive search procedure (GRASP) is used to produce the solutions. The design and implementation of the code are described in detail, and computational experiments are reported, illustrating solution quality as a function of running time.\n\nPostScript file of full paper\n\nGo back"
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https://astarmathsandphysics.com/ib-physics-notes/measurements-units-uncertainty-and-principles/1347-simplifying-assumptions.html | [
"## Simplifying Assumptions\n\nThe world is complex. Every situation has many possible factors to take into account. Often however many of these possible factors are irrelevant and can be ignored. For example, we would not expect the temperature to greatly influence the acceleration due to gravity so can ignore this factor. Even if a factor cannot be ignored, we can often gain insight into a phenomenon by ignoring it. A model can later be refined by taking more factors into account not making that simplifying assumption.\n\nThe table below lists some common assumptions.\n\n Assumption Examples Friction is negligible PulleyThe motion of a car down or up a slope No heat lost Heat being transferred from one body to another when both are insulated together Mass of string or spring is negligible Simple pendulumMass on stretched spring Resistor of ammeter is zero Electric circuits Resistance of voltmeter is infinite Electric circuits Internal resistance of battery is zero Electric circuits Conductor obeys Ohm's Law Electric circuits 100% efficiency TransformersProjectiles Gas is ideal Thermodynamics Collision is elastic Collisions. In fact perfectly elastic collsions are only possible between atoms and subatomic particles.",
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"https://astarmathsandphysics.com/component/jcomments/captcha/46736.html",
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https://discourse.pymc.io/t/how-can-i-terminate-pm-sample-when-a-convergence-criterion-is-met-e-g-r-hat-1-05/9704 | [
"# How can I terminate pm.sample when a convergence criterion is met (e.g. r_hat < 1.05)\n\nHi all,\n\nI am wondering if the documentation page for Sample Callbacks is valid in PyMC version 4. I have tried reproducing the first example which simply terminates the sampler when the number of samples exceeds 100 (even if the `draws` argument is set to a higher value.) However, PyMC gives me a TypeError since `trace` is a `NoneType` for some reason. For convenience I have included a self-contained minimum working example below.\n\nMinimum working example\n\n``````import numpy as np\nimport pymc as pm\nprint(f'PyMC v{pm.__version__}')\n\n# Generate artificial data\ntrue_mu = 0.\ntrue_sigma = 1.\ndata = np.random.normal(loc=true_mu, scale=true_sigma, size=1000)\n\n# Define custom callback (taken from pymc.io/projects/examples/en/latest/howto/sampling_callback.html)\ndef my_callback(trace, draw):\nif len(trace) >= 100:\nraise KeyboardInterrupt()\n\nwith pm.Model() as model:\nmu = pm.Flat(\"mu\")\nsigma = pm.Flat(\"sigma\")\nobs = pm.Normal(\"obs\", mu=mu, sigma=sigma, observed=data)\nidata = pm.sample(draws=1000, tune=1000, chains=4, callback=my_callback)\n``````\n\nOutput:\n\n``````Auto-assigning NUTS sampler...\nPyMC v4.0.0\nMultiprocess sampling (4 chains in 4 jobs)\nNUTS: [mu, sigma]\n\n0.01% [1/8000 00:00<00:11 Sampling 4 chains, 0 divergences]\n---------------------------------------------------------------------------\nTypeError Traceback (most recent call last)\nInput In , in <cell line: 31>()\n33 sigma = pm.Flat(\"sigma\")\n34 obs = pm.Normal(\"obs\", mu=mu, sigma=sigma, observed=data)\n---> 35 idata = pm.sample(draws=1000, tune=1000, chains=4, callback=my_callback)\n\nFile ~/miniconda3/lib/python3.9/site-packages/pymc/sampling.py:607, in sample(draws, step, init, n_init, initvals, trace, chain_idx, chains, cores, tune, progressbar, model, random_seed, discard_tuned_samples, compute_convergence_checks, callback, jitter_max_retries, return_inferencedata, idata_kwargs, mp_ctx, **kwargs)\n605 _print_step_hierarchy(step)\n606 try:\n--> 607 mtrace = _mp_sample(**sample_args, **parallel_args)\n608 except pickle.PickleError:\n609 _log.warning(\"Could not pickle model, sampling singlethreaded.\")\n\nFile ~/miniconda3/lib/python3.9/site-packages/pymc/sampling.py:1547, in _mp_sample(draws, tune, step, chains, cores, chain, random_seed, start, progressbar, trace, model, callback, discard_tuned_samples, mp_ctx, **kwargs)\n1546 if callback is not None:\n-> 1547 callback(trace=trace, draw=draw)\n1549 except ps.ParallelSamplingError as error:\n1550 strace = traces[error._chain - chain]\n\nInput In , in my_callback(trace, draw)\n8 def my_callback(trace, draw):\n----> 9 if len(trace) >= 100:\n10 raise KeyboardInterrupt()\n\nTypeError: object of type 'NoneType' has no len()\n``````\n\nWhy is trace NoneType?\n\nAdditional context: My end goal is to write a callback function which terminates the sampler when the Gelman-Rubin statistic satsifies `r_hat < 1.01`, which is the subject of the second example on the Sample Callback page."
]
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https://www.practically.com/studymaterial/blog/docs/class-6th/maths/rationally-part-1/ | [
"# rationally NCERT Questions\n\nNCERT TEXT BOOK EXERCISES\n\nEXERCISE-12.1\n\n1. There are 20 girls and 15 boys in a class.\n\n(a) What is the ratio of number of girls to the number of boys?\n\n(b) What is the ratio of number of girls to the total number of students in the class?\n\n1. Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of\n\n(a) Number of students liking football to number of students liking tennis.\n\n(b) Number of students liking cricket to total number of students.\n\n1. See the figure and find the ratio of",
null,
"(a) Number of triangles to the number of circles inside the rectangle.\n\n(b) Number of squares to all the figures inside the rectangle.\n\n(c) Number of circles to all the figures inside the rectangle.\n\n1. Distances travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Find the ratio of speed of Hamid to the speed of Akhtar.\n2. Fill in the following blanks:\n\n[Are these equivalent ratios?]\n\n1. Find the ratio of the following:\n\n(a) 81 to 108\n\n(b) 98 to 63\n\n(c) 33 km to 121 km\n\n(d) 30 minutes to 45 minutes\n\n1. Find the ratio of the following:\n\n(a) 30 minutes to 1.5 hours\n\n(b) 40 cm to 1.5 m\n\n(c) 55 paise to Re1\n\n(d) 500 mL to 2 litres\n\n1. In a year, Seema earns Rs 1, 50, 000 and saves Rs 50, 000. Find the ratio of\n\n(a) Money that Seema earns to the money she saves.\n\n(b) Money that she saves to the money she spends.\n\n1. There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students.\n2. In a college, out of 4320 students, 2300 are girls. Find the ratio of\n\n(a) Number of girls to the total number of students.\n\n(b) Number of boys to the number of girls.\n\n(c) Number of boys to the total number of students.\n\n1. Out of 1800 students in a school, 750 opted basketball, 800 opted cricket and remaining opted table tennis. If a student can opt only one game, find the ratio of\n\n(a) Number of students who opted basketball to the number of students who opted table tennis.\n\n(b) Number of students who opted cricket to the number of students opting basketball.\n\n(c) Number of students who opted basketball to the total number of students.\n\n1. Cost of a dozen pens is Rs 180 and cost of 8 ball pens is Rs 56. Find the ratio of the cost of a pen to the cost of a ball pen.\n2. Consider the statement: Ratio of breadth and length of a hall is 2 : 5. Complete the following table that shows some possible breadths and lengths of the hall.\n Breadth of the hall (in metres) 10 ? 40 Length of the hall (in metres) 25 50 ?\n1. Divide 20 pens between Sheela and Sangeeta in the ratio of 3:2.\n2. Mother wants to divide Rs 36 between her daughters Shreya and Bhoomika in the ratio of their ages. If age of Shreya is 15 years and age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get.\n3. Present age of father is 42 years and that of his son is 14 years. Find the ratio of\n\n(a) Present age of father to the present age of son.\n\n(b) Age of the father to the age of son, when son was 12 years old.\n\n(c) Age of father after 10 years to the age of son after 10 years.\n\n(d) Age of father to the age of son when father was 30 years old.\n\nEXERCISE-12.2\n\n1. Determine if the following are in proportion.\n\n(a) 15, 45, 40, 120 (b) 33, 121, 9, 96\n\n(c) 24, 28, 36, 48 (d) 32, 48, 70, 210\n\n(e) 4, 6, 8, 12 (f) 33, 44, 75, 100\n\n1. Write True (T) or False (F) against each of the following statements:\n\n(a) 16:24::20:30 (b) 21:6::35:10\n\n(c) 12:18::28:12 (d) 8:9::24:27\n\n(e) 5.2:3.9::3:4 (f) 0.9:0.36::10:4\n\n1. Are the following statements true?\n\n(a) 40 persons: 200 persons = Rs 15: Rs 75\n\n(b) 7.5 litres: 15 litres = 5 kg: 10 kg\n\n(c) 99 kg: 45 kg = Rs 44: Rs 20\n\n(d) 32 m: 64 m = 6 sec: 12 sec\n\n(e) 45 km: 60 km = 12 hours: 15 hours\n\n1. Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.\n\n(a) 25 cm: 1 m and Rs 40 : Rs 160\n\n(b) 39 litres: 65 litres and 6 bottles: 10 bottles\n\n(c) 2 kg: 80 kg and 25 g: 625 g\n\n(d) 200 mL: 2.5 litre and Rs 4: Rs 50\n\nEXERCISE-12.3\n\n1. If the cost of 7 m of cloth is Rs 294, find the cost of 5 m of cloth.\n2. Ekta earns Rs 1500 in 10 days. How much will she earn in 30 days?\n3. If it has rained 276 mm in the last 3 days, how many cm of rain will fall in one full week (7 days)? Assume that the rain continues to fall at the same rate.\n4. Cost of 5 kg of wheat is Rs 30.50.\n\n(a) What will be the cost of 8 kg of wheat?\n\n(b) What quantity of wheat can be purchased in Rs 61?\n\n1. The temperature dropped 15 degree Celsius in the last 30 days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next ten days?\n2. Shaina pays Rs 7500 as rent for 3 months. How much does she has to pay for a whole year, if the rent per month remains same?\n3. Cost of 4 dozens bananas is Rs 60. How many bananas can be purchased for Rs 12.50?\n\n1. The weight of 72 books is 9 kg. What is the weight of 40 such books?\n2. A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?\n3. Raju purchases 10 pens for Rs 150 and Manish buys 7 pens for Rs 84. Can you say who got the pens cheaper?\n4. Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over?"
]
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https://engineering.stackexchange.com/tags/finite-element-method/hot | [
"# Tag Info\n\n### What is the difference between FEA and CFD?\n\nCFD (computational fluid dynamics) includes any numerical method used to solve fluid flow problems. FEA (finite element analysis) is one numerical method for solving partial differential equations, ...\n\n### Does welding two metal parts together increases or decreases the strength locally there?\n\nRegarding the problem you have, without reviewing the results its difficult to make a definitive answer. However, if you are observing high loads near the weld, I think you should try to make ...\n\n### Does welding two metal parts together increases or decreases the strength locally there?\n\nYes, it increases and decreases strength in traditionally steels. Depends on what steel you have and what you mean by \"thin rods\", and the weld process , and the specific weld parameters. ...\nAccepted\n\n### FEA: What boundary conditions am I missing?\n\nAfter a long discussion with the OP in comments and chat, it emerged that the basic cause of the problem is the geometry of the structure. The curved \"beam\" has a length/diameter ratio of about 50,000:...\nAccepted\n\n### Rayleigh damping in Finite Element Models using beta-coefficient only\n\nThe report in your link explains it briefly in the sentence after your quote. The beta term models \"structural\" or \"hysteretic\" damping, which is described earlier in the report. The key fact about ...\nAccepted\n\n### How to model elastic support in FEM?\n\nBefore jumping into elastic support conditions, you have to realise that it's not a ground beam. For three reasons: With a minor axis width to depth ratio of 6ish, it's not far away from a square. ...\n\n### How do I use FEM to derive the torsional constant of an arbitrary shape?\n\nThis is a problem which is usually solved in books on elasticity theory. The underlying math is based on the solution to the Laplace PDE. If you do a Google search for Larry J. Segerlind's book \"...\nAccepted\n\n### ANSYS Workbench/Mechanical: Automatically export chart?\n\nThere are several options for doing this: If there are only a few values to be saved per design point, you could use 'output parameters'. If there is a lot of data to be saved, the report generator ...\nAccepted\n\n### How can I reduce the computation time required to simulate a brain tissue penetration model?\n\nGiven the subject matter my gut tells me to take the time to compute once you've triple checked your conditions. What material properties are you using? You can alter the mesh to decrease cpu usage as ...\n\nFrom a theoretical standpoint, the displacement gradient is equivalent to strain (assuming a structural problem). Numerically, you can obtain the derivate of a quantity through multiplication with ...\nAccepted\n\n### Are there other finite element types besides the usual?\n\nCertain geometries can benefit from polyhedral elements or elements with edge degrees of freedom. I can think of three main developments in that direction: 1) Voronoi cell finite elements, e.g., ...\nAccepted\n\n### How to properly validate transverse isotropic elasticity Finite Element Code\n\nDefinitions Before we can answer your question, let us look at two standard definitions of terms (from ASME Guide for Verification and Validation in Computational Solid Mechanics) : 1) Verification:...\n\n### Viscoelasticity and Hyperelastic model, history and difference\n\nLet's look at some (very rough) definitions: 1) Viscoelasticity = elastic behavior that changes if the rate of application of the load is changed or, if the load is kept fixed, the change in the ...\nAccepted\n\n### Viscoelasticity and Hyperelastic model, history and difference\n\nThe hyperelastic and viscoelastic material models are both constitutive relations that relate: Stress and strain, in the case of hyperelasticity. Stress, strain and strain rate, in the case of ...\nThe balance of linear momentum is $$\\nabla \\cdot \\boldsymbol{\\sigma} + \\rho \\mathbf{b} = \\rho \\mathbf{a}$$ where $\\boldsymbol{\\sigma}$ is the Cauchy stress, $\\rho$ is the mass density, $\\mathbf{b}$..."
]
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https://number.academy/5527 | [
"# Number 5527\n\nNumber 5,527 spell 🔊, write in words: five thousand, five hundred and twenty-seven . Ordinal number 5527th is said 🔊 and write: five thousand, five hundred and twenty-seventh. The meaning of number 5527 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 5527. What is 5527 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 5527.\n\n## What is 5,527 in other units\n\nThe decimal (Arabic) number 5527 converted to a Roman number is (V)DXXVII. Roman and decimal number conversions.\n The number 5527 converted to a Mayan number is",
null,
"Decimal and Mayan number conversions.\n\n#### Weight conversion\n\n5527 kilograms (kg) = 12184.8 pounds (lbs)\n5527 pounds (lbs) = 2507.0 kilograms (kg)\n\n#### Length conversion\n\n5527 kilometers (km) equals to 3435 miles (mi).\n5527 miles (mi) equals to 8895 kilometers (km).\n5527 meters (m) equals to 18133 feet (ft).\n5527 feet (ft) equals 1685 meters (m).\n5527 centimeters (cm) equals to 2176.0 inches (in).\n5527 inches (in) equals to 14038.6 centimeters (cm).\n\n#### Temperature conversion\n\n5527° Fahrenheit (°F) equals to 3052.8° Celsius (°C)\n5527° Celsius (°C) equals to 9980.6° Fahrenheit (°F)\n\n#### Power conversion\n\n5527 Horsepower (hp) equals to 4064.55 kilowatts (kW)\n5527 kilowatts (kW) equals to 7515.65 horsepower (hp)\n\n#### Time conversion\n\n(hours, minutes, seconds, days, weeks)\n5527 seconds equals to 1 hour, 32 minutes, 7 seconds\n5527 minutes equals to 3 days, 20 hours, 7 minutes\n\n### Zip codes 5527\n\n• Zip code 5527 COLONIA SANTA TERESA, MENDOZA, Argentina a map\n• Zip code 5527 LA PRIMAVERA (DPTO. GUAYMALLEN), MENDOZA, Argentina a map\n• Zip code 5527 LAGUNITA, MENDOZA, Argentina a map\nZip code areas 5527\n\n### Codes and images of the number 5527\n\nNumber 5527 morse code: ..... ..... ..--- --...\nSign language for number 5527:",
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"",
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"Number 5527 in braille:",
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"Images of the number\nImage (1) of the numberImage (2) of the number",
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"More images, other sizes, codes and colors ...\n\n#### Number 5527 infographic",
null,
"### Gregorian, Hebrew, Islamic, Persian and Buddhist year (calendar)\n\nGregorian year 5527 is Buddhist year 6070.\nBuddhist year 5527 is Gregorian year 4984 .\nGregorian year 5527 is Islamic year 5055 or 5057.\nIslamic year 5527 is Gregorian year 5983 or 5984.\nGregorian year 5527 is Persian year 4905 or 4906.\nPersian year 5527 is Gregorian 6148 or 6149.\nGregorian year 5527 is Hebrew year 9287 or 9288.\nHebrew year 5527 is Gregorian year 1767.\nThe Buddhist calendar is used in Sri Lanka, Cambodia, Laos, Thailand, and Burma. The Persian calendar is official in Iran and Afghanistan.\n\n## Share in social networks",
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"## Mathematics of no. 5527\n\n### Multiplications\n\n#### Multiplication table of 5527\n\n5527 multiplied by two equals 11054 (5527 x 2 = 11054).\n5527 multiplied by three equals 16581 (5527 x 3 = 16581).\n5527 multiplied by four equals 22108 (5527 x 4 = 22108).\n5527 multiplied by five equals 27635 (5527 x 5 = 27635).\n5527 multiplied by six equals 33162 (5527 x 6 = 33162).\n5527 multiplied by seven equals 38689 (5527 x 7 = 38689).\n5527 multiplied by eight equals 44216 (5527 x 8 = 44216).\n5527 multiplied by nine equals 49743 (5527 x 9 = 49743).\nshow multiplications by 6, 7, 8, 9 ...\n\n### Fractions: decimal fraction and common fraction\n\n#### Fraction table of 5527\n\nHalf of 5527 is 2763,5 (5527 / 2 = 2763,5 = 2763 1/2).\nOne third of 5527 is 1842,3333 (5527 / 3 = 1842,3333 = 1842 1/3).\nOne quarter of 5527 is 1381,75 (5527 / 4 = 1381,75 = 1381 3/4).\nOne fifth of 5527 is 1105,4 (5527 / 5 = 1105,4 = 1105 2/5).\nOne sixth of 5527 is 921,1667 (5527 / 6 = 921,1667 = 921 1/6).\nOne seventh of 5527 is 789,5714 (5527 / 7 = 789,5714 = 789 4/7).\nOne eighth of 5527 is 690,875 (5527 / 8 = 690,875 = 690 7/8).\nOne ninth of 5527 is 614,1111 (5527 / 9 = 614,1111 = 614 1/9).\nshow fractions by 6, 7, 8, 9 ...\n\n### Calculator\n\n 5527\n\n#### Is Prime?\n\nThe number 5527 is a prime number. The closest prime numbers are 5521, 5531.\n5527th prime number in order is 54331.\n\n#### Factorization and factors (dividers)\n\nThe prime factors of 5527\nPrime numbers have no prime factors less than themselves.\nThe factors of 5527 are 1 , 5527\nTotal factors 2.\nSum of factors 5528 (1).\n\n#### Prime factor tree\n\n5527 is a prime number.\n\n#### Powers\n\nThe second power of 55272 is 30.547.729.\nThe third power of 55273 is 168.837.298.183.\n\n#### Roots\n\nThe square root √5527 is 74,343796.\nThe cube root of 35527 is 17,680579.\n\n#### Logarithms\n\nThe natural logarithm of No. ln 5527 = loge 5527 = 8,6174.\nThe logarithm to base 10 of No. log10 5527 = 3,742489.\nThe Napierian logarithm of No. log1/e 5527 = -8,6174.\n\n### Trigonometric functions\n\nThe cosine of 5527 is -0,590981.\nThe sine of 5527 is -0,806686.\nThe tangent of 5527 is 1,364995.\n\n## Number 5527 in Computer Science\n\nCode typeCode value\nPIN 5527 It's recommendable to use 5527 as a password or PIN.\n5527 Number of bytes5.4KB\nUnix timeUnix time 5527 is equal to Thursday Jan. 1, 1970, 1:32:07 a.m. GMT\nIPv4, IPv6Number 5527 internet address in dotted format v4 0.0.21.151, v6 ::1597\n5527 Decimal = 1010110010111 Binary\n5527 Decimal = 21120201 Ternary\n5527 Decimal = 12627 Octal\n5527 Decimal = 1597 Hexadecimal (0x1597 hex)\n5527 BASE64NTUyNw==\n5527 MD5e41e164f7485ec4a28741a2d0ea41c74\n5527 SHA256670d1430eca8c1f5fc91dd0089e0e62f2409eea629d3a826fe2f6def428e57a1\n5527 SHA3848472e9fc2d279dea5a5a9de132a0e9f3952646eb1311ddfe4e29b6669336746fa44186c2cd5e5cdbf669d0b7c3c15008\nMore SHA codes related to the number 5527 ...\n\nIf you know something interesting about the 5527 number that you did not find on this page, do not hesitate to write us here.\n\n## Numerology 5527\n\n### The meaning of the number 5 (five), numerology 5\n\nCharacter frequency 5: 2\n\nThe number five (5) came to this world to achieve freedom. You need to apply discipline to find your inner freedom and open-mindedness. It is about a restless spirit in constant search of the truth that surrounds us. You need to accumulate as much information as possible to know what is happening in depth. Number 5 person is intelligent, selfish, curious and with great artistic ability. It is a symbol of freedom, independence, change, adaptation, movement, the search for new experiences, the traveling and adventurous spirit, but also of inconsistency and abuse of the senses.\nMore about the meaning of the number 5 (five), numerology 5 ...\n\n### The meaning of the number 7 (seven), numerology 7\n\nCharacter frequency 7: 1\n\nThe number 7 (seven) is the sign of the intellect, thought, psychic analysis, idealism and wisdom. This number first needs to gain self-confidence and to open his/her life and heart to experience trust and openness in the world. And then you can develop or balance the aspects of reflection, meditation, seeking knowledge and knowing.\nMore about the meaning of the number 7 (seven), numerology 7 ...\n\n### The meaning of the number 2 (two), numerology 2\n\nCharacter frequency 2: 1\n\nThe number two (2) needs above all to feel and to be. It represents the couple, duality, family, private and social life. He/she really enjoys home life and family gatherings. The number 2 denotes a sociable, hospitable, friendly, caring and affectionate person. It is the sign of empathy, cooperation, adaptability, consideration for others, super-sensitivity towards the needs of others.\n\nThe number 2 (two) is also the symbol of balance, togetherness and receptivity. He/she is a good partner, colleague or companion; he/she also plays a wonderful role as a referee or mediator. Number 2 person is modest, sincere, spiritually influenced and a good diplomat. It represents intuition and vulnerability.\nMore about the meaning of the number 2 (two), numerology 2 ...\n\n## Interesting facts about the number 5527\n\n### Asteroids\n\n• (5527) 1991 UQ3 is asteroid number 5527. It was discovered by S. Ueda; H. Kaneda from Kushiro on 10/31/1991.\n\n### Distances between cities\n\n• There is a 3,435 miles (5,527 km) direct distance between Ahmedabad (India) and Daegu (South Korea).\n• There is a 5,527 miles (8,894 km) direct distance between Antalya (Turkey) and Teresina (Brazil).\n• There is a 3,435 miles (5,527 km) direct distance between Bishkek (Kyrgyzstan) and Kawasaki (Japan).\n• There is a 5,527 miles (8,894 km) direct distance between Brisbane (Australia) and Hāora (India).\n• There is a 5,527 miles (8,894 km) direct distance between Cape Town (South Africa) and Mashhad (Iran).\n• There is a 5,527 miles (8,894 km) direct distance between Dallas (USA) and Freetown (Sierra Leone).\n• There is a 3,435 miles (5,527 km) direct distance between Damascus (Syria) and Chittagong (Bangladesh).\n• There is a 3,435 miles (5,527 km) direct distance between Depok (Indonesia) and Karachi (Pakistan).\n• There is a 5,527 miles (8,894 km) direct distance between Dhaka (Bangladesh) and Melbourne (Australia).\n• There is a 3,435 miles (5,527 km) direct distance between Dubai (United Arab Emirates) and Luoyang (China).\n• There is a 3,435 miles (5,527 km) direct distance between Chittagong (Bangladesh) and Voronezh (Russia).\n• There is a 3,435 miles (5,527 km) direct distance between Jakarta (Indonesia) and Multān (Pakistan).\n• There is a 3,435 miles (5,527 km) direct distance between Johannesburg (South Africa) and Monrovia (Liberia).\n• There is a 5,527 miles (8,894 km) direct distance between Köln (Germany) and Phoenix (USA).\n• There is a 5,527 miles (8,894 km) direct distance between Lusaka (Zambia) and Omsk (Russia).\n• There is a 3,435 miles (5,527 km) direct distance between Shiraz (Iran) and Taiyuan (China).\n• There is a 3,435 miles (5,527 km) direct distance between Tabrīz (Iran) and Xi’an (China).\n\n## Number 5,527 in other languages\n\nHow to say or write the number five thousand, five hundred and twenty-seven in Spanish, German, French and other languages. The character used as the thousands separator.\n Spanish: 🔊 (número 5.527) cinco mil quinientos veintisiete German: 🔊 (Anzahl 5.527) fünftausendfünfhundertsiebenundzwanzig French: 🔊 (nombre 5 527) cinq mille cinq cent vingt-sept Portuguese: 🔊 (número 5 527) cinco mil, quinhentos e vinte e sete Chinese: 🔊 (数 5 527) 五千五百二十七 Arabian: 🔊 (عدد 5,527) خمسة آلاف و خمسمائة و سبعة و عشرون Czech: 🔊 (číslo 5 527) pět tisíc pětset dvacet sedm Korean: 🔊 (번호 5,527) 오천오백이십칠 Danish: 🔊 (nummer 5 527) femtusinde og femhundrede og syvogtyve Hebrew: (מספר 5,527) חמש אלף חמש מאות עשרים ושבע Dutch: 🔊 (nummer 5 527) vijfduizendvijfhonderdzevenentwintig Japanese: 🔊 (数 5,527) 五千五百二十七 Indonesian: 🔊 (jumlah 5.527) lima ribu lima ratus dua puluh tujuh Italian: 🔊 (numero 5 527) cinquemilacinquecentoventisette Norwegian: 🔊 (nummer 5 527) fem tusen, fem hundre og tjue-syv Polish: 🔊 (liczba 5 527) pięć tysięcy pięćset dwadzieścia siedem Russian: 🔊 (номер 5 527) пять тысяч пятьсот двадцать семь Turkish: 🔊 (numara 5,527) beşbinbeşyüzyirmiyedi Thai: 🔊 (จำนวน 5 527) ห้าพันห้าร้อยยี่สิบเจ็ด Ukrainian: 🔊 (номер 5 527) п'ять тисяч п'ятсот двадцять сiм Vietnamese: 🔊 (con số 5.527) năm nghìn năm trăm hai mươi bảy Other languages ...\n\n## News to email\n\nPrivacy Policy.\n\n## Comment\n\nIf you know something interesting about the number 5527 or any natural number (positive integer) please write us here or on facebook."
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http://curious.astro.cornell.edu/people-and-astronomy/careers-in-astronomy/41-our-solar-system/the-earth/orbit/83-is-the-distance-from-the-earth-to-the-sun-changing-advanced?tmpl=component&print=1 | [
"## Is the distance from the Earth to the Sun changing? (Advanced)\n\nIs the distance from the Earth to the Sun increasing, and if so, by how much in kilometers per (Earth) year?\n\nFirst I should say that the Earth's orbit around the Sun is elliptical, not perfectly circular, so the Earth-Sun distance is changing as we speak just from the Earth traveling in its orbit around the Sun. See here for a discussion of that.\n\nIs the orbit itself changing? Well, there are some long-period oscillations, but those are very small, and don't imply that we're systematically moving towards or away from the Sun.\n\nThere is an effect which is making us move very slowly away from the Sun. That is the tidal interaction between the Sun and the Earth. This slows down the rotation of the Sun, and pushes the Earth farther away from the Sun. You can read about tides, as they relate to the Earth-Moon system here. The principle for the Sun-Earth system should be the same. But how big of an effect is this? It turns out that the yearly increase in the distance between the Earth and the Sun from this effect is only about one micrometer (a millionth of a meter, or a ten thousandth of a centimeter). So this is a very tiny effect.\n\nThere is another effect which is also small, but somewhat bigger than the tidal effect. The Sun is powered by nuclear fusion, which means the Sun is continuously transforming a small part of its mass into energy. As the mass of the Sun goes down, our orbit gets proportionally bigger. However, over the entire main sequence lifetime of the Sun (about 10 billion years), the Sun will only lose about 0.1% of its mass, which means that the Earth should move out by just ~150,000 km (small compared to the total Earth-Sun distance of ~150,000,000 km). If we assume that the Sun's rate of nuclear fusion today is the same as the average rate over those 10 billion years (a bold assumption, but it should give us a rough idea of the answer), then we're moving away from the Sun at the rate of ~1.5 cm (less than an inch) per year. I probably don't even need to mention that this is so small that we don't have to worry about freezing.",
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https://dzone.com/articles/a-polyglots-guide-to-multiple-dispatch | [
"{{announcement.body}}\n{{announcement.title}}\n\n# A Polyglot's Guide to Multiple Dispatch – Part 1\n\nDZone 's Guide to\n\n# A Polyglot's Guide to Multiple Dispatch – Part 1\n\n### What is multiple dispatch and what problems does it solve?\n\n· Performance Zone ·\nFree Resource\n\nComment (2)\n\nSave\n{{ articles.views | formatCount}} Views\n\nThis is the first article in a series dedicated to multiple dispatch, an advanced abstraction technique available to programmers out-of-the-box in some languages, and implementable in others. This first post in the series presents the technique and explains the problem it intends to solve. It uses C++ as the presentation language because C++ does not support multiple dispatch directly, but can be used to implement it in various ways. Showing how multiple dispatch is implemented in a language that doesn't support it natively is important, in my opinion, as it lets us understand the issue on a deeper level.\n\nFollow-up articles will keep focusing on multiple dispatch using other programming languages: Part 2 will show how to implement multiple dispatch in Python; Part 3 will use Common Lisp, where multiple dispatch comes built-in as part of a large and powerful object-oriented system called CLOS; Part 4 will use Clojure, a more modern attempt at a Lisp, where multiple dispatch is also built-in, but works somewhat differently.\n\n## Polymorphism, Single Dispatch, Multiple Dispatch\n\nThere are many kinds of polymorphism in programming. The kind we're talking about here is runtime subtype-based polymorphism, where behavior is chosen dynamically based on the runtime types of objects. More specifically, multiple dispatch is all about the runtime types of more than one object.\n\nThe best way to understand multiple dispatch is to first think about single dispatch. Single dispatch is what we usually refer to as \"runtime polymorphism\" in languages like C++ and Java . We have an object on which we call a method, and the actual method being called at runtime depends on the runtime type of the object. In C++ this is done with virtual functions:\n\n``````class Shape {\npublic:\nvirtual void ComputeArea() const = 0;\n};\n\nclass Rectangle : public Shape {\npublic:\nvirtual void ComputeArea() const {\nstd::cout << \"Rectangle: width times height\\n\";\n}\n};\n\nclass Ellipse : public Shape {\npublic:\nvirtual void ComputeArea() const {\nstd::cout << \"Ellipse: width times height times pi/4\\n\";\n}\n};\n\nint main(int argc, const char** argv) {\nstd::unique_ptr<Shape> pr(new Rectangle);\nstd::unique_ptr<Shape> pe(new Ellipse);\n\npr->ComputeArea(); // invokes Rectangle::ComputeArea\npe->ComputeArea(); // invokes Ellipse::ComputeArea\n\nreturn 0;\n}``````\n\nEven though both pr and pe are pointers to a Shape as far as the C++ compiler is concerned, the two calls toComputeArea get dispatched to different methods at runtime due to C++'s implementation of runtime polymorphism via virtual functions.\n\nNow, spend a few seconds thinking about the question: \"What is the dispatch done upon in the code sample above?\"\n\nIt's fairly obvious that the entity we dispatch upon is a pointer to Shape. We have pr and we call a method on it. The C++ compiler emits code for this call such that at runtime the right function is invoked. The decision which function to invoke is based upon examining a single object - what pr points to. Hence single dispatch.\n\nA natural extension of this idea is multiple dispatch, wherein the decision which function to call is based on the runtime types of multiple objects. Why is this useful? It's not a tool programmers reach for very often, but when it is appropriate, alternatives tend to be cumbersome and repetitive. A telling sign that multiple dispatch may be in order is when you have some operation that involves more than one class and there is no single obvious class where this operation belongs. Think of simulating a sound when a drumstick hits a drum. There are many kinds of drumsticks, and many kinds of drums; their combinations produce different sounds. Say we want to write a function (or family of functions) that determines which sound is produced. Should this function be a method of the Drum class or the DrumStick class? Forcing this decision is one of the follies of classical OOP, and multiple dispatch helps us solve it naturally without adding a kludge into our design.\n\nA simpler and more canonical example is computing intersections of shapes — maybe for computer graphics, or for simulation, or other use cases. A generic shape intersection computation can be complex to implement, but in many specific cases it's easy. For example, computing intersections of rectangles with rectangles is trivial; same for circles and ellipses; rectangles with triangles may be a tiny bit harder, but still much simpler than artibrary polygons, and so on.\n\nHow do we write code to handle all these cases? All in all, we just need an intersect function that takes two shapes and computes an intersection. This function may have a whole bunch of special cases inside for different combinations of shapes it knows how to do easily, before it resorts to some heavy-handed generic polygon intersection approach. Such code, however, would be gross to develop and maintain. Wouldn't it be nice if we could have:\n\n``````void Intersect(const Rectangle* r, const Ellipse* e) {\n// implement intersection of rectangle with ellipse\n}\n\nvoid Intersect(const Rectangle* r1, const Rectangle* r2) {\n// implement intersection of rectangle with another rectangle\n}\n\nvoid Intersect(const Shape* s1, const Shape* s2) {\n// implement interesction of two generic shapes\n}``````\n\nAnd then the call Intersect(some_shape, other_shape) would just magically dispatch to the right function? This capability is what's most often referred to by multiple dispatch in programming language parlance .\n\n## A Failed Attempt in C++\n\nYou may be tempted to come up with the following \"trivial\" solution in C++:\n\n``````class Shape {\npublic:\nvirtual std::string name() const {\nreturn typeid(*this).name();\n}\n};\n\nclass Rectangle : public Shape {};\n\nclass Ellipse : public Shape {};\n\nclass Triangle : public Shape {};\n\nvoid Intersect(const Rectangle* r, const Ellipse* e) {\nstd::cout << \"Rectangle x Ellipse [names r=\" << r->name()\n<< \", e=\" << e->name() << \"]\\n\";\n}\n\nvoid Intersect(const Rectangle* r1, const Rectangle* r2) {\nstd::cout << \"Rectangle x Rectangle [names r1=\" << r1->name()\n<< \", r2=\" << r2->name() << \"]\\n\";\n}\n\n// Fallback to shapes\nvoid Intersect(const Shape* s1, const Shape* s2) {\nstd::cout << \"Shape x Shape [names s1=\" << s1->name()\n<< \", s2=\" << s2->name() << \"]\\n\";\n}``````\n\nNow in main:\n\n``````Rectangle r1, r2;\nEllipse e;\nTriangle t;\n\nstd::cout << \"Static type dispatch\\n\";\nIntersect(&r1, &e);\nIntersect(&r1, &r2);\nIntersect(&r1, &t);``````\n\nWe'll see:\n\n``````Static type dispatch\nRectangle x Ellipse [names r=9Rectangle, e=7Ellipse]\nRectangle x Rectangle [names r1=9Rectangle, r2=9Rectangle]\nShape x Shape [names s1=9Rectangle, s2=8Triangle]``````\n\nNote how the intersections get dispatched to specialized functions when these exist and to a generic catch-allShape x Shape handler when there is no specialized function.\n\nSo that's it, multiple dispatch works out of the box? Not so fast... what we see here is just C++ function overloading in action. The compiler knows the static, compile-time types of the pointers passed to the Intersect calls, so it just emits the right call. Function overloading is great and useful, but this is not the general problem we're trying to solve. In a realistic code-base, you won't be passing pointers to concrete subclasses of Shape around. You are almost certainly going to be dealing with pointers to the Shape base class. Let's try to see how the code in the previous sample works with dynamic types:\n\n``````std::unique_ptr<Shape> pr1(new Rectangle);\nstd::unique_ptr<Shape> pr2(new Rectangle);\nstd::unique_ptr<Shape> pe(new Ellipse);\nstd::unique_ptr<Shape> pt(new Triangle);\n\nstd::cout << \"Dynamic type dispatch\\n\";\nIntersect(pr1.get(), pe.get());\nIntersect(pr1.get(), pr2.get());\nIntersect(pr1.get(), pt.get());``````\n\nPrints:\n\n``````Dynamic type dispatch\nShape x Shape [names s1=9Rectangle, s2=7Ellipse]\nShape x Shape [names s1=9Rectangle, s2=9Rectangle]\nShape x Shape [names s1=9Rectangle, s2=8Triangle]``````\n\nYeah... that's not good. All calls were dispatched to the generic Shape x Shape handler, even though the runtime types of the objects are different (see the names gathered from typeid). This is hardly surprising, because when the compiler sees Intersect(pr1.get(), pr2.get()), the static types for the two arguments are Shape* and Shape*. You could be forgiven for thinking that the compiler may invoke virtual dispatch here, but virtual dispatch in C++ doesn't work this way. It only works when a virtual method is called on a pointer to a base object, which is not what's happening here.\n\n## Multiple Dispatch in C++ With the Visitor Pattern\n\nI'll admit I'm calling this approach \"the visitor pattern\" only because this is how it's called elsewhere and because I don't have a better name for it. In fact, it's probably closer to an \"inverted\" visitor pattern, and in general the pattern name may obscure the code more than help. So forget about the name, and just study the code.\n\nThe last paragraph of the previous section ended with an important observation: virtual dispatch in C++ kicks in onlywhen a virtual method is called on a pointer to a base object. Let's leverage this idea to simulate double dispatch on our hierarchy of shapes. The plan is to arrange Intersect to hop through virtual dispatches on both its arguments to get to the right method for their runtime types.\n\nWe'll start by defining Shape like this:\n\n``````class Shape {\npublic:\nvirtual std::string name() const {\nreturn typeid(*this).name();\n}\n\n// Dispatcher that should be called by clients to intersect different shapes.\nvirtual void Intersect(const Shape*) const = 0;\n\n// Specific interesection methods implemented by subclasses. If subclass A\n// has a special way to intersect with subclass B, it should implement\n// InteresectWith(const B*).\nvirtual void IntersectWith(const Shape*) const {}\nvirtual void IntersectWith(const Rectangle*) const {}\nvirtual void IntersectWith(const Ellipse*) const {}\n};``````\n\nThe Intersect method is what the users of the code will invoke. To be able to make use of virtual dispatches, we are forced to turn a two-argument call Intersect(A*, B*) to a method call A->Intersect(B). The IntersectWithmethods are concrete implementations of intersections the code will dispatch to and should be implemented by subclasses on a case-per-case basis.\n\n``````class Rectangle : public Shape {\npublic:\nvirtual void Intersect(const Shape* s) const {\ns->IntersectWith(this);\n}\n\nvirtual void IntersectWith(const Shape* s) const {\nstd::cout << \"Rectangle x Shape [names this=\" << this->name()\n<< \", s=\" << s->name() << \"]\\n\";\n}\n\nvirtual void IntersectWith(const Rectangle* r) const {\nstd::cout << \"Rectangle x Rectangle [names this=\" << this->name()\n<< \", r=\" << r->name() << \"]\\n\";\n}\n};\n\nclass Ellipse : public Shape {\npublic:\nvirtual void Intersect(const Shape* s) const {\ns->IntersectWith(this);\n}\n\nvirtual void IntersectWith(const Rectangle* r) const {\nstd::cout << \"Ellipse x Rectangle [names this=\" << this->name()\n<< \", r=\" << r->name() << \"]\\n\";\n}\n};``````\n``````std::unique_ptr<Shape> pr1(new Rectangle);\nstd::unique_ptr<Shape> pr2(new Rectangle);\nstd::unique_ptr<Shape> pe(new Ellipse);\n\nstd::cout << \"Dynamic type dispatch\\n\";\npr1->Intersect(pe.get());\npr1->Intersect(pr2.get());``````\n\nWill now print\n\n``````Dynamic type dispatch\nEllipse x Rectangle [names this=7Ellipse, r=9Rectangle]\nRectangle x Rectangle [names this=9Rectangle, r=9Rectangle]``````\n\nSuccess! Even though we're dealing solely in pointers to Shape, the right intersections are computed. Why does this work?\n\nAs I've mentioned before, the key here is use C++'s virtual function dispatch capability twice. Let's trace through one execution to see what's going on. We have:\n\n``pr1->Intersect(pe.get());``\n\npr1 is a pointer to Shape, and Intersect is a virtual method. Therefore, the runtime type's Intersect is called here, which is Rectangle::Intersect. The argument passed into the method is another pointer to Shape which at runtime points to an Ellipse (pe). Rectangle::Intersect calls s->IntersectWith(this). The compiler sees that s is a Shape*, and IntersectWith is a virtual method, so this is another virtual dispatch. What gets called isEllipse::IntersectWith. But which overload of this method is called?\n\nThis is an extremely crucial point in the explanation, so please focus :-) Here is Rectangle::Intersect again:\n\n``````virtual void Intersect(const Shape* s) const {\ns->IntersectWith(this);\n}``````\n\ns->IntersectWith is called with this, which the compiler knows is a pointer to Rectangle, statically. If you wondered why I define Intersect in each subclass rather than doing it once in Shape, even though its code is exactly the same for each subclass, this is the reason. Had I defined it in Shape, the compiler would think the type of this is Shape* and would always dispatch to the IntersectWith(const Shape*) overload. Defining this method in each subclass helps the compiler leverage overloading to call the right method.\n\nWhat happens eventually is that the call pr1->Intersect(pe.get()) gets routed to Ellipse::IntersectWith(const Rectangle*), thanks to two virtual dispatches and one use of method overloading. The end result is double dispatch! \n\nBut wait a second, how did we end up with Ellipse::IntersectWith(Rectangle)? Shouldn'tpr1->Intersect(pe.get()) go to Rectangle::IntersectWith(Ellipse) instead? Well, yes and no. Yes because this is what you'd expect from how the call is syntactically structured. No because you almost certainly want double dispatches to be symmetric. I'll discuss this and other related issues in the next section.\n\n## Symmetry and Base-class Defaults\n\nWhen we come up with ways to do multiple dispatch, whether in C++ or in other languages, there are two aspects of the solution we should always keep in mind:\n\n1. Does it permit symmetry? In other words, does the order of objects dispatched upon matters? And if it doesn't, how much extra code is needed to express this fact.\n2. Does base-class default dispatch work as expected? Suppose we create a new subclass of Rectangle, calledSquare and we don't explicitly create an IntersectWith method for Square and Ellipse. Will the right thing happen and the intersection between a Rectangle and Ellipse be invoked when we ask forSquare x Ellipse? This is the right thing because this is what we've come to expect from class hierarchies in object-oriented languages.\n\nIn the visitor-based solution presented above, both aspects will work, though symmetry needs a bit of extra code. The full code sample is available here (and the accompanying .cpp file). It's conceptually similar to the code shown above, but with a bit more details. In particular, it implements symmetry between rectangle and ellipse intersections as follows:\n\n``````namespace {\n\n// All intersections between rectangles and ellipses dispatch here.\nvoid SymmetricIntersectRectangleEllipse(const Rectangle* r, const Ellipse* e) {\nstd::cout << \"IntersectRectangleEllipse [names r=\" << r->name()\n<< \", e=\" << e->name() << \"]\\n\";\n}\n}\n\nvoid Rectangle::IntersectWith(const Ellipse* e) const {\nSymmetricIntersectRectangleEllipse(this, e);\n}\n\nvoid Ellipse::IntersectWith(const Rectangle* r) const {\nSymmetricIntersectRectangleEllipse(r, this);\n}``````\n\nThis ensures that both rectangle->Intersect(ellipse) and ellipse->Intersect(rectangle) end up in the same function. As far as I know there's not way to do this automatically in the visitor approach, so a bit of extra coding is due when symmetry between subclasses is desired.\n\nNote also that this method doesn't force symmetry either. If some form of dispatch is order-dependent, it's easy to express.\n\n## The Problem With the Visitor-based Approach\n\nAlthough the visitor-based approach works, enables fairly clean client code, and is efficient (constant time - two virtual calls), there's a glaring issue with it that's apparent with the most cursory look at the code: it's very intrusive, and hence hard to maintain.\n\nImagine we want to add a new kind of shape - a HyperFrob. Suppose also that there's an efficient algorithm for intersecting a HyperFrob with an Ellipse. Ideally, we'd only have to write code for the new functionality:\n\n1. Define the new HyperFrob class deriving from Shape.\n2. Implement the generic HyperFrob x Shape intersection algorithm.\n3. Implement the specific HyperFrom x Ellipse algorithm.\n\nBut in reality, we're forced to modify the definition of the base class Shape to add an overload of IntersectWith forHyperFrob. Moreover, if we want intersections between HyperFrob and Ellipse to be symmetric (which we almost certainly do), we'll have to modify Ellipse as well to add the same overload.\n\nIf we don't control the Shape base class at all, we're in real trouble. This is an instance of the expression problem. I'll have more to say about the expression problem in a future post, but for now the Wikipedia link will have to do. It's not an easy problem to solve in C++, and the approaches to implement multiple dispatch should be judged by how flexible they are in this respect, along with the other considerations.\n\n## Multiple-dispatch in C++ By Brute-force\n\nThe visitor-based approach is kind-of clever, leveraging single virtual dispatch multiple times to simulate multiple dispatch. But if we go back to first principles for a moment, it becomes clear that there's a much more obvious solution to the problem - brute-force if-else checks. I mentioned this possibility early in the article and called it \"gross to develop and maintain\", but it makes sense to at least get a feel for how it would look:\n\n``````class Shape {\npublic:\nvirtual std::string name() const {\nreturn typeid(*this).name();\n}\n};\n\nclass Rectangle : public Shape {};\n\nclass Ellipse : public Shape {};\n\nclass Triangle : public Shape {};\n\nvoid Intersect(const Shape* s1, const Shape* s2) {\nif (const Rectangle* r1 = dynamic_cast<const Rectangle*>(s1)) {\nif (const Rectangle* r2 = dynamic_cast<const Rectangle*>(s2)) {\nstd::cout << \"Rectangle x Rectangle [names r1=\" << r1->name()\n<< \", r2=\" << r2->name() << \"]\\n\";\n} else if (const Ellipse* e2 = dynamic_cast<const Ellipse*>(s2)) {\nstd::cout << \"Rectangle x Ellipse [names r1=\" << r1->name()\n<< \", e2=\" << e2->name() << \"]\\n\";\n\n} else {\nstd::cout << \"Rectangle x Shape [names r1=\" << r1->name()\n<< \", s2=\" << s2->name() << \"]\\n\";\n}\n} else if (const Ellipse* e1 = dynamic_cast<const Ellipse*>(s1)) {\nif (const Ellipse* e2 = dynamic_cast<const Ellipse*>(s2)) {\nstd::cout << \"Ellipse x Ellipse [names e1=\" << e1->name()\n<< \", e2=\" << e2->name() << \"]\\n\";\n} else {\n// Handle other Ellipse x ... dispatches.\n}\n} else {\n// Handle Triangle s1\n}\n}``````\n\nOne thing is immediately noticeable: the intrusiveness issue of the visitor-based approach is completely solved. Obliterated! Intersect is now a stand-alone function that encapsulates the dispatch. If we add new kinds of shape, we only have to change Intersect, nothing else. Perfect... or is it?\n\nThe other immediately noticeable fact about this code is: holy cow, how long it is. I'm only showing a small snippet here, but the number of these if clauses grows as square of the number of subclasses. Imagine how this could look for 20 kinds of shapes. Moreover, Intersect is just one algorithm. We may have other \"multi methods\" - this travesty would have to be repeated for each algorithm.\n\nAnother, less obvious problem is that the code is somewhat brittle. Given a non-trivial inheritance hierarchy, we have to be very careful about the order of the if clauses, lest a parent class \"shadows\" all its subclasses by coming before them in the chain.\n\nIt's no wonder that one would be very reluctant to write all this code. In fact, smart folks came up with all kinds of ways to automate such if chains. If you're thinking - \"hey I could just store pairs of typeids in a map and dispatch upon that\" - congrats, you're in the right direction.\n\nOne of the most notable experts to tackle the beast is Andrei Alexandrescu, who dedicated chapter 11 of \"Modern C++ Design\" to this problem, implementing all kinds of automated solutions based on heavy template metaprogramming. It's a fairly impressive piece of work, presenting multiple approaches with different tradeoffs in terms of performance and intrusiveness. If you Google for Loki (his C++ template library) and look into the MultiMethods.h header you'll see it in all its glory - complete with type lists, traits, policies, and template templates. This is C++, and these are the abstractions the language provides for meta-programming - so take it or leave it :-) If you are seriously considering using multiple dispatch in your C++ code, Loki is well worth a look.\n\n## An Attempt for Standardization\n\nBy far the most interesting attempt to solve this problem came from Bjarne Stroustrup himself, who co-authored a paper with two of his students named \"Open Multi-Methods for C++\" . In this paper, the authors thoroughly review the problem and propose a C++ language extension that will implement it efficiently in the compiler.\n\nThe main idea is to let function arguments be potentially virtual, meaning that they perform dynamic dispatch and not just static overloading. So we could implement our intersection problem as follows:\n\n``````// This is not real C++: the syntax is based on the paper\n// \"Open Multi-Methods for C++\" and was only implemented experimentally.\n\n// Generic Shape x Shape intersection.\nvoid Intersect(virtual const Shape*, virtual const Shape*);\n\n// Interesection for Rectangle x Ellipse.\nvoid Intersect(virtual const Rectangle*, virtual const Ellipse*);``````\n\nNote how similar this is to the failed attempt to leverage overloading for multiple dispatch in the beginning of this article. All we add is the virtual keyword for arguments, and the dispatch turns from static to dynamic.\n\nUnfortunately, the proposal never made it into the standard (it was proposed as document number N2216).\n\n## Conclusions and Next Steps\n\nThis part in the series presented the multiple dispatch problem and demonstrated possible solutions in C++. Each solution has its advantages and issues, and choosing one depends on the exact needs of your project. C++ presents unique challenges in designing such high-level abstractions, because it's comparatively rigid and statically typed. Abstractions in C++ also tend to strive to being as cheap as possible in terms of runtime performance and memory consumption, which adds another dimension of complexity to the problem.\n\nIn the following parts of the series we'll examine how the same problem is solved in other, more dynamic and structurally flexible programming languages.\n\n As opposed to \"compile-time\" polymorphism which in C++ is done with overloaded functions and templates.\n\n More examples: You may have multiple event types handled by multiple handlers - mixing and matching them boils down to the same problem. Or in game code, you may have collision detection between different kinds of objects; or completely different battle scenarios depending on two kinds of units - knight vs. mage, mage vs. mage, knight vs. elf, or whatever. These examples sound like toys, but this is because realistic examples are often much more boring and more difficult to explain. Battles between mages and knights is more reasonable to discuss in an introductory article than different kinds of mathematical transforms applied to different kinds of nodes in a dataflow graph.\n\n To be more precise, this is a special case - double dispatch, where dispatch is done on two objects. I will mostly focus on double dispatch in this series, even though some of the languages and techniques presented support an arbitrary number of objects. In my experience, in 99% of the cases where multiple dispatch is useful, two objects are sufficient.\n\n I'll lament again that the \"visitor\" pattern is not a great name to apply here. An alternative way to talk about this approach is \"partial application\". With double dispatch, we route the call through two virtual method calls. The first of these can be seen to create a partially applied method that knows the dynamic type of one of its arguments, and what remains is to grab the other. This idea also extends naturally to multiple dispatch with more than 2 objects. As an exercise, try to figure out how to do triple dispatch using this technique.\n\nTopics:\npolymorphism ,abstractions\n\nComment (2)\n\nSave\n{{ articles.views | formatCount}} Views\n\nPublished at DZone with permission of Eli Bendersky . See the original article here.\n\nOpinions expressed by DZone contributors are their own."
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https://brilliant.org/practice/convert-cartesian-coordinates-to-polar/?subtopic=parametric-equations-calculus&chapter=polar-equations | [
"",
null,
"Calculus\n\n# Converting Cartesian Coordinates to Polar\n\nThe point $(5, 5\\sqrt{3})$ in Cartesian coordinates can be expressed as $(r, \\theta^{\\circ})$ in polar coordinates, where $r$ is a positive real number and $0 \\leq \\theta \\leq 180$. What is the value of $r+ \\theta$?\n\nThe point $(-21\\sqrt{3}, 21)$ in Cartesian coordinates can be expressed as $(r, \\theta)$ in polar coordinates, where r is a positive real number and $0^{\\circ} \\leq \\theta \\leq 180^{\\circ}$. If $\\theta$ is measured in degrees, what is the value of $r+ \\theta$?\n\nIf point $P$ is given in Cartesian coordinates as $P=(15, 15),$ what are the polar coordinates of $P?$\n\nThe point $(19, -19)$ in Cartesian coordinates can be expressed as $(q\\sqrt{2}, \\theta^{\\circ})$ in polar coordinates, where $q$ is a positive real number and $0 \\leq \\theta \\leq 360.$ What is the value of $q+ \\theta$?\n\nThe point $(-19\\sqrt{2}, 19\\sqrt{2})$ in Cartesian coordinates can be expressed as $(r, \\theta)$ in polar coordinates, where $r$ is a positive real number and $0^{\\circ} \\leq \\theta \\leq 180^{\\circ}$. If $\\theta$ is measured in degrees, what is the value of $r+\\theta$?\n\n×\n\nProblem Loading...\n\nNote Loading...\n\nSet Loading..."
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"https://ds055uzetaobb.cloudfront.net/brioche/chapter/Polar%20Equations-jQlloi.png",
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https://zh.wikipedia.org/zh-my/%E4%B8%89%E9%87%8D%E7%A7%AF | [
"# 三重积\n\n## 标量三重积\n\n### 定义\n\n$\\mathbf {a}$",
null,
"$\\mathbf {b}$",
null,
"$\\mathbf {c}$",
null,
"为三个向量，则标量三重积的定义为\n\n$\\mathbf {a} \\cdot (\\mathbf {b} \\times \\mathbf {c} )$",
null,
"### 特性\n\n$\\mathbf {a} =a_{1}\\mathbf {i} +a_{2}\\mathbf {j} +a_{3}\\mathbf {k}$",
null,
"$\\mathbf {b} =b_{1}\\mathbf {i} +b_{2}\\mathbf {j} +b_{3}\\mathbf {k}$",
null,
"$\\mathbf {c} =c_{1}\\mathbf {i} +c_{2}\\mathbf {j} +c_{3}\\mathbf {k}$",
null,
"，则有\n\n$\\mathbf {a} \\cdot (\\mathbf {b} \\times \\mathbf {c} )={\\begin{vmatrix}a_{1}&a_{2}&a_{3}\\\\b_{1}&b_{2}&b_{3}\\\\c_{1}&c_{2}&c_{3}\\\\\\end{vmatrix}}$",
null,
"$\\mathbf {a} \\cdot (\\mathbf {b} \\times \\mathbf {c} )$",
null,
"{\\begin{aligned}&=(a_{1}\\mathbf {i} +a_{2}\\mathbf {j} +a_{3}\\mathbf {k} )\\cdot {\\begin{vmatrix}\\mathbf {i} &\\mathbf {j} &\\mathbf {k} \\\\b_{1}&b_{2}&b_{3}\\\\c_{1}&c_{2}&c_{3}\\\\\\end{vmatrix}}\\\\&=(a_{1}\\mathbf {i} +a_{2}\\mathbf {j} +a_{3}\\mathbf {k} )\\cdot (\\mathbf {i} {\\begin{vmatrix}\\ b_{2}&b_{3}\\\\c_{2}&c_{3}\\\\\\end{vmatrix}}-\\mathbf {j} {\\begin{vmatrix}\\ b_{1}&b_{3}\\\\c_{1}&c_{3}\\\\\\end{vmatrix}}+\\mathbf {k} {\\begin{vmatrix}\\ b_{1}&b_{2}\\\\c_{1}&c_{2}\\\\\\end{vmatrix}})\\\\&=a_{1}{\\begin{vmatrix}\\ b_{2}&b_{3}\\\\c_{2}&c_{3}\\\\\\end{vmatrix}}-a_{2}{\\begin{vmatrix}\\ b_{1}&b_{3}\\\\c_{1}&c_{3}\\\\\\end{vmatrix}}+a_{3}{\\begin{vmatrix}\\ b_{1}&b_{2}\\\\c_{1}&c_{2}\\\\\\end{vmatrix}}\\\\&={\\begin{vmatrix}a_{1}&a_{2}&a_{3}\\\\b_{1}&b_{2}&b_{3}\\\\c_{1}&c_{2}&c_{3}\\\\\\end{vmatrix}}\\end{aligned}}",
null,
"$\\mathbf {a} \\cdot (\\mathbf {b} \\times \\mathbf {c} )=\\mathbf {b} \\cdot (\\mathbf {c} \\times \\mathbf {a} )=\\mathbf {c} \\cdot (\\mathbf {a} \\times \\mathbf {b} )$",
null,
"$\\mathbf {a} \\cdot (\\mathbf {b} \\times \\mathbf {c} )=-\\mathbf {a} \\cdot (\\mathbf {c} \\times \\mathbf {b} )$",
null,
"$\\mathbf {a} \\cdot (\\mathbf {b} \\times \\mathbf {c} )=-\\mathbf {b} \\cdot (\\mathbf {a} \\times \\mathbf {c} )$",
null,
"$\\mathbf {a} \\cdot (\\mathbf {b} \\times \\mathbf {c} )=-\\mathbf {c} \\cdot (\\mathbf {b} \\times \\mathbf {a} )$",
null,
"$\\mathbf {a} \\cdot (\\mathbf {a} \\times \\mathbf {b} )=\\mathbf {a} \\cdot (\\mathbf {b} \\times \\mathbf {a} )=\\mathbf {a} \\cdot (\\mathbf {b} \\times \\mathbf {b} )=\\mathbf {a} \\cdot (\\mathbf {a} \\times \\mathbf {a} )=0$",
null,
"### 其他记号\n\n$[\\mathbf {a} \\ \\mathbf {b} \\ \\mathbf {c} ]=\\mathbf {a} \\cdot (\\mathbf {b} \\times \\mathbf {c} )=(\\mathbf {a} \\times \\mathbf {b} )\\cdot \\mathbf {c}$",
null,
"### 几何意义\n\n$V=|\\mathbf {a} \\cdot (\\mathbf {b} \\times \\mathbf {c} )|=\\left|{\\begin{vmatrix}a_{1}&a_{2}&a_{3}\\\\b_{1}&b_{2}&b_{3}\\\\c_{1}&c_{2}&c_{3}\\end{vmatrix}}\\right|$",
null,
"$\\mathbf {b}$",
null,
"$\\mathbf {c}$",
null,
"来表示底面的边，则根据叉积的定义，底面的面积 $A$",
null,
"$A=|\\mathbf {b} ||\\mathbf {c} |\\sin \\theta =|\\mathbf {b} \\times \\mathbf {c} |$",
null,
"$h=|\\mathbf {a} |\\cos \\alpha$",
null,
"$\\cos \\alpha =\\pm \\cos \\beta =|\\cos \\beta |$",
null,
"$h=|\\mathbf {a} ||\\cos \\beta |$",
null,
"$V=Ah=|\\mathbf {a} ||\\mathbf {b} \\times \\mathbf {c} ||\\cos \\beta |$",
null,
"$V=|\\mathbf {a} \\cdot (\\mathbf {b} \\times \\mathbf {c} )|$",
null,
"## 向量三重积\n\n### 定义\n\n$\\mathbf {a} \\times (\\mathbf {b} \\times \\mathbf {c} )$",
null,
"$\\mathbf {a} \\times (\\mathbf {b} \\times \\mathbf {c} )\\neq (\\mathbf {a} \\times \\mathbf {b} )\\times \\mathbf {c}$",
null,
"### 特性\n\n$\\mathbf {a} \\times (\\mathbf {b} \\times \\mathbf {c} )=\\mathbf {b} (\\mathbf {a} \\cdot \\mathbf {c} )-\\mathbf {c} (\\mathbf {a} \\cdot \\mathbf {b} )$",
null,
"{\\begin{aligned}(\\mathbf {a} \\times \\mathbf {b} )\\times \\mathbf {c} &=-\\mathbf {c} \\times (\\mathbf {a} \\times \\mathbf {b} )\\\\&=-\\mathbf {a} (\\mathbf {c} \\cdot \\mathbf {b} )+\\mathbf {b} (\\mathbf {c} \\cdot \\mathbf {a} )\\end{aligned}}",
null,
"• 两个分项都带有三个向量 （$\\mathbf {a} ,\\mathbf {b} ,\\mathbf {c}$",
null,
"• 三重积一定是先做叉积的两向量之线性组合\n• 中间的向量所带的系数一定为正（此处为向量$\\mathbf {b}$",
null,
"### 证明\n\n{\\begin{aligned}(\\mathbf {u} \\times (\\mathbf {v} \\times \\mathbf {w} ))_{x}&=\\mathbf {u} _{y}(\\mathbf {v} _{x}\\mathbf {w} _{y}-\\mathbf {v} _{y}\\mathbf {w} _{x})-\\mathbf {u} _{z}(\\mathbf {v} _{z}\\mathbf {w} _{x}-\\mathbf {v} _{x}\\mathbf {w} _{z})\\\\&=\\mathbf {v} _{x}(\\mathbf {u} _{y}\\mathbf {w} _{y}+\\mathbf {u} _{z}\\mathbf {w} _{z})-\\mathbf {w} _{x}(\\mathbf {u} _{y}\\mathbf {v} _{y}+\\mathbf {u} _{z}\\mathbf {v} _{z})\\\\&=\\mathbf {v} _{x}(\\mathbf {u} _{x}\\mathbf {w} _{x}+\\mathbf {u} _{y}\\mathbf {w} _{y}+\\mathbf {u} _{z}\\mathbf {w} _{z})-\\mathbf {w} _{x}(\\mathbf {u} _{x}\\mathbf {v} _{x}+\\mathbf {u} _{y}\\mathbf {v} _{y}+\\mathbf {u} _{z}\\mathbf {v} _{z})\\\\&=(\\mathbf {u} \\cdot \\mathbf {w} )\\mathbf {v} _{x}-(\\mathbf {u} \\cdot \\mathbf {v} )\\mathbf {w} _{x}\\end{aligned}}",
null,
"{\\begin{aligned}(\\mathbf {u} \\times (\\mathbf {v} \\times \\mathbf {w} ))_{y}&=(\\mathbf {u} \\cdot \\mathbf {w} )\\mathbf {v} _{y}-(\\mathbf {u} \\cdot \\mathbf {v} )\\mathbf {w} _{y}\\\\(\\mathbf {u} \\times (\\mathbf {v} \\times \\mathbf {w} ))_{z}&=(\\mathbf {u} \\cdot \\mathbf {w} )\\mathbf {v} _{z}-(\\mathbf {u} \\cdot \\mathbf {v} )\\mathbf {w} _{z}\\end{aligned}}",
null,
"$\\mathbf {a} \\times (\\mathbf {b} \\times \\mathbf {c} )=\\mathbf {b} (\\mathbf {a} \\cdot \\mathbf {c} )-\\mathbf {c} (\\mathbf {a} \\cdot \\mathbf {b} )$",
null,
"$\\mathbf {a} \\times (\\mathbf {b} \\times \\mathbf {c} )\\;+\\mathbf {b} \\times (\\mathbf {c} \\times \\mathbf {a} )\\;+\\mathbf {c} \\times (\\mathbf {a} \\times \\mathbf {b} )=0$",
null,
"雅可比恒等式\n$(\\mathbf {a} \\times \\mathbf {b} )\\times \\mathbf {c} =\\mathbf {a} \\times (\\mathbf {b} \\times \\mathbf {c} )\\;-\\mathbf {b} \\times (\\mathbf {a} \\times \\mathbf {c} )$",
null,
"$\\nabla \\times (\\nabla \\times \\mathbf {f} )=\\nabla (\\nabla \\cdot \\mathbf {f} )-(\\nabla \\cdot \\nabla )\\mathbf {f}$",
null,
""
]
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https://trycolors.com/colors/87764B | [
"#87764B\n\nHex #87764B has hue angle of 43 degrees, value = 53 and saturation = 44. #87764B can be obtained by mixing 4 colors: 24% of RED, 24% of GREEN, 29% of WHITE, 24% of BLACK. Click \"ADJUST\" button to move #87764B to the mixer and play with it.\n\n#87764B\n24%\n4\nRED\n24%\n4\nGREEN\n29%\n5\nWHITE\n24%\n4\nBLACK\n\nMixing #87764B step by step\n\nThe diagram shows the process of mixing multiple colors step by step. Here you can see the mix of 4 drops of RED, 4 drops of GREEN, 5 drops of WHITE, 4 drops of BLACK.\n\n1\n=\n1\n1\n+\n1\n=\n2\n1\n+\n2\n=\n3\n1\n+\n3\n=\n4\n1\n+\n4\n=\n5\n1\n+\n5\n=\n6\n1\n+\n6\n=\n7\n1\n+\n7\n=\n8\n1\n+\n8\n=\n9\n1\n+\n9\n=\n10\n1\n+\n10\n=\n11\n1\n+\n11\n=\n12\n1\n+\n12\n=\n13\n1\n+\n13\n=\n14\n1\n+\n14\n=\n15\n1\n+\n15\n=\n16\n1\n+\n16\n=\n17\n\nColor #87764B conversion table\n\nHEX\n#87764B\n\nHSV\n43°, 44, 53\n\nHSL\n43°, 29, 41\n\nCIE Lab\n50.24, 0.25, 26.08\n\nRGB decimal\n135, 118, 75\n\nRGB percent\n52.9%, 46.3%, 29.4%\n\nCMYK\n0, 13, 44, 47\n\nColor name\n\nMix of color #87764B with water\n\nBelow you can see the model of the mix of #87764B with pure water. Labels indicate the transparency of the mixture.\n\n0%\n10%\n20%\n30%\n40%\n50%\n60%\n70%\n80%\n90%"
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http://book.caltech.edu/bookforum/printthread.php?s=897e22ac247c95c48aef598b32157e5c&t=4196 | [
"LFD Book Forum (http://book.caltech.edu/bookforum/index.php)\n- Chapter 1 - The Learning Problem (http://book.caltech.edu/bookforum/forumdisplay.php?f=108)\n\n vsthakur 08-10-2012 06:56 PM\n\nProblem 1.10 : Expected Off Training Error\n\nHi,\nIf I got it right, in a noiseless setting, for a fixed D, if all f are equally likely, the expected off-training-error of any hypothesis h is 0.5 (part d of problem 1.10, page 37) and hence any two algorithms are the same in terms of expected off training error (part e of the same problem).\n\nMy question is, does this not contradict the generalization by Hoeffding. Specifically, the following point is bothering me\n\nBy Hoeffding : Ein approaches Eout for larger number of hypothesis (i.e for small epsilon) as N grows sufficiently large. Which would imply that expected(Eout) should be approximately the same as expected(Ein) and not a constant (0.5).\n\nCan you please provide some insight on this, perhaps my comparison is erroneous.\n\nThanks.\n\n yaser 08-10-2012 08:25 PM\n\nRe: Problem 1.10 : Expected Off Training Error\n\nQuote:\n Originally Posted by vsthakur (Post 3952) Hi, If I got it right, in a noiseless setting, for a fixed D, if all f are equally likely, the expected off-training-error of any hypothesis h is 0.5 (part d of problem 1.10, page 37) and hence any two algorithms are the same in terms of expected off training error (part e of the same problem). My question is, does this not contradict the generalization by Hoeffding. Specifically, the following point is bothering me By Hoeffding : Ein approaches Eout for larger number of hypothesis (i.e for small epsilon) as N grows sufficiently large. Which would imply that expected(Eout) should be approximately the same as expected(Ein) and not a constant (0.5). Can you please provide some insight on this, perhaps my comparison is erroneous.\nThis is an important question, and I thank you for asking it. There is a subtle point that creates this impression of contradiction.\n\nOn face value, the statement \"all",
null,
"are equally likely\" sounds reasonable. Mathematically, it corresponds to trying to learn a randomly generated target function, and getting the average performance of this learning process. It should not be a surprise that we would get 0.5 under these circumstances, since almost all random functions are impossible to learn.\n\nIn terms of",
null,
"and",
null,
", Hoeffding inequality certainly holds for each of these random target functions, but",
null,
"itself will be close to 0.5 for almost all of these functions since they have no pattern to fit, so Hoeffding would indeed predict",
null,
"to be close to 0.5 on average.\n\nThis is why learning was decomposed into two separate questions in this chapter. In terms of these two questions, the one that \"fails\" in the random function approach is \"",
null,
"?\"\n\nLet me finally comment that treating \"all",
null,
"are equally likely\" as a plausible statement is a common trap. This issue is addressed in detail in the context of Bayesian learning in the following video segment:\n\nhttp://work.caltech.edu/library/182.html\n\n vsthakur 08-10-2012 10:56 PM\n\nRe: Problem 1.10 : Expected Off Training Error\n\nMy takeaway point : All f are equally likely corresponds to trying to learn a randomly generated target function\n\nThanks for the detailed explanation. The Bayesian Learning example highlights the ramifications of this assumption, very useful point indeed.\n\n All times are GMT -7. The time now is 12:23 PM."
]
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https://jira.mariadb.org/browse/MDEV-12313 | [
"",
null,
"# Histograms with equal-width bins in MariaDB\n\nXMLWordPrintable\n\n#### Details\n\n•",
null,
"Task\n• Status: Open\n•",
null,
"Major\n• Resolution: Unresolved\n• None\n\n#### Description\n\nHistograms with equal-width bins are easy to construct using samples. For this it's enough\nto look through the given sample set and for each value from it to figure out what bin this value can be placed in. Each bin requires only one counter.\nLet f be a column of a table with N rows and n be the number of samples by which the equal-width histogram of k bins for this column is constructed. Let after looking through all sample\nrows the counters created for the histogram bins contain numbers c,..,c[k]. Then\nm[i]= c[i]/n * 100 is the percentage of the rows whose values of f are expected to be in the interval\n\n ``` (max(f)-min(f))/k *(i-1), max(f)-min(f))/k *i-1). ```\n\nIt means that if the sample rows have been chosen randomly the expected number of rows with the values of f from this interval can be approximated by the number m[i]*/100 * N.\n\nTo collect such statistics it is suggested to use the following variant of the ANALYZE TABLE command:\n\n `ANALYZE FAST TABLE tbl [ WITH n ROWS ] [SAMPLING p PERCENTS ]` ``` PERSISTENT FOR COLUMNS (col1 [IN RANGE r] [WITH k INTERVALS],...) ```\n\nHere:\n\n• 'WITH n ROWS' provides an estimate for the number of rows in the table in the case when this estimate cannot be obtained from statistical data.\n• 'SAMPLING p PERCENTS' provides the percentage of sample rows to collect statistics.\nIf this is omitted the number is taken from the system variable samples_ratio.\n• 'IN RANGE r' sets the range of equal-width bins of the histogram built for the column col1. If this is omitted then and min and max values for the column can be read from statistical data\nthen the histogram is built for the range [min(col1), max(col1)]. Otherwise the range [MIN_type(col1), MAX_type(col1) is considered]. The values beyond the given range, if any, are also is taken into account in two additional bins.\n• WITH k INTERVALS says how many bins are included in the histogram. If it is omitted this value is taken from the system variable histogram_size.\n\n#### People",
null,
"Vicențiu Ciorbaru",
null,
"Igor Babaev"
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https://ncatlab.org/nlab/show/exceptional+spinors+and+division+algebras+--+table | [
"# nLab exceptional spinors and division algebras -- table\n\nexceptional spinors and real normed division algebras\n\nLorentzian\nspacetime\ndimension\n$\\phantom{AA}$spin groupnormed division algebra$\\,\\,$ brane scan entry\n$3 = 2+1$$Spin(2,1) \\simeq SL(2,\\mathbb{R})$$\\phantom{A}$ $\\mathbb{R}$ the real numberssuper 1-brane in 3d\n$4 = 3+1$$Spin(3,1) \\simeq SL(2, \\mathbb{C})$$\\phantom{A}$ $\\mathbb{C}$ the complex numberssuper 2-brane in 4d\n$6 = 5+1$$Spin(5,1) \\simeq$ SL(2,H)$\\phantom{A}$ $\\mathbb{H}$ the quaternionslittle string\n$10 = 9+1$Spin(9,1) ${\\simeq}$SL(2,O)$\\phantom{A}$ $\\mathbb{O}$ the octonionsheterotic/type II string\n\nLast revised on April 6, 2021 at 07:29:57. See the history of this page for a list of all contributions to it."
]
| [
null
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https://apcocoa.uni-passau.de/wiki/index.php?title=ApCoCoA-1:GLPK.IPCSolve&oldid=12028 | [
"# ApCoCoA-1:GLPK.IPCSolve\n\n## GLPK.IPCSolve\n\nSolves a system of polynomial equations over F_2 for one solution in F_2^n.\n\n### Syntax\n\n```GLPK.IPCSolve(F:LIST, QStrategy:INT, CStrategy:INT, MinMax:STRING)\n```\n\n### Description\n\nPlease note: The function(s) explained on this page is/are using the ApCoCoAServer. You will have to start the ApCoCoAServer in order to use it/them.\n\nThis function finds one solution in F_2^n of a system of polynomial equations over the field F_2. It uses Integer Polynomial Conversion (IPC) along with some strategies from propositional logic to model a mixed integer linear programming problem. Then the modelled mixed integer linear programming problem is solved using glpk.\n\n• @param F: A List containing the polynomials of the given system.\n\n• @param QStrategy: Strategy for quadratic substitution. 0 - Standard; 1 - Linear Partner; 2 - Double Linear Partner; 3 - Quadratic Partner;\n\n• @param CStrategy: Strategy for cubic substitution. 0 - Standard; and 1 - Quadratic Partner;\n\n• @param MinMax: Optimization direction i.e. minimization (\"Min\") or maximization (\"Max\").\n\n#### Example\n\n```Use Z/(2)[x[1..4]];\nF:=[\nxx + xx + xx + xx + x + x + 1,\nxx + xx + xx + xx + x + x + 1,\nxx + xx + xx + xx + x + x + 1,\nxx + xx + xx + xx + 1\n];\n\nQStrategy:=0;\nCStrategy:=0;\nMinMax:=<quotes>Max</quotes>;\n\n-- Then we compute the solution with\n\nGLPK.IPCSolve(F, QStrategy, CStrategy, MinMax);\n\n-- The result will be the following:\nModelling the system as a mixed integer programming problem.\nQStrategy: Standard, CStrategy: Standard.\nModel is ready to solve with GLPK...\n\nSolution Status: INTEGER OPTIMAL\nValue of objective function: 2\n\n[0, 1, 0, 1]\n-------------------------------\n```\n\n#### Example\n\n```Use S::=Z/(2)[x[1..5]];\nF:=[\nxx + xx + xx + x + x,\nxx + xx + xx + xx + xx + xx + x + x + x + 1,\nxx + xx + x + x + x,\nxx + xx + xx + x + x + x + x + 1,\nxx + xx + xx + xx + xx + x + x + x + x\n];\n\nQStrategy:=1;\nCStrategy:=0;\nMinMax:=<quotes>Max</quotes>;\n\n-- Then we compute the solution with\n\nGLPK.IPCSolve(F, QStrategy, CStrategy, MinMax);\n\n-- The result will be the following:\n\nModelling the system as a mixed integer programming problem.\nQStrategy: LinearPartner, CStrategy: Standard.\nModel is ready to solve with GLPK...\nSolution Status: INTEGER OPTIMAL\nValue of objective function: 4\n\n[1, 1, 1, 1, 0]\n-------------------------------\n```\n\n#### Example\n\n```Use ZZ/(2)[x[1..3]];\nF := [ xxx + xx + xx + x + x +1,\nxxx + xx + xx + x + x,\nxx + xx + x\n];\n\nQStrategy:=0;\nCStrategy:=1;\nMinMax:=<quotes>Max</quotes>;\n\n-- Then we compute the solution with\n\nGLPK.IPCSolve(F, QStrategy, CStrategy, MinMax);\n\n-- The result will be the following:\n\nModelling the system as a mixed integer programming problem.\nQStrategy: Standard, CStrategy: CubicParnterDegree2.\nModel is ready to solve with GLPK...\n\nSolution Status: INTEGER OPTIMAL\nValue of objective function: 1\n\n[0, 0, 1]\n-------------------------------\n```"
]
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https://mathstrek.blog/2013/01/21/topology-bases-and-subbases/ | [
"## Bases\n\nRecall that though a subring or ideal of a ring may be rather huge, it often suffices to specify just a few elements which will generate the subring or ideal. Likewise, in a topology, one can specify a few open sets and generate the rest via unions and finite intersections. We’ll expound upon that in what follows.\n\nFirst, note that when (Xd) is a metric space, a subset U of X is open if and only if it is a union of (possibly infinitely many) open balls. Indeed, since every open ball is open, so is a union of multiple open balls. Conversely, if U is open, each element",
null,
"$x\\in U$ is contained in some open ball",
null,
"$N(x,\\epsilon) \\subseteq U$ for some ε>0 which depends on x; hence U is the union of all these N(x, ε).\n\nIn other words, consider B = {N(x, ε) : x in X, ε>0}, the collection of all open balls; a subset of X is open if and only if it is a union of elements of B. Generalising this concept for topological spaces gives us the following definition.\n\nDefinition. A basis for a topology (X, T) is a collection of open sets,",
null,
"$B\\subseteq T$ such that every open subset U of X is a union",
null,
"$U=\\cup_i V_i$ of elements",
null,
"$V_i\\in B$. We shall call an element",
null,
"$V\\in B$ a basic open set.\n\nExamples\n\n1. As we saw above, the set B of open balls in a metric space (Xd) forms a basis of the induced topology.\n2. In particular, the set of open intervals (ab) in R forms a basis.\n3. In the discrete topology, the collection of singleton sets {x} forms a basis. In fact, if you take the collection of open balls {N(x, 1/2)} for the discrete metric, you get the same basis.\n4. Since the metrics dd1 and d on Rn give rise to the same topology, we can pick the set of open balls under any one metric to produce a basis.\n5. Exercise: find all possible bases of N under the right-order topology. [ Answer: there’s no non-trivial basis, i.e. any basis must include all open sets. ]",
null,
"Our next question is: suppose",
null,
"$B\\subseteq \\mathbf{P}(X)$ is some collection of subsets of X. Is it always the basis of some topology?\n\nIf it were, then clearly the resulting topology would be:",
null,
"$T=\\{\\cup_i V_i : \\{V_i\\}\\subseteq B\\}.$\n\nLet’s check if T satisfies the axioms of a topology.\n\n• Empty set and X : the empty set can be written as a union of an empty collection of",
null,
"$V_i$, so no problem there; for X, we need to specify that",
null,
"$\\cup_i V_i = X$.\n• Arbitrary union : this is obvious, since the union of sets",
null,
"$U_i := \\cup_j V_{ij}$ (where each",
null,
"$V_{ij} \\in B$) is also of the form",
null,
"$\\cup_{i, j} V_{ij}$ which lies in T.\n• Finite intersection : write",
null,
"$U_1 = \\cup_i V_{1i}$ and",
null,
"$U_2 = \\cup_j V_{2j}$, with each",
null,
"$V_{1i}, V_{2j}\\in B$. Since intersection is distributive over union, we get:",
null,
"$U_1\\cap U_2 = (\\cup_i V_{1i})\\cap (\\cup_j V_{2j}) = \\cup_{i,j} (V_{1i} \\cap V_{2j}).$\n\nFor this to be in T, a sufficient condition is that",
null,
"$V_1\\cap V_2\\in T$ for all",
null,
"$V_1, V_2\\in B$. On the other hand, this condition is obviously necessary since if",
null,
"$V_1, V_2\\in B$, we have",
null,
"$V_1, V_2\\in T\\implies V_1\\cap V_2 \\in T$. Thus we have proven:\n\nTheorem. A subset",
null,
"$B\\subseteq \\mathbf{P}(X)$ is a basis for some topology if and only if:\n\n• the union of all",
null,
"$V\\in B$ is the whole X; and\n• for any",
null,
"$V_1, V_2\\in B$, the intersection",
null,
"$V_1 \\cap V_2$ is a union of elements from B.\n\n## Application: Furstenberg’s Proof of the Infinitude of Primes\n\nWhile he was an undergraduate, Hillel Furstenberg found an innovative proof that there’re infinitely many prime numbers, via the concept of topology.\n\nTheorem. There are infinitely many prime numbers.\n\nProof (Furstenberg)\n\nDefine a topology on set of integers Z as follows. For integers a, b let V(a, b) be the set of all integers am+b for integer m. Then any intersection",
null,
"$V(a,b)\\cap V(c,d)$ corresponds to solutions of simultaneous linear congruences",
null,
"$x\\equiv b\\pmod a$,",
null,
"$x\\equiv d\\pmod c$. From elementary number theory, this intersection is either empty or V(e, f) for some integers e, f (where e = lcm(ac)). Since V(1, 0) = Z, {V(a, b)} forms a basis for some topology T on Z.\n\nSince the complement ZV(a, b) is a union of V(a, b’) for various b’, it is open, i.e. V(a, b) is both open and closed. Now, since any integer other than ±1 is divisible by some prime p, we have:",
null,
"$\\mathbf{Z} -\\{-1, +1\\} = \\cup_{p \\text{ prime}} V(p, 0).$\n\nIf there’re finitely many primes, then the RHS is a union of finitely many closed sets and is hence closed. Thus, {-1, +1} is open, which is ridiculous: since every basic open set V(a, b) is infinite, every open set must be infinite too. ♦",
null,
"## Subbases\n\nLet’s fix an underlying set X and consider various topologies on it. If",
null,
"$\\{T_i\\}$ is a collection of topologies on X, one easily checks that so is",
null,
"$T:=\\cap_i T_i$ where a subset of X is open in T if and only if it’s open in all Ti. In short, an intersection of topologies is still a topology; so given any subset",
null,
"$S\\subseteq \\mathbf{P}(X)$, one can consider the collection of all topologies containing S (this is a non-empty collection since it includes the discrete topology) and take their intersection.\n\nDefinition. Under the above definition, the resulting topology is called the topology generated by S. One also says that S is a subbasis for the topology T.\n\nPut in another way, T is the “smallest” topology on X containing S, in the following sense:\n\n•",
null,
"$S\\subseteq T$;\n• if T’ is any topology containing S, then",
null,
"$T\\subseteq T'$.\n\n[ This is similar to case where an arbitrary subset of a group or ring can be used to generate a subgroup or subring. ]\n\nTheorem. The topology T generated by subbasis S has, as basis,",
null,
"$B(S) := \\{W_1 \\cap W_2 \\cap \\ldots \\cap W_k : W_i\\in S\\} \\cup \\{X\\}.$\n\n[ Note: the additional term {X} may be superfluous if one interprets X as being the intersection of no Wi‘s. The more terms we intersect, the smaller the set becomes, so if we intersect no terms at all, we get the universal set X.]\n\nProof.\n\n• The intersection of any two elements of B(S) still lies in it, so B(S) is indeed a basis for some topology T’.\n• Since",
null,
"$S\\subseteq B(S)\\subseteq T'$, we have",
null,
"$T \\subseteq T'$ too, since T is the smallest topology containing S.\n• Finally, if",
null,
"$V=W_1\\cap \\ldots \\cap W_k \\in B(S)$, for some",
null,
"$W_i\\in S$, then since",
null,
"$W_i\\in T$, we have",
null,
"$V\\in T$ as well. Thus,",
null,
"$T'\\subseteq T$. ♦\n\nIn general, bases and subbases can be useful tools in topological proofs since to verify a property, it often only suffices to do it on basic open sets, or even subbasic open sets.\n\nSummary. We have:",
null,
"$\\text{Subbasis S} \\stackrel{\\footnotesize \\begin{matrix}\\text{take}\\\\ \\text{finite} \\\\ \\text{intersections}\\end{matrix}}{\\implies} \\text{Basis }B \\stackrel{\\footnotesize\\begin{matrix}\\text{take}\\\\ \\text{unions}\\end{matrix}}{\\implies} \\text{Topology }T.$\n\nExamples\n\n1. On R, the set of intervals of the form (-∞, b) and (a, ∞) forms a subbasis; indeed, the intersection gives (-∞, b) ∩ (a, ∞) = (ab) or the empty set.\n2. If |X|>2, then the collection of two-element subsets {ab} forms a subbasis since if abc are distinct, then {ab} ∩ {bc} = {a}.\n3. For the topology on Z in Furstenberg’s proof, the set of V(ab) for prime power a forms a subbasis by Chinese Remainder Theorem.\nThis entry was posted in Notes and tagged , , , , , , , . Bookmark the permalink."
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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https://www.geeksforgeeks.org/convert-given-binary-tree-to-doubly-linked-list-in-linear-time/ | [
"# Convert given Binary Tree to Doubly Linked List in Linear time\n\nGiven a Binary Tree (BT), convert it to a Doubly Linked List(DLL) In-Place. The left and right pointers in nodes are to be used as previous and next pointers respectively in converted DLL. The order of nodes in DLL must be same as Inorder of the given Binary Tree. The first node of Inorder traversal (left most node in BT) must be head node of the DLL.",
null,
"Below three different solutions have been discussed for this problem.\nConvert a given Binary Tree to Doubly Linked List | Set 1\nConvert a given Binary Tree to Doubly Linked List | Set 2\nConvert a given Binary Tree to Doubly Linked List | Set 3\n\nIn the following implementation, we traverse the tree in inorder fashion. We add nodes at the beginning of current linked list and update head of the list using pointer to head pointer. Since we insert at the beginning, we need to process leaves in reverse order. For reverse order, we first traverse the right subtree before the left subtree. i.e. do a reverse inorder traversal.\n\n## C++\n\n `// C++ program to convert a given Binary Tree to Doubly Linked List ` `#include ` ` ` `// Structure for tree and linked list ` `struct` `Node { ` ` ``int` `data; ` ` ``Node *left, *right; ` `}; ` ` ` `// Utility function for allocating node for Binary ` `// Tree. ` `Node* newNode(``int` `data) ` `{ ` ` ``Node* node = ``new` `Node; ` ` ``node->data = data; ` ` ``node->left = node->right = NULL; ` ` ``return` `node; ` `} ` ` ` `// A simple recursive function to convert a given ` `// Binary tree to Doubly Linked List ` `// root --> Root of Binary Tree ` `// head --> Pointer to head node of created doubly linked list ` `void` `BToDLL(Node* root, Node*& head) ` `{ ` ` ``// Base cases ` ` ``if` `(root == NULL) ` ` ``return``; ` ` ` ` ``// Recursively convert right subtree ` ` ``BToDLL(root->right, head); ` ` ` ` ``// insert root into DLL ` ` ``root->right = head; ` ` ` ` ``// Change left pointer of previous head ` ` ``if` `(head != NULL) ` ` ``head->left = root; ` ` ` ` ``// Change head of Doubly linked list ` ` ``head = root; ` ` ` ` ``// Recursively convert left subtree ` ` ``BToDLL(root->left, head); ` `} ` ` ` `// Utility function for printing double linked list. ` `void` `printList(Node* head) ` `{ ` ` ``printf``(``\"Extracted Double Linked list is:\\n\"``); ` ` ``while` `(head) { ` ` ``printf``(``\"%d \"``, head->data); ` ` ``head = head->right; ` ` ``} ` `} ` ` ` `// Driver program to test above function ` `int` `main() ` `{ ` ` ``/* Constructing below tree ` ` ``5 ` ` ``/ \\ ` ` ``3 6 ` ` ``/ \\ \\ ` ` ``1 4 8 ` ` ``/ \\ / \\ ` ` ``0 2 7 9 */` ` ``Node* root = newNode(5); ` ` ``root->left = newNode(3); ` ` ``root->right = newNode(6); ` ` ``root->left->left = newNode(1); ` ` ``root->left->right = newNode(4); ` ` ``root->right->right = newNode(8); ` ` ``root->left->left->left = newNode(0); ` ` ``root->left->left->right = newNode(2); ` ` ``root->right->right->left = newNode(7); ` ` ``root->right->right->right = newNode(9); ` ` ` ` ``Node* head = NULL; ` ` ``BToDLL(root, head); ` ` ` ` ``printList(head); ` ` ` ` ``return` `0; ` `} `\n\n## Java\n\n `// Java program to convert a given Binary Tree to Doubly Linked List ` ` ` `/* Structure for tree and Linked List */` `class` `Node { ` ` ``int` `data; ` ` ``Node left, right; ` ` ` ` ``public` `Node(``int` `data) ` ` ``{ ` ` ``this``.data = data; ` ` ``left = right = ``null``; ` ` ``} ` `} ` ` ` `class` `BinaryTree { ` ` ``// root --> Root of Binary Tree ` ` ``Node root; ` ` ` ` ``// head --> Pointer to head node of created doubly linked list ` ` ``Node head; ` ` ` ` ``// A simple recursive function to convert a given ` ` ``// Binary tree to Doubly Linked List ` ` ``void` `BToDLL(Node root) ` ` ``{ ` ` ``// Base cases ` ` ``if` `(root == ``null``) ` ` ``return``; ` ` ` ` ``// Recursively convert right subtree ` ` ``BToDLL(root.right); ` ` ` ` ``// insert root into DLL ` ` ``root.right = head; ` ` ` ` ``// Change left pointer of previous head ` ` ``if` `(head != ``null``) ` ` ``head.left = root; ` ` ` ` ``// Change head of Doubly linked list ` ` ``head = root; ` ` ` ` ``// Recursively convert left subtree ` ` ``BToDLL(root.left); ` ` ``} ` ` ` ` ``// Utility function for printing double linked list. ` ` ``void` `printList(Node head) ` ` ``{ ` ` ``System.out.println(``\"Extracted Double Linked List is : \"``); ` ` ``while` `(head != ``null``) { ` ` ``System.out.print(head.data + ``\" \"``); ` ` ``head = head.right; ` ` ``} ` ` ``} ` ` ` ` ``// Driver program to test the above functions ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``/* Constructing below tree ` ` ``5 ` ` ``/ \\ ` ` ``3 6 ` ` ``/ \\ \\ ` ` ``1 4 8 ` ` ``/ \\ / \\ ` ` ``0 2 7 9 */` ` ` ` ``BinaryTree tree = ``new` `BinaryTree(); ` ` ``tree.root = ``new` `Node(``5``); ` ` ``tree.root.left = ``new` `Node(``3``); ` ` ``tree.root.right = ``new` `Node(``6``); ` ` ``tree.root.left.right = ``new` `Node(``4``); ` ` ``tree.root.left.left = ``new` `Node(``1``); ` ` ``tree.root.right.right = ``new` `Node(``8``); ` ` ``tree.root.left.left.right = ``new` `Node(``2``); ` ` ``tree.root.left.left.left = ``new` `Node(``0``); ` ` ``tree.root.right.right.left = ``new` `Node(``7``); ` ` ``tree.root.right.right.right = ``new` `Node(``9``); ` ` ` ` ``tree.BToDLL(tree.root); ` ` ``tree.printList(tree.head); ` ` ``} ` `} ` ` ` `// This code has been contributed by Mayank Jaiswal(mayank_24) `\n\n## Python3\n\n `# Python3 program to convert a given Binary Tree to Doubly Linked List ` `class` `Node: ` ` ``def` `__init__(``self``, data): ` ` ``self``.data ``=` `data ` ` ``self``.left ``=` `self``.right ``=` `None` ` ` `class` `BinaryTree: ` ` ``# A simple recursive function to convert a given ` ` ``# Binary tree to Doubly Linked List ` ` ``# root --> Root of Binary Tree ` ` ``# head --> Pointer to head node of created doubly linked list ` ` ``root, head ``=` `None``, ``None` ` ` ` ``def` `BToDll(``self``, root: Node): ` ` ``if` `root ``is` `None``: ` ` ``return` ` ` ` ``# Recursively convert right subtree ` ` ``self``.BToDll(root.right) ` ` ` ` ``# Insert root into doubly linked list ` ` ``root.right ``=` `self``.head ` ` ` ` ``# Change left pointer of previous head ` ` ``if` `self``.head ``is` `not` `None``: ` ` ``self``.head.left ``=` `root ` ` ` ` ``# Change head of doubly linked list ` ` ``self``.head ``=` `root ` ` ` ` ``# Recursively convert left subtree ` ` ``self``.BToDll(root.left) ` ` ` ` ``@staticmethod` ` ``def` `print_list(head: Node): ` ` ``print``(``'Extracted Double Linked list is:'``) ` ` ``while` `head ``is` `not` `None``: ` ` ``print``(head.data, end ``=` `' '``) ` ` ``head ``=` `head.right ` ` ` `# Driver program to test above function ` `if` `__name__ ``=``=` `'__main__'``: ` ` ` ` ``\"\"\" ` ` ``Constructing below tree ` ` ``5 ` ` ``// \\\\ ` ` ``3 6 ` ` ``// \\\\ \\\\ ` ` ``1 4 8 ` ` ``// \\\\ // \\\\ ` ` ``0 2 7 9 ` ` ``\"\"\"` ` ``tree ``=` `BinaryTree() ` ` ``tree.root ``=` `Node(``5``) ` ` ``tree.root.left ``=` `Node(``3``) ` ` ``tree.root.right ``=` `Node(``6``) ` ` ``tree.root.left.left ``=` `Node(``1``) ` ` ``tree.root.left.right ``=` `Node(``4``) ` ` ``tree.root.right.right ``=` `Node(``8``) ` ` ``tree.root.left.left.left ``=` `Node(``0``) ` ` ``tree.root.left.left.right ``=` `Node(``2``) ` ` ``tree.root.right.right.left ``=` `Node(``7``) ` ` ``tree.root.right.right.right ``=` `Node(``9``) ` ` ` ` ``tree.BToDll(tree.root) ` ` ``tree.print_list(tree.head) ` ` ` `# This code is contributed by Rajat Srivastava `\n\n## C#\n\n `// C# program to convert a given Binary Tree to Doubly Linked List ` `using` `System; ` ` ` `/* Structure for tree and Linked List */` `public` `class` `Node { ` ` ``public` `int` `data; ` ` ``public` `Node left, right; ` ` ` ` ``public` `Node(``int` `data) ` ` ``{ ` ` ``this``.data = data; ` ` ``left = right = ``null``; ` ` ``} ` `} ` ` ` `class` `GFG { ` ` ``// root --> Root of Binary Tree ` ` ``public` `Node root; ` ` ` ` ``// head --> Pointer to head node of created doubly linked list ` ` ``public` `Node head; ` ` ` ` ``// A simple recursive function to convert a given ` ` ``// Binary tree to Doubly Linked List ` ` ``public` `virtual` `void` `BToDLL(Node root) ` ` ``{ ` ` ``// Base cases ` ` ``if` `(root == ``null``) ` ` ``return``; ` ` ` ` ``// Recursively convert right subtree ` ` ``BToDLL(root.right); ` ` ` ` ``// insert root into DLL ` ` ``root.right = head; ` ` ` ` ``// Change left pointer of previous head ` ` ``if` `(head != ``null``) ` ` ``head.left = root; ` ` ` ` ``// Change head of Doubly linked list ` ` ``head = root; ` ` ` ` ``// Recursively convert left subtree ` ` ``BToDLL(root.left); ` ` ``} ` ` ` ` ``// Utility function for printing double linked list. ` ` ``public` `virtual` `void` `printList(Node head) ` ` ``{ ` ` ``Console.WriteLine(``\"Extracted Double \"` ` ``+ ``\"Linked List is : \"``); ` ` ``while` `(head != ``null``) { ` ` ``Console.Write(head.data + ``\" \"``); ` ` ``head = head.right; ` ` ``} ` ` ``} ` ` ` ` ``// Driver program to test above function ` ` ``public` `static` `void` `Main(``string``[] args) ` ` ``{ ` ` ``/* Constructing below tree ` ` ``5 ` ` ``/ \\ ` ` ``3 6 ` ` ``/ \\ \\ ` ` ``1 4 8 ` ` ``/ \\ / \\ ` ` ``0 2 7 9 */` ` ` ` ``GFG tree = ``new` `GFG(); ` ` ``tree.root = ``new` `Node(5); ` ` ``tree.root.left = ``new` `Node(3); ` ` ``tree.root.right = ``new` `Node(6); ` ` ``tree.root.left.right = ``new` `Node(4); ` ` ``tree.root.left.left = ``new` `Node(1); ` ` ``tree.root.right.right = ``new` `Node(8); ` ` ``tree.root.left.left.right = ``new` `Node(2); ` ` ``tree.root.left.left.left = ``new` `Node(0); ` ` ``tree.root.right.right.left = ``new` `Node(7); ` ` ``tree.root.right.right.right = ``new` `Node(9); ` ` ` ` ``tree.BToDLL(tree.root); ` ` ``tree.printList(tree.head); ` ` ``} ` `} ` ` ` `// This code is contributed by Shrikant13 `\n\n## Javascript\n\n ` `\n\nOutput\n\n```Extracted Double Linked list is:\n0 1 2 3 4 5 6 7 8 9 ```\n\nTime Complexity: O(n), as the solution does a single traversal of given Binary Tree.\nAuxiliary Space: O(n)"
]
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"https://media.geeksforgeeks.org/wp-content/cdn-uploads/TreeToList.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.51810104,"math_prob":0.8703017,"size":6479,"snap":"2023-40-2023-50","text_gpt3_token_len":1908,"char_repetition_ratio":0.22872587,"word_repetition_ratio":0.16698474,"special_character_ratio":0.31223956,"punctuation_ratio":0.24700914,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99888295,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-29T18:56:26Z\",\"WARC-Record-ID\":\"<urn:uuid:0045f5e5-96f1-470f-b555-832aa0d2b7b1>\",\"Content-Length\":\"439425\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:417b3a66-902f-496d-9364-fe25b9aba528>\",\"WARC-Concurrent-To\":\"<urn:uuid:29d83d25-0b4f-4b8d-8a59-3700f73d69b2>\",\"WARC-IP-Address\":\"108.138.64.13\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/convert-given-binary-tree-to-doubly-linked-list-in-linear-time/\",\"WARC-Payload-Digest\":\"sha1:7TV5KNLAJBTUOHIHOAT36NO7KBFZOYHQ\",\"WARC-Block-Digest\":\"sha1:Y54VP5WO6D4BBP4TUDAHYZVWOGCGIG6X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100135.11_warc_CC-MAIN-20231129173017-20231129203017-00205.warc.gz\"}"} |
https://www.physicsforums.com/threads/linear-speed-and-rotational-quantity.242784/ | [
"# Linear speed and rotational quantity\n\nbaylorbelle\nWhat linear speed must a 6.0×10−2 hula hoop have if its total kinetic energy is to be 0.15 J ? Assume the hoop rolls on the ground without slipping.\n\nSo, i know that the formula for linear momentum is p=mv. however, the hoop is circular, meaning it has to have a rotational velocity (omega). I can't seem to figure out how the Joules play into any equation that helps link linear and rotational speeds. Anyone know of any formulas I can use for this problem?"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9525755,"math_prob":0.9002094,"size":794,"snap":"2023-14-2023-23","text_gpt3_token_len":195,"char_repetition_ratio":0.11392405,"word_repetition_ratio":0.7172414,"special_character_ratio":0.24307305,"punctuation_ratio":0.10714286,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9877342,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-20T08:44:17Z\",\"WARC-Record-ID\":\"<urn:uuid:19fb3035-f36b-4c6f-96c8-f2730f6a5ffa>\",\"Content-Length\":\"63908\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:de3071a9-eac5-4672-aa45-51ce41e92687>\",\"WARC-Concurrent-To\":\"<urn:uuid:6535410e-009e-4ade-9bc2-bc127d5c8413>\",\"WARC-IP-Address\":\"172.67.68.135\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/linear-speed-and-rotational-quantity.242784/\",\"WARC-Payload-Digest\":\"sha1:3RQGL3RIBJTWQJRUUISCDTFWNUBKZIYE\",\"WARC-Block-Digest\":\"sha1:VODC5EXKULQ25MT62TIUG2KOWPJWZ2HA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943471.24_warc_CC-MAIN-20230320083513-20230320113513-00793.warc.gz\"}"} |
https://affairscloud.com/quants-quiz-time-and-distance-set-7-boats/?amp | [
"",
null,
"Quants Questions : Time and Distance Set 7 – Boats\n\nHello Aspirants. Welcome to Online Quantitative Aptitude Section in AffairsCloud.com. Here we are creating question sample in Time and Distance- Boats and streams with Explanation, which is common for all the IBPS,SBI,SSC and other competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!\n\n1. A man can row upstream at 10 km/hr and downstream at 16 km/hr. Find the man’s rate in still water and the rate of current.\nA. 13 km/hr, 3 km/hr\nB. 10 km/hr, 2 km/hr\nC. 3 km/hr, 13 km/hr\nD. 15 km/hr, 5 km/hr\nA. 13 km/hr, 3 km/hr\nExplanation:\nman’s rate in still water = 1/2 (16+10)\nman’s rate in still water = 1/2 (16+10)\n\n2. A boat can row at 16 km/hr in still water and the speed of river is 10 km/hr. Find the speed of boat with the river and speed of boat against the river.\nA. 13 km/hr, 3 km/hr\nB. 15 km/hr, 5 km/hr\nC. 26 km/hr, 6 km/hr\nD. 6 km/hr, 26 km/hr\nC. 26 km/hr, 6 km/hr\nExplanation:\nSpeed with the river (downstream) = 16+10\nSpeed against the river (upstream) = 16-10\n\n3. A man goes downstream 60 km and upstream 20 km, taking 4 hrs each. What is the velocity of current?\nA. 4 km/hr\nB. 8 km/hr\nC. 6 km/hr\nD. 5 km/hr\nD. 5 km/hr\nExplanation:\nDownstream speed = 60/4 = 15 km/hr\nUpstream speed = 20/4 = 5 km/hr\nVelocity of stream = (15-5)/2 = 5 km/hr\n\n4. A man rows downstream 28 km and upstream 16 km, taking 5 hrs each time. What is the velocity of current?\nA. 4 km/hr\nB. 2.4 km/hr\nC. 1.2 km/hr\nD. 3 km/hr\nC. 1.2 km/hr\n\n5. A man can row 30 km upstream and 44 km downstream in 10 hrs. Also, he can row 40 km upstream and 55 km downstream in 13 hrs. Find the speed of the man in still water.\nA. 5 km/hr\nB. 8 km/hr\nC. 10 km/hr\nD. 12 km/hr\nB. 8 km/hr\nExplanation:\nLet upstream speed = x, downstream speed = y km/hr\nThen, 30/x + 44/y = 10 and 40/x + 55/y = 13\nPut 1/x = a, 1/y = b\nSolve the equations.\nA = 1/5, b = 1/11\nSo, x = 5, y = 11\nSpeed in still water = (5+11)/2 = 8\n\n6. A man can row 24 km upstream and 36 km downstream in 6 hrs. Also, he can row 36 km upstream and 24 km downstream in 6.5 hrs. Find the speed of the current.\nA. 2 km/hr\nB. 8 km/hr\nC. 10 km/hr\nD. 12 km/hr\nA. 2 km/hr\n\n7. A man can row 6 km/hr in still water. When the river is running at 2 km/hr, it takes him 1 ½ hr to row to a place and come back. How far is the place?\nA. 2.5 km\nB. 4 km\nC. 5 km\nD. 10 km\nB. 4 km\nExplanation:\nB is speed of boat in still water, R is speed of stream\nTime is total time taken for upstream and downstream\nDistance = time * [B^2 – R^2] / 2*B\n=3/2 * [6^2 – 2^2] / 2*6\n\n8. In a stream running at 2 km/hr, a motorboat goes 10 km upstream and back again to the starting point in 55 minutes. Find the speed (km/hr) of the motorboat in still water.\nA. 17\nB. 20\nC. 22\nD. 25\nC. 22\nExplanation:\nDistance = time * [B^2 – R^2] / 2*B\n10 =55/60 * [B^2 – 2^2] / 2*B\n\n9. A man can row a certain distance downstream in 2 hours and return the same distance in 6 hours. If the speed of current is 22 km/hr, find the speed of man in still water.\nA. 44km/hr\nB. 48 km/hr\nC. 50 km/hr\nD. 55 km/hr\nA. 44km/hr\nExplanation:\nUse:\nB = [tu + td] / [tu – td] * R\nB = [6+2] / [6-2] * 22\nB = 44\n\n10. A man can row 9 3/5 km/hr in still water and he finds that it takes him twice as much time to row up than as to row down the same distance in river. The speed (km/hr) of the current is\nA. 2\nB. 2 1/2\nC. 3 1/5\nD. 5"
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"data:image/svg+xml;base64,PHN2ZyBoZWlnaHQ9IjI4MiIgd2lkdGg9IjMzNiIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2ZXJzaW9uPSIxLjEiLz4=",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8894694,"math_prob":0.99287534,"size":3629,"snap":"2022-05-2022-21","text_gpt3_token_len":1330,"char_repetition_ratio":0.18896551,"word_repetition_ratio":0.110539846,"special_character_ratio":0.37779003,"punctuation_ratio":0.12855637,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98932993,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-20T18:12:29Z\",\"WARC-Record-ID\":\"<urn:uuid:f1525908-8a7d-43e4-b363-8f6b29197d73>\",\"Content-Length\":\"117112\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3cfa74b6-bb01-41ab-ab6c-434bf9b7ef96>\",\"WARC-Concurrent-To\":\"<urn:uuid:76dd2881-7463-44c7-99d5-9cdaf10f26ed>\",\"WARC-IP-Address\":\"104.26.5.172\",\"WARC-Target-URI\":\"https://affairscloud.com/quants-quiz-time-and-distance-set-7-boats/?amp\",\"WARC-Payload-Digest\":\"sha1:3GTMFKLNNETOPKUH66OE6U4GV2YM72SG\",\"WARC-Block-Digest\":\"sha1:LJU4THXFE5LS2JTUCTRK26LD4NBRXKW5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320302355.97_warc_CC-MAIN-20220120160411-20220120190411-00096.warc.gz\"}"} |
https://www.kdnuggets.com/2019/11/generalization-neural-networks.html | [
"# Generalization in Neural Networks\n\nWhen training a neural network in deep learning, its performance on processing new data is key. Improving the model's ability to generalize relies on preventing overfitting using these important methods.\n\nBy Harsha Bommana, Datakalp | Deep Learning Demystified.\n\nWhenever we train our own neural networks, we need to take care of something called the generalization of the neural network. This essentially means how good our model is at learning from the given data and applying the learnt information elsewhere.\n\nWhen training a neural network, there’s going to be some data that the neural network trains on, and there’s going to be some data reserved for checking the performance of the neural network. If the neural network performs well on the data which it has not trained on, we can say it has generalized well on the given data. Let’s understand this with an example.\n\nSuppose we are training a neural network which should tell us if a given image has a dog or not. Let’s assume we have several pictures of dogs, each dog belonging to a certain breed, and there are 12 total breeds within those pictures. I’m going to keep all the images of 10 breeds of dogs for training, and the remaining images of the 2 breeds will be kept aside for now.",
null,
"Dogs training testing data split.\n\nNow before going to the deep learning side of things, let’s look at this from a human perspective. Let’s consider a human being who has never seen a dog in their entire life (just for the sake of an example). Now we will show this human the 10 breeds of dogs and tell them that these are dogs. After this, if we show them the other 2 breeds, will they be able to tell that they are also dogs? Well hopefully they should, 10 breeds should be enough to understand and identify the unique features of a dog. This concept of learning from some data and correctly applying the gained knowledge on other data is called generalization.\n\nComing back to deep learning, our aim is to make the neural network learn as effectively from the given data as possible. If we successfully make the neural network understand that the other 2 breeds are also dogs, then we have trained a very general neural network, and it will perform really well in the real world.\n\nThis is actually easier said than done, and training a general neural network is one of the most frustrating tasks of a deep learning practitioner. This is because of a phenomenon in neural networks called overfitting. If the neural network trains on the 10 breeds of dogs and refuses to classify the other 2 breeds of dogs as dogs, then this neural network has overfitted on the training data. What this means is that the neural network has memorized those 10 breeds of dogs and considers only them to be dogs. Due to this, it fails to form a general understanding of what dogs look like. Combating this issue while training Neural Networks is what we are going to be looking at in this article.\n\nNow we don’t actually have the liberty to divide all our data on a basis like breed. Instead, we will simply split all the data. One part of the data, usually the bigger part (around 80–90%), will be used for training the model, and the rest will be used to test it. Our objective is to make sure that the performance on the testing data is around the same as the performance on the training data. We use metrics like loss and accuracy to measure this performance.\n\nThere are certain aspects of neural networks that we can control in order to prevent overfitting. Let’s go through them one by one. The first thing is the number of parameters.\n\n### Number of Parameters\n\nIn a neural network, the number of parameters essentially means the number of weights. This is going to be directly proportional to the number of layers and the number of neurons in each layer. The relationship between the number of parameters and overfitting is as follows: the more the parameters, the more the chance of overfitting. I’ll explain why.\n\nWe need to define our problem in terms of complexity. A very complex dataset would require a very complex function to successfully understand and represent it. Mathematically speaking, we can somewhat associate complexity with non-linearity. Let’s recall the neural network formula.",
null,
"Here, W1, W2, and W3 are the weight matrices of this neural network. Now what we need to pay attention to is the activation functions in the equation, which is applied to every layer. Because of these activation functions, each layer is nonlinearly connected with the next layer.\n\nThe output of the first layer is f(W_1*X) (Let it be L1), the output of the second layer is f(W_2*L1). As you can see here, because of the activation function (f), the output of the second layer has a nonlinear relationship with the first layer. So at the end of the neural network, the final value Y will have a certain degree of nonlinearity with respect to the input X depending on the number of layers in the neural network.\n\nThe more the number of layers, the more the number of activation functions disrupting the linearity between the layers, and hence the more the nonlinearity.\n\nBecause of this relationship, we can say that our neural network becomes more complex if it has more layers and more nodes in each layer. Hence we need to adjust our parameters based on the complexity of our data. There is no definite way of doing this except for repeated experimentation and comparing results.\n\nIn a given experiment, if the test score is much lower than the training score, then the model has overfit, and that means the neural network has too many parameters for the given data. This basically means that the neural network is too complex for the given data and needs to be simplified. If the test score is around the same as the training score, then the model has generalized, but this does not mean that we have reached the maximum potential of neural networks. If we increase the parameters the performance will increase, but it also might overfit. So we need to keep experimenting to optimize the number of parameters by balancing performance with generalization.\n\nWe need to match the neural network’s complexity with our data complexity. If the neural network is too complex, it will start memorizing the training data instead of having a general understanding of the data, hence causing overfitting.\n\nUsually how deep learning practitioners go about this is to first train a neural network with a sufficiently high number of parameters such that the model will overfit. So initially, we try to get a model that fits extremely well on the training data. Next we try and reduce the number of parameters iteratively until the model stops overfitting, this can be considered as an optimal neural network. Another technique that we can use to prevent overfitting is using dropout neurons.\n\n### Dropout Neurons\n\nIn neural networks, adding dropout neurons is one of the most popular and effective ways to reduce overfitting in neural networks. What happens in dropout is that essentially each neuron in the network has a certain probability of completely dropping out from the network. This means that at a particular instant, there will be certain neurons that will not be connected to any other neuron in the network. Here’s a visual example:",
null,
"At every instant during training, a different set of neurons will be dropped out in a random fashion. Hence we can say that at each instant we are effectively training a certain subset neural network that has fewer neurons than the original neural network. This subset neural network will change each and every time because of the random nature of the dropout neurons.\n\nWhat essentially happens here is that while we train a neural network with dropout neurons, we are basically training many smaller subset neural networks and since the weights are a part of the original neural network, the final weights of the neural network can be considered as an average of all the corresponding subset neural network weights. Here’s a basic visualization of what’s going on:",
null,
"This is how dropout neurons work in a neural network, but why does dropout prevent overfitting? There are two main reasons for this.\n\nThe first reason is that dropout neurons promote neuron independence. Because of the fact that the neurons surrounding a particular neuron may or may not exist during a certain instant, that neuron cannot rely on those neurons which surround it. Hence it will be forced to be more independent while training.\n\nThe second reason is that because of dropout, we are essentially training multiple smaller neural networks at once. In general, if we train multiple models and average their weights, the performance usually increases because of the accumulation of independent learnings of each neural network. However, this is an expensive process since we need to define multiple neural networks and train them individually. However, in the case of dropout, this does the same thing while we need only one neural network from which we are training multiple possible configurations of subset neural networks.\n\nTraining multiple neural networks and aggregating their learnings is called “ensembling” and usually boosts performance. Using dropout essentially does this with while having only 1 neural network.\n\nThe next technique for reducing overfitting is weight regularization.\n\n### Weight Regularization\n\nWhile training neural networks, there is a possibility that the value of certain weights can become very large. This happens because these weights are focusing on certain features in the training data, which is causing them to increase in value continuously throughout the training process. Because of this, the network overfits on the training data.\n\nWe don’t need the weight to continuously increase to capture a certain pattern. Instead, it’s fine if they have a value that is higher than the other weights on a relative basis. But during the training process while a neural network is trained on the data over multiple iterations, the weights have a possibility of constantly increasing in value till they become huge, which is unnecessary.\n\nOne of the other reasons why huge weights are bad for a neural network is because of the increased input-output variance. Basically when there is a huge weight in the network, it is very susceptible to small changes in the input, but the neural network should essentially output the same thing for similar inputs. When we have huge weights, even when we keep two separate data inputs, that are very similar, there is a chance that the outputs vastly differ. This causes many incorrect predictions to occur on the testing data, hence decreasing the generalization of the neural network.\n\nThe general rule of weights in neural networks is that the higher the weights in the neural network, the more complex the neural network. Because of this, neural networks having higher weights generally tend to overfit.\n\nSo basically, we need to limit the growth of the weights so that they don’t grow too much, but how exactly do we go about this? Neural networks try to minimize the loss while training, so we can try to include a part of the weights in that loss function so that weights are also minimized while training, but of course decreasing the loss in the first priority.\n\nThere are two methods of doing this called the L1 and L2 regularization. In L1 we take a small part of the sum of all the absolute values of the weights in the network. In L2, we take a small part of the sum of all the squared values of the weights in the network. We just add this expression to the overall loss function of the neural network. The equations are as follows:",
null,
"",
null,
"Here, lambda is a value that allows us to alter the extent of weight change. We basically just add the L1 or L2 terms to the loss function of the neural net so that the network will also try to minimize these terms. By adding L1 or L2 regularization, the network will limit the growth of its weights since the magnitude of the weights are a part of the loss function, and the network always tries to minimize the loss function. Let’s highlight some of the differences between L1 and L2.\n\nWith L1 regularization, while a weight is decreasing due to regularization, L1 tries to push it down completely to zero. Hence unimportant weights that aren’t contributing much to the neural network will eventually become zero. However, in the case of L2, since the square function becomes inversely proportional for values below 1, the weights aren’t pushed to zero, but they are pushed to small values. Hence the unimportant weights have much lower values than the rest.\n\nThat covers the important methods of preventing overfitting. In deep learning, we usually use a mix of these methods to improve the performance of our neural networks and to improve the generalization of our models.\n\nOriginal. Reposted with permission.\n\nRelated:",
null,
"Get the FREE ebook 'The Great Big Natural Language Processing Primer' and 'The Complete Collection of Data Science Cheat Sheets' along with the leading newsletter on Data Science, Machine Learning, AI & Analytics straight to your inbox.",
null,
"",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9489392,"math_prob":0.9519077,"size":12990,"snap":"2023-40-2023-50","text_gpt3_token_len":2574,"char_repetition_ratio":0.1858155,"word_repetition_ratio":0.031235853,"special_character_ratio":0.19314858,"punctuation_ratio":0.08469722,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9843203,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-07T23:55:22Z\",\"WARC-Record-ID\":\"<urn:uuid:8fc64a51-9415-419c-867c-1a29ecd8a60d>\",\"Content-Length\":\"130027\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:936ac514-bc77-4a1d-b53b-b994905fab1c>\",\"WARC-Concurrent-To\":\"<urn:uuid:285393ef-5b06-4f7d-86bb-8e95b8586e45>\",\"WARC-IP-Address\":\"104.26.3.64\",\"WARC-Target-URI\":\"https://www.kdnuggets.com/2019/11/generalization-neural-networks.html\",\"WARC-Payload-Digest\":\"sha1:EYWQLRL6JYQ3C76HWHXFDR3BEL4QTWHE\",\"WARC-Block-Digest\":\"sha1:HAS4NBVJK32I67WRD267JE6NNTBFPTLW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100705.19_warc_CC-MAIN-20231207221604-20231208011604-00106.warc.gz\"}"} |
https://mathoverflow.net/questions/403218/what-is-the-homotopy-category-of-the-sphere-spectrum | [
"# What is the homotopy category of the sphere spectrum?\n\nIs there a known explicit description of the abelian $$2$$-group $$\\mathsf{Ho}(\\mathbb{S})\\overset{\\mathrm{def}}{=}\\mathsf{Ho}(QS^0)\\cong\\Pi_{\\leq1}(QS^0)$$?\n\n• This question looks a bit cumbersome to me. Are not you asking about the fundamental groupoid of the infinite loop space of the sphere spectrum? Sep 5, 2021 at 8:55\n• @FernandoMuro Sorry, I was a bit confused.\n– Théo\nSep 5, 2021 at 19:08\n• The easiest description is that it is the stable quadratic module $\\mathbb{Z}\\otimes\\mathbb{Z}\\twoheadrightarrow\\mathbb{Z}/(2)\\stackrel{0}{\\rightarrow}\\mathbb{Z}$ Sep 6, 2021 at 8:31\n• In case you're not acquainted with stable quadratic modules, if you want to see it as a symmetric monoidal groupoid, then the object set is $\\mathbb{Z}$, the automorphism group of an object is $\\{\\pm1\\}$, there are no morphisms other than automorphisms, the tensor product is addition on objects and multiplication on morphisms, the associativity and unitarity constraints are identities, and the commutativity constraint $m+n\\rightarrow n+m$ is $(-1)^{mn}$. Sep 6, 2021 at 8:34\n• @FernandoMuro I had heard of stable quadratic modules before, but at the time I had trouble finding any reference to learn about them. This time however I found a paper you wrote pointing to Baues, so I can finally understand what they are now. Thank you!\n– Théo\nSep 6, 2021 at 22:55\n\nThis is the groupoid given by the 1-truncation $$\\tau_{\\leq 1}(QS^0)$$. This groupoid has $$\\mathbb Z$$-many objects (since $$\\pi_0^s = \\mathbb Z$$), and each one has automorphism group $$C_2$$ (since $$\\pi_1^s = C_2$$). The tensor product on objects is given by addition in $$\\mathbb Z$$, and on morphisms by addition in $$C_2$$. One way to see this is to consider the universal functor $$\\Sigma \\to QS^0$$ given by the Barratt-Priddy-Quillen theorem (i.e. the fact that $$K(\\Sigma) = QS^0$$; here $$\\Sigma$$ is the groupoid of finite sets with the disjoint union monoidal structure), and to postcompose with the truncation functor $$QS^0 \\to \\tau_{\\leq 1} (QS^0)$$; the fact that this functor is symmetric monoidal yields this description of the category. This perspective is discussed a bit more here.\nFrom the description I've given, I suppose it follows that $$\\tau_{\\leq 1} (QS^0)$$ splits symmetric monoidally as $$\\tau_{\\leq 1} (QS^0) = \\mathbb Z \\times BC_2$$, (where $$\\mathbb Z$$ is a discrete symmetric monoidal groupoid and $$BC_2$$ is a 1-object symmetric monoidal groupoid), which is maybe a little surprising. This is not to say that $$\\tau_{\\leq 1} \\mathbb S$$ splits..."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8654386,"math_prob":0.9988556,"size":3250,"snap":"2022-27-2022-33","text_gpt3_token_len":965,"char_repetition_ratio":0.124460876,"word_repetition_ratio":0.00814664,"special_character_ratio":0.28523076,"punctuation_ratio":0.094462544,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999894,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-02T16:53:40Z\",\"WARC-Record-ID\":\"<urn:uuid:a7807038-5f02-435b-b94e-39315c30b81d>\",\"Content-Length\":\"113477\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f7f48965-d9ee-43dc-a368-3d122027be32>\",\"WARC-Concurrent-To\":\"<urn:uuid:6ace472a-103a-4644-8d64-9561eb021d8d>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/403218/what-is-the-homotopy-category-of-the-sphere-spectrum\",\"WARC-Payload-Digest\":\"sha1:5BUFKLOPQZIAJQI6TYQZK6BH6JRYKLVN\",\"WARC-Block-Digest\":\"sha1:57ZDYBS7TYH63QIFUV4OC6MZILV5REOR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104189587.61_warc_CC-MAIN-20220702162147-20220702192147-00137.warc.gz\"}"} |
https://brilliant.org/problems/a-problem-by-mark-allen-facun/ | [
"Polynomial Function\n\nAlgebra Level 3\n\nA third-degree polynomial $P(x)$, satisfies $P(0)= -3$ and $P(1)= 4$. When $P(x)$ is divided by $x^{2}+ x+ 1$, the remainder is $2x -1$. What is the quotient when $P(x)$ is divided by $x^2 + x + 1$?\n\n×"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6933014,"math_prob":1.00001,"size":357,"snap":"2019-26-2019-30","text_gpt3_token_len":155,"char_repetition_ratio":0.16713881,"word_repetition_ratio":0.0,"special_character_ratio":0.44257703,"punctuation_ratio":0.15238096,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000099,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-25T00:39:13Z\",\"WARC-Record-ID\":\"<urn:uuid:b7d4d7ed-5823-4109-9199-b1da5edc9c50>\",\"Content-Length\":\"51239\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:66a82cd7-907d-4da6-9c0c-2a799b027fdc>\",\"WARC-Concurrent-To\":\"<urn:uuid:b99b016d-e0ca-4107-ac46-c4c85649e9c0>\",\"WARC-IP-Address\":\"104.20.35.242\",\"WARC-Target-URI\":\"https://brilliant.org/problems/a-problem-by-mark-allen-facun/\",\"WARC-Payload-Digest\":\"sha1:PTUV24US4EAMC4K726Z3OBTFF7JWWJIA\",\"WARC-Block-Digest\":\"sha1:GIUYTKQG5VO57GOLE2ECQSTLBDK7WQVH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999779.19_warc_CC-MAIN-20190624231501-20190625013501-00446.warc.gz\"}"} |
https://eudml.org/subject/MSC/20G15 | [
"Page 1 Next\n\n## Displaying 1 – 20 of 257\n\nShowing per page\n\n### $\\left(*\\right)$-groups and pseudo-bad groups.\n\nRevista Colombiana de Matemáticas\n\n### A Canonical Filtration for Certain Rational Modules.\n\nMathematische Zeitschrift\n\n### A classification of reductive linear groups.\n\nJournal of Lie Theory\n\n### A classification of the nilpotent triangular matrices\n\nCompositio Mathematica\n\n### A Filtration for Rational Modules.\n\nMathematische Zeitschrift\n\n### A modular compactification of the general linear group.\n\nDocumenta Mathematica\n\nSemigroup forum\n\n### A note on the characters of the projective modules for the infinitesimal subgroups of a semisimple algebraic group.\n\nMathematica Scandinavica\n\n### A Specialization Theorem for Certain Weyl Group Representations and an Application to the Green Polynomials of Unitary Groups.\n\nInventiones mathematicae\n\n### A Stiefel complex for the orthogonal group of a field.\n\nCommentarii mathematici Helvetici\n\n### Abstract homomorphisms of simple algebraic groups\n\nSéminaire Bourbaki\n\n### Algebraic groups with a distinguished conjugacy class.\n\nForum mathematicum\n\nSemigroup forum\n\n### Algebraic zip data.\n\nDocumenta Mathematica\n\n### An alternative proof of Scheiderer's theorem on the Hasse principle for principal homogeneous spaces.\n\nDocumenta Mathematica\n\n### Automorphism groups of polycyclic-by-finite groups and arithmetic groups\n\nPublications Mathématiques de l'IHÉS\n\nWe show that the outer automorphism group of a polycyclic-by-finite group is an arithmetic group. This result follows from a detailed structural analysis of the automorphism groups of such groups. We use an extended version of the theory of the algebraic hull functor initiated by Mostow. We thus make applicable refined methods from the theory of algebraic and arithmetic groups. We also construct examples of polycyclic-by-finite groups which have an automorphism group which does not contain an arithmetic...\n\n### Automorphismes et deformations des varietes de Borel\n\nInventiones mathematicae\n\n### Autour d'une conjecture de Serge Lang.\n\nInventiones mathematicae\n\n### Central extensions of reductive groups by ${K}_{2}$\n\nPublications Mathématiques de l'IHÉS\n\n### Classification des espaces homogènes sphériques\n\nCompositio Mathematica\n\nPage 1 Next"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9136915,"math_prob":0.464072,"size":542,"snap":"2022-40-2023-06","text_gpt3_token_len":106,"char_repetition_ratio":0.1394052,"word_repetition_ratio":0.0,"special_character_ratio":0.17527676,"punctuation_ratio":0.08421053,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9682668,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-02T03:19:02Z\",\"WARC-Record-ID\":\"<urn:uuid:484e65b8-1f54-4c64-8d08-2643d30e5261>\",\"Content-Length\":\"47476\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:722b0d34-cf4e-4491-9f58-8cd694298b94>\",\"WARC-Concurrent-To\":\"<urn:uuid:ddddf4dc-f590-4306-aa77-9e8d731eb04c>\",\"WARC-IP-Address\":\"213.135.60.110\",\"WARC-Target-URI\":\"https://eudml.org/subject/MSC/20G15\",\"WARC-Payload-Digest\":\"sha1:5AIDMRJAKSDL3XCUYZVTHYAIFQL7HCLR\",\"WARC-Block-Digest\":\"sha1:XPDTEVNRS3WZ3U2R3JV6MEZ33KBMVR25\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337244.17_warc_CC-MAIN-20221002021540-20221002051540-00738.warc.gz\"}"} |
https://dup4.cn/Leetcode/problems/404.sum-of-left-leaves/ | [
"# 404.sum-of-left-leaves\n\n## Statement\n\n• Difficulty: Easy\n• Tag: `树` `深度优先搜索` `广度优先搜索` `二叉树`",
null,
"``````输入: root = [3,9,20,null,null,15,7]\n\n``````\n\n``````输入: root = \n\n``````\n\n• 节点数在 `[1, 1000]` 范围内\n• `-1000 <= Node.val <= 1000`\n\n• Link: Sum of Left Leaves\n• Difficulty: Easy\n• Tag: `Tree` `Depth-First Search` `Breadth-First Search` `Binary Tree`\n\nGiven the `root` of a binary tree, return the sum of all left leaves.\n\nA leaf is a node with no children. A left leaf is a leaf that is the left child of another node.\n\nExample 1:",
null,
"``````Input: root = [3,9,20,null,null,15,7]\nOutput: 24\nExplanation: There are two left leaves in the binary tree, with values 9 and 15 respectively.\n``````\n\nExample 2:\n\n``````Input: root = \nOutput: 0\n``````\n\nConstraints:\n\n• The number of nodes in the tree is in the range `[1, 1000]`.\n• `-1000 <= Node.val <= 1000`\n\n## Solution\n\n``````# Definition for a binary tree node.\n# class TreeNode:\n# def __init__(self, val=0, left=None, right=None):\n# self.val = val\n# self.left = left\n# self.right = right\nfrom typing import Optional\n\nclass Solution:\ndef __init__(self):\nself.res = 0\n\ndef dfs(self, root: Optional[TreeNode], is_left: bool) -> None:\nif not root:\nreturn\n\nson = 0\nif root.left:\nson += 1\nself.dfs(root.left, True)\n\nif root.right:\nson += 1\nself.dfs(root.right, False)\n\nif son == 0 and is_left:\nself.res += root.val\n\ndef sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int:\nself.dfs(root, False)\nreturn self.res\n``````"
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"https://dup4.cn/Leetcode/problems/404.sum-of-left-leaves/problem-assets/https:--assets.leetcode.com-uploads-2021-04-08-leftsum-tree.jpg",
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"https://dup4.cn/Leetcode/problems/404.sum-of-left-leaves/problem-assets/https:--assets.leetcode.com-uploads-2021-04-08-leftsum-tree.jpg",
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http://7puzzleblog.com/29/ | [
"# day/dydd 29 at 7puzzleblog.com",
null,
"The Main Challenge\n\nWhat is the 7th 2-digit whole number that is both a multiple of 5 AND contains the letter ‘F’ when written in English?",
null,
"The 7puzzle Challenge\n\nThe playing board of the 7puzzle game is a 7-by-7 grid containing 49 different numbers, ranging from up to 84.\n\nThe 3rd & 6th rows contain the following fourteen numbers:\n\n5 12 13 18 20 25 33 36 42 45 49 56 66 80\n\nHow many pairs of numbers have a difference of 24?",
null,
"The Lagrange Challenge\n\nLagrange’s Four-Square Theorem states that every positive integer can be made by adding up to four square numbers.\n\nFor example, 7 can be made by 2²+1²+1²+1² (or 4+1+1+1).\n\nThere are TWO ways of making 29 when using Lagrange’s Theorem. Can you find them both?",
null,
"The Mathematically Possible Challenge\n\nUsing 26 and 11 once each, with + – × ÷ available, which are the only TWO numbers it is possible to make from the list below?\n\n5 10 15 20 25 30 35 40 45 50\n\n#5TimesTable\n\nThe Target Challenge\n\nCan you arrive at 29 by inserting 2, 3, 4 and 5 into the gaps on each line?\n\n• (◯+◯)×+◯ = 29 (there are two ways)\n• ◯×◯×◯+◯ = 29\n• ◯²+◯×◯–◯ = 29",
null,
"",
null,
"",
null,
""
]
| [
null,
"http://7puzzleblog.com/wp-content/uploads/2018/10/PaulRoundImage2-298x300.jpg",
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"http://7puzzleblog.com/wp-content/uploads/2017/06/7-newimage.jpg",
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"http://7puzzleblog.com/wp-content/uploads/2017/06/7-newimage.jpg",
null,
"http://7puzzleblog.com/wp-content/uploads/2017/06/7-newimage.jpg",
null,
"http://7puzzleblog.com/wp-content/uploads/2018/10/PaulRoundImage2-298x300.jpg",
null,
"http://7puzzleblog.com/wp-content/uploads/2017/06/7-newimage.jpg",
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"http://7puzzleblog.com/wp-content/uploads/2018/01/7puzzleblog2-dnp-1-e1515248090297-300x100.jpg",
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https://stacks.math.columbia.edu/tag/070Y | [
"Lemma 13.31.6. Let $\\mathcal{A}$ be an abelian category. Let $F : K(\\mathcal{A}) \\to \\mathcal{D}'$ be an exact functor of triangulated categories. Then $RF$ is defined at every complex in $K(\\mathcal{A})$ which is quasi-isomorphic to a K-injective complex. In fact, every K-injective complex computes $RF$.\n\nProof. By Lemma 13.14.4 it suffices to show that $RF$ is defined at a K-injective complex, i.e., it suffices to show a K-injective complex $I^\\bullet$ computes $RF$. Any quasi-isomorphism $I^\\bullet \\to N^\\bullet$ is a homotopy equivalence as it has an inverse by Lemma 13.31.2. Thus $I^\\bullet \\to I^\\bullet$ is a final object of $I^\\bullet /\\text{Qis}(\\mathcal{A})$ and we win. $\\square$\n\nThere are also:\n\n• 5 comment(s) on Section 13.31: K-injective complexes\n\nIn your comment you can use Markdown and LaTeX style mathematics (enclose it like $\\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar)."
]
| [
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https://astronu.jinr.ru/wiki/index.php?title=NOvA_results&oldid=2288 | [
"# Analysis 2019\n\nThe best-fit parameters of neutrino oscillation for each choice of",
null,
"$\\theta_{23}$ octant and hierarchy. The best-fit point is found for the normal hierarchy",
null,
"$\\theta_{23}$ in the upper octant. In this case:",
null,
"$\\Delta m_{32}^2 = +2.48^{=0.11}_{-0.06} \\times 10^{-3} eV^2/c^4$,",
null,
"$sin^2 \\theta_{23} = 0.56^{+0.04}_{-0.03}$,\n\nFailed to parse(unknown function '\\dalta'): \\dalta_{CP} = 0.0^{+1.3}_{-0.4} \\pi .\n\nOther results is showen by table:\n\n<\\tr> <\\tr> <\\tr> <\\tr>\n parameters NO, LO IH, UO IH, LO",
null,
"$\\Delta m_{32}^2/ (10^{-3} eV^2/c^4)$ +2.47 -2.54 -2.53",
null,
"$sin^2 \\theta_{23}$ 0.48 0.56 0.47 Failed to parse(unknown function '\\dalta'): \\dalta_{CP} 1.9 1.5 1.4 [/itex] +1.6",
null,
"$\\sigma$ +1.8",
null,
"$\\sigma$ +2.0",
null,
"$\\sigma$\n\n# Second Analysis\n\n## Numu disappearance results (Preliminary)\n\nBest fit in Normal Hierarchy",
null,
"$|\\Delta m_{32}^2 | = (2.67 \\pm 0.12 ) \\times 10^{-3} eV^2$",
null,
"$sin^2 \\theta_{23} = 0.40^{+0.03}_{-0.02} (0.63^{+0.02}_{-0.03})$\n\nBest fit in Inverted Hierarchy",
null,
"$|\\Delta m_{32}^2 | = (-2.71 \\pm 0.12 ) \\times 10^{-3} eV^2$",
null,
"$sin^2 \\theta_{23} = 0.40^{+0.03}_{-0.02} (0.62^{+0.02}_{-0.03})$\n\nMaximal mixing excluded at",
null,
"$2.5 \\sigma$"
]
| [
null,
"https://astronu.jinr.ru/wiki/images/math/8/b/9/8b9566a93731d3611bba489887c6f217.png ",
null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
"https://astronu.jinr.ru/wiki/images/math/0/c/3/0c3995eee8413bc525b707d3e5a03a02.png ",
null,
"https://astronu.jinr.ru/wiki/images/math/c/e/6/ce6e4fb02ebc80669728bc50b2d9e8e0.png ",
null
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http://fmjournal.org/article/382/10.11648.j.fm.20190501.13 | [
"Archive\nSpecial Issues",
null,
"Volume 5, Issue 1, June 2019, Page: 15-25\nA Class of Exact Solutions for a Variable Viscosity Flow with Body Force for Moderate Peclet Number ViaVon-Mises Coordinates\nMushtaq Ahmed, Department of Mathematics, University of Karachi, Karachi, Pakistan\nReceived: Mar. 13, 2019; Accepted: May 7, 2019; Published: Jun. 4, 2019\nAbstract\nThe objective of this article is to communicate a class of new exact solutions of the plane equation of momentum with body force, energy and continuity for moderate Peclet number in von-Mises coordinates. Viscosity of fluid is variable but its density and thermal conductivity are constant. The class characterizes the streamlines pattern through an equation relating two continuously differentiable functions and a function of stream function ψ. Applying the successive transformation technique, the basic equations are prepared for exact solutions. It finds exact solutions for class of flows for which the function of stream function varies linearly and exponentially. The linear case shows viscosity and temperature for moderate Peclet number for two variety of velocity profile. The first velocity profile fixes both the functions of characteristic equation whereas the second keeps one of them arbitrary. The exponential case finds that the temperature distribution, due to heat generation, remains constant for all Peclet numbers except at 4 where it follows a specific formula. There are streamlines, velocity components, viscosity and temperature distribution in presence of body force for a large number of the finite Peclet number.\nKeywords\nSuccessive Transformation Technique, Variable Viscosity Fluids, Navier-Stokes Equations with Body Force, Martin’s Coordinates, Von-MisesCoordinates\nMushtaq Ahmed, A Class of Exact Solutions for a Variable Viscosity Flow with Body Force for Moderate Peclet Number ViaVon-Mises Coordinates, Fluid Mechanics. Vol. 5, No. 1, 2019, pp. 15-25. doi: 10.11648/j.fm.20190501.13\nReference\n\nNaeem, R. K.; Exact solutions of flow equations of an incompressible fluid of variable viscosity via one – parameter group: The Arabian Journal for Science and Engineering, 1994, 19 (1), 111-114.\n\nNaeem, R. K.; Srfaraz, A. N.; Study of steady plane flows of an incompressible fluid of variable viscosity using Martin’s System: Journal of Applied Mechanics and Engineering, 1996, 1 (1), 397-433.\n\nNaeem, R. K.; Steady plane flows of an incompressible fluid of variable viscosity via Hodograph transformation method: Karachi University Journal of Sciences, 2003, 3 (1), 73-89.\n\nNaeem, R. K.; On plane flows of an incompressible fluid of variable viscosity: Quarterly Science Vision, 2007, 12 (1), 125-131.\n\nNaeem, R. K.; Mushtaq A.; A class of exact solutions to the fundamental equations for plane steady incompressible and variable viscosity fluid in the absence of body force: International Journal of Basic and Applied Sciences, 2015, 4 (4), 429-465. www.sciencepubco.com/index.php/IJBAS, doi: 10.14419/ijbas.v4i4.5064.\n\nGerbeau, J. -F.; Le Bris, C., A basic Remark on Some Navier-Stokes Equations With Body Forces: Applied Mathematics Letters, 2000, 13 (1), 107-112.\n\nGiga, Y.; Inui, K.; Mahalov; Matasui S.; Uniform local solvability for the Navier-Stokes equations with the Coriolis force: Method and application of Analysis, 2005, 12, 381-384.\n\nLandau L. D. and Lifshitz E. M.; Fluid Mechanics, Pergmaon Press, vol 6.\n\nMushtaq A., On Some Thermally Conducting Fluids: Ph. D Thesis, Department of Mathematics, University of Karachi, Pakistan, 2016.\n\nMushtaq A.; Naeem R. K.; S. Anwer Ali; A class of new exact solutions of Navier-Stokes equations with body force for viscous incompressible fluid,: International Journal of Applied Mathematical Research, 2018, 7 (1), 22-26. www.sciencepubco.com/index.php/IJAMR, doi: 10.14419/ijamr.v7i1.8836.\n\nMushtaq Ahmed, Waseem Ahmed Khan,: A Class of New Exact Solutions of the System ofPDEfor the plane motion of viscous incompressible fluids in the presence of body force,: International Journal of Applied Mathematical Research, 2018, 7 (2), 42-48. www.sciencepubco.com/index.php/IJAMR, doi: 10.14419/ijamr.v7i2.9694.\n\nMushtaq Ahmed, Waseem Ahmed Khan, S. M. Shad Ahsen: A Class of Exact Solutions of Equations for Plane Steady Motion of Incompressible Fluids of Variable viscosity in presence of Body Force: International Journal of Applied Mathematical Research, 2018, 7 (3), 77-81. www.sciencepubco.com/index.php/IJAMR, doi: 10.14419/ijamr.v7i2.12326.\n\nMushtaq Ahmed, (2018), A Class of New Exact Solution of equations for Motion of Variable Viscosity Fluid In presence of Body Force with Moderate Peclet number, International Journal of Fluid Mechanics and Thermal Sciences, 4 (4)429- www.sciencepublishingdroup.com/j/ijfmts doi: 10.11648/j.ijfmts.20180401.12.\n\nD. L. R. Oliver & K. J. De Witt, High Peclet number heat transfer from adroplet suspended in an electric field: Interior problem, Int. J. Heat Mass Transfer, vol. 36: 3153-3155, 1993.\n\nZ. G. Feng, E. E. Michaelides, Unsteady heat transfer from a spherical particle atfinite Peclet numbers, J. Fluids Eng. 118: 96-102, 1996.\n\nFayerweather Carl, Heat Transfer From a Droplet at Moderate Peclet Numbers with heat Generation. PhD. Thesis, U of Toledo, May 2007.\n\nMartin, M. H.; The flow of a viscous fluid I: Archive for Rational Mechanics and Analysis, 1971, 41 (4), 266-286.\n\nDaniel Zwillinger; Handbook of differential equations; Academic Press, Inc. (1989).\n\nWeatherburn C. E., Differential geometry of three Dimensions, Cambridge University Press, (1964).",
null,
""
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null,
"http://fmjournal.org/spgj/decorator/img/article/article_r2_c3.jpg",
null,
"http://fmjournal.org/spgj/decorator/img/home/home_r16_c13.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.69587356,"math_prob":0.61231863,"size":6282,"snap":"2019-51-2020-05","text_gpt3_token_len":1654,"char_repetition_ratio":0.10640331,"word_repetition_ratio":0.14141414,"special_character_ratio":0.26822668,"punctuation_ratio":0.23286605,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.959298,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,7,null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-11T08:57:12Z\",\"WARC-Record-ID\":\"<urn:uuid:5faa04e8-b969-433d-b535-254d69e68da8>\",\"Content-Length\":\"40476\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:60f8c0eb-ab2b-4415-9d15-39a1df885082>\",\"WARC-Concurrent-To\":\"<urn:uuid:4131444d-6db3-4d51-9ba7-ba9d17c8c6f5>\",\"WARC-IP-Address\":\"47.88.20.168\",\"WARC-Target-URI\":\"http://fmjournal.org/article/382/10.11648.j.fm.20190501.13\",\"WARC-Payload-Digest\":\"sha1:U3RE323VV6IP3JPS456DWFD26NHENOO3\",\"WARC-Block-Digest\":\"sha1:MOXBPOMZKN6KGSXNPZ527BF7QY24QITU\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540530452.95_warc_CC-MAIN-20191211074417-20191211102417-00543.warc.gz\"}"} |
https://www.dsprelated.com/showarticle/795.php | [
"# Fitting a Damped Sine Wave\n\nA damped sine wave is described by\n\n$$x_{(k)} = A \\cdot e^{\\alpha \\cdot k} \\cdot cos(\\omega \\cdot k + p) \\tag{1}$$\n\nwith frequency $\\omega$ , phase p , initial amplitude A and damping constant $\\alpha$ . The $x_{(k)}$ are the samples of the function at equally spaced points in time.\n\nWith $x_{(k)}$ given, one often has to find the unknown parameters of the function. This can be achieved for instance with nonlinear approximation or with DFT – methods.\n\nI present a method to find the unknown parameters which is based on linear algebra only.\n\nThe equation (1) is the solution to the difference equation\n$$x_{(k+2)} = 2 \\cdot \\sqrt b \\cdot cos(\\omega) \\cdot x_{(k+1)} - b \\cdot x_{(k)} \\tag{2}$$\n\nThis difference equation is a linear equation system for the unknown constants b and $2 \\cdot \\sqrt b \\cdot cos(\\omega)$.\n\nThe overdetermined linear system\n$$\\left( \\begin{matrix} x_{(k+2)} \\\\ x_{(k+3)} \\\\ . \\\\ . \\end{matrix} \\right) = \\left( \\begin{matrix} x_{(k+1)} & x_{(k )} \\\\ x_{(k+2)} & x_{(k+1)} \\\\ . & .\\\\ . & . \\end{matrix} \\right) \\cdot \\left( \\begin{matrix} 2 \\cdot \\sqrt b \\cdot cos(\\omega) \\\\ -b \\end{matrix} \\right) \\tag{3}$$is solved using linear regression analysis.\n\nThere is a pitfall in this calculation: For high oversampling rates, i.e. small resulting $\\omega$, the numerical results with additional noise are poor. Performance is best near half the Nyquist rate, which results in approximately 4 samples per sine wave. The $x_{(k)}$ can be rearranged accordingly without losing samples or accuracy, such as taking every 8th sample for the $x_{(k)},x_{(k+1)}, ...$ The resultant $\\omega$ has to be corrected by this integer factor.\n\nThe $b$ is related to $\\alpha$ by\n\n$$e^\\alpha\\ = \\sqrt b \\tag{4}$$\n\nFor the derivation of this relation see the appendix.\n\nSo the $\\omega$, $\\alpha$ and b are known.\n\nThe amplitude A and the phase $p$ are still unknown and calculated by\n'mixing with a quadrature carrier' :\n\nThe system of equations\n\n$$\\left( \\begin{matrix} x_{(0)} \\\\ x_{(1)} \\\\ . \\\\ . \\end{matrix} \\right) = \\left( \\begin{matrix} e^{\\alpha \\cdot 0} \\cdot \\sin{(\\omega \\cdot 0)} & e^{\\alpha \\cdot 0} \\cdot \\cos{(\\omega \\cdot 0)} \\\\ e^{\\alpha \\cdot 1} \\cdot \\sin{(\\omega \\cdot 1)} & e^{\\alpha \\cdot 1} \\cdot \\cos{(\\omega \\cdot 1)} \\\\ . & .\\\\ . & . \\end{matrix} \\right) \\cdot \\left( \\begin{matrix} u \\\\ v \\end{matrix} \\right) \\tag{5}$$\n\nis solved for u and v. With the u, v interpreted as the complex number $u+iv$ we obtain the amplitude and phase\n\n$$A=|u+iv|$$\n$$p=\\sphericalangle{(u-iv)}$$\n\nIn real setups the samples $x_{(k)}$ suffer from additional noise. It turns out that the results for frequency $\\omega$ and phase $p$ are very stable in noise. That is not the case for amplitude A nor for the damping factor $\\alpha$. These parameters are adjusted with a gradient iteration. Amplitude and damping are slightly varied until the best fit to the sampled data is achieved.\n\nI published a corresponding Matlab script in the 'Mathworks file exchange' system. You can find it by searching for 'matlab fit damped sine wave'. Download is free and you do not need a registration. I hope that this algorithm is useful and download numbers will rise.\n\nCheers\nDetlef\n\nAppendix\n\nReplacing $x_{(k)}$ in (2) with its definition in (1) yields\n\\begin{align} & A e^{\\alpha \\cdot (k+2)} \\cos {(\\omega (k+2) + p)} = \\\\ & \\\\ & 2 \\sqrt{b} \\cos{(\\omega) A e^{\\alpha(k+1)}} \\cdot \\cos{(\\omega (k+1)+p)} \\\\ & - b \\cdot A \\cdot e^{\\alpha \\cdot k} \\cdot \\cos{(\\omega k + p)} \\\\ \\end{align}\n\nDivision by $A \\cdot e^{\\alpha \\cdot k}$ and application of addition theorems for cos-functions gives\n\n\\begin{align} & e^{2 \\alpha} [\\cos{(\\omega k + p)} \\cos{(2 \\omega)}-\\sin{(\\omega k + p)} \\sin{(2 \\omega)} ] = \\\\ \\\\ & 2 e^{\\alpha}\\sqrt{b}\\cos{\\omega} [\\cos{(\\omega k + p)} \\cos{(\\omega)}-\\sin{(\\omega k + p)} \\sin{(\\omega)}] \\\\ & - b \\cos{(\\omega k+p)} \\\\ \\end{align}\n\nRearranging terms\n\n\\begin{align} 0=& & \\cos{(\\omega k + p)} [e^{2 \\cdot \\alpha} \\cos{(2 \\omega)}-2e^\\alpha \\sqrt b \\cos ^2 {\\omega} +b ]- \\\\ & & \\sin{(\\omega k + p)} [e^{2 \\cdot \\alpha} \\sin{(2 \\omega)}-2e^\\alpha \\sqrt b \\sin {\\omega}\\sin {\\omega} ]= \\\\ \\\\ 0= & & \\cos{(\\omega k + p)} [e^{2 \\cdot \\alpha} \\cos{(2 \\omega)}-2e^\\alpha \\sqrt b \\cos ^2 {\\omega} +b ]- \\\\ & & \\sin{(\\omega k + p)}\\sin{(2 \\omega)} \\cdot [e^{2 \\alpha } -e^{ \\alpha }\\sqrt b ] \\\\ \\end{align}\n\nThe last equation can only hold for any k if the expressions in the squared brackets equal 0:\n\n$$0= e^{2 \\alpha } -e^{ \\alpha }\\sqrt b$$\n$$\\Rightarrow e^{\\alpha } = \\sqrt b$$\n\n[ - ]\nComment by July 2, 2015",
null,
"That is how technical articles should be, straight into core subject, no messing about.\n\nKaz\n[ - ]\nComment by July 3, 2015",
null,
"What has ? got to do with oversampling as written in the sentence \"For high oversampling rates, i.e. small resulting ?\". Intuitively oversampling should give better results shouldn't it? Why is the fitting best for \"near half the Nyquist rate\"?\n[ - ]\nComment by July 4, 2015",
null,
"Hi jin_,\n\n$\\omega$ is a normalized frequency, it is related to the sample frequency. So for a given signal, if the sample rate gets higher, the $\\omega$ is going down. From the linear system of equations the $\\cos \\omega$ term is calculated. This function is flat for small $\\omega$, a small change in the $\\cos \\omega$ will result in a big change for $\\omega$. The situation ist best near $\\pi / 2$, because the $\\cos \\omega$ function ist steepest there, and this sample rate is corresponding to Nyquist / 2 . And yes, higher oversampling rate gives better results, you use all samples, but you have to rearrange the samples. Take a look at the Matlab-script.\n\nCheers\nDetlef\n[ - ]\nComment by September 21, 2015",
null,
"Hello Dr. Amberg. In my efforts to understand your blog I searched the MathWorks Exchange web page for 'matlab fit damped sine wave'. In doing so I received the \"Your search returned No results\" message. Just so you know, searching on 'Fit a damped sine wave' worked fine.\n\nRegards,\n[-Rick-]\n[ - ]\nComment by September 22, 2015",
null,
"Hi,\nthanks for the hint. For Google 'matlab fit damped sine wave', the MathWork Exchange link is the first hit, at least for my Google results. I hope, that your efforts to understand my blog were crowned with success. I'd be more than glad to give some additional clarifications.\n\nCheers\nDetlef\n\nTo post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments."
]
| [
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"https://s3.amazonaws.com/embeddedrelated/user/profilepictures/37480.jpg",
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"https://www.embeddedrelated.com/new/images/defaultavatar.jpg",
null,
"https://d23s79tivgl8me.cloudfront.net/user/profilepictures/25706.jpg",
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"https://d23s79tivgl8me.cloudfront.net/user/profilepictures/14446.jpg",
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"https://d23s79tivgl8me.cloudfront.net/user/profilepictures/25706.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.70949286,"math_prob":0.99971205,"size":5879,"snap":"2022-27-2022-33","text_gpt3_token_len":1799,"char_repetition_ratio":0.1518298,"word_repetition_ratio":0.08190091,"special_character_ratio":0.3340704,"punctuation_ratio":0.07438017,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99994874,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-06T19:05:16Z\",\"WARC-Record-ID\":\"<urn:uuid:64dd61e0-f9de-4d88-b7b3-b95ccbf288e0>\",\"Content-Length\":\"52488\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e88e2150-adab-4fbd-804a-66a2c2e6fbcc>\",\"WARC-Concurrent-To\":\"<urn:uuid:d4537f17-0522-4f21-a911-40e9e95d77b2>\",\"WARC-IP-Address\":\"69.16.201.59\",\"WARC-Target-URI\":\"https://www.dsprelated.com/showarticle/795.php\",\"WARC-Payload-Digest\":\"sha1:PJAQSBEDDGWI2HIYH2B3M7HMF467NWRU\",\"WARC-Block-Digest\":\"sha1:ARZQBPEPHVQOQ3JBQPOEYDM4OC3BWGO4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104676086.90_warc_CC-MAIN-20220706182237-20220706212237-00572.warc.gz\"}"} |
https://employment.blurtit.com/117999/discuss-the-marginal-productivity-theory-and-modern-theory-of-wages | [
"# Discuss The Marginal Productivity Theory And Modern Theory Of Wages.\n\nMarginal Productivity Theory: According to the marginal productivity theory of wages, under the conditions of perfect competition, every worker of same skill and efficiency will receive a wage equal to the value of the marginal product of labor. By the value of marginal product of labor is mean the net addition to the value of the total product of a firm when and additional unit of labor is added. In brief, imperfectly competitive product and input markets, a worker is paid the value of his marginal physical product. Modern Theory of Wages: The modern theory of wages is also known as the demand and supply theory. According to the modern economists, wages are determined by the interaction of the courses of demand for and supply of labor as in the case of a market equilibrium of an ordinary commodity. The marginal productivity theory is criticized by the modern economists on the following rounds. 1. It ignores the supply side of a factor of production. 2. It does not explain the real issue i.e. The determination of the price of a factor of production. Thus, the modern economist uses the forces of demand for and supplies of a factor of production its price.\nthanked the writer.\nState and explain the marginal productivity theory\nthanked the writer.\nMarginal productivity means the net addition or net subtraction caused in the total production by employing or withdrawing one unit of production. This theory is based on wrong assumptions that all the units of production are homogenous. For example all the workers ability and efficiency cannot be equal. It is also wrong to assume that the factors of production are close substitute for one another Labor cannot be a perfect substitute capital. This theory assumes that the reward of each factor of production is determined under the conditions of a perfect competition and full employment. While in actual world it is not possible.\n\nIt is also wrong to assume that the law of diminishing returns applies to the business organization it is very difficult to measure the marginal production and its value. In this theory demand factor has given much importance while the supply factor has been ignored. While in fact both have an equal importance. Each factor of production works in co-operation with other factors; if one is withdrawn it will disorganize the whole business and may cause a loss.\nthanked the writer.\nMarginal productivity theory:\n\nAccording to the theory, wages in perfect competition tend to wequal in marginal net product of a labor. It means the netaddition and net subtraction made to the value of total product of a firm when one unit is aded or withdrawn from it.\n\nWhen an enterprueneur employs a unit of labor, how much he pays to him as wages depend upon addition which he makes to the total revenue of the firm. If addition made to the total revenue by a labor is Rs 5000, the rate of wages will be equal to Rs 5000. the enterprenuer will not pay him more than the return which he contributes to the total production.\n\nAccording to this theory, the rate of wages paid to thelabor tends to be equal to the marginal net produc of the labor employed at the margin. All units of the labor are of the same rates. This theory assumes that all units of the factors are homogenoues. The factors used can be continously varied. Factors of production are mobile as between various uses. Theory is based on the law of diminishing marginal returns. The theory shows that when the employment increases the wages decline. It is difficult to calculate the marginal productivity of a factor in a most area. The theory assumes that the supply of factor is fixed. This theory is valid under the perfect competition.\n\nIf marginal productivity os a factor is more than price of it, there will be more demand for factorand remuneration or salary will rise up to the marginal productivity. If the marginal productivity s less than the price of a factor, so there will be less demand for factor and salary will come down to amrginal productivity\nthanked the writer.",
null,
""
]
| [
null,
"https://cf.blurtitcdn.com/var/avatar/thumb_default_avatar.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9474243,"math_prob":0.953717,"size":6018,"snap":"2022-40-2023-06","text_gpt3_token_len":1203,"char_repetition_ratio":0.18440306,"word_repetition_ratio":0.69473684,"special_character_ratio":0.19125955,"punctuation_ratio":0.0723104,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9856157,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-09T13:27:34Z\",\"WARC-Record-ID\":\"<urn:uuid:50ec0d30-4e13-4167-83fc-b50f3e073fbd>\",\"Content-Length\":\"66832\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a57c25a1-ef7b-4557-9b8b-79528b22bd94>\",\"WARC-Concurrent-To\":\"<urn:uuid:a8215b78-47b2-4e36-affc-8399a0ac4ea1>\",\"WARC-IP-Address\":\"104.25.49.23\",\"WARC-Target-URI\":\"https://employment.blurtit.com/117999/discuss-the-marginal-productivity-theory-and-modern-theory-of-wages\",\"WARC-Payload-Digest\":\"sha1:SVJ3XIPJPPYYZGHGIC3IBHSRD3WYBTYC\",\"WARC-Block-Digest\":\"sha1:COSYH3V3PMD6PEFZUNTV23D7ED27FJTZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499966.43_warc_CC-MAIN-20230209112510-20230209142510-00022.warc.gz\"}"} |
https://naucse.python.cz/2018/pyladies-en-prague/beginners-en/turtle/index/solutions/4/ | [
"Nauč se Python > Kurzy > Beginners course PyLadies > Functions > Turtle and loops\n\n# Turtle and loops – Řešení \n\nThe command `sum = sum + number` calculates the value of `sum + number`, adds the current number to the sum, and stores the result in the variable `sum`. The new value of `sum` will be used in the next loop.\n\nIn the beginning the sum is 0, and in the end the sum of our numbers will be printed.\n\nToto je stránka lekce z kurzu, který probíhá nebo proběhl naživo s instruktorem."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7089452,"math_prob":0.9714685,"size":287,"snap":"2021-43-2021-49","text_gpt3_token_len":73,"char_repetition_ratio":0.18727915,"word_repetition_ratio":0.0,"special_character_ratio":0.24738675,"punctuation_ratio":0.0952381,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.990773,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-16T09:29:47Z\",\"WARC-Record-ID\":\"<urn:uuid:4645e88d-6248-49b2-bccb-354dae20c5c3>\",\"Content-Length\":\"6054\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7a9135fb-5149-4f11-a17d-25eedc656dcd>\",\"WARC-Concurrent-To\":\"<urn:uuid:57c91738-741f-49ed-aa38-1c652cd18e76>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"https://naucse.python.cz/2018/pyladies-en-prague/beginners-en/turtle/index/solutions/4/\",\"WARC-Payload-Digest\":\"sha1:HJ6QBTFH63YYZN3LZ6J5L2KA3CGOFXAW\",\"WARC-Block-Digest\":\"sha1:RTOVXWKBYJBTTGAU6723Q4YNF7PKGIYD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323584554.98_warc_CC-MAIN-20211016074500-20211016104500-00412.warc.gz\"}"} |
https://annmethod.matlabprog.com/ | [
"# Matlab Help Unwrap\n\nFor all students, who have to face the time management issues while doing their matlab assignments, matlab help is a boon. It can be quite difficult to handle time when you have to spend it wisely. Matlab is a good solution to time wastage. With matlab help, students can work without any stress or strain. This will help them in saving time which can then be used for other important tasks.\n\nMany times, students have to do a huge chunk of work, which is of very high class and can really consume a lot of time. This might often lead to a feeling of stress. Matlab has the answer to all these problems. The support that it provides to the students while working on their matlab assignments is truly helpful. With help unwrap matlab assignments, students are able to work smarter.\n\nIn matlab, it is not necessary for the students to do a whole lot of work. There is no need for them to create complex structures like graphs, charts and others. All the work that needs to be done can be done in matlab and this helps in reducing the workload on the students. Matlab help makes it possible for the students to understand each and every aspect of matlab in a proper manner.\n\nOne can save a lot of time when he uses matlab help to do his work. It also gives a perfect view of what needs to be done and the final result. The matlab application will let you know how much is left to be done by you. This is very helpful for students who always do a lot of work. The matlab help enables them to see all the curves in the curves site link and will guide them accordingly. It takes no stress on your part and you can easily work in the comfort of your home.\n\nAnother major benefit of matlab help is that it can easily prepare and complete matrices of various shapes as required. This can be used for classroom assignments and for publishing information online. This makes matlab a good choice for all students, since matrices can be prepared as per your convenience. You can prepare as many assignments as you want without fear of getting stuck at the end of the assignment.\n\nMatlab also helps make you more organized and less problematic. As you do the work, you can go through the matlab help to check if there is any error or if there is something wrong with the mathematically related functions. If you feel that there is some problem then you can immediately find out and correct them. In matlab, there are no complicated procedures; all processes are straight forward and easy to handle.\n\nThe matlab help that you get online will help you in saving a lot of time. Instead of spending hours at a computer, you can use matlab help on your laptop. You can also do your work while on the move and without being tied to the desk. The advantage of this is that you can check out matlab assignments on the go. Once you are done with the task you can go ahead with it and finish all your work. It will not only save you time, but also a lot of effort.\n\nMatlab help provides you with all the facilities that are required to run a spreadsheet application. These include cell references, worksheet navigation, paste and drop functionality, sorting options, sorting option and lot more. With matlab you can even create beautiful graphics and impress your audiences. Thus, it is the best option to solve the equations and solve problems associated with complex mathematical worksheets. You can even print charts and graphs from matlab without any hassles."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.96512616,"math_prob":0.5865949,"size":3432,"snap":"2021-31-2021-39","text_gpt3_token_len":690,"char_repetition_ratio":0.1470245,"word_repetition_ratio":0.0,"special_character_ratio":0.19900933,"punctuation_ratio":0.08479532,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9512772,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-31T15:39:36Z\",\"WARC-Record-ID\":\"<urn:uuid:dcc69b9b-6dce-46b1-87a7-3685db1f37d8>\",\"Content-Length\":\"17943\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:238e2d96-0ea4-4f41-a887-8cd8ec91aacd>\",\"WARC-Concurrent-To\":\"<urn:uuid:129c52be-d572-4858-9750-d468b9f6d519>\",\"WARC-IP-Address\":\"104.21.32.226\",\"WARC-Target-URI\":\"https://annmethod.matlabprog.com/\",\"WARC-Payload-Digest\":\"sha1:QRZFL3VQYYTH347IE3TKZOU4NW6ZKT25\",\"WARC-Block-Digest\":\"sha1:5YSFHTJM6GPYXV6BCRPUH6CS42AKJSDB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154089.68_warc_CC-MAIN-20210731141123-20210731171123-00713.warc.gz\"}"} |
https://geniebook.com/tuition/primary-2/maths/picture-graphs-1 | [
"",
null,
"Study P2 Mathematics Picture Graphs 1 - Geniebook",
null,
"",
null,
"",
null,
"# Picture Graphs 1\n\n## Topic Recap:\n\nLook at the picture below.",
null,
"We can arrange the fruits above and put them into a picture graph as shown below. The picture graph allows us to count and compare the fruits in a more systematic way.",
null,
"Example:\n\n1. How many apples are there?\n\n$$5$$ apples\n\n1. The number of __________ is the least.\n\nSolution:\n\nThere are $$5$$ apples, $$6$$ oranges and $$3$$ watermelons.\n\nThe number of watermelons is the least.\n\nwatermelons\n\n1. How many fruits are there altogether?\n\nSolution:\n\n$$5 + 6 + 3 = 14$$\n\n$$14$$ fruits\n\nThe picture below shows the number of bread rolls Jane and Jack have.",
null,
"After counting, we know that:\n\nJane have $$12$$ bread rolls.\n\nJack have $$18$$ bread rolls.\n\nWe can make a picture graph to represent the number of bread rolls Jack and Jane have individually. Instead of drawing each bread roll repeatedly, we have a simpler way to represent the given data.\n\nSince every plate has $$3$$ bread rolls, we can represent every $$3$$ bread rolls using a symbol. It saves us the trouble of drawing so many bread rolls! Therefore, the picture graph representing the given information is as below:",
null,
"While using a picture graph, we need to find out how many items each symbol represents. Picture graphs are useful especially when we have large numbers of items that we want to compare.\n\nExample:\n\nThe picture graph below shows the number of slices of pizza ‘The Pizza Stall’ sold last week. Fill in the blanks based on the picture graph below:",
null,
"1. The Pizza Stall sold the least number of slices of pizza on __________.\n2. The Pizza Stall sold the same number of slices of pizza on __________ and __________.\n3. The Pizza Stall sold __________ slices of pizza altogether.\n\nSolution:\n\nNumber of slices of pizza sold on Monday\n\n\\begin{align} &= 5 \\times 5 \\\\[2ex] &= 25 \\end{align}\n\nNumber of slices of pizza sold on Tuesday\n\n\\begin{align} &= 2 \\times 5 \\\\[2ex] &= 10 \\end{align}\n\nNumber of slices of pizza sold on Wednesday\n\n\\begin{align} &= 4 \\times 5 \\\\[2ex] &= 20 \\end{align}\n\nNumber of slices of pizza sold on Thursday\n\n\\begin{align} &= 3 \\times 5 \\\\[2ex] &= 15 \\end{align}\n\nNumber of slices of pizza sold on Friday\n\n\\begin{align} &= 5 \\times 5 \\\\[2ex] &= 25 \\end{align}\n\n1. The least number of slices of pizzas sold was $$10$$.\n\nSo, the Pizza Stall sold the least number of slices of pizza on Tuesday.\n\nTuesday\n\n1. The Pizza Stall sold $$25$$ slices of pizza on Monday and $$25$$ slices of pizza on Friday.\n\nSo, the Pizza Stall sold the same number of slices of pizza on Monday and Friday.\n\nMonday and Friday\n\n1. Total number of slices of pizza sold\n\n\\begin{align} &= 25 + 10 + 20 + 15 + 25 \\\\[2ex] &= 95 \\end{align}\n\n$$95$$ slices\n\nQuestion 1:\n\nThe picture graph below shows the number of burritos $$3$$ boys sold during a food fair.",
null,
"1. Who sold the most number of burritos?\n\n1. Mark\n2. Jason\n3. Samuel\n\n(3) Samuel\n\n1. How many burritos did Mark sell?\n\n1. $$5$$\n2. $$7$$\n3. $$10$$\n4. $$12$$\n\nSolution:\n\nNumber of burritos Mark sell\n\n\\begin{align} &= 5 \\times 2 \\\\[2ex] &= 10 \\end{align}\n\n(3) 10\n\n1. How many burritos did the boys sell altogether?\n\n1. $$17$$\n2. $$24$$\n3. $$32$$\n4. $$34$$\n\nSolution:\n\nNumber of burritos Mark sold $$= 10$$\n\nNumber of burritos Jason sold\n\n\\begin{align} &= 4 \\times 2 \\\\[2ex] &= 8 \\end{align}\n\nNumber of burritos Samuel sold\n\n\\begin{align} &= 8 \\times 2 \\\\[2ex] &= 16 \\end{align}\n\nTotal number of burritos sold\n\n\\begin{align} &= 10 + 8 + 16 \\\\[2ex] &= 34 \\end{align}\n\n(4) $$34$$\n\nQuestion 2:\n\nThe picture graph shows the number of stamps collected by 4 students.",
null,
"1. How many stamps did Sam collect?\n\n1. $$8$$\n2. $$11$$\n3. $$16$$\n4. $$24$$\n\nSolution:\n\nNumber of stamps Sam collected\n\n\\begin{align} &= 8 × 3 \\\\[2ex] &= 24 \\end{align}\n\n(4) $$24$$\n\n1. How many stamps did Fabian and Kris collect?\n\n1. $$9$$\n2. $$18$$\n3. $$27$$\n4. $$46$$\n\nSolution:\n\nNumber of stamps Fabian sold\n\n\\begin{align} &= 4 \\times 3 \\\\[2ex] &=12 \\end{align}\n\nNumber of stamps Kris sold\n\n\\begin{align} &= 5 \\times 3 \\\\[2ex] &= 15 \\end{align}\n\nThe total number of stamps collected by Fabian and Kris\n\n\\begin{align} &= 12 + 15 \\\\[2ex] &= 27 \\end{align}\n\n(3) $$27$$\n\nQuestion 3:\n\nThe picture below shows the number of pet dogs in $$5$$ housing estates.",
null,
"1. Which $$2$$ estates have the same number of pet dogs?\n\n1. Cherry Town and Peach Villa\n2. Peach Villa and Apricot Ville\n3. Cherry Town and Berry Town\n\n(2) Peach Villa and Apricot Ville\n\n1. How many pet dogs are there in Kiwi Bay?\n\n1. $$6$$\n2. $$2$$\n3. $$8$$\n4. $$4$$\n\nSolution:\n\nNumber of dogs in Kiwi Bay\n\n\\begin{align} &= 2 \\times 4 \\\\[2ex] &= 8 \\end{align}\n\n(3) $$8$$\n\n1. How many pet dogs are there in the five estates altogether?\n\n1. $$17$$\n2. $$34$$\n3. $$68$$\n4. $$70$$\n\nSolution:\n\nNumber of pet dogs in Cherry Town\n\n\\begin{align} &= 4 × 4\\\\[2ex] &= 16 \\end{align}\n\nNumber of pet dogs in Peach Ville\n\n\\begin{align} &= 3 \\times 4\\\\[2ex] &= 12 \\end{align}\n\nNumber of pet dogs in Apricot Ville\n\n\\begin{align} &= 3 \\times 4\\\\[2ex] &= 12 \\end{align}\n\nNumber of pet dogs in Berry Town\n\n\\begin{align} &= 5 \\times 4 \\\\[2ex] &= 20 \\end{align}\n\nNumber of pet dogs in Kiwi Bay\n\n\\begin{align} &= 2 \\times 4 \\\\[2ex] &= 8 \\end{align}\n\nTotal number of pet dogs\n\n\\begin{align} &= 16 + 12 + 12 + 20 + 8\\\\[2ex] &= 68 \\end{align}\n\n(3) $$68$$\n\nQuestion 4:\n\nThe picture graph shows the number of marbles collected by $$4$$ children.",
null,
"1. How many more marbles did Amy collect than Randy?\n\n1. $$10$$\n2. $$15$$\n3. $$20$$\n4. $$25$$\n\nSolution:\n\nNumber of marbles collected by Amy\n\n\\begin{align} &= 5 \\times 5 \\\\[2ex] &= 25 \\end{align}\n\nNumber of marbles collected by Randy\n\n\\begin{align} &= 2 \\times 5 \\\\[2ex] &= 10 \\end{align}\n\nDifference in marbles between Amy and Randy\n\n\\begin{align} &= 25 - 10 \\\\[2ex] &= 15 \\end{align}\n\n(2) $$15$$\n\n1. Who collected twice as many marbles than Randy?\n\n1. Charles\n2. Jun Xi\n3. Amy\n\nSolution:\n\nNumber of marbles Randy collected $$= 10$$\n\nNumber of marbles that is twice of $$10$$\n\n\\begin{align} &= 2 \\times 10 \\\\[2ex] &= 20 \\end{align}\n\nNumber of marbles collected by Amy $$= 25$$\n\nNumber of marbles collected by Jun Xi\n\n\\begin{align} &= 3 \\times 5 \\\\[2ex] &= 15 \\end{align}\n\nNumber of marbles collected by Charles\n\n\\begin{align} &= 4 \\times 5 \\\\[2ex] &= 20 \\end{align}\n\nSo, Charles collected twice as many marbles as Randy.\n\n(1) Charles\n\nQuestion 5:\n\nThe picture graph represents the number of hotdogs sold by $$3$$ children.",
null,
"1. Given that the 3 children sold 80 hotdogs in all, how many hotdogs did Sara sell?\n\n1. $$20$$\n2. $$30$$\n3. $$35$$\n4. $$40$$\n\nSolution:\n\nNumber of hotdogs sold by Molly\n\n\\begin{align} &= 3 \\times 5 \\\\[2ex] &= 15 \\end{align}\n\nNumber of hotdogs sold by Betty\n\n\\begin{align} &= 7 \\times 5 \\\\[2ex] &= 35 \\end{align}\n\nNumber of hotdogs sold by Molly and Betty\n\n\\begin{align} &= 15 + 35 \\\\[2ex] &= 50 \\end{align}\n\nTotal hotdogs sold $$= 80$$\n\nNumber of hotdogs sold by Sara\n\n\\begin{align} &= 80 - 50 \\\\[2ex] &= 30 \\end{align}\n\n(2) $$30$$\n\nContinue Learning\nNumbers To 1000 Multiplication And Division 1\nMultiplication And Division 2 Addition And Subtraction 1\nAddition And Subtraction 2 Fractions 1\nLength 1 Mass 1\nVolume 1 Money 1\nTime 1 Shapes And Patterns\nPicture Graphs 1 Model Drawing 1\nModel Drawing 4",
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https://www.engpaper.com/cse/quantum-computer-algorithm.html | [
"# quantum computer algorithm\n\na quantum algorithm is an algorithm which runs on a realistic model of quantum computation, the most commonly used model being the quantum circuit model of computation\n\nGeneralization and partial demonstration of an entanglement based deutsch-jozsa like algorithm using a 5-qubit quantum computer\n\nDeutsch-Jozsa algorithm is one of the first examples of quantum algorithm which is exponentially faster than any possible deterministic classical algorithm that solves the same problem. A special case of Deutsch-Jozsa problem is Deutsch problem ; given that a The authors aim is to present how to construct circuits, built of known quantum gates, solving some decision making problems. A decision making algorithm shows how to solve a binary problem with use of quantum computer by constructing quantum operators\n\nSimulation of Grovers Algorithm Quantum Search in a Classical Computer\n\nThe rapid progress of computer science has been accompanied by a corresponding evolution of computation, from classical computation to quantum computation. As quantum computing is on its way to becoming an established discipline of computing science, much\n\nFast algorithm for efficient simulation of quantum algorithm gates on classical computer\n\nThe general approach for quantum algorithm simulation on classical computer is introduced. Efficient fast algorithm for simulation of Grovers quantum search algorithm in unsorted database is presented. Comparison with common quantum algorithm simulation approach is\n\nA Quantum Based Algorithm for Computer Network Routing\n\nThe aim of this study, is to design and implement a quantum algorithm for network routing by exploiting the massive parallelism existing in the quantum environment and to deal with the demands of continuous growing of the internet. This algorithm is compared according to the\n\nAPPLYING QUANTUM COMPUTER FOR THE REALIZATION OF SPSA ALGORITHM\n\nThe estimates of the algorithm of the simultaneously perturbation stochastic approximation (SPSA) are critical against the requirement of the simultaneous generation of a large amount of a random value realization that have to be independent from each other, that is a hard\n\nImplementation of the Deutsch-Josza Algorithm on an ion-trap quantum computer\n\nPage 1. Implementation of the Deutsch- Josza Algorithm on an ion-trap quantum computer Przemyslaw Kotara HU Berlin (Ger) Tijl Schmelzer U Ghent (Be) S. Gulde et Al. Page 2. Outline Theory: Problem/Motivation The algorithm Quantum Circuit Deutsch algorithm\n\nalgorithm to solve and extended Deutsch problem with an implementation on an NMR quantum computer (Analytical Study of Quantum Information and Related\n\nQuantum computation is arapidly growing field of research. Aquantum computer uses aset of tw0-state systems as quantum bits (qubits), and executes acomputation by asequence ofunitary transformations on them; the desired useful information is extracted by\n\nSimulation of the statical and dynamical noise on Grovers search algorithm implemented in a Ising nuclear spin chain quantum computer\n\nWe consider Grovers quantum search algorithm on a model quantum computer implemented on a chain of four nuclear spins with first and second neighbor Ising interactions. Noise is introduced into the system in terms of random fluctuations of the\n\nAn extention of Grovers quantum search algorithm (Foundations of Computer Science)\n\nQuantum computers and their algorithm are now widely studied. We consider the prob-lem of making a superposition with an arbitrarilyspecified amplitudes. By modifying Grovers quantum search algorithm , we show that this can be done in time $O (\\sqrt {N})$ with two\n\nFast Algorithm and HW Design for Efficient Intelligent Computation of Main Quantum Algorithm Fuzzy Operators on Classical Computer . Part"
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https://answers.everydaycalculation.com/divide-fractions/2-18-divided-by-1-10 | [
"Solutions by everydaycalculation.com\n\n## Divide 2/18 with 1/10\n\n2/18 ÷ 1/10 is 10/9.\n\n#### Steps for dividing fractions\n\n1. Find the reciprocal of the divisor\nReciprocal of 1/10: 10/1\n2. Now, multiply it with the dividend\nSo, 2/18 ÷ 1/10 = 2/18 × 10/1\n3. = 2 × 10/18 × 1 = 20/18\n4. After reducing the fraction, the answer is 10/9\n5. In mixed form: 11/9\n\nMathStep (Works offline)",
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http://nd.ics.org.ru/authors_nd/detail/451388-andrei_starostin | [
"# Andrei Starostin\n\nul. 2-ya Baumanskaya 5, Moscow, 105005 Russia\nBauman Moscow State Technical University\n\n## Publications:\n\n Kuzenov V. V., Ryzhkov S. V., Starostin A. V. Development of a Mathematical Model and the Numerical Solution Method in a Combined Impact Scheme for MIF Target 2020, Vol. 16, no. 2, pp. 325-341 Abstract This work is devoted to the theoretical calculation of the processes of compression and energy release in the target by a combined action of a system of pulsed jets and intense laser radiation using a magnetic inertial plasma confinement method. A mathematical model, a numerical method, and a computational algorithm are developed to describe plasma-physical processes occurring in various types of high-temperature installations with high density. The results of the calculation of the hybrid effect of intensive energy flows on a cylindrical target are presented. The main gas-dynamic and radiative parameters of the compressed target plasma are found. Keywords: computer simulation, magnetized target, mathematical modeling, numerical algorithm Citation: Kuzenov V. V., Ryzhkov S. V., Starostin A. V., Development of a Mathematical Model and the Numerical Solution Method in a Combined Impact Scheme for MIF Target, Rus. J. Nonlin. Dyn., 2020, Vol. 16, no. 2, pp. 325-341 DOI:10.20537/nd200207"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.83406514,"math_prob":0.7916039,"size":977,"snap":"2023-40-2023-50","text_gpt3_token_len":215,"char_repetition_ratio":0.10483042,"word_repetition_ratio":0.0,"special_character_ratio":0.20368475,"punctuation_ratio":0.13559322,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9598856,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T04:21:19Z\",\"WARC-Record-ID\":\"<urn:uuid:68e90659-97d4-4272-96cc-45900a81badf>\",\"Content-Length\":\"17847\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:37657796-b70d-4fec-9bb6-b2f57c707d56>\",\"WARC-Concurrent-To\":\"<urn:uuid:0af44e1a-59a5-4242-96af-0820355a425a>\",\"WARC-IP-Address\":\"188.43.29.91\",\"WARC-Target-URI\":\"http://nd.ics.org.ru/authors_nd/detail/451388-andrei_starostin\",\"WARC-Payload-Digest\":\"sha1:LHPYV4WXA5JK33Z5BDSSSXFCA6R2XO3B\",\"WARC-Block-Digest\":\"sha1:U26TSLWENBLYSWLM6TY3IY7OJZFJXO5G\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506479.32_warc_CC-MAIN-20230923030601-20230923060601-00779.warc.gz\"}"} |
https://www.freemathhelp.com/forum/threads/simple-interest.129325/ | [
"# Simple interest\n\n#### Pens8675309\n\n##### New member\nCompare the number of years it will take to double money at 5% interest, with simple interest and compounded annually. What is the difference in the number of years? (Round your answer to the nearest tenth , if necessary)\n\n#### jonah2.0\n\n##### Full Member\nBeer soaked request follows.\nCompare the number of years it will take to double money at 5% interest, with simple interest and compounded annually. What is the difference in the number of years? (Round your answer to the nearest tenth , if necessary)\nPlease show us what you have tried and exactly where you are stuck.\n\nPlease follow the rules of posting in this forum, as enunciated at:\n\nPlease share your work/thoughts about this problem.\n\n#### HallsofIvy\n\n##### Elite Member\nIf A is the initial amount invested at 5% per year simple interest, the first year it earns 0.5A interest, the second year it earns 0,5A, etc., That is every year the account earns 0.5A interest so in n years the interest earned will be 0.5nA. In order that the account have doubled the interest earned must be equal to the amount originally in the account: 0.5nA= A. Solve that for n.\n\nWith interest compounded annually, the first year it earns 0.5A interest but that interest is add to the account so the interest the second year is 0.5(A+ 0.5A)= 0.5A+ 0.5^2A= A(0.5+0.5^2). The third year the interest earned is 0.5(A+A(0.5+ 0.5^2)= A(0.5+ 0.5+0.5^2+ 0.5^3). The nth year it would be A(0.5+ 0.5^2+ 0.5^3+ ...+ 0.5^n). Including the initial amount in the account, A, that is A(1+0.5+0.5^2+ 0.5^3+ ... + 0.5^n). That sum is a \"geometric sum\", a sum of the for a+ ar+ ar^2+....+ar^n for some a and r. Here a= 1 and r= 0.5.\n\nWhoever gave you this problem clearly expects you to know how to sum a geometric series. If you don't, or if you need a review, look at Finite Geometric Sum - Bing images .\n\nLast edited:"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8883372,"math_prob":0.98196423,"size":1094,"snap":"2021-21-2021-25","text_gpt3_token_len":384,"char_repetition_ratio":0.17798164,"word_repetition_ratio":0.020408163,"special_character_ratio":0.35374773,"punctuation_ratio":0.1967742,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99892414,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-12T13:39:30Z\",\"WARC-Record-ID\":\"<urn:uuid:d1485b50-dd44-4ae6-b5e1-2a12a43e0005>\",\"Content-Length\":\"47311\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:328bce3e-3ba9-402b-bb82-23cd8a075d3c>\",\"WARC-Concurrent-To\":\"<urn:uuid:739ba26b-7f08-478a-a75f-f657dce1083e>\",\"WARC-IP-Address\":\"50.28.58.203\",\"WARC-Target-URI\":\"https://www.freemathhelp.com/forum/threads/simple-interest.129325/\",\"WARC-Payload-Digest\":\"sha1:SLCM6E3J5HP7K4JZHYLTATVV5BHNK3V2\",\"WARC-Block-Digest\":\"sha1:OQT7752OECAYWMOC362S2HTJOMPK4BSQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487584018.1_warc_CC-MAIN-20210612132637-20210612162637-00020.warc.gz\"}"} |
https://formulasearchengine.com/wiki/Winding_number | [
"# Winding number\n\nThe term winding number may also refer to the rotation number of an iterated map.\n\nIn mathematics, the winding number of a closed curve in the plane around a given point is an integer representing the total number of times that curve travels counterclockwise around the point. The winding number depends on the orientation of the curve, and is negative if the curve travels around the point clockwise.\n\nWinding numbers are fundamental objects of study in algebraic topology, and they play an important role in vector calculus, complex analysis, geometric topology, differential geometry, and physics, including string theory.\n\n## Intuitive description",
null,
"An object traveling along the red curve makes two counterclockwise turns around the person at the origin.\n\nSuppose we are given a closed, oriented curve in the xy plane. We can imagine the curve as the path of motion of some object, with the orientation indicating the direction in which the object moves. Then the winding number of the curve is equal to the total number of counterclockwise turns that the object makes around the origin.\n\nWhen counting the total number of turns, counterclockwise motion counts as positive, while clockwise motion counts as negative. For example, if the object first circles the origin four times counterclockwise, and then circles the origin once clockwise, then the total winding number of the curve is three.\n\nUsing this scheme, a curve that does not travel around the origin at all has winding number zero, while a curve that travels clockwise around the origin has negative winding number. Therefore, the winding number of a curve may be any integer. The following pictures show curves with winding numbers between −2 and 3:\n\n## Formal definition\n\nA curve in the xy plane can be defined by parametric equations:\n\n$x=x(t)\\quad {\\text{and}}\\quad y=y(t)\\qquad {\\text{for }}0\\leq t\\leq 1.$",
null,
"If we think of the parameter t as time, then these equations specify the motion of an object in the plane between t = 0 and t = 1. The path of this motion is a curve as long as the functions x(t) and y(t) are continuous. This curve is closed as long as the position of the object is the same at t = 0 and t = 1.\n\nWe can define the winding number of such a curve using the polar coordinate system. Assuming the curve does not pass through the origin, we can rewrite the parametric equations in polar form:\n\n$r=r(t)\\quad {\\text{and}}\\quad \\theta =\\theta (t)\\qquad {\\text{for }}0\\leq t\\leq 1.$",
null,
"The functions r(t) and θ(t) are required to be continuous, with r > 0. Because the initial and final positions are the same, θ(0) and θ(1) must differ by an integer multiple of 2π. This integer is the winding number:\n\n${\\text{winding number}}={\\frac {\\theta (1)-\\theta (0)}{2\\pi }}.$",
null,
"This defines the winding number of a curve around the origin in the xy plane. By translating the coordinate system, we can extend this definition to include winding numbers around any point p.\n\n## Alternative definitions\n\nWinding number is often defined in different ways in various parts of mathematics. All of the definitions below are equivalent to the one given above:\n\n### Differential geometry\n\nIn differential geometry, parametric equations are usually assumed to be differentiable (or at least piecewise differentiable). In this case, the polar coordinate θ is related to the rectangular coordinates x and y by the equation:\n\n$d\\theta ={\\frac {1}{r^{2}}}\\left(x\\,dy-y\\,dx\\right)\\quad {\\text{where }}r^{2}=x^{2}+y^{2}.$",
null,
"By the fundamental theorem of calculus, the total change in θ is equal to the integral of . We can therefore express the winding number of a differentiable curve as a line integral:\n\n${\\text{winding number}}={\\frac {1}{2\\pi }}\\oint _{C}\\,{\\frac {x}{r^{2}}}\\,dy-{\\frac {y}{r^{2}}}\\,dx.$",
null,
"The one-form (defined on the complement of the origin) is closed but not exact, and it generates the first de Rham cohomology group of the punctured plane. In particular, if ω is any closed differentiable one-form defined on the complement of the origin, then the integral of ω along closed loops gives a multiple of the winding number.\n\n### Complex analysis\n\nIn complex analysis, the winding number of a closed curve C in the complex plane can be expressed in terms of the complex coordinate z = x + iy. Specifically, if we write z = re, then\n\n$dz=e^{i\\theta }dr+ire^{i\\theta }d\\theta \\!\\,$",
null,
"and therefore\n\n${\\frac {dz}{z}}\\;=\\;{\\frac {dr}{r}}+i\\,d\\theta \\;=\\;d[\\ln r]+i\\,d\\theta .$",
null,
"The total change in ln(r) is zero, and thus the integral of dz ⁄ z is equal to i multiplied by the total change in θ. Therefore:\n\n${\\text{winding number}}={\\frac {1}{2\\pi i}}\\oint _{C}{\\frac {dz}{z}}.$",
null,
"More generally, the winding number of C around any complex number a is given by\n\n${\\frac {1}{2\\pi i}}\\oint _{C}{\\frac {dz}{z-a}}.$",
null,
"This is a special case of the famous Cauchy integral formula. Winding numbers play a very important role throughout complex analysis (c.f. the statement of the residue theorem).\n\n### Topology\n\nIn topology, the winding number is an alternate term for the degree of a continuous mapping. In physics, winding numbers are frequently called topological quantum numbers. In both cases, the same concept applies.\n\nThe above example of a curve winding around a point has a simple topological interpretation. The complement of a point in the plane is homotopy equivalent to the circle, such that maps from the circle to itself are really all that need to be considered. It can be shown that each such map can be continuously deformed to (is homotopic to) one of the standard maps $S^{1}\\to S^{1}:s\\mapsto s^{n}$",
null,
", where multiplication in the circle is defined by identifying it with the complex unit circle. The set of homotopy classes of maps from a circle to a topological space form a group, which is called the first homotopy group or fundamental group of that space. The fundamental group of the circle is the group of the integers, Z; and the winding number of a complex curve is just its homotopy class.\n\nMaps from the 3-sphere to itself are also classified by an integer which is also called the winding number or sometimes Pontryagin index.\n\n### Polygons\n\nIn polygons, the winding number is referred to as the polygon density. For convex polygons, and more generally simple polygons (not self-intersecting), the density is 1, by the Jordan curve theorem. By contrast, for a regular star polygon {p/q}, the density is q.\n\n## Turning number\n\nOne can also consider the winding number of the path with respect to the tangent of the path itself. As a path followed through time, this would be the winding number with respect to the origin of the velocity vector. In this case the example illustrated on the right has a winding number of 4 (or −4), because the small loop is counted.\n\nThis is only defined for immersed paths (i.e., for differentiable paths with nowhere vanishing derivatives), and is the degree of the tangential Gauss map.\n\nThis is called the turning number, and can be computed as the total curvature divided by 2π.\n\n## Winding number and Heisenberg ferromagnet equations\n\nFinally, note that the winding number is closely related with the (2 + 1)-dimensional continuous Heisenberg ferromagnet equations and its integrable extensions: the Ishimori equation etc. Solutions of the last equations are classified by the winding number or topological charge (topological invariant and/or topological quantum number)."
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"https://upload.wikimedia.org/wikipedia/commons/a/ac/Winding_Number_Animation_Small.gif",
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"https://wikimedia.org/api/rest_v1/media/math/render/svg/614638a1775bf07e13bd8a981a3a11944d38b4b4",
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"https://wikimedia.org/api/rest_v1/media/math/render/svg/22622b242fc6f3c7a6ae9ffb8431d4282bf30edd",
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https://uk.mathworks.com/help/matlab/math/choose-an-ode-solver.html | [
"## Choose an ODE Solver\n\n### Ordinary Differential Equations\n\nAn ordinary differential equation (ODE) contains one or more derivatives of a dependent variable, y, with respect to a single independent variable, t, usually referred to as time. The notation used here for representing derivatives of y with respect to t is $y\\text{'}$ for a first derivative, $y\\text{'}\\text{'}$ for a second derivative, and so on. The order of the ODE is equal to the highest-order derivative of y that appears in the equation.\n\nFor example, this is a second order ODE:\n\n`$y\\text{'}\\text{'}=9y$`\n\nIn an initial value problem, the ODE is solved by starting from an initial state. Using the initial condition, ${y}_{0}$, as well as a period of time over which the answer is to be obtained, $\\left({t}_{0},{t}_{f}\\right)$, the solution is obtained iteratively. At each step the solver applies a particular algorithm to the results of previous steps. At the first such step, the initial condition provides the necessary information that allows the integration to proceed. The final result is that the ODE solver returns a vector of time steps $t=\\left[{t}_{0},{t}_{1},{t}_{2},...,{t}_{f}\\right]$ as well as the corresponding solution at each step $y=\\left[{y}_{0},{y}_{1},{y}_{2},...,{y}_{f}\\right]$.\n\n### Types of ODEs\n\nThe ODE solvers in MATLAB® solve these types of first-order ODEs:\n\n• Explicit ODEs of the form $y\\text{'}=f\\left(t,y\\right)$.\n\n• Linearly implicit ODEs of the form $M\\left(t,y\\right)y\\text{'}=f\\left(t,y\\right)$, where $M\\left(t,y\\right)$ is a nonsingular mass matrix. The mass matrix can be time- or state-dependent, or it can be a constant matrix. Linearly implicit ODEs involve linear combinations of the first derivative of y, which are encoded in the mass matrix.\n\nLinearly implicit ODEs can always be transformed to an explicit form, $y\\text{'}={M}^{-1}\\left(t,y\\right)f\\left(t,y\\right)$. However, specifying the mass matrix directly to the ODE solver avoids this transformation, which is inconvenient and can be computationally expensive.\n\n• If some components of $y\\text{'}$ are missing, then the equations are called differential algebraic equations, or DAEs, and the system of DAEs contains some algebraic variables. Algebraic variables are dependent variables whose derivatives do not appear in the equations. A system of DAEs can be rewritten as an equivalent system of first-order ODEs by taking derivatives of the equations to eliminate the algebraic variables. The number of derivatives needed to rewrite a DAE as an ODE is called the differential index. The `ode15s` and `ode23t` solvers can solve index-1 DAEs.\n\n• Fully implicit ODEs of the form $f\\left(t,y,y\\text{'}\\right)=0$. Fully implicit ODEs cannot be rewritten in an explicit form, and might also contain some algebraic variables. The `ode15i` solver is designed for fully implicit problems, including index-1 DAEs.\n\nYou can supply additional information to the solver for some types of problems by using the `odeset` function to create an options structure.\n\n### Systems of ODEs\n\nYou can specify any number of coupled ODE equations to solve, and in principle the number of equations is only limited by available computer memory. If the system of equations has n equations,\n\n`$\\left(\\begin{array}{c}y{\\text{'}}_{1}\\\\ y{\\text{'}}_{2}\\\\ ⋮\\\\ y{\\text{'}}_{n}\\end{array}\\right)=\\left(\\begin{array}{c}{f}_{1}\\left(t,{y}_{1},{y}_{2},...,{y}_{n}\\right)\\\\ {f}_{2}\\left(t,{y}_{1},{y}_{2},...,{y}_{n}\\right)\\\\ ⋮\\\\ {f}_{n}\\left(t,{y}_{1},{y}_{2},...,{y}_{n}\\right)\\end{array}\\right),$`\n\nthen the function that encodes the equations returns a vector with n elements, corresponding to the values for $y{\\text{'}}_{1},\\text{\\hspace{0.17em}}y{\\text{'}}_{2},\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\dots \\text{\\hspace{0.17em}},\\text{\\hspace{0.17em}}y{\\text{'}}_{n}$. For example, consider the system of two equations\n\n`$\\left\\{\\begin{array}{l}y{\\text{'}}_{1}={y}_{2}\\\\ y{\\text{'}}_{2}={y}_{1}\\text{\\hspace{0.17em}}{y}_{2}-2\\text{\\hspace{0.17em}}.\\end{array}$`\n\nA function that encodes these equations is\n\n```function dy = myODE(t,y) dy(1) = y(2); dy(2) = y(1)*y(2)-2; end```\n\n### Higher-Order ODEs\n\nThe MATLAB ODE solvers only solve first-order equations. You must rewrite higher-order ODEs as an equivalent system of first-order equations using the generic substitutions\n\n`$\\begin{array}{l}{y}_{1}=y\\\\ {y}_{2}=y\\text{'}\\\\ {y}_{3}=y\\text{'}\\text{'}\\\\ \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}⋮\\\\ {y}_{n}={y}^{\\left(n-1\\right)}.\\end{array}$`\n\nThe result of these substitutions is a system of n first-order equations\n\n`$\\left\\{\\begin{array}{l}y{\\text{'}}_{1}={y}_{2}\\\\ y{\\text{'}}_{2}={y}_{3}\\\\ \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}⋮\\\\ y{\\text{'}}_{n}=f\\left(t,{y}_{1},{y}_{2},...,{y}_{n}\\right).\\end{array}$`\n\nFor example, consider the third-order ODE\n\n`$y\\text{'}\\text{'}\\text{'}-y\\text{'}\\text{'}y+1=0.$`\n\nUsing the substitutions\n\n`$\\begin{array}{l}{y}_{1}=y\\\\ {y}_{2}=y\\text{'}\\\\ {y}_{3}=y\\text{'}\\text{'}\\end{array}$`\n\nresults in the equivalent first-order system\n\n`$\\left\\{\\begin{array}{l}y{\\text{'}}_{1}={y}_{2}\\\\ y{\\text{'}}_{2}={y}_{3}\\\\ y{\\text{'}}_{3}={y}_{1}\\text{\\hspace{0.17em}}{y}_{3}-1.\\end{array}$`\n\nThe code for this system of equations is then\n\n```function dydt = f(t,y) dydt(1) = y(2); dydt(2) = y(3); dydt(3) = y(1)*y(3)-1; end```\n\n### Complex ODEs\n\nConsider the complex ODE equation\n\n`$y\\text{'}=f\\left(t,y\\right)\\text{\\hspace{0.17em}},$`\n\nwhere $y={y}_{1}+i{y}_{2}$. To solve it, separate the real and imaginary parts into different solution components, then recombine the results at the end. Conceptually, this looks like\n\n`$\\begin{array}{l}yv=\\left[\\text{Real}\\left(y\\right)\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{Imag}\\left(y\\right)\\right]\\\\ fv=\\left[\\text{Real}\\left(f\\left(t,y\\right)\\right)\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{Imag}\\left(f\\left(t,y\\right)\\right)\\right]\\text{\\hspace{0.17em}}.\\end{array}$`\n\nFor example, if the ODE is $y\\text{'}=yt+2i$, then you can represent the equation using the function file:\n\n```function f = complexf(t,y) f = y.*t + 2*i; end```\n\nThen, the code to separate the real and imaginary parts is\n\n```function fv = imaginaryODE(t,yv) % Construct y from the real and imaginary components y = yv(1) + i*yv(2); % Evaluate the function yp = complexf(t,y); % Return real and imaginary in separate components fv = [real(yp); imag(yp)]; end ```\n\nWhen you run a solver to obtain the solution, the initial condition `y0` is also separated into real and imaginary parts to provide an initial condition for each solution component.\n\n```y0 = 1+i; yv0 = [real(y0); imag(y0)]; tspan = [0 2]; [t,yv] = ode45(@imaginaryODE, tspan, yv0); ```\n\nOnce you obtain the solution, combine the real and imaginary components together to obtain the final result.\n\n`y = yv(:,1) + i*yv(:,2);`\n\n### Basic Solver Selection\n\n`ode45` performs well with most ODE problems and should generally be your first choice of solver. However, `ode23`, `ode78`, `ode89` and `ode113` can be more efficient than `ode45` for problems with looser or tighter accuracy requirements.\n\nSome ODE problems exhibit stiffness, or difficulty in evaluation. Stiffness is a term that defies a precise definition, but in general, stiffness occurs when there is a difference in scaling somewhere in the problem. For example, if an ODE has two solution components that vary on drastically different time scales, then the equation might be stiff. You can identify a problem as stiff if nonstiff solvers (such as `ode45`) are unable to solve the problem or are extremely slow. If you observe that a nonstiff solver is very slow, try using a stiff solver such as `ode15s` instead. When using a stiff solver, you can improve reliability and efficiency by supplying the Jacobian matrix or its sparsity pattern.\n\nThis table provides general guidelines on when to use each of the different solvers.\n\nSolverProblem TypeAccuracyWhen to Use\n`ode45`NonstiffMedium\n\nMost of the time. `ode45` should be the first solver you try.\n\n`ode23`Low\n\n`ode23` can be more efficient than `ode45` at problems with crude tolerances, or in the presence of moderate stiffness.\n\n`ode113`Low to High\n\n`ode113` can be more efficient than `ode45` at problems with stringent error tolerances, or when the ODE function is expensive to evaluate.\n\n`ode78`High\n\n`ode78` can be more efficient than `ode45` at problems with smooth solutions that have high accuracy requirements.\n\n`ode89`High\n\n`ode89` can be more efficient than `ode78` on very smooth problems, when integrating over long time intervals, or when tolerances are especially tight.\n\n`ode15s`StiffLow to Medium\n\nTry `ode15s` when `ode45` fails or is inefficient and you suspect that the problem is stiff. Also use `ode15s` when solving differential algebraic equations (DAEs).\n\n`ode23s`Low\n\n`ode23s` can be more efficient than `ode15s` at problems with crude error tolerances. It can solve some stiff problems for which `ode15s` is not effective.\n\n`ode23s` computes the Jacobian in each step, so it is beneficial to provide the Jacobian via `odeset` to maximize efficiency and accuracy.\n\nIf there is a mass matrix, it must be constant.\n\n`ode23t`Low\n\nUse `ode23t` if the problem is only moderately stiff and you need a solution without numerical damping.\n\n`ode23t` can solve differential algebraic equations (DAEs).\n\n`ode23tb`Low\n\nLike `ode23s`, the `ode23tb` solver might be more efficient than `ode15s` at problems with crude error tolerances.\n\n`ode15i`Fully implicitLow\n\nUse `ode15i` for fully implicit problems f(t,y,y’) = 0 and for differential algebraic equations (DAEs) of index 1.\n\nFor details and further recommendations about when to use each solver, see .\n\n### Summary of ODE Examples and Files\n\nThere are several example files available that serve as excellent starting points for most ODE problems. To run the Differential Equations Examples app, which lets you easily explore and run examples, type\n\n`odeexamples`\n\nTo open an individual example file for editing, type\n\n`edit exampleFileName.m`\n\nTo run an example, type\n\n`exampleFileName`\n\nThis table contains a list of the available ODE and DAE example files as well as the solvers and options they use. Links are included for the subset of examples that are also published directly in the documentation.\n\n`amp1dae``ode23t`\n• `'Mass'`\n\nStiff DAE — electrical circuit with constant, singular mass matrix\n\nSolve Stiff Transistor Differential Algebraic Equation\n`ballode``ode23`\n• `'Events'`\n\n• `'OutputFcn'`\n\n• `'OutputSel'`\n\n• `'Refine'`\n\n• `'InitialStep'`\n\n• `'MaxStep'`\n\nSimple event location — bouncing ball\n\nODE Event Location\n`batonode``ode45`\n• `'Mass'`\n\nODE with time- and state-dependent mass matrix — motion of a baton\n\nSolve Equations of Motion for Baton Thrown into Air\n`brussode``ode15s`\n• `'JPattern'`\n\n• `'Vectorized'`\n\nStiff large problem — diffusion in a chemical reaction (the Brusselator)\n\nSolve Stiff ODEs\n`burgersode``ode15s`\n• `'Mass'`\n\n• `'MStateDependence'`\n\n• `'JPattern'`\n\n• `'MvPattern'`\n\n• `'RelTol'`\n\n• `'AbsTol'`\n\nODE with strongly state-dependent mass matrix — Burgers' equation solved using a moving mesh technique\n\nSolve ODE with Strongly State-Dependent Mass Matrix\n`fem1ode``ode15s`\n• `'Mass'`\n\n• `'MStateDependence'`\n\n• `'Jacobian'`\n\nStiff problem with a time-dependent mass matrix — finite element method\n\n`fem2ode``ode23s`\n• `'Mass'`\n\nStiff problem with a constant mass matrix — finite element method\n\n`hb1ode``ode15s`\n\nStiff ODE problem solved on a very long interval — Robertson chemical reaction\n\n`hb1dae``ode15s`\n• `'Mass'`\n\n• `'RelTol'`\n\n• `'AbsTol'`\n\n• `'Vectorized'`\n\nStiff, linearly implicit DAE from a conservation law — Robertson chemical reaction\n\nSolve Robertson Problem as Semi-Explicit Differential Algebraic Equations (DAEs)\n`ihb1dae``ode15i`\n• `'RelTol'`\n\n• `'AbsTol'`\n\n• `'Jacobian'`\n\nStiff, fully implicit DAE — Robertson chemical reaction\n\nSolve Robertson Problem as Implicit Differential Algebraic Equations (DAEs)\n`iburgersode``ode15i`\n• `'RelTol'`\n\n• `'AbsTol'`\n\n• `'Jacobian'`\n\n• `'JPattern'`\n\nImplicit ODE system — Burgers’ equation\n\n`kneeode``ode15s`\n• `'NonNegative'`\n\nThe “knee problem” with nonnegativity constraints\n\nNonnegative ODE Solution\n`orbitode``ode45`\n• `'RelTol'`\n\n• `'AbsTol'`\n\n• `'Events'`\n\n• `'OutputFcn'`\n\nAdvanced event location — restricted three body problem\n\nODE Event Location\n`rigidode``ode45`\n\nNonstiff problem — Euler equations of a rigid body without external forces\n\nSolve Nonstiff ODEs\n`vdpode``ode15s`\n• `'Jacobian'`\n\nParameterizable van der Pol equation (stiff for large μ)\n\nSolve Stiff ODEs\n\n Shampine, L. F. and M. K. Gordon, Computer Solution of Ordinary Differential Equations: the Initial Value Problem, W. H. Freeman, San Francisco, 1975.\n\n Forsythe, G., M. Malcolm, and C. Moler, Computer Methods for Mathematical Computations, Prentice-Hall, New Jersey, 1977.\n\n Kahaner, D., C. Moler, and S. Nash, Numerical Methods and Software, Prentice-Hall, New Jersey, 1989.\n\n Shampine, L. F., Numerical Solution of Ordinary Differential Equations, Chapman & Hall, New York, 1994.\n\n Shampine, L. F. and M. W. Reichelt, “The MATLAB ODE Suite,” SIAM Journal on Scientific Computing, Vol. 18, 1997, pp. 1–22.\n\n Shampine, L. F., Gladwell, I. and S. Thompson, Solving ODEs with MATLAB, Cambridge University Press, Cambridge UK, 2003."
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https://www.studystack.com/flashcard-2224348 | [
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"why\n\nMake sure to remember your password. If you forget it there is no way for StudyStack to send you a reset link. You would need to create a new account.\n\nDon't know\nKnow\nremaining cards\nSave\n0:01\nEmbed Code - If you would like this activity on your web page, copy the script below and paste it into your web page.\n\nNormal Size Small Size show me how\n\n# Waves 9-13\n\n### Chapters 9-13: Waves, Sound, EM, Color, mirrors, lenses\n\nAll waves carry _____. Energy\nWaves that need a medium to transfer energy are called _____. Mechanical\nWaves that do not need a medium to travel are _____. What kind of wave? Electromagnetic, travel in transverse waves\nThe lowest point on a transverse wave is a _____. trough\nThe highest point on a transverse wave is a _____. crest\nSound waves require a _____ and travel in ____ waves. medium; compressional/longitudinal\nThe most energetic electromagnetic waves are the _____. Gamma Rays\n_____occurs when light waves strike an object and bounce off. Reflection\nEnergy in a wave is measured by its _____. Amplitude\nThe speed of sound travels fastest in _____, then in _____, and then in ______. Solids, liquids, gases\nIf the frequency in a wave doubles, what happens to the wavelength? gets shorter by half\nSounds are produced by _______ particles. Vibrating\nLow frequency in an electromagnetic waves means that it has a _______ wavelength. long\nThe color in the visible light spectrum with the highest frequency is _____. Purple (Violet)\nColors that we can see are _____(Reflected or absorbed). Reflected\nWhat type of material does not allow light to pass through? Give an example. Opaque, example is a coffee mug\nWhat type of material does allow light to pass through? Give an example. Transparent, wine glass\nA smooth surface that reflects light to form an image is a _____. Mirror\nLight is refracted and spread out by a ____ lens. Concave\nAreas of a compressional wave that are spread out are called _____. Rarefactions\nLight is refracted and forms a focal point in a _____ lens. Used to correct farsighted vision. Convex\nIf S=d/t, how long will it take a person to travel a distance of 10km at an average speed of 5 km/hr? t=d/S so t=2 hr\nIf a=(Vf-Vi)/t, what is the acceleration of a car that speeds up from 30 m/s to 50 m/s in 5 seconds? (50-30)/5 so a=4m/s2\nThe _______ is when you or the source of a sound is moving closer and it causes the sound waves to compress and the pitch gets higher. Doppler Effect\nIf W=F*d, how much work is done when you push a cart 6 meters using 20 Newtons of force? W=20 * 6, then W=120 Joules\nIf I=V/R, then how much current is flowing in a circuit with a 20-Volt battery and has a resistance of 20 ohms. I=1 Amp\nConvert 0.79 meters to centimeters. 79 centimeters\nConvert 750 millimeters to meters. 0.75 meters.\nIf V=frequency*wavelength, then what is the speed of a wave with a frequency of 22 Hz and a wavelength of 65cm? First, 65 cm equals 0.65m. Then, 22*.65= 14.3 m/s\nCreated by: kbordelon2"
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https://republicofsouthossetia.org/question/what-is-the-length-of-th-diagonal-of-a-10-cm-by-15-cm-rectangle-to-the-nearest-whole-number-15925804-69/ | [
"What is the length of th diagonal of a 10 cm by 15 cm rectangle, to the nearest whole number\n\nQuestion\n\nWhat is the length of th diagonal of a 10 cm by 15 cm rectangle, to the nearest whole number\n\nin progress 0\n2 weeks 2022-01-12T12:43:12+00:00 1 Answer 0 views 0\n\nThe diagonal measures 5\n\n13 cm\n\nStep-by-step explanation:\n\na\n2 + b\n2 = c\n2\n\n10\n2\n+ 15\n2 = c\n2\n\n100 + 225 = c\n2\n\n325 = c\n\n5\n\n3\n= c"
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https://greprepclub.com/forum/a-circular-region-has-circumference-c-inches-and-area-k-squa-14900.html?sort_by_oldest=true | [
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"It is currently 26 Feb 2020, 11:37",
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"### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we’ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.",
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"# A circular region has circumference c inches and area k squa",
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"Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\nFounder",
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"Joined: 18 Apr 2015\nPosts: 9739\nFollowers: 202\n\nKudos [?]: 2385 , given: 9227\n\nA circular region has circumference c inches and area k squa [#permalink]\nExpert's post",
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"based on 26 sessions\nA circular region has circumference c inches and area k square inches. If $$c = 3k$$, what is the radius of the circle in inches?\n\nA. $$\\frac{\\sqrt{2}}{3}$$\n\nB. $$\\sqrt{\\frac{2}{3}}$$\n\nC. $$\\frac{2}{3}$$\n\nD. $$\\frac{4 \\pi}{9}$$\n\nE. $$\\frac{2\\pi}{3}$$\n[Reveal] Spoiler: OA\n\n_________________\n\nNeed Practice? 20 Free GRE Quant Tests available for free with 20 Kudos\nGRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests.",
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"Joined: 20 Apr 2016\nPosts: 1129\nWE: Engineering (Energy and Utilities)\nFollowers: 18\n\nKudos [?]: 1042 , given: 231\n\nRe: A circular region has circumference c inches and area k squa [#permalink]\n1\nKUDOS\nCarcass wrote:\nA circular region has circumference c inches and area k square inches. If $$c = 3k$$, what is the radius of the circle in inches?\n\nA. $$\\frac{\\sqrt{2}}{3}$$\n\nB. $$\\sqrt{\\frac{2}{3}}$$\n\nC. $$\\frac{2}{3}$$\n\nD. $$\\frac{4 \\pi}{9}$$\n\nE. $$\\frac{2\\pi}{3}$$\n\nWe know,\n\nCircumference of circle =$$2*\\pi*{Radius}$$\n\nand Area of the circle = $$\\pi*{Radius}^2$$\n\nSince,$$C = 3k$$\n\nor $$2*\\pi*Radius = 3 *\\pi*{Radius}^2$$\n\nor $$Radius = \\frac{2}{3}$$\n_________________\n\nIf you found this post useful, please let me know by pressing the Kudos Button\n\nRules for Posting\n\nGot 20 Kudos? You can get Free GRE Prep Club Tests\n\nGRE Prep Club Members of the Month:TOP 10 members of the month with highest kudos receive access to 3 months GRE Prep Club tests",
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"Re: A circular region has circumference c inches and area k squa [#permalink] 07 Sep 2019, 18:48\nDisplay posts from previous: Sort by\n\n# A circular region has circumference c inches and area k squa",
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https://huffydeb2003.savingadvice.com/2009/05/20/another-penny_51145/ | [
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Array\n(\n => 106.197.31.240\n)\n\n => Array\n(\n => 122.168.179.251\n)\n\n => Array\n(\n => 39.37.167.126\n)\n\n => Array\n(\n => 171.48.8.115\n)\n\n => Array\n(\n => 157.44.152.14\n)\n\n => Array\n(\n => 103.77.43.219\n)\n\n => Array\n(\n => 122.161.49.38\n)\n\n => Array\n(\n => 122.161.52.83\n)\n\n => Array\n(\n => 122.173.108.210\n)\n\n => Array\n(\n => 60.254.109.92\n)\n\n => Array\n(\n => 103.57.85.75\n)\n\n => Array\n(\n => 106.0.58.36\n)\n\n => Array\n(\n => 122.161.49.212\n)\n\n => Array\n(\n => 27.255.182.159\n)\n\n => Array\n(\n => 116.75.230.159\n)\n\n => Array\n(\n => 122.173.152.133\n)\n\n => Array\n(\n => 129.0.79.247\n)\n\n => Array\n(\n => 223.228.163.44\n)\n\n => Array\n(\n => 103.168.78.82\n)\n\n => Array\n(\n => 39.59.67.124\n)\n\n => Array\n(\n => 182.69.19.120\n)\n\n => Array\n(\n => 196.202.236.195\n)\n\n => Array\n(\n => 137.59.225.206\n)\n\n => Array\n(\n => 143.110.209.194\n)\n\n => Array\n(\n => 117.201.233.91\n)\n\n => Array\n(\n => 37.120.150.107\n)\n\n => Array\n(\n => 58.65.222.10\n)\n\n => Array\n(\n => 202.47.43.86\n)\n\n => Array\n(\n => 106.206.223.234\n)\n\n => Array\n(\n => 5.195.153.158\n)\n\n => Array\n(\n => 223.227.127.243\n)\n\n => Array\n(\n => 103.165.12.222\n)\n\n => Array\n(\n => 49.36.185.189\n)\n\n => Array\n(\n => 59.96.92.57\n)\n\n => Array\n(\n => 203.194.104.235\n)\n\n => Array\n(\n => 122.177.72.33\n)\n\n => Array\n(\n => 106.213.126.40\n)\n\n => Array\n(\n => 45.127.232.69\n)\n\n => Array\n(\n => 156.146.59.39\n)\n\n => Array\n(\n => 103.21.184.11\n)\n\n => Array\n(\n => 106.212.47.59\n)\n\n => Array\n(\n => 182.179.137.235\n)\n\n => Array\n(\n => 49.36.178.154\n)\n\n => Array\n(\n => 171.48.7.128\n)\n\n => Array\n(\n => 119.160.57.96\n)\n\n => Array\n(\n => 197.210.79.92\n)\n\n => Array\n(\n => 36.255.45.87\n)\n\n => Array\n(\n => 47.31.219.47\n)\n\n => Array\n(\n => 122.161.51.160\n)\n\n => Array\n(\n => 103.217.123.129\n)\n\n => Array\n(\n => 59.153.16.12\n)\n\n => Array\n(\n => 103.92.43.226\n)\n\n => Array\n(\n => 47.31.139.139\n)\n\n => Array\n(\n => 210.2.140.18\n)\n\n => Array\n(\n => 106.210.33.219\n)\n\n => Array\n(\n => 175.107.203.34\n)\n\n => Array\n(\n => 146.196.32.144\n)\n\n => Array\n(\n => 103.12.133.121\n)\n\n => Array\n(\n => 103.59.208.182\n)\n\n => Array\n(\n => 157.37.190.232\n)\n\n => Array\n(\n => 106.195.35.201\n)\n\n => Array\n(\n => 27.122.14.83\n)\n\n => Array\n(\n => 194.193.44.5\n)\n\n => Array\n(\n => 5.62.43.245\n)\n\n => Array\n(\n => 103.53.80.50\n)\n\n => Array\n(\n => 47.29.142.233\n)\n\n => Array\n(\n => 154.6.20.63\n)\n\n => Array\n(\n => 173.245.203.128\n)\n\n => Array\n(\n => 103.77.43.231\n)\n\n => Array\n(\n => 5.107.166.235\n)\n\n => Array\n(\n => 106.212.44.123\n)\n\n => Array\n(\n => 157.41.60.93\n)\n\n => Array\n(\n => 27.58.179.79\n)\n\n => Array\n(\n => 157.37.167.144\n)\n\n => Array\n(\n => 119.160.57.115\n)\n\n => Array\n(\n => 122.161.53.224\n)\n\n => Array\n(\n => 49.36.233.51\n)\n\n => Array\n(\n => 101.0.32.8\n)\n\n => Array\n(\n => 119.160.103.158\n)\n\n => Array\n(\n => 122.177.79.115\n)\n\n => Array\n(\n => 107.181.166.27\n)\n\n => Array\n(\n => 183.6.0.125\n)\n\n => Array\n(\n => 49.36.186.0\n)\n\n => Array\n(\n => 202.181.5.4\n)\n\n => Array\n(\n => 45.118.165.144\n)\n\n => Array\n(\n => 171.96.157.133\n)\n\n => Array\n(\n => 222.252.51.163\n)\n\n => Array\n(\n => 103.81.215.162\n)\n\n => Array\n(\n => 110.225.93.208\n)\n\n => Array\n(\n => 122.161.48.200\n)\n\n => Array\n(\n => 119.63.138.173\n)\n\n => Array\n(\n => 202.83.58.208\n)\n\n => Array\n(\n => 122.161.53.101\n)\n\n => Array\n(\n => 137.97.95.21\n)\n\n => Array\n(\n => 112.204.167.123\n)\n\n => Array\n(\n => 122.180.21.151\n)\n\n => Array\n(\n => 103.120.44.108\n)\n\n => Array\n(\n => 49.37.220.174\n)\n\n => Array\n(\n => 1.55.255.124\n)\n\n => Array\n(\n => 23.227.140.173\n)\n\n => Array\n(\n => 43.248.153.110\n)\n\n => Array\n(\n => 106.214.93.101\n)\n\n => Array\n(\n => 103.83.149.36\n)\n\n => Array\n(\n => 103.217.123.57\n)\n\n => Array\n(\n => 193.9.113.119\n)\n\n => Array\n(\n => 14.182.57.204\n)\n\n => Array\n(\n => 117.201.231.0\n)\n\n => Array\n(\n => 14.99.198.186\n)\n\n => Array\n(\n => 36.255.44.204\n)\n\n => Array\n(\n => 103.160.236.42\n)\n\n => Array\n(\n => 31.202.16.116\n)\n\n => Array\n(\n => 223.239.49.201\n)\n\n => Array\n(\n => 122.161.102.149\n)\n\n => Array\n(\n => 117.196.123.184\n)\n\n => Array\n(\n => 49.205.112.105\n)\n\n => Array\n(\n => 103.244.176.201\n)\n\n => Array\n(\n => 95.216.15.219\n)\n\n => Array\n(\n => 103.107.196.174\n)\n\n => Array\n(\n => 203.190.34.65\n)\n\n => Array\n(\n => 23.227.140.182\n)\n\n => Array\n(\n => 171.79.74.74\n)\n\n => Array\n(\n => 106.206.223.244\n)\n\n => Array\n(\n => 180.151.28.140\n)\n\n => Array\n(\n => 165.225.124.114\n)\n\n => Array\n(\n => 106.206.223.252\n)\n\n => Array\n(\n => 39.62.23.38\n)\n\n => Array\n(\n => 112.211.252.33\n)\n\n => Array\n(\n => 146.70.66.242\n)\n\n => Array\n(\n => 222.252.51.38\n)\n\n => Array\n(\n => 122.162.151.223\n)\n\n => Array\n(\n => 180.178.154.100\n)\n\n => Array\n(\n => 180.94.33.94\n)\n\n => Array\n(\n => 205.164.130.82\n)\n\n => Array\n(\n => 117.196.114.167\n)\n\n => Array\n(\n => 43.224.0.189\n)\n\n => Array\n(\n => 154.6.20.59\n)\n\n => Array\n(\n => 122.161.131.67\n)\n\n => Array\n(\n => 70.68.68.159\n)\n\n => Array\n(\n => 103.125.130.200\n)\n\n => Array\n(\n => 43.242.176.147\n)\n\n => Array\n(\n => 129.0.102.29\n)\n\n => Array\n(\n => 182.64.180.32\n)\n\n => Array\n(\n => 110.93.250.196\n)\n\n => Array\n(\n => 139.135.57.197\n)\n\n => Array\n(\n => 157.33.219.2\n)\n\n => Array\n(\n => 205.253.123.239\n)\n\n => Array\n(\n => 122.177.66.119\n)\n\n => Array\n(\n => 182.64.105.252\n)\n\n => Array\n(\n => 14.97.111.154\n)\n\n => Array\n(\n => 146.196.35.35\n)\n\n => Array\n(\n => 103.167.162.205\n)\n\n => Array\n(\n => 37.111.130.245\n)\n\n => Array\n(\n => 49.228.51.196\n)\n\n => Array\n(\n => 157.39.148.205\n)\n\n => Array\n(\n => 129.0.102.28\n)\n\n => Array\n(\n => 103.82.191.229\n)\n\n => Array\n(\n => 194.104.23.140\n)\n\n => Array\n(\n => 49.205.193.252\n)\n\n => Array\n(\n => 222.252.33.119\n)\n\n => Array\n(\n => 173.255.132.114\n)\n\n => Array\n(\n => 182.64.148.162\n)\n\n => Array\n(\n => 175.176.87.8\n)\n\n => Array\n(\n => 5.62.57.6\n)\n\n => Array\n(\n => 119.160.96.229\n)\n\n => Array\n(\n => 49.205.180.226\n)\n\n => Array\n(\n => 95.142.120.59\n)\n\n => Array\n(\n => 183.82.116.204\n)\n\n => Array\n(\n => 202.89.69.186\n)\n\n => Array\n(\n => 39.48.165.36\n)\n\n => Array\n(\n => 192.140.149.81\n)\n\n => Array\n(\n => 198.16.70.28\n)\n\n => Array\n(\n => 103.25.250.236\n)\n\n => Array\n(\n => 106.76.202.244\n)\n\n => Array\n(\n => 47.8.8.165\n)\n\n => Array\n(\n => 202.5.145.213\n)\n\n => Array\n(\n => 106.212.188.243\n)\n\n => Array\n(\n => 106.215.89.2\n)\n\n => Array\n(\n => 119.82.83.148\n)\n\n => Array\n(\n => 123.24.164.245\n)\n\n => Array\n(\n => 187.67.51.106\n)\n\n => Array\n(\n => 117.196.119.95\n)\n\n => Array\n(\n => 95.142.120.66\n)\n\n => Array\n(\n => 156.146.59.35\n)\n\n => Array\n(\n => 49.205.213.148\n)\n\n => Array\n(\n => 111.223.27.206\n)\n\n => Array\n(\n => 49.205.212.86\n)\n\n => Array\n(\n => 103.77.42.103\n)\n\n => Array\n(\n => 110.227.62.25\n)\n\n => Array\n(\n => 122.179.54.140\n)\n\n => Array\n(\n => 157.39.239.81\n)\n\n => Array\n(\n => 138.128.27.234\n)\n\n => Array\n(\n => 103.244.176.194\n)\n\n => Array\n(\n => 130.105.10.127\n)\n\n => Array\n(\n => 103.116.250.191\n)\n\n => Array\n(\n => 122.180.186.6\n)\n\n => Array\n(\n => 101.53.228.52\n)\n\n => Array\n(\n => 39.57.138.90\n)\n\n => Array\n(\n => 197.156.137.165\n)\n\n => Array\n(\n => 49.37.155.78\n)\n\n => Array\n(\n => 39.59.81.32\n)\n\n => Array\n(\n => 45.127.44.78\n)\n\n => Array\n(\n => 103.58.155.83\n)\n\n => Array\n(\n => 175.107.220.20\n)\n\n => Array\n(\n => 14.255.9.197\n)\n\n => Array\n(\n => 103.55.63.146\n)\n\n => Array\n(\n => 49.205.138.81\n)\n\n => Array\n(\n => 45.35.222.243\n)\n\n => Array\n(\n => 203.190.34.57\n)\n\n => Array\n(\n => 205.253.121.11\n)\n\n => Array\n(\n => 154.72.171.177\n)\n\n => Array\n(\n => 39.52.203.37\n)\n\n => Array\n(\n => 122.161.52.2\n)\n\n => Array\n(\n => 82.145.41.170\n)\n\n => Array\n(\n => 103.217.123.33\n)\n\n => Array\n(\n => 103.150.238.100\n)\n\n => Array\n(\n => 125.99.11.182\n)\n\n => Array\n(\n => 103.217.178.70\n)\n\n => Array\n(\n => 197.210.227.95\n)\n\n => Array\n(\n => 116.75.212.153\n)\n\n => Array\n(\n => 212.102.42.202\n)\n\n => Array\n(\n => 49.34.177.147\n)\n\n => Array\n(\n => 173.242.123.110\n)\n\n => Array\n(\n => 49.36.35.254\n)\n\n => Array\n(\n => 202.47.59.82\n)\n\n => Array\n(\n => 157.42.197.119\n)\n\n => Array\n(\n => 103.99.196.250\n)\n\n => Array\n(\n => 119.155.228.244\n)\n\n => Array\n(\n => 130.105.160.170\n)\n\n => Array\n(\n => 78.132.235.189\n)\n\n => Array\n(\n => 202.142.186.114\n)\n\n => Array\n(\n => 115.99.156.136\n)\n\n => Array\n(\n => 14.162.166.254\n)\n\n => Array\n(\n => 157.39.133.205\n)\n\n => Array\n(\n => 103.196.139.157\n)\n\n => Array\n(\n => 139.99.159.20\n)\n\n => Array\n(\n => 175.176.87.42\n)\n\n => Array\n(\n => 103.46.202.244\n)\n\n => Array\n(\n => 175.176.87.16\n)\n\n => Array\n(\n => 49.156.85.55\n)\n\n => Array\n(\n => 157.39.101.65\n)\n\n => Array\n(\n => 124.253.195.93\n)\n\n => Array\n(\n => 110.227.59.8\n)\n\n => Array\n(\n => 157.50.50.6\n)\n\n => Array\n(\n => 95.142.120.25\n)\n\n => Array\n(\n => 49.36.186.141\n)\n\n => Array\n(\n => 110.227.54.161\n)\n\n => Array\n(\n => 88.117.62.180\n)\n\n => Array\n(\n => 110.227.57.8\n)\n\n => Array\n(\n => 106.200.36.21\n)\n\n => Array\n(\n => 202.131.143.247\n)\n\n => Array\n(\n => 103.46.202.4\n)\n\n => Array\n(\n => 122.177.78.217\n)\n\n => Array\n(\n => 124.253.195.201\n)\n\n => Array\n(\n => 27.58.17.91\n)\n\n => Array\n(\n => 223.228.143.162\n)\n\n => Array\n(\n => 119.160.96.233\n)\n\n => Array\n(\n => 49.156.69.213\n)\n\n => Array\n(\n => 41.80.97.54\n)\n\n => Array\n(\n => 122.176.207.193\n)\n\n => Array\n(\n => 45.118.156.6\n)\n\n => Array\n(\n => 157.39.154.210\n)\n\n => Array\n(\n => 103.48.197.173\n)\n\n => Array\n(\n => 103.46.202.98\n)\n\n => Array\n(\n => 157.43.214.102\n)\n\n => Array\n(\n => 180.191.125.73\n)\n\n)\n```\nAnother penny!: Purple Flower's Finance Blog\n Layout: Blue and Brown (Default) Author's Creation\n Home > Another penny!\n\n# Another penny!\n\nMay 20th, 2009 at 12:25 am\n\nI found a penny at the Wal-mart checkout area again. A little more extra income!!\n\n+++\n\nI feel like I'm getting sick...sore throat and generally just weak/tired. Hopefully I can stop it from progressing any further, but we shall see. I think I will have a hot cup of lemon tea now.\n\n### 0 Responses to “Another penny!”\n\n(Note: If you were logged in, we could automatically fill in these fields for you.)\n Name: * Email: Will not be published. Subscribe: Notify me of additional comments to this entry. URL: Verification: * Please spell out the number 4. [ Why? ]\n\nvB Code: You can use these tags: [b] [i] [u] [url] [email]"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9293072,"math_prob":0.9994655,"size":485,"snap":"2021-43-2021-49","text_gpt3_token_len":123,"char_repetition_ratio":0.11018711,"word_repetition_ratio":0.38372093,"special_character_ratio":0.25979382,"punctuation_ratio":0.16666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99976784,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-03T05:02:12Z\",\"WARC-Record-ID\":\"<urn:uuid:90f11e9b-0fa7-46f6-a2a3-d22fe81dae43>\",\"Content-Length\":\"310229\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c2eb1180-470e-44a0-b0db-7736b503e850>\",\"WARC-Concurrent-To\":\"<urn:uuid:2876a945-aa9a-421f-9d47-965d5f24af49>\",\"WARC-IP-Address\":\"173.231.200.26\",\"WARC-Target-URI\":\"https://huffydeb2003.savingadvice.com/2009/05/20/another-penny_51145/\",\"WARC-Payload-Digest\":\"sha1:GN5TWB5KDQWY5MTIQBNCG4AAEE7CSFJG\",\"WARC-Block-Digest\":\"sha1:QDEEV5G4QUX4LYRXGQ5LKZDRIIM6CZCS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362589.37_warc_CC-MAIN-20211203030522-20211203060522-00363.warc.gz\"}"} |
https://www.tutorialsrack.com/articles/205/how-to-get-the-year-month-and-day-from-a-datetime-in-python | [
";\n\n# How to Get the year, month, and day from a datetime in Python\n\n###### Tutorialsrack 29/03/2020 Programs Python\n\nIn this article, we will learn how to get or extract the year, month and day from a DateTime in python.\n\nExample: In this example, we will get the date part from the DateTime\n\n##### Example\n``````# How to Get the year, month, and day of a datetime in Python\n\n# Import Module\nimport datetime\n\nnow = datetime.datetime.now()\n\nprint (\"\\nCurrent date and time using instance attributes:\")\nprint (\"\\nCurrent year: \", now.year)\n# Output ==> Current year: 2020\n\nprint (\"\\nCurrent month: \", now.month)\n#Output ==> Current month: 3\n\nprint (\"\\nCurrent day: \", now.day)\n# Output ==> Current day: 29\n\nprint (\"\\nCurrent hour: \", now.hour)\n# Output ==> Current hour: 16\n\nprint (\"\\nCurrent minute: \", now.minute)\n# Output ==> Current minute: 5\n\nprint (\"\\nCurrent second: \", now.second)\n# Output ==> Current second: 44\n\nprint (\"\\nCurrent microsecond: \", now.microsecond)\n# Output ==> Current microsecond: 405009\n``````\n\nI hope this article will help you to understand how to get or extract the year, month and day from a DateTime in python."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.76143545,"math_prob":0.49145794,"size":1090,"snap":"2020-34-2020-40","text_gpt3_token_len":265,"char_repetition_ratio":0.19429098,"word_repetition_ratio":0.12429378,"special_character_ratio":0.2880734,"punctuation_ratio":0.20476191,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95847607,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-27T17:26:09Z\",\"WARC-Record-ID\":\"<urn:uuid:79651049-5af1-443b-931b-8a5d696f48e1>\",\"Content-Length\":\"25014\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:100587de-854c-48bd-848e-a09e2558b83c>\",\"WARC-Concurrent-To\":\"<urn:uuid:d7c93271-5ddd-4272-8ddc-80f01b501d2c>\",\"WARC-IP-Address\":\"103.20.215.198\",\"WARC-Target-URI\":\"https://www.tutorialsrack.com/articles/205/how-to-get-the-year-month-and-day-from-a-datetime-in-python\",\"WARC-Payload-Digest\":\"sha1:XWB56D7PH5ESZLNY5CICGPCTRRJNAGDJ\",\"WARC-Block-Digest\":\"sha1:IFYHXTMK7POYF56LXI2RZK2MS2P3IZT3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400283990.75_warc_CC-MAIN-20200927152349-20200927182349-00338.warc.gz\"}"} |
http://access.feld.cvut.cz/rservice.php?akce=tisk&cisloclanku=2007030003 | [
"# Time Precision of Navigation Systems\n\nAutor: E. Kozlowska <kozloe1(at)fel.cvut.cz>, Pracoviště: České vysoké učení technické v Praze, FEL, Téma: Lokalizace polohy, Vydáno dne: 20. 03. 2007\n\nThis paper will present advantages and potential possibilities of satellite navigation systems, especially with the emphasis on what makes them so precise in showing time.\n\nSatellite navigation is a type of radio navigation that use radio waves from artificial satellites in order to determine points and moving receivers location, along with their movement parameters, in any place on the Earth. The most popular system of satellite navigation is GPS (Global Positioning System).\n\nPrinciple of operation\nSatellite navigation system principle of operation is based on measurements of distance between four satellites and receiver, using the formula:",
null,
"(1)\n\nRadio waves information about local time is constantly broadcasts by each satellite; time is counted down by four cesium-beam atomic clocks. To compute the distance from satellite, GPS receiver measure time, which passed between sending and receiving the signal; difference between these two values is multiply by speed of light, which is speed of radio wave propagation. When the distance from first satellite is calculate, it is known, that the GPS receiver’s location is somewhere on sphere’s surface with satellite as a center and radius equal to the distance, Figure 1.a). By knowing distance from second satellite, it is possible to reduce this space to a circle, which is interception point of both spheres, Figure 1.b). Distance from third satellite to receiver makes possible to narrow down search area to two points, where one for being an impossible result can be rejected, Figure 1.c). In theory those data would be sufficient to determine precise distance, but the GPS receiver is equipped with quartz clocks, which are not as accurate as those very expensive atomic clocks on board of satellites with accuracy of 1 nanosecond. That is why the signal from fourth satellite is being use to synchronize quartz clock in receiver with satellite’s atomic clock .\n\n a) b)",
null,
"",
null,
"c)",
null,
"Figure 1. Positioning.\n\nSatellite broadcast two microwave signals: C/A code and P-code.\n\nCoarse Acquisition (C/A-code)\nIt is a basic signal for civil GPS receivers on the L1 frequency (1575,42 MHz), which carries navigation message and SPS code signal (Standard Positioning Service). The C/A code is a repeating 1 MHz Pseudo Random Noise (PRN) Code. L1 carrier signal is modulated by this code. The C/A code is repeated every one millisecond (1023 bits) and each satellite broadcast different PRN one, so by this they can be identify. Because C/A code is designed for civil use, it is less accurate then P-code, and it is possible to jam it (broadcast false signal to mislead the receiver about its real position). C/A code has one advantage over P-code, it is faster to gain it from satellite and compute first position.\n\nPrecision (P-code)\nIt is signal for the Precise Positioning Service (PPS), broadcasted on L2 frequency (1227,60 MHz). It is dedicated to be used by U.S. military because of it highly precise location information. Also it is used by PPS receivers to calculate ionospheric delays. The P-code modulates L1 and L2 carrier phase. It is a very long 10 MHz PRN code. To be free of jamming and spoofing its encrypted form is broadcasted, Y-code. To receive it a special receiver is needed, with cryptographic keys.\n\nIt is 50 Hz signal, which modulates L1-C/A code signal. It contains data about satellite orbit, clock corrections and other parameters.\n\nRelativity\nGPS was first technology were Einstein theory of relativity found its application. After sending first satellite, result was not as accurate as everybody expected; difference between real position and determined by GPS was more that 11 km. It became obvious that clocks on deck of satellite count time a little bit different than on earth. After first try it was necessary to provide each satellite with relativity correction. We distinguish two theories of relativity: General and Special. General Relativity (GR) predicts that clocks in a stronger gravitational field will tick at a slower rate. Special Relativity (SR) predicts that moving clocks will appear to tick slower than non-moving ones. Because Earth’s spin rate determines its shape, these two effects are not independent, and it is therefore not entirely coincidental that the effects exactly cancel. The cancellation is not general, however. The special theory, in addition to claiming, that the frequency received is a function of the relative velocity, also claims that the speed of light is isotropic relative to the (observer) receiver; what is more, the GPS system uses the earth-centered non-rotating frame and also assumes the speed of light is isotropic in that frame. The nominal velocity of a GPS satellite with respect to an earth-centered non-rotating frame is about 3,87 kilometers per second. Using this frame, the computed clock rates should slow by:",
null,
"(2)\n\nwhere\n\n• f is satellite clock frequency,\n• f0 is clocks on Earth frequency,\n• v is speed of satellite relatively to Earth and is speed of light.\n\nUsing this expression, one obtains a frequency decrease of 8,32 parts in 1011 for GPS satellites. General Relativity experiments show that the gravitational potential affects the rate at which clocks run. In order to demonstrate these effects without excessive use of mathematics, scale factor “s” was define, slightly less than one, which is used to multiply or scale the parameter of interest. This scale factor is a direct function of the gravitational potential, and can be computed from it. The lower the gravitational potential the smaller the scale factor becomes. The scale factor is defined as,",
null,
"(3)\n\nwhere\n\n• G is Newton's gravitational constant,\n• M is the mass causing the gravitational potential,\n• r is the distance from the center of the gravitational potential.\n\nThe clocks run slower (measured time appears dilated) as compared to the rate at which they would run if they were located external to the gravitational field. The comparative clock rate is given in terms of the scale factor, s, defined above as,",
null,
"(4)\n\nwhere\n\n• fs is the rate the clock would run if it were external to the gravitational potential.\n\nThere is a change in the clock frequency of the GPS satellite clocks at the time of their launch. The change in the gravitational potential at the surface of the earth to the gravitational potential at the satellite orbital height causes an increase in the average rate at which the clock runs of 5,311 parts in 1010. As stated above, the speed of the GPS satellites in orbit causes a clock frequency decrease of 8,32 parts in 1011. These two effects combine to give a net increase in frequency of 4,479 parts in 1010. These two frequency-biasing effects and the additional small mean effects of the earth oblateness, sun and moon are compensated before launch by setting the frequency low by 4,45 parts in 1010 . Further, each GPS receiver has built into it a microcomputer that (among other things) performs the necessary relativistic calculation when it is determining the user’s position.\n\nAccuracy\nThe position accuracy is primarily dependent on the satellite position and signal delay. Several factors exist that has effect on accuracy:\n\nIonosphere delays\nThe satellite signal slows as it passes through the atmosphere. The GPS system uses a built-in model that calculates an average amount of delay to partially correct for this type of error. The ionospheric time delay is given by:",
null,
"(5)\n\nwhere TEC is the total number of electrons, called the Total Electron Content, along the path from the transmitter to the receiver, c is the velocity of light in meters per second, and f is the carrier frequency in Hz. TEC common definition is a number of electrons in a unit cross-section column of 1 m2 area along the path and range from 1016 electrons per m2 to 1019 electrons per m2. This leads to the delay of 54 ns, for 1,575 GHz C/A carrier frequency for GPS satellite system and for a TEC of 1018 electrons per m2. Obviously, the TEC parameter is significant for GPS system .\n\nTroposphere delays\nWhen time signal is transferred between ground stations by use of common-view satellite, one records the time of arrival of the signal and computes the time of transmission by subtracting the propagation time. By dividing the range to the satellite by the velocity of light a the propagation time is found. Nevertheless, moisture and oxygen in the troposphere have an effect on the signal velocity of propagation, which has effect on computing time of transmission and as a result, the time transfer. The geometry, the latitude, the pressure, and the temperature, all those factors has influence on this effect and may fluctuate in error from 3 ns to 300 ns. That is why a reasonable models are employed and use of high elevation angels. It should lower the differential delay between two sited to 10 ns .\n\nSignal multipath\nThis kind of error occurs when the GPS signal is reflected off objects such as tall buildings or large rock surfaces before it reaches the receiver. This increases the travel time of the signal.\n\nA receiver is not provide with clock as accurate as the atomic clocks onboard the GPS satellites. For that reason, it may have very slight timing errors.\n\nOrbital errors\nReported location of satellite, known also ephemeris, can be inaccurate, which can lead to form an error.\n\nNumber of satellites visible\nThe best accuracy receiver can get by seeing large number of GPS satellites. Buildings, terrain, electronic interference, or sometimes even dense foliage can block signal reception, causing position errors or possibly no position reading at all. GPS units typically will not work indoors, underwater or underground.\n\nThis refers to the relative position of the satellites at any given time. Ideal satellite geometry exists when the satellites are located at wide angles relative to each other. Poor geometry results when the satellites are located in a line or in a tight grouping .\n\nConclusion\nUsing today’s state of the art global navigation satellite systems (GNSS), you can pinpoint your location anywhere on earth with an accuracy of less than fifteen meters. Currently, the only system available to the general public is the American Global Positioning System, which has been fully operational since mid-1994. The upcoming European competitor, Galileo, promises to improve accuracy. This made it possible to find your way through a city, solely based on the data gathered by your receiver from space. Already nowadays GNSS will find extremely useful applications, thanks to its precision in measuring the time, such as replacing seeing-eye dogs and guiding motor vehicles.\n\nAcknowledgement\nThis work has been supported by the grant No. FT TA2/0475 of the Ministry of Industry and Trade of the Czech Republic.\n\nReferences\n http://pl.wikipedia.org/wiki/Nawigacja_satelitarna, 13.03.2007\n LAMPARSKI J.: Navstar GPS od teorii do praktyki, Wydawnictwo Uniwersytetu Warmińsko-Mazurskiego, Olsztyn 2001\n NARKIEWICZ J.: GPS Globalny System Pozycyjny, Wydawnictwa Komunikacji i Łączności, Warszawa 2003"
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https://djalil.chafai.net/blog/page/3/?s=julia | [
"# 11 search results for \"julia\"\n\nThis micro post is about automatic differentiation, a programming technique allowing to compute automatically the derivative of a function from its source code. This technique, in its reverse mode, is attributed to Seppo Linnainmaa who introduced the idea in his Master thesis in 1970.\n\nHow to compute the derivative of a numerical function? If we have a symbolic mathematical expression for it, then we can use the rules of differential calculus, and this can be automated: this is known as symbolic differentiation. If the function is a black box, then we can compute numerically the differential ratio by using evaluations for very close arguments: this is known as numerical differentiation. But actually, if we have the source code of the function, one could follow the code and create an augmented code in parallel that computes the effect of each elementary operation on the derivative: this is known as automatic differentiation. Automatic differentiation can be implemented by using object programming and operators overloading. Automatic differentiation can be in trouble in the presence of conditionals.\n\nNowadays automatic differentiation is fashionable, partly because of the development of deep machine learning and multi-layered neural networks. For more, take a look at the Wikipedia article. If you like the Julia programming language, you may take a look at JuliaDiff.",
null,
"Recently, a friend of mine, Arnaud Guyader, discovered by accident during a teaching session that the geometric distribution generator of GNU R, implemented in the rgeom function, is sub-optimal. The command help rgeom says rgeom uses the derivation as an exponential mixture of Poissons, see Devroye, L. (1986) Non-Uniform Random Variate Generation. Springer-Verlag, New York. Page 480. Moreover, page 480 of this book says nothing about the geometric distribution.\n\n/* Source code of rgeom.c in latest version of GNU R (2014-11-15).\n* Mathlib : A C Library of Special Functions\n* Copyright (C) 1998 Ross Ihaka and the R Core Team.\n* Copyright (C) 2000 The R Core Team\n*\n* This program is free software; you can redistribute it and/or modify\n* the Free Software Foundation; either version 2 of the License, or\n* (at your option) any later version.\n*\n* This program is distributed in the hope that it will be useful,\n* but WITHOUT ANY WARRANTY; without even the implied warranty of\n* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the\n* GNU General Public License for more details.\n*\n* You should have received a copy of the GNU General Public License\n* along with this program; if not, a copy is available at\n*\n* SYNOPSIS\n*\n* #include <Rmath.h>\n* double rgeom(double p);\n*\n* DESCRIPTION\n*\n* Random variates from the geometric distribution.\n*\n* NOTES\n*\n* We generate lambda as exponential with scale parameter\n* p / (1 – p). Return a Poisson deviate with mean lambda.\n*\n* REFERENCE\n*\n* Devroye, L. (1986).\n* Non-Uniform Random Variate Generation.\n* New York: Springer-Verlag.\n* Page 480.\n*/\n\n#include “nmath.h”\n\ndouble rgeom(double p)\n{\nif (!R_FINITE(p) || p <= 0 || p > 1) ML_ERR_return_NAN;\n\nreturn rpois(exp_rand() * ((1 – p) / p));\n}\n\nThe fact that the geometric distribution is an exponential mixture of Poisson is a special case of the more general fact that the negative binomial distribution, i.e. the convolution power of the geometric distribution also known as the Pascal distribution, is a gamma mixture of Poisson. Namely if ${X\\sim\\mathrm{Gamma}(n,\\ell)}$ with ${\\ell=p/(1-p)}$, and if ${Y|X\\sim\\mathrm{Poisson}(X)}$ then ${Y\\sim\\mathrm{NegativeBinomial}(n,p)}$, since for every ${k\\geq n}$, ${\\mathbb{P}(Y=k)}$ writes\n\n$\\int_0^\\infty\\!e^{-\\lambda} \\frac{\\lambda^k}{k!} \\frac{\\ell^n\\lambda^{n-1}}{\\Gamma(n)}e^{-\\lambda\\ell}\\,d\\lambda =\\frac{\\ell^n}{k!\\Gamma(n)} \\int_0^\\infty\\!\\lambda^{n+k-1}e^{-\\lambda(\\ell+1)}\\,d\\lambda =\\frac{\\Gamma(n+k)}{k!\\Gamma(n)}(1-p)^kp^n.$\n\nFor ${n=1}$ we recover the geometric distribution. This mixture representation is useful for the simulation of negative binomial distributions with high integer parameter, but is not the best choice for the simulation of the special case of the geometric distribution.\n\nThe simulation of the geometric distribution is better done by using the so-called inversion method (generalized inverse of the cdf), which boils down to take the integer part of the standard exponential distribution generator: ceil(log(rand())/log(1-p)) since if ${X\\sim\\mathrm{Exponential}(\\lambda)}$ with ${\\lambda=-\\log(1-p)}$ then\n\n$\\mathbb{P}(k\\leq X\\leq k+1) =e^{-k\\lambda}-e^{-(k+1)\\lambda} =e^{-\\lambda k}(1-e^{-\\lambda}) =(1-p)^kp.$\n\nPatch. Here is a possible patch to the rgeom.c of GNU R.\n\n31,32c31\n< * We generate lambda as exponential with scale parameter\n< * p / (1 – p). Return a Poisson deviate with mean lambda.\n\n> * returns 1+ integer part of exponential with scale parameter -1 / log(1 – p).\n39c38\n< * Page 480.\n\n> * Section X.2 page 498.\n48c47\n< return rpois(exp_rand() * ((1 – p) / p));\n\n> return ceil(log(rand()) / log(1 – p));\n\nFurther reading. The simulation of the geometric distribution is discussed in Section X.2 page 498 of the book of Devroye: the inversion method is considered, but no words on the exponential mixture of Poisson. The simulation of the negative binomial distribution is discussed in Section X.4.7 page 543, and the Gamma mixture of Poisson is considered.\n\nOther software. The random generator for the geometric distribution proposed presently by the GNU Scientific Library, by the GNU Octave, and by the Scilab software are based on the inversion method. The Distributions.jl package for the Julia software also uses the inversion method. The good news with open source software is that even if the help does not mention the algorithm that is used, you can always check the source code.\n\nThis post is about the discrete Dirichlet problem and Gaussian free field, linked with the random walk on ${\\mathbb{Z}^d}$. Discreteness allows to go to the concepts with minimal abstraction. The Gaussian free field (GFF) is an important Gaussian object which appears, like most Gaussian objects, as a limiting object in many models of Statistical Physics.\n\nSymmetric Nearest Neighbors random walk. The symmetric nearest neighbors random walk on ${\\mathbb{Z}^d}$ is the sequence of random variables ${X={(X_n)}_{n\\geq0}}$ defined by the linear recursion\n\n$X_{n+1}=X_n+\\xi_{n+1}=X_0+\\xi_1+\\cdots+\\xi_{n+1}$\n\nwhere ${{(\\xi)}_{n\\geq1}}$ is a sequence of independent and identically distributed random variables on ${\\mathbb{Z}^d}$, independent of ${X_0}$, and uniformly distributed on the discrete ${\\ell^1}$ sphere: ${\\{\\pm e_1,\\ldots,\\pm e_d\\}}$ where ${e_1,\\ldots,e_d}$ is the canonical basis of ${\\mathbb{R}^d}$. The term symmetric comes from the fact that in each of the ${d}$ dimensions, both directions are equally probable in the law of the increments.\n\nThe sequence ${X}$ is a homogeneous Markov chain with state space ${\\mathbb{Z}^d}$ since\n\n$\\begin{array}{rcl} \\mathbb{P}(X_{n+1}=x_{n+1}\\,|\\,X_n=x_n,\\ldots,X_0=x_0) &=&\\mathbb{P}(X_{n+1}=x_{n+1}\\,|\\,X_n=x_n)\\\\ &=&\\mathbb{P}(X_1=x_{n+1}\\,|\\,X_0=x_n)\\\\ &=&\\mathbb{P}(\\xi_1=x_{n+1}-x_n). \\end{array}$\n\nIts transition kernel ${P:\\mathbb{Z}^d\\times\\mathbb{Z}^d\\rightarrow[0,1]}$ is given for every ${x,y\\in\\mathbb{Z}^d}$ by\n\n$P(x,y) = \\mathbb{P}(X_{n+1}=y\\,|\\,X_n=x) =\\frac{\\mathbf{1}_{\\{\\left|x-y\\right|_1=1\\}}}{2d} \\quad\\mbox{where}\\quad \\left|x-y\\right|_1=\\sum_{k=1}^d\\left|x_k-y_k\\right|.$\n\nIt is an infinite Markov transition matrix. It acts on a bounded function ${f:\\mathbb{Z}^d\\rightarrow\\mathbb{R}}$ as\n\n$(Pf)(x)=\\sum_{y\\in\\mathbb{Z}^d}P(x,y)f(y)=\\mathbb{E}(f(X_1)\\,|\\,X_0=x).$\n\nThe sequence ${X}$ is also a martingale for the filtration ${{(\\mathcal{F}_n)}_{n\\geq0}}$ defined by ${\\mathcal{F}_0=\\sigma(X_0)}$ and ${\\mathcal{F}_{n+1}=\\sigma(X_0,\\xi_1,\\ldots,\\xi_{n+1})}$. Indeed, by measurability and independence,\n\n$\\mathbb{E}(X_{n+1}\\,|\\,\\mathcal{F}_n) =X_n+\\mathbb{E}(\\xi_{n+1}\\,|\\,\\mathcal{F}_n) =X_n+\\mathbb{E}(\\xi_{n+1}) =X_n.$\n\nDirichlet problem. For any bounded function ${f:\\mathbb{Z}^d\\rightarrow\\mathbb{R}}$ and any ${x\\in\\mathbb{Z}^d}$,\n\n$\\mathbb{E}(f(X_{n+1})\\,|\\,X_n=x)-f(x) =(Pf)(x)-f(x) =(\\Delta f)(x) \\quad\\mbox{where}\\quad \\Delta=P-I.$\n\nHere ${I}$ is the identity operator ${I(x,y)=\\mathbf{1}_{\\{x=y\\}}}$. The operator\n\n$\\Delta =P-I$\n\nis the generator of the symmetric nearest neighbors random walk. It is a discrete Laplace operator, which computes the difference between the value at a point and the mean on its neighbors for the ${\\ell^1}$ distance:\n\n$(\\Delta f)(x) =\\left\\{\\frac{1}{2d}\\sum_{y:\\left|y-x\\right|_1=1}f(y)\\right\\}-f(x) =\\frac{1}{2d}\\sum_{y:\\left|y-x\\right|_1=1}(f(y)-f(x)).$\n\nWe say that ${f}$ is harmonic on ${A\\subset\\mathbb{Z}^d}$ when ${\\Delta f=0}$ on ${A}$, which means that on every point of ${A}$, the value of ${f}$ is equal to the mean of the values of ${f}$ on the ${2d}$ neighbors of ${x}$. The operator ${\\Delta}$ is local in the sense that ${(\\Delta f)(x)}$ depends only on the value of ${f}$ at ${x}$ and its nearest neighbors. Consequently, the value of ${\\Delta f}$ on ${A}$ depends only on the values of ${f}$ on\n\n$\\bar{A}:=A\\cup\\partial\\!A$\n\nwhere\n\n$\\partial\\!A:=\\{y\\not\\in A:\\exists x\\in A,\\left|x-y\\right|_1=1\\}$\n\nis the exterior boundary of ${A}$. In this context, the Dirichlet problem consists in finding a function which is harmonic on ${A}$ and for which the value on ${\\partial\\!A}$ is prescribed. This problem is actually a linear algebra problem for which the result below provides a probabilistic expression of the solution, based on the stopping time\n\n$\\tau_{\\partial\\!A}:=\\inf\\{n\\geq0:X_n\\in\\partial\\!A\\}.$\n\nIn Physics, the quantity ${f(x)}$ models typically the temperature at location ${x}$, while the harmonicity of ${f}$ on ${A}$ and the boundary condition on ${\\partial\\!A}$ express the thermal equilibrium and a thermostatic boundary, respectively.\n\nTheorem 1 (Dirichlet problem) Let ${\\varnothing\\neq A\\subset\\mathbb{Z}^d}$ be finite. Then for any ${x\\in A}$,\n\n$\\mathbb{P}_x(\\tau_{\\partial\\!A}<\\infty)=1.$\n\nMoreover, for any ${g:\\partial\\!A\\rightarrow\\mathbb{R}}$, the function ${f:\\bar{A}\\rightarrow\\mathbb{R}}$ defined for any ${x\\in\\bar{A}}$ by\n\n$f(x)=\\mathbb{E}_x(g(X_{\\tau_{\\partial\\!A}}))$\n\nis the unique solution of the system\n\n$\\left\\{ \\begin{array}{rl} f=g &\\mbox{on } \\partial\\!A,\\\\ \\Delta f=0&\\mbox{on } A. \\end{array} \\right.$\n\nWhen ${d=1}$ we recover the function which appears in the gambler ruin problem. Recall that the image of a Markov chain by a function which is harmonic for the generator of the chain is a martingale (discrete Itô formula !), and this explains a posteriori the probabilistic formula for the solution of the Dirichlet problem, thanks to the Doob optional stopping theorem.\n\nThe quantity ${f(x)=\\mathbb{E}_x(g(\\tau_{\\partial\\!A}))=\\sum_{y\\in\\partial\\!A}g(y)\\mathbb{P}_x(\\tau_{\\partial\\!A}=y)}$ is the mean of ${g}$ for the law ${\\mu_x}$ on ${\\partial\\!A}$ called the harmonic measure defined for any ${y\\in\\partial\\!A}$ by\n\n$\\mu_x(y):=\\mathbb{P}_x(X_{\\tau_{\\partial\\!A}}=y)$\n\nProof: The property ${\\mathbb{P}_x(\\tau_{\\partial\\!A}<\\infty)=1}$ follows from the Central Limit Theorem or by using conditioning which gives a geometric upper bound for the tail of ${\\tau_{\\partial\\!A}}$.\n\nLet us check that the proposed function ${f}$ is a solution. For any ${x\\in\\partial\\!A}$, we have ${\\tau_{\\partial\\!A}=0}$ on ${\\{X_0=x\\}}$ and thus ${f=g}$ on ${\\partial\\!A}$. Let us show now that ${\\Delta f=0}$ on ${A}$. We first reduce the problem by linearity to the case ${g=\\mathbf{1}_{\\{z\\}}}$ with ${z\\in\\partial\\!A}$. Next, we write, for any ${y\\in\\bar{A}}$,\n\n$\\begin{array}{rcl} f(y) &=&\\mathbb{P}_y(X_{\\tau_{\\partial\\!A}}=z)\\\\ &=&\\sum_{n=0}^\\infty\\mathbb{P}_y(X_n=z,\\tau_{\\partial\\!A}=n)\\\\ &=&\\mathbf{1}_{y=z}+\\mathbf{1}_{y\\in A}\\sum_{n=1}^\\infty\\sum_{x_1,\\ldots,x_{n-1}\\in A}P(y,x_1)P(x_1,x_2)\\cdots P(x_{n-1},z). \\end{array}$\n\nOn the other hand, since ${f=0}$ on ${\\partial\\!A\\setminus\\{z\\}}$ and since ${\\Delta}$ is local, we have, for any ${x\\in A}$,\n\n$(P f)(x) =\\sum_{y\\in\\mathbb{Z}^d}P(x,y)f(y) =P(x,z)f(z)+\\sum_{y\\in A}P(x,y)f(y).$\n\nAs a consequence, for any ${x\\in A}$,\n\n$\\begin{array}{rcl} (P f)(x) &=&P(x,z)f(z)+\\sum_{n=1}^\\infty\\sum_{y,x_1,\\ldots,x_{n-1}\\in A} P(x,y)P(y,x_1)\\cdots P(x_{n-1},z)\\\\ &=&P(x,z)f(z)+(f(x)-(\\mathbf{1}_{x=z}+P(x,z))), \\end{array}$\n\nwhich is equal to ${f(x)}$ since ${\\mathbf{1}_{x=z}=0}$ and ${f(z)=1}$. Hence ${Pf=f}$ on ${A}$, in other words ${\\Delta f=0}$ on ${A}$.\n\nTo establish the uniqueness of the solution, we first reduce the problem by linearity to show that ${f=0}$ is the unique solution when ${g=0}$. Next, if ${f:\\bar{A}\\rightarrow\\mathbb{R}}$ is harmonic on ${A}$, the interpretation of ${\\Delta f(x)}$ as a difference with the mean on the nearest neighbors allows to show that both the minimum and the maximum of ${f}$ on ${\\bar{A}}$ are (at least) necessarily achieved on the boundary ${\\partial\\!A}$. But since ${f}$ is vanishes on ${\\partial\\!A}$, it follows that ${f}$ vanishes on ${\\bar{A}}$. ☐\n\nDirichlet problem and Green function. Here is a generalization of Theorem 1. When ${h=0}$ we recover Theorem 1.\n\nTheorem 2 (Dirichlet problem and Green function) If ${\\varnothing\\neq A\\subset\\mathbb{Z}^d}$ is finite then for any ${g:\\partial\\!A\\rightarrow\\mathbb{R}}$ and ${h:A\\rightarrow\\mathbb{R}}$, the function ${f:\\bar{A}\\rightarrow\\mathbb{R}}$ defined for any ${x\\in\\bar{A}}$ by\n\n$f(x)=\\mathbb{E}_x\\left(g(X_{\\tau_{\\partial\\!A}})+\\sum_{n=0}^{\\tau_{\\partial\\!A}-1}h(X_n)\\right)$\n\nis the unique solution of\n\n$\\left\\{ \\begin{array}{rl} f=g&\\mbox{on } \\partial\\!A,\\\\ \\Delta f=-h&\\mbox{on } A. \\end{array} \\right.$\n\nWhen ${d=1}$ and ${h=1}$ then we recover a function which appears in the analysis of the gambler ruin problem.\n\nFor any ${x\\in A}$ we have\n\n$f(x) =\\sum_{y\\in\\partial\\!A}g(y)\\mathbb{P}_x(\\tau_{\\partial\\!A}=y)+\\sum_{y\\in A}h(y)G_A(x,y)$\n\nwhere ${G_A(x,y)}$ is the average number of passages at ${y}$ starting from ${x}$ and before escaping from ${A}$:\n\n$G_A(x,y):=\\mathbb{E}_x\\left\\{\\sum_{n=0}^{\\tau_{\\partial\\!A}-1}\\mathbf{1}_{X_n=y}\\right\\} =\\sum_{n=0}^\\infty\\mathbb{P}_x(X_n=y,n<\\tau_{\\partial\\!A}).$\n\nWe say that ${G_A}$ is the Green function of the symmetric nearest neighbors random walk on ${A}$ killed at the boundary ${\\partial\\!A}$. It is the inverse of the restriction ${-\\Delta_A}$ of ${-\\Delta}$ to functions on ${\\bar{A}}$ vanishing on ${\\partial\\!A}$:\n\n$G_A=-\\Delta_A^{-1}.$\n\nIndeed, when ${g=0}$ and ${h=\\mathbf{1}_{\\{y\\}}}$ we get ${f(x)=G_A(x,y)}$ and thus ${\\Delta_AG_A=-I_A}$.\n\nProof: Thanks to Theorem 1 it suffices by linearity to check that ${f(x)=\\mathbf{1}_{x\\in A}G_A(x,z)}$ is a solution when ${g=0}$ and ${h=\\mathbf{1}_{\\{z\\}}}$ with ${z\\in A}$. Now for any ${x\\in A}$,\n\n$\\begin{array}{rcl} f(x) &=&\\mathbf{1}_{\\{x=z\\}}+\\sum_{n=1}^\\infty\\mathbb{P}_x(X_n=z,n<\\tau_{\\partial\\!A})\\\\ &=&\\mathbf{1}_{\\{x=z\\}}+\\sum_{y:\\left|x-y\\right|_1=1}\\sum_{n=1}^\\infty\\mathbb{P}(X_n=z,n<\\tau_{\\partial\\!A}\\,|\\,X_1=y)P(x,y)\\\\ &=&\\mathbf{1}_{\\{x=z\\}}+\\sum_{u:\\left|x-y\\right|_1=1}f(y)P(x,y) \\end{array}$\n\nthanks to the Markov property. We have ${f=h+Pf}$ on ${A}$, in other words ${\\Delta f=-h}$ on ${A}$. ☐\n\nBeyond the symmetric nearest neighbors random walk. The proofs of Theorem 1 and Theorem 2 remain valid for asymmetric nearest neighbors random walks on ${\\mathbb{Z}^d}$, provided that we replace the generator ${\\Delta}$ of the symmetric nearest neighbors random walk by the generator ${L:=P-I}$, which is still a local operator: ${\\left|x-y\\right|>1\\Rightarrow L(x,y)=P(x,y)=0}$. One may even go beyond this framework by adapting the notion of boundary: ${\\{y\\not\\in A:\\exists x\\in A, L(x,y)>0\\}}$.\n\nGaussian free field. The Gaussian free field is model of Gaussian random interface for which the covariance matrix is the Green function of the discrete Laplace operator. More precisely, let ${\\varnothing\\neq A\\subset\\mathbb{Z}^d}$ be a finite set. An interface is a height function ${f:\\bar{A}\\rightarrow\\mathbb{R}}$ which associates to each site ${x\\in\\bar{A}}$ a height ${f(x)}$, also called spin in the context of Statistical Physics. For simplicity, we impose a zero boundary condition: ${f=0}$ on the boundary ${\\partial\\!A}$ of ${A}$.\n\nLet ${\\mathcal{F}_A}$ be the set of interfaces ${f}$ on ${\\bar{A}}$ vanishing at the boundary ${\\partial\\!A}$, which can be identified with ${\\mathbb{R}^A}$. The energy ${H_A(f)}$ of the interface ${f\\in\\mathcal{F}_A}$ is defined by\n\n$H_A(f)=\\frac{1}{4d}\\sum_{\\substack{\\{x,y\\}\\subset\\bar{A}\\\\\\left|x-y\\right|_1=1}}(f(x)-f(y))^2,$\n\nwhere ${f=0}$ on ${\\partial\\!A}$. The flatter is the interface, the smaller is the energy ${H_A(f)}$. Denoting ${\\left<u,v\\right>_A:=\\sum_{x\\in A}u(x)v(x)}$, we get, for any ${f\\in\\mathcal{F}_A}$,\n\n$H_A(f) =\\frac{1}{4d}\\sum_{x\\in A}\\sum_{\\substack{y\\in\\bar{A}\\\\\\left|x-y\\right|_1=1}}(f(x)-f(y))f(x) =\\frac{1}{2}\\left<-\\Delta f,f\\right>_A.$\n\nSince ${H_A(0)=0}$ and ${H_A(f)=0}$ give ${f=0}$, the quadratic form ${H_A}$ is not singular and we can define the Gaussian law ${Q_A}$ on ${\\mathcal{F}_A}$ which favors low energies:\n\n$Q_A(df)=\\frac{1}{Z_A}e^{-H_A(f)}\\,df \\quad\\mbox{where}\\quad Z_A:=\\int_{\\mathcal{F}_A}\\!e^{-H_A(f)}\\,df.$\n\nThis Gaussian law, called the Gaussian free field, is characterized by its mean ${m_A:A\\rightarrow\\mathbb{R}}$ and its covariance matrix ${C_A:A\\times A\\rightarrow\\mathbb{R}}$, given for any ${x,y\\in A}$ by\n\n$m_A(x):=\\int\\!f_x\\,Q_A(df)=0$\n\nand\n\n$C_A(x,y):=\\int\\!f_xf_y\\,Q_A(df)-m_A(x)m_A(y)=-\\Delta_A^{-1}(x,y)=G_A(x,y),$\n\nwhere ${f_x}$ denotes the coordinate map ${f_x:f\\in\\mathcal{F}_A\\mapsto f(x)\\in\\mathbb{R}}$.\n\nSimulation of the Gaussian free field. The simulation of the Gaussian free field can be done provided that we know how to compute a square root of ${G_A}$ in the sense of quadratic forms, using for instance a Cholesky factorization or diagonalization. Let us consider the square case:\n\n$A=L(0,1)^d\\cap\\mathbb{Z}^d=\\{1,\\ldots,L-1\\}^d.$\n\nThis gives ${\\bar{A}=\\{0,1,\\ldots,L\\}^d}$. In this case, one can compute the eigenvectors and eigenvalues of ${\\Delta_A}$ and deduce the ones of ${G_A=-\\Delta_A^{-1}}$. In fact, the continuous Laplace operator ${\\Delta}$ on ${[0,1]^d\\in\\mathbb{R}^n}$, with Dirichlet boundary conditions, when defined on the Sobolev space ${\\mathrm{H}_0^2([0,1]^d)}$, is a symmetric operator with eigenvectors ${\\{e_n:n\\in\\{1,2,\\ldots\\}^d\\}}$ and eigenvalues ${\\{\\lambda_n:n\\in\\{1,2,\\ldots\\}^d\\}}$ given by\n\n$e_n(t):=2^{d/2}\\prod_{i=1}^d\\sin(\\pi n_i t_i), \\quad\\mbox{and}\\quad \\lambda_n:=-\\pi^2\\left|n\\right|_2^2 =-\\pi^2(n_1^2+\\cdots+n_d^2).$\n\nSimilarly, the discrete Laplace operator ${\\Delta_A}$ with Dirichlet boundary conditions on this ${A}$ has eigenvectors ${\\{e^L_n:n\\in A\\}}$ and eigenvalues ${\\{\\lambda_n^L:n\\in A\\}}$ given for any ${k\\in A}$ by\n\n$e^L_n(k):=e_n\\left(\\frac{k}{L}\\right) \\quad\\mbox{and}\\quad \\lambda_n^L:=-2\\,\\sum_{i=1}^d \\sin^2\\left(\\frac{\\pi n_i}{2L}\\right).$\n\nThe vectors ${e_n^L}$ vanish on the boundary ${\\partial\\!A}$. It follows that for any ${x,y\\in A}$,\n\n$\\Delta_A(x,y)=\\sum_{n\\in A}\\lambda_n^L\\left<e^L_n,f_x\\right>\\left<e^L_n,f_y\\right> =\\sum_{n\\in A}\\lambda_n^L e^L_n(x)e^L_n(y)$\n\nand\n\n$G_A(x,y)=-\\sum_{n\\in A}\\left(\\lambda_n^L\\right)^{-1} e^L_n(x)e^L_n(y).$\n\nA possible matrix square root ${\\sqrt{G_A}}$ of ${G_A}$ is given for any ${x,y\\in A}$ by\n\n$\\sqrt{G_A}(x,y)=-\\sum_{n\\in A}\\left(\\lambda_n^L\\right)^{-1/2} e^L_n(x)e^L_n(y).$\n\nNow if ${Z={(Z_y)}_{y\\in A}}$ are independent and identically distributed Gaussian random variables with zero mean and unit variance, seen as a random vector of ${\\mathbb{R}^A}$, then the random vector\n\n$\\sqrt{G_A}Z = \\left(\\sum_{y\\in A}\\sqrt{G_A}(x,y)Z_y\\right)_{x\\in A} =\\left(\\sum_{y\\in A}\\sum_{n\\in A}\\left(-\\lambda_n^L\\right)^{-1/2}e_n^L(x)e_n^L(y)Z_y\\right)_{x\\in A}$\n\nis distributed according to the Gaussian free field ${Q_A}$.\n\nContinuous analogue. The scaling limit ${\\varepsilon\\mathbb{Z}^d\\rightarrow\\mathbb{R}^d}$ leads to natural continuous objects (in time and space): the simple random walk becomes the standard Brownian Motion via the Central Limit Theorem, the discrete Laplace operator becomes the Laplace second order differential operator via a Taylor formula, while the discrete Gaussian free field becomes the continuous Gaussian free field (its covariance is the Green function of the Laplace operator). The continuous object are less elementary since their require much more analytic and probabilistic machinery, but in the mean time, they provide differential calculus, a tool which is problematic in discrete settings due in particular to the lack of chain rule.\n\nNote. The probabilistic approach to the Dirichlet problem goes back at least to Shizuo Kakutani (1911 – 2004), who studied around 1944 the continuous version with Brownian Motion. This post is the English translation of an excerpt from a forthcoming book on stochastic models, written, in French, together with my old friend Florent Malrieu.\n\n#\n# load within julia using include(\"dirichlet.jl\")\n#\nnx, ny = 25, 25\nf = zeros(nx,ny)\nf[1,:], f[nx,:], f[:,1], f[:,ny] = 2, 2, -2, -2\nC = zeros(nx,ny)\nN = 5000\n#\nfor x = 1:nx\nfor y = 1:ny\nfor k = 1:N\nxx = x\nyy = y\nwhile ((xx > 1) && (xx < nx) && (yy > 1) && (yy < ny))\nb = rand(2)\nif (b < .5)\nif (b < .5)\nxx -= 1\nelse\nxx += 1\nend\nelseif (b < .5)\nyy -= 1\nelse\nyy += 1\nend\nend # while\nC[x,y] += f[xx,yy]\nend # for k\nend # for y\nend # for x\ndraw(PNG(\"dirichlet.png\", 10cm, 10cm), spy(C/N))\n\n#\n# load within julia using include(\"gff.jl\")\n#\nusing Compose, Gadfly, Cairo # for 2D graphics\n#using PyPlot # for 3D graphics (Gadfly does not provide 3D graphics)\n#\n# dimension 1\n#\nL = 500\nZ = randn(L-1)\nlambda = 2*sin(pi*[1:L-1]/(2*L)).^2\nevec = sqrt(2)*sin(pi*kron([1:L-1],[1:L-1]')/L)\nGFF1 = zeros(L-1)\nfor x = 1:L-1\nfor y = 1:L-1\nfor n = 1:L-1\nGFF1[x] += (lambda[n])^(-1/2)*evec[n,x]*evec[n,y]*Z[y]\nend\nend\nend\np = plot(x = [1:L-1], y = GFF1,\nGeom.line,\nGuide.xlabel(\"\"),\nGuide.ylabel(\"\"),\n#Guide.title(\"GFF\"),\nTheme(default_color=color(\"black\")))\ndraw(PNG(\"gff1.png\", 10cm, 10cm), p)\n#\n# dimension 2\n#\nL = 20\nZ = randn(L-1,L-1)\nAUX = kron(ones(L-1)',[1:L-1])\nlambda = 2*sin(pi*AUX/(2*L)).^2 + 2*sin(pi*AUX'/(2*L)).^2\nevec = zeros(L-1,L-1,L-1,L-1) # n1,n2,k1,k2 : e_n^L(k)\nfor n1 = 1:L-1\nfor n2 = 1:L-1\nAUX = kron(ones(L-1)',[1:L-1])\nevec[n1,n2,:,:] = reshape(2*sin(pi*n1*AUX/L).*sin(pi*n2*AUX'/L),(1,1,L-1,L-1))\nend\nend\nGFF2 = zeros(L-1,L-1)\nfor x1 = 1:L-1\nfor x2 = 1:L-1\nfor y1 = 1:L-1\nfor y2 = 1:L-1\nfor n1 = 1:L-1\nfor n2 = 1:L-1\naux = (lambda[n1,n2])^(-1/2)\naux *= evec[n1,n2,x1,x2]\naux *= evec[n1,n2,y1,y2]\naux *= Z[y1,y2]\nGFF2[x1,x2] += aux\nend\nend\nend\nend\nend\nend\n# graphics with PyPlot\n#surf(GFF2)\n#savefig(\"gff2.svg\")"
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https://pakaccountants.com/excel-formulas/value/ | [
"## VALUE\n\nChanges given text string (that should have been a number) to a number.\n\nSyntax of Excel Value formula:\n=VALUE(text)\n\nIn words:\n=Change to numbers (this text)\n\n#### Examples of Excel Value formula\n\n=VALUE(“21 December 2015”) will return 42359. Once formatted in date format it will be 12/21/2015 and can be used in formula now as normal."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8415182,"math_prob":0.857497,"size":374,"snap":"2019-26-2019-30","text_gpt3_token_len":96,"char_repetition_ratio":0.14054054,"word_repetition_ratio":0.3,"special_character_ratio":0.28074867,"punctuation_ratio":0.08108108,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95937425,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-23T09:19:51Z\",\"WARC-Record-ID\":\"<urn:uuid:20692c0b-bc66-4d2d-9ad9-89bff00f71ad>\",\"Content-Length\":\"70491\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e820a3cc-d615-42a4-96af-290ee49ec409>\",\"WARC-Concurrent-To\":\"<urn:uuid:98f31662-4861-454d-a88d-d307944dd124>\",\"WARC-IP-Address\":\"199.168.186.34\",\"WARC-Target-URI\":\"https://pakaccountants.com/excel-formulas/value/\",\"WARC-Payload-Digest\":\"sha1:AB6AI5ID4PPTEHDCVQ2C44H543JKUQPL\",\"WARC-Block-Digest\":\"sha1:R65QOY7DTB7BKMQHWBVVVRC7YU6WUCJV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195529175.83_warc_CC-MAIN-20190723085031-20190723111031-00157.warc.gz\"}"} |
https://calculatorsonline.org/rounding-numbers/what-is-2438-rounded-to-the-nearest-whole | [
"# 2438 rounded to the nearest whole\n\nHere you will see step by step solution to round of the 2438 to the nearest whole. What is 2438 rounded to the nearest whole? 2438 rounded to the nearest whole is 2438, check the explanation that how to rounding the 2438 to nearest whole.\n\n## Answer: Rounded 2438 to nearest whole is\n\n= 2438\n\n### How to round 2438 to the nearest whole?\n\nTo round of the number 2438 simply find the digit at whole place, then look to the right digit next to the whole place, if this number is greater than or equal to 5 (5, 6, 7, 8, 9) round up or if number is less then 5 (0, 1, 2, 3, 4) round down whole(8) number. Remove all the digits to the right of the rounding digit.\n\n#### Solution for decimal number 2438 to the nearest whole\n\nGiven number is => 2438\n\n• Number at whole place is = 8\n• 2438 = 8\n• Now we need to find the digit at the right side of whole place = 0\n• 0 is smaller than 5, whole place digit 8 we don't need to change it. It will remain unchanged.\n• In 2438, after rounding digit 8, remove all the numbers in the right side, then rewrite the numbers.\n• Final Conclusion: 2438\n\nHence, the 2438 to the nearest whole is 2438."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8715178,"math_prob":0.9930762,"size":1136,"snap":"2023-14-2023-23","text_gpt3_token_len":324,"char_repetition_ratio":0.23233216,"word_repetition_ratio":0.04888889,"special_character_ratio":0.3177817,"punctuation_ratio":0.1124498,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96370184,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-30T00:37:40Z\",\"WARC-Record-ID\":\"<urn:uuid:77a42e27-d15e-41de-bc50-e599505656a8>\",\"Content-Length\":\"18151\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:af2ca557-9a19-43d7-8dee-766a7dcd4326>\",\"WARC-Concurrent-To\":\"<urn:uuid:bd00bff2-5ac4-4145-a5b1-6ed845ce34a1>\",\"WARC-IP-Address\":\"104.21.85.191\",\"WARC-Target-URI\":\"https://calculatorsonline.org/rounding-numbers/what-is-2438-rounded-to-the-nearest-whole\",\"WARC-Payload-Digest\":\"sha1:PRL3TNQ5V5G262ZQ5ARGIQFZMFWB2QO6\",\"WARC-Block-Digest\":\"sha1:PBBRSPTOBGYC2FIZSRWA3QR452O2IDJK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224644915.48_warc_CC-MAIN-20230530000715-20230530030715-00717.warc.gz\"}"} |
https://kr.mathworks.com/matlabcentral/cody/problems/1058-subtract-two-positive-numbers/solutions/1239109 | [
"Cody\n\n# Problem 1058. Subtract two positive numbers\n\nSolution 1239109\n\nSubmitted on 25 Jul 2017 by Augusto Mazzei\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\na = '0205'; b = '8120'; y_correct = 7915; A = fread(fopen('yourSubtract.m')); assert(~any(regexpi(char(A)','str2num|str2double'))); assert(isequal(yourSubtract(a,b),y_correct))\n\nans = 7915\n\n2 Pass\na = '12589'; b = '78956'; y_correct = 66367; assert(isequal(yourSubtract(a,b),y_correct))\n\nans = 66367"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5124457,"math_prob":0.92536885,"size":624,"snap":"2019-51-2020-05","text_gpt3_token_len":191,"char_repetition_ratio":0.12903225,"word_repetition_ratio":0.0,"special_character_ratio":0.36698717,"punctuation_ratio":0.14529915,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99620545,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-24T19:42:02Z\",\"WARC-Record-ID\":\"<urn:uuid:8e5c8dd7-1ead-44e9-bba2-88be6dbc4c86>\",\"Content-Length\":\"71754\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4a011897-d6b8-41ba-a5b3-bda3a1228d01>\",\"WARC-Concurrent-To\":\"<urn:uuid:ae664560-3659-4570-90ba-b744bf6a782d>\",\"WARC-IP-Address\":\"104.110.193.39\",\"WARC-Target-URI\":\"https://kr.mathworks.com/matlabcentral/cody/problems/1058-subtract-two-positive-numbers/solutions/1239109\",\"WARC-Payload-Digest\":\"sha1:GFHHDXW53JUX3XLTPHYFOKZ53OXNM5E5\",\"WARC-Block-Digest\":\"sha1:6Y5YNK6ZR5RJIN36XOQEHN47OEBLKNLP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250625097.75_warc_CC-MAIN-20200124191133-20200124220133-00305.warc.gz\"}"} |
https://www.zhygear.com/dynamic-model-of-planetary-gear-train-of-shearer-cutting-part/ | [
"# Dynamic model of planetary gear train of shearer cutting part\n\nAccording to Newton’s second law, the dynamic equation of the system is established as follows\n\nWhere I (I = R, s, 1, 2,…) , n) is the moment of inertia of component I around its center of mass; MP is the mass of a single planetary gear, KSN is the time-varying meshing stiffness of the sun gear and the nth planetary gear; Krn is the time-varying meshing stiffness of the inner ring gear and the nth planetary gear; TC and TS are the external torques of the planet carrier and the sun gear respectively, and the positive and negative values are judged by the right hand rule.\n\nThe formula can be written in matrix form\n\nThe matrix and vector in the formula are 3 + N dimensions, and the specific meaning of each symbol is as follows:\n\n1) Q is the generalized coordinate vector of the system\n\n2) M is a mass matrix and a diagonal matrix\n\n3) C is the damping matrix, using proportional damping\n\nWhere β is the proportional damping coefficient and is a constant.\n\n4) KB is the support stiffness matrix and a diagonal matrix\n\n5) Km is the meshing stiffness matrix, in which the non-zero elements are\n\n6) T is the external excitation torque vector\n\nIt should be noted that although the displacement projection of the sun gear and the inner gear ring relative to the nth planetary gear is related to the rotation direction of the sun gear (rot value), the meshing stiffness matrix of the planetary gear train does not change with the change of the rotation direction of the sun gear.",
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"https://www.zhygear.com/wp-content/themes/luminescence-lite/images/post-shadow.png",
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https://eng_rus_building.academic.ru/2052/direction_cosine_matrix | [
"# direction cosine matrix\n\n\ndirection cosine matrix\nn\nматрица направляющих косинусов\n\nАнгло-русский строительный словарь. — М.: Русский Язык. . 1995.\n\n### Смотреть что такое \"direction cosine matrix\" в других словарях:\n\n• Matrix - получить на Академике рабочий купон на скидку Летуаль или выгодно matrix купить с бесплатной доставкой на распродаже в Летуаль\n\n• Orthogonal matrix — In linear algebra, an orthogonal matrix (less commonly called orthonormal matrix), is a square matrix with real entries whose columns and rows are orthogonal unit vectors (i.e., orthonormal vectors). Equivalently, a matrix Q is orthogonal if… … Wikipedia\n\n• Lambert's cosine law — See also: Lambertian reflectance In optics, Lambert s cosine law says that the radiant intensity observed from a Lambertian surface or a Lambertian radiator is directly proportional to the cosine of the angle θ between the observer s line of… … Wikipedia\n\n• Euclidean vector — This article is about the vectors mainly used in physics and engineering to represent directed quantities. For mathematical vectors in general, see Vector (mathematics and physics). For other uses, see vector. Illustration of a vector … Wikipedia\n\n• Rotation representation (mathematics) — In geometry a rotation representation expresses the orientation of an object (or coordinate frame) relative to a coordinate reference frame. This concept extends to classical mechanics where rotational (or angular) kinematics is the science of… … Wikipedia\n\n• Flight dynamics — is the science of air and space vehicle orientation and control in three dimensions. The three critical flight dynamics parameters are the angles of rotation in three dimensions about the vehicle s center of mass, known as pitch , roll and yaw… … Wikipedia\n\n• Rigid body — Classical mechanics Newton s Second Law History of classical mechanics … Wikipedia\n\n• Gimbal — A gimbal is a pivoted support that allows the rotation of an object about a single axis. A set of two gimbals, one mounted on the other with pivot axes orthogonal, may be used to allow an object mounted on the innermost gimbal to remain vertical… … Wikipedia\n\n• DCM — may refer to: Technology: Data Center Manager Deep chlorophyll maximum, subsurface maximum in the concentration of chlorophyll Dichloromethane, a common solvent in organic chemistry Digital Clock Manager, programmed clock signal transformation… … Wikipedia\n\n• DCM — 1. Design Compliance Matrix Contributor: MSFC 2. Direction Cosine Matrix Contributor: MSFC 3. Display and Control Module (Space Shuttle) Contributor: CASI 4. Display Control Module Contributor: MSFC 5. Division Computing Management… … NASA Acronyms\n\n• DCM — • Digital Circuit Multiplication • Dark Cold Matter Dunkelmaterie Astrophysik • Digital Carrier Module • DECOM Control Memory NASA • Display & Control Module NASA • Direction Cosine Matrix NASA … Acronyms\n\n• DCM — Digital Circuit Multiplication Dark Cold Matter Dunkelmaterie {Astrophysik} Digital Carrier Module DECOM Control Memory ( > NASA Acronym List ) Display & Control Module ( > NASA Acronym List ) Direction Cosine Matrix … Acronyms von A bis Z"
]
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http://www.vpuzzles.com/find-the-missing-number-math-puzzle-game-answer/ | [
"# find the missing number math puzzle game answer\n\nTo answer to this math puzzle game is 105 as explained below the picture\n\n#### Left Block Answer Explanation\n\nIn each block Step 1 : Multiply Opposite Diagonal Values and add them together as (6 x 3) + (6 x 3) = 36 Step 2: Add to this 36 number we get in step 1 to top middle digit so 36 + 6 = 42 Step 3 : Subtract from this 42 number we get in step 2 by middle digit which is 3 so 42 – 3 = 39\n\n#### Right Block Answer Explanation\n\nStep 1 : Multiply Opposite Diagonal Values and add them together as (4 x 2) + (4 x 2) = 16 Step 2: Add to this 16 number we get in step 1 to top middle digit so 16 + 4 = 20 Step 3 : Subtract from this 20 number we get in step 2 by middle digit which is 2 so 20 – 2 = 18\n\n#### Top Block Answer Explanation\n\nStep 1 : Multiply Opposite Diagonal Values and add them together as (10 x 5) + (10 x 5) = 100 Step 2: Add to this 100 number we get in step 1 to top middle digit so 100 + 10 = 110 Step 3 : Subtract from this 110 number we get in step 2 by middle digit which is 5 so 110 – 5 = 105\n\nSubscribe\nNotify of",
null,
"1 Comment\nInline Feedbacks",
null,
"Jeandre\n3 years ago\n\nI used two steps….\n\nAll i needed was the top value and bottom value once.\n\nMultiply the top value with itself and add the bottom value by that answer.\nExample2:\n\n4×4=16\n16+2=18\n\n6×6=36\n36+3=39\n\n10×10=100\n100+5=105"
]
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"http://1.gravatar.com/avatar/",
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"http://2.gravatar.com/avatar/53124557e9da72e77fc57bbf6d829bd5",
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http://math.portonvictor.org/2013/06/29/funcoids-history/ | [
"# The history of discovery of funcoids\n\nIn my book I introduce funcoids as a generalization of proximity spaces. This is the most natural way to introduce funcoids, but it was not the actual way I’ve discovered them.\n\nThe first thing discovered equivalent to funcoids was a function",
null,
"$\\Delta$ (generalizing a topological space) which I defined to get a set as argument and return a filter, subject to several equalities, after the removal of superfluous equalities it has become:\n\n•",
null,
"$\\Delta\\varnothing = \\varnothing$;\n•",
null,
"$\\Delta(I\\cup J) = \\Delta I\\cup\\Delta J$.\n\n(Here",
null,
"$\\cup$ and",
null,
"$\\cap$ may mean the join and meet on the lattice of filters ordered reverse to set-theoretic inclusion of filters,",
null,
"$\\varnothing$ also denotes the improper filter.)\n\nThen somehow (I don’t remember how exactly) I managed that this",
null,
"$\\Delta$ can be reverted (like modern notion of reverse funcoid).\n\nSomehow I have managed to define composition of funcoids.\n\nAfterward I defined funcoids in the following cumbersome way:\n\nFuncoids is a set of objects",
null,
"$f$ such that it is unambiguously defined by the values",
null,
"$\\langle f\\rangle$ and",
null,
"$\\langle f^{-1}\\rangle$ of functions from filters to filters such that",
null,
"$\\mathcal{Y}\\cap\\langle f\\rangle\\mathcal{X}\\ne\\varnothing \\Leftrightarrow \\mathcal{X}\\cap\\langle f^{-1}\\rangle\\mathcal{Y}\\ne\\varnothing$.\n\nLater I have come to the simple idea that instead of this cumbersome definition can be replaced with simply defining a funcoid as a pair of functions from filters to filters.\n\nAn other anachronism: Initially I considered funcoids as a special case of reloids conforming to the above formula. (I don’t remember whether I had an exact definition for a reloid to be a funcoid.) Later funcoids have become an independent kind of objects, rather than a special case of reloids.\n\nFinally, I’ve replaced the funcoid on a single “universal” set with funcoids between two sets",
null,
"$A$ and",
null,
"$B$, so forming a category of funcoids.\n\nI have not told the history of filter objects (where principal filters were equated with corresponding sets, such as",
null,
"$\\varnothing$ was also the improper filter) and then removing this notion in regard of simple reverse-ordered lattice of filters with order, meets, and joins denoted differently than set-theoretic subset, intersection, and union, not to make mess between these.\n\nFinally: If I would know the notion of proximity spaces at the time when I wrote the function",
null,
"$\\Delta$ would it prevent me to discover funcoids (counting that proximity spaces is what I need and thus not continuing the research further)?\n\n1.",
null,
"A.K. Devaraj says:\n1.",
null,
"porton says:"
]
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"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
null,
"http://2.gravatar.com/avatar/b46210678da98727bbbbc7ac094409e7",
null,
"http://0.gravatar.com/avatar/fa49a1c90a6af7544d764c5b22ff780d",
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https://www.geoteknikk.com/2015/08/introduction-to-nonlinear-finite.html | [
"# Introduction to Nonlinear Finite Element Analysis\n\nNam-Ho Kim ... 430 pages - Publisher: Springer; (November, 2014) ... Language: English - ISBN-10: 1441917454 - ISBN-13: 978-1441917454 ...\n\nThis book introduces the key concepts of nonlinear finite element analysis procedures. The book explains the fundamental theories of the field and provides instructions on how to apply the concepts to solving practical engineering problems. Instead of covering many nonlinear problems, the book focuses on three representative problems: nonlinear elasticity, elastoplasticity, and contact problems. The book is written independent of any particular software, but tutorials and examples using four commercial programs are included as appendices: ANSYS, NASTRAN, ABAQUS, and MATLAB. In particular, the MATLAB program includes all source codes so that students can develop their own material models, or different algorithms. This book also: Presents clear explanations of nonlinear finite element analysis for elasticity, elastoplasticity, and contact problems. + Includes many informative examples of nonlinear analyses so that students can clearly understand the nonlinear theory. + Offers practical applications of FEM to engineering analysis, providing a balance between theory and practice.\n\nIntroduction to Nonlinear Finite Element Analysis, Nam-Ho Kim"
]
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https://salve.digication.com/meganwalton/Introduction_Background/published | [
"DRAFT: This module has unpublished changes.\n\nLab 1: Determining the Empirical Formula of a Compound\n\nPurpose: The purpose of this lab is to determine the empirical formula of a compound. In order to do this, the empirical formula, which is the simplest whole number ratio of elements in a compound is found by looking at the stoichiometric ratio of atoms in a specific compound.\n\nQuestion: What is the stoichiometric composition of an ionic compound?\n\nBackground:\n\nThe empirical formula of a compound is the simplest whole number ratio of the elements in the compound. The stoichiometric ratio of the different types of atoms in a compound is given by that ratio of whole numbers. One mole of a substance has the number of moles of different types of atoms in the same ratio as the stoichiometic ratio.\n\nFor example, the formula reflects a 2:1 stoichiometric ratio between hydrogen and oxygen atoms. Also, one mole of molecules has a 2:1 molar ratio between the total number of hydrogen and oxygen atoms. That is, there are 2 moles of hydrogen and 1 mole of oxygen atoms in 1 mole of molecule.\n\nThe empirical formula can be determined experimentally if a compound can be synthesized from its elements. This process requires three steps: determining the mass of each element in the compound; calculating the number of moles of each element in the sample; and expressing the molar ratio of each element as the smallest whole number.\n\nMolecular oxygen is very reactive, whether in pure form or in a mixture such as air. The most abundant component of air, nitrogen, is relatively unreactive. An element reacting with oxygen forms an oxide. For example, magnesium and oxygen form magnesium oxide:",
null,
"The reaction of nitrogen with an element forms a nitride. Because oxygen has a greater reactivity than nitrogen, the oxide is more likely to form.\n\nIn this experiment, you will burn magnesium (Mg) in air to form magnesium oxide and magnesium nitride. The magnesium nitride will be converted to magnesium hydroxide and ammonia by adding water. Upon heating, the magnesium hydroxide will be converted to magnesium oxide with the release of water vapor.",
null,
"",
null,
"",
null,
"DRAFT: This module has unpublished changes."
]
| [
null,
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https://lammps.sandia.gov/doc/fix_gravity.html | [
"# fix gravity/kk command\n\n## Syntax\n\nfix ID group gravity magnitude style args\n\n• ID, group are documented in fix command\n\n• gravity = style name of this fix command\n\n• magnitude = size of acceleration (force/mass units)\n\n• magnitude can be a variable (see below)\n\n• style = chute or spherical or gradient or vector\n\nchute args = angle\nangle = angle in +x away from -z or -y axis in 3d/2d (in degrees)\nangle can be a variable (see below)\nspherical args = phi theta\nphi = azimuthal angle from +x axis (in degrees)\ntheta = angle from +z or +y axis in 3d/2d (in degrees)\nphi or theta can be a variable (see below)\nvector args = x y z\nx y z = vector direction to apply the acceleration\nx or y or z can be a variable (see below)\n\n## Examples\n\nfix 1 all gravity 1.0 chute 24.0\nfix 1 all gravity v_increase chute 24.0\nfix 1 all gravity 1.0 spherical 0.0 -180.0\nfix 1 all gravity 10.0 spherical v_phi v_theta\nfix 1 all gravity 100.0 vector 1 1 0\n\n\n## Description\n\nImpose an additional acceleration on each particle in the group. This fix is typically used with granular systems to include a “gravity” term acting on the macroscopic particles. More generally, it can represent any kind of driving field, e.g. a pressure gradient inducing a Poiseuille flow in a fluid. Note that this fix operates differently than the fix addforce command. The addforce fix adds the same force to each atom, independent of its mass. This command imparts the same acceleration to each atom (force/mass).\n\nThe magnitude of the acceleration is specified in force/mass units. For granular systems (LJ units) this is typically 1.0. See the units command for details.\n\nStyle chute is typically used for simulations of chute flow where the specified angle is the chute angle, with flow occurring in the +x direction. For 3d systems, the tilt is away from the z axis; for 2d systems, the tilt is away from the y axis.\n\nStyle spherical allows an arbitrary 3d direction to be specified for the acceleration vector. Phi and theta are defined in the usual spherical coordinates. Thus for acceleration acting in the -z direction, theta would be 180.0 (or -180.0). Theta = 90.0 and phi = -90.0 would mean acceleration acts in the -y direction. For 2d systems, phi is ignored and theta is an angle in the xy plane where theta = 0.0 is the y-axis.\n\nStyle vector imposes an acceleration in the vector direction given by (x,y,z). Only the direction of the vector is important; it’s length is ignored. For 2d systems, the z component is ignored.\n\nAny of the quantities magnitude, angle, phi, theta, x, y, z which define the gravitational magnitude and direction, can be specified as an equal-style variable. If the value is a variable, it should be specified as v_name, where name is the variable name. In this case, the variable will be evaluated each timestep, and its value used to determine the quantity. You should insure that the variable calculates a result in the appropriate units, e.g. force/mass or degrees.\n\nEqual-style variables can specify formulas with various mathematical functions, and include thermo_style command keywords for the simulation box parameters and timestep and elapsed time. Thus it is easy to specify a time-dependent gravitational field.\n\nStyles with a gpu, intel, kk, omp, or opt suffix are functionally the same as the corresponding style without the suffix. They have been optimized to run faster, depending on your available hardware, as discussed on the Speed packages doc page. The accelerated styles take the same arguments and should produce the same results, except for round-off and precision issues.\n\nThese accelerated styles are part of the GPU, USER-INTEL, KOKKOS, USER-OMP and OPT packages, respectively. They are only enabled if LAMMPS was built with those packages. See the Build package doc page for more info.\n\nYou can specify the accelerated styles explicitly in your input script by including their suffix, or you can use the -suffix command-line switch when you invoke LAMMPS, or you can use the suffix command in your input script.\n\nSee the Speed packages doc page for more instructions on how to use the accelerated styles effectively.\n\nRestart, fix_modify, output, run start/stop, minimize info:"
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https://mathoverflow.net/questions/143144/intuition-behind-the-ricci-flow/149341 | [
"# Intuition behind the ricci flow\n\nI hope you don't shoot me for this question. I try to understand among other things the Ricci flow. However I have no idea of the intuition behind the definition. So my questions is:\n\nWhat is the intuition behind the Ricci flow?\n\nI would appreciate any illustrative example.\n\n• Hamilton states in his original paper that the Ricci tensor is analogous to a Laplacian of the metric, and therefore has similarities to a parabolic PDE. If one wants to find a flow on metrics, this is the simplest symmetric 2-tensor to write down in terms of the curvature. One can modify it by other terms, like scalar curvature times the metric, but this is not weakly parabolic, or the traceless Ricci tensor, but this just rescales the Ricci flow. In fact, Hamilton said when he originally defined the flow, he didn't know the correct sign until he tried proving local existence. – Ian Agol Sep 25 '13 at 16:11\n• I wrote a non-technical article on Ricci flow for the Princeton Companion to Mathematics : terrytao.files.wordpress.com/2008/03/ricci1.pdf – Terry Tao Sep 26 '13 at 17:13\n\nSee\n\nJ. Hyam Rubinstein and Robert Sinclair. \"Visualizing Ricci Flow of Manifolds of Revolution\", Experimental Mathematics v. 14 n. 3, pp. 257–384. (Journal link)",
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"(Image from that paper, via Wikipedia)\n\n• I was about to post the same thing, I think this is very illustrative. – Abhimanyu Pallavi Sudhir Nov 19 '13 at 16:05\n• The 2-dimensional case is special for the Ricci flow. For $n\\geq 3$, we shall have neck-pinch on the thin parts. To have a correct picture in mind is necessary, otherwise we will be confused that \"why cigar cannot occur on $(S^2,g(t))$ while Bryant soliton do occur on $(S^3,g(t))$.\" – Chih-Wei Chen Apr 12 '14 at 2:53\n\nI've combined two answers into one (out of order).\n\nSecond answer: some historical intuition. This is only a partial answer. Assume that you have stumbled upon the equation $\\frac{\\partial}{\\partial t}g=-2\\operatorname{Ric}$ and that you are interested in whether you can use it to deform metrics to better metrics on closed manifolds.\n\nPDE intuition. The first question is that of short time existence given a $C^{\\infty}$ initial metric $g_{0}$. So one linearizes the operator $g\\mapsto-2\\operatorname{Ric}_{g}$ and computes its symbol and finds that it is weakly elliptic. In fact $\\operatorname{Ric}_{\\varphi^{\\ast}g} =\\varphi^{\\ast}\\operatorname{Ric}_{g}$, accounts for the kernel of the symbol. By breaking the diffeomorphism invariance of the Ricci flow in the right way, DeTurck simplified Hamilton's proof of short time existence by obtaining an equivalent equation for the metric which linearizes to a heat-type equation.\n\nTo see if the metric gets better, one computes the evolutions of geometric quantities associated to $g(t)$. If $Q=Q[g]$ is such a quantity, assuming the variation $\\frac{\\partial}{\\partial t}g=-2\\operatorname{Ric}$ of the metric, one computes the corresponding variation $\\frac{\\partial Q}{\\partial t}$. One immediately sees heat-type equations everywhere. For example, the scalar curvature evolves by $\\frac{\\partial R}{\\partial t}=\\Delta R+2|\\operatorname{Ric}|^{2}$. Since the global method of the maximum principle relies on local calculations, it applies to closed manifolds. So $R_{\\min }(t)=\\min_{x\\in M}R(x,t)$ is nondecreasing. One finds other examples of Ricci flow preferring positive curvature over negative curvature. Basically, any polynomial of the curvature and its covariant derivatives, whether it be a function or more generally a tensor, satisfies a heat-type equation. E.g., derivative of curvature estimates follow from the maximum principle.\n\nHaving obtained some control of the metric as it evolves, one then aims to prove convergence. In dimension two, this is always possible after rescaling to normalize the volume to be constant. Generally, an Einstein metric shrinks, is stationary, or expands according to whether $R$ is positive, zero, or negative, respectively.\n\nQuantities satisfying heat-type equations. The full curvature tensor $\\operatorname{Rm}$ satisfies an equation of the form $\\frac{\\partial }{\\partial t}\\operatorname{Rm}=\\Delta\\operatorname{Rm}+q(\\operatorname{Rm})$, where $q$ is a quadratic polynomial. Since $\\operatorname{Rm}$ is a symmetric bilinear form on the vector space $\\wedge^{2}T_{x}^{\\ast}M$ at each point $x$, we have the notion of nonnegativity of $\\operatorname{Rm}$. Since $q(\\operatorname{Rm})$ satisfies a property sufficient for the maximum principle for systems to be applied, $\\operatorname{Rm}\\geq0$ is preserved under the Ricci flow. Generally, we can analyze the behavior of $\\operatorname{Rm}$ by the maximum principle under various hypotheses.\n\nGeometric application. In particular, when $n=3$ and $\\operatorname{Ric} _{g_{0}}>0$, we have $\\pi_{1}(M)=0$ and hence the universal cover $\\tilde{M}$ is a homotopy $3$-sphere. Encouraged by this, Hamilton proved that the solution to the normalized Ricci flow exists for all time and converges to a constant positive sectional curvature metric; thus $M$ is diffeomorphic to a spherical space form. The main gonzo estimate is $\\frac{|\\operatorname{Ric}% -\\frac{R}{3}g|^{2}}{R^{2}}\\leq CR^{-\\delta}$ for some $C$ and $\\delta>0$. Intuitively, we expect $R\\rightarrow\\infty$ and hence $\\operatorname{Ric} -\\frac{R}{3}g\\rightarrow0$.\n\nSingularities. Schoen and Yau proved that if an orientable $M^{3}$ admits a metric with $R>0$, then it is a connected sum of quotients of homotopy $3$-spheres and $S^{2}\\times S^{1}$'s.\\ Yau proposed to Hamilton that in this case Ricci flow should be able to produce surgeries to obtain a connected sum of spherical space forms and $S^{2}\\times S^{1}$'s. One first sees, that by the strong maximum principle, the universal cover of singularities of the Ricci flow often split as products of $\\mathbb{R}$ with a solution on a surface. This is a motivation to study the Ricci flow on surfaces, to rule out the formation of the cigar soliton.\n\nInspired by his corresponding results for the curve shortening flow and the Ricci flow on surfaces, Hamilton proved that the Li-Yau differential Harnack method extends to the Ricci flow assuming $\\operatorname{Rm}\\geq0$. Since $3$-dimensional singularity models have $\\operatorname{Rm}\\geq0$, Hamilton was able to classify certain singularities as steady gradient Ricci solitons and cylinders. Provided the Little Loop Lemma is true, or more aptly, no local collapsing is true, for finite time singular solutions one obtains round cylinder $S^{2}\\times\\mathbb{R}$ limits unless one is one a spherical space form. At this point, one can begin to believe that Ricci flow does indeed perform the surgeries that Yau proposed.\n\nFirst answer: Ricci flow as a heat-type equation. Remark about Hamilton's statement: \"The Ricci flow is the heat equation for metrics\". (The following is a well known calculation.) The Ricci tensor is given in local coordinates by \\begin{align*} -2R_{jk} & =-2\\left( \\partial_{q}\\Gamma_{jk}^{q}-\\partial_{j}\\Gamma_{qk}% ^{q}+\\Gamma_{jk}^{p}\\Gamma_{qp}^{q}-\\Gamma_{qk}^{p}\\Gamma_{jp}^{q}\\right) \\\\ & =-g^{qr}\\partial_{q}\\left( \\partial_{j}g_{kr}+\\partial_{k}g_{jr}% -\\partial_{r}g_{jk}\\right) +g^{qr}\\partial_{j}\\left( \\partial_{q}% g_{kr}+\\partial_{k}g_{qr}-\\partial_{r}g_{qk}\\right) \\\\ & \\quad\\;+\\left( g^{-1}\\right) ^{\\ast2}\\ast\\left( \\partial g\\right) ^{\\ast2}\\\\ & =\\Delta\\left( g_{jk}\\right) -g^{qr}\\left( \\partial_{q}\\partial_{j}% g_{kr}+\\partial_{q}\\partial_{k}g_{jr}-\\partial_{j}\\partial_{k}g_{qr}\\right) +\\left( g^{-1}\\right) ^{\\ast2}\\ast\\left( \\partial g\\right) ^{\\ast2}\\\\ & =\\Delta\\left( g_{jk}\\right) -g_{k\\ell}\\partial_{j}\\left( g^{qr}% \\Gamma_{qr}^{\\ell}\\right) -g_{j\\ell}\\partial_{k}\\left( g^{qr}\\Gamma _{qr}^{\\ell}\\right) +\\left( g^{-1}\\right) ^{\\ast2}\\ast\\left( \\partial g\\right) ^{\\ast2}. \\end{align*} In harmonic coordinates $\\{x^{i}\\},$ $0=g^{ij}\\Gamma_{ij}^{k},$ so then $-2R_{jk}=\\Delta\\left( g_{jk}\\right) +Q\\left( g^{-1},\\partial g\\right) ,$ where $Q$ is quadratic in both arguments. (From line to line, various terms are absorbed in the lower order quadratic term.)\n\nIn normal coordinates $\\{x^{i}\\}$ centered at $p$ we have $g_{ij}\\left( x\\right) =\\delta_{ij}-\\frac{1}{3}R_{i\\ell mj}\\left( p\\right) x^{\\ell}% x^{m}+O\\left( r^{3}\\right)$, where $r=d\\left( x,p\\right) =(\\sum_{i}% (x^{i})^{2})^{1/2}$. Then $\\Delta\\left( g_{ij}\\right) \\left( p\\right) =-\\frac{2}{3}R_{ij}\\left( p\\right)$. Note that $\\partial_{i}g_{jk}\\left( p\\right) =0$.\n\nHamilton likes to joke that when he first wrote down the Ricci flow equation, he wrote: $\\frac{\\partial}{\\partial t}g_{ij} = 2 R_{ij}$, having a preference for positivity over negativity.\n\nDecember 13, 2013. Answer to Qfwfq's question. $\\partial g$ denotes a nonspecific factor of the form $\\partial_{i}g_{jk}$. $\\ast$ denotes a product, possibly together with contractions (summing over a pair of repeated indices, one upper and one lower). For example, \\begin{align*} 2\\partial_{i}\\Gamma_{jk}^{\\ell} & =\\partial_{i}(g^{\\ell m}(\\partial_{j} g_{km}+\\partial_{k}g_{jm}-\\partial_{m}g_{jk}))\\\\ & =g^{\\ell m}(\\partial_{i}\\partial_{j}g_{km}+\\partial_{i}\\partial_{k} g_{jm}-\\partial_{i}\\partial_{m}g_{jk})\\\\ & \\quad-g^{\\ell p}g^{qm}\\partial_{i}g_{pq}(\\partial_{j}g_{km}+\\partial _{k}g_{jm}-\\partial_{m}g_{jk})\\\\ & =g^{\\ell m}(\\partial_{i}\\partial_{j}g_{km}+\\partial_{i}\\partial_{k} g_{jm}-\\partial_{i}\\partial_{m}g_{jk})+(g^{-1})^{\\ast2}\\ast(\\partial g)^{\\ast2}. \\end{align*}\n\n• What is the \" $∗$ \" that appears in the above calculations? (I guess it's the symmetric product of symmetric tensors, right?) – Qfwfq Dec 14 '13 at 0:41\n• And, what is $\\partial g$? (sorry for spamming with notational questions) – Qfwfq Dec 14 '13 at 0:44\n\nThe original idea behind the Ricci flow equation, namely $$\\frac{\\partial g}{\\partial t}=-2Ric (g)$$ was to deform 'rough' or 'uneven' metrics to try to obtain more uniform ones. Heuristically, we can see that regions where $Ric>0$ tend to contract under the flow, thus making curvature even more positive (think for example of a sphere with the usual metric; this has constant positive curvature, so the metric will contract uniformly. In this case the sphere shrinks, thus making curvature even larger, until it finally becomes a point). In contrast, regions with $Ric>0$ tend to expand, thereby reducing the curvature. As is obvious from the case of the sphere, in general the volume $V(t)=\\int d\\mu(g)$ changes with time. We can renormalise the flow to make volume constant in time (incidentally, this renormalized Ricci flow was the original flow that Hamilton used in his 1982 JDG paper). This yields, at least in principle, a method to try to deform a given metric into a 'more uniform one' (i.e. one with constant sectional curvature, Ricci flat, Einstein, etc.). In practice this is not always possible because sometimes certain regions of the manifold shrink to fast and the curvature goes to infinity in finite time. This is called a singularity. These singularities in the flow are important because they give us information about the underlying topology and the geometry of the initial manifold.\n\nElaborating a little bit more on Prof. Agol's comment, another way to get some feel for the Ricci flow is to think of it as a heat equation. In normal coordinates, the Ricci tensor can be expressed as $-2R_{ij}=\\Delta (g_{ij})+2Q(g,\\partial g)$, (ie. the second term only dependes on $g$ and its first derivative). On these coordinates the Ricci flow looks like $$\\frac{\\partial g_{ij}}{\\partial t}=\\Delta (g_{ij})-2Q_{ij},$$ which looks suspiciously similar to the usual heat equation (it is not exactly a heat equation, though!). From this we can see that regions where curvature is big and positive tend to shrink (because the $g_{ij}$ have a value smaller than average in such a region; hence $\\Delta (g_{ij})<0$), whereas regions with very negative curvature tend to expand (in this case $\\Delta (g_{ij})>0$).\n\nFinally, it is worth saying something about the asympotic behaviour of the flow. In the paper cited before, Hamilton showed that a simply connected 3-manifold $(M,g(0))$ with $Ric(0)>0$ converges eventually to a manifold of positive sectional curvature (more precisely, $g(t)$ converges in some sense to a metric $g_\\infty$ of constant sectional curvature; the convergence is a rather technical issue so I will not elaborate more). In general, provided that we do not encounter a singularity in the way, the Ricci flow converges to something called Ricci solitons, which are generalizations of the more well known Einstein metrics.\n\nDetails about the asymptotic behaviour of the flow are endless, but I would recommend to have a look at section 3 of Hamilton's 1995 paper, The formation of singularities in the Ricci flow, for lots of intuitive examples. Also chapter 3 in Peter Toppings excellent Lectures on the Ricci Flow provides an introduction to the maximum principle, which is a great tool to understand this evolution equation.\n\n• Thank you very much, especially for reference of Peter Toppings lectures. – Oliver Straser Sep 30 '13 at 9:34\n• A typo there, it should be \"regions with Ric>0 tend to expand\", but I don't have enough rep to edit it. – John Harvey Sep 3 '15 at 12:16\n\nThis YouTube video is the intuition behind the Ricci flow in 1-dimensional and discussed by James Isenberg from the University of Oregon."
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https://www.dataunitconverter.com/kilobit-per-hour-to-zebibit-per-minute | [
"# kbit/Hour to Zibit/Minute Converter - CONVERT Kilobits Per Hour to Zebibits Per Minute\n\nInput Kilobits Per Hour - and press Enter\nkbit/Hour\n\nSec\nMin\nHr\nDay\nSec\nMin\nHr\nDay\nQuickly and accurately convert between Kilobits Per Hour and Zebibits Per Minute with our free online tool. Learn about the conversion formula and calculation steps. Get precise results and save time with DataUnitConverter.\n\n## Recent Conversions\n\nHistory Empty ! No Recent Conversions.\n\n## How to use Kilobits Per Hour to Zebibits Per Minute Converter\n\nkbit/Hour to Zibit/Minute Calculator Tool convert the data transfer rate from Kilobits Per Hour to Zebibits Per Minute.\nIt is very easy to use, just follow the below steps.\n\n• Type the value in kbit/Hour input box and click CONVERT button or simply hit ENTER key.\n• The calculator will process the conversion with the highest accuracy and display the result.\n• Use the Copy button to copy the result to clipboard.\n• Click on the Swap⇄ button to reverse the conversion direction.\n\nYou can also change the source and target units in the drop-downs and quickly navigate to an entirely different conversion. Alternatively, switch to Data Size Converter for calculating the data storage size.\n\nIf you are looking to convert from one number system to another, such as binary, decimal, octal, or hexadecimal, try out the Number Base Converters.\n\n## kbit/Hour to Zibit/Minute Formula and Manual Conversion Steps\n\nKilobit and Zebibit are units of digital information used to measure storage capacity and data transfer rate. Kilobit is a decimal standard unit where as Zebibit is binary. One Kilobit is equal to 1000 bits. One Zebibit is equal to 1024^7 bits. There are 1,180,591,620,717,411,303.424 Kilobits in one Zebibit. - view the difference between both units",
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"Source Data UnitTarget Data Unit\nKilobit (kbit)\nEqual to 1000 bits\n(Decimal Unit)\nZebibit (Zibit)\nEqual to 1024^7 bits\n(Binary Unit)\n\nThe formula of converting the Kilobits Per Hour to Zebibits Per Minute is represented as follows :\n\nZibit/Minute = kbit/Hour x 1000 / 10247 / 60\n\nNote : Here we are converting the units between different standards. The source unit Kilobit is Decimal where as the target unit Zebibit is Binary. In such scenario, first we need to convert the source unit to the basic unit - Bit - multiply with 1000, and then convert to target unit by dividing with 1024^7 .\n\nNow let us apply the above formula and see how to manually convert Kilobits Per Hour (kbit/Hour) to Zebibits Per Minute (Zibit/Minute). We can further simplify the formula to ease the calculation.\n\nFORMULA\n\nZebibits Per Minute = Kilobits Per Hour x 1000 / 10247 / 60\n\nSTEP 1\n\nZebibits Per Minute = Kilobits Per Hour x 1000 / (1024x1024x1024x1024x1024x1024x1024) / 60\n\nSTEP 2\n\nZebibits Per Minute = Kilobits Per Hour x 1000 / 1180591620717411303424 / 60\n\nSTEP 3\n\nZebibits Per Minute = Kilobits Per Hour x 0.0000000000000000008470329472543003390683 / 60\n\nExample : If we apply the above Formula and steps, conversion from 10 kbit/Hour to Zibit/Minute, will be processed as below.\n\n1. = 10 x 1000 / 10247 / 60\n2. = 10 x 1000 / (1024x1024x1024x1024x1024x1024x1024) / 60\n3. = 10 x 1000 / 1180591620717411303424 / 60\n4. = 10 x 0.0000000000000000008470329472543003390683 / 60\n5. = 0.000000000000000000141172157875716723178\n6. i.e. 10 kbit/Hour is equal to 0.000000000000000000141172157875716723178 Zibit/Minute.\n\n(Result rounded off to 40 decimal positions.)\n\nYou can use above formula and steps to convert Kilobits Per Hour to Zebibits Per Minute using any of the programming language such as Java, Python or Powershell.\n\n#### Definition : Kilobit\n\nA Kilobit (kb or kbit) is a decimal unit of digital information that is equal to 1000 bits. It is commonly used to express data transfer speeds, such as the speed of an internet connection and to measure the size of a file. In the context of data storage and memory, the binary-based unit of Kibibit (Kibit) is used instead.\n\n#### Definition : Zebibit\n\nA Zebibit (Zib or Zibit) is a binary unit of digital information that is equal to 1,180,591,620,717,411,303,424 bits and is defined by the International Electro technical Commission(IEC). The prefix \"zebi\" is derived from the binary number system and it is used to distinguish it from the decimal-based \"zettabit\" (Zb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.\n\n### Excel Formula to convert from kbit/Hour to Zibit/Minute\n\nApply the formula as shown below to convert from Kilobits Per Hour to Zebibits Per Minute.\n\nABC\n1Kilobits Per Hour (kbit/Hour)Zebibits Per Minute (Zibit/Minute)\n21=A2 * 0.0000000000000000008470329472543003390683 * 0.0166666666666666666666666666666666666666\n3\n\nDownload - Excel Template for Kilobits Per Hour to Zebibits Per Minute Conversion\n\nIf you want to perform bulk conversion locally in your system, then download and make use of above Excel template.\n\n### Python Code for kbit/Hour to Zibit/Minute Conversion\n\nYou can use below code to convert any value in Kilobits Per Hour to Zebibits Per Minute in Python.\n\nkilobitsPerHour = int(input(\"Enter Kilobits Per Hour: \"))\nzebibitsPerMinute = kilobitsPerHour * 1000 / (1024*1024*1024*1024*1024*1024*1024) / 60\nprint(\"{} Kilobits Per Hour = {} Zebibits Per Minute\".format(kilobitsPerHour,zebibitsPerMinute))\n\nThe first line of code will prompt the user to enter the Kilobits Per Hour as an input. The value of Zebibits Per Minute is calculated on the next line, and the code in third line will display the result."
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https://www.physicsforums.com/search/11188950/?c[users]=Crosshash&o=date | [
"# Search results for query: *\n\n1. ### A How do Magnetic Gradiometers work?\n\nI'm a bit of a layman when it comes to this sort of thing, but from the literature I've read, it looks like most gradiometers utilise two magnetometers. Much of my understanding stems from this paper: http://library.seg.org/doi/abs/10.1190/1.2164759 but from the fairly limited articles on...\n2. ### A How do Magnetic Gradiometers work?\n\nWhenever I've encountered measuring magnetic fields with magnetometers, they've typically been very large magnetic fields (up to a few T). Magnetic fields fall off rapidly with distance and so I generally figured that measuring weak fields at distances on the order of cm (let alone m) was very...\n3. ### Dispersion relation for the free electron model\n\nI'm trying to get my head around what this means exactly. I've plotted the graph to help verse me with the functions that I've derived. From the free electron model, the wavefunctions are treated as planewaves of the form \\psi_\\mathbf{k}(\\mathbf{r}) = e^{i\\mathbf{k}\\cdot\\mathbf{r}} Due to...\n4. ### Drawing Feynman Diagrams\n\nYeah, okay, so I could use a photon or a Z for that interaction. So would it be accurate to say both diagrams would contribute to the probability amplitude?\n5. ### Drawing Feynman Diagrams\n\nHey there, this isn't a homework / coursework question so I didn't think it should be posted in that section. I can't seem to find a source which explicitly tells you how to determine which bosons are used in reactions. I understand how to work out if a reaction is legal or not by using charge...\n6. ### Cv as a function of temperature for ionic solid and metal\n\nWould it be okay to bump this question? I have a similar question and I can't find these plots anywhere.\n7. ### Finishing my Bsc in Physics & Computer Science, not sure what to do now.\n\nHello everyone. I'm a third year Physics and Computer science student (predicted a 1st class - studying in the UK) and I'd like some advice on what to do in the future. I apologise in advance if this OP turns into a wall of text but I have tried to do some research prior to this post. I feel...\n8. ### Quick Braket notation question\n\nAh crud, I completely forgot that's how you write Expectation value. so, just to confirm I have a grip on this, \\langle x \\rangle = \\langle \\psi \\vert {x} \\vert \\psi \\rangle = \\int_{-\\infty}^{\\infty} \\psi^* x \\psi dx Is that right? Assuming the limits are from infinity to minus infinity...\n9. ### Quick Braket notation question\n\nI'm a complete noob with Braket and I've only just started getting to grips with it. For completeness' sake though (from the book I'm currently reading), I can't seem to find a definition for: \\langle J_z \\rangle Would this just be the \"magnitude\" of J_z? Thanks\n10. ### How do you get good at statistical physics?\n\nThank you for the reply. I'll give these a read over christmas and see how things go. I'll still check this thread to see if somebody else can contribute.\n11. ### How do you get good at statistical physics?\n\nHello there, I'm a second year physics student who like most, has exams around the start of the next year and as such, have started revising for my exams. The term has introduced new physics I wasn't initially familiar with such as Quantum mechanics and advanced differential calculus. Another...\n12. ### Is homework help plagiarism?\n\nYeah, I'll pay him a visit with the rules (which I've just printed) on Monday. Thanks for the words. I'd like to see this thread stay open just so I can see what other people have to say on this topic. Thanks.\n13. ### Is homework help plagiarism?\n\nGuess I'll have to ask him next time I see him. I suppose the rules being obeyed make sense though.\n14. ### Is homework help plagiarism?\n\nI usually find myself asking a question here if I get stuck on a question. Is this plagiarism?\n15. ### Maxwell Boltzmann Distribution Question\n\nOk, so \\bigg[{-e^{-\\frac {mv^2}{kT}}\\bigg]_0^t will give me \\lim_{t \\to \\infty}({-e^{-\\frac {mt^2}{kT}} + 1) yeah? Sorry if this is annoying, I've never actually done something like this which seems a bit strange considering a question requires it. I appreciate the help.\n16. ### Maxwell Boltzmann Distribution Question\n\nI thought so. This might seem stupid, but I really don't know how to evaluate the integral when one of the limits is \\infty. Could you shed some light on that please?\n17. ### Maxwell Boltzmann Distribution Question\n\nok, so you're saying N = \\frac {Ckt}{2m} \\bigg[{-e^{-\\frac {mv^2}{kt}}\\bigg]_0^\\infty which yeah, makes sense. But do you want me to rearrange it to make C the subject while not evaluating the integral?\n18. ### Maxwell Boltzmann Distribution Question\n\nHello everyone Homework Statement The equivalent of the Maxwell-Boltzman distribution for a two-dimensional gas is P(v) = Cv e^-\\frac {mv^2}{kt} Determine C so that \\int_0^\\infty P(v)dv = N Homework Equations Not really sure The Attempt at a Solution I wasn't really sure how to tackle...\n19. ### Rotational Energy Deriving\n\nThanks for the reply CompuChip, sorry for this late reply. I took onboard what you said, perhaps you could verify my answer. I have v = \\omega r I = \\int dV \\rho r^{2} dm = \\rho dV I = \\int \\dfrac {dm}{\\rho} \\rho r^{2} I = \\int dm r^{2} I = r^{2} \\int dm I = r^{2}m m = \\dfrac...\n20. ### Rotational Energy Deriving\n\nHello, I think I've got the right idea on how to perform this question but I just need a little bit of help. Homework Statement Show that for a rigid body rotating with angular velocity \\omega the energy of rotation may be written as: E = \\dfrac {1}{2}I\\omega^{2} where the moment of...\n21. ### Determining whether a matrix function is linear?\n\nOk, I think I understand, so if I take the first question and use the criteria f(cx) = cf(x). a) \\begin{pmatrix} a & b\\\\ c & d\\\\ \\end{pmatrix} \\begin{pmatrix} x\\\\ y\\\\ \\end{pmatrix} = \\begin{pmatrix} x - y \\\\ x\\\\ \\end{pmatrix} I get \\begin{pmatrix} ax + by\\\\ cx + dy\\\\ \\end{pmatrix}...\n22. ### Determining whether a matrix function is linear?\n\nHomework Statement Which of the following functions is linear? Give reasons if they are not linear. If they are linear, give the corresponding matrix. Homework Equations a) R \\begin{pmatrix} x\\\\ y\\\\ \\end{pmatrix} = \\begin{pmatrix} x - y \\\\ x\\\\ \\end{pmatrix} b)...\n23. ### Satellite Slingshot manoeuvre question\n\nan unfortunate bump :(\n24. ### Satellite Slingshot manoeuvre question\n\nI've already attempted to work everything out by the way but my answers seem a little off. Homework Statement A satellite performs a sling shot manoeuvre around a planet. The mass of the planet is 6.00 * 10^{24} kg and the mass of the satellite is 7.00 * 10^{2} kg. The satellite approaches...\n25. ### A regular expression and a Windows permission (cacls) question\n\nHello, firstly, what does the regular expression \"\\\\+\" do? Would that literally match up \"\\+\". Could I possibly get some recommendations on some software to try regular expressions with as well actually please? and secondly, I've been fiddling around with the cacls command and I've noticed...\n26. ### Solving Ideal Gas Q: Temps & Press at A,B,C,D\n\nSo basically, if I can find out what K is from the Adiabatic condition, then I should be able to calculate the values of P and V for the other points? Except I don't have a point which has both P and V values.\n27. ### Solving Ideal Gas Q: Temps & Press at A,B,C,D\n\nHomework Statement 1) Consider one-mole of gas in a heat engine undergoing the Otto Cycle a) The gas absorbs heat, at constant volume between 120'C and 300'C b) The gas expands adiabatically from V1 to V2 = 5V1 c) The gas cools, at constant volume to Td at point D where the pressure is 1At d)...\n28. ### Vector equation questions\n\nThankyou both very much for the help Yeah, I completely understand where my problem is, I'll definitely do this. Thanks again.\n29. ### Vector equation questions\n\nah, I see the error of my ways, ok: OP(t) = OA + (t)(BA) = (1-t)OA + (t)OB (1-t)OA = (5 -5t, 0, 1-t) (t)OB = (7t, 4t, 7t) OP(t) = (5 -5t, 0, 1-t) + (7t, 4t, 7t) OP(t) = (5 + 2t, 4t, 1 + 6t) So that's the answer? I think I'll have to do more research on this regardless?\n30. ### Vector equation questions\n\nOk, well, I'm really not sure what I'm doing but I did this: OP(t) = OA + (t)(BA) = (1-t)OA + (t)OB (1-t)(5,0,1) + (t)(7,4,7) (-6t + 6) + (7t + 4t + 7t) 12t + 6 OP(t) = 12t + 6 Surely there's more to it then that?\n31. ### Vector equation questions\n\nThanks a lot Yeah, just got back, I'll take a stab at this now and post my result :)\n32. ### Vector equation questions\n\nI was hoping I could be linked to some resources to find the answer actually. Equation of a plane takes the form of Ax + By + Cz + D = 0 Point = (1, 0, 1) Normal point = (i + j - k) = (1, 1, -1) Point Normal Form = (1, 1, -1) x ((x, y, z) - (1, 0, 1) = 1(x -1) + 1(y-0) -1(z-1) = x...\n33. ### Vector equation questions\n\nHello guys, I've got two questions here which I'm really unsure about on how to actually tackle. As a result, maybe it would be better if I could possibly be linked to some resources where I could read up on how to actually solve them. Homework Statement 1. Find the equation of the plane in 3...\n34. ### Simple complex number question\n\nHomework Statement 3. Write down the (x+iy) form for the complex numbers with the following modulus and argument (in radians): (a) Modulus 1, argument pi (b) Modulus 3, argument -pi/3 (c)Modulus 7, argument -4 Homework Equations Modulus = ((a)^2 + (b)^2)^1/2 Arg = the angle...\n35. ### Gravitational Fields question\n\nah cool, ok, thanks for the help everyone.\n36. ### Gravitational Fields question\n\ngod damn it. Ok. Weight at sea level = 980N weight at summit = mg g = 6.67*10^-11 * 6.0 * 10^24 / 6410,000² g = 9.74 100 * 9.74 = 974 980 - 974 = 6N so the difference in weight is 6N? Is that correct? Can't believe my entire mistake was using the wrong mass.\n37. ### Gravitational Fields question\n\ng = GM/r² gr²/M = G 9.8 * 6400,000² / 100 = G G = 4.01408*10^12 gr² / G = M 9.8 * (6400,000 + 10,000)² / 4.01408*10^12 = M M = 100.313 Weight = Mg 983.06 - 980 = 3.06N So the difference in weight is 3.06N? The only problem is that she gained weight at the summit, she didn't lose it. So...\n38. ### Gravitational Fields question\n\nOk, so at sea level, the lady weighs 100*9.8 = 980N so I need to find the value for g at the summit of the mountain. I'm still confused with this constant business though, on wikipedia, it says the units of G are N(m/kg)². So do i need to square the mass? g = GM/r² g = 6.67*10^-11 * 100² /...\n39. ### Gravitational Fields question\n\nthanks for the reply. But actually, I think you've lost me even more :)\n40. ### Gravitational Fields question\n\n[SOLVED] Gravitational Fields question Homework Statement 1. Mount Everest is approximately 10 km high. how much less would a mountaineer of mass 100kg (including backpack) weigh at its summit, compared to her weight at sea level? Would this difference be measureable with bathroom scales...\n41. ### Circle equation question\n\nThanks for the replies, I managed to reach sqr29 using a different method though. x² + y² - 8x - 4y = 9 so x² + y² - 8x - 4y - 9 = 0 and the equation of a circle is (x - a)² + (x - b)² = r² expand out x² + y² - 2ax - 2by + a² + b² - r² equating coeficients -2a = -8 a = 4 -2b = -4 b =...\n42. ### Circle equation question\n\noops, fix'd (i hope :) )\n43. ### Circle equation question\n\nHomework Statement A circle has equation x² + y² - 8x - 4y = 9 (i) Show that the centre of this circle is (4,2) and find the radius of the circle. Homework Equations Circle equation = (x-a)² + (x-b)² = r² The Attempt at a Solution Well, if the centre of the circle is (4,2). Then the...\n44. ### Solving Fish Refraction Problem Using Law of Refraction\n\nThink about it, what is a perfectly flat surface of water with no waves a metaphoric equivalent to?\n45. ### Momentum of a snooker ball\n\nargh! you're kidding me? I've been stuck on this question for almost 30 minutes and it's that simple!? I thought it deflects off the side at a right angle :P I get the right answer now, thankyou!\n46. ### Momentum of a snooker ball\n\nHomework Statement A snooker ball of mass 0.350kg hits the side of a snooker table at right angles, and bounces off also at right angles. If its speed before collision is 2.8ms^-1 and its speed after is 2.5ms^-1, calculate the change in its momentum. (The answer to the question is not 0.105kg...\n47. ### Expansion of exponentials (very easy)\n\naha, thanks for the help. Just got a bit confused with the powers for a moment it seems.\n48. ### Expansion of exponentials (very easy)\n\nYeah, i tried using foil. Foil if I'm not mistaken would grant me. (e^x + e^-x) (e^x + e^-x) (e^x * e^x) + (e^x * e^-x) + (e^-x * e^x) + (e^-x * e^-x) (e^2x) + (1) + (1) + (e^-2x) (e^2x) + (2) + (e^-2x) Is that right?\n49. ### Expansion of exponentials (very easy)\n\nSorry, the +2. I can reach the e^2x + e^-2x but I can't reach the \"+2\" which means the method I'm using is incorrect. I haven't touched this stuff in a while so I'm just trying to re-remember it.\n50. ### Expansion of exponentials (very easy)\n\nSorry, I'm having a severe mental block here, sorry if this is laughable. Question Expand (e^x + e^-x)^2 I've got the answer but I'm just stumped at how to reach it, I don't know where the 2 comes from: Answer e^2x + 2 + e^-2x Thanks (and sorry again)"
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https://researchers.mq.edu.au/en/publications/complexity-mapping-of-a-photonic-integrated-circuit-laser-using-a | [
"# Complexity mapping of a photonic integrated circuit laser using a correlation-dimension-based approach\n\nChristopher J. McMahon, Joshua P. Toomey, Apostolos Argyris, Deb M. Kane\n\nResearch output: Contribution to journalArticleResearchpeer-review\n\n### Abstract\n\nQuantifying complexity from experimental time series generated by nonlinear systems, including laser systems, remains a challenge. Methods that are based on entropy, such as permutation entropy (PE), have proven to be useful tools for the relative measure of time series complexity. However, the numerical value of PE is not readily linked to a specific type of dynamical output. Thus, the quest to calculate quantitatively meaningful fractal dimension values, such as the correlation dimension (CD), from experimental signals, is still important. A protocol for calculating minimum gradient values and their spread, an integral part of CD analysis, is used here. Minimum gradient values with small spread are presented as approximate CD values. Here-in we report mapping these values, derived from analyzing experimental time series, obtained from a 4-section photonic integrated circuit laser (PICL) across a large parameter space. The PICL is an integrated form of a semiconductor laser subject to controllable optical feedback system. The minimum gradient/approximate CD mapping shows it has some qualitatively different map regions in its dynamics as compared to a free-space-based equivalent system. We show that the minimum gradient values give insight into the dynamics even when approximate CD values cannot be determined. The agreement between the qualitative features of permutation entropy mapping and minimum gradient/approximate CD value mapping provides further support for this. Regions of time series with close to periodic and quasi-periodic dynamics are identifiable using minimum gradient value maps.\n\nLanguage English 086202 1-7 7 Laser Physics 29 8 10.1088/1555-6611/ab27bb Published - Jun 2019\n\n### Fingerprint\n\nPhotonics\nintegrated circuits\nIntegrated circuits\nTime series\nEntropy\nphotonics\nLasers\npermutations\nlasers\nentropy\nOptical feedback\nFractal dimension\nSemiconductor lasers\nNonlinear systems\nnonlinear systems\nfractals\nsemiconductor lasers\noutput\n\n### Keywords\n\n• Chaos\n• Complex systems\n• Complexity\n• Integrated optoelectronic devices\n• Nonlinear dynamics\n• Semiconductor lasers\n\n### Cite this\n\nMcMahon, Christopher J. ; Toomey, Joshua P. ; Argyris, Apostolos ; Kane, Deb M. / Complexity mapping of a photonic integrated circuit laser using a correlation-dimension-based approach. In: Laser Physics. 2019 ; Vol. 29, No. 8. pp. 1-7.\n@article{273348d57d034d0faf8ff1d8e9512c1f,\ntitle = \"Complexity mapping of a photonic integrated circuit laser using a correlation-dimension-based approach\",\nabstract = \"Quantifying complexity from experimental time series generated by nonlinear systems, including laser systems, remains a challenge. Methods that are based on entropy, such as permutation entropy (PE), have proven to be useful tools for the relative measure of time series complexity. However, the numerical value of PE is not readily linked to a specific type of dynamical output. Thus, the quest to calculate quantitatively meaningful fractal dimension values, such as the correlation dimension (CD), from experimental signals, is still important. A protocol for calculating minimum gradient values and their spread, an integral part of CD analysis, is used here. Minimum gradient values with small spread are presented as approximate CD values. Here-in we report mapping these values, derived from analyzing experimental time series, obtained from a 4-section photonic integrated circuit laser (PICL) across a large parameter space. The PICL is an integrated form of a semiconductor laser subject to controllable optical feedback system. The minimum gradient/approximate CD mapping shows it has some qualitatively different map regions in its dynamics as compared to a free-space-based equivalent system. We show that the minimum gradient values give insight into the dynamics even when approximate CD values cannot be determined. The agreement between the qualitative features of permutation entropy mapping and minimum gradient/approximate CD value mapping provides further support for this. Regions of time series with close to periodic and quasi-periodic dynamics are identifiable using minimum gradient value maps.\",\nkeywords = \"Chaos, Complex systems, Complexity, Integrated optoelectronic devices, Nonlinear dynamics, Semiconductor lasers\",\nauthor = \"McMahon, {Christopher J.} and Toomey, {Joshua P.} and Apostolos Argyris and Kane, {Deb M.}\",\nyear = \"2019\",\nmonth = \"6\",\ndoi = \"10.1088/1555-6611/ab27bb\",\nlanguage = \"English\",\nvolume = \"29\",\npages = \"1--7\",\njournal = \"Laser Physics\",\nissn = \"1054-660X\",\npublisher = \"Maik Nauka-Interperiodica Publishing\",\nnumber = \"8\",\n\n}\n\nComplexity mapping of a photonic integrated circuit laser using a correlation-dimension-based approach. / McMahon, Christopher J.; Toomey, Joshua P.; Argyris, Apostolos; Kane, Deb M.\n\nIn: Laser Physics, Vol. 29, No. 8, 086202, 06.2019, p. 1-7.\n\nResearch output: Contribution to journalArticleResearchpeer-review\n\nTY - JOUR\n\nT1 - Complexity mapping of a photonic integrated circuit laser using a correlation-dimension-based approach\n\nAU - McMahon, Christopher J.\n\nAU - Toomey, Joshua P.\n\nAU - Argyris, Apostolos\n\nAU - Kane, Deb M.\n\nPY - 2019/6\n\nY1 - 2019/6\n\nN2 - Quantifying complexity from experimental time series generated by nonlinear systems, including laser systems, remains a challenge. Methods that are based on entropy, such as permutation entropy (PE), have proven to be useful tools for the relative measure of time series complexity. However, the numerical value of PE is not readily linked to a specific type of dynamical output. Thus, the quest to calculate quantitatively meaningful fractal dimension values, such as the correlation dimension (CD), from experimental signals, is still important. A protocol for calculating minimum gradient values and their spread, an integral part of CD analysis, is used here. Minimum gradient values with small spread are presented as approximate CD values. Here-in we report mapping these values, derived from analyzing experimental time series, obtained from a 4-section photonic integrated circuit laser (PICL) across a large parameter space. The PICL is an integrated form of a semiconductor laser subject to controllable optical feedback system. The minimum gradient/approximate CD mapping shows it has some qualitatively different map regions in its dynamics as compared to a free-space-based equivalent system. We show that the minimum gradient values give insight into the dynamics even when approximate CD values cannot be determined. The agreement between the qualitative features of permutation entropy mapping and minimum gradient/approximate CD value mapping provides further support for this. Regions of time series with close to periodic and quasi-periodic dynamics are identifiable using minimum gradient value maps.\n\nAB - Quantifying complexity from experimental time series generated by nonlinear systems, including laser systems, remains a challenge. Methods that are based on entropy, such as permutation entropy (PE), have proven to be useful tools for the relative measure of time series complexity. However, the numerical value of PE is not readily linked to a specific type of dynamical output. Thus, the quest to calculate quantitatively meaningful fractal dimension values, such as the correlation dimension (CD), from experimental signals, is still important. A protocol for calculating minimum gradient values and their spread, an integral part of CD analysis, is used here. Minimum gradient values with small spread are presented as approximate CD values. Here-in we report mapping these values, derived from analyzing experimental time series, obtained from a 4-section photonic integrated circuit laser (PICL) across a large parameter space. The PICL is an integrated form of a semiconductor laser subject to controllable optical feedback system. The minimum gradient/approximate CD mapping shows it has some qualitatively different map regions in its dynamics as compared to a free-space-based equivalent system. We show that the minimum gradient values give insight into the dynamics even when approximate CD values cannot be determined. The agreement between the qualitative features of permutation entropy mapping and minimum gradient/approximate CD value mapping provides further support for this. Regions of time series with close to periodic and quasi-periodic dynamics are identifiable using minimum gradient value maps.\n\nKW - Chaos\n\nKW - Complex systems\n\nKW - Complexity\n\nKW - Integrated optoelectronic devices\n\nKW - Nonlinear dynamics\n\nKW - Semiconductor lasers\n\nUR - http://www.scopus.com/inward/record.url?scp=85070818815&partnerID=8YFLogxK\n\nUR - http://purl.org/au-research/grants/arc/LP100100312\n\nU2 - 10.1088/1555-6611/ab27bb\n\nDO - 10.1088/1555-6611/ab27bb\n\nM3 - Article\n\nVL - 29\n\nSP - 1\n\nEP - 7\n\nJO - Laser Physics\n\nT2 - Laser Physics\n\nJF - Laser Physics\n\nSN - 1054-660X\n\nIS - 8\n\nM1 - 086202\n\nER -"
]
| [
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https://quantumfinancier.wordpress.com/2010/05/ | [
"Volatility Autocorrelation in Different Markets\n\nFor an introduction to autocorrelation see this previous post: First Order Autocorrelation as a Moderator of Daily MR.\n\nIn the previous post, I looked at how autocorrelation indicated MR performance, in this post, I observed the autocorrelation of another moderator of daily follow-through; volatility. Intuitively, it makes sense for volatility to be highly autocorrelated. VIX spikes are a good example of it. Let’s look at how the autocorrelation of volatility behave during bull/bear markets as defined by the 200/50 days moving averages crossover.\n\nFor this test, I looked at the first through fifth order autocorrelation of the 21 day rolling standard deviation of SPY returns, dividing them into percentiles (using the percent rank function with a 252 days lookback period) and looked at the average and the standard deviation of the percentiles.\n\nAs I thought before beginning the test, the volatility of returns is highly autocorrelated. However I thought that a bigger difference would exist between bull and bear markets. One thing to notice in the table below is that the autocorrelation is on average greater when the 50 day MA is greater than 200 day MA for all orders. The same observation holds true regarding the standard deviation.",
null,
"Trading wise, this information a no direct value by itself, however it puts in numbers a well defined market mechanics. We know that volatility is easier to predict than returns, and the level of autocorrelation is greatly responsible for this fact. We also know that daily mean-reversion strategies perform better during high volatility regimes. Thus we can use the volatility autocorrelation as an indicator of the suitability of our strategy. For example, during a high volatility regime with high autocorrelation we can expect our strategy to perform well and vice-versa if we change the contingency. For a very interesting read about the predictability of volatility and its application to trading, I recommend this recent article winning the NAAIM 2010 Wagner award from Tony Cooper. http://naaim.org/files/2010/1st_Place_Tony_Cooper_abstract.pdf\n\nQF\n\nDrivers of MR Performance\n\nA lot of the daily mean reversion strategies discussed in the blogosphere are designed for equity index ETFs or equity index mutual funds. Indexes being a group of individual stocks, if we could determine which stocks will drive the index returns or in this case, the behavior, we could adapt a strategy to profit from the prevailing market paradigm as indicated by the drivers of the index.\n\nFor example if the stocks affecting the index the most are exhibit mostly mean reverting behavior, we can expect the index to exhibit the same behavior. The same conclusion holds for trending behavior. The important point here is that indices are not an entity in themselves; they are made to reflect a certain sector/type of stocks/area etc. To illustrate the concept, I performed a simple test using the popular unbounded DV 2.\n\nI used the Nasdaq 100 and its related ETF QQQQ to perform this test. First I needed a way to quantitatively identify the drivers of the market. For this purpose, I used a weighted r-squared. Many of you know that the r-square under its other name: the coefficient of determination. The statistic provides an indication of the level in which a series is predicted by another (i.e. the goodness of fit). To obtain it you can just elevate the correlation coefficient to the power 2. For this test I looked at the r-squared only for the 21 previous days on a rolling basis. Then I looked at the volume of each individual component of the index relative to the total volume of all the components, and then weighted the r-squared with the proportion of the volume of this particular stock.\n\nIn a more rigorous form:",
null,
"After this computation for every stock at every period n, I take the top 10 weighted r-squared stocks and compute their DV2 values, average it and trade the QQQQ using the signal given by the DV2 (long/short from 0). The results below compare this strategy, buy & hold QQQQ, and DV2 on the QQQQ data.",
null,
"",
null,
"As you can see, the results are quite similar and as expected, highly correlated. It shows that in an index formed with 100 stocks, looking at only a select group with a big influence on the returns can help determine the MR performance of the index itself.\n\nQF\n\nIntroduction of SVM Learning in Strategies\n\nThis post is derived from a comment I received my last post on probability density function for an adaptive RSI strategy. pinner made the following observation:\n\n“Alternatively you could regress the returns against the set of 6mo & 1yr RSI points as a means to determine the best decision. While this approach probably requires more historical data, it also affords more detailed metrics from which to make each decision.”\n\nI am personally not a big fan of simple regression model for my trading decisions. However, I think there is something in the concept. Such a setup is also a good place to use a common traditional machine learning technique; the support vector machine (SVM).\n\nFrom wikipedia: given a set of training examples, each marked as belonging to one of two categories, an SVM training algorithm builds a model that predicts whether a new example falls into one category or the other. Intuitively, an SVM model is a representation of the examples as points in space, mapped so that the examples of the separate categories are divided by a clear gap that is as wide as possible. New examples are then mapped into that same space and predicted to belong to a category based on which side of the gap they fall on.\n\nIn concrete terms, the training algorithm separate the data into up days and down days then look at the predictors value. It then creates a set of rules dividing the data; these rules minimize the classification error while also maximizing the margin of safety, thus giving the model more room, resulting (hopefully) in a greater accuracy. Based on this set of rules, the algorithm classifies new data in a category. Note that this is not a numeric prediction (i.e. next day return should be xx%) it is a binary factor prediction, thus allowing us to derive a probability along with the prediction. It comes in handy when we want a confidence based system.\n\nThis is a nice technique, but it has its drawbacks. First of all, it is a parameter dependant technique. The effectiveness of the svm is mostly determined by two parameters. It is usually recommended to test different values for the pair and to retain the pair that performs best during cross-validation for the model. This can become quite computationally annoying. Without getting to technical (or into details), if we want a non-linear classification algorithm, we have to choose the type of kernel function we want; there is several.\n\nI just wanted to throw the idea out there, since pinner’s suggestion was a good one. If readers are interested to try it out, I encourage you to send me the results, I would be happy to post them. Alternatively, I might post some results depending on the interest. I just wanted to introduce support vector machine and its potential application when developing strategies.\n\nQF\n\nUsing Probability Density as an Adaptive Mechanism\n\nI will take a pause of the Time Machine series for now while I work on it some more and prepare future posts. Today I will follow up on my post on return distributions and show a simplistic way to include it in an adaptive strategy.\n\nIt has been discussed quite a lot on the blogosphere that a strategies with fixed parameters are inferior to adaptive strategies. For example the simplest daily MR strategy I can think of is probably RSI 2 50/50, but this strategy did not always worked and I certainly don’t expect it to keep working forever. Furthermore, the most profitable lookback parameter for RSI also varies in time. This is where return distribution is useful. From it, we can derive the probability density function and use that to create an adaptive mechanism.\n\nJust a little background on probability density function; from wiki: “density of a continuous random variable is a function that describes the relative likelihood for this random variable to occur at a given point in the observation space.” In plain language; the probability of a certain event happening. I recommend using your favorite statistical software to do so, unless you want to be doing integrals for a long time!\n\nFor this test, I took SPY data, computed RSI values for different lookback periods (2 to 30), and then looked at the results for each strategy. For a rolling period of 1 year and 6 months I looked at the probability densities of returns for every strategy looking specifically at the probability of returns greater than zero (this can be changed to a higher threshold, just want to keep it basic for this). I then traded the strategy that had the highest combination of 1 year and 6 months values. That way, the capital is allocated to the strategy with the parameter generating the highest probability of positive returns as measured by the probability density function. I compared the strategy, RSI 2 50/50 and buy and hold.",
null,
"The results are not particularly impressive, the point of the article was to illustrate the concept as simply as possible. I believe that there is ways to make this particular strategy more robust; as a starter to take a shorter time frame for the lookback period to make it more sensitive to recent market data or introducing a weighting scheme to weight more recent data. I will let the reader experiment with it, I would be happy to post results if you care to share. Even though the results are not spectacular, the strategy seems to adapt to the different waves in the market and allocate the capital to a more appropriate parameter length for the RSI depending on the current market paradigm.\n\nQF\n\n(Part 4) Time Machine Test – Commodities\n\nResults on a commodities basket.\n\nAs you can see, the algorithm adapts to different classes of commodities and outperforms most of the buy and hold returns. But I want to emphasis that this concept alone is not something I would trade as is (not robust enough). The results are in no way good enough to rely on out-of-sample.\n\nI found that it is much harder for the algorithm to find strategies significant enough to trade on for long periods on these assets. For equities, there is on average 7 different active strategies (i.e. the level of significance is high enough) at the same time. The number is much less with commodities and currencies, the lack of diversification between strategies certainly hurts performance when compared to equity indices. It also add on more exposure to a given strategy adding a lot of volatility to the returns as showed by the numbers above.\n\nQF\n\n(Part 3) Time Machine Test – Currencies\n\nEdit: This is a repost of the previous version of the post for currencies and commodities. When adapting my code for Datastream data, I made an error, the code was peeking which explains the ridiculously straight equity curve. Here is the corrected version of the post. I sincerely apologize to readers for the inconvenience. I also want to assure you that the previous results on equity indices are correct; I had a couple colleagues take a look at it to confirm that there were no bugs left.\n\nFirst thing I notice looking at these results is difference between the usefulness of run analysis on currencies. It makes sense since currencies are related in good part to macroeconomics factors. It is also much harder for the algorithm to find significant strategies, the time in the market for the strategy is much less than with equities.. Because of this, it becomes harder to squeeze out alpha out of the strategy with a high confidence level in the analysis.\n\nQF\n\nBlog Beauty Pageant\n\nJust a quick post to tell you not to be surprised if I change the theme of my blog, or at the very least the font size, during the day, some reader have brought to my attention the difficulty to read posts. I will be experimenting with WordPress’ themes and HTML today to find a solution. It is made harder by the fact that I don’t really want to take a blog theme already used by the quant blogosphere big players (a real shame since they are pretty nice). Regardless, expect a post testing the time machine algorithm on different asset classes; commodities, currencies later today.\nQF"
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"https://quantumfinancier.files.wordpress.com/2010/05/volatility-autocorrelation-in-different-markets-results.png",
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"https://quantumfinancier.files.wordpress.com/2010/05/drivers-of-mr-formulas.png",
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"https://quantumfinancier.files.wordpress.com/2010/05/drivers-of-mr.png",
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"https://quantumfinancier.files.wordpress.com/2010/05/drivers-of-mr-res1.png",
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"https://quantumfinancier.files.wordpress.com/2010/05/density-adative-results.png",
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https://file.scirp.org/Html/5-7900259_35558.htm | [
"Decoupling Control Research on Test System of Hydraulic Drive Unit of Quadruped Robot Based on Diagonal Matrix Method\n\nIntelligent Control and Automation\nVol. 4 No. 3 (2013) , Article ID: 35558 , 6 pages DOI:10.4236/ica.2013.43033\n\nDecoupling Control Research on Test System of Hydraulic Drive Unit of Quadruped Robot Based on Diagonal Matrix Method*\n\nLingxiao Quan1,2, Wei Zhang1, Bin Yu1, Liang Ha1\n\n1College of Mechanical Engineering, Yanshan University, Qinhuangdao, China\n\n2Engineering Research Center of Advanced Forging & Stamping Technology and Science, Yanshan University, Qinhuangdao, China\n\nEmail: [email protected]\n\nCopyright © 2013 Lingxiao Quan et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.\n\nReceived March 26, 2013; revised April 26, 2013; accepted May 7, 2013\n\nKeywords: Position Control System; Force Control System; Diagonal Matrix Method; Decoupling Control\n\nABSTRACT\n\nThe mathematical model of hydraulic drive unit of quadruped robot was built in this paper. According to the coupling characteristics between position control system and force control system, the decoupling control strategy was realized based on diagonal matrix method in AMESimÒ. The results of simulation show that using diagonal matrix method can achieve the decoupling control effectively and it can achieve the decoupling control more effectively with the method of not offset pole-zero in the S coordinate. This research can provide theoretical basis for the application of test system of hydraulic drive unit.\n\n1. Introduction\n\nHydraulic quadruped bionic robot is a walking machinery with the high power weight ratio, fast response and good dynamic performance, which uses the valve controlling hydraulic cylinder system as its drive unit. The overall performance of the quadruped bionic robot is determined by the performance of its hydraulic driving unit. Therefore, it is necessary to test the performance of the hydraulic driving unit which was designed for the quadruped bionic robot [1-5]. A dedicated test system for the hydraulic drive unit of the quadruped bionic robot was built in this paper. The system consists of the subsystem of the hydraulic drive unit and the subsystem of the load simulation unit. The piston rod of the hydraulic cylinders of the two subsystems is rigidly connected. When the input signal of one part makes a sudden change, its output will influence the other part as a load interference and the surplus force is produced. The presence of surplus force seriously affects the precision and dynamic quality of the load simulation. In order to improve the control performance of the test system, it is necessary to eliminate or suppress the mutual interference between the two subsystems [6-12].\n\nIn this paper, according to the characteristics of hydraulic drive unit test system, a decoupling control strategy based on the diagonal matrix method was used to control the double inputs and double outputs electrohydraulic servo system [13,14].\n\n2. Composition and Mathematical Model of Test System\n\n2.1. Composition of Test System\n\nTest system is shown in Figure 1. The left part is the hydraulic drive unit which is a position control subsystem, the other part is the load simulation unit which is a force control subsystem. The two parts are rigidly connected together by the force sensor.\n\n2.2. Mathematical Model of Test System\n\nMaking the following assumptions to the test system:\n\n1) The oil is incompressible; 2) The oil supply pressure of the servo valve is constant; 3) The oil temperature has no effect on the system performance; 4) Regarding the servo valve as a second-order shock link. Considering the three basic equations of the valve controlling cylinder",
null,
"Figure 1. Composition diagram of test system.\n\nsystem: the flow equation of servo valve, the continuity equation of the valve controlling cylinder and the force balance equation of the piston, the open-loop transfer functions of the hydraulic drive unit and the load simulation unit are obtained.",
null,
"(1)",
null,
"(2)\n\nIn this paper, the subsystem of the hydraulic drive unit and the subsystem of the load simulation unit use two identical systems with servo valve controlling symmetrical cylinder. The parameters of two subsystems are same. They are shown in Table 1.\n\nAccording to Formulae (1) and (2), the block diagram of test system is obtained. It is shown in Figure 2.\n\nThe top half of Figure 2 is the hydraulic drive unit, the other half is the load simulation unit.",
null,
"and",
null,
"are coupling loops.\n\n3. Decoupling Control Strategy of Test System\n\n3.1. Research on the Decoupling Theory\n\nThe test system of hydraulic drive unit is a double inputs and double outputs coupled system. It belongs to V standard coupled system which was proposed by Mesarovic. The output of each object is influenced not only by the input of its own channel, but also by the other output through the middle channel. In order to eliminate the mutual effect of the position control channel and the force control channel, diagonal matrix method, which was proposed and developed by Boksenbom, is used to achieve the decoupling control in this paper. The algo-",
null,
"Table 1. Parameters of test system.\n\nrithm of this method is concise, easy to implement and suited for the linear time-invariant systems [15-18].\n\nThe schematic diagram of decoupling control based on diagonal matrix method is shown in Figure 3. G1 is the transfer function of servo valve of the position channel and G3 is the transfer function of servo valve of the Force channel. G2 is the transfer function of servo cylinder in the Position channel and G4 is the transfer function of servo cylinder the force channel. G5 is the coupling function of the Position channel and G6 is the coupling function of the Force channel. E1 is the decoupling transfer function from displacement channel to the Force channel. E2 is the decoupling transfer function from force channel to the position channel.\n\nThe transfer function of the decoupling link is obtained:",
null,
"(3)\n\nCombined with the transfer function of the test system, the transfer function of test system with decoupling link which is shown in Figure 4 is obtained.",
null,
"Figure 2. Block diagram of test system.",
null,
"Figure 3. The schematic diagram of decoupling control.",
null,
"Figure 4. Transfer function of test system with decoupling link.\n\n3.2. Research on the Decoupling Simulation\n\nIn this paper, take below situation as the research object. The output power of the force control system is constant, the output displacement of the position control system is changed. By comparing the influence degree of position fluctuation to force control, the availability of the decoupling control is validated.\n\nAt first, the AMESimÒ simulation model of test system is built, which is shown in Figure 5. The simulation parameters are shown in Table 2. Set the initial force of the force control system as 7 kN. Start the simulation at 0.5 s with a step displacement signal as the input of the position control system. The simulation results found that when the PID parameters of the position control system are P = 1, I = 0, D = 0 and the PID parameters of the force control system are P = 7 × 10−7, I = 1 × 104, D =",
null,
"Figure 5. AMESimÒ simulation model of test system.",
null,
"Table 2. Parameters of AMESimÒ simulation model.\n\n0, the control effects of the position control system and the force control system are good. The simulation result is shown in Figure 6.\n\nIt can be seen in Figure 6 that the force fluctuation of the force control system is drastic before taking the decoupling control strategy. Its peak reaches 18 kN and it is 2.57 times as much as the given force. The output force increases to the maximum and then decreases and tends to be stable finally. The adjustment time is about 0.2 s. It can be seen that the position fluctuation has influenced the force control system. After taking the decoupling control strategy, the force fluctuation decreases. Its peak reaches 9 kN and it is 1.29 times as much as the given force. The adjustment time reduces to 0.05 s. The output force decreases at first and then increases. The influence of position fluctuation to force control decreases with the decoupling control strategy. But the effect is not ideal enough.\n\n3.3. The Optimization of Decoupling Control\n\n3.3.1. Analysis of the Fluctuation of the Output Force\n\nThe essence of diagonal decoupling is adding a compensation link in the system to counteract the effects of in-",
null,
"Figure 6. The influence of position fluctuation to force control.\n\nterference loop. The output force of the force channel is controlled by its input, and it is also influenced by the interference channel. It is necessary to add a compensating link to counteract its influence. Diagonal matrix method has been applied to obtain the transfer function from the input of the position channel to force channel which named E1. The block diagram from u1 to F is shown in Figure 7.\n\n3.3.2. The Analysis of the Simulation\n\nThe AMESimÒ simulation model of output force from position channel to force channel was built as shown in Figure 8. Start the simulation with step signal valued 25 mm as its input. The simulation result is shown in Figure 9.\n\nIt can be seen in Figure 9 that the output force of interference channel is equal to the output force of the compensation channel in the opposite direction. The decoupling link is able to counteract the influence of position fluctuation to force control. But the actual output force fluctuates at the moment of adding the step signal. This is the reason why the decoupling effect is not ideal in the Figure 6. The reason is that the response speed of compensation channel is faster than that of the interference channel.\n\nIt can be seen in Figure 7 that the integral element and differentiation element in the interference channel make its response slow. While there is no integral element and differentiation element in the compensation channel. So the integral element and differentiation element in the decoupling link are added and the AMESimÒ simulation model is built as shown in the . The simulation result is shown in .\n\nIt can be seen in that adding the integral element and differentiation element in the decoupling link keeps the response of the interference channel synchronizing with the response of the compensation channel and solves the problem that the output force fluctuates.",
null,
"Figure 7. The block diagram from u1 to F.",
null,
"Figure 8. AMESimÒ simulation model of output force from position channel to force channel.",
null,
"Figure 9. The figure of output force from position channel to force channel.",
null,
". AMESimÒ simulation model of output force from position channel to force channel.\n\nBased on the conclusion above, the integral element and differentiation element are added in the test system. The simulation result is shown in .\n\nThe simulation result shows that position fluctuation has no effect on force control. The decoupling effect is very good.",
null,
". The figure of output force from position channel to force channel.",
null,
". The influence of position fluctuation to force control with decoupling link.\n\n3.3.3. Frequency Response of the Test System\n\nThe frequency domain diagram of closed loop from position channel to force channel is shown in .\n\nIt can be seen in that the gain from input displacement to output force is as high as 120 dB. It shows that the position fluctuation has a great influence on force control. The gain of low-frequency period declines with the decoupling control while the gain of high frequency period is still high with the decoupling control. The fluctuation still existes while the influence of position fluctuation to force control decreases. Adding the",
null,
". Frequency domain diagram of closed loop from position channel to force channel.\n\nintegral element and differentiation element in the decoupling link makes the gain from input displacement to output force under 0 dB. The influence of position fluctuation to force control is eliminated. Above results are the same with results obtained from time domain analysis.\n\n4. Conclusions\n\nThe mathematical model of hydraulic driving unit of quadruped bionic robot was built in this paper. Simulation analysis of decoupling control strategy was done and the conclusions are reached as follows:\n\n1) The diagonal matrix method can effectively eliminate the mutual interference between the position control and the force control.\n\n2) The diagonal matrix method cannot completely eliminate the mutual interference between the channels. In order to get better decoupling effect, it is necessary to add the integral element and differentiation element in the compensation link to keep symmetrical and equal with the interference link.\n\nREFERENCES\n\n1. L. H. Ding, R. X. Wang, H. S. Feng and J. Li, “Brief Analysis of BigDog Quadruped Robot,” China Mechanical Engineering, Vol. 23, No. 5, 2012, pp. 505-514.\n2. “BigDog—The Most Advanced Rough-Terrain Robot on Earth”. http://www.bostondynamics.com/robot_bigdog.html\n3. R. Playter, M. Buehler and M. Raibert, “BigDog,” Proceedings of SPIE 6230, Unmanned Systems Technology VIII, 62302O, Orlando, 9 May 2006.\n4. M. Buehler, R. Playter and M. Raibert, “Robots Step Out-Side,” International Symposium of Adaptive Motion of Animal and Machines, Ilmenau, September 2005, pp. 1-4.\n5. M. Raibert, K. Blankepoor and G. Nelson, “Big-Dog, the Rough-Terrain Quadruped Robot,” Proceeings of the 17th World Congress of the International Federation of Automatic Control, Seoul, 2008, pp. 6-9.\n6. Q. Hua, “Study on the Key Technology of Electro-Hydraulic Load Simulator,” Beijing University of Aeronautics and Astronautics, Beijing, 2001.\n7. Y. H. Li, “Research on Method to Reject to Extraneous Moment of Load Simulator,” Machine & Hydraulics, No. 2, 1999, pp. 27-30.\n8. Z. X. Jiao, Q. Hua and X. D. Wang, “Estimation for Performance of Load Simulator,” Chinese Journal of Mechanical Engineering, Vol. 11, No. 38, 2002, pp. 26-29.\n9. Z. X. Jiao, Q. Hua and X. D. Wang, “Hybrid Control on the Electro-Hydraulic Load Simulator,” Chinese Journal of Mechanical Engineering, Vol. 12, No. 38, 2002, pp. 34-38.\n10. J. X. Gao, Q. Hua and Z. X. Jiao, “The Redundant Force of Electro-Hydraulic Loading System and the Comparison of Various Kinds of Compensation Methods,” Hydraulics Pneumatics & Seals, No. 5, 2003, pp. 1-6.\n11. H. Hu, C. H. Han, J. G. Liu and Y. Wei, “On Multivariable Feedback Decoupling Control Systems,” Control Engineering of China, Vol. 11, No. 6, 2004, pp. 500-502.\n12. N. Cherouvim and E. Papadopoulos, “Speed and Height Control for a Special Class of Running Quadruped Robots,” 2008 IEEE International Conference on Robotics and Automation, Pasadena, 19-23 May 2008, pp. 825-830. doi:10.1109/ROBOT.2008.4543307\n13. Q. G. Wang, “Decoupling with Internal Stability for Unity Output Feedback Systems,” Automatica, Vol. 28, No. 2, 1992, pp. 411-415.\n14. Q. G. Wang and Y. S. Yang, “Transfer function Matrix Approach to Decoupling Problem with Stability,” Systems & Control Letters, Vol. 47, No. 2, 2002, pp. 103- 110.\n15. L. H. Liang, “Research on the Electro-Hydraulic Load Simulator of Fin Stabilizer,” Harbin Engineering University, Harbin, 2003.\n16. L. H. Liang, Q. Liu and L. L. Zhao, “Research on Feedforward Compensation Decoupling Control for the Electro-Hydraulic Load Simulator of Fin Stabilizer,” China Mechanical Engineering, Vol. 4, 2007, pp. 439-411.\n17. L. B. Xian, G. Yang, X. Y. Fu and B. R. Li, “Study on the Position/Force Control of an Electro-Hydraulic Actuator Based on AMEsim/Simulink,” Machine Tool & Hydraulics, Vol. 35, No. 5, 2007, pp. 205-207.\n18. L. He, “Study on Control and Energy Saving Characteristics of Hydraulic Secondary Regulated Inertia Loading System,” Yanshan University, Qinhuangdao, 2009.\n\nNOTES\n\n*This work was supported by the foundation item “Key Basic Research of Hebei Province (12962147D)”."
]
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https://smcint.com/transformer-turns-ratio-2/ | [
"# Transformer Turns Ratio\n\nThe transformer turns ratio is the number of turns of the primary winding divided by the number of turns of the secondary coil. The transformer turns ratio provides the expected operation of the transformer and the corresponding voltage required on the secondary winding.\n\nIf it is required a secondary voltage smaller than the primary voltage – step-down transformer- the number of turns on the secondary must be lower than in primary, and for step-up transformers is the other way round; when the transformer turns ratio step-down the voltage, it steps-up the current and vice versa, so that the voltage and current ratio of an ideal transformer is directly related to the turns ratio.\n\n## The transformer turns ratio in real transformers\n\nUnfortunately, transformers are not ideal, and in a real transformer the voltage or current ratio may be not equal to the physical transformer turns ratio, due to the different electrical losses such as in the transformer iron core (hysteresis and eddy current losses) and copper losses (due to the electrical resistance of the primary and secondary windings); therefore, the manufacturers try to design the transformers in a way that minimize those losses, for getting a maximum efficiency at full load, higher to 95% of power transformation, and thus providing a voltage ratio which differs as maximum in 5% to the transformer turns ratio.\n\nBut since the transformers suffers different stresses and changes in their life, electrical and mechanical, the proper transformer turns ratio must be verified before placing it in service and during the different maintenance schedules, which is the target of the transformer turns ratio testers; therefore, the ratio measured with the different transformer turns ratio test sets (TTR equipment) includes the losses normally found in the transformer, which result in a ratio different to the physical turns but reflects the real voltage ratio expected by the manufacturer and the user, or the true electrical transformer turns ratio.\n\n## Transformer turns ratio in power and instrument transformers, different testing solutions and methods\n\nThe transformer turns ratio test helps to identify problems such as shorted-turns or open-turns, incorrect connections, tap changer and internal magnetic core problems, etc; the different transformer turns ratio test sets are specifically designed for the application depending on the type of transformer and the ratio to measure.\n\nIn power transformers the transformer turns ratio test is made through voltage injection, phase by phase and tap by tap, measuring the corresponding voltage ratio for the related windings, which is compared with the expected ratio of the nameplate; since in the 3-phase power transformers it is required to keep in mind the connection group, in some configurations the transformer turns ratio must be calculated from the measured voltage ratio with some conversion formulas; the magnetizing current must be also kept at a minimum, by using low voltages in the injection, and so reducing the voltage drop in the primary winding impedance which could be a main source of error; in brief, the power transformer turns ratio test equipment must have a dedicated design, with special performance characteristics and accuracy, for the procedure and range required, and with special arrangements for the 3-phase transformers that facilitate that type of test.\n\nIn the case of the current transformers for instrumentation, the transformer turns ratio is defined as a ratio of the rated primary current to the rated secondary current at specified burden, so that a different injection range and type is required in the suitable current transformer turns ratio test sets; those must provide the high rated currents required and must have a high power to meet the different testing situations; the power required for the same nominal current range will not be the same whether the transformer is close or situated very high at a distance of many meters.",
null,
"Fortunately, the small size and weight of the Raptor System allows it to be placed closer to the transformer under test and reduce the test leads length and the power required; furthermore, the modularity of the Raptor provides the capacity to increase the power of the system, when required, adding current slave units, in a very simple and quick way, thanks to the self-detecting infrared technology and the no need to interconnect the units; up to three slaves can be added for up to 15,000 A and up to 18-kVA total injection power, to cover the different situations in transformer turns ratio testing.\n\nThe Raptor System provides several test templates and functions for transformer turns ratio measurement in current transformers, voltage transformers and power transformers, with the required current and voltage injection and corresponding integrated measurement capability. The current transformer turns ratio test can be performed through current or voltage injection, being always preferable the current method, but the Raptor includes both for those cases in which is not possible to inject primary current in the CT; the Raptor also includes templates for the current transformer turns ratio of Rogowski and low power CTs; the VT and PT ratio is performed through voltage injection, which range is extended if the system is complemented with the optional Raptor high voltage slave.\n\nThe Raptor is a multifunctional substation test system, including transformer tests for the ratio, burden, polarity, knee point and withstand voltage, and many other primary injection applications for testing breakers, reclosers, relays, ground grid, resistance, etc.\n\nAs usual with EuroSMC test equipment, simplicity and ease-of-use are common to all solutions for transformer turns ratio tests."
]
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"https://smcint.com/wp-content/uploads/2016/01/prueba-de-relacion-de-transformacion-raptor-217x300.jpg",
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https://www.soulzfitness.com/profile/frddsonrandalie/profile | [
"## Profile\n\nJoin date: May 12, 2022",
null,
"Radio Comercial is an online radio that aims to present all kinds of music and cultural information that gives it a special character. Most of its programs are done by a group of professionals and loyal listeners.Fold generation of Gaussian beam Fold generation of Gaussian beam (also known as Gaussian beam folding) is an optical effect resulting in the formation of higher-order (non-paraxial) Gaussian beams. When a Gaussian beam propagates through a region of high refractive index the wave front of the beam becomes \"flattened\" as a result of the curvature of the beam. When the local refractive index is low, the curvature of the wave front may become so small that it does not significantly change the beam profile. However, as the local refractive index becomes higher and higher, the curvature of the wave front eventually grows to become of the same order of magnitude as the Gaussian beam diameter. In this case, the beam profile becomes \"folded\" around the beam waist. Gaussian beam folding may be analyzed using an approximate plane-wave approach, starting from the paraxial wave equation for a Gaussian beam: where is the transverse wave vector, is the wavelength, and the real part of is set to 1 for brevity. The quantity is the Gouy phase and the quantity is the curvature of the wavefront. The local divergence angle is set to 0. In this approximation, the beam propagates through a region of refractive index. The paraxial wave equation may be solved by separating the real and imaginary parts: where the complex conjugate has been neglected for brevity. The equation for the transverse wave vector is then where the complex wave vector is related to the transverse wave number by The solution to the equation for is found by separating variables and integrating over the transverse coordinate: where the angle is the transverse coordinate, and is a radial coordinate related to the beam radius. The integral may be evaluated using the error function and the incomplete gamma function to give where is the Wronskian. The transverse wave number is related to the radial coordinate by and this implies The integral may"
]
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"https://static.wixstatic.com/media/a894bb_56196071f0624c5aae109046727e3dfd~mv2.png/v1/fit/w_256,h_256,al_c,q_5,enc_auto/file.png",
null
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https://www.biologic.net/documents/eis-optimizations-electrochemistry-battery-application-note-23/ | [
"20 min read",
null,
"",
null,
"",
null,
"# EIS measurements on Li-ion batteries EC-Lab® software parameters adjustment (EIS optimizations) Battery – Application Note 23\n\nLatest updated: May 5, 2023\n\nAbstract\n\nThis application note explains many software and hardware considerations that need to be taken into account when setting up an Electrochemical Impedance Spectroscopy experiment on low impedance systems, in particular battery cells. From a hardware point of view, separating the contact of current-carrying leads and voltage sensing leads is one important parameter. Another one is cable length and/or cable extensions by the user. On the software side, importance of choosing appropriate excitation amplitude (and its definition), pre-delays, averaging of multiple excitation cycles and drift correction are discussed.\n\n## Introduction\n\nTo obtain significant EIS plots, without experiencing noise or other issues, experimental parameters should be chosen carefully. Users should also pay attention to the definition of each parameter and test the effects of each one on the results.\nFurthermore, all electrochemistry software is designed differently, and users must adapt parameters according to their own particular software interface.\nThe aim of this note is to help the user obtain good experimental results with EC-Lab® software. To this end, a detailed description of some key points (connections, cable length, experimental parameters etc.) is explained below.\n\n## Experimental Part\n\nIn the following application note, experiments were carried out on a Li-ion battery with a nominal capacity of 10 Ah. Experiments were executed at the Eoc potential, i.e. 3.3 V.\n\nNote that experiments were obtained with EC-Lab® software in potentiostatic mode.\n\n### Connection\n\nTwo kinds of two electrodes connections can be considered:\n\n1) The “two-point” connection with CA2+Ref1 together on the positive electrode and CA1+Ref2+Ref3 together on the negative electrode:",
null,
"2) The “five-point” connection by separating each cable on each electrode, i.e. CA2 and Ref1 on the positive electrode and CA1, Ref2, and Ref3 on the negative electrode:",
null,
"This connection is also recommended for ac-curate potential measurements.\nAs shown in Figure 1, we can notice a shift of + 2.5 mΩ between the EIS diagram obtained with the banana plugs connected together (1) comparing with the banana plugs connected separately (2).",
null,
"###### Figure 1: Comparison of EIS diagrams obtained with two-point connection (green line and markers) and five-point connection (red line and markers).\n\nThis example shows that for systems with a small resistance, the connection is very important and can significantly influence results. This is why it is necessary to minimize the value of stray capacitance, stray inductance or resistance generated by the connection. Each connection lead should be separated and connected as close as possible to the electrochemical system to reduce the intrinsic influence of the connection.\n\n### Cable Length\n\nLengthened cables are sometimes deemed necessary for certain configurations. However, this should be avoided wherever possible, as extended cables can impact negatively on result quality.\nWe used a standard cable (red EIS curve) and a cable of 10 m length (blue EIS curve) for these experiments. Note, that to avoid oscillations of the potentiostat, resistors were added to the reference plugs on the 10 m cable.\nAs shown in Figure 2, a slight difference can be noticed between the EIS curves obtained with the standard and long cables. Depending on the system studied, EIS measurements with long cables should be analyzed with caution.",
null,
"### Experimental Parameters\n\nPEIS technique experiments (Figure 3).\n\nExcitation amplitude\nThe first parameter to determine is the value of the potential excitation Va. Note that before the 9.56 EC-Lab® version, excitation amplitude was defined as Va (sinus amplitude). Equivalence between Va, Vpp (peak to peak amplitude) and VRMS is defined by the relationship:\n\n$$V_a=\\frac{1}{2}V_{pp}=\\sqrt2V_{RMS} \\tag{1}$$\n\nThis parameter value has to be chosen considering the current amplitude |I| and the potential amplitude |E| values. The |I| and |E| are the amplitude applied around the DC level of current or around the DC level of potential. It is the AC amplitude. In EC-Lab® software, DC levels of current or of potential are called I and E. The Va value has to be determined in such a way to ensure the system is linear in order to obtain significant EIS results .",
null,
"###### Figure 3: “Parameters Settings” window of PEIS experiment.\n\nFirstly, a Va value of 0.5 mV was chosen. The obtained EIS diagram is given in Figure 4.",
null,
"###### Figure 4: EIS diagram obtained with the following experimental conditions: Va = 0.5 mV, pw = 0, Na = 1 and no drift correction.\n\nConsidering the poor quality of this EIS diagram, we should analyze current and potential amplitude values, which are significant for EIS measurements (Figure 5 & Figure 6).\nIndeed, the value, which is the open circuit potential of the battery, is not significant for the EIS measurement, but as expected, this value is maintained during the measurement.\nThe value of |Ewe|, which is the modulus of the applied potential modulation is very small, only 0.5 mV (Figure 5). Considering the specifications of the instrument in impedance, signals smaller than 1 mV present the same level as the noise. At this point, in this measurement, the value of potential amplitude (|Ewe|) is included in the measurement noise that explains the poor quality of the EIS diagram.",
null,
"###### Figure 5: Change of potential amplitude IEweI with frequency.\n\nThe value of |I| is small but in good agreement with the accuracy of the instrument (Figure 6).",
null,
"###### Figure 6: Change of current amplitude III with frequency.\n\nIn this experiment, the critical parameter is the amplitude of potential. Note however, that the current and/or potential can be critical parameters depending on the studied system.\nTo improve the EIS diagram, the Va value has to be changed. Considering the previous results, it seems that an increase of the Va value could help to obtain a good quality EIS diagram.\nFigure 7 gives the EIS diagram obtained with 10 mV as Va value.",
null,
"###### Figure 7: EIS diagram obtained with the following experimental conditions: Va = 10 mV, pw = 0, Na = 1 and no drift correction.\n\nThis time, the value of the potential amplitude (|Ewe|) is significant and in agreement with the accuracy of the instrument.\nThe influence of the increase of the Va value can clearly be seen on the comparison between EIS diagrams obtained with Va = 0.5 mV and with Va = 10 mV as shown in Figure 9.",
null,
"",
null,
"###### Figure 8: Top: Change of potential amplitude |Ewe| with frequency. Bottom: Change of current amplitude |I| with frequency.",
null,
"###### Figure 9: Comparison of EIS diagrams obtained with a potential excitation Va of 0.5 mV (purple curve) and of 10 mV (red curve).\n\nThe pw value\nThe pw box offers the user the ability to add a delay before the measurement at each frequency. This delay is defined as a fraction of the period. In other words, this delay offers users the ability to let the system come out of a transient period resulting from the frequency transition and back to a steady-state sinusoidal behavior. This parameter is important especially for systems with a large time constant.\nThe aim of this part of the measurement is to show the effect of pw value on the EIS diagram in comparison to a well-defined diagram (for Va = 10 mV). Figure 10 shows a comparison between two EIS diagrams obtained with two values of pw (0 and 1) and Va = 0.5 mV and one diagram obtained with Va = 10 mV and pw = 0.",
null,
"###### Figure 10: Comparison of EIS diagrams obtained with Va = 0.5 mV, Na = 1 with no drift correction and a value of pw equal to 0 (purple curve) or 1 (blue curve) with EIS diagram obtained with Va = 10 mV (red curve).\n\nFirstly, we can see that the diagram obtained with the value of pw = 1 is less noisy than the one obtained with pw = 0. Moreover, this diagram obtained with pw = 1 is almost superimposed with the “correct” EIS diagram obtained with Va = 10 mV.\nThis result means that it is possible to compensate slightly for a noisy shape of an EIS diagram just by increasing the pw value and without disturbing the cell much. This result is in agreement with a high time constant of the system. Of course, this increases the experiment time. For example in this experiment for a value of pw of 0, a time of 6 s is needed for one scan, whereas when the value of pw is 1 a time of 13 s is needed for one scan.\n\nThe Na value\nNa is the number of repetitions of measurements at each frequency, and then an average is carried out for each frequency. This means that the noise is reduced following the mathematical law .\nIn the following section, two values of Na were tested 1 and 36, the other parameters are equal: Va = 0.5 mV, pw = 0, no drift correction, separate connection. The full frequency sweep is repeated 15 times to show the data point dispersion.\nFigure 11 shows the results obtained with Na = 1. It is possible to see that for each frequency the recorded points are not superimposed. It is clearly visible on the enlargement.\nFigure 12 shows the experiment done with Na = 36. Obviously, for each frequency, recorded points are superimposed, meaning that the coordinates of each recorded point are not more affected by the surrounding noise.",
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"",
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"###### Figure 11: Repetition of EIS diagrams obtained with the following experimental conditions: Va = 0.5 mV, pw = 0, Na = 1 and no drift correction. Top: complete diagram, bottom: enlargement.\n\nDrift correction\nThe drift correction tool is especially dedicated to systems with very long relaxation times. Indeed, the unsteadiness of systems can induce a slight deviation on the obtained impedance graphs compared with a theoretical one. This tool is explained extensively in reference .\n\n## Conclusion\n\nGreat care should be taken to ensure optimized EIS results are obtained. Indeed, each parameter that is not well defined can have a huge influence on the final result as demonstrated by this application note. Therefore, before carrying out a measurement, experimental conditions have to be defined with care, in agreement with the studied system and the software in use. A good compromise between significant results and acceptable experiment times must be found.\nThis note gives an example of measurements made for on Li-ion batteries; however, the same precautions can be used with other systems.",
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"###### Figure 12: Repetition of EIS diagrams obtained with the following experimental conditions: Va = 0.5 mV, pw = 0, Na = 36 and no drift correction. Top: complete diagram, bottom: enlargement.\n\nData files can be found in :\nC:\\Users\\xxx\\Documents\\EC-Lab\\Data\\Samples\\Corrosion\\battery 10Ah_Va=xmV_pw=x_Na=x_\ndrift?_connection?_repetition?_12\n\n## References\n\n1. Application Note #5 “Precautions for good impedance measurements\n2. Application Note #9 “Linear vs. non linear systems in impedance measurements\n3. Application Note #17 ”Drift correction in electrochemical impedance measurements\n\nRevised in 08/2019\n\n## Related products\n\nWork smarter. Not harder.\n\nTech-tips, theory, latest functionality, new products & more.\n\nNo thanks!"
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https://www.cfm.brown.edu/people/dobrush/am33/Mathematica/taylor.html | [
"# Preface\n\nThis is a tutorial made solely for the purpose of education and it was designed for students taking Applied Math 0330. It is primarily for students who have very little experience or have never used Mathematica before and would like to learn more of the basics for this computer algebra system. As a friendly reminder, don't forget to clear variables in use and/or the kernel.\n\nFinally, the commands in this tutorial are all written in bold black font, while Mathematica output is in normal font. This means that you can copy and paste all commands into Mathematica, change the parameters and run them. You, as the user, are free to use the scripts for your needs to learn the Mathematica program, and have the right to distribute this tutorial and refer to this tutorial as long as this tutorial is accredited appropriately.\n\nThe most important single result in numerical computations is Taylor's theorem, which we now state below.\n\nTheorem: Let f(x) have n+1 continuous derivatives on the closed interval [a.b] for some positive n, and let $$x, x_0 \\in [a,b] .$$ Then\n\n$f(x) = p_n + R_n (x)$\nfor\n$p_n (x) = f(x_0 ) + f' (x_0 ) \\,(x-x_0 ) + \\cdots + \\frac{f^{(n)} (x_0 )}{n!} \\, (x- x_0 )^n = \\sum_{k=0}^n \\frac{(x- x_0 )^k}{k!} \\, f^{(k)} (x_0 )$\nand\n$R_n (x) = \\frac{1}{n!} \\int_{x_0}^x (x-t)^n (t)\\, {\\text d}t = \\frac{(x- x_0 )^{n+1}}{(n+1)!} \\,f^{(n+1)} (\\xi_x ) ,$\nwhere ξ is a point between x and x0. ■\n\nBrook Taylor (1685--1731) was educated at St. John's College of Cambridge University (England), from which he graduated in 1709. He wrote two very significant books, ‘Methodus incrementorum directa et inversa’ and ‘Linear Perspective,’ which were published in 1715. He was the one to invent ‘Integration of Parts’ and also a series called the ‘Taylor’s Expansion.’ The Taylor’s Theorem is based on the letter written by him to Machin in which he tells about the origination of this idea. Although it appears that he did not appreciate its importance and he certainly did not bother with a proof of Taylor's theorem, which was discovered and published in 1694 by Johann Bernoulli (1667--1748). Taylor acknowledged that his work was based on that of Newton and Kepler. However, Johann Bernoulli claimed his priority in discovering both of his main results (Taylor's expansion and integration by parts).\n\n# Polynomial Approximations\n\nThe Taylor series method is of general applicability, and it is a standard to which we compare the accuracy of the various other numerical methods for solving an initial value problem for ordinary differential equations:\n$y' = f(x,y ) , \\quad y(x_0 ) = y_0 .$\nIn what follows, it will be assumed that the given slope function f(x,y) has as many derivatives in a neighborhood of the initial point as needed. For simplicity, we consider the set of uniformly distributed grid points starting from x0: $$x_n = x_0 + n\\,h , \\quad n=0,1,2,\\ldots .$$\n\nTheorem: Suppose that a function y(t) has m+1 continuous derivatives on the interval [a,b] containing grid points $$\\{ x_n \\}_{n\\ge 0} .$$ Assume that y(t) has a Taylor series expansion of order m about fixed value $$x_n \\in [a,b] :$$\n\n$y(x_n +h) = y(x_n ) + h\\, \\Phi_m \\left( x_n, y(x_n )\\right) + O\\left( h^{m+1} \\right) ,$\nwhere the increment function is\n$\\Phi_m \\left( x_n, y(x_n )\\right) = \\sum_{j=1}^m \\frac{y^{(j)} (x_n )}{j!} \\, h^{j-1}$\nand\n$y^{(k)} (x) = \\left( \\frac{\\partial}{\\partial x} + f\\,\\frac{\\partial}{\\partial y} \\right)^{k-1}\\, f\\left( x, y(x) \\right) , \\qquad k=0,1,2,\\ldots .$\nIn particular,\n\\begin{align*} y' (x) &= f(x,y(x)) , \\\\ y'' (x) &= f_x + f_y \\, y' = f_x + f\\, f_y , \\\\ y^{(3)} (x)&= f_{xx} + 2\\, f_{xy} y' + f_y y'' + f_{yy} \\left( y' \\right)^2 \\\\ &= f_{xx} + 2\\,f_{xy}\\,f + f_{yy} f^2 + f_y \\left( f_x + f\\, f_y \\right) , \\\\ y^{(4)} (x)&= f_{xxx} + 3\\, f_{xxy} y' +3\\, f_{xyy} \\left( y' \\right)^2 +3\\, f_{xy} y'' + f_y y''' +3\\, f_{yy} y' y'' + f_{yyy} \\left( y' \\right)^3 \\\\ &= f_{xxx} + 3\\, f_{xxy} \\,f +3\\, f_{xyy} \\left( f \\right)^2 +3\\, f_{xy} \\left( f_x + f\\, f_y \\right) + f_{yy} f\\left( f_x + f\\, f_y \\right) + f_y y''' . \\qquad ■ \\end{align*}\n\nThe approximate numerical solution to the initial value problem $$y' = f(x,y ) , \\quad y(x_0 ) = y_0$$ over the interval [a,b] is derived by using the Taylor series expansion on each subinterval $$[ x_n , x_{n+1}] .$$ The general step for Taylor's method of order m is\n\n$y_{n+1} = y_n + h\\,y' (x_n ) + \\frac{h^2}{2} \\,y'' (x_n ) + \\cdots + \\frac{h^m}{m!} \\, y^{(m)} (x_n ).$\nThe Taylor method of order m has the property that the final global error is of the order $$O\\left( h^{m+1} \\right) .$$ It is possible theoretically to choose m as large as necessary to make this error as small as desired. However, in practice, we usually compute two sets of approximations using step sizes h and h/2 and compare the results.\n\n## II. Second Order Polynomial Approximation\n\nSince the first order Taylor series approximation is identical with Euler’s method, we start with the second order one:\n\n$y_{n+1} = y_n + h\\,f (x_n , y_n ) + \\frac{h^2}{2} \\left[ f_x (x_n , y_n ) + f(x_n , y_n )\\, f_y (x_n , y_n ) \\right] = y_n + h\\, \\Phi_2 (h) ,$\nwhere the increment function Φ2 is just adding the second order differential deviation to the next term in the same equation used for the first order Taylor expansion. We show how it works in the following examples.\n\nExample: Consider the initial value problem $$y’=1/(2x-3y+5), \\quad y(0)=1$$ and let us try to find out y(1).\n\nTo achieve this, we pick up a fixed step length h = 0.1, which means that we need to do 10 iterations to reach the end point x = 1.\n\nClear[y]\nh:=0.1;\nf1[x_, y_] := 1 / (2 x - 3 y + 5)\nfx[x_, y_] := D[f1[a, b], a] /. {a -> x, b -> y}\nfy[x_, y_] := D[f1[a, b], b] /. {a -> x, b -> y}\nf2[x_, y_] := fx[x, y] + fy[x, y] * f1[x, y]\ny = 1;\nDo[y[n + 1] = y[n] + h * f1[h n, y[n]] + 1 / 2 * h^2 * f2[h n, y[n]], {n, 0, 9}]\ny\nOut= 1.43529\n\nAt each step in the sequence, the expansion requires 3 addition, 1 subtraction, 3 multiplication, and 1 division operations. Since there are 8 steps to this sequence, in order to obtain the desired approximation, the code performs 80 operations, which would be hectic work to do by hand. The approximation is y10 = 1.435290196. This algorithm requires at least 22 operations per step, which means that the entire sequence requires 220 steps. However, this is not significant since Mathematica does all the work.\n\nTo check the answer, we write a subroutine to evaluate numerically the initial value problem for arbitrary slope function, initial conditions, and step size:\n\ntaylor2[f_, {x_, x0_, xn_}, {y_, y0_}, stepsize_] :=\nBlock[{xold = x0, yold = y0, sollist = {{x0, y0}}, h},\nh = N[stepsize];\nDo[xnew = xold + h;\nfold = f /. {x -> xold, y -> yold};\nPhi]2 = fold + (h/2)*((D[f, x] /. {x -> xold, y -> yold}) + (D[f, y] /. {x -> xold, y -> yold})*(f /. {x -> xold, y -> yold})); ynew = yold + h*\\[Phi]2; sollist = Append[sollist, {xnew, ynew}]; xold = xnew; yold = ynew, {(xn - x0)/stepsize}]; Return[sollist]] Now we apply this subroutine to our slope function: taylor2[1/(2*x - 3*y + 5), {x, 0, 1}, {y, 1}, .1] // Flatten Out= {0, 1, 0.1, 1.04938, 0.2, 1.09747, 0.3, 1.14427, 0.4, 1.18976, 0.5, 1.23393, 0.6, 1.27678, 0.7, 1.31833, 0.8, 1.35858, 0.9, 1.39756, 1., 1.43529} Using Mathematica command NDSolve, we check the obtained value: sol = NDSolve[{y'[x] == 1/(2*x - 3*y[x] + 5), y == 1}, y[x], {x, 0, 2}] Out= {{y[x] -> InterpolatingFunction[{{0., 2.}}<>][x]}} z = sol /. x -> sol /. x -> 1.0 Out= {{y[{{y[1.] -> 1.43529}}] -> {{InterpolatingFunction[0.,2.}},<>][y[1.] -> 1.43529]}}}} Example: Let us take a look at the initial value problem $$y’=(x^2 -y^2) \\sin (y), \\quad y(0)=1 ,$$ for which it is impossible to find the actual solution. First we find two derivatives: \\begin{align*} y' &= f(x,y) = \\left( x^2 - y^2 \\right) \\sin (y) , \\\\ y'' &= 2\\left( x- y\\, y' \\right) \\sin (y) + \\left( x^2 - y^2 \\right) y' \\, \\cos (y) . \\end{align*} We check the formulas with Mathematica: f[x_, y_] = (x^2 + y^2)*Sin[y] f2[x_,y_] = D[f[x,y],x] + D[f[x,y],y] f[x,y] Then we ask Mathematica to do all calculations y = 1; x = 0; h = 0.1; n = Floor[(1 - 0)/h] Solution = taylor2[(x^2 + y^2)*Sin[y], {x, 0, 1}, {y, 1}, .1] Out= {0, 1, 0.1, 1.0935, 0.2, 1.21486, 0.3, 1.37875, 0.4, 1.60729, 0.5, 1.92811, 0.6, 2.34585, 0.7, 2.75325, 0.8, 2.97922, 0.9, 3.06953, 1., \\ 3.10788} We compare the second order Taylor series approximation y10 = 3.10788 with those obtained with the aid of NDSolve (one needs to clear variables before invoking function NDSolve): sol = NDSolve[{y'[x] == (x^2 + y[x]^2)*Sin[y[x]], y == 1}, y[x], {x, 0, 2}] z = sol /. x -> sol /. x -> 1.0 Out= {{y[{{y[1.] -> 3.11958}} So we see that Taylor's approximation gives two correct significant digits. Now we use another approach involving NestList command. f[x_,y_]= (x^2 + y^2)*Sin[y] f2[x_,y_] = D[f[x,y],x] + D[f[x,y],y] f[x,y] h = 0.1 euler2[{x_, y_}] = {x + h, N[y + h*f[x, y] + (h^2/2)*f2[x, y]]} Expand[euler2[{x,y}]] NestList[euler2, {0, 1}, 10] // Flatten Out= {0, 1, 0.1, 1.0935, 0.2, 1.21486, 0.3, 1.37875, 0.4, 1.60729, 0.5, 1.92811, 0.6, 2.34585, 0.7, 2.75325, 0.8, 2.97922, 0.9, 3.06953, 1., 3.10788} Here is the direct application of polynomial approximation with many loops. f[x_,y_]= (x^2 + y^2)*Sin[y] f1[x_,y_]=D[f[x,y],x]+f[x,y]*D[f[x,y],y] y = 1; h =0.1; a = 0; b = 1; n = Floor[(b-a)/h]; Do[x[i] = 0.0 + (i-1)*h, {i,1,n+1}] Do[{k1=f1[x[i],y[i]], y[i+1] = y[i] + h*f[x[i],y[i]] +h*h/2*k1}, {i,1,n}] Do[Print[x[i], \" \", y[i]],{i,1,n+1}] ## III. Third Order Polynomial Approximation The third order Taylor approximation is adding a third order differential deviation to the equation for the 2nd order expansion. \\[ y_{n+1} = y_n + h\\,f (x_n , y_n ) + \\frac{h^2}{2} \\ y'' (_n ) + \\frac{h^3}{3!}\\, y''' (x_n ) = y_n + h\\, \\Phi_3 (h) ,\nwhere the increment function Φ3 adds just one term to Φ2. We will show how it works in two examples that were considered previously.\n\nExample: We reconsider the IVP: $$y’=1/(2x-3y+5), \\quad y(0)=1 .$$ To use the third order polynomial approximation, we choose a uniform grid with constant step length h = 0.1.\n\nHere is the syntax for the third order Taylor approximation (which takes too long to execute):\n\nClear[y]\nf1[x_, y_] := 1 / (2 x - 3 y + 5)\nfx[x_, y_] := D[f1[a, b], a] /. {a -> x, b -> y}\nfy[x_, y_] := D[f1[a, b], b] /. {a -> x, b -> y}\nf2[x_, y_] := fx[x, y] + fy[x, y] * f1[x, y]\nfxx[x_, y_] := D[fx[a, b], a] /. {a -> x, b -> y}\nfxy[x_, y_] := D[fx[a, b], b] /. {a -> x, b -> y}\nfyy[x_, y_] := D[fy[a, b], b] /. {a -> x, b -> y}\nf3[x_, y_] := fxx[x, y] + 2 * fxy[x, y] * f1[x, y] +\nfyy[x, y] * (f1[x, y])^2 + fx[x, y] * fy[x, y] + (fy[x, y])^2 * f1[x, y]\ny = 1;\nDo[y[n + 1] = y[n] + h * f1[h n, y[n]] +\n1 / 2 * h^2 * f2[h n, y[n]] + 1 / 6 * h^3 * f3[h n, y[n]], {n, 0, 9}]\ny\nOut= 1.43528\nThe approximation for this value is y10=1.435283295 while the true value is 1.4352866048707.\n\nThe total number of numerical operations here is 420. It is obvious at this point why using mathematical programs such as Mathematica is a desired approach for such problems.\n\nExample: We reconsider the IVP: $$y’=(x^2 -y^2) \\sin (y), \\quad y(0)=1 .$$ To use the third order polynomial approximation, we choose a uniform grid with constant step length h = 0.1.\n\ntaylor3[f_, {x_, x0_, xn_}, {y_, y0_}, stepsize_] :=\nBlock[{xold = x0, yold = y0, sollist = {{x0, y0}}, h},\nh = N[stepsize];\nDo[xnew = xold + h;\nfold = f /. {x -> xold, y -> yold};\n\\[Phi]2 =\nfold + (h/2)*((D[f, x] /. {x -> xold,\ny -> yold}) + (D[f, y] /. {x -> xold,\ny -> yold})*(f /. {x -> xold, y -> yold}));\n\\[Alpha] = (D[f, {x, 2}] /. {x -> xold, y -> yold}) +\n2*(D[f, x, y] /. {x -> xold, y -> yold})*(f /. {x -> xold,\ny -> yold}) + (D[f, {y, 2}] /. {x -> xold,\ny -> yold})*(f /. {x -> xold, y -> yold})^2 + (D[f,\nx] /. {x -> xold, y -> yold})*(D[f, y] /. {x -> xold,\ny -> yold}) + (D[f, y] /. {x -> xold,\ny -> yold})^2*(f /. {x -> xold, y -> yold});\nynew = yold + h*\\[Phi]2 + h^3/3!*\\[Alpha];\nsollist = Append[sollist, {xnew, ynew}];\nxold = xnew;\nyold = ynew, {(xn - x0)/stepsize}];\nReturn[sollist]]\n\nSolution = taylor3[(x^2 + y^2)*Sin[y], {x, 0, 1}, {y, 1}, .1]\nOut= {{0, 1}, {0.1, 1.09483}, {0.2, 1.21862}, {0.3, 1.38695}, {0.4, 1.62303}, {0.5, 1.95295},\n{0.6, 2.36595}, {0.7, 2.73957}, {0.8, 2.9658}, {0.9, 3.07578}, {1., 3.12003}}\n\nAnother example of code (with the same output):\n\nf[x_,y_]=(x^2 + y^2)*Sin[y]\nf1[x_,y_]=D[f[x,y],x]+f[x,y]*D[f[x,y],y]\nf2[x_,y_]=D[f[x,y],x,x]+2*f[x,y]*D[f[x,y],x,y] + D[f[x,y],y,y]*f[x,y]*f[x,y] + D[f[x,y],x]*D[f[x,y],y] +f[x,y]*(D[f[x,y],y])^2\ny = 1; h =0.1;\na = 0; b = 1; n = Floor[(b-a)/h];\nDo[x[i] = 0.0 + (i-1)*h, {i,1,n+1}]\n\nDo[{k1=f1[x[i],y[i]],\nk2=f2[x[i],y[i]],\ny[i+1] = y[i] + h*f[x[i],y[i]] +h*h/2*k1 + k2/6*h^3}, {i,1,n}]\nDo[Print[x[i], \" \", y[i]],{i,1,n+1}]"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.83131754,"math_prob":0.99989593,"size":9281,"snap":"2021-31-2021-39","text_gpt3_token_len":3273,"char_repetition_ratio":0.1238547,"word_repetition_ratio":0.27878788,"special_character_ratio":0.39887944,"punctuation_ratio":0.20175055,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99999046,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-31T22:43:48Z\",\"WARC-Record-ID\":\"<urn:uuid:fcd7d03c-aaae-4bae-8166-fbe7feeb824f>\",\"Content-Length\":\"32954\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c366890d-6e61-47c5-85c8-83e776a76227>\",\"WARC-Concurrent-To\":\"<urn:uuid:8b29b22d-a2aa-4fd6-ad9b-46b4ea3771df>\",\"WARC-IP-Address\":\"128.148.254.15\",\"WARC-Target-URI\":\"https://www.cfm.brown.edu/people/dobrush/am33/Mathematica/taylor.html\",\"WARC-Payload-Digest\":\"sha1:MJ5OHWAGNO24D25NI4B5OCPHWYBUK5J2\",\"WARC-Block-Digest\":\"sha1:OJ6OX526SRQG3RSXIBZOHXIQ2BN2C4FZ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154126.73_warc_CC-MAIN-20210731203400-20210731233400-00636.warc.gz\"}"} |
https://www.fincover.com/banking/loan/emi-calculator/ | [
"",
null,
"Personal Loan Car Loan Home Loan Business Loan Bike Loan Gold Loan\n\nChart by Visualizer\n\nChart by Visualizer\n\nChart by Visualizer\n\nChart by Visualizer\n\nChart by Visualizer\n\nChart by Visualizer\n\n## How EMI Calculator Works\n\nYou can calculate your EMI with the help of the EMI Calculator. All you have to do is to input details like Tenure, Principal, and Interest rate to get your EMI.\n\n#### Principal\n\nThe loan amount borrowed. You should enter the loan amount you borrowed in the space provided for Principal.\n\n#### Rate of Interest\n\nThe borrower lends the loan amount at a rate of interest. You should compare interest rates offered by different lenders for a particular type of loan before you apply for it.\n\n#### Tenure\n\nThe time given to repay the loan amount that you borrowed is referred to as tenure. Loan Tenure depends on the type of loan that you avail and the borrower.",
null,
"Calculate your Monthly EMIs in seconds\n\n# EMI Calculator\n\n#### What is an EMI?\n\nAn Equated Monthly Instalment, or EMI, is a bundled payment of interest on outstanding loan and repayment of part of the principal so that the same amount is paid every month for the three-year period. Your EMI will work out to about Rs 16,000 a month for three years. The beauty is the major portion of the Rs 16,000 will be interest in month 1, while in month 36, the major portion will be principal.\n\n### What are the advantages of EMI?\n\nEMIs are incredibly convenient as you are servicing the interest and repaying the principal methodically, and your monthly outgo is still stable and predictable.\nYou won’t be faced with random demands for due interest or the nearly impossible task of repaying a lump sum suddenly.\nEMIs help smoother access to financial products like home loans and car loans.\n\n## What is Loan EMI Calculator?\n\nEMI calculator is a tool that can calculate your monthly EMIs. With it you can quickly figure out your monthly instalments and plan your budget accordingly. Calculating your EMIs manually is cumbersome and complicated, while an EMI calculator can determine your EMIs within a few seconds.\n\nWhile applying for a loan, you can use this EMI calculator to keep in mind the monthly payment you can afford to make as it gives a clear breakdown of your repayment schedule.\n\nIn the case of loans with floating interest rates, EMIs will vary. Fincover’s online EMI Calculator will help you calculate your EMIs for Personal Loan, Car Loan, and Home Loan with different tenures so that you can choose the most appropriate loan.\n\n## Formula for calculating the EMI\n\nIt's the formula used by the EMI Loan Calculator to give you your EMIs in a matter of seconds.\n\n• EMI = [P x R x (1+R) ^N]/[(1+R) ^N-1]\n• P = Principal Loan amount\n• R = Rate of Interest\n• N = Loan Tenure\n\nWith this formula, the EMI Loan Calculator gives you your EMIs in a matter of seconds.",
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"https://www.facebook.com/tr",
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"https://www.fincover.com/wp-content/uploads/2020/11/emi.png",
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"https://www.fincover.com/wp-content/uploads/2020/11/emi-cal.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.93133754,"math_prob":0.8897997,"size":2195,"snap":"2023-40-2023-50","text_gpt3_token_len":505,"char_repetition_ratio":0.15107258,"word_repetition_ratio":0.020253165,"special_character_ratio":0.2164009,"punctuation_ratio":0.08275862,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95121473,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-01T08:12:11Z\",\"WARC-Record-ID\":\"<urn:uuid:4fe6bc76-6777-443a-84f7-b12790a24f85>\",\"Content-Length\":\"255982\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7b3b8128-de60-42f3-a392-f61b1b21ba10>\",\"WARC-Concurrent-To\":\"<urn:uuid:c46981cb-a4e3-4e9d-95da-7693ac383be1>\",\"WARC-IP-Address\":\"13.232.51.71\",\"WARC-Target-URI\":\"https://www.fincover.com/banking/loan/emi-calculator/\",\"WARC-Payload-Digest\":\"sha1:OLAX6SLUKTOEE4Y2M33Y4QE3KCI6XGZ5\",\"WARC-Block-Digest\":\"sha1:KZFZU4MTYRD7PP25CJEDBISVGOEERDRU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510810.46_warc_CC-MAIN-20231001073649-20231001103649-00380.warc.gz\"}"} |
https://nineplanets.org/percentage-calculator/what-is-percentage-difference-from-2856-to-2208/ | [
"# What is the percentage increase/decrease from 2856 to 2208?\n\n## Quickly work out the percentage increase or decrease from 2856 to 2208 in this step-by-step percentage calculator tutorial. (Spoiler alert: it's -22.69%!)\n\nSo you want to work out the percentage increase or decrease from 2856 to 2208? Fear not, intrepid math seeker! Today, we will guide you through the calculation so you can figure out how to work out the increase or decrease in any numbers as a percentage. Onwards!\n\nIn a rush and just need to know the answer? The percentage decrease from 2856 to 2208 is -22.69%.\n\nWhat is the % change from to\n\n## Percentage increase/decrease from 2856 to 2208?\n\nAn increase or decrease percentage of two numbers can be very useful. Let's say you are a shop that sold 2856 t-shirts in January, and then sold 2208 t-shirts in February. What is the percentage increase or decrease there? Knowing the answer allows you to compare and track numbers to look for trends or reasons for the change.\n\nWorking out a percentage increase or decrease between two numbers is pretty simple. The resulting number (the second input) is 2208 and what we need to do first is subtract the old number, 2856, from it:\n\n2208 - 2856 = -648\n\nOnce we've done that we need to divide the result, -648, by the original number, 2856. We do this because we need to compare the difference between the new number and the original:\n\n-648 / 2856 = -0.22689075630252\n\nWe now have our answer in decimal format. How do we get this into percentage format? Multiply -0.22689075630252 by 100? Ding ding ding! We have a winner:\n\n-0.22689075630252 x 100 = -22.69%\n\nWe're done! You just successfully calculated the percentage difference from 2856 to 2208. You can now go forth and use this method to work out and calculate the increase/decrease in percentage of any numbers.\n\nHead back to the percentage calculator to work out any more calculations you need to make or be brave and give it a go by hand. Hopefully this article has shown you that it's easier than you might think!"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.92250544,"math_prob":0.9134881,"size":1742,"snap":"2020-45-2020-50","text_gpt3_token_len":413,"char_repetition_ratio":0.16685846,"word_repetition_ratio":0.006535948,"special_character_ratio":0.2738232,"punctuation_ratio":0.10511363,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9918265,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-22T13:09:29Z\",\"WARC-Record-ID\":\"<urn:uuid:ccedeb40-8866-481f-b4e1-c23f7f5d3dd3>\",\"Content-Length\":\"26296\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:140cf71a-0138-4839-bcb4-6db383157d90>\",\"WARC-Concurrent-To\":\"<urn:uuid:fc865a8e-c434-4bea-a556-aa0ab743a1ae>\",\"WARC-IP-Address\":\"172.67.196.14\",\"WARC-Target-URI\":\"https://nineplanets.org/percentage-calculator/what-is-percentage-difference-from-2856-to-2208/\",\"WARC-Payload-Digest\":\"sha1:KB6NFFUEEI5R3LOGLVQB67PFNC5IEMGS\",\"WARC-Block-Digest\":\"sha1:FZXSXMQBNCZWL4RNXS243SJZSTWVPNNK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107879537.28_warc_CC-MAIN-20201022111909-20201022141909-00297.warc.gz\"}"} |
https://chemistry.stackexchange.com/questions/7955/which-ions-are-accounted-for-in-total-and-net-ionic-equations | [
"# Which ions are accounted for in total and net ionic equations?\n\nThere are a number of things I don't understand about ionic equations.\n\nFirst off, when you do net ionic equations, is it correct that you're only focusing on precipitates? The way I understood it, you put \"no reaction\" if there are no solid or gaseous products. To clarify, if all of the products of the reaction are aqueous, we look at it as if no reaction had taken place.\n\nThen I went and looked in my book, and that made it seem like all products that weren't covalent were spectator ions, i.e., covalent molecules were the only ones you note. Now I'm even more confused.\n\n• Welcome to Chemistry.SE! You are not the first one who has struggled with spectator ions, complete and net ionic equations. Please feel free to use the search function of this site. In addition, this video on net ionic equations might be helpful. Jan 22, 2014 at 6:21\n\nAfter reading into the matter (as a non-native English speaker, I've never heard of total and net ionic equations) here goes my small little guide to ionic equations:\n\n## Total Ionic Equations\n\nA total ionic equation is where you start with a balanced equation and write all the ionic compounds in their ionic form. For example, in the following reaction $$\\ce{2 Na3PO4 + 3 CaCl2 -> 6 NaCl + Ca3(PO4)2 v},$$ one of the products precipitates (noted by the downwards pointing arrow). So we will not write it in ionic form for the total ionic equation, which is the following: $$\\ce{6 Na+ + 2PO4^3- + 3Ca^2+ + 6 Cl- -> 6 Na+ + 6 Cl- + Ca3(PO4)2 v}.$$\n\n## Net Ionic Equations\n\nNet ionic equations can be derived from total ionic equations: We just eliminate the compounds that appear both on the right and on the left side of the reaction arrow. Thus we get for the above total ionic equation: $$\\ce{2 PO4^3- + 3 Ca^2+ -> Ca3(PO4)2 v}$$\n\nThe ions that were eliminated are termed spectator ions, because they do not participate in any notable chemical reaction.\n\nIf something is still unclear, maybe have a go at Kristy M. Bailey's website (Oklahoma City Community College), which explains it pretty neatly."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.92219263,"math_prob":0.98620385,"size":1190,"snap":"2022-40-2023-06","text_gpt3_token_len":326,"char_repetition_ratio":0.15092748,"word_repetition_ratio":0.0,"special_character_ratio":0.26638657,"punctuation_ratio":0.07826087,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98186606,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-25T20:11:10Z\",\"WARC-Record-ID\":\"<urn:uuid:953b3fd5-b496-474f-b74e-c618b5ed211a>\",\"Content-Length\":\"229085\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fec436f3-d515-4f69-be11-9cd75b0a25db>\",\"WARC-Concurrent-To\":\"<urn:uuid:536b7741-9d72-4bf3-b3ad-1c13a52689c1>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://chemistry.stackexchange.com/questions/7955/which-ions-are-accounted-for-in-total-and-net-ionic-equations\",\"WARC-Payload-Digest\":\"sha1:LQIADGUM5K6HVQS4R7HG3Z7RAVCMTNRT\",\"WARC-Block-Digest\":\"sha1:LIVF7MQYDYA2SLINACC6SQOUJ374D7XY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030334596.27_warc_CC-MAIN-20220925193816-20220925223816-00693.warc.gz\"}"} |
https://link.springer.com/article/10.1134%2FS0040577919100027 | [
"Theoretical and Mathematical Physics\n\n, Volume 201, Issue 1, pp 1426–1441\n\n# Solution Space Monodromy of a Special Double Confluent Heun Equation and Its Applications\n\nArticle\n\n## Abstract\n\nWe consider three linear operators determining automorphisms of the solution space of a special double confluent Heun equation of positive integer order (L-operators). We propose a new method for describing properties of the solution space of this equation based on using eigenfunctions of one of the L-operators, called the universal L-operator. We construct composition laws for L-operators and establish their relation to the monodromy transformation of the solution space of the special double confluent Heun equation. We find four functionals quadratic in eigenfunctions of the universal automorphism; they have a property with respect to the considered equation analogous to the property of the first integral. Based on them, we construct matrix representations of the L-operators and also the monodromy operator. We give a method for extending solutions of the special double confluent Heun equation from the subset Re z > 0 of a complex plane to a maximum domain on which the solution exists. As an example of its application to the RSJ model theory of overdamped Josephson junctions, we give the explicit form of the transformation of the phase difference function induced by the monodromy of the solution space of the special double confluent Heun equation and propose a way to continue this function from a half-period interval to any given interval in the domain of the function using only algebraic transformations.\n\n## Keywords\n\ndouble confluent Heun equation solution space automorphism monodromy composition law matrix representation solution continuation RSJ model of Josephson junction\n\n## References\n\n1. 1.\nR. Foote, “Geometry of the Prytz planimeter,” Rep. Math. Phys., 42, 249–271 (1998); arXiv:math/9808070v1 (1998).\n2. 2.\nM. Levi and S. Tabachnikov, “On bicycle tire tracks geometry, hatchet planimeter, Menzin’s conjecture, and oscillation of unicycle tracks,” Experiment. Math., 18, 173–186 (2009).\n3. 3.\nR. L. Foote, M. Levi, and S. Tabachnikov, “Tractrices, bicycle tire tracks, hatchet planimeters, and a 100-year-old conjecture,” Amer. Math. Monthly, 120, 199–216 (2013).\n4. 4.\nG. Bor, M. Levi, R. Perline, and S. Tabachnikov, “Tire tracks and integrable curve evolution,” arXiv: 1705.06314v3 [math.DG] (2017).Google Scholar\n5. 5.\nJ. Guckenheimer and Yu. S. Ilyashenko, “The duck and the devil: Canards on the staircase,” Moscow Math. J., 1, 27–47 (2001).\n6. 6.\nV. A. Kleptsyn, O. L. Romaskevich, and I. V. Shchurov, “Josephson effect and fast–slow systems [in Russian],” Nanostrukt. Matem. Fiz. i Modelir., 8, 31–46 (2013).Google Scholar\n7. 7.\nW. C. Stewart, “Current–voltage characteristics of Josephson junctions,” Appl. Phys. Lett., 12, 277–280 (1968).\n8. 8.\nD. E. McCumber, “Effect of ac impedance on dc voltage–current characteristics of superconductor weak-link junctions,” J. Appl. Phys., 39, 3113–3118 (1968).\n9. 9.\nP. Mangin and R. Kahn, Superconductivity: An Introduction, Springer, New York (2017).\n10. 10.\nV. M. Buchstaber, O. V. Karpov, and S. I. Tertychnyi, “Rotation number quantization effect,” Theor. Math. Phys., 162, 211–221 (2010).\n11. 11.\nYu. S. Ilyashenko, D. A. Ryzhov, and D. A. Filimonov, “Phase-lock effect for equations modeling resistively shunted Josephson junctions and for their perturbations,” Funct. Anal. Appl., 45, 192–203 (2011).\n12. 12.\nA. Glutsyuk and L. Rybnikov, “On families of differential equations on two-torus with all phase-lock areas,” Nonlinearity, 30, 61–72 (2017).\n13. 13.\nG. V. Osipov and A. V. Polovinkin, Synchronization by an External Periodic Action [in Russian], Nizhny Novgorod State Univ. Press, Nizhny Novgorod (2005).Google Scholar\n14. 14.\nV. M. Buchstaber and A. A. Glutsyuk, “Josephson effect, Arnold tongues, and double confluent Heun equations,” Talk at Intl. Conf. “Contemporary Mathematics,” dedicated to the 80th birthday of V. I. Arnold, Higher School of Economics, Skolkovo Inst. of Science and Technology, Steklov Math. Inst., Moscow, 18–23 December 2017 (2017).Google Scholar\n15. 15.\nS. I. Tertychniy, “Long-term behavior of solutions to the equation ø+sin ø = f with periodic f and the modeling of dynamics of overdamped Josephson junctions: Unlectured notes,” arXiv:math-ph/0512058v1 (2005).Google Scholar\n16. 16.\nV. M. Buchstaber and S. I. Tertychnyi, “Dynamical systems on a torus with identity Poincaré map which are associated with the Josephson effect,” Russian Math. Surveys, 69, 383–385 (2014).\n17. 17.\nS. I. Tertychniy, “The interrelation of the special double confluent Heun equation and the equation of RSJ model of Josephson junction revisited,” arXiv:1811.03971v1 [math-ph] (2018).Google Scholar\n18. 18.\nD. Schmidt and G. Wolf, “Double confluent Heun equation,” in: Heun’s Diffrential Equations (A. Ronveaux, ed.), Oxford Univ. Press, Oxford (1995), pp. 129–188.Google Scholar\n19. 19.\nS. Yu. Slavyanov and W. Lay, Special Functions: A Unified Theory Based on Singularities, Oxford Univ. Press, Oxford (2000).\n20. 20.\nThe Heun Project, “Heun functions, their generalizations and applications: Bibliography,” http://theheunproject.org/bibliography.html (2017).Google Scholar\n21. 21.\nM. Horta¸csu, “Heun functions and some of their applications in physics,” Adv. High Energy Phys., 2018, 8621573 (2018); arXiv:1101.0471v11 [math-ph] (2011).Google Scholar\n22. 22.\nV. M. Buchstaber and A. A. Glutsyuk, “On phase-lock areas in a model of Josephson effect and double confluent Heun equations,” Talk at Intl. Conf. “Real and Complex Dynamical Systems,” dedicated to the 75th anniversary of Yu. S. Il’yashenko, Steklov Math. Inst., Moscow, 30 November 2018 (2018).\n23. 23.\nV. M. Buchstaber and S. I. Tertychnyi, “Explicit solution family for the equation of the resistively shunted Josephson junction model,” Theor. Math. Phys., 176, 965–986 (2013).\n24. 24.\nA. A. Glutsyuk, “On constrictions of phase-lock areas in model of overdamped Josephson effect and transition matrix of the double-confluent Heun equation,” J. Dyn. Control. Syst., 25, 323–349 (2019).\n25. 25.\nV. M. Buchstaber and S. I. Tertychnyi, “Holomorphic solutions of the double confluent Heun equation associated with the RSJ model of the Josephson junction,” Theor. Math. Phys., 182, 329–355 (2015).\n26. 26.\nV. M. Buchstaber and A. A. Glutsyuk, “On determinants of modified Bessel functions and entire solutions of double confluent Heun equations,” Nonlinearity, 29, 3857–3870 (2016); arXiv:1509.01725v4 [math.DS] (2015).\n27. 27.\nY. Bibilo, “Josephson effect and isomonodromic deformations,” arXiv:1805.11759v2 [math.CA] (2018).Google Scholar\n28. 28.\nA. A. Glutsyuk, V. A. Kleptsyn, D. A. Filimonov, and I. V. Shchurov, “On the adjacency quantization in an equation modeling the Josephson effect,” Funct. Anal. Appl., 48, 272–285 (2014).\n29. 29.\nV. M. Buchstaber and A. A. Glutsyuk, “On monodromy eigenfunctions of Heun equations and boundaries of phase-lock areas in a model of overdamped Josephson effect,” Proc. Steklov Inst. Math., 297, 50–89 (2017).\n30. 30.\nA. A. Salatic and S. Yu. Slavyanov, “Antiquantization of the double confluent Heun equation: The Teukolsky equation,” Russ. J. Nonlinear Dyn., 15, 79–85 (2019).\n31. 31.\nS. Yu. Slavyanov, “Isomonodromic deformations of Heun and Painlevé equations,” Theor. Math. Phys., 123, 744–753 (2000).\n32. 32.\nS. Yu. Slavyanov and O. L. Stesik, “Antiquantization of deformed Heun-class equations,” Theor. Math. Phys., 186, 118–125 (2016).\n33. 33.\nS. I. Tertychniy, “Square root of the monodromy map for the equation of RSJ model of Josephson junction,” arXiv:1901.01103v3 [math.CA] (2019).Google Scholar\n34. 34.\nV. M. Buchstaber and S. I. Tertychnyi, “Automorphisms of the solution spaces of special double-confluent Heun equations,” Funct. Anal. Appl., 50, 176–192 (2016).\n35. 35.\nV. M. Buchstaber and S. I. Tertychnyi, “Representations of the Klein group determined by quadruples of polynomials associated with the double confluent Heun equation,” Math. Notes, 103, 357–371 (2018)."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.73270833,"math_prob":0.87651914,"size":8872,"snap":"2019-43-2019-47","text_gpt3_token_len":2570,"char_repetition_ratio":0.1865133,"word_repetition_ratio":0.0558799,"special_character_ratio":0.27434626,"punctuation_ratio":0.26126125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9860362,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-19T10:39:48Z\",\"WARC-Record-ID\":\"<urn:uuid:3770a9f6-d3db-4005-aca3-5db6629752c5>\",\"Content-Length\":\"104471\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8b453e06-be7c-4a7b-9f43-4dbd73071a10>\",\"WARC-Concurrent-To\":\"<urn:uuid:b2961097-07b4-445f-9de0-f7c00f623b5b>\",\"WARC-IP-Address\":\"151.101.248.95\",\"WARC-Target-URI\":\"https://link.springer.com/article/10.1134%2FS0040577919100027\",\"WARC-Payload-Digest\":\"sha1:6RPQJTMTOHXTST24DFIYRQZNXVZG3RZX\",\"WARC-Block-Digest\":\"sha1:MAGCCPVZLYBUW652FTQOXSFWHWYOH5NJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670135.29_warc_CC-MAIN-20191119093744-20191119121744-00479.warc.gz\"}"} |
https://uk.mathworks.com/matlabcentral/answers/539623-find-max-value-position | [
"# Find max value position\n\n10 views (last 30 days)\nGonzalo Guerrero on 1 Jun 2020\nHi,\nI am trying to sort out a problem where I got 210 rows in 1 column, which variable is called 'Force1' and I need to find the position of the max value. I tried the next code:\nx=find(max(Force1))\nThe actual value is in row 170, however, matlab gives me x=1.\nThen I tried to apply this one:\nx=find(max(Force1))\nand result is x= to 210 rows where first one is the max value and then 0\nThank you\nGonzalo Gomez Guerrero\n\nKALYAN ACHARJYA on 1 Jun 2020\nEdited: KALYAN ACHARJYA on 1 Jun 2020\nx=find(Force1==max(Force1))\n\nmadhan ravi on 1 Jun 2020\nWhy is an additional transpose needed?\nKALYAN ACHARJYA on 1 Jun 2020"
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https://cran.case.edu/web/packages/breakDown/vignettes/break_lm.html | [
"# breakDown plots for the linear models\n\n#### 2020-04-04\n\nHere we will use the wine quality data (archive.ics.uci.edu/ml/machine-learning-databases/wine-quality/winequality-white.csv) to present the breakDown package for lm models.\n\nlibrary(\"breakDown\")\n#> fixed.acidity volatile.acidity citric.acid residual.sugar chlorides\n#> 1 7.0 0.27 0.36 20.7 0.045\n#> 2 6.3 0.30 0.34 1.6 0.049\n#> 3 8.1 0.28 0.40 6.9 0.050\n#> free.sulfur.dioxide total.sulfur.dioxide density pH sulphates alcohol\n#> 1 45 170 1.0010 3.00 0.45 8.8\n#> 2 14 132 0.9940 3.30 0.49 9.5\n#> 3 30 97 0.9951 3.26 0.44 10.1\n#> quality\n#> 1 6\n#> 2 6\n#> 3 6\n\nNow let’s create a liner model for quality.\n\nmodel <- lm(quality ~ fixed.acidity + volatile.acidity + citric.acid + residual.sugar + chlorides + free.sulfur.dioxide + total.sulfur.dioxide + density + pH + sulphates + alcohol,\ndata = wine)\n\nThe common goodness-of-fit parameteres for lm model are R^2, adjusted R^2, AIC or BIC coefficients.\n\nsummary(model)$r.squared #> 0.2818704 summary(model)$adj.r.squared\n#> 0.2802536\nBIC(model)\n#> 11197.94\n\nThey assess the overall quality of fit. But how to understand the factors that drive predictions for a single observation?\n\nWith the breakDown package!\n\nlibrary(breakDown)\nlibrary(ggplot2)\n\nnew_observation <- wine[1,]\nbr <- broken(model, new_observation)\nbr\n#> contribution\n#> (Intercept) 5.878\n#> residual.sugar = 20.7 1.166\n#> density = 1.001 -1.048\n#> alcohol = 8.8 -0.332\n#> pH = 3 -0.129\n#> free.sulfur.dioxide = 45 0.036\n#> sulphates = 0.45 -0.025\n#> volatile.acidity = 0.27 0.015\n#> fixed.acidity = 7 0.010\n#> total.sulfur.dioxide = 170 -0.009\n#> citric.acid = 0.36 0.001\n#> chlorides = 0.045 0.000\n#> final_prognosis 5.563\n#> baseline: 0\n# different roundings\nprint(br, digits = 2, rounding_function = signif)\n#> contribution\n#> (Intercept) 5.90000\n#> residual.sugar = 20.7 1.20000\n#> density = 1.001 -1.00000\n#> alcohol = 8.8 -0.33000\n#> pH = 3 -0.13000\n#> free.sulfur.dioxide = 45 0.03600\n#> sulphates = 0.45 -0.02500\n#> volatile.acidity = 0.27 0.01500\n#> fixed.acidity = 7 0.00950\n#> total.sulfur.dioxide = 170 -0.00900\n#> citric.acid = 0.36 0.00057\n#> chlorides = 0.045 0.00019\n#> final_prognosis 5.60000\n#> baseline: 0\nprint(br, digits = 6, rounding_function = round)\n#> contribution\n#> (Intercept) 5.877909\n#> residual.sugar = 20.7 1.165904\n#> density = 1.001 -1.047875\n#> alcohol = 8.8 -0.331669\n#> pH = 3 -0.129216\n#> free.sulfur.dioxide = 45 0.036178\n#> sulphates = 0.45 -0.025162\n#> volatile.acidity = 0.27 0.015355\n#> fixed.acidity = 7 0.009514\n#> total.sulfur.dioxide = 170 -0.009041\n#> citric.acid = 0.36 0.000570\n#> chlorides = 0.045 0.000191\n#> final_prognosis 5.562658\n#> baseline: 0\nplot(br) + ggtitle(\"breakDown plot for predicted quality of a wine\")",
null,
"Use the baseline argument to set the origin of plots.\n\nbr <- broken(model, new_observation, baseline = \"Intercept\")\nbr\n#> contribution\n#> residual.sugar = 20.7 1.166\n#> density = 1.001 -1.048\n#> alcohol = 8.8 -0.332\n#> pH = 3 -0.129\n#> free.sulfur.dioxide = 45 0.036\n#> sulphates = 0.45 -0.025\n#> volatile.acidity = 0.27 0.015\n#> fixed.acidity = 7 0.010\n#> total.sulfur.dioxide = 170 -0.009\n#> citric.acid = 0.36 0.001\n#> chlorides = 0.045 0.000\n#> final_prognosis -0.315\n#> baseline: 5.877909\nplot(br) + ggtitle(\"breakDown plot for predicted quality of a wine\")",
null,
"Works for interactions as well\n\nmodel <- lm(quality ~ (alcohol + density + residual.sugar)^2,\ndata = wine)\nnew_observation <- wine[1,]\n\nbr <- broken(model, new_observation, baseline = \"Intercept\")\nbr\n#> contribution\n#> alcohol = 8.8 -5.546\n#> density:residual.sugar = 1.001:20.7 5.422\n#> alcohol:density = 8.8:1.001 4.882\n#> residual.sugar = 20.7 -3.791\n#> alcohol:residual.sugar = 8.8:20.7 -0.705\n#> density = 1.001 -0.401\n#> final_prognosis -0.139\n#> baseline: 5.877909\nplot(br) + ggtitle(\"breakDown plot for predicted quality of a wine\")",
null,
""
]
| [
null,
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",
null,
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",
null,
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",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.5161946,"math_prob":0.951936,"size":3800,"snap":"2021-04-2021-17","text_gpt3_token_len":1543,"char_repetition_ratio":0.12671232,"word_repetition_ratio":0.2573913,"special_character_ratio":0.4686842,"punctuation_ratio":0.2552204,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9551555,"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\":\"2021-01-20T10:56:23Z\",\"WARC-Record-ID\":\"<urn:uuid:1b05e4f8-8606-4532-a83a-242c78e9549a>\",\"Content-Length\":\"202349\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:01621a99-7f4b-4e45-9896-e92022d05f65>\",\"WARC-Concurrent-To\":\"<urn:uuid:f4c27333-4297-4386-b272-9f0f895b949b>\",\"WARC-IP-Address\":\"129.22.208.161\",\"WARC-Target-URI\":\"https://cran.case.edu/web/packages/breakDown/vignettes/break_lm.html\",\"WARC-Payload-Digest\":\"sha1:3QOZNKPP7SWUKBGU5ULOUAHJNMDKT44X\",\"WARC-Block-Digest\":\"sha1:NKFSB46Y4TIRIEH5QLALME65NQNG5VDU\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703519984.9_warc_CC-MAIN-20210120085204-20210120115204-00706.warc.gz\"}"} |
https://www.math.ku.dk/english/research/conferences/2010_and_before/localfs/ | [
"# Local methods in fusion systems\n\n## December 15-19, 2008",
null,
"Organizer: Antonio Díaz Ramos\n\nExternal Participants:\n\n• Nadia Mazza (University of Lancaster)\n• Sejong Park (Oberwolfach Institute)\n• Radu Stancu (Universite de Picardie)\n\nThis is a workshop on fusion systems. There will be very few talks (1 per day) and plenty of time to interact. The interest is to make further progress in generalizing results from group theory to the fusion system setting. Nevertheless, the talks are not restricted to this topic, and they cover several topics related to fusion systems.\n\nProgramme\n\n Mon 15 13:15-14:15 Aud. 2 Antonio Díaz Introduction to fusion systems Tue 16 13:15-14:15 Aud. 2 Sejong Park Control of fusion and transfer Wed 17 10:15-11:15 Aud. 9 Nadia Mazza Oliver's conjecture Thu 18 13:15-14:15 Aud. 9 Radu Stancu Characteristic bisets and fusion Fri 19 10:15-11:15 Aud. 8 Adam Glesser Sparse fusion systems\n\nOverview of the topics:\n\n• Introduction to fusion systems: this will be an introduction to fusion systems theory with emphasis on the homotopy theoretical point of view. Main open problems of the theory will be described. Analogies with group theory will be discussed (recommended reading: [BLO04]).\n• Sparse fusion systems: In studying fusion systems, one is inexorably led towards proving statements by considering a minimal counterexample and showing that the accompanying fusion system is constrained, hence modeled by a finite group. Commonly, the fusion system involved has very few subfusion systems. In an extreme case, we call the fusion system sparse. In this talk, we will give some basic properties of sparse fusion systems, allowing us to streamline the proof of a result of Kessar and Linckelmann as well as to give a proof of a new result based on an unpublished lemma of Navarro.\n• Oliver's conjecture: In the proof of the Martino-Priddy conjecture in odd characteristic Bob Oliver defines a certain characteristic subgroup in any finite p-group (Definition 3.1 in [O04]). He then conjectures that this subgroup always contains the Thompson subgroup that is built up using the elementary abelian subgroups of maximal order (Conjecture 3.9 in [O04]). In fact, if it were true, then it would provide another proof of the Martino-Priddy conjecture in odd characteristic. But a proof of Oliver's conjecture is still awaiting, and only few cases have been shown. In this talk, we will present a survey of the conjecture and an update of the results known so far (recommended reading: [O04,GHL]).\n• Control of fusion and transfer: We review the notion of control of fusion and transfer for finite groups and see how it can be generalized for (saturated) fusion systems. Then we discuss recent generalizations of classical control of fusion/transfer theorems in local group theory to fusion systems - notably Glauberman-Thompson p-nilpotency theorem and Glauberman's ZJ- theorem [KL], Thompson's p-nilpotency theorem [DGMP1], Stellmacher's characteristic 2-subgroup theorem [OS], and Glauberman's K-group theorem [KL] [DGMP2]. Finally we discuss a new generalization of Yoshida's theorem on control of transfer to fusion systems.\n• Characteristic bisets and fusion: To a saturated fusion system on a finite $p$-group $S$ Broto, Levi and Oliver associate a characteristic $S$-$S$ biset having stability properties with respect to the fusion system. I'll explain that the existence of such characteristic biset implies the saturation of the fusion system. A characteristic biset associated to a saturated fusion system also satisfies some Frobenius reciprocity type properties (that I'll introduce and discuss). Again we have a converse statement saying that if a characteristic biset of a fusion system satisfies Frobenius reciprocity then the fusion system is saturated. This is joint work with Kari Ragnarsson.\n• Generalizing results from group theory to fusion systems: Alperin's fusion theorem for finite groups is the most central result which has successfully been generalized to the fusion system setting. Other such results are Frobenius' theorem on normal p-complements, Glauberman and Thompson's p-nilpotency criterion and Glauberman's ZJ-theorem.\n\nWe will use the visitors office 4.01 for informal meetings. The discussion topics in these meeting will be all of the above and possibly:\n\n• Yoshida's theorem\n• Category of elementary abelian p-subgroups\n• Burnside ring of fusion systems\n\nBibliography:\n\n• [BLO04] C. Broto, R. Levi, R. Oliver, The theory of p-local groups: a survey, Homotopy theory: relations with algebraic geometry, group cohomology, and algebraic K-theory, 51--84, Contemp. Math., 346, Amer. Math. Soc., Providence, RI, 2004.\n• [GHL] D.J. Green, L. Hethelyi, and M. Lilienthal, On Oliver's p-group conjecture, preprint.\n• [DGMP1] A. Diaz, A. Glesser, N. Mazza, S. Park, Glauberman and Thompson's theorems for fusion systems, Proc. Amer. Math. Soc., 137 (2009), 495-503.\n• [DGMP2] A. Diaz, A. Glesser, N. Mazza, S. Park, Control of transfer\nand weak closure in fusion systems, preprint (2008), preprint.\n• [KL] R. Kessar, M. Linckelmann, ZJ-theorem for fusion systems, Trans. Amer. Math. Soc., 360 (2008), 3093--3206.\n• [O04] R. Oliver, Equivalences of classifying spaces completed at odd primes, Math. Proc. Cambridge Philos. Soc., vol. 137 (2004), number 2, 321-347.\n• [OS] S. Onofrei, R. Stancu, A characteristic subgroup for fusion systems, preprint (2008), preprint."
]
| [
null,
"https://www.math.ku.dk/english/images/lillehavfrue.xxs.jpg",
null
]
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https://dannagifford.com/category/sharing-is-caring/ | [
"# R and ggplot2: Make zero print as “0”\n\nThe aesthetics of R printing zeros with superfluous decimal places (e.g. “0.00” instead of “0”) has always bothered me. I also want to keep the number of decimal places for the other ticks consistent (e.g. “0, 0.5, 1.0,” not “0, 0.5, 1”). I could get prettyNum() to fix the “0” instead of “0.0”, but then it was also giving me “1” instead of “1.0”.\n\nIt’s possible to get the desired output, for an arbitrary number of decimal places, without resorting to manually defining the break labels. We first define a function to format the labels how we want, using formatC(). We then supply it to the “labels” argument of scale_*_continuous, which takes the original numeric values calculated for the breaks, and applies the function to them.\n\n```# Make zeros print as \"0\" always\nprettyZero <- function(l){\nmax.decimals = max(nchar(str_extract(l, \"\\\\.[0-9]+\")), na.rm = T)-1\nlnew = formatC(l, replace.zero = T, zero.print = \"0\",\ndigits = max.decimals, format = \"f\", preserve.width=T)\nreturn(lnew)\n}\n\n# Try it out: compare x-axis and y-axis\nsomedata = tibble(x = runif(n = 100)-0.5, y = runif(n = 100)-0.5)\nggplot(data = somedata, aes(x, y)) +\ngeom_point() +\nscale_y_continuous(labels = prettyZero)\n```\n\n# A tidyverse approach to manipulating BMG Labtech OMGEA and stacker output\n\nHere is some R code for dealing with ASCII files as output by BMG MARS analysis",
null,
"software, specifically to make use of data spread across different plates as a result of using the BMG Microplate Stacker.\n\nI had a few problems with the stacker jamming when we first got it, but this was solved by (1) lubricating the magazines with a silicon spray (food-grade, not sure if that matters), (2) periodic cleaning with a degreaser, and (3) making the first step of any protocol a ‘restack’ all plates command, so that the plates are positioned more accurately. Since doing these steps, I’ve had minimal problems. In general, I like BMG plate readers for the exact reason most people don’t: the scripting language. Although it isn’t the most sophisticated, the scripting language is powerful enough to design some fairly complex protocols. Combined with the stacker, which can hold either 25 or 50 microplates depending on which magazine set you have, it’s useful for running high-throughput phenotype assays.\n\n```require(lubridate)\nrequire(tidyverse)\nrequire(growthcurver)\n\n# Some functions for importing BMG-style CSV files in a tidy way\nBMGtime = function(otime){\n# A function to convert BMG's default time format to fractional hours (can also set in MARS)\nmissing.s=grep(\"[0-9] s\",otime, value=F,invert=T)\nmissing.m=grep(\"[0-9] min\",otime, value=F,invert=T)\nmissing.h=grep(\"[0-9] h\",otime, value=F,invert=T)\notime[missing.s]=sub(\"\\$\",\" 0 s\", otime[missing.s])\notime[missing.h]=sub(\"^\",\"0 h \", otime[missing.h])\notime[missing.m]=sub(\"h\", \"h 0 min \", otime[missing.m])\notime=period_to_seconds(hms(gsub(\"[a-z ]+\",\":\", otime)))/3600\nreturn(otime)\n}\n\n# BMG MARS CSV exports have a trailing comma, so this drops the empty last column\n\n# Read a single BMG Mars wide-format time series table into long format\nselect(-contains(\"Blank corrected\"))\nndescriptors = ifelse(\"Group\"%in%names(growthdata), 4, 3)\nheader = tolower(gsub(\"Well \", \"\", names(growthdata)[1:ndescriptors]))\ntime = growthdata %>% select(-(1:ndescriptors)) %>% slice(1) %>% t()\nif(grepl(\"[a-z]\", time)) time = BMGtime(time)\ngrowthdata = growthdata %>%\nslice(2:n()) %>%\ntype_convert() %>%\nreturn(growthdata)\n}\n# Read multiple files from the same experiment (resulting from a stacker run)\nreadStack = function(path=NULL, pattern, parse.name=F, prefix=NULL, suffix=NULL, into=NULL, sep=NULL, ...){\nstackdata = data_frame(id = list.files(pattern=pattern, path=path, ...), data = map2(id, path, ~readBMG(paste0(.y,\"/\",.x))))\nif(parse.name){\nif(!is.null(prefix)){\nstackdata = stackdata %>%\nmutate(id = gsub(prefix, \"\", id))\n}\nif(!is.null(suffix)){\nstackdata = stackdata %>%\nmutate(id = gsub(suffix, \"\", id))\n}\nstackdata = stackdata %>% separate(id, sep=sep, into=into)\n}\nstackdata = stackdata %>% unnest(data) %>% type_convert()\nreturn(stackdata)}\n\n```\n\n# Getting BMG Labtech CSV files into R\n\nHere is some R code for reading wide-format BMG MARS-style time series CSV files into long-format in R, for either single data files (readBMG) or multiple data files from the same experiment, e.g. using the BMG Microplate stacker (readStack).\n\n# Re-blog: Convert Accuri .c6 flow cytometry files to .fcs\n\naccuri2fcs: convert proprietary Accuri C6 files to .fcs for batch processing. Could be useful!\n\n# Re-blog: How to peer review\n\nInformative article on how to peer review as a junior scientist. Worth a read!"
]
| [
null,
"https://dannagifford.files.wordpress.com/2018/10/csm_microplate-stacker_317840ffa5.png",
null
]
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https://help.geogebra.org/topic/extend-point-definition-with-an-checkbox | [
"# Extend point definition with an checkbox\n\nRumburak shared this question 5 years ago\n\nAn point A move on two segments.\n\nPoint definition : Point[{f, g}]\n\nWhen I click an checkbox I want to extend point definition with more one line :\n\nNew Point definition : Point[{f, g,h}]\n\nhow can I do?",
null,
"1\n\nIf the name of your checkbox is \"c\", use\n\n1. A=If[c , Point[{f, g,h}], Point[{f, g] ]",
null,
"2\n\nIf you want it to be moveable you should use use\n\nR2=Point[If[c , {f, g,h}, {f, g} ]]",
null,
"1\n\nYes, I thought that it is easier to add. The problem is that in reality there are many points and segments and the textbox \"Definition\" is too small ...\n\nThanks for help",
null,
"2\n\nIf you have 100 segments and the segments are the content of spreadsheet column \"A\",\n\nyou will get the set of these 100 segments with the simple and short input\n\n1. A1:A100",
null,
"1\n\nYes, work fine.\n\nCan you tell me how to do a cells union ?\n\nex: A1:A3 and A5:A10",
null,
"1\n\nPoint[If[¬(o ∨ p ∨ q), {B1:B3}]] work fine\n\nBut I need\n\n{B1:B3} AND {A1:A12}",
null,
"",
null,
"2\n\nJoin[{B1:B3, A1:A12}]",
null,
"1\n\nYes. Is perfect !\n\nThank you very much !",
null,
"",
null,
"1\n\nPoint A move on segments depending chexbox df1,dc1,nf1 :\n\nPoint[If[(df1 ∨ dc1 ∨ nf1) ≟ false, Join[{B1:B12}], df1, Join[{B1:B12, A1:A12}], dc1, Join[{B1:B12, A16:A19}], nf1, Join[{B1:B12, A13:A15}]]]\n\nPoint B :\n\nPoint[If[(df2 ∨ dc2 ∨ nf2) ≟ false, Join[{B1:B12}], df2, Join[{B1:B12, A1:A12}], dc2, Join[{B1:B12, A16:A19}], nf2, Join[{B1:B12, A13:A15}]]]\n\nPoint C :\n\nPoint[{a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11, a12, A1:A20}]\n\nFirst time everything is ok.\n\nAfter a while points A and B moves only in one segment but point C is ok.\n\nAfter close and open Geogebra all is ok. For now.\n\nThat is a big problem.\n\nIn each checbox (set of three chexbox) because I want only one to be active I have :\n\nOn Update script :\n\ndc1=false\n\nnf1=false",
null,
"1\n\nSorry, it is my mistake.\n\nI replaced dc1=false with SetValue[dc1,false] and work fine allways."
]
| [
null,
"https://accounts.geogebra.org/api/avatar.php",
null,
"https://accounts.geogebra.org/api/avatar.php",
null,
"https://accounts.geogebra.org/api/avatar.php",
null,
"https://accounts.geogebra.org/api/avatar.php",
null,
"https://accounts.geogebra.org/api/avatar.php",
null,
"https://accounts.geogebra.org/api/avatar.php",
null,
"https://help.geogebra.org/public/avatars/default-avatar.svg",
null,
"https://accounts.geogebra.org/api/avatar.php",
null,
"https://accounts.geogebra.org/api/avatar.php",
null,
"https://help.geogebra.org/public/avatars/default-avatar.svg",
null,
"https://accounts.geogebra.org/api/avatar.php",
null,
"https://accounts.geogebra.org/api/avatar.php",
null
]
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http://www.expertsmind.com/questions/what-is-called-when-a-star-connected-intermediate-exchange-30166208.aspx | [
"## What is called when a star connected intermediate exchange, Computer Engineering\n\nAssignment Help:\n\nA star connected intermediate exchange is known as a?\n\nA star connected intermediate exchange is termed as a Hub exchange.\n\n#### Explain about interrupt cycle, Q. Explain about Interrupt Cycle? On com...\n\nQ. Explain about Interrupt Cycle? On completion of execute cycle the current instruction execution gets completed. At this point a test is made to conclude whether any enabled\n\n#### Explain multiple instruction and single data stream (misd), Multiple Instru...\n\nMultiple Instruction and Single Data stream (MISD): In this type of organization multiple processing elements are ordered under the control of multiple control units. Every contro\n\n#### K-nearest neighbor for text classification, Assignment 2: K-nearest neighbo...\n\nAssignment 2: K-nearest neighbor for text classification. The goal of text classification is to identify the topic for a piece of text (news article, web-blog, etc.). Text clas\n\n#### Define race condition, Define race condition. When several process acc...\n\nDefine race condition. When several process access and manipulate similar data concurrently, then the outcome of the implementation depends on particular order in which the ac\n\n#### Page directive include, What is the difference between, page directive incl...\n\nWhat is the difference between, page directive include, action tag include? Ans) One difference is whereas using the include page directive, in translation time it is making t\n\n#### Baic electrical, Ask question #Minimumstate and explain thevenins theorem a...\n\nAsk question #Minimumstate and explain thevenins theorem as applicable to electrical circuits 100 words accepted#\n\n#### Explain working of counters, Q. Explain working of Counters? A counter ...\n\nQ. Explain working of Counters? A counter is a register that goes through a predetermined sequence of states when clock pulse is applied. In principle value of counters is incr\n\n#### Solve out linear equations, Assume that you have been asked to solve proble...\n\nAssume that you have been asked to solve problem with exact area constraints, the area error being no more than 1% for each department. What are the linear equations you would nee\n\n#### What is arithmetic and logic unit, What is Arithmetic and Logic Unit Ar...\n\nWhat is Arithmetic and Logic Unit Arithmetic and Logic Unit: The ALU is the 'core' of any processor. It implements all arithmetic operations (addition, multiplication, subtract\n\n#### Self knowledge - characteristics of an experts system, An experts system ...\n\nAn experts system has knowledge that lets it reason about its own operations plus a structure that simplifies this reasoning process. For example if an expert system",
null,
"",
null,
""
]
| [
null,
"http://www.expertsmind.com/questions/CaptchaImage.axd",
null,
"http://www.expertsmind.com/prostyles/images/3.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9131189,"math_prob":0.84279937,"size":2417,"snap":"2022-05-2022-21","text_gpt3_token_len":486,"char_repetition_ratio":0.07915458,"word_repetition_ratio":0.010840109,"special_character_ratio":0.19817956,"punctuation_ratio":0.1072261,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96026677,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-25T08:10:30Z\",\"WARC-Record-ID\":\"<urn:uuid:de2fb5c1-0355-466c-9f3b-c7bacf8a626d>\",\"Content-Length\":\"65536\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3799f67d-0362-4152-98e4-0e986568d246>\",\"WARC-Concurrent-To\":\"<urn:uuid:b910dd8f-a13f-45d9-ad57-b6b0afc0a0d2>\",\"WARC-IP-Address\":\"198.38.85.49\",\"WARC-Target-URI\":\"http://www.expertsmind.com/questions/what-is-called-when-a-star-connected-intermediate-exchange-30166208.aspx\",\"WARC-Payload-Digest\":\"sha1:3MXOUXQFR5Q3G2RECYMHJKECZQOHIQWB\",\"WARC-Block-Digest\":\"sha1:NYDXV7TCRLDNB6V6EYMTGPINIYA4HN3M\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662580803.75_warc_CC-MAIN-20220525054507-20220525084507-00513.warc.gz\"}"} |
https://www2.nagare.or.jp/mm/2008/taniguchi-ishiwatari/crossmode_en.htm | [
"## 5. Characteristic of neutral modes whose dispersion curves have intersection with other neutral modes",
null,
"",
null,
"",
null,
"### 5.1 The objective of this section\n\nIn dispersion relations obtained by TI2006 for large value of E, dispersion curves of some neutral modes intersect nearby dispersion curves of other neutral modes. An example of such intersection is shown in figure 6. Dispersion curves indicated by blue arrows, superimposed on dispersion curves of eastward (westward) inertial-gravity modes, intersect each other. In the followings, the modes whose dispersion curves intersect nearby curves are referred to as crossing modes. Crossing modes have following characteristics.\n\n• Intersection with dispersion curves which have pseudomomentum with opposite sign does not cause instability. Continuous modes have negative pseudomomentum (Iga, 1999a), and eastward crossing modes indicated by blue arrows in figure 6 (a) and (b) have positive pseudomomentum (Iga, 1999b). Although instability is expected to occur by intersection of dispersion curves of these modes, any unstable mode does not appear (the result does not change in calculations with ten times higher resolution. Figures are not shown).\n• Phase speed of crossing modes does not have asymptote of c=2.5 for large value of E. Phase speed of all equatorial waves in shear flow approach velocity of basic flow at dynamic equator in the limit of large E (Clark and Haynes, 1996). However, crossing modes have phase speed beyond the asymptotic value. Eastward crossing modes indicated by blue arrows in figure 6 have phase speed smaller than 2.5 for large wavenumber, and westward crossing modes have phase speed larger than 2.5 .\n\nTI2006 did not mention those crossing modes. In this section, we examine crossing modes, and show their horizontal structures and existing condition.\n\nFigure 6: Dispersion curves for (a) log E=+2.70, (b) log E=+4.00. Blue arrows indicate the region in which dispersion curves of crossing modes exist. Other symbols in the figures are same as table 1. Click above figures to show larger figures.\n\n### 5.2 Existence condition for crossing modes\n\nWith calculations for various meridional width of computational domain, it is shown that the existence of computational domain outside the inertially unstable region causes crossing modes. Figure 7 shows a series of results in which computational domains are enlarged gradually from the case only with the inertially unstable region. The figure shows dispersion curves for large value of E, in which crossing modes, if they exist, can be observed clearly. The left panel of figure 7 shows the result for the case only with the inertially unstable region. In this case, no crossing mode emerges. The middle panel of figure 7 shows dispersion curves for the case in which two grid points exist on the north of the inertially unstable region. In this case, there exist two eastward crossing modes and two westward crossing modes. For the case of the right panel of figure 7, five grid points exist on the north of the inertially unstable region, and there exist five eastward crossing modes and five westward crossing modes. The results of other cases with various computational domain show that adding a grid point outside the inertially unstable region produces an eastward crossing mode, an westward crossing mode, and a continuous mode (Appendix B). On the other hand, crossing modes do not appear for the cases with only the inertially unstable region (0 ≤ y ≤ 1) regardless the number of grid points (Appendix C).\n\nFigure 7: Dispersion curves for various meridional width of computational domain. (Left) 0.0 < y < 1.0, (Middle) 0.0 < y < 1.2, (Right) 0.0 < y < 1.4. The figures for log E=+7.50 are shown. Click each figure to show larger figure. Click [Animation] to show dispersion curves for various values of E. Refer to the caption of table 1 for the meanings of the symbols.\n\n### 5.3 Horizontal structures of crossing modes\n\nFigure 8 shows the result for log E=7.50 obtained with computational domain of 0.0 < y < 1.2. In this case, dispersion curves of crossing modes remain outside dispersion curves of continuous modes. The horizontal structure of the eastward crossing mode is shown in the left panel of figure 8. This figure shows that the crossing mode has large amplitude in the region of y > 1; outside the inertially unstable region. It is also confirmed that westward crossing mode has large amplitude in the region of y > 1 (see Appendix D). For smaller values of E, although dispersion curves of crossing modes do not intersect with other curves, crossing modes have large amplitude in the outside the inertially unstable region (see Appendix D). When computational domain exists on the north (south) of the inertially unstable region, both of eastward and westward crossing modes have large amplitude on the north (south) of the inertially unstable region. It is considered that crossing modes are gravity wave like modes existing in the channel region outside the inertially unstable region. Crossing modes are considered to be gravity wave like modes on a mid-latitude β-plane rather than equatorial wave modes.",
null,
"Figure 8: Horizontal structure (left) and dispersion curves (right) of crossing modes for log E=7.50, k=0.15 obtained with computational domain of 0 < y < 1.2. The position of the mode shown in left panel is indicated by green filled circles in right panel. Contours and vectors in the left panels indicate φ and the velocity field, respectively. Contour intervals (left) are 1.0 × 10-6. Refer to the caption of table 1 for the meanings of symbols in right panels. Click figure to show larger figure."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.90334874,"math_prob":0.947641,"size":4441,"snap":"2020-45-2020-50","text_gpt3_token_len":963,"char_repetition_ratio":0.18210502,"word_repetition_ratio":0.12742382,"special_character_ratio":0.21414097,"punctuation_ratio":0.10666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9658135,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-25T19:19:18Z\",\"WARC-Record-ID\":\"<urn:uuid:44c8c53d-0824-4791-8654-aeef06f664bc>\",\"Content-Length\":\"10254\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bb3f8ad1-bfc1-49f4-8a8a-9158d434cb73>\",\"WARC-Concurrent-To\":\"<urn:uuid:f9e67b81-a0c9-468b-ab4e-dc1e52b531eb>\",\"WARC-IP-Address\":\"118.27.95.151\",\"WARC-Target-URI\":\"https://www2.nagare.or.jp/mm/2008/taniguchi-ishiwatari/crossmode_en.htm\",\"WARC-Payload-Digest\":\"sha1:MTWLSJBY32MS5KPNZQFSMMFDBIMTQYNN\",\"WARC-Block-Digest\":\"sha1:NETFLUZ5B6FEO6NW3ANLVGG3M7PVXKP5\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107889651.52_warc_CC-MAIN-20201025183844-20201025213844-00012.warc.gz\"}"} |
https://gullele.com/tag/algorithm-2/ | [
"find longest word in the sentence\n\nFind the longest word in the sentence\n\nI am using javascript for this question.\n\nIt can be done in a lot of ways. The straight forward way could be the fastest probably. You will loop over the array of the words – you can split it with space – ” “.\n\nThen just compare the value of the word length with maximum value and overwrite when it is higher – you know what I mean..\n\nI will show the single liner for this though\n\nvar longestWord = \"Guess what the longest word in the dictionary is\".split(\" \").sort(function(a, b){\nreturn b.length - a.length;\n});\n\nLet’s go through what is going on..\n\nThe first part \"Guess what the longest word in the dictionary is\" is the given example sentence we are trying to find the longest word for.\n\nThen we split the string into array of words using the split(\" \") function. Now we have array to work with.\n\nThen we sorted the array with sorter function passed to it. Sorter function in this one is anonymous function that we pass to it.\n\nFinally we just took the first element of the sorted array which is sorted by the length of the word.\n\nAlso with the same concept, we have have it using reduce function;\n\n\"I still am having the wonderfulness of this thing \".split(\" \").reduce((word1, word2) => word2.length > word1.length ? word2 : word1 );\n\nThat is it!\n\nGet maximum occurring character\n\nFind maximum occurring character from the given string\n\nMax occurring character is the most frequent interview whiteboard question. Ok, there are so many of ways to handle this question and this is just one of them.\n\nI have showed it with 6 characters as starting point and use those as reference. But you can extend that given object to hold as many characters as you want.\nContinue reading Get maximum occurring character\n\nUsing stack data structure to implement queue in javascript\n\nThis is one way of implementing of the stack to be reused as queue using two stacks.\n\nYou may complain you are abusing the memory. But we are not copying the data on both stacks we are using the stacks for switching values only\n\nclass Queue {\nconstructor() {\nthis.inputStack = [];\nthis.outputStack = [];\n}\n\nenqueue(item) {\nthis.inputStack.push(item);\n}\n\ndequeue(item) {\nthis.outputStack = [];\nif (this.inputStack.length > 0) {\nwhile (this.inputStack.length > 0) {\nthis.outputStack.push(this.inputStack.pop());\n}\n}\nif (this.outputStack.length > 0) {\nconsole.log(this.outputStack.pop());\nthis.inputStack = [];\nwhile (this.outputStack.length > 0) {\nthis.inputStack.push(this.outputStack.pop());\n}\n}\n}\n\nlistIn() {\nlet i=0;\nwhile(i < this.inputStack.length) {\nconsole.log(this.inputStack[i]);\ni++;\n}\n}\nlistOut() {\nlet i=0;\nwhile(i < this.outputStack.length) {\nconsole.log(this.outputStack[i]);\ni++;\n}\n}\n}\n\n*There is part left out for improvement and refactoring. But, can you see a way to use only one while rather than two whiles and make it a bit efficient?\n\nJava solution for checking anagram strings – tell if phrases are anagrams\n\nAnagram Algorithm in Java solving if two strings are anagrams or not\n\nThis is the java solution for checking if the given two strings are anagrams or not. For the examples of anagrams and the approach I used and analysis of the code, see analysis of anagram algorithm\n\nFind missing numbers from billion number lists with limited memory\n\nAnagram algorithm in Java solution\n\npackage algorithm;\n\nimport java.util.Arrays;\n\n/**\n* Determine if the given two strings are anagrams or not\n* A solution in java programming language\n* @author https://gullele.com\n*\n* Analysis - this would run in o(nlogn) for the sorting part and all other others would be in o(n)\n* Hence it is o(nlogn) + o(n) ==> o(nlogn) assuming worst case sorting.\n*/\npublic class Anagram implements AnagramFinder {\n\npublic static void main(String string[]) {\nString string1 = \"Debit Card \";\n\nAnagramFinder anagramFinder = new Anagram();\nif (anagramFinder.areAnagrams(string1, string2)) {\nSystem.out.println(\"Anagrams\");\n} else {\nSystem.out.println(\"Not Anagrams\");\n}\n}\n\npublic Anagram() {\n\n}\n\n@Override\n/**\n* Verify if the given words are anagrams or not\n*/\npublic boolean areAnagrams(String s1, String s2) {\nif (s1 == null || s2 == null) { //I don't think we would assume null is anagram at all..\nreturn false;\n}\n\n//get rid of the spaces\ns1 = s1.replaceAll(\"\\\\s+\", \"\").toLowerCase();\ns2 = s2.replaceAll(\"\\\\s+\", \"\").toLowerCase();\n\n//no need to proceed if the length is not the same. assumed the anagrams are the same in length\nif (s1.length() != s2.length()) {\nreturn false;\n}\n\n//then order the string in characters\nchar[] ordereds1 = sortChars(createCharArray(s1)); //o(nlogn)\nchar[] ordereds2 = sortChars(createCharArray(s2)); //o(nlogn)\n\n//first thing first, if the size is not the same, then we are done\nint index = 0;\nwhile (index < ordereds1.length) { //this would run o(n)\nif (ordereds1[index] != ordereds2[index]) {\nreturn false;\n}\nindex++;\n}\n\nreturn true;\n}\n\n/**\n* Takes the string and converts it to characters\n* @param string\n* @return\n*/\nprivate char[] createCharArray(String string) {\nreturn new char[string.length()];\n}\n\n/**\n* if I have to implement it I can use quick sort and make it at least n(logn)\n* @param chars\n* @return array\n*/\nprivate char[] sortChars(char[] chars) {\nArrays.sort(chars);\nreturn chars;\n}\n}\n\nThe above anagram algorithm solution is in java but it can be performed in any programming language.\n\nCheck if two strings are anagrams or not\n\nAre the two strings or phrases anagrams or not algorithm\n\nFirst thing first, What is anagram?\n\nAnagrams are phrases/words of which one can be created from the other with rearrangement of characters.\n\nExamples of anagram strings:\n\nGeorge Bush and He Bugs Gore\n\nNow, given two strings, how would do find if the strings are anagrams or not.\n\nThe approach I used is a follows:\n\n1. remove the space out of the strings\n2. compare the length of strings, if they don’t match, they ain’t anagrams\n3. sort the characters of each string\n4. check character by character and see if they match till the end\n\nThis is approach I used with o(nlogn) because of the sorting part. All the others can be done in o(n)\n\nA java solution for anagram algorithm\n\nFind K Complementary numbers from array Java implementation\n\nFind K complementary numbers from the given array\n\nThis is another approach to the problem that I have done it here. On that post, a good deal of visitors pointed out that it is actually o(n*n) not o(n) as I claimed.\n\nYes, the naive usage of Hashmap for holding the numbers has soared the performance and I have changed the approach as follows.\n\nTo remind the k complementary problem and its solution, you are given array of numbers and a number k of which you are going to find numbers that give k complementary. The following is an example of it.\n\nThe problem is, given numbers like 7, 1, 5, 6, 9, 3, 11, -1 and given number 10, write a script that would print numbers that would add to 10. In the example, it would be like\n7 and 3, 1 and 9, -1 and 11.\n\npackage algorithm;\n\nimport java.util.ArrayList;\nimport java.util.List;\n\n/**\n* Algorithm to find the pairs making the K complementary in O(n) complexity\n*\n* @author http://gullele.com\n*/\npublic class KComplementary2 {\n\npublic static void main(String[] args) {\nKComplementary2 kcomp = new KComplementary2();\nint[] numbers = new int[]{7, 1, 5, 6, 9, 3, 11, -1};\n\nfor (Integer number : kcomp.getKComplementaryPairs(10, numbers)) {\nSystem.out.println(\" Pairs are \"+ number + \" and \" + (10-number));\n}\n}\n\npublic KComplementary2() {}\n\n/**\n* An algorithm to find the pair from the given array that would sum up the given K\n*\n* @note - the algorithm would be done in o(n)+o(nlogn). First it will run through the whole\n* numbers and creates a temporary list of pairs in HashMap with\n* (value, sum-value).\n* @param sum\n* @param listOfIntegers\n* @return\n*/\npublic List getKComplementaryPairs(int sum, int[] listOfIntegers) {\n\n/*\n* The algorithm works using front and last pointers on ascendingly sorted array. The front would be\n* instantiated with 0 and last with length-1; if the arr[front]+arr[last] == sum, then pick\n* the numbers and add them to the pair array.\n* if their sum is greater than sum, it means time to check the second higher number that is lower\n* than the current highest number. And the reverse would hold true if the sum is less than the sum\n* time to move to the next higher number from the lower side.\n*/\nif (listOfIntegers == null || listOfIntegers.length == 0) {\nreturn null;\n}\n\n//quick sort the array\nquickSort(0, listOfIntegers.length-1, listOfIntegers);\n\nint[] sortedArray = listOfIntegers;\n\n//holder for the complementary pairs\nList pairs = new ArrayList();\nint frontPointer = 0;\nint lastPointer = sortedArray.length-1;\n\nwhile (frontPointer < lastPointer) {\nint currentSum = sortedArray[frontPointer] + sortedArray[lastPointer];\nif (currentSum == sum) {\n/*\n* Since sum is found, increment front and decrement last pointer\n* Only one number is required to be hold, the other can be found\n* from sum-number since complementary\n*/\nfrontPointer++;\nlastPointer--;\n} else if (currentSum > sum) {\nlastPointer--;\n} else {\nfrontPointer++;\n}\n}\n\nreturn pairs;\n}\n\n/**\n* sort the numbers. I have used quick sort here. QuickSort is nlogn in average case\n* well, in worst case it still would be n**2 though :(\n* @param numbers\n* @return sorted array numbers.\n*/\npublic void quickSort(int lowerIndex, int higherIndex, int[] numbers) {\n/*\n* Recursively inarray sort. Start from arbitrary value to compare from and recursively sort its\n* left and right.\n*/\nint pivot = lowerIndex+(higherIndex-lowerIndex)/2;\nint lower = lowerIndex;\nint higher = higherIndex;\n\nwhile (lower < higher) {\nwhile (numbers[lower] < numbers[pivot]) {\nlower++;\n}\nwhile (numbers[higher] > numbers[pivot]) {\nhigher--;\n}\n\n//swap those needed to be on the left on those on the right.\nif (lower <= higher) {\nint temp = numbers[lower];\nnumbers[lower] = numbers[higher];\nnumbers[higher] = temp;\nlower++;\nhigher--;\n}\n}\nif (lowerIndex < higher) {\nquickSort(lowerIndex, higher, numbers);\n}\n\nif (lower < higherIndex) {\nquickSort(lower, higherIndex, numbers);\n}\n}\n}\n\nLove algorithms? See how you would solve the following\n\nCan you find the three numbers that would sum to T from given array?\n\nFrom a file of billion numbers, find missing numbers with limited memory\n\nCheck if there are three numbers a, b, c giving a total T from array A\n\nI got this question while helping a friend on the course work. It is relatively simple question. But the way how it is approached can make a difference on efficiency.\n\nThe question is, given an array of numbers, you are to find if there are three numbers that would total the given number T.\n\nIf done in a very naive way, it can soar to o(n^3) like having three loops and checking the sum inside the third loop.. well.. this is a no no..\n\nI have approached it in a log n ( for sorting) and n for (searching) approach..\n\npackage algorithm;\n\nimport java.util.Arrays;\n\n/**\n* Given an array of integers, find if there are three numbers that would sum up to the\n* number T\n*\n* @author http://gullele.com\n*\n*/\npublic static void main(String[] args) {\nint[] test = new int[]{1,3,4,5,10,12, 18};\n\nint[] response = summt.findThreeNumbers(test, 29);\n\nif (response.length > 1) {\nfor(int num : response) {\nSystem.out.println(num);\n}\n} else {\nSystem.out.println(\":( Couldn't find those three gems\");\n}\n\n}\n\npublic int[] findThreeNumbers(int[] nums, int t) {\n\nint[] indexes = new int;\nif (nums.length == 0 || nums.length <= 2) {\nreturn indexes;\n}\n\n//for primitive this would be quick sort so we have nlogn\nArrays.sort(nums);\n\nint minIndex =0;\nint maxIndex = nums.length-1;\nint current = 1;\nwhile (minIndex != maxIndex) {\nif (nums[minIndex] + nums[maxIndex] + nums[current] == t) {\nint[] summingNumbers = new int;\nsummingNumbers = nums[minIndex];\nsummingNumbers = nums[current];\nsummingNumbers = nums[maxIndex];\nreturn summingNumbers;\n}\n\nint lookingFor = t-(nums[minIndex] + nums[maxIndex]);\n//if the number being sought is beyond the max, then jack up the min index\nif (lookingFor >= nums[maxIndex]) {\nminIndex++;\ncurrent = minIndex + 1;\n} else if (nums[minIndex] + nums[maxIndex] + nums[current] < t) {\ncurrent++;\n} else {\nmaxIndex--;\ncurrent = minIndex + 1;\n}\n\n}\n\nreturn indexes;\n}\n}\n\nbinary tree problems with solution\n\ntoday is saturday and was catching upon some blogs. Then come across some binary tree algorithms and it reminded me the famous stanford binary tree questions.. decided to work on those and got to the mid of it..\nActually the questions get harder as you go further.. the first 7 are relatively easy..\nI will continue to work on them and post my solution here.\nthe solutions are given on the this page but, it is advisable to work them out without looking a the solution..\n\nhere it is!\n\n#include\n#include\n/*\n* @author http://gullele.com\n* binary tree problems and solutions.\n*/\nstruct Node\n{\nint number;\nstruct Node *left;\nstruct Node *right;\n};\n\nstruct Node *createNode(int value);\nint hasPathSum(struct Node *head, int sum);\nint main()\n{\nstruct Node *left=createNode(8);\nstruct Node *right=createNode(15);\nstruct Node *right1=createNode(5);\nstruct Node *right2=createNode(1);\nleft->left=right1;\nright1->left=right2;\n\nstruct Node *two=createNode(2);\nstruct Node *seven=createNode(7);\nstruct Node *eleven=createNode(11);\nstruct Node *lfour=createNode(4);\nstruct Node *thirteen=createNode(13);\nstruct Node *eight=createNode(8);\nstruct Node *rfour=createNode(4);\nstruct Node *one=createNode(1);\n\nlfour->left=eleven;\neleven->left=seven;\neleven->right=two;\neight->left=thirteen;\neight->right=rfour;\nrfour->right=one;\n\nprintf(\"Size of the tree is %d n\", countNode(head));\nprintf(\"Max depth is %d n\", maxDepth(head)-1);\nprintf(\"Minimum Value is %d n\", minValue(head));\nprintf(\"n\");\nif(hasSum)\nprintf(\"it has sum\");\nelse\nprintf(\"it does not has sum\");\nreturn 0;\n}\n\n/**\n* Takes the head of the binary tree and counts how many children are there in the tree\n* it will recursively count the left and right nodes to come to the conclusion\n*/\n{\n{\nreturn 0;\n}\n}\n\n/**\n* Finds the maximum depth of the tree\n*\n*/\n{\nint maxLength = 0;\n{\nreturn 0;\n}\nelse\n{\nint leftMax = 1 + maxDepth(head->left);\nint rightMax = 1 + maxDepth(head->right);\nif (leftMax > rightMax)\n{\nreturn leftMax;\n}\nreturn rightMax;\n}\n}\n\n/**\n* Works on the Binary Search Tree - since on the BST, for sure the left child is always lesser in value.\n*/\n{\nreturn 0; //might not be valid answer here\nwhile(current->left!=NULL)\n{\ncurrent=current->left;\n}\nreturn current->number;\n}\n\n/**\n* Prints the value of the BST\n* this is inorder traversal\n*/\n{\n{\nreturn;\n}\nelse\n{\n}\n}\n\nint hasPathSum(struct Node *head, int sum)\n{\nint localsum = 0;\n{\nreturn 0;\n}\n{\n}\nelse\n{\n}\n}\n/**\n* Post order traversal version of the tree traversal\n*/\n{\n{\nreturn;\n}\nelse\n{\n}\n}\n\n/**\n* Create new node\n*/\nstruct Node *createNode(int value)\n{\nstruct Node *newNode=malloc(sizeof(struct Node));\nnewNode->number=value;\nnewNode->left=NULL;\nnewNode->right=NULL;\nreturn newNode;\n}\n\nKadane’s algorithm in C – Dynamic Programming\n\nHow do we find a keys that hold maximum consecutive values in sum from the given array?\nThe usual approach could be O(n^2).\n\nBut Kadanes algorithm can perform this same problem in a linear time.\nhere is the implementation using C\n\n#include <stdio.h>\n\n/*\n* @author http://gullele.com\n*\n*/\nvoid main()\n{\nint nums[] = {-1,2,3,-9,8,7,2};\nint start_index;\nint end_index;\nint sum = maximum_consequential_sum(nums, &start_index, &end_index);\nprintf(\"maximum sum is %d n\",sum );\nprintf(\"maximum index starts at %d n\", start_index);\nprintf(\"maximum index ends at %d n\", end_index);\n}\n/*\n* Get the maximum subsequent sum from the given array\n*/\nint maximum_consequential_sum(int* nums, int*start_index, int*end_index)\n{\nint max_start_index = 0, max_end_index = 0;\nint max_sum = 0;\nint cur_index, cur_max_sum = 0;\nfor(cur_index = 0; cur_index <= max_sum ; cur_index++)\n{\ncur_max_sum += nums[cur_index];\nif (cur_max_sum > max_sum)\n{\nmax_sum = cur_max_sum;\nmax_end_index = cur_index;\n}\nif (cur_max_sum < 0)\n{\ncur_max_sum = 0 ;\nmax_start_index = cur_index + 1;\n}\n}\n*start_index = max_start_index;\n*end_index = max_end_index;\nreturn max_sum;\n}\n\nThe above code is, of course, just for illustration purpose. Make sure to check\nfor empty array and possible division by zero stuff..\n\nTo run this on Linux\n\n1. Save the above file as kadanes.c\n2. go to terminal and locate to the file"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.81019443,"math_prob":0.96818817,"size":4156,"snap":"2022-05-2022-21","text_gpt3_token_len":1054,"char_repetition_ratio":0.15149325,"word_repetition_ratio":0.0058224164,"special_character_ratio":0.27141482,"punctuation_ratio":0.14438502,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99556464,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-19T13:39:01Z\",\"WARC-Record-ID\":\"<urn:uuid:1ed80c09-bc15-4cb8-8e53-de98ec0c9b89>\",\"Content-Length\":\"91151\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:21451495-c937-4371-b8e6-957e0c75900a>\",\"WARC-Concurrent-To\":\"<urn:uuid:ceb5aac3-046c-4e2f-9307-a220367ea00d>\",\"WARC-IP-Address\":\"74.208.236.164\",\"WARC-Target-URI\":\"https://gullele.com/tag/algorithm-2/\",\"WARC-Payload-Digest\":\"sha1:Q6TW7NIKCGUJIUZZIJXHDOFUSZORLNO4\",\"WARC-Block-Digest\":\"sha1:ANCVGZJWXS6GQNNDGJL5TP5M3N7C6TTQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301341.12_warc_CC-MAIN-20220119125003-20220119155003-00380.warc.gz\"}"} |
https://nl.mathworks.com/help/symbolic/sin.html | [
"# sin\n\nSymbolic sine function\n\n## Syntax\n\n``sin(X)``\n\n## Description\n\nexample\n\n``sin(X)` returns the sine function of `X`.`\n\n## Examples\n\n### Sine Function for Numeric and Symbolic Arguments\n\nDepending on its arguments, `sin` returns floating-point or exact symbolic results.\n\nCompute the sine function for these numbers. Because these numbers are not symbolic objects, `sin` returns floating-point results.\n\n`A = sin([-2, -pi, pi/6, 5*pi/7, 11])`\n```A = -0.9093 -0.0000 0.5000 0.7818 -1.0000```\n\nCompute the sine function for the numbers converted to symbolic objects. For many symbolic (exact) numbers, `sin` returns unresolved symbolic calls.\n\n`symA = sin(sym([-2, -pi, pi/6, 5*pi/7, 11]))`\n```symA = [ -sin(2), 0, 1/2, sin((2*pi)/7), sin(11)]```\n\nUse `vpa` to approximate symbolic results with floating-point numbers:\n\n`vpa(symA)`\n```ans = [ -0.90929742682568169539601986591174,... 0,... 0.5,... 0.78183148246802980870844452667406,... -0.99999020655070345705156489902552]```\n\n### Plot Sine Function\n\nPlot the sine function on the interval from $-4\\pi$ to $4\\pi$.\n\n```syms x fplot(sin(x),[-4*pi 4*pi]) grid on```",
null,
"### Handle Expressions Containing Sine Function\n\nMany functions, such as `diff`, `int`, `taylor`, and `rewrite`, can handle expressions containing `sin`.\n\nFind the first and second derivatives of the sine function:\n\n```syms x diff(sin(x), x) diff(sin(x), x, x)```\n```ans = cos(x) ans = -sin(x)```\n\nFind the indefinite integral of the sine function:\n\n`int(sin(x), x)`\n```ans = -cos(x)```\n\nFind the Taylor series expansion of `sin(x)`:\n\n`taylor(sin(x), x)`\n```ans = x^5/120 - x^3/6 + x```\n\nRewrite the sine function in terms of the exponential function:\n\n`rewrite(sin(x), 'exp')`\n```ans = (exp(-x*1i)*1i)/2 - (exp(x*1i)*1i)/2```\n\n### Evaluate Units with `sin` Function\n\n`sin` numerically evaluates these units automatically: `radian`, `degree`, `arcmin`, `arcsec`, and `revolution`.\n\nShow this behavior by finding the sine of `x` degrees and `2` radians.\n\n```u = symunit; syms x f = [x*u.degree 2*u.radian]; sinf = sin(f)```\n```sinf = [ sin((pi*x)/180), sin(2)]```\n\nYou can calculate `sinf` by substituting for `x` using `subs` and then using `double` or `vpa`.\n\n## Input Arguments\n\ncollapse all\n\nInput, specified as a symbolic number, scalar variable, matrix variable (since R2021a), expression, or function, or as a vector or matrix of symbolic numbers, scalar variables, expressions, or functions.\n\ncollapse all\n\n### Sine Function\n\nThe sine of an angle, α, defined with reference to a right angled triangle is",
null,
"The sine of a complex argument, α, is\n\n`$\\mathrm{sin}\\left(\\alpha \\right)=\\frac{{e}^{i\\alpha }-{e}^{-i\\alpha }}{2i}\\text{\\hspace{0.17em}}.$`"
]
| [
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"https://nl.mathworks.com/help/examples/symbolic/win64/SinPlotTheSineFunctionExample_01.png",
null,
"https://nl.mathworks.com/help/symbolic/definition_sine.png",
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]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.5802431,"math_prob":0.9979505,"size":1758,"snap":"2021-43-2021-49","text_gpt3_token_len":565,"char_repetition_ratio":0.14139111,"word_repetition_ratio":0.00754717,"special_character_ratio":0.37144482,"punctuation_ratio":0.20157067,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99987626,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-27T14:35:09Z\",\"WARC-Record-ID\":\"<urn:uuid:edff764c-3e39-4830-96e2-13c78b76b4de>\",\"Content-Length\":\"82386\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1b8144f6-e6dd-46e9-a6d2-908402966edf>\",\"WARC-Concurrent-To\":\"<urn:uuid:52e5d3f1-5050-477b-bb73-205439397b01>\",\"WARC-IP-Address\":\"184.25.188.167\",\"WARC-Target-URI\":\"https://nl.mathworks.com/help/symbolic/sin.html\",\"WARC-Payload-Digest\":\"sha1:6PX4NRIJ2AN5BLZWP737YKWABUJC6SSQ\",\"WARC-Block-Digest\":\"sha1:AE5JEN7EOB43R4H7THY43ZUIKKBLLQMG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358189.36_warc_CC-MAIN-20211127133237-20211127163237-00018.warc.gz\"}"} |
https://proficientwritershub.com/circle-has-equation-x2y24x-4y-10-graph-circle-using-center-hk-and-raduis-r-find-intercepts-i/ | [
"# circle has equation x2y24x 4y 10 graph circle using center hk and raduis r find intercepts i\n\na circle has the equation x^2+y^2^4x 4y 1=0 graph the circle using center (h,k) and raduis r. find the intercepts, if any of the graph. then find the x intercepts?\n\n##### Looking for a similar assignment? Our writers will offer you original work free from plagiarism. We follow the assignment instructions to the letter and always deliver on time. Be assured of a quality paper that will raise your grade. Order now and Get a 15% Discount! Use Coupon Code \"Newclient\"",
null,
""
]
| [
null,
"https://proficientwritershub.com/wp-content/uploads/2019/10/order-now.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.577095,"math_prob":0.9371601,"size":650,"snap":"2023-40-2023-50","text_gpt3_token_len":212,"char_repetition_ratio":0.119195044,"word_repetition_ratio":0.35820895,"special_character_ratio":0.35384616,"punctuation_ratio":0.115384616,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96591884,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-01T01:56:35Z\",\"WARC-Record-ID\":\"<urn:uuid:ca2b0f37-c9f9-4eef-8bdc-8dafa5e84cd0>\",\"Content-Length\":\"61261\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3d565806-3f77-487f-922d-cb8b3cc674fa>\",\"WARC-Concurrent-To\":\"<urn:uuid:02727f5d-e5d0-467a-886a-fdd3843986be>\",\"WARC-IP-Address\":\"198.54.125.202\",\"WARC-Target-URI\":\"https://proficientwritershub.com/circle-has-equation-x2y24x-4y-10-graph-circle-using-center-hk-and-raduis-r-find-intercepts-i/\",\"WARC-Payload-Digest\":\"sha1:MQVPWKQ4CQGA54DFOVQNBAURJKTLU5UA\",\"WARC-Block-Digest\":\"sha1:H7SENOLCM43BHME2ZBUND56SXZIREXGP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510734.55_warc_CC-MAIN-20231001005750-20231001035750-00826.warc.gz\"}"} |
https://pyglow.github.io/userguide.html | [
"# User Guide¶\n\n## Standard Models¶\n\nclass `network.``Network`(input_shape, device, gpu, track_dynamics=False)\n\nBase class for all neural network modules.\n\nYour network should also subclass this class.\n\nParameters: input_shape (tuple) – input tensor shape device (torch.device or int) – GPU or CPU for training purposes gpu (bool) – true if GPU is enabled on the system, false otherwise track_dynamics (bool) – tracks the NN dynamics during training (stores input-output for every intermediate layer)\n`input_shape`\n\ntuple – input tensor shape\n\n`layer_list`\n\niterable – an iterable of pytorch `torch.nn.modules.container.Sequential` type layers\n\n`num_layers`\n\nint – number of layers in the model\n\n`is_gpu`\n\nbool – true if GPU is enabled on the system, false otherwise\n\n`device`\n\ntorch.device or intGPU or CPU for training purposes\n\n`track_dynamics`\n\nbool – tracks the NN dynamics during training (stores input-output for every intermediate layer)\n\n`criterion`\n\ncallable – loss function for the model\n\n`optimizer`\n\ntorch.optim.Optimizer – optimizer for training the model\n\n`metrics`\n\nstr – metric to be used for evaluating performance of the model\n\n`add`(layer_obj)\n\nAdds specified (by the layer_obj argument) layer to the model.\n\nParameters: layer_obj (glow.Layer) – object of specific layer to be added\n`attach_evaluator`(evaluator_obj)\n\nAttaches an evaluator with the model which will get evaluated at every pass of batch and obtain information plane coordinates according to defined criterion in the ‘evaluator_obj’.\n\nIt appends the ‘evaluator_obj’ to the list evaluator_list which contains all the attached evaluators with the model.\n\nParameters: evaluator_obj (glow.information_bottleneck.Estimator) – evaluator object with has criterion defined which will get evaluated for the dynamics of the training process\n`compile`(optimizer='SGD', loss='cross_entropy', metrics=[], learning_rate=0.001, momentum=0.95, **kwargs)\n\nCompile the model with attaching optimizer and loss function to the model.\n\nParameters: optimizer (torch.optim.Optimizer) – optimizer to be used during training process loss (loss) – loss function for back-propagation metrics (list) – list of all performance metric which needs to be evaluated in validation pass learning_rate (float, optional) – learning rate for gradient descent step (default: 0.001) momentum (float, optional) – momentum for different variants of optimizers (default: 0.95)\n`fit`(x_train, y_train, batch_size, num_epochs, validation_split=0.2, show_plot=False)\n\nFits the dataset passed as numpy array (Keras like pipeline) in the arguments.\n\nParameters: x_train (numpy.ndarray) – training input dataset y_train (numpy.ndarray) – training ground-truth labels batch_size (int) – batch size of one batch num_epochs (int) – number of epochs for training validation_split (float, optional) – proportion of the total dataset to be used for validation (default: 0.2) show_plot (bool, optional) – if true plots the training loss (red), validation loss (blue) vs epochs (default: True)\n`fit_generator`(train_loader, val_loader, num_epochs, show_plot=False)\n\nFits the dataset by taking data-loader as argument.\n\nParameters: num_epochs (int) – number of epochs for training train_loader (torch.utils.data.DataLoader) – training dataset (with already processed batches) val_loader (torch.utils.data.DataLoader) – validation dataset (with already processed batches) show_plot (bool, optional) – if true plots the training loss (red), validation loss (blue) vs epochs (default: True)\n`forward`(x)\n\nMethod for defining forward pass through the model.\n\nThis method needs to be overridden by your implementation contain logic of the forward pass through your model.\n\nParameters: x (torch.Tensor) – input tensor to the model tuple containing: (torch.Tensor): output tensor of the network (iterable): list of hidden layer outputs for dynamics tracking purposes (tuple)\n\n### Sequential¶\n\nclass `network.``Sequential`(input_shape, gpu=False)\n\nKeras like Sequential model.\n\nParameters: input_shape (tuple) – input tensor shape gpu (bool, optional) – if true then PyGlow will attempt to use GPU, for false CPU will be used (default: False)\n\n### IBSequential¶\n\nclass `network.``IBSequential`(input_shape, gpu=False, track_dynamics=False, save_dynamics=False)\n\nKeras like Sequential model with extended more sophisticated Information Bottleneck functionalities for analysing the dynamics of training.\n\nParameters: input_shape (tuple) – input tensor shape gpu (bool, optional) – if true then PyGlow will attempt to use GPU, for false CPU will be used (default: False) track_dynamics (bool) – if true then will track the input-hidden-output dynamics segment and will allow evaluator to attach to the model, for false no track for dynamics is kept save_dynamics (bool, optional) – if true then saves the whole training process dynamics into a distributed file (for efficiency)\n`evaluator_list`\n\niterable – list of `glow.information_bottleneck.Estimator` instances which stores the evaluators for the model\n\n`evaluated_dynamics`\n\niterable – list of evaluated dynamics segment information coordinates for intermediate layer for each evaluator averaged over batch for each epoch\n\nShape:\nevaluator_list has shape (N, E, L, 2) where:\n• N: Number of epochs\n• E: Number of evaluators\n• L: Number of layers with parameters (Flatten and Dropout excluded)\n\nand last dimension is equal to 2 which stores 2-D information plane coordinates\n\n## ‘Models without Back-prop’¶\n\nclass `hsic.``HSIC`(input_shape, device, gpu, **kwargs)\n\nThe HSIC Bottelneck: Deep Learning without backpropagation.\n\nBase class for all HSIC network.\n\nYour HSIC network should also subclass this class.\n\nParameters: input_shape (tuple) – input tensor shape device (torch.device, optional) – gpu (bool) – true if GPU is enabled on the system, false otherwise\n`add`(layer_obj, loss_criterion=None, regularize_coeff=0)\n\nAdds specified layer and loss criterion to the model which will be used for measuring objective between layer’s current representation and optimal representation (representation with minimum IB-based objective).\n\nParameters: layer_obj (glow.Layer) – object of specific layer to be added loss_criterion (glow.information_bottleneck.Estimator) – loss function for the layer which is added regularize_coeff (float) – regularization coefficient between generalization and compression of IB type objective\n`compile`(loss_criterion=None, optimizer='SGD', regularize_coeff=100, learning_rate=0.001, momentum=0.95, **kwargs)\n\nCompile the HSIC network with loss criterion (criterion objective used as loss function for intermediate representations).\n\nAll the layers which did not have any criterion passed as argument at the time of ‘.add’ of layer automatically takes this loss criterion.\n\nParameters: loss_criterion (glow.information_bottleneck.Estimator) – criterion function which is an instance of `glow.information_bottleneck.Estimator` optimizer (torch.optim.Optimizer) – optimizer to be used during training process for all the layers regularize_coeff (float) – trade-off parameter between generalization and compression according to IB-based theory learning_rate (float, optional) – learning rate for gradient descent step (default: 0.001) momentum (float, optional) – momentum for different variants of optimizers (default: 0.95)\n`forward`(x)\n\nMethod for defining forward pass through the model.\n\nThis method needs to be overridden by your implementation which contains the logic of the forward pass through your model.\n\nParameters: x (torch.Tensor) – input tensor to the model list of hidden layer outputs (objects of type `torch.Tensor`) which are detached from their previous layer’s gradients (iterable)\n`pre_training_loop`(num_epochs, train_loader, val_loader)\n\nPre training phase in which hidden representations are learned using HSIC training paradigm.\n\n`sequential_forward`(x)\n\nSequentially calculate the output taking HSIC network as sequential feedforward neural network and is equivalent to forward pass in standard models.\n\nParameters: x (torch.Tensor) – input tensor to the network output of the sequential feedforward network (torch.Tensor)\n\n### HSICSequential¶\n\nclass `hsic.``HSICSequential`(input_shape, gpu=False, **kwargs)\n\nBase implementation for HSIC networks.\n\nThis class forms instances for multi-model sigma network as given in the paper https://arxiv.org/abs/1908.01580 .\n\nParameters: input_shape (tuple) – input tensor shape gpu (bool, optional) – if true then PyGlow will attempt to use GPU, for false CPU will be used (default: False)\n\n## Layers¶\n\nclass `layer.``Layer`(*args)\n\nBase class for all layer modules.\n\nYour layer should also subclass this class.\n\n`forward`(x)\n\nForward method overrides PyTorch forward method and contains the logic for the forward pass through the custom layer defined.\n\nParameters: x (torch.Tensor) – input tensor to the layer output tensor of the layer y (torch.Tensor)\n`set_input`(input_shape)\n\nTakes input_shape and demands user to define a variable self.output_shape which stores the output shape of the custom layer.\n\nParameters: input_shape (tuple) – input shape of the tensor which the layer expects to receive\n\n### Core¶\n\nclass `core.``Dense`(output_dim, activation=None)\n\nBases: `glow.layer.Layer`\n\nClass for full connected dense layer.\n\nParameters: output_dim (int) – output dimension of the dense layer activation (str) – activation function to be used for the layer (default: None)\nclass `core.``Dropout`(prob)\n\nBases: `glow.layer.Layer`\n\nClass for dropout layer - regularization using noise stablity of output.\n\nParameters: prob (float) – probability with which neurons in the previous layer is dropped\nclass `core.``Flatten`\n\nBases: `glow.layer.Layer`\n\nClass for flattening the input shape.\n\n### Convolutional¶\n\nclass `convolutional.``Conv1d`(filters, kernel_size, stride, padding=0, dilation=1, activation=None, **kwargs)\n\nBases: `convolutional._Conv`\n\nConvolutional layer of rank 1.\n\nParameters: filters (int) – number of filters for the layer kernel_size (int) – size of kernel to be used for convolutional operation stride (int) – stride for the kernel in convolutional operations padding (int, optional) – padding for the image to handle edges while convoluting (default: 0) dilation (int, optional) – dilation for the convolutional operation (default: 1) activation (str) – activation function to be used for the layer (default: None)\nclass `convolutional.``Conv2d`(filters, kernel_size, stride, padding=0, dilation=1, activation=None, **kwargs)\n\nBases: `convolutional._Conv`\n\nConvolutional layer of rank 2.\n\nParameters: filters (int) – number of filters for the layer kernel_size (int) – size of kernel to be used for convolutional operation stride (int) – stride for the kernel in convolutional operations padding (int, optional) – padding for the image to handle edges while convoluting (default: 0) dilation (int, optional) – dilation for the convolutional operation (default: 1) activation (str) – activation function to be used for the layer (default: None)\nclass `convolutional.``Conv3d`(filters, kernel_size, stride, padding=0, dilation=1, activation=None, **kwargs)\n\nBases: `convolutional._Conv`\n\nConvolutional layer of rank 3.\n\nParameters: filters (int) – number of filters for the layer kernel_size (int) – size of kernel to be used for convolutional operation stride (int) – stride for the kernel in convolutional operations padding (int, optional) – padding for the image to handle edges while convoluting (default: 0) dilation (int, optional) – dilation for the convolutional operation (default: 1) activation (str) – activation function to be used for the layer (default: None)\n\n### Normalization¶\n\nclass `normalization.``BatchNorm1d`(eps=1e-05, momentum=0.1)\n\nBases: `normalization._BatchNorm`\n\n1-D batch normalization layer.\n\nParameters: eps (float) – a value added to the denominator for numerical stability (default: 1e-5) momentum (float) – the value used for the running_mean and running_var computation. Can be set to None for cumulative moving average (i.e. simple average) (default: 0.1)\nclass `normalization.``BatchNorm2d`(eps=1e-05, momentum=0.1)\n\nBases: `normalization._BatchNorm`\n\n2-D batch normalization layer.\n\nParameters: eps (float) – a value added to the denominator for numerical stability (default: 1e-5) momentum (float) – the value used for the running_mean and running_var computation. Can be set to None for cumulative moving average (i.e. simple average) (default: 0.1)\nclass `normalization.``BatchNorm3d`(eps=1e-05, momentum=0.1)\n\nBases: `normalization._BatchNorm`\n\n3-D batch normalization layer.\n\nParameters: eps (float) – a value added to the denominator for numerical stability (default: 1e-5) momentum (float) – the value used for the running_mean and running_var computation. Can be set to None for cumulative moving average (i.e. simple average) (default: 0.1)\n\n### Pooling¶\n\n#### 1-D¶\n\nclass `pooling.``MaxPool1d`(kernel_size, stride, padding=0, dilation=1)\n\nBases: `pooling._Pooling1d`\n\n1-D max pooling layer.\n\nParameters: kernel_size (int) – size of kernel to be used for pooling operation stride (int) – stride for the kernel in pooling operations padding (int, optional) – padding for the image to handle edges while pooling (default: 0) dilation (int, optional) – dilation for the pooling operation (default: 1)\nclass `pooling.``AvgPool1d`(kernel_size, stride, padding=0)\n\nBases: `pooling._Pooling1d`\n\n1-D average pooling layer.\n\nParameters: kernel_size (int) – size of kernel to be used for pooling operation stride (int) – stride for the kernel in pooling operations padding (int, optional) – padding for the image to handle edges while pooling (default: 0)\n\n#### 2-D¶\n\nclass `pooling.``MaxPool2d`(kernel_size, stride, padding=0, dilation=1)\n\nBases: `pooling._Pooling2d`\n\n2-D max pooling layer.\n\nParameters: kernel_size (int) – size of kernel to be used for pooling operation stride (int) – stride for the kernel in pooling operations padding (int, optional) – padding for the image to handle edges while pooling (default: 0) dilation (int, optional) – dilation for the pooling operation (default: 1)\nclass `pooling.``AvgPool2d`(kernel_size, stride, padding=0)\n\nBases: `pooling._Pooling2d`\n\n2-D average pooling layer.\n\nParameters: kernel_size (int) – size of kernel to be used for pooling operation stride (int) – stride for the kernel in pooling operations padding (int, optional) – padding for the image to handle edges while pooling (default: 0)\n\n#### 3-D¶\n\nclass `pooling.``MaxPool3d`(kernel_size, stride, padding=0, dilation=1)\n\nBases: `pooling._Pooling3d`\n\n3-D max pooling layer.\n\nParameters: kernel_size (int) – size of kernel to be used for pooling operation stride (int) – stride for the kernel in pooling operations padding (int, optional) – padding for the image to handle edges while pooling (default: 0) dilation (int, optional) – dilation for the pooling operation (default: 1)\nclass `pooling.``AvgPool3d`(kernel_size, stride, padding=0)\n\nBases: `pooling._Pooling3d`\n\n3-D average pooling layer.\n\nParameters: kernel_size (int) – size of kernel to be used for pooling operation stride (int) – stride for the kernel in pooling operations padding (int, optional) – padding for the image to handle edges while pooling (default: 0)\n\n### HSIC¶\n\nclass `hsic_output.``HSICoutput`(output_dim, activation='softmax')\n\nBases: `glow.layers.core.Dense`\n\nClass for HSIC sigma network output layer. This class extends functionalities of `glow.layers.Dense` with more robust features to serve for HSIC sigma network purposes.\n\nParameters: output_dim (int) – output dimension of the HSIC output layer used after pre-training phase activation (str, optional) – activation function to be used for the layer (default: softmax)\n\n## Information Bottleneck¶\n\n### Estimator¶\n\nclass `estimator.``Estimator`(gpu, **kwargs)\n\nBase class for all the estimator modules.\n\nYour estimator should also subclass this class.\n\nThis Class is for implementing functionalities to estimate different dependence criterion in information theory like mutual information etc. These methods are further used in analysing training dyanmics of different architechures.\n\nParameters: gpu (bool) – if true then all the computation is carried on GPU else on CPU **kwargs – the keyword that stores parameters for the estimators\n`criterion`(x, y)\n\nDefines the criterion of the estimator for example EDGE algorithm have mutual information as its criterion. Generally criterion is some kind of dependence or independence measure between x and y. In the context of information theory most widely used criterion is mutual information between the two arguments.\n\nParameters: x (torch.Tensor) – first random variable y (torch.Tensor) – second random variable calculated criterion of the two random variables ‘x’ and ‘y’ (torch.Tensor)\n`eval_dynamics_segment`(dynamics_segment)\n\nProcess smallest segment of dynamics and calculate coordinates using the defined criterion.\n\nParameters: dynamics_segment (iterable) – smallest segment of the dynamics of a batch containing input, hidden layer output and label in form of `torch.Tensor` objects list of calculated coordinates according to the criterion with length equal to ‘len(dynamics_segment)-2’ (iterable)\nclass `estimator.``HSIC`(kernel, gpu=True, **kwargs)\n\nClass for estimating Hilbert-Schmidt Independence Criterion as done in paper “The HSIC Bottleneck: Deep Learning without Back-Propagation”.\n\nParameters: kernel (str) – kernel which is used for calculating K matrix in HSIC criterion gpu (bool) – if true then all the computation is carried on GPU else on CPU **kwargs – the keyword that stores parameters for HSIC criterion\n`criterion`(x, y)\n\nDefines the HSIC criterion.\n\n## Preprocessing¶\n\nclass `data_generator.``DataGenerator`\n\nclass for implementing data generators and loaders.\n\n`prepare_numpy_data`(x_train, y_train, batch_size, validation_split)\n\nConverts numpy type dataset into PyTorch data-loader type dataset.\n\nParameters: x_train (numpy.ndarray) – training input dataset y_train (numpy.ndarray) – training ground-truth labels batch_size (int) – batch size of a single batch validation_split (float): proportion of the total dataset which is used for validation contains training data-loader with processed batches val_loader (torch.utils.data.DataLoader): contains validation data-loader with processed batches train_loader (torch.utils.data.DataLoader)\n`set_dataset`(X, y, batch_size, validation_split=0.2)\n\nConverts raw dataset into processed batched dataset loaders for training and validation.\n\nParameters: X (torch.Tensor) – input dataset y (torch.Tensor) – labels batch_size (int) – batch size of a single batch validation_split (float): proportion of the total dataset which is used for validation contains training data-loader with processed batches validation_dataset (torch.utils.data.DataLoader): contains validation data-loader with processed batches train_dataset (torch.utils.data.DataLoader)"
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https://yutsumura.com/is-the-derivative-linear-transformation-diagonalizable/ | [
"# Is the Derivative Linear Transformation Diagonalizable?",
null,
"## Problem 690\n\nLet $\\mathrm{P}_2$ denote the vector space of polynomials of degree $2$ or less, and let $T : \\mathrm{P}_2 \\rightarrow \\mathrm{P}_2$ be the derivative linear transformation, defined by\n$T( ax^2 + bx + c ) = 2ax + b .$\n\nIs $T$ diagonalizable? If so, find a diagonal matrix which represents $T$. If not, explain why not.",
null,
"Add to solve later\n\n## Solution.\n\nThe standard basis of the vector space $\\mathrm{P}_2$ is the set $B = \\{ 1 , x , x^2 \\}$. The matrix representing $T$ with respect to this basis is\n$[T]_B = \\begin{bmatrix} 0 & 1 & 0 \\\\ 0 & 0 & 2 \\\\ 0 & 0 & 0 \\end{bmatrix} .$\n\nThe characteristic polynomial of this matrix is\n$\\det ( [T]_B – \\lambda I ) = \\begin{vmatrix} -\\lambda & 1 & 0 \\\\ 0 & -\\lambda & 2 \\\\ 0 & 0 & -\\lambda \\end{vmatrix} = \\, – \\lambda^3 .$ We see that the only eigenvalue of $T$ is $0$ with algebraic multiplicity $3$.\n\nOn the other hand, a polynomial $f(x)$ satisfies $T(f)(x) = 0$ if and only if $f(x) = c$ is a constant. The null space of $T$ is spanned by the single constant polynomial $\\mathbb{1}(x) = 1$, and thus is one-dimensional. This means that the geometric multiplicity of the eigenvalue $0$ is only $1$.\n\nBecause the geometric multiplicity of $0$ is less than the algebraic multiplicity, the map $T$ is defective, and thus not diagonalizable.",
null,
"Add to solve later\n\n### More from my site\n\n#### You may also like...\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.\n\n###### More in Linear Algebra",
null,
"##### Dot Product, Lengths, and Distances of Complex Vectors\n\nFor this problem, use the complex vectors \\[ \\mathbf{w}_1 = \\begin{bmatrix} 1 + i \\\\ 1 - i \\\\ 0...\n\nClose"
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https://www.clutchprep.com/chemistry/balancing-redox-reaction | [
"Ch.18 - ElectrochemistryWorksheetSee all chapters\n\n# Balancing Redox Reaction\n\nSee all sections\nSections\nRedox Reaction\nBalancing Redox Reaction\nThe Nernst Equation\nGalvanic Cell\nBatteries and Electricity\nNernst Equation (IGNORE)\n###### Balancing Oxidation-Reduction Reactions\n\nConcept #1: Balancing Redox Reactions\n\nTranscript\n\nHey guys. We're going to continue with our discussion of redox reactions by now learning how to balance them whether they're in acidic solutions or basic solutions. So, we know the ins and outs of redox reactions, we know oxidation, we know the reduction from the past videos, but now if our goal is to learn how to balance then when they're in acidic solutions or basic solutions and to do that we're going to follow a different set of rules, they almost go hand-in-hand together except when we get to a basic solution we have to add an additional step. So, first let's to balance it out in an acidic solution. So, here going to say step one, we have to write the equation, the full redox equation first into two half reactions and we'll see exactly how do we do this but basically we look for elements different from oxygen or hydrogen first on either sides of the arrows and bring those down, separating them into two separate equations, once we do that we balance out these elements that are not oxygen or hydrogen's first, once that's done, we can move on to step three, we have to balance our oxygens by adding molecules of water to the opposite side so that both sides have the same number of oxygen, what we do next after that is we have to balance out hydrogen by adding H plus. Now, here comes the tricky part after, we do this, we have to balance out the overall charge of that half reaction by adding electrons to the more positive side and we have to make sure that when we do this, the electrons in both half reactions must match and if they don't, that means we're going to have to multiply either one of our half reactions or maybe even both. So, we're going to say electrons from both half-reactions must match, if not that means we have to balance either one of half reactions or both of them again. Alright, after we do that, once our electrons match up in both half reactions, we can then combine those two half reactions back into our full redox reaction and this time we're going to cross out intermediates and remember intermediates are species that look alike except one will be a reactant and one will be a product, those will cancel out member, if they're on the same side, let's say they're both reactants of products, they don't cancel out at all they actually add up together. So, remember intermediates ones are reactant ones are product, they look alike they'll cancel out, once we've done this an acidic solution, basic solution is basically almost all the same rules except for one additional rule at the end. So, for imbalancing in basic solutions, we follow rules, and once we've accomplished this, we're going to, we're going to balance H plus by adding OH minus ions to both sides of the chemical reaction, when we get to basic solutions, we'll see quickly, what this really means. So, remember it's imperative that you guys first remember the basic rules that we went over, because this tells us what's being reduced and what's being oxidized, remember if you've been reduced you're the oxidizing agent but if you've been oxidized you're the reducing agent, mastering that helps you to move on to this aspect of redox reactions whereas to balance them out and remember you have to remember all the guidelines all the steps for either acidic or basic in order to get the full credit for that question. So, hopefully you guys can remember these rules and will apply them on the next series of videos on how to balance redox reactions.\n\nConcept #2: Balancing Acidic Redox Reactions\n\nTranscript\n\nHey guys. In this new video we're going to put to practice some of the rules that we learn to balance redox reactions in acidic solution. So, let's take a look at this first one, here we need to balance out this redox reaction in an acidic solution. So, let's go over the rules that we know, first we're going to break this up into two half reactions. So, we look for the elements different from oxygen or hydrogen first, here we have an N here, here we have another N right here. So, we're going to do is we're going to bring them both down and say that represents our first half reaction. So, we're going to say NO2- gives us NO3-, here we have Mn and, here we have Mn. So, those will also give us another half reaction MnO4- gives us Mn2+. Alright, next we have to balance out elements different from oxygen or hydrogen. So, what we need to see is on the first half reaction, we only have one nitrogen. So, we don't have to worry about balancing they're already balanced, on the other side we have one manganese on each side. So, we don't need to balance that either. Now, we move on to balancing out oxygens by adding water, on this side we have two oxygens, on this side we have three. So, we need to put one mole of water, or one, yeah one mole of water on the left side. So, now that both sides have three, let's look at the other half reaction, here this has four but the other side has none. So, we have to add four moles of water. Now, we balanced out oxygen. Now, it's time to balance out H+. So, looking back on the left half reaction, we're going to say we have two H's right here, but we have none on the product side. So, we have to add two H+ to this side, then looking at the other half reaction, we have 4 times 2, we have 8 H's there. So, we have to add eight H+ here. Now we, this is the tricky part, we have to balance that overall charge. So, we're going to say here is we're going to say water is neutral, there's no charge on this. So, we're going to say it's charge is 0, but nitrite ion NO2- has a negative one charge. So, we're going to say overall this side is negative one, let's look at the other side, here we have H plus for H, but there's two of them so that really counts as plus 2 and then we're going to say our nitrate ion NO3- has a minus one charge. So, that's minus 1. So, overall the charge on this side is plus 1, remember, we have to add electrons to the more positive side so that both sides have the same exact charge. So, we're going to add electrons to the plus 1 side.\n\nNow, we need that plus 1 to become a negative 1 just like on the left side. So, how many electrons do we need to add? we need to add two electrons, because adding two electrons is equivalent to adding a negative 2. So, here we're going to say plus 1 minus 2 equals minus 1. So, both sides are now negative 1, let's go to the other side, here we have eight plusses. So, that's plus 8, we have a negative one here. So, that's minus 1. So, this side is plus 7, here on the other side water again is neutral has no charge. So, it's 0, here we have plus two. So, this side is plus 2 overall. Now, we need to add electrons to the more positive side. So, we're going to add them to the left side because it's plus 7, we need both sides to be the same exact charge. So, how many electrons do I need to add two plus seven so that drops down to plus two, we're going to have to add five electrons, because remember, each electron is negative. So, adding five electrons is equivalent to doing negative 5 plus 7 minus 5 equals plus 2, both sides are now plus 2. Now, finally, we have to check to make sure our electrons are equal, here we have two electrons, but here we have five, they're not the same number. So, we're going to do here is we have to multiply both half reactions by a number so that they both have equal number of electrons so the common multiple between two and five is 10. So, we multiply this whole thing here, times 5 and we multiply this whole thing times 1, and remember, we're doing this because we need to have the same number of electrons on in both half reactions, if we don't then it can't be a valid way of balancing it in acidic solution. Alright, so everything gets multiplied by five. So, we're going to have five NO2- plus 5 waters, gives me 5 and a 3- plus 10 H+ plus 10 electrons and what you should remember here, just remember this for later on, we have electrons as products, if you have electrons as products then this is an oxidation. So, remember, if you have electron this product it's an oxidation, let's look at the other one, everyone is getting multiplied by two. So, this would be 2 MnO4- plus 8 times 2, 16 H+, plus 10 electrons gives me 2 Mn2+, plus 8 H2O, and look we have electrons in the second one as reactants. So, electrons are reactance, reactance means it's reduction, okay? So, remember, if your electrons are products it's oxidation, but if your electrons are reactants its reduction. Alright, so we have to make sure that the electrons cancel out and they do, one's a product ones are reacting, they're intermediates, let's see who else can be intermediates, we can say that here, these 10 H+ that are products, all of them get canceled out by 10 from here, leaving us with 6 left, we're going to say all 5 of these waters that are reactants get cancelled out by 5 from here, leaving us with 3 left, and it looks like those are our only intermediates that we have, everything left we bring it down. So, what we have left at the end, we're going to have 2 MnO4- plus 5 NO2- plus 6 H+, gives me 5 NO3- plus 2Mn 2+ plus 3H2O, so that would be our final answer when balancing this redox reaction in acidic solution, I know it's a lot of moves, a lot of steps to remember, but remember, go at it slowly if you're getting lost, I think the worst part of this whole thing is really looking at the overall charge, but just remember, if you don't see the compound with any type of charge it's 0, if it has a coefficient in front of it that would multiply the charge, whatever it is, making it bigger, and as long as you can get those basics down you'll know how to add the number of electrons you need. So, both sides have the same overall charge and again remember, your electrons in both half reactions must match up, if they don't then it's done incorrectly. Now, that we've done this one I want you guys to attempt to do the next one on your own. So, it's the same basic concept. So, we're going over the same rules, break it up into half reactions first and from there we take it on to do the rest of the steps to balance it in acidic solution, if you get lost it's, okay? Just click on the explanation button and a video of me will show you how best to approach this next problem, good luck guys.\n\nPractice: Balance the following reaction in an acidic solution.\n\nCl2(g) + S2O32-(aq) ----> Cl-(aq) + SO42-(aq)\n\nConcept #3: Balancing Basic Redox Reactions\n\nTranscript\n\nHey guys in this new video we're finally going to look at how do we balance the redox reaction in a basic solution. So, let's take a look at the first example. So, here we're going to have to balance out this massive reaction in a basic solution. So, let's just follow the rules that we know first, when following an acidic solution because remember, the first six rules are the same for basic solutions. So, here we have Mo and here, we have Mo. So, let's bring that down as its own half reaction and here, we have Br and here, we have Br, let's bring those down as their own half reaction next, we have to make sure we balance out elements that are different from oxygen or hydrogen. So, here we have three Mo's but over here we only have one. So, we're going to throw a three in front of that and then on the other side both bromines are just one. So, we don't have to worry about anything there, next we have to balance out oxygens by adding water, on this side, we have nine oxygens on the other side we have none. So, we have to add nine waters here we have four oxygens but on the other side we have none. So, we're going to add four waters, next we're going to balance out H plus. So, here we have nine times two that gives us 18. So, we're going to put 18 H plus on this side and then here, we have 4 times 2, that's 8. So, we have to add 8 h plus to this side. Now, for the tricky part overall charge. Now, the overall charge here, we have 18 times plus 1. So, that's plus 18, minus 3, it'll be plus 15 overall on this side, on the other side Mo and water are both neutral, we don't show any charges for them. So, overall it's 0 on this side then on the other side we have, water is neutral again, but we have a negative with the Br. So, this side is negative 1 overall, then we're going to say, we have 8 times plus 1. So, that's plus 8 and then here this is a minus 2. So, here this is plus 6 on this side. So, we need to balance out the overall charge by adding electrons to the more positive side. So, on this side, we're going to add 15 electrons so that both sides are 0 and then here, plus 6, we're going to add how many electrons we need to get to negative 1, we need to add 7 electrons. Now, we're going to say the electrons definitely don't match. So, we have to think of a number we can multiply them both by so that they both match. So, we have to do here is we have to multiply this 1 by 7 and we multiply this 1 by 15 because the common number between them is 105. So, now we're going to have really big numbers we have to deal with. So, we're going to have 7 Mo3 O9 3 minus plus, we have 18 times 7. So, it's 126 h plus, plus 105 electrons, gives me 45 Mo plus, we're going to have 15 I'm actually, sorry, we're multiplying by 7, I was looking at the 15. So, it's 7 times 3, which is 21 and then we have 7 times 9, which is 63, okay? So, remember to multiplied by 7, I was looking at the 15 for a moment but it's times 7. Now, we look at the other one. So, here we are going to have 15 multiplying every one. So, we're going to have 15 Br minus plus 60 H2O gives me 15 Br O4 2 minus, plus, we're going to have 15 times 8, which is 120 plus 105 electrons. So, there's a lot of big numbers going on here.\n\nNow, we have to cross on intermediates. So, who looks like but ones are reacting one is a product, all 120 of these h plus cancel out with 120 from here leaving us with 6, all 105 electrons of course have to cancel out, all 60 of these waters cancel out with 60 from here, leaving us with 3, bring down everything. So, at the end we're just going to have a balanced redox reaction in an acidic solution but we need it in a basic solution. So, we have to do one additional step. So, we said we have to balance out H plus by adding OH minus to both sides. So, how many H pluses do we have, we have 6 here. So, that means we have to add 6 OH minus here, plus 6 OH minus here and remember, what happens when you have H plus and OH minus together, they're opposite charges so they're going to combine to give us water. So, 6OH minus 6H plus combined to give us six waters but we can have water on both sides of the equation. So, what's going to happen here they're going to cancel each other out. So, all three of these would cancel out with 3 from here, leaving us with three waters. So, we'd say the final answer, we'll just bring down everyone, okay? So, we get this for our final answer, so that last step is a little bit different from what we're custom to seeing but that's the last step you'd have to use in order to bounce it in a basic solution. So, look at how many H+ you have left add OH- to both sides equal to that number. Remember, H+ OH- together gives us water and then you have to cancel out the waters from both sides of the reactant and products, whoever is left is whoever's left, at the end nothing happens to this OH- on this side. So, we just bring it down, so that would be our final answer. Now, that you guys have seen this I want you guys to attempt the next one and I'll give you guys some help on the last one because it's different than all the others, here the common element found in both is xenon. So, I'll actually help you out here, when it's a xenon. So, XeO2 gives me H2Xe and XeO2 gives me XeO4, since xenon is found in both xenon will be part of both half reactions xenon O2. So, now that I set that up for you just follow the rules that we've learned from all the other examples in order to balance it first in an acidic solution, soak it in a basic solution, good luck guys.\n\nPractice: Balance the following reaction in a basic solution.\n\nXeO2 (aq) ----> H2Xe (aq) + XeO4 (aq)"
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https://sqlserverfast.com/blog/hugo/2006/09/the-prime-number-challenge-great-waste-of-time/ | [
"# The prime number challenge – great waste of time!\n\nNo sane person would even consider using SQL Server to construct a list of prime numbers. So just to prove that I’m not sane (as if there could be any doubt!), this post will be about finding prime numbers.\n\nFirst a bit of history. Ward Pond wrote about efficient ways to populate a table with one million GUIDs. I posted a comment with a slightly more efficient algorithm. And that was quickly followed by a new post from Ward, tweaking my syntax even further. And that’s when Denis the SQL Menace lived true to his name by posting this comment:\n\n“How about the next challenge is to return all 78498 prime numbers between 1 and 1000000?”\n\nNow of course, this is a silly challenge. Not because prime numbers are silly, mind you. They are very useful to mathematicians, and many encryption algorithms wouldn’t even be possible without (large) prime numbers. The silly part is using SQL Server, a data manipulation tool, to calculate prime numbers. If you really need them, code a quick algorithm in a C++ program. Or buy a ready-made list with the first million or so prime numbers. So I attempted to resist the challenge.\n\nAlas – the flesh is weak. So when I saw Ward’s reply to Denis’ challenge, I was no longer able to resist temptation. After all, Ward’s attempt is not only interesting – it is also very long, and apparently (with an estimated completion time of 1 to 2 days!!) not very efficient. I decided that I should be able to outperform that.\n\nMy assumptions are that a table of numbers is already available, and that this table holds at least all numbers from 1 to 1,000,000. (Mine holds numbers from 1 to 5,764,801), and that the challenge is to create and populate a table with the prime numbers from 1 to 1,000,000, in as little speed as possible. Displaying the prime numbers is not part of the challenge. For testing purposes, I replaced the upper limit of 1,000,000 with a variable @Limit, and I set this to a mere 10,000. That saves me a lot of idle waiting time!\n\nAs a first attempt, I decided to play dumb. Just use one single set-based query that holds the definition of prime number. Here’s the SQL:\n\nDROP TABLE dbo.Primes\n\ngo\n\nCREATE TABLE dbo.Primes\n\n(Prime int NOT NULL PRIMARY KEY)\n\ngo\n\nDECLARE @Start datetime, @End datetime\n\nSET @Start = CURRENT_TIMESTAMP\n\nDECLARE @Limit int\n\nSET @Limit = 10000\n\nINSERT INTO dbo.Primes (Prime)\n\nSELECT n1.Number\n\nFROM dbo.Numbers AS n1\n\nWHERE n1.Number > 1\n\nAND n1.Number < @Limit\n\nAND NOT EXISTS\n\n(SELECT\n\nFROM dbo.Numbers AS n2\n\nWHERE n2.Number > 1\n\nAND n2.Number < n1.Number\n\nAND n1.Number % n2.Number = 0)\n\nSET @End = CURRENT_TIMESTAMP\n\nSELECT @Start AS Start_time, @End AS End_time,\n\nDATEDIFF(ms, @Start, @End) AS Duration,\n\nCOUNT() AS Primes_found, @Limit AS Limit\n\nFROM dbo.Primes\n\ngo\n\n–select * from dbo.Primes\n\ngo\n\nThis ran in 1,530 ms on my test system. (And in case you ask – I also tested the equivalent query with LEFT JOIN; that took 11,466 ms, so I quickly discarded it). With @Limit set to 20,000 and 40,000, execution times were 5,263 and 18,703 ms – so each time we double @Limit, execution time grows with a factor 3.5. Using this factor, I can estimate an execution time of one to two hours.\n\nThat’s a lot better than the one to two days Ward estimates for his version – but not quite fast enough for me. So I decided to try to convert the Sieve of Eratosthenes to T-SQL. This algorithm is known to be both simple and fast for getting a list of prime numbers. Here’s my first attempt:\n\nDROP TABLE dbo.Primes\n\ngo\n\nCREATE TABLE dbo.Primes\n\n(Prime int NOT NULL PRIMARY KEY)\n\ngo\n\nDECLARE @Start datetime, @End datetime\n\nSET @Start = CURRENT_TIMESTAMP\n\nDECLARE @Limit int\n\nSET @Limit = 10000\n\n— Initial fill of sieve;\n\n— filter out the even numbers right from the start.\n\nINSERT INTO dbo.Primes (Prime)\n\nSELECT Number\n\nFROM dbo.Numbers\n\nWHERE (Number % 2 <> 0 OR Number = 2)\n\nAND Number <> 1\n\nAND Number <= @Limit\n\n— Set @Current to 2, since multiples of 2 have already been processed\n\nDECLARE @Current int\n\nSET @Current = 2\n\nWHILE @Current < SQRT(@Limit)\n\nBEGIN\n\n— Find next prime to process\n\nSET @Current =\n\n(SELECT TOP (1) Prime\n\nFROM dbo.Primes\n\nWHERE Prime > @Current\n\nORDER BY Prime)\n\nDELETE FROM dbo.Primes\n\nWHERE Prime IN (SELECT n.Number @Current\n\nFROM dbo.Numbers AS n\n\nWHERE n.Number >= @Current\n\nAND n.Number <= @Limit / @Current)\n\nEND\n\nSET @End = CURRENT_TIMESTAMP\n\nSELECT @Start AS Start_time, @End AS End_time,\n\nDATEDIFF(ms, @Start, @End) AS Duration,\n\nCOUNT() AS Primes_found, @Limit AS Limit\n\nFROM dbo.Primes\n\ngo\n\n–select * from dbo.Primes\n\nThe time that Eratosthenes takes to find the prime numbers up to 10,000 is 7,750 ms – much longer than the previous version. My only hope was that the execution time would not increase with a factor of 3.5 when doubling @Limit – and indeed, it didn’t. With @Limit set to 20,000 and 40,000, execution times were 31,126 and 124,923 ms, so the factor has gone up to 4. With @Limit set to 1,000,000, I expect an execution time of 15 to 20 hours.\n\nTime to ditch the sieve? No, not at all. Time to make use of the fact that SQL Server prefers to process whole sets at a time. Let’s look at the algorithm in more detail – after the initial INSERT that fills the sieve and removes the multiples of 2, it finds 3 as the next prime number and removes multiples of 3. It starts removing at 9 (3 squared) – so we can be pretty sure that numbers below 9 that have not yet been removed will never be removed anymore. Why process them one at a time? Why not process them all at once? That’s what the algorithm below does – on the first pass of the WHILE loop, it takes the last processed number (2), finds the first prime after that (3), then uses the square of that number (9) to define the range of numbers in the sieve that are now guaranteed to be primes. It then removes multiples of all primes in that range. And after that, it repeats the operation, this time removing multiples in the range between 11 (first prime after 9) and 121 (11 square). Let’s see how this affects performance.\n\nDROP TABLE dbo.Primes\n\ngo\n\nCREATE TABLE dbo.Primes\n\n(Prime int NOT NULL PRIMARY KEY)\n\ngo\n\nDECLARE @Start datetime, @End datetime\n\nSET @Start = CURRENT_TIMESTAMP\n\nDECLARE @Limit int\n\nSET @Limit = 10000\n\n— Initial fill of sieve;\n\n— filter out the even numbers right from the start.\n\nINSERT INTO dbo.Primes (Prime)\n\nSELECT Number\n\nFROM dbo.Numbers\n\nWHERE (Number % 2 <> 0 OR Number = 2)\n\nAND Number <> 1\n\nAND Number <= @Limit\n\n— Set @Last to 2, since multiples of 2 have already been processed\n\nDECLARE @First int, @Last int\n\nSET @Last = 2\n\nWHILE @Last < SQRT(@Limit)\n\nBEGIN\n\n— Find next prime as start of next range\n\nSET @First =\n\n(SELECT TOP (1) Prime\n\nFROM dbo.Primes\n\nWHERE Prime > @Last\n\nORDER BY Prime)\n\n— Range to process ends at square of starting point\n\nSET @Last = @First @First\n\nDELETE FROM dbo.Primes\n\nWHERE Prime IN (SELECT n.Number * p.Prime\n\nFROM dbo.Primes AS p\n\nINNER JOIN dbo.Numbers AS n\n\nON n.Number >= p.Prime\n\nAND n.Number <= @Limit / p.Prime\n\nWHERE p.Prime >= @First\n\nAND p.Prime < @Last)\n\nEND\n\nSET @End = CURRENT_TIMESTAMP\n\nSELECT @Start AS Start_time, @End AS End_time,\n\nDATEDIFF(ms, @Start, @End) AS Duration,\n\nCOUNT() AS Primes_found, @Limit AS Limit\n\nFROM dbo.Primes\n\ngo\n\n–select * from dbo.Primes\n\nThe time taken for the 1,229 primes between 1 and 10,000? A mere 266 ms!! With execution times like that, I saw no need to rely on extrapolation – I set @Limit to 1,000,000, hit the Execute button, sat back – and get the following result after 19 seconds:\n\nStart_time End_time Duration Primes_found Limit\n\n———————– ———————– ———– ———— ———–\n\n2006-09-24 00:42:22.780 2006-09-24 00:42:41.750 18970 78498 1000000\n\nFrom 1-2 days to just under 20 seconds – and this time, I didn’t even have to add an index!\n\nFinally, to top things off, I tried one more thing. I have often read that SQL Server won’t optimize an IN clause as well as an EXISTS clause, especially if the subquery after IN returns a lot of rows – which is definitely the case here. So I rewrote the DELETE statement in the heart of the WHILE loop to read like this:\n\nDELETE FROM dbo.Primes\n\nWHERE EXISTS\n\n(SELECT *\n\nFROM dbo.Primes AS p\n\nINNER JOIN dbo.Numbers AS n\n\nON n.Number >= p.Prime\n\nAND n.Number <= @Limit / p.Prime\n\nWHERE p.Prime >= @First\n\nAND p.Prime < @Last\n\nAND Primes.Prime = n.Number * p.Prime)\n\nAnd here are the results:\n\nStart_time End_time Duration Primes_found Limit\n\n———————– ———————– ———– ———— ———–\n\n2006-09-24 00:47:42.797 2006-09-24 00:48:01.903 19106 78498 1000000\n\nWhich just goes to prove that you shouldn’t believe everything you read, I guess. <g>\n\n#### Related Posts\n\n•",
null,
"Denis The SQL Menace\nSeptember 24, 2006 00:19\n\nHugo, let me know how many seconds (yes seconds) this version takes\n\nSET NOCOUNT ON\n\nDECLARE @i INT\n\n— Create a 10-digit table\nDECLARE @D TABLE (N INT)\nINSERT INTO @D (N)\nSELECT 0 UNION ALL\nSELECT 1 UNION ALL\nSELECT 2 UNION ALL\nSELECT 3 UNION ALL\nSELECT 4\n\nINSERT INTO @D (N)\nSELECT N+5 FROM @D\n\n— build a small sieve table between 2 and 1000\nDECLARE @T TABLE (N INT)\nINSERT INTO @T( N )\nSELECT 1+A.N+10(B.N+10C.N)\nFROM @D A, @D B, @D C\n\nDELETE FROM @T WHERE N = 1\n\nSET @I = 2\nWHILE @I <= SQRT(1000) BEGIN DELETE FROM @T WHERE N % @I = 0 AND N > @I\nSET @I = @I + 1\nEND\n\n— Create large table between 1001 and 1000000\nSELECT A+10(B+10(C+10(D+10(E+ 10F)))) AS N\nINTO #P\nFROM\n( SELECT A.N AS A, B.N AS B, C.N AS C, D.N AS D, E.N AS E, F.N AS F\nFROM @D A, @D B, @D C, @D D, @D E, @D F\nWHERE A.N in (1, 3, 7, 9) — Not divisible by 2 or 5\n) blah\nWHERE (A+B+C+D+E+F) % 3 <> 0 — Or 3\nAND (A+3\nB+2C-D-3E-2F) % 7 <> 0 — Or 7\nAND (B-A+D-C+F-E) % 11 <> 0 — Or 11\nAND D|E|F <> 0 — Don’t include the first 1000 numbers,\n–we already have these in the small sieve table\nUNION ALL SELECT 1000000\n\n— sieve the big table with smaller one\nSELECT @I = 2\nWHILE @I IS NOT NULL\nBEGIN\nDELETE FROM #P WHERE N% @I = 0\nSELECT @I = MIN(N) FROM @T WHERE N > @I\nEND\n\n— add primes up to 1000\nINSERT INTO #P SELECT N FROM @T\n\n— Here are the results\n–78498 rows\nSELECT\nFROM #P ORDER BY 1\n\ndrop table #P\ngo\n\n•",
null,
"Ward Pond\nSeptember 24, 2006 06:02\n\nSix seconds on my laptop, Denis.. pretty cool!\n\n•",
null,
"Path Enumeration using Prime number products\nSeptember 25, 2006 08:10\n•",
null,
"Lonedog\nOctober 18, 2006 21:54\n\nWHERE (A+B+C+D+E+F) % 3 <> 0 — Or 3\nAND (A+3B+2C-D-3E-2F) % 7 <> 0 — Or 7\nAND (B-A+D-C+F-E) % 11 <> 0 — Or 11\n\nIs there a mathematical reference for any of the lines?\n\n•",
null,
"Hugo Kornelis\nOctober 23, 2006 22:36\n\nHi Lonedog,\n\nI understand less than half of it myself, but I think that http://en.wikipedia.org/wiki/Divisibility_rule should be the reference you want.\n\nI must admit the the (A+B+C+D+E+F) % 3 <> 0 is the only part I understand <g>.\n\n•",
null,
"Ward Pond\nJanuary 23, 2007 06:40\n\nThis is the modulus operator, which returns the remainder of a division by the operand. So, if one Denis’ modulus operations returns 0, the calculated expression is evenly divisible by the operand. This is a very performant way of finding factors.\n\n•",
null,
"RBArryYoung\nFebruary 27, 2009 08:03\n\nWhat he is doing is what we used to cal \"digital factoring\" (before that came to mean something else), wherein he is using the decimal digits of the number to calculate \"digital roots\" for various primes which shortcut methods for determining their remainder modulo that prime.\n\nThese are tricks that are used by mathemagicians, etc. to do various mental calculations.\n\n•",
null,
"RBArryYoung\nFebruary 28, 2009 08:37\n\nHere is what I would use. It runs about 15-20% faster on my system than Denis’s (of course, Ive had an extra three years too):\n\nIf Not (object_id(‘tempdb..#Sieve2’) is Null) Drop table #Sieve2\n\nIf Not (object_id(‘tempdb..#Sieve3’) is Null) Drop table #Sieve3\n\nIf Not (object_id(‘tempdb..#Candidates1M’) is Null) Drop table #Candidates1M\n\nIf Not (object_id(‘tempdb..#Candidates1Ma’) is Null) Drop table #Candidates1Ma\n\n;WITH Primes7to36 as (\n\n` Select 7 as p`\n\n``` Union ALL Select 11 Union ALL Select 13 Union ALL Select 17 Union ALL Select 19 Union ALL Select 23 Union ALL Select 29 ```\n\n```Union ALL Select 31) ```\n\nSelect p\n\nInto #Sieve2\n\nFrom Primes7to36\n\n;WITH Base30 as (Select 1 as rem\n\n`Union Select 7 as p`\n\n``` Union ALL Select 11 Union ALL Select 13 Union ALL Select 17 Union ALL Select 19 Union ALL Select 23 ```\n\n```Union ALL Select 29) ```\n\n, Numbers10E4 as (Select TOP 10000\n\n`ROW_NUMBER() Over(Order by id) as Num`\n\n``` ```\n\n```From master.sys.syscolumns) ```\n\n, Numbers34 as (Select Top 34 Num\n\n```From Numbers10E4) ```\n\n, Candidates1000 as (\n\n`SELECT rem+30*Num as Cand`\n\n``` From Base30 ```\n\n``` Cross Join Numbers34) ```\n\n, Primes7to1000 as (\n\n` Select p From #sieve2 --Primes7to36`\n\n``` Where p<>7 and p<>11 UNION ALL Select Cand as p From Candidates1000 Where Cand <= 1000 And Not Exists(Select * From #Sieve2 Where (Cand % p)=0)) ```\n\nSELECT p\n\nInto #Sieve3\n\nFrom Primes7to1000\n\n;WITH Base30 as (Select 1 as rem\n\n`Union Select 7 as p`\n\n``` Union ALL Select 11 Union ALL Select 13 Union ALL Select 17 Union ALL Select 19 Union ALL Select 23 ```\n\n```Union ALL Select 29) ```\n\n, Base90 as (Select rem as rem From Base30\n\n`Union ALL Select rem+30 From Base30`\n\n``` ```\n\n```Union ALL Select rem+60 From Base30) ```\n\n, Numbers11120 as (Select TOP 11120\n\n`ROW_NUMBER() Over(Order by id) as Num`\n\n``` ```\n\n```From master.sys.syscolumns) ```\n\n, Candidates1M as (\n\n`Select rem+(90*Num) as Cand`\n\n``` From Base90 ```\n\n``` Cross Join Numbers11120) ```\n\n, Cand2_1M as (\n\n`Select Cand`\n\n``` From Candidates1M Where Cand <= 1000000 And (Cand % 7) <> 0 ```\n\n``` And (Cand % 11) <> 0 ```\n\n)\n\nSelect Cand\n\nInto #Candidates1M\n\nFrom Cand2_1m\n\n;WITH Cand2a as (\n\n`Select Cand as p`\n\n``` From #Candidates1M Where Cand <= 1000000 And Not Exists(Select * From #Sieve3 Where (Cand % p)=0 ) ```\n\n)\n\nSelect p as Cand\n\nInto #Candidates1Ma\n\nFrom Cand2a\n\n;WITH FilterPrimes as (\n\n` Select 2 as p`\n\n``` Union ALL Select 3 Union ALL Select 5 Union ALL Select 7 ```\n\n```Union ALL Select 11 ```\n\n)\n\n, PrimesLE1M as ( Select p From FilterPrimes\n\n`UNION ALL Select p From #Sieve3`\n\n``` UNION ALL Select Cand as p ```\n\n``` From #Candidates1Ma ```\n\n)\n\nSelect p\n\nFrom PrimesLE1M\n\n–=======\n\n— RBarryYoung\n\n•",
null,
"Stan Hudecek\nMarch 7, 2009 13:46\n\nalso wish to prove my (in)sanity…takes about 0.7 sec on my notebook. (no access to real server at the moment) trying to do it all in one sql statement as well. seemed to run faster with a pre-populated a table of ints with an index rather then the row_number, but this seems fairly fast to me.\n\nselect 2 [num] union select 3 union select 5 union\n\nselect num\n\nfrom\n\n(\n\n`select num,min(num%den) [div]`\n\n``` from ( select a.id [num], b.id [den] from ( select top 10000 row_number() over(order by id) [id] from master.sys.syscolumns --from (select top 40000 c1.id from master.sys.syscolumns c1 cross join master.sys.syscolumns c2) [c] ) [a] join ( select top 100 row_number() over(order by id) [id] from master.sys.syscolumns ) [b] on ( a.id % 2 &lt;&gt; 0 and a.id % 3 &lt;&gt; 0 and a.id % 5 &lt;&gt; 0 and b.id &gt; 1 and b.id &lt;= sqrt( a.id) ) ) [base] group by num ```\n\n```having min(num%den)>0 ```\n\n) [cols]\n\n•",
null,
"Jeff Moden\nFebruary 25, 2017 09:12\n\nHi Hugo,\n\nI tried your code on a million rows and it drove all 4 processors on my poor little I5 laptop into the stops. I halted it after 10 minutes.\n\nThe only things I can think of for such a difference between what you got for performance and what I got would be number of CPUs, type of disk, and, of course, how the Numbers table you used was built. Here’s how I built that. Please let me know what kind of box you were using and whether or not I correctly duplicated your Numbers table. Thanks.\n\nCREATE TABLE dbo.Numbers (Number INT PRIMARY KEY CLUSTERED);\n\nGO\n\nINSERT INTO dbo.Numbers WITH(TABLOCK)\n\n(Number)\n\nSELECT TOP 5764801\n\nNumber = ROW_NUMBER()OVER(ORDER BY (SELECT NULL))\n\nFROM sys.all_columns ac1\n\nCROSS JOIN sys.all_columns ac2\n\nOPTION (RECOMPILE)\n\n;"
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http://re.mcalls.cn/qspevdu_t1002001/ | [
"• 防诈骗中心\n• 客服中心 |\n• 网站导航 |\n• 设为主页 |\n• 加入收藏\n• 您当前位置： 首页>产品库>工业品>照明工业\n相关分类：\n• 湖南\n• 长沙市\n• 常德市\n• 郴州市\n• 衡阳市\n• 怀化市\n• 娄底市\n• 邵阳市\n• 湘潭市\n• 湘西土家族苗族自治州\n• 益阳市\n• 永州市\n• 岳阳市\n• 张家界市\n• 株洲市\n• 山西\n• 长治市\n• 大同市\n• 晋城市\n• 晋中市\n• 临汾市\n• 吕梁市\n• 朔州市\n• 太原市\n• 忻州市\n• 阳泉市\n• 运城市\n• 安徽\n• 安庆市\n• 蚌埠市\n• 亳州市\n• 巢湖市\n• 池州市\n• 滁州市\n• 阜阳市\n• 合肥市\n• 淮北市\n• 淮南市\n• 黄山市\n• 六安市\n• 马鞍山市\n• 宿州市\n• 铜陵市\n• 芜湖市\n• 宣城市\n• 广西\n• 百色市\n• 北海市\n• 崇左市\n• 防城港市\n• 贵港市\n• 桂林市\n• 河池市\n• 贺州市\n• 来宾市\n• 柳州市\n• 南宁市\n• 钦州市\n• 梧州市\n• 玉林市\n• 河南\n• 安阳市\n• 鹤壁市\n• 焦作市\n• 开封市\n• 洛阳市\n• 漯河市\n• 南阳市\n• 平顶山市\n• 濮阳市\n• 三门峡市\n• 商丘市\n• 新乡市\n• 信阳市\n• 许昌市\n• 郑州市\n• 周口市\n• 驻马店市\n• 吉林\n• 白城市\n• 白山市\n• 长春市\n• 吉林市\n• 辽源市\n• 四平市\n• 松原市\n• 通化市\n• 延边朝鲜族自治州\n• 广东\n• 潮州市\n• 东莞市\n• 佛山市\n• 广州市\n• 河源市\n• 惠州市\n• 江门市\n• 揭阳市\n• 茂名市\n• 梅州市\n• 清远市\n• 汕头市\n• 汕尾市\n• 韶关市\n• 深圳市\n• 阳江市\n• 云浮市\n• 湛江市\n• 肇庆市\n• 中山市\n• 珠海市\n• 辽宁\n• 鞍山市\n• 本溪市\n• 朝阳市\n• 大连市\n• 丹东市\n• 抚顺市\n• 阜新市\n• 葫芦岛市\n• 锦州市\n• 辽阳市\n• 盘锦市\n• 沈阳市\n• 铁岭市\n• 营口市\n• 湖北\n• 鄂州市\n• 恩施土家族苗族自治州\n• 黄冈市\n• 黄石市\n• 荆门市\n• 荆州市\n• 直辖行政单位\n• 十堰市\n• 随州市\n• 武汉市\n• 咸宁市\n• 襄阳市\n• 孝感市\n• 宜昌市\n• 江西\n• 抚州市\n• 赣州市\n• 吉安市\n• 景德镇市\n• 九江市\n• 南昌市\n• 萍乡市\n• 上饶市\n• 新余市\n• 宜春市\n• 鹰潭市\n• 浙江\n• 杭州市\n• 湖州市\n• 嘉兴市\n• 金华市\n• 丽水市\n• 宁波市\n• 衢州市\n• 绍兴市\n• 台州市\n• 温州市\n• 舟山市\n• 青海\n• 果洛藏族自治州\n• 海北藏族自治州\n• 海东地区\n• 海南藏族自治州\n• 海西蒙古族藏族自治州\n• 黄南藏族自治州\n• 西宁市\n• 玉树藏族自治州\n• 甘肃\n• 白银市\n• 定西市\n• 甘南藏族自治州\n• 嘉峪关市\n• 金昌市\n• 酒泉市\n• 兰州市\n• 临夏回族自治州\n• 陇南市\n• 平凉市\n• 庆阳市\n• 天水市\n• 武威市\n• 张掖市\n• 贵州\n• 安顺市\n• 毕节市\n• 贵阳市\n• 六盘水市\n• 黔东南苗族侗族自治州\n• 黔南布依族苗族自治州\n• 黔西南布依族苗族自治州\n• 铜仁地区\n• 遵义市\n• 陕西\n• 安康市\n• 宝鸡市\n• 汉中市\n• 商洛市\n• 铜川市\n• 渭南市\n• 西安市\n• 咸阳市\n• 延安市\n• 榆林市\n• 西藏\n• 阿里地区\n• 昌都地区\n• 拉萨市\n• 林芝地区\n• 那曲地区\n• 日喀则地区\n• 山南地区\n• 宁夏\n• 固原市\n• 石嘴山市\n• 吴忠市\n• 银川市\n• 中卫市\n• 福建\n• 福州市\n• 龙岩市\n• 南平市\n• 宁德市\n• 莆田市\n• 泉州市\n• 三明市\n• 厦门市\n• 漳州市\n• 内蒙古\n• 阿拉善盟\n• 巴彦淖尔市\n• 包头市\n• 赤峰市\n• 鄂尔多斯市\n• 呼和浩特市\n• 呼伦贝尔市\n• 通辽市\n• 乌海市\n• 乌兰察布市\n• 锡林郭勒盟\n• 兴安盟\n• 云南\n• 保山市\n• 楚雄彝族自治州\n• 大理白族自治州\n• 德宏傣族景颇族自治州\n• 迪庆藏族自治州\n• 红河哈尼族彝族自治州\n• 昆明市\n• 丽江市\n• 临沧市\n• 怒江傈僳族自治州\n• 曲靖市\n• 思茅市\n• 文山壮族苗族自治州\n• 西双版纳傣族自治州\n• 玉溪市\n• 昭通市\n• 新疆\n• 阿克苏地区\n• 阿勒泰地区\n• 巴音郭楞蒙古自治州\n• 博尔塔拉蒙古自治州\n• 昌吉回族自治州\n• 哈密地区\n• 和田地区\n• 喀什地区\n• 克拉玛依市\n• 克孜勒苏柯尔克孜自治州\n• 直辖行政单位\n• 塔城地区\n• 吐鲁番地区\n• 乌鲁木齐市\n• 伊犁哈萨克自治州\n• 黑龙江\n• 大庆市\n• 大兴安岭地区\n• 哈尔滨市\n• 鹤岗市\n• 黑河市\n• 鸡西市\n• 佳木斯市\n• 牡丹江市\n• 七台河市\n• 齐齐哈尔市\n• 双鸭山市\n• 绥化市\n• 伊春市\n• 香港\n• 香港\n• 九龙\n• 新界\n• 澳门\n• 澳门\n• 其它地区\n• 台湾\n• 台中市\n• 台南市\n• 高雄市\n• 台北市\n• 基隆市\n• 嘉义市\n•",
null,
"新款led箭头灯 车载作业转向灯\n\n品牌:骧虎\n\n出厂地:忻城县(城关镇)\n\n报价：面议\n\n杭州骧虎交通科技有限公司\n\n主营：太阳能爆闪灯、太阳能警示灯、太阳能移动红绿灯、太阳能仿真警车、太阳能回转灯、太阳...\n\n•",
null,
"太阳能移动红绿灯（三联箭头灯）\n\n品牌:骧虎\n\n出厂地:忻城县(城关镇)\n\n报价：面议\n\n杭州骧虎交通科技有限公司\n\n主营：太阳能爆闪灯、太阳能警示灯、太阳能移动红绿灯、太阳能仿真警车、太阳能回转灯、太阳...\n\n•",
null,
"到哪买彩灯亮化比较好-吴忠彩灯亮化厂家\n\n品牌:巧工匠,,\n\n出厂地:柳南区\n\n报价：面议\n\n宁夏巧工匠智能科技发展有限公司\n\n主营：宁夏护栏,宁夏不锈钢岗亭,宁夏电动伸缩门,宁夏立体停车库,宁夏交通市政护栏\n\n•",
null,
"优良的郴州信号灯在哪有卖_永兴交通信号灯\n\n品牌:众申,众申交通设施,众申交通\n\n出厂地:天峨县(六排镇)\n\n报价：面议\n\n郴州众申交通设施开发有限公司\n\n黄金会员：",
null,
"主营：交通标线,标牌,电子警察,信号灯,地下停车场\n\n•",
null,
"报价：面议\n\n肇庆市端州区景豪装饰设计工程有限公司\n\n主营：商铺装修工程,宣传栏工程,庆典活动策划,室内装修工程,室外装修工程\n\n•",
null,
"甩卖led防水电源-想买优惠的LED防水电源就来中孚光电\n\n品牌:中孚光电,中孚,Turefull\n\n出厂地:合山市\n\n报价：面议\n\n深圳市中孚光电科技有限公司\n\n黄金会员：",
null,
"主营：LED隧道灯电源,LED工矿灯电源,LED泛光灯电源,LED投光灯电源,LED面...\n\n•",
null,
"要买郴州信号灯当选众申交通设施|资兴信号灯\n\n品牌:众申,众申交通设施,众申交通\n\n出厂地:天峨县(六排镇)\n\n报价：面议\n\n郴州众申交通设施开发有限公司\n\n黄金会员：",
null,
"主营：交通标线,标牌,电子警察,信号灯,地下停车场\n\n•",
null,
"太阳能路障警示灯 多功能路锥警示灯批发\n\n品牌:骧虎\n\n出厂地:忻城县(城关镇)\n\n报价：面议\n\n杭州骧虎交通科技有限公司\n\n主营：太阳能爆闪灯、太阳能警示灯、太阳能移动红绿灯、太阳能仿真警车、太阳能回转灯、太阳...\n\n•",
null,
"太阳能标志牌 led交通标志牌 太阳能限速标志牌\n\n品牌:骧虎\n\n出厂地:忻城县(城关镇)\n\n报价：面议\n\n杭州骧虎交通科技有限公司\n\n主营：太阳能爆闪灯、太阳能警示灯、太阳能移动红绿灯、太阳能仿真警车、太阳能回转灯、太阳...\n\n•",
null,
"报价：面议\n\n肇庆市端州区景豪装饰设计工程有限公司\n\n主营：商铺装修工程,宣传栏工程,庆典活动策划,室内装修工程,室外装修工程\n\n• 没有找到合适的供应商？您可以发布采购信息\n\n没有找到满足要求的供应商？您可以搜索 照明工业批发 照明工业公司 照明工业厂\n\n### 最新入驻厂家\n\n相关产品:\nled箭头导向灯 太阳能移动红绿灯 宁夏护栏 郴州信号灯 广告招牌LED发光字全彩显示屏 甩卖led防水电源 郴州信号灯 多功能路锥警示灯 限速标志牌 广告招牌LED发光字全彩显示屏"
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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