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https://www.emathhelp.net/standard-gravity-to-feet-per-second-squared/	[ "# Standard gravity to feet per second squared\n\nThis free conversion calculator will convert standard gravity to feet per second squared, i.e. g_0 to fps^2. Correct conversion between different measurement scales.\n\nConvert", null, "The formula is $A_{fps^2} = \\frac{196133}{6096} A_{g_0}$, where $A_{g_0} = 15$.\n\nTherefore, $A_{fps^2} = \\frac{980665}{2032}$.\n\nAnswer: $15 g_{0} = \\frac{980665}{2032} fps^{2}\\approx 482.610728346456693 fps^{2}$." ]	[ null, "https://www.emathhelp.net/static/assets/images/flip_horizontal-512.51112b4c0ae4.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.71544695,"math_prob":0.9999286,"size":436,"snap":"2021-43-2021-49","text_gpt3_token_len":139,"char_repetition_ratio":0.11805555,"word_repetition_ratio":0.040816326,"special_character_ratio":0.44954127,"punctuation_ratio":0.15,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999497,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-26T20:16:55Z\",\"WARC-Record-ID\":\"<urn:uuid:0ec8de64-8c28-4b4b-848c-76d658ce6f75>\",\"Content-Length\":\"19666\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:416b3f7c-9c80-4b07-a98d-686cd60ad8c0>\",\"WARC-Concurrent-To\":\"<urn:uuid:a3155ad8-9b44-4097-9f77-1595327170e4>\",\"WARC-IP-Address\":\"104.248.238.17\",\"WARC-Target-URI\":\"https://www.emathhelp.net/standard-gravity-to-feet-per-second-squared/\",\"WARC-Payload-Digest\":\"sha1:KGKVVICCW3YCI5SLWTWACXJZRDFIMKFJ\",\"WARC-Block-Digest\":\"sha1:G2K4YNZWEGKEYDR7XH7OD5IT46ZEM5B7\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587926.9_warc_CC-MAIN-20211026200738-20211026230738-00494.warc.gz\"}"}
http://journals.tubitak.gov.tr/physics/abstract.htm?id=8038	[ "Optical Pulse distortion in Fibonacci-class Multilayer Stacks\n\nAuthors: A ROSTAMI, S. MATLOUB\n\nAbstract: Optical pulse distortion and propagation through quasi-periodic structures generally and especially for Fibonacci-class, is investigated analytically and simulated numerically. In this analysis, the transfer matrix method (TMM) for distortion evaluation is used. The simulated results for Fibonacci-class quasi-periodic structures and pure periodic multilayer stacks are compared. We show that the Fibonacci-class quasi-periodic structure has a large dispersion coefficient with respect to a similar case in the periodic structure. So, quasi-periodic structures will destroy the incident pulse shape for smaller stack length than a periodic case in a similar situation. Finally, using output power, the distortion limit can be estimated and the maximum number of layers with acceptable distortion can be determined. Also, we have calculated the second order (D) and third order (B) dispersion coefficients for Fibonacci-class quasi-periodic structures around 1.55 \\mu m for dispersion compensation purposes.\n\nKeywords: Pulse distortion, dispersion, quasi-periodic structures, and Fibonacci-class.\n\nFull Text: PDF" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86081445,"math_prob":0.9103946,"size":1151,"snap":"2019-43-2019-47","text_gpt3_token_len":215,"char_repetition_ratio":0.163034,"word_repetition_ratio":0.0,"special_character_ratio":0.17202432,"punctuation_ratio":0.12631579,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9785674,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-17T15:44:14Z\",\"WARC-Record-ID\":\"<urn:uuid:97c86302-9763-44ce-a774-ceb9788023c4>\",\"Content-Length\":\"9871\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e37be967-3337-4ec7-9288-7887b144f938>\",\"WARC-Concurrent-To\":\"<urn:uuid:fa454179-6c86-488f-bc7f-35f5ea182b87>\",\"WARC-IP-Address\":\"193.140.81.167\",\"WARC-Target-URI\":\"http://journals.tubitak.gov.tr/physics/abstract.htm?id=8038\",\"WARC-Payload-Digest\":\"sha1:SEHHZ4WRHV7KA2LOPAPHVIGSXPJ3JEBT\",\"WARC-Block-Digest\":\"sha1:VN4WIO4E5PYT3EXBB3C2JXO27WBEJTSZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986675409.61_warc_CC-MAIN-20191017145741-20191017173241-00325.warc.gz\"}"}
https://murray.cds.caltech.edu/index.php?title=Lecture_7.1_bugs&diff=next&oldid=6870	[ "# Lecture 7.1 bugs: Difference between revisions\n\nNotes and bugs from Lecture 7.1.\n\nThe Nyquist D contour is defined as being traversed in the counter-clockwise direction. On Page 5, if there were arrows on the \"D\" countour, they would show the tracing starting at the origin and moving downard toward $-j\\infty$", null, ", traversing around toward $+j\\infty$", null, "at $\|s\|=R$", null, "where <amsmath>R \\approx \\infty</amsmath>, and then traversing back to the origin. Then the Nyquist mapping (the second figure on P. 5) shows the value of $L(s)$", null, "in the complex plane as $s$", null, "takes on all of the values of the \"D\" contour.\n\nBug 1: With that convention, which is the one used in the book, the Nyquist mapping would have arrows in the opposite direction as shown. They are different because that plot was generated in MATLAB, which uses a convention that the traverse is performed in the opposite direction. So you should mentally reverse the arrows in all Nyquist plots shown in this lecture.\n\nWith this convention established, then the Nyquist theorem works with N=#counter-clockwise encirclements of -1, as stated in the notes. As stated in lecture, amnyquist() from the book's web page will draw the nyquist diagrams with the arrows in the convention used in the course and by mathematicians.\n\nBug 2: On page 13, N is #CCW encirclements, not #CW encirclements (as above).\n\nBug 3: The Nyquist diagrams on P. 10 and 13 show a unit circle referring to unity gain magnitude. But they are not drawn to scale according to the axes on the figures -- the width is correct, but they should be squished vertically (along the imaginary axis) by a factor of 2.\n\n--Fuller 16:33, 12 November 2007 (PST)" ]	[ null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/573bfe3c126f3fe23878015480ac5c9477599e1e", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/46f74816c863a4c2f9496bb95e87ddb0e33bab04", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/ddb70cb0e3fe7ee693a881e2cef01ed392929dfe", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/0a39f3622b00851baab207a8d42bc94243366839", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/01d131dfd7673938b947072a13a9744fe997e632", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94490117,"math_prob":0.92548203,"size":3400,"snap":"2023-40-2023-50","text_gpt3_token_len":834,"char_repetition_ratio":0.14163722,"word_repetition_ratio":0.65957445,"special_character_ratio":0.23382352,"punctuation_ratio":0.10329341,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9806448,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,5,null,5,null,5,null,8,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-06T11:08:29Z\",\"WARC-Record-ID\":\"<urn:uuid:dc17de29-f57f-4b8a-88c7-f44067d23742>\",\"Content-Length\":\"45216\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a1e52bc0-47ed-4b24-80bf-3d6686594031>\",\"WARC-Concurrent-To\":\"<urn:uuid:2f6d9313-9537-4c0b-84e9-a268cc0c3f58>\",\"WARC-IP-Address\":\"52.10.80.185\",\"WARC-Target-URI\":\"https://murray.cds.caltech.edu/index.php?title=Lecture_7.1_bugs&diff=next&oldid=6870\",\"WARC-Payload-Digest\":\"sha1:YWLAK6XYY3Z32VEPTJD6WXN4U4GZZIBM\",\"WARC-Block-Digest\":\"sha1:HWOG3YEVSMHXP62LECCIY6PLZQ4WUNWH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100593.71_warc_CC-MAIN-20231206095331-20231206125331-00225.warc.gz\"}"}
https://blog.learnsignal.com/blog/what-does-volatile-mean	[ "Volatility\nis a statistical measure of a security's or market index's return dispersion. The more the volatility, the riskier the security is in most circumstances. A \"volatile\" market, for example, is one in which the stock market rises and falls by more than 1% over a long time.\n\nThere are several ways to measure volatility, including beta coefficients, option pricing models, and standard deviations of returns.\n\nExample of Volatility:\n\nThe formula for daily volatility is computed by finding out the square root of a daily stock price variance.\n\nDaily Volatility Formula is represented as,\n\nDaily Volatility formula = √Variance\n\nFurther, the annualised volatility formula is calculated by multiplying the daily volatility by a square root of 252.\n\nWhy is Volatility important?\n\nVolatility is a measure of risk, while volatility measures variability. It assists investors in assessing the risk they take when purchasing a specific asset and determining if the purchase will be worthwhile.\n\nTopics: ACCA, CIMA, CPD, AAT, FRM", null, "•", null, "### Central Counterparties\n\n•", null, "### Country Risk\n\n•", null, "### Convexity Formula\n\n•", null, "### Clean and Dirty Price\n\n•", null, "### Basic Indicator Approach\n\n•", null, "### ACCA Exam Fees 2022: Everything You Need to Know\n\n•", null, "### ACCA Exam Dates for 2022\n\n•", null, "### How Difficult is Passing ACCA? An Honest Review\n\n•", null, "### 7 of the Best YouTube Channels for Accounting Students\n\n•", null, "" ]	[ null, "https://no-cache.hubspot.com/cta/default/6067029/3fa0b031-c01b-450d-a12f-926decf77688.png", null, "https://blog.learnsignal.com/hubfs/Central%20Counterparties.jpg", null, "https://blog.learnsignal.com/hubfs/Country%20Risk.jpg", null, "https://blog.learnsignal.com/hubfs/Convexity%20Formula.jpg", null, "https://blog.learnsignal.com/hubfs/Clean%20and%20Dirty%20Price.jpg", null, "https://blog.learnsignal.com/hubfs/Basic%20Indicator%20Approach.jpg", null, "https://blog.learnsignal.com/hubfs/shutterstock_1079701271.jpg", null, "https://blog.learnsignal.com/hubfs/Untitled%20design%20%282%29-3.png", null, "https://blog.learnsignal.com/hubfs/How-difficult-is-the-ACCA-An-honest-review.png", null, "https://blog.learnsignal.com/hubfs/apps-blur-button-close-up-267350.jpg", null, "https://blog.learnsignal.com/hubfs/social-suggested-images/blog.learnsignal.comhubfsaccounting-alone-application-938965.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9361671,"math_prob":0.87198657,"size":985,"snap":"2022-27-2022-33","text_gpt3_token_len":191,"char_repetition_ratio":0.16921508,"word_repetition_ratio":0.0,"special_character_ratio":0.18477157,"punctuation_ratio":0.11111111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97575086,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22],"im_url_duplicate_count":[null,8,null,8,null,8,null,8,null,8,null,8,null,7,null,7,null,7,null,7,null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-18T08:29:42Z\",\"WARC-Record-ID\":\"<urn:uuid:090f6186-865c-4735-87fc-b7e6374cf0ba>\",\"Content-Length\":\"48196\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fb9a23d7-ba00-4bc9-bd98-9c5ec3a1489a>\",\"WARC-Concurrent-To\":\"<urn:uuid:84a57e37-4fae-4088-bed4-e02d295838c9>\",\"WARC-IP-Address\":\"199.60.103.2\",\"WARC-Target-URI\":\"https://blog.learnsignal.com/blog/what-does-volatile-mean\",\"WARC-Payload-Digest\":\"sha1:72E247JEGQVKJDOJ3U4NGZYSCKUA4WYO\",\"WARC-Block-Digest\":\"sha1:6IDPGGHOT2KBZI5RCKNNHY6NJBLJ2W36\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573172.64_warc_CC-MAIN-20220818063910-20220818093910-00074.warc.gz\"}"}
https://file.scirp.org/xml/85751.xml	[ "We present an iterative scheme for solving Poisson’s equation in 2D. Using finite differences, we discretize the equation into a Sylvester system, AU + UB = F, involving tridiagonal matrices A and B. The iterations occur on this Sylvester system directly after introducing a deflation-type parameter that enables optimized convergence. Analytical bounds are obtained on the spectral radii of the iteration matrices. Our method is comparable to Successive Over-Relaxation (SOR) and amenable to compact programming via vector/array operations. It can also be implemented within a multigrid framework with considerable improvement in performance as shown herein.\n\nPoisson’s Equation Sylvester System Multigrid\n1. Introduction\n\nPoisson’s equation ∇ 2 u = f , an elliptic partial differential equation , was first published in 1813 in the Bulletin de la Société Philomatique by Siméon-Denis Poisson. The equation has since found wide utility in applications such as electrostatics , fluid dynamics , theoretical physics , and engineering . Due to its expansive applicability in the natural sciences, analytic and efficient approximate solution methods have been sought for nearly two centuries. Analytic solutions to Poisson’s equation are unlikely in most scientific applications because the forcing or boundary conditions on the system cannot be explicitly represented by means of elementary functions. For this reason, numerical approximations have been developed, dating back to the Jacobi method in 1845. The linear systems arising from these numerical approximations are solved either directly, using methods like Gaussian elimination, or iteratively. To this day, there are applications solved by direct solvers and others solved by iterative solvers, depending largely on the structure and size of the matrices involved in the computation. The 1950s and 1960s saw an enormous interest in relaxation type methods, prompted by the studies on optimal relaxation and work by Young, Varga, Southwell, Frankel and others. The books by Varga and Young give a comprehensive guide to iterative methods used in the 1960s and 1970s, and have remained the handbooks used by academics and practitioners alike .\n\nThe Problem Description\n\nThe Poisson equation on a rectangular domain is given by\n\n∇ 2 u = f in Ω = { x , y \| 0 ≤ x ≤ a , 0 ≤ y ≤ b } (1)\n\nwhere u = u ( x , y ) is to be solved in the 2D domain Ω , and f ( x , y ) is the forcing function. Typical boundary conditions for this equation are either Dirichlet, where the value of f ( x , y ) is specified on the boundary, or Neumann, where the value of the normal derivative is specified on the boundary. These are given mathematically as,\n\nu = g D or ∂ u / ∂ n ≡ n ^ ⋅ ∇ u = g N on ∂ Ω , (2)\n\nwhere n ^ is the outward unit normal along ∂ Ω and g D and g N are the function values specified by Dirichlet or Neumann boundary conditions. It is also possible to have mixed boundary conditions along the boundary, where some edges have Dirichlet and some have Neumann, so long as the problem is well-posed. Furthermore, edges could be subject to Robin boundary conditions of the form c 1 u + c 2 ∂ u / ∂ n = g R . Any numerical scheme designed to solve the Poisson equation should be robust in its ability to incorporate any form of boundary condition into the solver. A detailed discussion of boundary condition implementation is given in the Appendix. Discretizing (1) using central differences with equal grid size Δ x = Δ y = h leads to an M × N rectangular array of unknown U, such that U i , j ≈ u ( x i , y j ) (assuming that a and b are both integer multiples of h). This discretization leads to a linear system of the form A U + U B = F , the Sylvester equation, which can be solved either directly or iteratively. The direct method utilizes the Kronecker product approach , given by\n\nK u = f where K = kron ( I , A ) + kron ( B T , I ) (3)\n\nwhere u and f are appropriately ordered M N × 1 column vectors obtained from the M × N arrays U and F, and K is a sparse M N × M N matrix. The Kronecker product, kron ( P , Q ) , of any two matrices P and Q is a partitioned matrix whose ijth partition contains matrix Q multiplied by component p i j of P. Due to the potentially large size of the system given in (3), direct solvers are not the preferred solution approach. Specifically addressed here is an iterative approach to solving the Sylvester equation,\n\nA U int + U int B = F int , (4)\n\nwhere U int is the m × n array of interior unknowns (not including the known boundary values when Dirichlet boundary conditions are given) with m = M − 2 and n = N − 2 . Operator matrices A and B are given by the O ( h 2 ) finite difference approximation to the second derivative,\n\n1 h 2 [ − 2 1 1 − 2 1 ⋱ ⋱ ⋱ 1 − 2 1 1 − 2 ] . (5)\n\nFor an array of unknowns U int ∈ ℝ m × n , the operator matrices are of dimension A ∈ ℝ m × m , and B ∈ ℝ n × n . The matrix structure of U should be modified such that the first index i corresponds to the x-direction, and the second index j corresponds to the y-direction. With this orientation, multiplying U int by the matrix A on the left approximates the second derivative in the x-direction, and multiplying U int by the matrix B on the right approximates the second derivative in the y-direction.\n\n2. Sylvester Iterative Scheme\n\nExamining (4) it might seem natural to move one term to the right-hand side of the equation to achieve an iterative scheme such as:\n\nA U + U B = F ⇒ A U = F − U B\n\n⇒ U k + 1 = A − 1 F − A − 1 U k B , (6)\n\nHowever, this scheme diverges, and an alternative approach is required to iterate on the Sylvester system. An appropriate method is to break up the iterative scheme into two “half-steps’’ as follows\n\n1) First half-step: U k → U \n\nA U = F − U B ⇒ ( A − α I + α I ) U = F − U B\n\n⇒ ( A − α I ) U = F − U k ( B + α I ) (7)\n\n2) Second half-step: U * → U k + 1\n\nU B = F − A U ⇒ U ( B − β I + β I ) = F − A U\n\n⇒ U k + 1 ( B − β I ) = F − ( A + β I ) U * (8)\n\nwhere U * is some intermediate solution between steps k and k + 1 . Rearranging, this leads to\n\n( I − 1 α A ) U * = U k ( I + 1 α B ) − 1 α F , U k + 1 ( I − 1 β B ) = ( I + 1 β A ) U * − 1 β F . (9)\n\nThe iterative scheme (9) is similar to the Alternating Direction Implicit (ADI) formulation , where Poisson’s equation is reformulated to have pseudo-time dependency,\n\nd U d t = A U + U B − F , (10)\n\nwhich achieves the solution to Equation (4) when it reaches steady-state. This method is separated into two half-steps, the first time step going from time k → k + 1 / 2 treating the x-direction implicitly and the y-direction explicitly. The second time step then goes from time k + 1 / 2 → k + 1 , treating the y-direction implicitly and the x-direction explicitly. The two half-steps are,\n\nU k + 1 / 2 − U k Δ t / 2 = 1 h 2 [ A U k + 1 / 2 + U k B ] , U k + 1 − U k + 1 / 2 Δ t / 2 = 1 h 2 [ A U k + 1 / 2 + U k + 1 B ] , (11)\n\n( I − Δ t 2 A ) U k + 1 / 2 = U k ( I + Δ t 2 B ) − Δ t 2 F , U k + 1 ( I − Δ t 2 B ) = ( I + Δ t 2 A ) U k + 1 / 2 − Δ t 2 F . (12)\n\nThis iteration procedure looks nearly identical to our Sylvester iterations given in (9) with Δ t / 2 replaced by the unknown parameters 1/α and 1/β. However in our formulation, there is no pseudo-time dependency introduced. Instead, the eigenvalues of our operator matrices A and B are deflated to produce an iterative scheme that optimally converges, and finding the values of the parameters α and β becomes an optimization problem.\n\nConvergence\n\nAfter the Sylvester Equation (4) is modified into the iterative system (9), the iterative scheme can be written as a single step by substituting the expression for the intermediate solution U * into the second step of the iterative process; this yields the single update equation for U k + 1 given by\n\nU k + 1 = ( I + 1 β A ) ( I − 1 α A ) − 1 U k ( I + 1 α B ) ( I − 1 β B ) − 1 − [ 1 α ( I + 1 β A ) ( I − 1 α A ) − 1 − 1 β I ] F ( I − 1 β B ) − 1 . (13)\n\nAssuming that an exact solution U exact exists that exactly satisfies the linear system (4), i.e. A U exact + U exact B = F , we define the error between the kth iteration and the exact solution as\n\nE k ≡ U k − U exact . (14)\n\nFinding an update equation for the error is done by subtracting the error at the kth step from the error at the ( k + 1 ) s t step, noting that the expressions involving the forcing F disappear, we arrive at\n\nE k + 1 = P E k Q , (15)\n\nwhere the matrices P and Q are given by\n\nP = ( I + 1 β A ) ( I − 1 α A ) − 1 , Q = ( I + 1 α B ) ( I − 1 β B ) − 1 . (16)\n\nDenoting the m eigenvalues of A ∈ ℝ m × m by λ k A , and the n eigenvalues of B ∈ ℝ n × n by λ k B , the corresponding deflated eigenvalues of the iteration matrices P and Q are\n\nλ k P = 1 + ( λ k A / β ) 1 − ( λ k A / α ) = α β ( β + λ k A α − λ k A ) , k = 1 , 2 , ⋯ , m , λ k Q = 1 + ( λ k B / α ) 1 − ( λ k B / β ) = β α ( α + λ k B β − λ k B ) , k = 1 , 2 , ⋯ , n . (17)\n\nA sufficient condition for convergence of the iterative process is achieved if the spectral radii of both iteration matrices P and Q are less than one,\n\nρ ( P ) ≡ max \| λ k P \| < 1 and ρ ( Q ) ≡ max \| λ k Q \| < 1. (18)\n\nThe error at each consecutive iteration is decreased by the product of ρ ( P ) and ρ ( Q ) ,\n\nE k + 1 = P E k Q , ‖ E k + 1 ‖ = ρ ( P ) ρ ( Q ) ‖ E k ‖ , ‖ E k + 1 ‖ = ( ρ ( P ) ρ ( Q ) ) k ‖ E 0 ‖ , (19)\n\nwhere E 0 is the initial error. Often in practical applications, the exact solution is not known, so the error E k cannot be computed directly. In this case, the preferred measure in iterative schemes is given by the residual, which measures the difference of the left and right hand sides of the linear system being solved. This will be further discussed in the Results section.\n\n3. Finding Optimal Parameters α and β\n\nFinding α and β is an optimization problem for achieving the fastest convergence rate of the Sylvester iterative scheme (9). Given the operator matrices A and B and their respective eigenvalues λ k A and λ k B , it seems feasible to find optimal values of α and β to minimize the spectral radii of the iteration matrices P and Q given in Equations (17). From (15) the error E k + 1 is found by multiplying by P on the left, and Q on the right, thus the convergence is governed by the spectral radii of both P and Q.\n\nFigure 1 shows the eigenvalues of P and Q for arbitrary m and n, given by Equation (17), plotted vs. the eigenvalues of A and B for some parameters α and β. It can be seen that as λ A or λ B get large in magnitude, the values of λ P or λ Q approach − α / β and − β / α , respectively. This implies that if α ≠ β ,\n\nthe high frequency eigenvalues of P and Q, in magnitude, will be greater than one, thus convergence condition (18) will not be satisfied. This provides the restriction for convergence that,\n\nα = β . (20)\n\nThis optimal value of α = β will henceforth be called α * . It is important to note that the operator A ∈ ℝ m × m or B ∈ ℝ n × n with the larger dimension\n\nl ≡ m a x ( m , n ) , (21)\n\nhas a larger range of eigenvalues. Figure 1 shows m > n , (i.e. l = m ) so it can be seen that min ( λ A ) < min ( λ B ) and max ( λ A ) > max ( λ B ) . This property of the eigenvalues is important when calculating an expression for c, which will soon prove to be a highly useful parameter for an adaptive approach to smoothing. Letting α = β = α * in (17) gives the following expression for the eigenvalues of P and Q:\n\nλ k P = α * + λ k A α * − λ k A , k = 1 , 2 , ⋯ , m , λ k Q = α * + λ k B α * − λ k B , k = 1 , 2 , ⋯ , n . (22)\n\nFinding the optimal parameter α * is done by considering the error reduction of Sylvester iterations on an arbitrary initial condition U0. Assume that U0 can be decomposed into its constituent error (Fourier) modes, ranging from low frequency (smooth) to high frequency (oscillatory) modes. Given that U0 contains error modes of all frequencies, the most conservative method would be to choose α * such that the spectral radii ρ ( P ) and ρ ( Q ) are minimized over the full range of frequencies. This ensures that all modes of error are efficiently relaxed, and convergence is governed by the product of spectral radii.\n\nReferring to the lower curve in Figure 2, the conservative method of\n\ndetermining α * would be to set L min * = L max * and according to (22),\n\n− ( α * + λ min A α * − λ min A ) = ( α * + λ max A α * − λ max A ) . (23)\n\nNoting that eigenvalues for all dimensions collapse onto the curves shown in Figure 2, this conservative approach “locks in’’ the value of the larger operator’s spectral radius, thus providing an upper bound for convergence. Figure 2 shows m > n , so ρ ( P ) > ρ ( Q ) , so convergence will be limited by ρ ( P ) . Equation (23) can then be solved for α * giving\n\nα * = \| min ( λ min A , λ min B ) \| × \| max ( λ max A , λ max B ) \| , (24)\n\nwhere absolute values are introduced as a reminder that λ A , λ B are negative. This value of the parameter α * most uniformly smooths all frequencies for any arbitrary U 0 containing all frequency modes of error. It can be seen that the spectral radii of P and Q shown in Figure 2 occur at either endpoint, and the minimum amplitude occurs near the intersection of the curve with the axis. Varying the parameter α * in (22) controls the intersection point, and thus creates an effective “optimal smoothing region’’, denoted R smooth ∗ . Modes of error associated with this optimal smoothing region will be damped fastest, which makes Sylvester iterations highly adaptive in nature. This adaptive nature of Sylvester iterations lends itself nicely to a multigrid formulation.\n\nThe Sylvester multigrid formulation is based on the philosophy that most iterative schemes, including Sylvester iterations, relax high frequency modes fastest, leaving low frequency components relatively unchanged . On all grids traversed by a multigrid V-cycle, the high frequency modes are eliminated fastest by finding the optimal parameter value α mg ∗ such that L min mg = L mid mg , as shown in the upper curve of Figure 2. The height L mid mg is essentially the distance above the axis associated with the approximate “middle’’ eigenvalue in the range of λ A or λ B . This equality gives the following optimal parameter value for the Sylvester multigrid method\n\nα mg ∗ = \| m i n ( λ min A , λ min B ) \| × \| m i n ( λ mid A , λ mid B ) \| , (25)\n\nwhere, if m ≠ n , the minimum (i.e., most negative) middle eigenvalue λ mid ≡ ( λ min + λ max ) / 2 is chosen to shrink the optimal smoothing region R smooth mg such that high frequencies are smoothed most effectively. This choice of optimal parameter can be observed in Figure 2 to drastically decrease the magnitude of λ P and λ Q associated with high frequencies which significantly enhance relaxation in accordance with the multigrid philosophy.\n\nTo find analytical expressions for α * and α mg ∗ , it is necessary to have values for λ A and λ B . For Dirichlet boundary conditions, analytical expressions for λ A and λ B are derived below, but for Neumann boundary conditions numerical approaches are necessary to find λ A and λ B . The operator matrices A and B are each of tridiagonal form,\n\n[ d 0 d 1 d 1 d 0 d 1 ⋱ ⋱ ⋱ d 1 d 0 d 1 d 1 d 0 ] ∈ ℝ p × p . (26)\n\nTridiagonal matrices with constant diagonals, such as A and B for Dirichlet boundary conditions, have analytical expressions for their eigenvalues given by\n\nλ k = d 0 + 2 d 1 cos ( k π p + 1 ) , k = 1 , 2 , ⋯ , p (27)\n\nwhere p is the arbitrary dimension of the matrix . Neumann boundary conditions alter the upper and lower diagonals of A or B, thus there is no analytical form of eigenvalues for Neumann boundary conditions. Using (5) and (27) gives the following analytic form of the eigenvalues of the tridiagonal matrices A and B,\n\nλ k = 2 h 2 ( − 1 + cos ( k π p + 1 ) ) , k = 1 , 2 , ⋯ , p , (28)\n\nwhich achieves minimum and maximum values given by\n\nλ min = 2 h 2 ( − 1 + cos ( p π p + 1 ) ) and λ max = 2 h 2 ( − 1 + cos ( π p + 1 ) ) , (29)\n\nrespectively. Using (24), (25), and (29) the analytic expressions for optimal parameters for both conservative and multigrid approaches are given by\n\nα * = 2 h 2 ( − 1 + cos ( l π l + 1 ) ) ( − 1 + cos ( π l + 1 ) ) , α mg ∗ = 2 h 2 ( − 1 + cos ( l π l + 1 ) ) ( − 2 + cos ( π l + 1 ) + cos ( l π l + 1 ) ) , (30)\n\nwhere again l ≡ m a x ( m , n ) . Having expressions for α * and α mg ∗ allows λ P and λ Q to be found analytically using (22) which subsequently allows the spectral radii of the iteration matrices P and Q to be calculated. Knowing the spectral radii of the iteration matrices P and Q is highly advantageous, as it allows for an analysis of the Sylvester iterative scheme.\n\n4. Analysis\n\nThe analysis of standard Sylvester iterations can be performed and describes the error reduction with each consecutive iteration using (19). Having the optimal parameters given by (30) and eigenvalues of P and Q in (22), the spectral radii can be calculated to be\n\nρ ( P ) = − 1 + cos ( m π m + 1 ) + ( − 1 + cos ( l π l + 1 ) ) ( − 1 + cos ( π l + 1 ) ) − 1 + cos ( m π m + 1 ) − ( − 1 + cos ( l π l + 1 ) ) ( − 1 + cos ( π l + 1 ) ) , ρ ( Q ) = − 1 + cos ( n π n + 1 ) + ( − 1 + cos ( l π l + 1 ) ) ( − 1 + cos ( π l + 1 ) ) − 1 + cos ( n π n + 1 ) − ( − 1 + cos ( l π l + 1 ) ) ( − 1 + cos ( π l + 1 ) ) . (31)\n\nRewriting the last expression of (19), we see that\n\n‖ E k ‖ ‖ E 0 ‖ ~ ( ρ ( P ) ρ ( Q ) ) k . (32)\n\nIf we want to reduce our error to ‖ E k ‖ ~ ϵ ‖ E 0 ‖ and we wish to know how many iterations it will take to achieve this error reduction, using (32) we set ( ρ ( P ) ρ ( Q ) ) k ~ ϵ , and solving for k, we find it will take\n\nk ~ l o g ( ϵ ) l o g ( ρ ( P ) ρ ( Q ) ) (33)\n\niterations to reduce the error by ϵ . Here log can be with respect to any base, as long as the same one is used in both the numerator and denominator; e.g., the natural log can be used. Recall that the exact solution U exact of (4) is only an approximate solution of the differential Equation (1) we are actually solving. Due to this, we can only expect accuracy of the truncation error of the approximation. With an O ( h 2 ) method, U i , j exact differs from U ( x i , y j ) on the order of h 2 so we cannot achieve better accuracy than this no matter how well we solve the linear system. Thus, it is practical to take ϵ to be something proportional to the expected global error, e.g. ϵ = C h 2 for some fixed C .\n\nTo calculate the order of work required asymptotically as h → 0 , (i.e. m → ∞ ) using (33) and our choice for ϵ , we see that\n\nk ~ l o g ( C ) + 2 l o g ( h ) l o g ( ρ ( P ) ρ ( Q ) ) . (34)\n\nThe expressions for ρ ( P ) and ρ ( Q ) in (31) contain several cosine terms which can be Taylor expanded about different values. Cosines with arguments like π x can be expanded about x = 1 or x = 0 depending on the form of x, namely\n\nc o s ( π x ) ~ − 1 + π 2 2 ( x − 1 ) 2 + O ( ( x − 1 ) 3 ) for x ≈ 1, c o s ( π x ) ~ 1 − π 2 2 x 2 + O ( x 3 ) for x ≈ 0, (35)\n\nwhere, from (31), the form of x is something like m / ( m + 1 ) or 1 / ( m + 1 ) , which clearly approach one or zero, respectively, in the limit that m → ∞ . Using these expansions, along with the fact that 1 / ( 1 − x ) ~ 1 + x + O ( x 2 ) for x ≪ 1 to simplify the spectral radii, we arrive at the following\n\nρ ( P ) ~ ρ ( Q ) ~ 1 − π l + 1 + 1 4 ( π l + 1 ) 2 , (36)\n\nwhen m , n ≫ 1 . Since h = 1 / ( m + 1 ) , (34) combined with (36) gives the following order of work needed for convergence to within ϵ ~ C h 2 :\n\nk ~ − 2 l o g ( m + 1 ) 2 l o g ( 1 − π l + 1 ) ~ l π l o g ( m ) , (37)\n\nwhere only linear terms are used from (36), and the latter simplified expression can be deduced by using the property that l o g ( 1 + x ) ~ x + O ( x 2 ) for x ≪ 1 . Note that when m = n , the order of work for Sylvester iterations is k ~ ( m / π ) l o g ( m ) , which is comparable to the work necessary for the Successive Over-Relaxation (SOR) algorithm to solve Poisson’s equation . This will be our basis for comparison in the Results section for standard Sylvester iterations.\n\n5. Results\n\nProblems solved by Sylvester iterations can, in general, be written shorthand as L U = F , where L is a linear operator. In the case of Poisson’s equation, L is the Laplacian operator. As an error measure, the discrete ‖ ⋅ ‖ 2 norm of the residual, r ≡ F − L U , can be measured at each iteration. This number provides the stopping criterion for our iterative schemes, namely the iterations are run until\n\n‖ r ( k ) ‖ = ‖ F − L U ( k ) ‖ < tol × ‖ r ( 0 ) ‖ , (38)\n\nwhere r ( 0 ) is the initial residual, and tol is the tolerance. The tolerance is set to machine precision tol ~ 10 − 16 to illustrate the asymptotic convergence rate,\n\nq ( k ) = ‖ r ( k ) ‖ ‖ r ( k − 1 ) ‖ , (39)\n\nhowever, in practice, the discretization error O ( h 2 ) is the best accuracy that can be expected. These numerical results were run using MATLAB on a 1.5 GHz Mac PowerPC G4. The model problem that is solved is given by\n\n∇ 2 u = − 2 [ y 2 ( 1 − 6 x 2 ) ( 1 − y 2 ) + x 2 ( 1 − 6 y 2 ) ( 1 − x 2 ) ] in Ω , u = 0 on ∂ Ω , (40)\n\nwhere Ω = { x , y \| 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 } , and whose exact solution\n\nu exact ( x , y ) = ( x 2 − x 4 ) ( y 4 − y 2 ) , (41)\n\nis known so errors can be computed . This model problem is used to show performance of both standard and multigrid Sylvester iterations. In all cases, the initial guess U ( 0 ) of the iterative scheme is a normalized random array and can be assumed to contain all modes of error.\n\n5.1. Standard Sylvester Iterations\n\nFor comparison, standard Sylvester iterations were tested against Successive Over-Relaxation (SOR) with Chebyshev acceleration (see e.g., ). In SOR with Chebyshev acceleration, one uses odd-even ordering of the grid and changes the relaxation parameter ω at each half-step, which converges to the optimal relaxation parameter. The results are shown in Table 1. It is important to note that SOR iterations involve no matrix inversions, whereas Sylvester iterations do, thus the CPU time measure might not be an appropriate gauge for this particular comparison. From these results, it is clear that standard Sylvester iterations are comparable to SOR, and in most cases, converge to within tolerance in fewer iterations than SOR. An unfortunate artifact of standard iterative schemes is that as system size increases, so does the spectral radii governing the convergence. This can be observed in Table 1, as asymptotic convergence rates for each method q sylvester and q sor steadily increase, thus requiring higher numbers of iterations to solve within tolerance. The number of Sylvester iterations required to converge is consistent with the predicted number of iterations given by Equation (33) letting ϵ = tol = 10 − 16 .\n\n5.2. Multigrid Sylvester Iterations\n\nIn multigrid Sylvester iterations, the performance of the V ( ν 1 , ν 2 ) -cycle using Sylvester iterations is compared to that using the traditional Gauss-Seidel (GS)\n\nSylvester iterations vs. successive over-relaxation (SOR)\nSystem SizeSylvesterSORSylvesterSOR\nM × NIterationsIterationsq sylvesterq sorCPU TimeCPU Time\n16 × 1684930.6380.6740.0590.072\n32 × 321731850.8060.8190.4740.612\n64 × 643523680.8990.9054.5915.082\n128 × 1287097350.9490.95143.1348.28\n256 × 256147314690.9750.975744.5786.7\nMultigrid sylvester iterations vs. multigrid gauss-seidel (GS) iterations\nSystem SizeSylvesterGSSylvesterGS\nM × NV(2,1)-cyclesV(2,1)-cyclesq sylvester mgq gs mgCPU TimeCPU Time\n16 × 1612140.0550.0671.531.66\n32 × 3213150.0620.0782.072.16\n64 × 6413150.0660.0823.292.94\n128 × 12813150.0680.0836.344.87\n256 × 25613150.0680.08324.115.4\n512 × 51213150.0690.083151.8102.3\n\niterations. The parameter ν 1 represents the number of smoothing iterations done on each level of the downward branch of the V-cycle, while ν 2 represents the number done on the upward branch. In practice, common choices are ν = ν 1 + ν 2 ≤ 3 , so our performance is based on the V ( 2,1 ) -cycle . In each case, the V-cycle descends to the coarsest grid having gridwidth h 0 = 1 / 2 , and in the Sylvester implementation, the value of α mg ∗ is calculated to smooth high frequencies most effectively on each grid traversed by the cycle. The results are shown in Table 2. It can be seen that the asymptotic convergence rates q sylvester mg and q gs mg reach steady values independent of the gridwidth h. This is characteristic of multigrid methods, and enables the optimality of the multigrid method. It is clear when comparing the CPU times of the Sylvester multigrid formulation in Table 2 with standard Sylvester iterations in Table 1 that the multigrid framework is substantially faster (e.g., 30 times faster than standard iterations for a grid of size 256 × 256 ). It can also be seen that the asymptotic convergence rates are such that q sylvester mg < q gs mg , thus convergence is met in fewer V ( 2,1 ) cycles using Sylvester smoothing versus Gauss-Seidel smoothing.\n\n6. Conclusion\n\nSylvester iterations provide an alternative iterative scheme to solve Poisson’s equation that is comparable to SOR in the number of iterations necessary to converge, namely converging to discretization accuracy within k ~ ( m / π ) l o g ( m ) iterations. The true benefit of the Sylvester iterations, however, comes from its adaptive ability to smooth any range of error frequencies, thus being a perfect candidate for smoothing in a multigrid framework. Multigrid V ( 2,1 ) -cycles using Sylvester smoothing have an asymptotic convergence rate of q sylvester mg = 0.069 (versus q gs mg = 0.083 for Gauss-Seidel smoothing) and indicate significant improvement in efficiency over standard Sylvester iterations.\n\nCite this paper\n\nFranklin, M.B. and Nadim, A. (2018) A Poisson Solver Based on Iterations on a Sylvester System. Applied Mathematics, 9, 749-763. https://doi.org/10.4236/am.2018.96052\n\nAppendix\n\n1) Boundary condition implementation\n\nSolving the Poisson equation using Sylvester iterations lends itself nicely to boundary condition implementation. Dirichlet boundary conditions of the form\n\nu ( 0 , y ) = u 1 ( y ) , u ( a , y ) = u 2 ( y ) , u ( x , 0 ) = u 3 ( x ) , u ( x , b ) = u 4 ( x ) , (1)\n\nwhere u 1 , u 2 , u 3 and u 4 are functions describing the edges of U, can be implemented as follows. The unknown values in the Sylvester system given in Equation (4) are the m × n array of interior values, where m = M − 2 , n = N − 2 . It is possible to incorporate Dirichlet boundary conditions directly into this interior system by examining the partitioned matrix product, for example AU, given by\n\nwith UB taking an analogous partitioned form. Multiplying through by h 2 associated with the operator matrices A and B, the partitioned Sylvester system for internal unknowns gives\n\nA L ⋅ u 1 T + ( A int ) ( U int ) + A R ⋅ u 2 T + u 3 ⋅ B T T + ( U int ) ( B int ) + u 4 ⋅ B B T = ( h 2 ) ( F int ) , (3)\n\nwhere all matrix-vector products are m × n outer products. Note that the product AU incorporates Dirichlet boundary conditions in the x-direction, and UB incorporates Dirichlet boundary conditions in the y-direction. Combining the partitioned systems incorporating both A and B matrix multiplications and boundary conditions yields\n\n( A int ) ( U int ) = ( h 2 ) ( F int ) − ( A L ⋅ u 1 T + A R ⋅ u 2 T ) − ( u 3 ⋅ B T T + u 4 ⋅ B B T ) , (4)\n\nwhich is an m × n linear system for U int .\n\nFor Neumann boundary conditions, the edge at which the condition is imposed becomes part of the internal unknowns in the Sylvester system. As an example, consider a Neumann boundary condition given by\n\n∂ u ∂ x = g ( y ) on x = 0. (5)\n\nStaying within the finite difference formulation of derivatives and letting g ( y j ) ≡ g j , this condition can be discretized and approximated with the O ( h 2 ) central difference approximation, which yields\n\n( ∂ u ∂ x ) i , j ≈ U i + 1 , j − U i − 1 , j 2 h = g j for i = 0 , 0 ≤ j ≤ N . (6)\n\nFor a Neumann condition along the edge x = 0 , the row vector u 1 T described in (2) becomes a part of the internal array of unknowns U int . In order to implement this finite difference on the edge, we need to introduce a ghost layer with index i = − 1 , and pair Equation (6) to the second derivative operator AU for i = 0 . This gives\n\nU 1 , j − U − 1 , j 2 h = g j ⇒ U − 1 , j = U 1 , j − 2 h g j , ( ∂ 2 u ∂ x 2 ) 0 , j ≈ ( U 1 , j − 2 h g j ) − 2 U 0 , j + U 1 , j h 2 = 2 U 1 , j − 2 U 0 , j h 2 − 2 g j h , (7)\n\nwhich leads to the following partitioned form of AU,\n\nwhere the additional term − 2 g j / h is taken to the right hand side of the Sylvester system such that F 0 , j int = F 0 , j int + 2 g j / h for 0 < j < N . Comparing (8) to (2), the size of U int changes from m × n to ( m + 1 ) × n , and A 0,1 is changed from 1 to 2 (shown boldface in (8)), and the right hand side is slightly modified along that edge. Similarly, any edge with a Neumann condition can be handled in this fashion. It is clear that both Dirichlet and Neumann boundary conditions are very simple to implement in the Sylvester iteration method, and only slightly modify the structure of the arrays involved.\n\nReferencesDouglass, C., Hasse, G. and Langer, U. (2003) A Tutorial on Elliptic PDE Solvers and Their Parallelization. Society for Industrial and Applied Math, Philadelphia. https://doi.org/10.1137/1.9780898718171Feig, M., Onufriev, A., Lee, M.S., Im, W., Case, D.A. and Brooks III, C.L. (2003) Performance Comparison of Generalized Born and Poisson Methods in the Calculation of Electrostatic Solvation Energies for Protein Structures. Journal of Computational Chemistry, 25, 265-284. https://doi.org/10.1002/jcc.10378Ravoux, J.F., Nadim, A. and Haj-Hariri, H. (2003) An Embedding Method for Bluff Body Flows: Interactions of Two Side-by-Side Cylinder Wakes. Theoretical and Computational Fluid Dynamics, 16, 433-466. https://doi.org/10.1007/s00162-003-0090-4Trellakis, A., Galick, A.T., Pacelli, A. and Ravaioli, U. (1997) Iteration Scheme for the Solution of the Two-Dimensional Schrodinger-Poisson Equations in Quantum Structures. Journal of Applied Physics, 81, 7880-7884. https://doi.org/10.1063/1.365396Saraniti, M., Rein, A., Zandler, G., Vogl, P. and Lugli, P. (1996) An Efficient Multigrid Poisson Solver for Device Simulations. IEEE Transactions on Computer-Aided Design of Integrated Circuits and Systems, 15, 141-150. https://doi.org/10.1109/43.486661Varga, R.S. (1962) Matrix Iterative Analysis. Prentice-Hall, New Jersey.Young, D.M. (1971) Iterative Solution of Large Linear Systems. Academic Press, New York.Brezinski, C. and Wuytack, L. (2001) Numerical Analysis: Historical Developments in the 20th Century. Elsevier Science B.V., Netherlands.Van Loan, C.F. (2000) The Ubiquitous Kronecker Product. Journal of Computational and Applied Mathematics, 123, 85-100. https://doi.org/10.1016/S0377-0427(00)00393-9Peaceman, D.W. and Rachford, H.H. (1955) The Numerical Solution of Parabolic and Elliptic Differential Equations. Journal of the Society for Industrial and Applied Mathematics, 3, 28-41. https://doi.org/10.1137/0103003Briggs, W.L., Henson, V.E. and McCormick, S.F. (2000) A Multigrid Tutorial. 2nd Edition, Society for Industrial and Applied Math, Philadelphia. https://doi.org/10.1137/1.9780898719505LeVeque, R.J. (2007) Finite Difference Methods for Ordinary and Partial Differential Equations: Steady-State and Time-Dependent Problems. Society for Industrial and Applied Math, Philadelphia. https://doi.org/10.1137/1.9780898717839Press, W., Vetterling, W., Teukolsky, S. and Flannery, B. (1992) Numerical Recipes in Fortran. 2nd Edition, Cambridge University Press, New York.Trottenberg, U., Oosterlee, C. and Schuller, A. (2001) Multigrid. 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https://tech.apdaga.com/2019/07/high-bias-and-variance-problem-in.html	[ "# High Bias and Variance problem in Machine Learning [Cause & Solution]", null, "These days everybody talks about Machine Learning and Artificial Intelligence. More and more people started studying Machine Learning.\n\nFor beginners, Studying Machine Learning starts with understanding Linear and Logistic Regression.\n\nWhile implementing Linear or Logistic Regression, we mainly face two types of problems:\n\nRecommended Machine Learning Courses:\n\n## What is a high bias problem in machine learning?\n\nWhen our model (hypothesis function) fits poorly with the training data trend, then we can say our model is underfitting. This is also known as high bias problem.", null, "Figure 1: Underfitted\n• High bias problem is mainly caused due to:\n\n• If the hypothesis function is too simple.\n\n• If the hypothesis function uses very few features.\n\n• Solution for high bias problem :\n\n• Adding more features to the hypothesis function might solve the high bias problem.\n\n• If new features are not available, we gen create new features by combining two or more existing features or by taking a square, cube, etc of the existing feature.\n\n• If your model is underfitting (high bias), then getting more data for training will NOT help.\n\n• Adding new features will solve the problem of high bias, but if you add too many new features then your model will lead to overfitting also known as high variance.\n\n## What is a high variance problem in machine learning?\n\nUnlike high bias (underfitting) problem, When our model (hypothesis function) fits very well with the training data but doesn't work well with the new data, we can say our model is overfitting. This is also known as high variance problem.", null, "Figure 2: Overfitted\n\n• High variance problem is mainly caused due to:\n\n• If the hypothesis function is too complex.\n\n• If the hypothesis function uses too many features.\n\nThe higher-order polynomial in hypothesis function creates unnecessary curves and angles in the model which is unrelated to data.\n\n• Solution for high variance problem :\n\n• Either reduce the unnecessary features from the hypothesis function.\n\n• Otherwise, keep all the features in the hypothesis function but reduce the magnitude of the higher-order features (terms). This method is known as Regularization. Regularization is the most widely used method to solve the problem of high variance or overfitting.\n\n• Regularization works well when we have a lot of slightly useful features.\n\n• Lambda (λ) is the regularization parameter.", null, "Equation 1: Linear regression with regularization\n\n• Increasing the value of λ will solve the Overfitting (High Variance) problem.\n\n• Decreasing the value of λ will solve the Underfitting (High Bias) problem.\n\n• Selecting the correct/optimum value of λ will give you a balanced\n\n• If your model is overfitting (high variance), getting more data for training will help.\n\n## Summary :\n\n• Too simple or very few features in hypothesis function will cause high bias (underfitting) problem. Adding new features will solve it but adding too many features might lead to high variance (overfitting) problem. You can apply regularization to your model to solve the high variance problem.\n\n• Getting more training data will help in case of high variance (overfitting) but it won't help in high bias (underfitting) situation.\n\n--------------------------------------------------------------------------------" ]	[ null, "https://1.bp.blogspot.com/-2yIb8dXdTsc/XS92H6CcYDI/AAAAAAAAavY/8gc1-HYFZPIjFq9kyeRrC59Buf-Ll3HCwCLcBGAs/s320/High%2BBias%2BVariance%2Bin%2BMachine%2BLearning-min.png", null, "https://1.bp.blogspot.com/-4pdwpnOfmKI/XSl35ObuzvI/AAAAAAAAatI/P72GaJWOYXQxiB8vMxFSRln4D0e5CW8VgCLcBGAs/s400/Underfitted-min.jpg", null, "https://1.bp.blogspot.com/-QP9uD2gKMsc/XSl4PcRgSmI/AAAAAAAAatQ/xikRgaxJJg0jYYZAAHWPJNZc8RA9mvWiQCLcBGAs/s400/Overfitted-min.jpg", null, "https://1.bp.blogspot.com/-nktouDdej1Y/XSl4hd4Iy9I/AAAAAAAAatY/X-L85JMtZYI2dNAebEt89ECsPLLpKL7_QCLcBGAs/s400/Equation-min.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90216386,"math_prob":0.90845555,"size":3771,"snap":"2023-40-2023-50","text_gpt3_token_len":784,"char_repetition_ratio":0.16803823,"word_repetition_ratio":0.09933775,"special_character_ratio":0.2198356,"punctuation_ratio":0.10664606,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.978943,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,6,null,6,null,6,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-04T19:57:14Z\",\"WARC-Record-ID\":\"<urn:uuid:d778e83d-6aad-4e57-8438-2da1a10b9a69>\",\"Content-Length\":\"230399\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e1610744-76e9-46cd-9274-e8fd09e683f2>\",\"WARC-Concurrent-To\":\"<urn:uuid:128fd12c-3ae4-4038-8fc2-4237d950ffb0>\",\"WARC-IP-Address\":\"142.251.163.121\",\"WARC-Target-URI\":\"https://tech.apdaga.com/2019/07/high-bias-and-variance-problem-in.html\",\"WARC-Payload-Digest\":\"sha1:IQDLTNDTHEHNNTIVF67HQCGJNIBG4YQW\",\"WARC-Block-Digest\":\"sha1:F2FMLIWDGINYJGRVP5MXYOL4GE5MO25A\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511406.34_warc_CC-MAIN-20231004184208-20231004214208-00814.warc.gz\"}"}
https://www.supplyden.com/janitorial/waste-receptacles-material-handling/transport-trucks/c1_897_1051/	[ "# Transport Trucks\n\nCategories\nWaste Receptacles & Material Handling\n51008AG\nEACH\n\\$815.26\n51008BK\nEACH\n\\$815.26\n51008BL\nEACH\n\\$815.26\n51008BT\nEACH\n\\$815.26\n51008GY\nEACH\n\\$815.26\n51008GR\nEACH\n\\$815.26\n51008L\nEACH\n\\$815.26\n51008N\nEACH\n\\$815.26\n51008O\nEACH\n\\$815.26\n51008R\nEACH\n\\$815.26\n51008SS\nEACH\n\\$815.26\n51008VG\nEACH\n\\$815.26\n51008W\nEACH\n\\$815.26\n51008Y\nEACH\n\\$815.26\n51009AG\nEACH\n\\$1,059.80\n51009BK\nEACH\n\\$1,059.80\n51009BL\nEACH\n\\$1,059.80\n51009BT\nEACH\n\\$1,059.80\n51009GY\nEACH\n\\$1,059.80\n51009GR\nEACH\n\\$1,059.80\n51009L\nEACH\n\\$1,059.80\n51009N\nEACH\n\\$1,059.80\n51009O\nEACH\n\\$1,059.80\n51009R\nEACH\n\\$1,059.80\n51009SS\nEACH\n\\$1,059.80\n51009VG\nEACH\n\\$1,059.80\n51009W\nEACH\n\\$1,059.80\n51009Y\nEACH\n\\$1,059.80\n\nDisplaying 1 to 28 (of 28 products)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.500929,"math_prob":0.9679655,"size":4017,"snap":"2021-04-2021-17","text_gpt3_token_len":1741,"char_repetition_ratio":0.24370795,"word_repetition_ratio":0.66819745,"special_character_ratio":0.4386358,"punctuation_ratio":0.10584678,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95884854,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-19T01:40:42Z\",\"WARC-Record-ID\":\"<urn:uuid:ef5f329e-bbd4-41c2-8692-4c35744c3aea>\",\"Content-Length\":\"877686\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5bfb57be-535e-4b4b-a8d9-63e1a2e282c2>\",\"WARC-Concurrent-To\":\"<urn:uuid:ea5a64e5-3e60-462c-82f6-4c4963a746ef>\",\"WARC-IP-Address\":\"3.19.100.13\",\"WARC-Target-URI\":\"https://www.supplyden.com/janitorial/waste-receptacles-material-handling/transport-trucks/c1_897_1051/\",\"WARC-Payload-Digest\":\"sha1:E6DXKLEZRN7RELE56FKYT3I2VJQ257AM\",\"WARC-Block-Digest\":\"sha1:7YTJ4H7S6WVT2YK45DUZT6ZM2LO3IXYA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703517559.41_warc_CC-MAIN-20210119011203-20210119041203-00561.warc.gz\"}"}
https://avesis.hacettepe.edu.tr/yayin/3610b3de-8861-4b07-96b5-bdadddf12d8b/numerical-solution-to-the-optimization-problem-in-sampling	[ "## NUMERICAL SOLUTION TO THE OPTIMIZATION PROBLEM IN SAMPLING\n\nPAKISTAN JOURNAL OF STATISTICS, vol.30, no.2, pp.285-289, 2014 (Journal Indexed in SCI)", null, "", null, "• Publication Type: Article / Article\n• Volume: 30 Issue: 2\n• Publication Date: 2014\n• Title of Journal : PAKISTAN JOURNAL OF STATISTICS\n• Page Numbers: pp.285-289\n\n#### Abstract\n\nThe sampling literature contains many examples of estimators of population parameters. In the case of generalization of these estimators, estimation of optimum values is essential. For the optimum estimator, the mean square error equations are minimized with respect to unknown parameters and nonlinear constraints. In this paper an attempt is made to get these optimum estimators of parameters in stratified random sampling using genetic algorithms (GAs). Numerical examples are given to illustrate the algorithms. The results show that the genetic algorithm is more efficient than classical ratio type estimator in stratified random sampling." ]	[ null, "https://avesis.hacettepe.edu.tr/Content/images/integrations/small/integrationtype_2.png", null, "https://avesis.hacettepe.edu.tr/Content/images/integrations/small/integrationtype_1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7678626,"math_prob":0.8090142,"size":732,"snap":"2021-31-2021-39","text_gpt3_token_len":143,"char_repetition_ratio":0.125,"word_repetition_ratio":0.0,"special_character_ratio":0.18715847,"punctuation_ratio":0.12096774,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95834607,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-03T07:34:13Z\",\"WARC-Record-ID\":\"<urn:uuid:255cc86c-52fe-4fe6-aca2-4837d9b05bf5>\",\"Content-Length\":\"32108\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c342452e-351e-4ffc-b409-281adeb7a881>\",\"WARC-Concurrent-To\":\"<urn:uuid:a98c809d-8791-4d5d-98bc-9b33e2477e2f>\",\"WARC-IP-Address\":\"193.140.229.23\",\"WARC-Target-URI\":\"https://avesis.hacettepe.edu.tr/yayin/3610b3de-8861-4b07-96b5-bdadddf12d8b/numerical-solution-to-the-optimization-problem-in-sampling\",\"WARC-Payload-Digest\":\"sha1:J3Z2LTYQWW5NJW6GPATPLHCPD3V4ETLG\",\"WARC-Block-Digest\":\"sha1:X4LXDPSPHGEMSOMAXOD62BKJMB7MNUXZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154432.2_warc_CC-MAIN-20210803061431-20210803091431-00211.warc.gz\"}"}
https://www.univerkov.com/suppose-that-the-diameter-of-the-earth-has-become-2-times-larger-but-its-mass-remains-the-same/	[ "# Suppose that the diameter of the Earth has become 2 times larger, but its mass remains the same.\n\nSuppose that the diameter of the Earth has become 2 times larger, but its mass remains the same. Under these conditions, the force acting from the Earth on a person who is on its surface will be …\n\nProblem data: D2 (considered diameter of the Earth) = 2D1 (real diameter); Mz (mass of the Earth) = const.\n\n1) Ratio of gravitational accelerations: n = g2 / g1 = (G * Mz / R2 ^ 2) / (G * Mz / R1 ^ 2) = R1 ^ 2 / R2 ^ 2 = (0.5D1) ^ 2 / ( 0.5D2) ^ 2 = D1 ^ 2 / D2 ^ 2 = D1 ^ 2 / (2D1) ^ 2 = D1 ^ 2 / 4D) ^ 2 = 1/4.\n\n2) Change in the force of attraction: n = Ft2 / Ft1 = mh * g2 / (mh * g1) = g2 / g1 = 1/4.\n\nAnswer: The force acting from the side of the Earth will decrease by 4 times.", null, "One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities." ]	[ null, "https://www.univerkov.com/01.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93270963,"math_prob":0.9940686,"size":1111,"snap":"2023-40-2023-50","text_gpt3_token_len":345,"char_repetition_ratio":0.13279133,"word_repetition_ratio":0.16738197,"special_character_ratio":0.34563455,"punctuation_ratio":0.10454545,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98296285,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-03T07:45:24Z\",\"WARC-Record-ID\":\"<urn:uuid:0e89b616-8d6c-4559-a618-23c43c9d46ec>\",\"Content-Length\":\"22700\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:049bceb1-a0b6-4378-be96-0fc469ce7059>\",\"WARC-Concurrent-To\":\"<urn:uuid:4871a459-ea1b-465d-8cf7-043f6061b3c8>\",\"WARC-IP-Address\":\"37.143.13.208\",\"WARC-Target-URI\":\"https://www.univerkov.com/suppose-that-the-diameter-of-the-earth-has-become-2-times-larger-but-its-mass-remains-the-same/\",\"WARC-Payload-Digest\":\"sha1:J6F2GTLNZPUMHHN4JOZVQCQXJXAMU6VS\",\"WARC-Block-Digest\":\"sha1:UJRG3352VKUGPRDYY6XQO26YSARHHK4L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100489.16_warc_CC-MAIN-20231203062445-20231203092445-00533.warc.gz\"}"}
https://answers.everydaycalculation.com/compare-fractions/1-2-and-9-3	[ "Solutions by everydaycalculation.com\n\n## Compare 1/2 and 9/3\n\n1st number: 1/2, 2nd number: 3 0/3\n\n1/2 is smaller than 9/3\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 2 and 3 is 6\n\nNext, find the equivalent fraction of both fractional numbers with denominator 6\n2. For the 1st fraction, since 2 × 3 = 6,\n1/2 = 1 × 3/2 × 3 = 3/6\n3. Likewise, for the 2nd fraction, since 3 × 2 = 6,\n9/3 = 9 × 2/3 × 2 = 18/6\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 3/6 < 18/6 or 1/2 < 9/3\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]	[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8537638,"math_prob":0.9986555,"size":431,"snap":"2022-05-2022-21","text_gpt3_token_len":201,"char_repetition_ratio":0.27166277,"word_repetition_ratio":0.0,"special_character_ratio":0.45939675,"punctuation_ratio":0.05785124,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99583644,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-19T12:34:01Z\",\"WARC-Record-ID\":\"<urn:uuid:f8e3b649-69d6-487b-9c0a-8b4f40ae19f3>\",\"Content-Length\":\"8287\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:67fd7fe2-3747-4417-8446-c0e5bd4c33f8>\",\"WARC-Concurrent-To\":\"<urn:uuid:d0d4902c-e9b9-4165-9aec-d6a21725c81a>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/compare-fractions/1-2-and-9-3\",\"WARC-Payload-Digest\":\"sha1:CA4HPNFXIKWYQLFZMIAITH4DXTLRTLW2\",\"WARC-Block-Digest\":\"sha1:5NZ6J27MK7MV4BTUWDILXMWCWDYCOD2P\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662527626.15_warc_CC-MAIN-20220519105247-20220519135247-00591.warc.gz\"}"}
https://spsstools.net/en/syntax/syntax-index/parse-or-flag-data/splita-string-variable-into-plaintiff-and-defendant-portions/	[ "``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25``` ```DATA LIST FIXED /a 1-80 (A). BEGIN DATA STAPF V US BUNTEN VS CUMBERLAND T ATTY PIPER V. USA MCCAMMON JR VS. US BOARD OF PAROLES END DATA. LIST. STRING part1 part2(A80). DO IF INDEX(a,\" VS. \")>0. + COMPUTE part1=SUBSTR(a,1,INDEX(a,\" VS. \")). + COMPUTE part2=SUBSTR(a,INDEX(a,\" VS. \")+5). ELSE IF INDEX(a,\" VS \")>0. + COMPUTE part1=SUBSTR(a,1,INDEX(a,\" VS \")). + COMPUTE part2=SUBSTR(a,INDEX(a,\" VS \")+4). ELSE IF INDEX(a,\" V. \")>0. + COMPUTE part1=SUBSTR(a,1,INDEX(a,\" V. \")). + COMPUTE part2=SUBSTR(a,INDEX(a,\" V. \")+4). ELSE IF INDEX(a,\" V \")>0. + COMPUTE part1=SUBSTR(a,1,INDEX(a,\" V \")). + COMPUTE part2=SUBSTR(a,INDEX(a,\" V \")+3). END IF. EXECUTE. ```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.53475094,"math_prob":0.99599123,"size":778,"snap":"2021-21-2021-25","text_gpt3_token_len":317,"char_repetition_ratio":0.24547803,"word_repetition_ratio":0.0,"special_character_ratio":0.43573266,"punctuation_ratio":0.24038461,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9978837,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-15T05:44:46Z\",\"WARC-Record-ID\":\"<urn:uuid:7fd69910-3e69-4534-9ae5-7838efe3689c>\",\"Content-Length\":\"14249\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:359cfd3b-3a02-4214-babe-14a5621625c6>\",\"WARC-Concurrent-To\":\"<urn:uuid:89013332-9ea1-4f4e-8809-1d1fcdee1f0d>\",\"WARC-IP-Address\":\"78.108.88.70\",\"WARC-Target-URI\":\"https://spsstools.net/en/syntax/syntax-index/parse-or-flag-data/splita-string-variable-into-plaintiff-and-defendant-portions/\",\"WARC-Payload-Digest\":\"sha1:WKVKQQXIAVZWQ7NEQUXDIPKGJG5QSH7N\",\"WARC-Block-Digest\":\"sha1:F5OTJUXA6YDLWF4QD77DWTWZS5C54HC2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487617599.15_warc_CC-MAIN-20210615053457-20210615083457-00332.warc.gz\"}"}
https://books.google.com.ag/books?id=vLkIAAAAQAAJ&hl=de&lr=	[ "# Stewart's guide-book to the queen's scholarship and certificate examinations, and to public elementary schools and training schools\n\n### Was andere dazu sagen -Rezension schreiben\n\nEs wurden keine Rezensionen gefunden.\n\n### Beliebte Passagen\n\nSeite 94 - If a straight line meets two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles...\nSeite 11 - At length the freshening western blast Aside the shroud of battle cast; And, first, the ridge of mingled spears Above the brightening cloud appears; And in the smoke the pennons flew , As in the storm the white sea-mew.\nSeite 95 - If, from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than, the other two sides of the triangle, but shall contain a greater angle. Let...\nSeite 98 - In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side on which, when produced, the perpendicular falls, and the straight line intercepted without the triangle, between the perpendicular and the obtuse angle.\nSeite 97 - If a straight line be divided into any two parts, the square on the whole line is equal to the squares on the two parts, together with twice the rectangle contained by the parts.\nSeite 102 - To read a short paragraph from a book not confined to words of one syllable.\nSeite 100 - A tangent to a circle is perpendicular to the radius drawn to the point of contact.\nSeite 97 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.\nSeite 94 - Parallel straight lines are such as are in the same plane, and which being produced ever so far both ways, do not meet.\nSeite 100 - If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles which this line makes with the line touching the circle shall be equal to the angles which are in the alternate segments of the circle." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.78037167,"math_prob":0.96299475,"size":3496,"snap":"2021-43-2021-49","text_gpt3_token_len":753,"char_repetition_ratio":0.12829325,"word_repetition_ratio":0.06239168,"special_character_ratio":0.1999428,"punctuation_ratio":0.08548387,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97155625,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-04T12:58:27Z\",\"WARC-Record-ID\":\"<urn:uuid:bc526f88-107a-4ee9-957b-e2b310c28f76>\",\"Content-Length\":\"59735\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c10b8ba9-5821-4ed0-8a4d-04afbaaac340>\",\"WARC-Concurrent-To\":\"<urn:uuid:07535363-fb8d-4cb9-80d9-dd093dcb0935>\",\"WARC-IP-Address\":\"142.250.65.78\",\"WARC-Target-URI\":\"https://books.google.com.ag/books?id=vLkIAAAAQAAJ&hl=de&lr=\",\"WARC-Payload-Digest\":\"sha1:X6XSRVRUO3ALJABHXRBMRCSN2WGI7DO3\",\"WARC-Block-Digest\":\"sha1:OSCPJCXVCUEMN7T3475UHBUGOWKOZCLW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362992.98_warc_CC-MAIN-20211204124328-20211204154328-00484.warc.gz\"}"}
https://socratic.org/questions/how-do-you-simplify-i-33	[ "# How do you simplify i^-33?\n\nMay 27, 2015\n\nAlgebraically you could do this like this: ${i}^{- 33} = \\frac{1}{{i}^{33}} = \\frac{1}{{i}^{32} \\cdot i} = \\frac{1}{{\\left({i}^{2}\\right)}^{16} \\cdot i} = \\frac{1}{{\\left(- 1\\right)}^{16} \\cdot i} =$\n$= \\frac{1}{1 \\cdot i} = \\frac{1}{i} \\cdot \\frac{i}{i} = \\frac{i}{{i}^{2}} = \\frac{i}{- 1} = - i$\n\nIf you have to use the trigonometric form it goes like this:\n${i}^{- 33} = {\\left[1 \\cdot \\left(\\cos \\left(\\frac{\\pi}{2}\\right) + i \\sin \\left(\\frac{\\pi}{2}\\right)\\right)\\right]}^{- 33} =$\n$= {1}^{- 33} \\cdot \\left(\\cos \\left(\\frac{- 33 \\pi}{2}\\right) + i \\sin \\left(\\frac{- 33 \\pi}{2}\\right)\\right) =$\n$= \\cos \\left(\\frac{33 \\pi}{2}\\right) - i \\sin \\left(\\frac{33 \\pi}{2}\\right) =$\n$= \\cos \\left(8 \\cdot 2 \\pi + \\frac{\\pi}{2}\\right) - i \\sin \\left(8 \\cdot 2 \\pi + \\frac{\\pi}{2}\\right) =$\n$= \\cos \\left(\\frac{\\pi}{2}\\right) - i \\sin \\left(\\frac{\\pi}{2}\\right) =$\n$= 0 - i \\cdot 1 = - i$\nWhat I used here is the De Moivre's formula and some basic properties of $\\sin$ and $\\cos$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7926441,"math_prob":1.0000098,"size":355,"snap":"2021-31-2021-39","text_gpt3_token_len":89,"char_repetition_ratio":0.116809115,"word_repetition_ratio":0.0,"special_character_ratio":0.23661971,"punctuation_ratio":0.05882353,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000092,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-22T18:33:40Z\",\"WARC-Record-ID\":\"<urn:uuid:9c73b20b-4e9a-4c0d-90d0-5579fcef70a4>\",\"Content-Length\":\"33029\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d8ab4b31-93d9-439d-88be-15863c928638>\",\"WARC-Concurrent-To\":\"<urn:uuid:82d33221-7c43-418d-8a79-8d9cead6f5c3>\",\"WARC-IP-Address\":\"216.239.32.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-simplify-i-33\",\"WARC-Payload-Digest\":\"sha1:OKQ6CPW4E5Z2RU4LRJAPEWDPVD6TEROO\",\"WARC-Block-Digest\":\"sha1:CEX4KTHM6UJS5FJR6V4FLEPCLWZFO3XV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057371.69_warc_CC-MAIN-20210922163121-20210922193121-00200.warc.gz\"}"}
https://www.lynxbee.com/c-program-for-finding-remainder-from-floating-point-division-using-fmod/	[ "# C program for finding remainder from floating point division using fmod\n\nAs we have seen in “C program for using modulo operator, finding if number is dividable and print remainder.” if we want to use modulo operator, we have to use either integer or typecast the value with “int” and result also was integer.\n\nNow, what if we have to do float operations and want to identify whether certain float number is completely divisible by another number or not and if not what could be the remainder. For this we have to use math libraries function “fmod” and link the same library during the compilation.\n\n` \\$ vim remainder_from_float_division.c `\n```#include <stdio.h>\n#include <math.h> /* fmod */\n\nint main () {\nfloat i = fmod (15.0,3.0);\nprintf(\"i is %f\\n\", i);\n\nif (i==0) {\nprintf(\"number is divisible by 3\\n\");\n} else {\nprintf(\"number is not divisible by 3\\n\");\n}\n\nprintf ( \"fmod of 5.3 / 2 is %f\\n\", fmod (5.3,2) );\nprintf ( \"fmod of 18.5 / 4.2 is %f\\n\", fmod (18.5,4.2) );\nreturn 0;\n}\n```\n\nNote: we are passing addtional “-lm” to link the math library during compilation.\nIf you don’t link the same, you may get an error like,\n\n```/tmp/ccUbRn5m.o: In function `main':\nremainder_from_float_division.c:(.text+0x2c): undefined reference to `fmod'\n```\n` \\$ gcc -o remainder_from_float_division remainder_from_float_division.c -lm `\n``` \\$ ./remainder_from_float_division\ni is 0.000000\nnumber is divisible by 3\nfmod of 5.3 / 2 is 1.300000\nfmod of 18.5 / 4.2 is 1.700000\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7670877,"math_prob":0.9808565,"size":1357,"snap":"2019-43-2019-47","text_gpt3_token_len":401,"char_repetition_ratio":0.1308204,"word_repetition_ratio":0.018018018,"special_character_ratio":0.31982315,"punctuation_ratio":0.17182131,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9935008,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-12T23:38:18Z\",\"WARC-Record-ID\":\"<urn:uuid:d98784ee-237f-48cb-baa3-92fa8fd5ae13>\",\"Content-Length\":\"79197\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fcda194b-f388-4819-afbd-3fc85f7e373a>\",\"WARC-Concurrent-To\":\"<urn:uuid:bd4dbd64-b937-4db3-8ace-9754ce02fe97>\",\"WARC-IP-Address\":\"104.27.130.51\",\"WARC-Target-URI\":\"https://www.lynxbee.com/c-program-for-finding-remainder-from-floating-point-division-using-fmod/\",\"WARC-Payload-Digest\":\"sha1:WVSGPYTWITNIX4YLSVZA5XD6HBQHGQKO\",\"WARC-Block-Digest\":\"sha1:2KAOHSHK3ICHFQMVD7J7YMBVFNJFVVMR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496665809.73_warc_CC-MAIN-20191112230002-20191113014002-00118.warc.gz\"}"}
https://fj.ics.keio.ac.jp/publication/en/archives/590	[ "# What does the Riemann sphere’s axis stand for?: Understanding the big picture of how mathematical functions behave\n\nAtsushi MiyazawaMasanori NakayamaIssei Fujishiro\n\nin Proceedings of SIGGRAPH Asia 2017 Symposium on Education,\nBangkok, Thailand, Octorber 23―27, 2017 [doi: 10.1145/3134368.3139212]\n\nAbstract\n\nTo visualize the behavior of complex functions, we need four dimensions. We raise the question that has been neglected: “What does the Riemann sphere’s axis stand for?” The answer can be obtained by setting the immersive environment at the sphere’s origin, which is always undefined in projective geometry. By using our immersive math environment, we visually confirmed that a continuous mapping exists between a complex projective line (i.e., a complexified circle) and the Riemann sphere; therefore, we can call them homeomorphic. Such a visualization idiom seems to be applicable to other learning themes, and it also helps us more easily visualize even higher-dimensional math objects.\n\nBy using the same method that we used to construct a real projective line with the equation RP^1 = R^1 ⋃ R^0, we can then create a new, fourth form of the projective plane (the first form is often referred to as a cross-cap, the second as a “Boy’s surface,” and the third as a “Roman surface”), according to the equation RP^2 = R^2 ⋃ RP^1, as shown in the above figure. This new form looks almost like a normal spherical surface, except that an infinite number of self-intersections occur at both poles of the sphere.Therefore, we can draw some elementary math functions that we are familiar with on this new projective plane. The figure shown below illustrates an example of visualizing y = cos x and y = cos 1/x. Note that both are drawn in the same shape as a whole only by differences in whether they are oscillating at infinity or oscillating at the origin.\n\nPublication page in 2017 is here" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94757324,"math_prob":0.89764017,"size":1563,"snap":"2021-43-2021-49","text_gpt3_token_len":339,"char_repetition_ratio":0.1045542,"word_repetition_ratio":0.0,"special_character_ratio":0.20921305,"punctuation_ratio":0.09271523,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98879355,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-04T10:13:18Z\",\"WARC-Record-ID\":\"<urn:uuid:cc79ae64-2283-4711-938d-933e945ac264>\",\"Content-Length\":\"77394\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c56b4c14-7669-4cba-a783-7f4f22553503>\",\"WARC-Concurrent-To\":\"<urn:uuid:c2f7b14e-be18-4874-9a59-47deb62a8735>\",\"WARC-IP-Address\":\"54.250.100.90\",\"WARC-Target-URI\":\"https://fj.ics.keio.ac.jp/publication/en/archives/590\",\"WARC-Payload-Digest\":\"sha1:72JXFQ4ZYIG22FW2DB3QSS4XRTL6K5LI\",\"WARC-Block-Digest\":\"sha1:DVVYJHNMH4QF7V72H5YWEHWZPDXXUS3T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362969.51_warc_CC-MAIN-20211204094103-20211204124103-00213.warc.gz\"}"}
https://sidekick.teachinglearningcollaborative.org/Home/ViewLesson?lessonId=70	[ "Caged Mice Problem\n\nOAA/NBT\nOA\n\nOverview\n\nMathematics Content:\n\nUse place value understanding and properties of operations to add and subtract\n\nWork with addition and subtraction equations\n\nRepresent and solve problems involving addition and subtraction\n\nEstimated Time: Several Class Periods\n\nDuring this lesson students will work as mathematicians to help talk and question through the problem solving process with a partner. Students will work with a problem involving caged mice and different ways they can be organized in two cages. Students will explore a variety of organizational strategies to solve the problem, represent their thinking and summarize their solutions." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9452379,"math_prob":0.7048386,"size":629,"snap":"2022-05-2022-21","text_gpt3_token_len":106,"char_repetition_ratio":0.1264,"word_repetition_ratio":0.0,"special_character_ratio":0.15421304,"punctuation_ratio":0.06122449,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96935636,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-25T05:17:35Z\",\"WARC-Record-ID\":\"<urn:uuid:7fa9f4a8-2163-4cff-8adf-84304b4dd613>\",\"Content-Length\":\"10161\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9c462cd6-3e81-4907-b78b-8b96da1dd969>\",\"WARC-Concurrent-To\":\"<urn:uuid:5adeb4d4-ca4e-4f65-9647-1482727bb76d>\",\"WARC-IP-Address\":\"40.123.45.47\",\"WARC-Target-URI\":\"https://sidekick.teachinglearningcollaborative.org/Home/ViewLesson?lessonId=70\",\"WARC-Payload-Digest\":\"sha1:RTQG3GRNXCLOFB36VKUGQKX5JV6BEZYA\",\"WARC-Block-Digest\":\"sha1:KLHL5A2DG4IC7HHX4PUELNKVFQ5I4IAY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304760.30_warc_CC-MAIN-20220125035839-20220125065839-00225.warc.gz\"}"}
https://www.emanueleferonato.com/2008/08/19/create-a-flash-game-like-pixelfield/	[ "# Create a Flash game like PixelField\n\nI spent some time on PixelField and I liked the way you control the “player” and the overall concept.", null, "So here I am ready to clone it for Tony Pa’s pleasure…\n\nIn this first step I’ll show you how to control the player.\n\nIn this version you control a square (in the original one it was a triangle, so don’t say I did’t add something to the original concept… I added a side…) that will move wherever you click the mouse.\n\nAttached to the square there are four little squares that will follow the main square with an elastic effect.\n\nYou will find the code familiar if you read Controlling a ball like in Flash Elasticity game tutorial… the four satellites move using this engine.\n\n```// friction and speed_scale\n// playing with these variables will affect gameplay\nfriction = 0.95;\nspeed_scale = 0.05;\n// movieclip where I will draw the four elastics\n_root.createEmptyMovieClip(\"drawing\", _root.getNextHighestDepth());\n// big square, the \"player\"\n_root.attachMovie(\"big_square\", \"big_square\", _root.getNextHighestDepth(), {_x:250, _y:200});\n//attaching the four satellites\nfor (x=0; x<=3; x++) {\nsq = _root.attachMovie(\"square\", \"square_\"+x, _root.getNextHighestDepth());\n// set their initial speeds to zero\nsq.xspeed = 0;\nsq.yspeed = 0;\n}\n// main function, to be executed at every frame\n_root.onEnterFrame = function() {\n// rotate the big square\nbig_square._rotation += 2;\nfor (x=0; x<=3; x++) {\n// determining where SHOULD be the x-th satellite without elasticity\nshould_be_x = big_square._x+80Math.cos((big_square._rotation+45+90x)0.0174532925);\nshould_be_y = big_square._y+80Math.sin((big_square._rotation+45+90x)0.0174532925);\n// determining the distance between the point where it SOULD BE and where it IS\ndist_x = (should_be_x-_root[\"square_\"+x]._x)speed_scale;\ndist_y = (should_be_y-_root[\"square_\"+x]._y)speed_scale;\n// adding elasticity... refer to http://www.emanueleferonato.com/2007/09/01/controlling-a-ball-like-in-flash-elasticity-game-tutorial/\n_root[\"square_\"+x].xspeed += dist_x;\n_root[\"square_\"+x].yspeed += dist_y;\n_root[\"square_\"+x].xspeed = friction;\n_root[\"square_\"+x].yspeed = friction;\n_root[\"square_\"+x]._x += _root[\"square_\"+x].xspeed;\n_root[\"square_\"+x]._y += _root[\"square_\"+x].yspeed;\n_root[\"square_\"+x]._rotation = big_square._rotation;\n}\n// drawing the elastics\ndrawing.clear();\ndrawing.lineStyle(1, 0x000000, 50);\ndrawing.moveTo(big_square._x, big_square._y);\ndrawing.lineTo(square_0._x, square_0._y);\ndrawing.moveTo(big_square._x, big_square._y);\ndrawing.lineTo(square_1._x, square_1._y);\ndrawing.moveTo(big_square._x, big_square._y);\ndrawing.lineTo(square_2._x, square_2._y);\ndrawing.moveTo(big_square._x, big_square._y);\ndrawing.lineTo(square_3._x, square_3._y);\n};\n// moving the big square if the player pressed mouse button\n_root.onMouseDown = function() {\nbig_square._x = _root._xmouse;\nbig_square._y = _root._ymouse;\n};```\n\nEnjoy... press the mouse and see what happens\n\n214 GAME PROTOTYPES EXPLAINED WITH SOURCE CODE\n// 1+2=3\n// 10000000\n// 2 Cars\n// 2048\n// Avoider\n// Ballz\n// Block it\n// Blockage\n// Bloons\n// Boids\n// Bombuzal\n// Breakout\n// Bricks\n// Columns\n// CubesOut\n// Dots\n// DROP'd\n// Dudeski\n// Eskiv\n// Filler\n// Fling\n// Globe\n// HookPod\n// Hundreds\n// InkTd\n// Iromeku\n// Lumines\n// Magick\n// MagOrMin\n// Maze\n// Memdot\n// Nano War\n// Nodes\n// o:anquan\n// Ononmin\n// Pacco\n// Phyballs\n// Platform\n// Poker\n// Pool\n// Poux\n// Pudi\n// qomp\n// Racing\n// Renju\n// SameGame\n// Security\n// Sling\n// Slingy\n// Sokoban\n// Splitter\n// Sproing\n// Stack\n// Stringy\n// Sudoku\n// Tetris\n// Threes\n// Toony\n// Turn\n// TwinSpin\n// vvvvvv\n// Wordle\n// Worms\n// Yanga\n// Zhed\n// zNumbers" ]	[ null, "https://www.emanueleferonato.com/images/pixelfield.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6487665,"math_prob":0.97004676,"size":2939,"snap":"2022-27-2022-33","text_gpt3_token_len":795,"char_repetition_ratio":0.19250426,"word_repetition_ratio":0.005882353,"special_character_ratio":0.30758762,"punctuation_ratio":0.23407917,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9957191,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-04T15:30:13Z\",\"WARC-Record-ID\":\"<urn:uuid:cb97add4-b98b-40d5-b41e-5a95c62311c3>\",\"Content-Length\":\"63266\",\"Content-Type\":\"application/http; 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https://socratic.org/questions/how-do-you-simplify-2-3-1	[ "How do you simplify (2/3) ^-1?\n\nJun 10, 2016\n\n$1 \\cdot \\left(\\frac{3}{2}\\right) = 1 \\left(\\frac{1}{2}\\right)$\n\nExplanation:\n\nTo simplify the ${\\left(\\frac{2}{3}\\right)}^{-} 1$\n\nWe first eliminate the negative exponent by placing the term in the denominator.\n\n$\\frac{1}{\\frac{2}{3}} ^ 1$ = $\\frac{1}{\\frac{2}{3}}$\n\nNow invert the fraction from the denominator and multiply by the numerator.\n\n$1 \\cdot \\left(\\frac{3}{2}\\right) = 1 \\left(\\frac{1}{2}\\right)$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7159944,"math_prob":1.0000054,"size":368,"snap":"2022-05-2022-21","text_gpt3_token_len":84,"char_repetition_ratio":0.115384616,"word_repetition_ratio":0.0,"special_character_ratio":0.23097827,"punctuation_ratio":0.07575758,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999815,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-24T13:19:16Z\",\"WARC-Record-ID\":\"<urn:uuid:72065d0d-963e-4eb5-b807-2fb529d9c95a>\",\"Content-Length\":\"32450\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c92dd032-e7a7-4e4f-b1f0-25a0e15bd29b>\",\"WARC-Concurrent-To\":\"<urn:uuid:34a51a33-7c03-405e-9501-de4bfbfd4a19>\",\"WARC-IP-Address\":\"216.239.34.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-simplify-2-3-1\",\"WARC-Payload-Digest\":\"sha1:N3QCVFZWEIMC6BTZALO6NSY7MLUWRFQV\",\"WARC-Block-Digest\":\"sha1:54CSBKBNCYCOEUNPLUBA5XGGPF7DGAVQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304570.90_warc_CC-MAIN-20220124124654-20220124154654-00136.warc.gz\"}"}
http://www.alcyone.com/max/reference/maths/derivatives.html	[ "Mathematics reference\nRules for differentiation\n 18Ma5 MathRef\nEssential rules for differentiation.\n\nLegend.\n• a and n are constants,\n• u and v are functions of x,\n• d is the differential operator.\nBasic.\n(d/dx) (a u) = a du/dx equation 1\n(d/dx) (u +- v) = du/dx +- dv/dx equation 2\n(d/dx) (u v) = u dv/dx + du/dx v equation 3\n(d/dx) (u/v) = (v du/dx - u dv/dx)/v2 equation 4\n(d/dx) a = 0 equation 5\n(d/dx) x = 1 equation 6\n(d/dx) xn = n xn - 1 equation 7\n(d/dx) x1/2 = (1/2) x-1/2 equation 8\n(d/dx) \|x\| = x/\|x\|, x != 0 equation 9\n(d/dx) ex = ex equation 10\n(d/dx) ln x = 1/x equation 11\nTrigonometry.\n(d/dx) sin x = cos x equation 12\n(d/dx) cos x = -sin x equation 13\n(d/dx) tan x = sec2 x equation 14\n(d/dx) cot x = -csc2 x equation 15\n(d/dx) sec x = sec x tan x equation 16\n(d/dx) csc x = -csc x cot x equation 17\n(d/dx) arcsin x = 1/(1 - x2)1/2 equation 18\n(d/dx) arccos x = -1/(1 - x2)1/2 equation 19\n(d/dx) arctan x = 1/(1 + x2) equation 20\n(d/dx) arccot x = -1/(1 + x2) equation 21\n(d/dx) arcsec x = 1/[\|x\| (x2 - 1)1/2] equation 22\n(d/dx) arccsc x = -1/[\|x\| (x2 - 1)1/2] equation 23\nHyperbolic trigonometry.\n(d/dx) sinh x = cosh x equation 24\n(d/dx) cosh x = sinh x equation 25\n(d/dx) tanh x = sech2 x equation 26\n(d/dx) coth x = -csch2 x equation 27\n(d/dx) sech x = -sech x tanh x equation 28\n(d/dx) csch x = -csch x coth x equation 29\n(d/dx) arcsinh x = 1/(x2 + 1)1/2 equation 30\n(d/dx) arccosh x = 1/(x2 - 1)1/2 equation 31\n(d/dx) arctanh x = 1/(1 - x2) equation 32\n(d/dx) arccoth x = 1/(1 - x2) equation 33\n(d/dx) arcsech x = -1/[x (1 - x2)1/2] equation 34\n(d/dx) arccsch x = -1/[\|x\| (1 + x2)1/2] equation 35\nErik Max Francis -- TOP\nWelcome to my homepage.\n 0e\nReference -- UP\nA technical reference.\n 8Re\nMathematics reference -- UP\nA mathematics reference for students and teachers.\n 18Ma\nMathematics reference: Limits -- PREVIOUS\nProperties of limits.\n 18Ma4\nMathematics reference: Rules for integration -- NEXT\nEssential rules for integration.\n 18Ma6\nContents of Erik Max Francis' homepages -- CONTENTS\nEverything in my homepages.\n 1In1\nFeedback -- FEEDBACK\nHow to send feedback on these pages to the author.\n 1In5\nAbout Erik Max Francis -- PERSONAL" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.50286716,"math_prob":1.0000082,"size":2563,"snap":"2019-13-2019-22","text_gpt3_token_len":1083,"char_repetition_ratio":0.31027746,"word_repetition_ratio":0.013864818,"special_character_ratio":0.4541553,"punctuation_ratio":0.05263158,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999813,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-25T22:47:26Z\",\"WARC-Record-ID\":\"<urn:uuid:3400793a-e185-40ed-92bf-badadbfb94e5>\",\"Content-Length\":\"15812\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:53a8faf3-d469-4b14-9677-4ea434b8d0b3>\",\"WARC-Concurrent-To\":\"<urn:uuid:b06ddb1f-c5da-4aa9-9780-dff7281b8a9b>\",\"WARC-IP-Address\":\"216.218.218.4\",\"WARC-Target-URI\":\"http://www.alcyone.com/max/reference/maths/derivatives.html\",\"WARC-Payload-Digest\":\"sha1:EYHWYAOJZX7DLVH2652SM23P7EAMZYJB\",\"WARC-Block-Digest\":\"sha1:7C6MI45PRJBVU7BIXSMLHTKFQU3YXITW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912204461.23_warc_CC-MAIN-20190325214331-20190326000331-00559.warc.gz\"}"}
https://blogs.mathworks.com/loren/2007/03/01/creating-sparse-finite-element-matrices-in-matlab/	[ "# Creating Sparse Finite-Element Matrices in MATLAB\n\nI'm pleased to introduce Tim Davis as this week's guest blogger. Tim is a professor at the University of Florida, and is the\nauthor or co-author of many of our sparse matrix functions (lu, chol, much of sparse backslash, ordering methods such as amd\nand colamd, and other functions such as etree and symbfact). He is also the author of a recent book, Direct Methods for Sparse Linear Systems, published by SIAM, where more details of MATLAB sparse matrices are discussed ( http://www.cise.ufl.edu/~davis ).\n\n### MATLAB is Slow? Only If You Use It Incorrectly\n\nFrom time to time, I hear comments such as \"MATLAB is slow for large finite-element problems.\" When I look at the details,\nthe problem is typically due to an overuse of A(i,j)= ... when creating the sparse matrix (assembling the finite elements). This can be seen in typical user's code, MATLAB code in\nbooks on the topic, and even in MATLAB itself. The problem is widespread. MATLAB can be very fast for finite-element problems,\nbut not if it's used incorrectly.\n\n### How Not to Create a Finite-Element Matrix\n\nA good example of what not to do can be found in the wathen.m function, in MATLAB. A=gallery('wathen',200,200) takes a huge amount of time; a very minor modification cuts the run time drastically. I'm not intending to single out this\none function for critique; this is a very common issue that I see over and over again. This function is built into MATLAB,\nwhich makes it an accessible example. It was first written when MATLAB did not support sparse matrices, and was modified\nonly slightly to exploit sparsity. It was never meant for generating large sparse finite-element matrices. You can see the\nentire function with the command:\n\n type private/wathen.m\n\nBelow is an excerpt of the relevant parts of wathen.m. The function wathen1.m creates a finite-element matrix of an nx-by-ny mesh. Each i loop creates a single 8-by-8 finite-element matrix, and adds it into A.\n\ntype wathen1\ntic\nA = wathen1 (200,200) ;\ntoc\nfunction A = wathen1 (nx,ny)\nrand('state',0)\ne1 = [6 -6 2 -8;-6 32 -6 20;2 -6 6 -6;-8 20 -6 32];\ne2 = [3 -8 2 -6;-8 16 -8 20;2 -8 3 -8;-6 20 -8 16];\ne = [e1 e2; e2' e1]/45;\nn = 3nxny+2nx+2ny+1;\nA = sparse(n,n);\nRHO = 100rand(nx,ny);\nnn = zeros(8,1);\nfor j=1:ny\nfor i=1:nx\nnn(1) = 3jnx+2i+2j+1;\nnn(2) = nn(1)-1;\nnn(3) = nn(2)-1;\nnn(4) = (3j-1)nx+2j+i-1;\nnn(5) = 3(j-1)nx+2i+2j-3;\nnn(6) = nn(5)+1;\nnn(7) = nn(6)+1;\nnn(8) = nn(4)+1;\nem = eRHO(i,j);\nfor krow=1:8\nfor kcol=1:8\nA(nn(krow),nn(kcol)) = A(nn(krow),nn(kcol))+em(krow,kcol);\nend\nend\nend\nend\n\nElapsed time is 305.832709 seconds.\n\n\n### What's Wrong with It?\n\nThe above code is fine for generating a modest sized matrix, but the A(i,j) = ... statement is quite slow when A is large and sparse, particularly when i and j are also scalars. The inner two for-loops can be vectorized so that i and j are vectors of length 8. Each A(i,j) = ... statement is assembling an entire finite-element matrix into A. However, this leads to very minimal improvement in run time.\n\ntype wathen1b\ntic\nA1b = wathen1b (200,200) ;\ntoc\nfunction A = wathen1b (nx,ny)\nrand('state',0)\ne1 = [6 -6 2 -8;-6 32 -6 20;2 -6 6 -6;-8 20 -6 32];\ne2 = [3 -8 2 -6;-8 16 -8 20;2 -8 3 -8;-6 20 -8 16];\ne = [e1 e2; e2' e1]/45;\nn = 3nxny+2nx+2ny+1;\nA = sparse(n,n);\nRHO = 100rand(nx,ny);\nnn = zeros(8,1);\nfor j=1:ny\nfor i=1:nx\nnn(1) = 3jnx+2i+2j+1;\nnn(2) = nn(1)-1;\nnn(3) = nn(2)-1;\nnn(4) = (3j-1)nx+2j+i-1;\nnn(5) = 3(j-1)nx+2i+2j-3;\nnn(6) = nn(5)+1;\nnn(7) = nn(6)+1;\nnn(8) = nn(4)+1;\nem = eRHO(i,j);\nA (nn,nn) = A (nn,nn) + em ;\nend\nend\n\nElapsed time is 282.945593 seconds.\n\ndisp (norm (A-A1b,1))\n 0\n\n\n### How MATLAB Stores Sparse Matrices\n\nTo understand why the above examples are so slow, you need to understand how MATLAB stores its sparse matrices. An n-by-n\nMATLAB sparse matrix is stored as three arrays; I'll call them p, i, and x. These three arrays are not directly accessible from M, but they can be accessed by a mexFunction. The nonzero entries in\ncolumn j are stored in i(p(j):p(j+1)-1) and x(p(j):p(j+1)-1), where x holds the numerical values and i holds the corresponding row indices. Below is a very small example. First, I create a full matrix and convert it into a sparse one. This is only so that you can easily see the matrix C and how it's stored in sparse form. You should never create a sparse matrix this way, except for tiny examples.\n\nC = [\n4.5 0.0 3.2 0.0\n3.1 2.9 0.0 0.9\n0.0 1.7 3.0 0.0\n3.5 0.4 0.0 1.0 ] ;\nC = sparse (C)\nC =\n(1,1) 4.5000\n(2,1) 3.1000\n(4,1) 3.5000\n(2,2) 2.9000\n(3,2) 1.7000\n(4,2) 0.4000\n(1,3) 3.2000\n(3,3) 3.0000\n(2,4) 0.9000\n(4,4) 1.0000\n\n\nNotice that the nonzero entries in C are stored in column order, with sorted row indices. The internal p, i, and x arrays can be reconstructed as follows. The find(C) statement returns a list of \"triplets,\" where the kth triplet is i(k), j(k), and x(k). This specifies that C(i(k),j(k)) is equal to x(k). Next, find(diff(...)) constructs the column pointer array p (this only works if there are no all-zero columns in the matrix).\n\n[i j x] = find (C) ;\nn = size(C,2) ;\np = find (diff ([0 ; j ; n+1])) ;\nfor col = 1:n\nfprintf ('column %d:\\n k row index value\\n', col) ;\ndisp ([(p(col):p(col+1)-1)' i(p(col):p(col+1)-1) x(p(col):p(col+1)-1)])\nend\ncolumn 1:\nk row index value\n1.0000 1.0000 4.5000\n2.0000 2.0000 3.1000\n3.0000 4.0000 3.5000\ncolumn 2:\nk row index value\n4.0000 2.0000 2.9000\n5.0000 3.0000 1.7000\n6.0000 4.0000 0.4000\ncolumn 3:\nk row index value\n7.0000 1.0000 3.2000\n8.0000 3.0000 3.0000\ncolumn 4:\nk row index value\n9.0000 2.0000 0.9000\n10.0000 4.0000 1.0000\n\n\nNow consider what happens when one entry is added to C:\n\nC(3,1) = 42 ;\n\n[i j x] = find (C) ;\nn = size(C,2) ;\np = find (diff ([0 ; j ; n+1])) ;\nfor col = 1:n\nfprintf ('column %d:\\n k row index value\\n', col) ;\ndisp ([(p(col):p(col+1)-1)' i(p(col):p(col+1)-1) x(p(col):p(col+1)-1)])\nend\ncolumn 1:\nk row index value\n1.0000 1.0000 4.5000\n2.0000 2.0000 3.1000\n3.0000 3.0000 42.0000\n4.0000 4.0000 3.5000\ncolumn 2:\nk row index value\n5.0000 2.0000 2.9000\n6.0000 3.0000 1.7000\n7.0000 4.0000 0.4000\ncolumn 3:\nk row index value\n8.0000 1.0000 3.2000\n9.0000 3.0000 3.0000\ncolumn 4:\nk row index value\n10.0000 2.0000 0.9000\n11.0000 4.0000 1.0000\n\n\nand you can see that nearly every entry in C has been moved down by one in the i and x arrays. In general, the single statement C(3,1)=42 takes time proportional to the number of entries in matrix. Thus, looping nnz(A) times over the statement A(i,j)=A(i,j)+... takes time proportional to nnz(A)^2.\n\n### A Better Way to Create a Finite-Element Matrix\n\nThe version below is only slightly different. It could be improved, but I left it nearly the same to illustrate how simple\nit is to write fast MATLAB code to solve this problem, via a minor tweak. The idea is to create a list of triplets, and let\nMATLAB convert them into a sparse matrix all at once. If there are duplicates (which a finite-element matrix always has)\nthe duplicates are summed, which is exactly what you want when assembling a finite-element matrix. In MATLAB 7.3 (R2006b),\nsparse uses quicksort, which takes nnz(A)log(nnz(A)) time. This is slower than it could be (a linear-time bucket sort can be used, taking essentially nnz(A) time). However, it's still much faster than nnz(A)^2. For this matrix, nnz(A) is about 1.9 million.\n\ntype wathen2.m\ntic\nA2 = wathen2 (200,200) ;\ntoc\nfunction A = wathen2 (nx,ny)\nrand('state',0)\ne1 = [6 -6 2 -8;-6 32 -6 20;2 -6 6 -6;-8 20 -6 32];\ne2 = [3 -8 2 -6;-8 16 -8 20;2 -8 3 -8;-6 20 -8 16];\ne = [e1 e2; e2' e1]/45;\nn = 3nxny+2nx+2ny+1;\nntriplets = nxny64 ;\nI = zeros (ntriplets, 1) ;\nJ = zeros (ntriplets, 1) ;\nX = zeros (ntriplets, 1) ;\nntriplets = 0 ;\nRHO = 100rand(nx,ny);\nnn = zeros(8,1);\nfor j=1:ny\nfor i=1:nx\nnn(1) = 3jnx+2i+2j+1;\nnn(2) = nn(1)-1;\nnn(3) = nn(2)-1;\nnn(4) = (3j-1)nx+2j+i-1;\nnn(5) = 3(j-1)nx+2i+2j-3;\nnn(6) = nn(5)+1;\nnn(7) = nn(6)+1;\nnn(8) = nn(4)+1;\nem = eRHO(i,j);\nfor krow=1:8\nfor kcol=1:8\nntriplets = ntriplets + 1 ;\nI (ntriplets) = nn (krow) ;\nJ (ntriplets) = nn (kcol) ;\nX (ntriplets) = em (krow,kcol) ;\nend\nend\nend\nend\nA = sparse (I,J,X,n,n) ;\n\nElapsed time is 1.594073 seconds.\n\ndisp (norm (A-A2,1))\n 1.4211e-014\n\n\nIf you do not know how many entries your matrix will have, you may not be able to preallocate the I, J, and X arrays, as done in wathen2.m. In that case, start them at a reasonable size (anything larger than zero will do) and add this code to the innermost loop,\njust after ntriplets is incremented:\n\n len = length (X) ;\nif (ntriplets > len)\nI (2len) = 0 ;\nJ (2len) = 0 ;\nX (2*len) = 0 ;\nend\n\nand when done, use sparse(I(1:ntriplets),J(1:ntriplets),X(1:ntriplets),n,n).\n\n### Moral: Do Not Abuse A(i,j)=... for Sparse A; Use sparse Instead\n\nAvoid statements such as\n\nC(4,2) = C(4,2) + 42 ;\n\nin a loop. Create a list of triplets (i,j,x) and use sparse instead. This advice holds for any sparse matrix, not just finite-element ones.\n\n### Try a Faster Sparse Function\n\nCHOLMOD includes a sparse2 mexFunction which is a replacement for sparse. It uses a linear-time bucket sort. The MATLAB 7.3 (R2006b) sparse accounts for about 3/4ths the total run time of wathen2.m. For this matrix sparse2 in CHOLMOD is about 10 times faster than the MATLAB sparse. CHOLMOD can be found in the SuiteSparse package, in MATLAB Central.\n\nIf you would like to see a short and concise C mexFunction implementation of the method used by sparse2, take a look at CSparse, which was written for a concise textbook-style presentation. The cs_sparse mexFunction uses cs_compress.c, to convert the triplets to a compressed column form of A', cs_dupl.c to sum up duplicate entries, and then cs_transpose.c to transpose the result (which also sorts the columns).\n\n### When Fast is Not Fast Enough...\n\nEven with this dramatic improvement in constructing the matrix A, MATLAB could still use additional features for faster construction of sparse finite-element matrices. Constructing the\nmatrix should be much faster than x=A\\b, since chol is doing about 700 times more work as sparse for this matrix (1.3 billion flops, vs 1.9 million nonzeros in A). The run times are not that different, however:\n\ntic\nA = wathen2 (200,200) ;\ntoc\nb = rand (size(A,1),1) ;\ntic\nx=A\\b ;\ntoc\nElapsed time is 1.720397 seconds.\nElapsed time is 3.125791 seconds.\n\n\nPublished with MATLAB® 7.4\n\n\|" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7623933,"math_prob":0.99585754,"size":10747,"snap":"2020-45-2020-50","text_gpt3_token_len":3735,"char_repetition_ratio":0.1211952,"word_repetition_ratio":0.22204472,"special_character_ratio":0.3786173,"punctuation_ratio":0.17893218,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9985094,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-31T08:03:55Z\",\"WARC-Record-ID\":\"<urn:uuid:9bb8eb93-69c9-4a9a-8fd0-01e3728e0320>\",\"Content-Length\":\"186243\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7f1e9a10-adb4-495e-a724-18a7ca7211b6>\",\"WARC-Concurrent-To\":\"<urn:uuid:fb1f8e5e-c156-4734-a967-fcf24d06cbc4>\",\"WARC-IP-Address\":\"23.212.144.59\",\"WARC-Target-URI\":\"https://blogs.mathworks.com/loren/2007/03/01/creating-sparse-finite-element-matrices-in-matlab/\",\"WARC-Payload-Digest\":\"sha1:QMDMS5BQXZHJA4YOFD2DXQD7S3UVEJKI\",\"WARC-Block-Digest\":\"sha1:FHCPZNQYXAFX3H6LMGLEI57ZKFYGGF4U\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107916776.80_warc_CC-MAIN-20201031062721-20201031092721-00429.warc.gz\"}"}
https://de.mathworks.com/matlabcentral/cody/problems/42917-nth-roots-of-unity/solutions/1497697	[ "Cody\n\n# Problem 42917. Nth roots of unity\n\nSolution 1497697\n\nSubmitted on 19 Apr 2018 by J. S. Kowontan\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\nn = 5; y_correct = -0.467800202134647; assert( abs(your_fcn_name(n)-y_correct) < .0001)\n\n2 Pass\nn = 50; y_correct = -2.151544927902936 - 0.430301217000093i assert( abs(your_fcn_name(n)-y_correct) < .0001)\n\ny_correct = -2.1515 - 0.4303i\n\n3 Pass\nn = 7; y_correct = -0.435928596902380 assert( abs(your_fcn_name(n)-y_correct) < .0001)\n\ny_correct = -0.4359\n\n4 Pass\nn = 70; y_correct = -3.031653804728051 - 0.430301217000095i assert( abs(your_fcn_name(n)-y_correct) < .0001)\n\ny_correct = -3.0317 - 0.4303i\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.50387007,"math_prob":0.8762687,"size":789,"snap":"2020-45-2020-50","text_gpt3_token_len":286,"char_repetition_ratio":0.16942675,"word_repetition_ratio":0.054054055,"special_character_ratio":0.4917617,"punctuation_ratio":0.16058394,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9843867,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-25T11:14:02Z\",\"WARC-Record-ID\":\"<urn:uuid:000a7ebc-2d93-4b73-9df6-4df7062244f9>\",\"Content-Length\":\"80343\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f475cf36-4bf4-4b57-91c4-de28fb94414c>\",\"WARC-Concurrent-To\":\"<urn:uuid:f5d0c715-0c4d-4297-9494-414d467feb46>\",\"WARC-IP-Address\":\"184.25.198.13\",\"WARC-Target-URI\":\"https://de.mathworks.com/matlabcentral/cody/problems/42917-nth-roots-of-unity/solutions/1497697\",\"WARC-Payload-Digest\":\"sha1:QXQ7THLSOJLLRTP2GRFO2TYRGRRQEMLQ\",\"WARC-Block-Digest\":\"sha1:SDGLCW63KBYLJWKJAH4DCIM6INBQNNF2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141182776.11_warc_CC-MAIN-20201125100409-20201125130409-00020.warc.gz\"}"}
http://www.fetsystem.com/mathematics/factors-13	[ " What are the factors of 13? \| Foreign Educator Teaching System", null, "", null, "# What are the factors of 13?\n\nFactors of 13\nFactors of a number are the numbers which give the same number when they are multiplied.\n1×13=13\n-1x-13=13\nFactors of 13 are 1,13,-1,-13\n\nRating 3.00 out of 5\n" ]	[ null, "http://www.fetsystem.com/wp-content/themes/fetsysbeta/images/logo.png", null, "http://www.fetsystem.com/wp-content/themes/fetsysbeta/images/share.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8721242,"math_prob":0.9929358,"size":214,"snap":"2019-35-2019-39","text_gpt3_token_len":65,"char_repetition_ratio":0.17619048,"word_repetition_ratio":0.0,"special_character_ratio":0.3317757,"punctuation_ratio":0.08695652,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9932852,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-23T20:03:02Z\",\"WARC-Record-ID\":\"<urn:uuid:764a72e5-c2e1-4e89-a9c6-c4bddcd46597>\",\"Content-Length\":\"48459\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6ac3cc44-cb03-4fe0-b0f3-1c6f04618a08>\",\"WARC-Concurrent-To\":\"<urn:uuid:3ad3e2f3-c3af-42ca-8600-b65e659d0ae3>\",\"WARC-IP-Address\":\"184.107.235.178\",\"WARC-Target-URI\":\"http://www.fetsystem.com/mathematics/factors-13\",\"WARC-Payload-Digest\":\"sha1:JOV7KD2LJC5L3DUKNVOPFZDLUJ36MTAT\",\"WARC-Block-Digest\":\"sha1:C4PQLD4GCDW47OJHEHHA3TT7HKQZ5VKZ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514578201.99_warc_CC-MAIN-20190923193125-20190923215125-00548.warc.gz\"}"}
http://bulletin.kpi.ua/article/view/151759	[ "# Program of Simplification of High-Level Polynomial at the Example of Simplification of Wilson’s Formula\n\n## Authors\n\n• Artem M. Yevtushenko Igor Sikorsky Kyiv Polytechnic Institute, Ukraine\n• Yuri D. Shcherbashin Igor Sikorsky Kyiv Polytechnic Institute, Ukraine\n\n## Keywords:\n\nHigh-level polynomial, Wilson’s formula, Experiment planning method, Least squares method, Polynomial of the 2nd degree\n\n## Abstract\n\nBackground. Since many formulas have the form of high-level polynomials and their use leads to a large number of computations, which slows down the speed of obtaining results, the technology of simplification of high-level polynomials is considered.\n\nObjective. The aim of the paper is to obtain a technology for the simplification of high-level polynomials based on the application of the theory of experiment planning to the Wilson’s formula.\n\nMethods. To simplify high-level polynomials, combination and consistent application of experiment planning and least squares methods are proposed. For the field of input values, matrix of rotatable central composite plan Box of second order for three factors is constructed. To the constructed matrix the least squares method was applied, by which the coefficients of the simplified formula can be found. The resulting simplified formula will have the form of a polynomial of the 2nd degree.\n\nResults. The Wilson’s formula, which has the form of a polynomial of degree 4, is simplified to the form of a polynomial of degree 2. Having broken down the entire definition domain for Wilson's formula on the parts and constructed a simplified formula for a particular part, we obtained a result that, using the simplified formula, one can calculate the speed of sound almost 25 times faster than using the Wilson's formula, with only a slight deviation in the results.\n\nConclusions. When simplifying polynomials of high degree, the reduction of the ranges of input parameters is decisive for obtaining a satisfactory deviation between the calculated values. The proposed approach to simplifying the formulas worked quite well on the example of Wilson's formula. It can also be used to simplify other formulas that have the form of high-level polynomials. One of the options for further use of the results of this work is the creation of a technology that would enable the parallel calculation of the sound speed based on the simplified formulas obtained for each of the parts to which the definition area for the Wilson's formula is divided.\n\n## Author Biographies\n\n### Artem M. Yevtushenko, Igor Sikorsky Kyiv Polytechnic Institute\n\nАртем Михайлович Євтушенко\n\n### Yuri D. Shcherbashin, Igor Sikorsky Kyiv Polytechnic Institute\n\nЮрій Дмитрович Щербашин\n\n## References\n\nN.B. Vargaftik, Handbook on the Thermophysical Properties of Gases and Hydroxides. Moscow, SU: Nauka, 1972, 721 p.\n\nV.I. Babiy, Problems and Prospects of Measuring the Speed of Sound in the Ocean. Sevastopol, Ukraine: Scientific and Production Center “EKOSI-Hidrofizika”, 2009, 142 p.\n\nC.C. Leroy, “A new equation for the accurate calculation of sound speed in all oceans”, J. Acoust. Soc. Am., vol. 124, no. 5, pp. 2774–2782, 2008. doi: 10.1121/1.2988296\n\nW.D. Wilson, “Equation for the speed of sound in sea water”, J. Acoust. Soc. Am., vol. 32, no. 10, p. 1357, 1960. doi: 10.1121/1.1907913\n\nYu.P. Adler, Introduction to Experiment Planning. Moscow, SU: Metalurgia, 1968, 155 p.\n\nV.I. Asaturyan, The Theory of Experiment Planning. Moscow, SU: Radio i Sviaz’, 1983.\n\nV.V. Nalimov, Theory of the Experiment. Moscow, SU: Nauka, 1971.\n\nYu.V. Linnik. The Method of least Squares and the Fundamentals of the Mathematical-Statistical Theory of Processing Observations, 2nd ed. Leningrad, SU: Fizmatgiz, 1962.\n\nL.I. Turchak, Fundamentals of Numerical Methods. Moscow, Russia: Fizmatlit, 2005, 304 p." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8885354,"math_prob":0.86219084,"size":4548,"snap":"2022-40-2023-06","text_gpt3_token_len":1044,"char_repetition_ratio":0.13556337,"word_repetition_ratio":0.029112082,"special_character_ratio":0.22581354,"punctuation_ratio":0.17508417,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9901705,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-02T10:18:47Z\",\"WARC-Record-ID\":\"<urn:uuid:1b2d34d8-1681-43a6-997b-235eb204712d>\",\"Content-Length\":\"38398\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dea5a5d8-99c4-4663-a63c-182633a81406>\",\"WARC-Concurrent-To\":\"<urn:uuid:92095cf1-464f-4336-a312-603186cea593>\",\"WARC-IP-Address\":\"212.111.212.230\",\"WARC-Target-URI\":\"http://bulletin.kpi.ua/article/view/151759\",\"WARC-Payload-Digest\":\"sha1:AX2Y5L4YV2B4YBQYOVOIFLTWW6LWQVTL\",\"WARC-Block-Digest\":\"sha1:KFRHX3VMO236RPEVVJVR45G6FLMMALPD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337307.81_warc_CC-MAIN-20221002083954-20221002113954-00604.warc.gz\"}"}
https://ginga.readthedocs.io/en/stable/api/ginga.canvas.CanvasObject.CanvasObjectBase.html	[ "# CanvasObjectBase\n\nclass ginga.canvas.CanvasObject.CanvasObjectBase(*kwdargs)[source]\n\nBases: `Callbacks`\n\nThis is the abstract base class for a CanvasObject. A CanvasObject is an item that can be placed on a Ginga canvas.\n\nThis class defines common methods used by all such objects.\n\nMethods Summary\n\n `calc_dual_scale_from_pt`(pt, detail) `calc_radius`(viewer, p1, p2) `calc_rotation_from_pt`(pt, detail) `calc_scale_from_pt`(pt, detail) `calc_vertexes`(start_cx, start_cy, end_cx, end_cy) `canvascoords`(viewer, data_x, data_y[, center]) `contains`(x, y) For backward compatibility. `contains_arr`(x_arr, y_arr) For backward compatibility. `contains_pts`(points) `convert_mapper`(tomap) Converts our object from using one coordinate map to another. `copy`([share]) `draw_arrowhead`(cr, x1, y1, x2, y2) `draw_caps`(cr, cap, points[, radius]) `draw_edit`(cr, viewer) `get_bbox`([points]) Get bounding box of this object. Return the geometric average of points as data_points. `get_cpoints`(viewer[, points, no_rotate]) `get_data`(args) `get_data_points`([points]) Points returned are in data coordinates. Returns 3 edit control points for editing this object: a move point, a scale point and a rotate point. Get the set of points that is used to draw the object. `get_pt`(viewer, points, pt[, canvas_radius]) Takes an array of points `points` and a target point `pt`. `initialize`(canvas, viewer, logger) `move_delta`(xoff, yoff) For backward compatibility. `move_delta_pt`(off_pt) `move_to`(xdst, ydst) For backward compatibility. `move_to_pt`(dst_pt) `point_within_line`(points, p_start, p_stop, ...) `point_within_radius`(points, pt, canvas_radius) Points `points` and point `pt` are in data coordinates. `rerotate_by_deg`(thetas, detail) `rescale_by`(scale_x, scale_y, detail) For backward compatibility. `rescale_by_factors`(factors, detail) `rotate`(theta_deg[, xoff, yoff]) For backward compatibility. `rotate_by`(theta_deg) For backward compatibility. `rotate_by_deg`(thetas) `rotate_deg`(thetas, offset) `scale_by`(scale_x, scale_y) For backward compatibility. `scale_by_factors`(factors) `scale_font`(viewer) `select_contains`(viewer, x, y) For backward compatibility. `select_contains_pt`(viewer, pt) `set_data`(*kwdargs) `set_data_points`(points) Input `points` must be in data coordinates, will be converted to the coordinate space of the object and stored. `set_point_by_index`(i, pt) `setup_edit`(detail) subclass should override as necessary. `swapxy`(x1, y1, x2, y2) This method called when changes are made to the parameters. `use_coordmap`(mapobj) `within_line`(viewer, points, p_start, p_stop, ...) Points `points` and line endpoints `p_start`, `p_stop` are in data coordinates. `within_radius`(viewer, points, pt, canvas_radius) Points `points` and point `pt` are in data coordinates.\n\nMethods Documentation\n\ncalc_dual_scale_from_pt(pt, detail)[source]\ncalc_rotation_from_pt(pt, detail)[source]\ncalc_scale_from_pt(pt, detail)[source]\ncalc_vertexes(start_cx, start_cy, end_cx, end_cy, arrow_length=10, arrow_degrees=0.35)[source]\ncanvascoords(viewer, data_x, data_y, center=None)[source]\ncontains(x, y)[source]\n\nFor backward compatibility. TO BE DEPRECATED–DO NOT USE. Use contains_pt() instead.\n\ncontains_arr(x_arr, y_arr)[source]\n\nFor backward compatibility. TO BE DEPRECATED–DO NOT USE. Use contains_pts() instead.\n\ncontains_pt(pt)[source]\ncontains_pts(points)[source]\nconvert_mapper(tomap)[source]\n\nConverts our object from using one coordinate map to another.\n\nNOTE: In some cases this only approximately preserves the equivalent point values when transforming between coordinate spaces.\n\ncopy(share=[])[source]\ndraw_edit(cr, viewer)[source]\nget_bbox(points=None)[source]\n\nGet bounding box of this object.\n\nReturns\n(p1, p2, p3, p4): a 4-tuple of the points in data coordinates,\nbeginning with the lower-left and proceeding counter-clockwise.\nget_center_pt()[source]\n\nReturn the geometric average of points as data_points.\n\nget_cpoints(viewer, points=None, no_rotate=False)[source]\nget_data(args)[source]\nget_data_points(points=None)[source]\n\nPoints returned are in data coordinates.\n\nget_llur()[source]\nget_move_scale_rotate_pts(viewer)[source]\n\nReturns 3 edit control points for editing this object: a move point, a scale point and a rotate point. These points are all in data coordinates.\n\nget_num_points()[source]\nget_point_by_index(i)[source]\nget_points()[source]\n\nGet the set of points that is used to draw the object.\n\nPoints are returned in data coordinates.\n\nTakes an array of points `points` and a target point `pt`. Returns the first index of the point that is within the radius of the target point. If none of the points are within the radius, returns None.\n\nget_reference_pt()[source]\ninitialize(canvas, viewer, logger)[source]\nis_compound()[source]\nmove_delta(xoff, yoff)[source]\n\nFor backward compatibility. TO BE DEPRECATED–DO NOT USE. Use move_delta_pt instead.\n\nmove_delta_pt(off_pt)[source]\nmove_to(xdst, ydst)[source]\n\nFor backward compatibility. TO BE DEPRECATED–DO NOT USE. Use move_to_pt() instead.\n\nmove_to_pt(dst_pt)[source]\n\nPoints `points` and point `pt` are in data coordinates. Return True for points within the circle defined by a center at point `pt` and within canvas_radius.\n\nrerotate_by_deg(thetas, detail)[source]\nrescale_by(scale_x, scale_y, detail)[source]\n\nFor backward compatibility. TO BE DEPRECATED–DO NOT USE. Use rescale_by_factors() instead.\n\nrescale_by_factors(factors, detail)[source]\nrotate(theta_deg, xoff=0, yoff=0)[source]\n\nFor backward compatibility. TO BE DEPRECATED–DO NOT USE. Use rotate_deg() instead.\n\nrotate_by(theta_deg)[source]\n\nFor backward compatibility. TO BE DEPRECATED–DO NOT USE. Use rotate_by_deg() instead.\n\nrotate_by_deg(thetas)[source]\nrotate_deg(thetas, offset)[source]\nscale_by(scale_x, scale_y)[source]\n\nFor backward compatibility. TO BE DEPRECATED–DO NOT USE. Use scale_by_factors() instead.\n\nscale_by_factors(factors)[source]\nscale_font(viewer)[source]\nselect_contains(viewer, x, y)[source]\n\nFor backward compatibility. TO BE DEPRECATED–DO NOT USE. Use select_contains_pt() instead.\n\nselect_contains_pt(viewer, pt)[source]\nset_data(**kwdargs)[source]\nset_data_points(points)[source]\n\nInput `points` must be in data coordinates, will be converted to the coordinate space of the object and stored.\n\nset_point_by_index(i, pt)[source]\nsetup_edit(detail)[source]\n\nsubclass should override as necessary.\n\nswapxy(x1, y1, x2, y2)[source]\nsync_state()[source]\n\nThis method called when changes are made to the parameters. subclasses should override if they need any special state handling.\n\nuse_coordmap(mapobj)[source]\nPoints `points` and line endpoints `p_start`, `p_stop` are in data coordinates. Return True for points within the line defined by a line from p_start to p_end and within `canvas_radius`. The distance between points is scaled by the viewer’s canvas scale.\nPoints `points` and point `pt` are in data coordinates. Return True for points within the circle defined by a center at point `pt` and within canvas_radius. The distance between points is scaled by the canvas scale." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.61097044,"math_prob":0.94354,"size":5686,"snap":"2022-27-2022-33","text_gpt3_token_len":1367,"char_repetition_ratio":0.16772263,"word_repetition_ratio":0.28075254,"special_character_ratio":0.22810411,"punctuation_ratio":0.18534032,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97144586,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-17T16:50:40Z\",\"WARC-Record-ID\":\"<urn:uuid:97a9bacf-6aa5-43f0-938e-a65666d609eb>\",\"Content-Length\":\"82509\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:22a8d6e7-c244-4531-97db-dc2da138b136>\",\"WARC-Concurrent-To\":\"<urn:uuid:ea54290b-7020-4879-b6cb-da141fcfa5a8>\",\"WARC-IP-Address\":\"104.17.33.82\",\"WARC-Target-URI\":\"https://ginga.readthedocs.io/en/stable/api/ginga.canvas.CanvasObject.CanvasObjectBase.html\",\"WARC-Payload-Digest\":\"sha1:MWNCWIW4UFNJYAUHSELNHZOECZVX2LZT\",\"WARC-Block-Digest\":\"sha1:CCOTTTW4BFOGGWDB5UGPWVOLNDEGKPHJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573029.81_warc_CC-MAIN-20220817153027-20220817183027-00332.warc.gz\"}"}
https://www.thelearningpoint.net/home/mathematics/rhombus/rhombus-area-perimeter-diagonal-symmetry-62	[ "The Learning Point‎ > ‎Mathematics‎ > ‎Rhombus‎ > ‎\n\n### Area of Rhombus of side 62 and Geometric properties like symmetry,perimeter of rhombus,diagonals of rhombus", null, "Area of a rhombus with side 62 units and base angle 60 degrees = side x side x sine of baseAngle = 62 x 62 x 0.866 = 3328.904 square units Perimeter of a rhombus with side 62 units = 4 * side = 4 * 62 = 248 units Length of the two Diagonals of a rhombus with side 62 units = 2 * length of side * sin(BaseAngle/2.0) and 2 * length of side cos(BaseAngle/2.0) = 2 62 * sine of 30.0 degrees and 2 * 62 * cosine of 30.0 degrees = 62.0 and 107.39 units Height of the Rhombus = Area of Rhombus / (length of side) = 53.69 units The order of symmetry of a rhombus is 2 (number of times the shape co-incides with itself during a 360 degree rotation). The symmetry group is dihedral. And it is isotoxal or edge-transitive in nature, though these are concepts and details which some of you might encounter later. The number of axes of symmetry of a rhombus is 2: the two diagonals In general, there are multiple formulas to compute the area of rhombus. Depending on information available you may use either of them. Area of a rhombus = (square of side) x sine(base Angle) = product of diagonals/2 = base x height To understand more about the geometric features and properties of a rhombus, formulas related to mensuration etc. you might find it useful to read the properties of a Rhombus tutorial over here. Many of these concepts are a part of the Grade 9 and 10 Mathematics syllabus of the UK GCSE curriculum, Common Core Standards in the US, ICSE/CBSE syllabus in India. Many of these concepts are a part of the Grade 9 and 10 Mathematics syllabus of the UK GCSE curriculum, Common Core Standards in the US, ICSE/CBSE/SSC syllabus in Indian high schools. You may check out our free and printable worksheets for Common Core and GCSE. Geometric Properties of a rhombus of side 57 units. Geometric Properties of a rhombus of side 58 units. Geometric Properties of a rhombus of side 59 units. Geometric Properties of a rhombus of side 60 units. Geometric Properties of a rhombus of side 61 units. Geometric Properties of a rhombus of side 62 units. Geometric Properties of a rhombus of side 63 units. Geometric Properties of a rhombus of side 64 units. Geometric Properties of a rhombus of side 65 units. Geometric Properties of a rhombus of side 66 units. Geometric Properties of a rhombus of side 67 units." ]	[ null, "http://thelearningpoint.net/home/mathematics/rhombus/Rhombus.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81358206,"math_prob":0.98157036,"size":2246,"snap":"2021-31-2021-39","text_gpt3_token_len":610,"char_repetition_ratio":0.23505798,"word_repetition_ratio":0.25365853,"special_character_ratio":0.2600178,"punctuation_ratio":0.08520179,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9953114,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-05T10:51:31Z\",\"WARC-Record-ID\":\"<urn:uuid:a207a508-6b30-435e-8947-835c95a33b34>\",\"Content-Length\":\"104937\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:89bfb874-78d1-4a5c-ba27-56c09acbb3a6>\",\"WARC-Concurrent-To\":\"<urn:uuid:c8c5b2ae-f6d4-41d8-82c1-8bab3f81d67e>\",\"WARC-IP-Address\":\"104.21.91.36\",\"WARC-Target-URI\":\"https://www.thelearningpoint.net/home/mathematics/rhombus/rhombus-area-perimeter-diagonal-symmetry-62\",\"WARC-Payload-Digest\":\"sha1:DMHFDRJF6HCSPRMHHVZB3T4NHLL2QPG6\",\"WARC-Block-Digest\":\"sha1:EZLVW4YUQYT5Z7ZUI6AQ3HK4HFVF7US5\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046155529.97_warc_CC-MAIN-20210805095314-20210805125314-00224.warc.gz\"}"}
https://pubs2.ttiedu.com/course_outline?tid=4	[ "# 108 Mechanical and Structural Theory\n\n Purchase Options Printed Material, Expedited Shipping 108 \\$180.00 Printed Material, Standard Shipping 108 \\$165.00 Online Text Material 108 \\$100.00\n\nFor whom intended This course is intended for individuals whose primary formal training is not in the field of mechanical or structural engineering. Mechanical and structural considerations are fundamental to almost every technical activity, and all technical personnel have to deal, at least to some extent, with some aspects of mechanical engineering. A basic understanding of mechanical principles is essential to better perform their main function.", null, "Objectives To help participants to understand basic mechanical and structural concepts and terminology. It is not an in-depth mechanical engineering course but rather a course aimed at individuals who require an intensive review of basic principals, without the assumption of any prior knowledge of the topic. The course is fast paced and as non-mathematical as possible.\n\nBrief Course Description The course covers basic concepts of mechanical theory, starting with basic mathematics and conversion factors. To fully comprehend sinusoidal and non-sinusoidal waveforms, a basic understanding of complex algebra is required. The instructor reviews this topic as it applies to mechanical technology.\n\nThe instructor next introduces the basics of mechanical and structural theory, such as measurement of mass, displacement, acceleration and velocity, before moving into somewhat greater depth on dynamics theory. Single and multiple degree-of-freedom systems are considered, in regard to spring stiffness, dynamic properties of different materials, natural frequency and damping.\n\nThe Rayleigh and Dunkerley methods of calculating the first natural frequency of systems are briefly considered, with examples. Forced vibration and loading effects are also included in the dynamics theory section.\n\nMoving on to structural design fundamentals, the instructor addresses the concepts of stress and strain; moment of inertia and the torsional shape factor. Useful formulas are provided for calculating stiffness and stress, also tables for determining moments of inertia and torsional shape factors. The instructor discusses the dynamic characteristics of structural elements such as compression members, flanges and beams. Finally, the course provides useful tables and formulas for the calculation of beam stiffness and resonant frequency, as well as resonances of plates and columns.\n\nRelated Courses TTi Course 310, Mechanical Design for Product Reliability, which is available as an OnDemand Complete Internet Course, contains all the theory contained in this course, plus advanced examples and exercises in applying the theory.\n\nDiploma Programs This may be used as an optional course for any Specialist Diploma Program.\n\nPrerequisites There are no definite prerequisites. However, this course is aimed toward individuals involved in various technical fields. An understanding of basic algebra will be useful.\n\nText Each student will receive 180 days access to the on-line electronic course workbook. Renewals and printed textbooks are available for an additional fee\n\nCourse Hours, Certificate and CEUs Class hours/days for on-site courses can vary from 14-35 hours over 2-5 days as requested by our clients. Upon successful course completion, each participant receives a certificate of completion and one Continuing Education Unit (CEU) for every ten class hours.\n\nClick for a printable course outline (pdf).\n\n## Course Outline\n\n### Chapter 1 - Review of Mathematics\n\n Purchase Options OnDemand Short Topics \\$25.00\n• Math Reference\n• Fractions, Powers, Exponents and Roots\n• Mathematical symbols, Conversions\n• Calculator exponential notation\n• Useful Constants, Trigonometric ratios\n• Scientific and Engineering Notation\n• Greek Letter Symbols\n• Algebra\n• Constants and Variables, Equations, Terms\n• Proportionality\n• Exponential, Linear and Quadratic Equations\n• Graphs\n• Geometry and Trigonometry\n• Arcs and Circles, Radians, Angles, Chords\n• Vectors\n• Angular Frequency, Phase Difference, Critical Functions\n• Linear Coordinates of a Rotating Vector\n• Complex Algebra\n• Rectangular and Polar Coordinates, Vector in Complex Plane\n• Complex Number operations (Addition, Subtraction, Multiplication and Division)\n• Calculus\n• Differentiation, Curve Drawing in Differential Calculus\n• Integration; Calculating the area under a curve\n• Differential Equations\n\n### Chapter 2 - Introduction to Vibration\n\n Purchase Options OnDemand Short Topics \\$25.00\n• Design and Testing for Vibration and Shock\n• Rotational Unbalance Example—Automobile engine\n• Vibration and Shock Examples\n• Natural Frequency (Resonance)\n• Forcing Frequency\n• Prolonged Excitation of Natural Frequency\n• Tacoma Narrows Bridge\n• Natural and Resonant Frequencies\n• Effects of Shock and Vibration\n• Dynamic Inputs\n• Fragility\n• Effect of Failures\n\n### Chapter 3 - Introduction to Mechanical Terms and Material Properties\n\n Purchase Options OnDemand Short Topics \\$25.00\n• Laws of Motion\n• Weight, Mass and Gravity\n• Units\n• Specific Weight and Density\n• Definitions of Common Mechanical Terms\n• Friction and Wear\n• Work, Power\n• Energy\n• Engineering Materials: Metals\n• Stress and Strain\n• Definition of Stress\n• Shear Stress and Tensile Stress\n• Examples of Stress:\n• Simple Tension or Compression\n• Pure Shear\n• Strain\n• Shear Strain\n• Elasticity—Definitions and Laws\n• Stress-Strain Relationship\n• Tensile Strength\n• Non-linear Elastic and Anelastic Solid Material\n• Non-Elastic (or Plastic) Load-Extension (F/u) Curves\n• Torque (T)\n• Tangential Acceleration (at )\n• The Mass Moment of Inertia (IM)\n• Mass Moments of Inertia\n• The Area Moment of Inertia (IA)\n• Relative Stiffness (k) of Structural Members\n• Shearing Torques/Twisting Torques\n• Torsional Stiffness of an Open Cross-Section\n\n### Chapter 4 - Application of Vibration Theory\n\n Purchase Options OnDemand Short Topics \\$25.00\n• Principles of Analysis\n• Fundamentals of Dynamics\n• Stiffness\n• Mass\n• A Simple Dynamic System\n• Degrees of Freedom\n• Single-Degree-of-Freedom (SDoF)\n• Undamped Vibrations—Single Degree of Freedom Systems\n• Sinusoidal Waveform\n• SDoF—Sinusoidal Relationships\n• Relationship between Displacement, Velocity and Acceleration\n• Sinusoidal Relationships\n• Undamped Vibrations in SDoF Systems\n• Natural Frequency\n• Decaying Sinusoidal Vibration\n• Undamped MDoF System Vibration\n• Dimensionless Ratio Graph for 2DoF Systems\n• Complex Systems\n• Spring Stiffness, in Parallel and in Series\n• Multi-Degree-of-Freedom (MDoF) Modeling\n• Rayleigh’s Method\n• Dunkerley’s Method\n• Forced Vibration for SDoF System\n• Transmissibility\n• Plotting Permissibility vs. Frequency Ratio\n• Isolation\n\n### Chapter 5 - Materials and Beams\n\n Purchase Options OnDemand Short Topics \\$25.00\n• Material Selection in Engineering Design\n• Overall Material Properties\n• Design-Limiting Material Properties\n• Deriving Application-Specific Material Properties\n• Properties of Materials\n• Dynamic Response\n• Simple Beam\n• Bending Moment\n• Elastic Deflection of a Simply Supported Beam (with center load)\n• Sandwich Structures\n• Bending Strengths\n• Structural Beams\n• Bending Stiffness of a Beam Kb\n• Finding Stiffness of a Composite Beam\n• Beam Instability—Twisting\n• Compression Member Instability\n• Instability of Flanges\n• Flange Buckling\n• Structure Buckling\n• Resonant Frequencies of Flanges\n\n### Chapter 6 - Frequency and Stiffness Considerations\n\n Purchase Options OnDemand Short Topics \\$25.00\n• Designing for Stiffness vs. Strength\n• Frequency Oscillation of a Rod\n• Natural Frequency of a Simply Supported Beam\n• Natural Frequency of a Cantilever\n• Effective Mass\n• Effective Mass of a Beam: Example\n• Natural Frequency of Simply Supported Plate\n• Beam Formulas\n• Stiffness of Gussets—with End load\n• Effective Stiffness of Gusset\n• Beam Formulas\n• Plate Frequency Equation\n• Plate Frequency Parameters\n• Column Resonance\n• Axial Resonance\n• Example: Determining Stress in a Loaded Beam\n\n### Appendix C - Dynamic Force and Motion\n\n• Weight, Specific Weight and Density\n• Relative Density or Specific Gravity\n• Common Units of Force—Comparison\n• SDoF — Sinusoidal Relationships\n• Calculating Peak X, V and A\n• Undamped Vibrations in Single Degree of Freedom Systems\n• Calculating Natural Frequency\n• Calculating Stiffness\n• Example—Damped Resonant System\n\n### Award of Certificates for sucessful completion\n\nClick for a printable course outline (pdf).\n\nRevised 6/7/2018" ]	[ null, "https://pubs2.ttiedu.com/sites/default/files/108.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84041995,"math_prob":0.6686922,"size":8224,"snap":"2022-27-2022-33","text_gpt3_token_len":1788,"char_repetition_ratio":0.118856445,"word_repetition_ratio":0.011484823,"special_character_ratio":0.18871595,"punctuation_ratio":0.069630876,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9687934,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-27T17:45:11Z\",\"WARC-Record-ID\":\"<urn:uuid:1e570c52-9a01-4485-9c55-f6cb727bd2c6>\",\"Content-Length\":\"23594\",\"Content-Type\":\"application/http; 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https://askainow.com/challenge/preview/cpid/884	[ "#### Print the size of the largest subset of distinct integers such that no two...\n\nGiven a set of distinct integers, this function will print the size of the largest subset such that no two integers in the subset sum to more than k. For example, given the input [ 1, 2, 3, 4, 5] and k = 6, the output would be 3.\n\n#### Problem\n\n```Given a set of distinct integers, print the size of the largest subset such that no two integers in the subset sum to more than k.\nExample input:\nnums = [1, 2, 3, 4, 5], k = 6\nExample output:\n3```\n\n#### Solution\n\nThis solution is in Python. Our A.I. can create solutions in multiple languages.\n```# Solution:\n# The optimal solution is to sort the array and then iterate through the array, keeping track of the largest subset.\n# The reason this is optimal is because the time complexity is O(n log n) because of the sorting, and O(n) for the iteration.\n# The space complexity is O(1) because we are not storing any data.\n\ndef largest_subset(nums, k):\nnums.sort()\nlargest_subset = 0\nfor i in range(len(nums)):\nif nums[i] > k:\nbreak\nlargest_subset += 1\nreturn largest_subset\n\nprint(largest_subset([1, 2, 3, 4, 5], 6))```\n\nA.I. Evaluation of the Solution\n\nThe candidate's solution is correct and demonstrates a level of completeness. The approach is sound and the time and space complexity are both optimal.\n\nEvaluated at: 2022-11-19 00:16:56" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87162566,"math_prob":0.98614424,"size":415,"snap":"2023-40-2023-50","text_gpt3_token_len":110,"char_repetition_ratio":0.10705596,"word_repetition_ratio":0.0,"special_character_ratio":0.2819277,"punctuation_ratio":0.15053764,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9986114,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-05T03:38:37Z\",\"WARC-Record-ID\":\"<urn:uuid:c8c2f887-0aa8-4049-b414-2f2a16b383c1>\",\"Content-Length\":\"10465\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e5e87ce4-9018-4854-9c4d-75775bd85c52>\",\"WARC-Concurrent-To\":\"<urn:uuid:c711c37b-d6ce-4e4d-99a9-bf5536bf4953>\",\"WARC-IP-Address\":\"20.118.56.3\",\"WARC-Target-URI\":\"https://askainow.com/challenge/preview/cpid/884\",\"WARC-Payload-Digest\":\"sha1:ATZDY3TFFOYEH2EAUIUMIRHAROS3LS3D\",\"WARC-Block-Digest\":\"sha1:LVG4UUBO624SXF557VVOWKRBAECH33LU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511717.69_warc_CC-MAIN-20231005012006-20231005042006-00739.warc.gz\"}"}
https://www.jobilize.com/physics-ap/course/28-3-length-contraction-special-relativity-by-openstax?qcr=www.quizover.com	[ "# 28.3 Length contraction\n\n Page 1 / 5\n• Describe proper length.\n• Calculate length contraction.\n• Explain why we don’t notice these effects at everyday scales.", null, "People might describe distances differently, but at relativistic speeds, the distances really are different. (credit: Corey Leopold, Flickr)\n\nHave you ever driven on a road that seems like it goes on forever? If you look ahead, you might say you have about 10 km left to go. Another traveler might say the road ahead looks like it’s about 15 km long. If you both measured the road, however, you would agree. Traveling at everyday speeds, the distance you both measure would be the same. You will read in this section, however, that this is not true at relativistic speeds. Close to the speed of light, distances measured are not the same when measured by different observers.\n\n## Proper length\n\nOne thing all observers agree upon is relative speed. Even though clocks measure different elapsed times for the same process, they still agree that relative speed, which is distance divided by elapsed time, is the same. This implies that distance, too, depends on the observer’s relative motion. If two observers see different times, then they must also see different distances for relative speed to be the same to each of them.\n\nThe muon discussed in [link] illustrates this concept. To an observer on the Earth, the muon travels at $0.950c$ for $7.05\\phantom{\\rule{0.25em}{0ex}}\\mu s$ from the time it is produced until it decays. Thus it travels a distance\n\n${L}_{0}=v\\Delta t=\\left(0.950\\right)\\left(3.00×{\\text{10}}^{8}\\phantom{\\rule{0.25em}{0ex}}\\text{m/s}\\right)\\left(7.05×{\\text{10}}^{-6}\\phantom{\\rule{0.25em}{0ex}}\\text{s}\\right)=2.01\\phantom{\\rule{0.25em}{0ex}}\\text{km}$\n\nrelative to the Earth. In the muon’s frame of reference, its lifetime is only $2.20\\phantom{\\rule{0.25em}{0ex}}\\mu s$ . It has enough time to travel only\n\n$L=v\\Delta {t}_{0}=\\left(0\\text{.}\\text{950}\\right)\\left(3\\text{.}\\text{00}×{\\text{10}}^{8}\\phantom{\\rule{0.25em}{0ex}}\\text{m/s}\\right)\\left(2\\text{.}\\text{20}×{\\text{10}}^{-6}\\phantom{\\rule{0.25em}{0ex}}\\text{s}\\right)=0\\text{.627 km}.$\n\nThe distance between the same two events (production and decay of a muon) depends on who measures it and how they are moving relative to it.\n\n## Proper length\n\nProper length ${L}_{0}$ is the distance between two points measured by an observer who is at rest relative to both of the points.\n\nThe Earth-bound observer measures the proper length ${L}_{0}$ , because the points at which the muon is produced and decays are stationary relative to the Earth. To the muon, the Earth, air, and clouds are moving, and so the distance $L$ it sees is not the proper length.", null, "(a) The Earth-bound observer sees the muon travel 2.01 km between clouds. (b) The muon sees itself travel the same path, but only a distance of 0.627 km. The Earth, air, and clouds are moving relative to the muon in its frame, and all appear to have smaller lengths along the direction of travel.\n\n## Length contraction\n\nTo develop an equation relating distances measured by different observers, we note that the velocity relative to the Earth-bound observer in our muon example is given by\n\n$v=\\frac{{L}_{0}}{\\Delta t}.$\n\nThe time relative to the Earth-bound observer is $\\Delta t$ , since the object being timed is moving relative to this observer. The velocity relative to the moving observer is given by\n\n$v=\\frac{L}{\\Delta {t}_{0}}.$\n\nThe moving observer travels with the muon and therefore observes the proper time $\\Delta {t}_{0}$ . The two velocities are identical; thus,\n\n$\\frac{{L}_{0}}{\\Delta t}=\\frac{L}{\\Delta {t}_{0}}.$\n\nWe know that $\\Delta t=\\gamma \\Delta {t}_{0}$ . Substituting this equation into the relationship above gives\n\nthe meaning of phrase in physics\nis the meaning of phrase in physics\nChovwe\nwrite an expression for a plane progressive wave moving from left to right along x axis and having amplitude 0.02m, frequency of 650Hz and speed if 680ms-¹\nhow does a model differ from a theory\nwhat is vector quantity\nVector quality have both direction and magnitude, such as Force, displacement, acceleration and etc.\nBesmellah\nIs the force attractive or repulsive between the hot and neutral lines hung from power poles? Why?\nwhat's electromagnetic induction\nelectromagnetic induction is a process in which conductor is put in a particular position and magnetic field keeps varying.\nLukman\nwow great\nSalaudeen\nwhat is mutual induction?\nje\nmutual induction can be define as the current flowing in one coil that induces a voltage in an adjacent coil.\nJohnson\nhow to undergo polarization\nshow that a particle moving under the influence of an attractive force mu/y³ towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v²k² and distance uk²/√u-vk as origin\nshow that a particle moving under the influence of an attractive force mu/y^3 towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v^2k^2 and distance uk^2/√u-k as origin\nNo idea.... Are you even sure this question exist?\nMavis\nI can't even understand the question\nyes it was an assignment question \"^\"represent raise to power pls\nGabriel\nGabriel\nAn engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10kg . Pendulum 2 has a bob with a mass of 100 kg . Describe how the motion of the pendula will differ if the bobs are both displaced by 12º .\nno ideas\nAugstine\nif u at an angle of 12 degrees their period will be same so as their velocity, that means they both move simultaneously since both both hovers at same length meaning they have the same length\nModern cars are made of materials that make them collapsible upon collision. Explain using physics concept (Force and impulse), how these car designs help with the safety of passengers.\ncalculate the force due to surface tension required to support a column liquid in a capillary tube 5mm. If the capillary tube is dipped into a beaker of water\nfind the time required for a train Half a Kilometre long to cross a bridge almost kilometre long racing at 100km/h\nmethod of polarization\nAjayi\nWhat is atomic number?\nThe number of protons in the nucleus of an atom\nDeborah\ntype of thermodynamics\noxygen gas contained in a ccylinder of volume has a temp of 300k and pressure 2.5×10Nm\n\n#### Get Jobilize Job Search Mobile App in your pocket Now!", null, "By", null, "By OpenStax", null, "By Brenna Fike", null, "By Tamsin Knox", null, "By OpenStax", null, "By Courntey Hub", null, "By Madison Christian", null, "By Kimberly Nichols", null, "By Madison Christian", null, "By Richley Crapo", null, "By Abby Sharp" ]	[ null, "https://www.jobilize.com/ocw/mirror/col11406/m42535/Figure_29_03_01a.jpg", null, "https://www.jobilize.com/ocw/mirror/col11406/m42535/Figure_29_03_02a.jpg", null, "https://www.jobilize.com/ocw/mirror/course-thumbs/col11406-course-thumb.png;jsessionid=ztdK73uCzVkLs19oBvY715RHJUVSgRmYjciRgpEY.web04", null, "https://www.jobilize.com/quiz/thumb/biology-ch-01-the-study-of-life-mcq-quiz-openstax-college.png;jsessionid=ztdK73uCzVkLs19oBvY715RHJUVSgRmYjciRgpEY.web04", null, "https://www.jobilize.com/quiz/thumb/social-media-quiz-by-brenna-fike.png;jsessionid=ztdK73uCzVkLs19oBvY715RHJUVSgRmYjciRgpEY.web04", null, "https://www.jobilize.com/quiz/thumb/pathophysiology-mcq-questions-1064704.png;jsessionid=ztdK73uCzVkLs19oBvY715RHJUVSgRmYjciRgpEY.web04", null, "https://www.jobilize.com/quiz/thumb/human-body-anatomy-physiology-essay-quiz.png;jsessionid=ztdK73uCzVkLs19oBvY715RHJUVSgRmYjciRgpEY.web04", null, "https://farm9.staticflickr.com/8796/17127874488_9ce17b2f7d_t.jpg", null, "https://farm9.staticflickr.com/8621/16637604532_eb208bd3d3_t.jpg", null, "https://www.jobilize.com/quiz/thumb/neuropsychology-midterm-test-by-kimberly-nichols.png;jsessionid=ztdK73uCzVkLs19oBvY715RHJUVSgRmYjciRgpEY.web04", null, "https://www.jobilize.com/quiz/thumb/microbiology-chapter-10-11-unit-3-test-3-by-madison.png;jsessionid=ztdK73uCzVkLs19oBvY715RHJUVSgRmYjciRgpEY.web04", null, "https://www.jobilize.com/quiz/thumb/cultural-anthropology-definition-by-prof-richley-crapo-utah-usu.png;jsessionid=ztdK73uCzVkLs19oBvY715RHJUVSgRmYjciRgpEY.web04", null, "https://farm4.staticflickr.com/3839/14360304974_f9cdb740f9_t.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9487487,"math_prob":0.9879754,"size":2507,"snap":"2021-21-2021-25","text_gpt3_token_len":528,"char_repetition_ratio":0.16580103,"word_repetition_ratio":0.020361992,"special_character_ratio":0.21180694,"punctuation_ratio":0.0985325,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9931231,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26],"im_url_duplicate_count":[null,5,null,5,null,1,null,1,null,1,null,1,null,1,null,null,null,null,null,1,null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-11T17:29:25Z\",\"WARC-Record-ID\":\"<urn:uuid:c6aad207-3d88-4daa-8b7b-c6db5abc95f9>\",\"Content-Length\":\"117609\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:80f80aa4-c1a6-447c-9318-6c08997d58dc>\",\"WARC-Concurrent-To\":\"<urn:uuid:ef0e3e43-7927-4104-83bd-ca1f50b24e83>\",\"WARC-IP-Address\":\"207.38.87.179\",\"WARC-Target-URI\":\"https://www.jobilize.com/physics-ap/course/28-3-length-contraction-special-relativity-by-openstax?qcr=www.quizover.com\",\"WARC-Payload-Digest\":\"sha1:DABS36H2SPTSFJRHXR4KXOI54SSJ5NCP\",\"WARC-Block-Digest\":\"sha1:GIASSZ7YPHWZDQ22US5K7SPBF656NFZS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991648.10_warc_CC-MAIN-20210511153555-20210511183555-00579.warc.gz\"}"}
https://numberworld.info/1349006080	[ "Number 1349006080\n\nProperties of number 1349006080\n\nCross Sum:\nFactorization:\n2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 5 * 19 * 55469\nDivisors:\nCount of divisors:\nSum of divisors:\nPrime number?\nNo\nFibonacci number?\nNo\nBell Number?\nNo\nCatalan Number?\nNo\nBase 3 (Ternary):\nBase 4 (Quaternary):\nBase 5 (Quintal):\nBase 8 (Octal):\nBase 32:\n186gco0\nsin(1349006080)\n-0.6226854914944\ncos(1349006080)\n0.7824722223072\ntan(1349006080)\n-0.79579245593965\nln(1349006080)\n21.022633921204\nlg(1349006080)\n9.1300139070511\nsqrt(1349006080)\n36728.818113302\nSquare(1349006080)\n\nNumber Look Up\n\nLook Up\n\n1349006080 (one billion three hundred forty-nine million six thousand eighty) is a very special figure. The cross sum of 1349006080 is 31. If you factorisate 1349006080 you will get these result 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 5 * 19 * 55469. The number 1349006080 has 72 divisors ( 1, 2, 4, 5, 8, 10, 16, 19, 20, 32, 38, 40, 64, 76, 80, 95, 128, 152, 160, 190, 256, 304, 320, 380, 608, 640, 760, 1216, 1280, 1520, 2432, 3040, 4864, 6080, 12160, 24320, 55469, 110938, 221876, 277345, 443752, 554690, 887504, 1053911, 1109380, 1775008, 2107822, 2218760, 3550016, 4215644, 4437520, 5269555, 7100032, 8431288, 8875040, 10539110, 14200064, 16862576, 17750080, 21078220, 33725152, 35500160, 42156440, 67450304, 71000320, 84312880, 134900608, 168625760, 269801216, 337251520, 674503040, 1349006080 ) whith a sum of 3401420400. 1349006080 is not a prime number. The number 1349006080 is not a fibonacci number. The figure 1349006080 is not a Bell Number. The figure 1349006080 is not a Catalan Number. The convertion of 1349006080 to base 2 (Binary) is 1010000011010000011001100000000. The convertion of 1349006080 to base 3 (Ternary) is 10111000101121110011. The convertion of 1349006080 to base 4 (Quaternary) is 1100122003030000. The convertion of 1349006080 to base 5 (Quintal) is 10230321143310. The convertion of 1349006080 to base 8 (Octal) is 12032031400. The convertion of 1349006080 to base 16 (Hexadecimal) is 50683300. The convertion of 1349006080 to base 32 is 186gco0. The sine of 1349006080 is -0.6226854914944. The cosine of 1349006080 is 0.7824722223072. The tangent of the figure 1349006080 is -0.79579245593965. The root of 1349006080 is 36728.818113302.\nIf you square 1349006080 you will get the following result 1819817403876966400. The natural logarithm of 1349006080 is 21.022633921204 and the decimal logarithm is 9.1300139070511. You should now know that 1349006080 is very impressive number!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6135349,"math_prob":0.9798265,"size":3278,"snap":"2022-05-2022-21","text_gpt3_token_len":1357,"char_repetition_ratio":0.16921197,"word_repetition_ratio":0.48842105,"special_character_ratio":0.6705308,"punctuation_ratio":0.28654972,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99317074,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-20T17:58:26Z\",\"WARC-Record-ID\":\"<urn:uuid:be3f7981-502f-493d-a0b9-5fd54b44d213>\",\"Content-Length\":\"17937\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:df11d7a1-edb6-47c5-ac87-46f5660a07bd>\",\"WARC-Concurrent-To\":\"<urn:uuid:a3790b38-2db2-4142-b197-5bb17a593327>\",\"WARC-IP-Address\":\"176.9.140.13\",\"WARC-Target-URI\":\"https://numberworld.info/1349006080\",\"WARC-Payload-Digest\":\"sha1:DW3NMN6OURIUBXW65AVT746L7PWRBTN6\",\"WARC-Block-Digest\":\"sha1:KWYRZXJA4RZZBXN3HW2JWQJYY65MSI5L\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320302355.97_warc_CC-MAIN-20220120160411-20220120190411-00293.warc.gz\"}"}
http://waytoohuman.com/laws-of-exponents-worksheet/	[ "", null, "Worksheets\n\nLaws Of Exponents Worksheet\n\nMixed exponent rules all positive a math worksheet freemath more versions httpwww drills comalgebra phpexponentrul. Algebra 1 unit 7 exponent rules worksheet 2 simplify each math each. Worksheets laws of exponents cricmag free worksheet thedanks for everyone worksheet. Worksheets laws of exponents cricmag free worksheet thedanks for everyone exponent all download. Exponents worksheets.", null, "Mixed exponent rules all positive a math worksheet freemath more versions httpwww drills comalgebra phpexponentrul", null, "Algebra 1 unit 7 exponent rules worksheet 2 simplify each math each", null, "Worksheets laws of exponents cricmag free worksheet thedanks for everyone worksheet", null, "Worksheets laws of exponents cricmag free worksheet thedanks for everyone exponent all download", null, "Exponents worksheets", null, "Worksheets laws of exponents cricmag free worksheet thedanks for everyone exponent all download and share worksheets", null, "Free exponents worksheets addsubtractmultiplydivide powers bases are both positive and negative integers", null, "Laws of exponent worksheets for all download and share free on bonlacfoods com", null, "Endearing pre algebra worksheets negative exponents for worksheet and multiplication mytourvn", null, "Multiplying exponents all positive a the math worksheet page 2", null, "Related Posts\n\nGrade 8 Math Worksheets", null, "" ]	[ null, "https://i.pinimg.com/originals/a0/d9/08/a0d908005948a05a996a9a0b4a79df7b.png", null, "https://i.pinimg.com/originals/14/30/af/1430af98a6592020745e233ace0b290a.jpg", null, "https://i.pinimg.com/originals/a0/d9/08/a0d908005948a05a996a9a0b4a79df7b.png", null, "http://bonlacfoods.com/images/laws-of-exponent-worksheets/laws-of-exponent-worksheets-19.jpg", null, "http://bonlacfoods.com/images/practice-worksheet-for-law-of-exponents/practice-worksheet-for-law-of-exponents-3.jpg", null, "https://www.dadsworksheets.com/worksheets/exponents/simple-exponents-with-hints-v1.jpg", null, "https://homeshealth.info/wp-content/uploads/2018/02/interesting-fun-worksheets-for-exponents-about-exponents-rules-worksheet-free-worksheets-library-of-fun-worksheets-for-exponents.jpg", null, "https://www.homeschoolmath.net/worksheets/examples/exponents_worksheet_9.gif", null, "http://bonlacfoods.com/images/laws-of-exponent-worksheets/laws-of-exponent-worksheets-2.jpg", null, "https://homeshealth.info/wp-content/uploads/2018/02/endearing-pre-algebra-worksheets-negative-exponents-for-worksheet-exponents-and-multiplication-worksheet-mytourvn-of-pre-algebra-worksheets-negative", null, "https://www.math-drills.com/algebra/images/algebra_exponent_rules_basic_multiplying_samebase_positive_001_pin2.jpg", null, "http://gigidiaries.com/wp-content/uploads/2018/01/grade-math-worksheets-laws-algebraic-free-for-8th-library-integersble.jpg", null, "http://www.2ndgradeworksheets.net/ccss2oa1/ccss2oa1wordproblems1d1.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.62346613,"math_prob":0.6999392,"size":1380,"snap":"2019-43-2019-47","text_gpt3_token_len":286,"char_repetition_ratio":0.26380813,"word_repetition_ratio":0.38297874,"special_character_ratio":0.15144928,"punctuation_ratio":0.025773196,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99556774,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26],"im_url_duplicate_count":[null,null,null,null,null,null,null,4,null,1,null,null,null,3,null,null,null,1,null,3,null,10,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-20T23:46:49Z\",\"WARC-Record-ID\":\"<urn:uuid:a1f20198-7e21-4eda-b3ca-2fc8673eef17>\",\"Content-Length\":\"18799\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c2065b56-2501-47ca-bb42-ac55f82873e4>\",\"WARC-Concurrent-To\":\"<urn:uuid:1a5ace38-073c-468e-81fc-e598ea951670>\",\"WARC-IP-Address\":\"104.27.143.115\",\"WARC-Target-URI\":\"http://waytoohuman.com/laws-of-exponents-worksheet/\",\"WARC-Payload-Digest\":\"sha1:DCOKITIXK7AQFSL6N2SCQWPNGGUTKALI\",\"WARC-Block-Digest\":\"sha1:WMXS3F2AN5KEWOVWDNQIYVKQPBUER4XE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987750110.78_warc_CC-MAIN-20191020233245-20191021020745-00441.warc.gz\"}"}
https://www.mentorforbankexams.com/2017/01/quantitative-aptitude-notes-averages.html	[ "## Quantitative Aptitude Notes - Averages\n\n### Quantitative Aptitude Notes - Averages\n\nWhat is Average?\nThe result obtained by adding several quantities together and then dividing this total by the number of quantities is called Average. The main term of average is equal sharing of a value among all, where it may share persons or things. We obtain the average of a number using formula that is sum of observations divide by Number of observations.\nAverage = Sum of Quantities/Number of quantities =>S = A × N\nImportant Points to Remember:\na) If all the numbers are increased by 'a’, then their average will also be increased by 'a’.\nb) If all the numbers are decreased by ‘a’, then their average will also be decreased by 'a’.\nc) If all the numbers are multiplied by 'a’, then their average will also be multiplied by 'a'.\nd) If all the numbers are divided by 'a', then their average will also be divided by 'a’.\nImportant Formulae on Averages:", null, "", null, "", null, "### Memory Based Example Problems base on various types\n\nType 1: Relationship between Averages and Numbers\n1. Find the average of the following set of scores 216, 463, 154, 605,446, 336\nSolution:\nS = 216 + 463 + 154 + 605 + 446 + 336 = 2220; N = 6\nA = S/N = 2220/6 = 370\n2. The average of four consecutive even numbers A, B, C and D respectively is 55. What is the product of A and C?\nSolution:\nThe average of A, B, C and D = Average of B and C\nBut B and C are consecutive even numbers. Their average will be equal to the odd number in between them (Which is 55)\n=> B = 55 – 1 = 54; C = 55 + 1 = 56\n=> A = B – 2 = 52\n=> A × C = 52 × 56 = 2912\n3. Average of four consecutive odd numbers is 106. What is the third number in ascending order?\nSolution:\nA, B, C and D be the four consecutive odd numbers in ascending order\nTheir average = Average of B and C = the even number between B and C = 106\n=> B = 106 – 1 = 105; C = 106 + 1 = 107\nTherefore, the third number in ascending order = C = 107\n4. The average of five positive integers is 55.8 if average of first two integers is 4 and the average of fourth and fifth integers is 69.5. Then find the third integer.\nSolution:\n55.8 × 5 = 279; 49 × 2 = 98; 69.5 × 2 = 139\nTherefore third number = 279 – 98 – 139 = 42\nType 2: Questions based on Partial Average\n1. In a college, 16 girls has the average age is 18 years and 14 boys has the average age 17 years. What would be the average age of entire college?\nSolution:\n16 girls has the average age is 18 years (16 × 18) = 288\n4 boys has the average age 17 years (14 × 17) = 238\nAverage age = 288 + 238/30 = 526/30 = 17.54\n2. The average salary of 25 employee in a company per month is Rs.6000. If the manager’s salary also added then the average is increases by Rs.500. What would be the salary of Manager?\nSolution:\nAverage salary of employee in company = 6000\nWhen added one member salary 25 + 1 = 26 = 6500\nSo, (26 x 6500) – (25 x 6000) = 169000 – 150000 = 19000\n3. The average wages of a worker during a fortnight comprising 15 consecutive working days was Rs.90 per day. During the first 7 days, his average wages was Rs.87/day and the average wages during the last 7 days was Rs.92 /day. What was his wage on the 8th day?\nSolution:\nThe total wages earned during the 15 days that the worker worked =15×90= Rs.1350\nThe total wages earned during the first 7 days = 7×87 = Rs. 609\nThe total wages earned during the last 7 days = 7×92 = Rs. 644\nTotal wages earned during the 15 days = wages during first 7 days + wage on 8thday + wages during the last 7 days.\n1350=609+1350=609+ wage on 8th day +644\nWage on 8thday = 1350 – 609 – 644 = Rs. 97\n4. 40% of the employees in a factory are workers. All the remaining employees are executives. The annual income of each worker is Rs. 390. The annual income of each executive is Rs. 420. What is the average annual income of all the employees in the factory together?\nSolution:\nLet the number of employees be x.\n40% of employees are workers = 2x/5; Number of executives = 3x/5\nThe annual income of each worker is Rs. 390.\nHence, the total annual income of all the workers together = 2x/5 × 390 = 156x\nAlso, the annual income of each executive is Rs. 420.\nHence, the total income of all the executives together = 3x/5 × 420 = 252x\nHence, the total income of the employees = 156x + 252x = 408x\nThe average income of all the employees together equals = 408x/x = 408\n5. The average annual income of Ramesh and Suresh is Rs. 3800. The average annual income of Suresh and Pratap was Rs. 4800, and the average annual incomes of Pratap and Ramesh was Rs. 5800. What is the average of the incomes of the three?\nSolution:\nAverage of Ramesh and Suresh (R + S)/2 = 3800 => Total income (R + S) = 3800 x 2 = 7600\nAverage of Surech and Pratap (S +P)/2= 4800 => Total (S + P) = 4800 x 2 = 9600\nAverage of Pratap and Ramesh (P + R)/2 = 5800 => Total (P + R) = 5800 x 2 = 11600\nTotal of three (2R + 2P + 2S) = 7600 + 9600 + 11600 = 28800 => R + P + S = 14400\nTherefore average = 14400/3 = 4800\n6. How many students are there in the class who has an average age of 15?\nI. The ratio between the boys and girls is 4:3, total ages of whole class is 420.\nII. The ratio between the average age of girls and boys is 13:15.\nSolution:\nFrom statement I, it is not cleared about the total strength of students.\nFrom statement II, only the ratio between the ages of boys and girls is given, which is inadequate.\nOn combining we get that total number of student is multiple of 7 but we do not get the total number, so both the statement I and II are not sufficient to answer the question.\n7. On a school’s Annual day sweets were to be distributed amongst 112 children. But on that particular day, 32 children were absent. Thus, the remaining children got extra 6 sweets. How many sweets did each child originally supposed to get?\nSolution:\nChildren attended the function = 112 – 32 = 80\nTotal extra sweets = 80 * 6 =480 due to absence of children.\nOriginal share of each child = 480 / 32 = 15.\nAlternative Method:\n112x = ( 112-32) * ( x+6) => 32x = 680 => x= 6 80/ 32 = 15\n8. Arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of the remaining 55% is:\nSolution:\nRemaining students = 100% - 20% - 25% = 55%\nLet remaining students got a mean score of x marks.\nThen, 20% of 80 + 25% of 31+ 55% of x = 52 => 16+ 7.75 + 55% of x = 52\nTherefore, 55% of x = 52 – 16 – 7.75 = 28.25 => x = 2825/100 * 100/55 = 565/11 = 51.4% (approx.)\nType 3: Questions based on Replacement/ without Replacement\n1. When a student weighing 45 kg left a class, the average weight of the remaining 59 students increased by 200 g. What is the average weight of the remaining 59 students?\nSolution:\nLet the average weight of the 59 students be A.\nTherefore, the total weight of the 59 of them will be 59 A.\nThe questions states that when the weight of this student who left is added, the total weight of the class= 59A+ 45\nWhen this student is also included, the average weight decreases by 0.2 kg.\n59A+45/60=A−0.259A+45=60A−1245+12=60A−59AA=57\n2. There were 35students in a hostel. Due to the admission of 7 new students the expenses of the mess were increased by Rs.42 per day while the average expenditure per head diminished by Re.1. What was the original expenditure of the mess?\nSolution:\nLet the original average expenditure be Rs.x.\nThen 42(x – 1) – 35x = 42 => 7x = 84 => x = 12\nTherefore original expenditure = Rs.(35 x 12) = Rs.420\n3. The average age of 40 students of a class is 18 years. When 20 new students are admitted to the same class, the average age of the students of the class is increased by six months. The average age of newly admitted students is:\nSolution:\nTotal age of 40 students = 40 * 18 = 720\nTotal age of 60 students = 60 * 18.5 = 1110\nTotal age of 20 new students = 1110-720=390\nAverage age of 20 new students = 390/20 = 19 years 6 months.\nAlternative Method: Average age of 60 students = 18 years 6 months\nTotal increase in age of 40 students = 40 * 6 = 20 years\nIncrease in age of new students = 20/20= 1year\nTherefore, Avg. of new 20 students = 18 years 6 months + 1 year = 19 years 6 months.\nType 4: Questions based on Mistaken Average\n1. The average of 8 observations was 25.5. It was noticed later that two of those observations were wrongly taken. One observation was 14 more than the original value and the other observation was wrongly taken as 31 instead of 13. What will be the correct average of those 8 observations?\nSolution:\nLet correct average =x\nThen, correct total =8x\nObtained total =8×25.5=204\n204−14 − (31−13) = 8x => x=21.5\n2. The arithmetic mean of 100 numbers was computed as 89.05. It was later found that two numbers 92 and 83 have been misreads 192 and 33 respectively. What is the correct arithmetic mean of the numbers?\nSolution:\nSum of marks were wrongly increased by = (192 + 33) * (92+83) = 50\nAverage was wrongly increased by 50 / 100 = 0.5\nTherefore, correct mean = 89.05 – 0.5 = 88.55\n3. In an examination, the average marks of all the students calculated to be 58 marks. It was later found that marks of 60 students were wrongly written as 70 instead of 50. If the corrected average is 55, find the total no. Of students who took the exam.\nSolution:\nTotal decrease in marks = 60 * (70-50) = 1200\nDecrease in average = 58 – 55 = 3.\nTherefore, No. of students = 1200 / 3 = 400\nType 5: Questions based on Cricket\n1. A cricketer has completed 10 innings and his average is 21.5 runs. How many runs must be make in his next innings so as to rise is average to 24?\nSolution:\nTotal of 10 innings = 21.5 x 10 = 215\nSuppose he needs a score of x in 11th innings then average in 11 innings = (215 + x)/11\n(215 + x)/11 = 24 => x = (24 x 11) – 215 = 264 – 215 = 49\n2. In a cricket team, the average age of eleven players in 28 years. What is the age of the captain?\nI. The captain is eleven years older than the youngest player.\nII. The average age of 10 players, other than the captain is 27.3 years.\nIII. Leaving aside the captain and the youngest player, the average ages of three groups of three players each are 25 years, 28 years and 30 years respectively.\nSolution:\nTotal age of 11 players = (28 x 11) years = 308 years.\nI. C = Y + 11 C - Y = 11 .... (i)\nII. Total age of 10 players (excluding captain) = (27.3 x 10) years = 273 years.\nAge of captain = (308 - 273) years = 35 years.\nThus, C = 35. .... (ii)\nFrom (i) and (ii), we get Y = 24\nIII. Total age of 9 players = [ (25 x 3) + (28 x 3) + (30 x 3)] years = 249 years.\nC + Y = (308 - 249) = 59 .... (iii)\nFrom (i) and (iii), we get C = 35.\nThus, II alone gives the answer.\nAlso, I and III together give the answer.\n3. A cricketer had a certain average of runs for his 64th innings. In his 65th inning, he is bowled out for no score on his part. This brings down his average by 2 runs. His new average of runs is:\nSolution:\nLet original average be x. Then 64x = 65(x-2) => x = 130\nTherefore, New average = 130-2 = 128 runs.\nAlternative Method: Decrease in average = 2 runs\nTotal decrease in 64 innings = 642 = 128 runs => New average = 0+ 128 = 128 runs.\n4. The batting average of a cricket player for 64 innings in 62 runs. His highest score exceeds his lowest score by 180 runs. Excluding these two innings, the average of the remaining innings becomes 60 runs. His highest score is\nSolution:\nTotal runs of the two innings = 2 62 + 62 * (64-62)= 124+ 124 = 248\nHighest score – Lowest score = 180 runs => Highest Score = (240 + 180) / 2 = 214 runs\n5. Vinod's bowling average till yesterday was 19.2 . Today he took 7 more wickets and conceded 84 runs, there by his average decreased by 0.2 . How many wickets had he taken till yesterday?\nSolution:\nBowling average = (Total runs conceded) / (Number of wickets taken)\nbowling average till yesterday = 19.2\nLet total runs conceded till yesterday =r\ntotal wickets taken till yesterday =w\nr / w = 19.2 => r = 19.2w ------(1)\nbowling average till today = 19.2 - 0.2 = 19\nTotal wickets taken till today = (w+7)\nTotal runs conceded till today = (r+84)\n(r+84)/(w+7) = 19 --------(2)\nUsing (1) and (2)\n(19.2w + 84)/(w + 7) = 19 => 19.2w + 84 = 19w + 133 => 0.2w = 49 => w = 245\nType 6: Miscellaneous\n1. How many candidates were interviewed everyday by the panel A out of the three panels A, B and C?\nI. The three panels on average interview 15 candidates every day.\nII. Out of a total of 45 candidates interviewed everyday by the three panels, the number of candidates interviewed by panel A is more by 2 than the candidates interviewed by panel c and is more by 1 than the candidates interviewed by panel B.\nSolution:\nI. Total candidates interviewed by 3 panels = (15 x 3) = 45.\nII. Let x candidates be interviewed by C.\nNumber of candidates interviewed by A = (x + 2).\nNumber of candidates interviewed by B = (x + 1).\nTherefore x + (x + 2) + (x + 1) = 45 =>3x = 42 =>x = 14\nHence II alone sufficient while I alone not sufficient to answer\n2. The average weight of a class of 25 students is 30kg. The average weight of girls is 5 kg more than that of boys. If the class teacher's weight, which is between 64kg and 106 kg is included, the average weight of the male members of the class equals that of the female members. What is the number of girls in the class, if the average weight of the boys(in kg) is an integer?\nSolution:\nLet average weight of boys =x kg\nThen, average weight of girls =(x+5)kg\nLet number of boys = n\nThen, number of girls = (25- n)\nnx+(25- n)(x+5)=25×30 => 5x - n=125 (1)\nLet class teacher's weight =y\n(nx+y)/(n+1)=(x+5) => y = 5n+x+5\nFrom (1), x = (125+n) / 5\nFrom (2), x=y - 5n - 5\nTherefore,\n( 125+n)/ 5 = y - 5n - 5 => n = (5y - 150) / 26 (3)\nGiven, 64 ≤ y ≤ 106\ny=64 n ≥ 7 y=106 => n ≤ 14\nTherefore, 7 ≤ n ≤ 14\nFrom (3), y = (26n + 150)/5 = 26n/5 + 30\nOnly for n=10, y is an integer ( Since xx is an integer, from (2), it is clear that y is also an integer)\nThus we get n=10\nTherefore, number of girls in the class =25 - n = 15\n3. A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. The average age of the family is\nSolution:\nTotal age of the grandparents = 67 × 2\nTotal age of the parents = 35 × 2\nTotal age of the grandchildren = 6 × 3\nAverage age of the family = ((67×2) + (35×2) + (6×3))/7 = (134+70+18)/7 = 222/7 = 31 5/7\n4. Average temperature from 9th to 16th of a month is 30° C and that from 10th to 17th is 31°C. What is the temperature on 17th , if temperature on 9th is 35° C ?\nSolution:\nDifference between temperature on 9th and 17th = 8 * (31° C - 30° C) = 8° C\nTemperature for 8 days including 17th is more than that of 8 days including 9th.\nTherefore, Temperature on 17th is more than the temperature on 9th.\nTherefore, Temperature on 17th = Temperature on 9th + Difference = 35° C + 8° C\n5. Some students planned a trip and estimated their total expenses to be Rs. 500. However, 5 of them could not go for the trip and as a result average expenditure of the remaining students is increased by Rs. 5. How many students have gone for trip?\nSolution:\nLet the total no. of students be x.\nThen, 500/x = 500/(x – 5) – 5\n500/x + 5 = 500/(x – 5) => (500 + 5x)/x = 500/(x – 5)\n(500 + 5x) (x – 5) = 500x => 500x + 5x2 – 2500 – 25x = 500x\n5x2 – 25x – 2500 = 0 => x = 25, - 20. We cannot take negative value. So, x = 25.\n6. A ship 40 km from shore springs a leak which admits 3 ¾ quintals of water in 12 minutes. 60quintals would suffice to sink the ship, but its pump can throw out 12 quintals of water in one hour. Find the average rate of sailing so that it may reach the shore just it begins to sink.\nSolution:\nIn 12 min. leak admits = 15/4 quintals\nIn one hour leak admits = 15/4 * 60/12 =75/4 quintals\nIn one hour pumps throw out = 12 quintals\nWater left in the ship in one hour = 75/4 – 12 =27/4 quintals\n27/4 quintals of water is left in the ship in 1 hour\n60 quintals of water is left in the ship in = 1604 / 27 = 80/9\nNow in 80/9hrs ship runs = 40 km\n1 hr the ship runs = 40 * 9 / 80 =4.5 km/hr" ]	[ null, "https://1.bp.blogspot.com/-UoIg6YyFGQs/WIIdfhEkceI/AAAAAAAABcQ/wfmLXTjG7QQ1ST9_QMzY4RLzyxYZMuk9QCLcB/s640/1.png", null, "https://1.bp.blogspot.com/-Mgum_n09HA8/WIIdmqft2YI/AAAAAAAABcU/PJJK5g63C-kZsk95xlieyOj4xj62TJ7TACEw/s640/2.png", null, "https://3.bp.blogspot.com/-5lIURWQU1uI/WIIdpPMF1II/AAAAAAAABcY/BHxYoFkqFI0atdCOIxnDpeBbncpNuZfqQCEw/s640/3.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9396699,"math_prob":0.9981587,"size":16056,"snap":"2019-51-2020-05","text_gpt3_token_len":5103,"char_repetition_ratio":0.15711437,"word_repetition_ratio":0.042909313,"special_character_ratio":0.36852267,"punctuation_ratio":0.11059506,"nsfw_num_words":2,"has_unicode_error":false,"math_prob_llama3":0.9996327,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-13T13:27:47Z\",\"WARC-Record-ID\":\"<urn:uuid:4402ee81-8b99-4f19-8c24-b221017bae32>\",\"Content-Length\":\"298364\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8a6c558d-f772-4ca6-928e-42dc4e1294d0>\",\"WARC-Concurrent-To\":\"<urn:uuid:02265bc8-e023-41a8-a698-c7586e44412a>\",\"WARC-IP-Address\":\"172.217.15.115\",\"WARC-Target-URI\":\"https://www.mentorforbankexams.com/2017/01/quantitative-aptitude-notes-averages.html\",\"WARC-Payload-Digest\":\"sha1:MNMCUH3YMYCMR7TFSW6NPXREYXPTZJYO\",\"WARC-Block-Digest\":\"sha1:HJIGUKLFOXJXTUARMOEGGPMDTSU7B2QM\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540555616.2_warc_CC-MAIN-20191213122716-20191213150716-00542.warc.gz\"}"}
https://deepai.org/publication/removing-the-curse-of-superefficiency-an-effective-strategy-for-distributed-computing-in-isotonic-regression	[ "DeepAI\n\n# Removing the Curse of Superefficiency: an Effective Strategy For Distributed Computing in Isotonic Regression\n\nWe propose a strategy for computing the isotonic least-squares estimate of a monotone function in a general regression setting where the data are distributed across different servers and the observations across servers, though independent, can come from heterogeneous sub-populations, thereby violating the identically distributed assumption. Our strategy fixes the super-efficiency phenomenon observed in prior work on distributed computing in the isotonic regression framework, where averaging several isotonic estimates (each computed at a local server) on a central server produces super-efficient estimates that do not replicate the properties of the global isotonic estimator, i.e. the isotonic estimate that would be constructed by transferring all the data to a single server. The new estimator proposed in this paper works by smoothing the data on each local server, communicating the smoothed summaries to the central server, and then computing an isotonic estimate at the central server, and is shown to replicate the asymptotic properties of the global estimator, and also overcome the super-efficiency phenomenon exhibited by earlier estimators. For data on N observations, the new estimator can be constructed by transferring data just over order N^1/3 across servers [as compared to transferring data of order N to compute the global isotonic estimator], and requires the same order of computing time as the global estimator.\n\n02/04/2021\n\n### Improved Communication Efficiency for Distributed Mean Estimation with Side Information\n\nIn this paper, we consider the distributed mean estimation problem where...\n10/14/2020\n\n### Robust covariance estimation for distributed principal component analysis\n\nPrincipal component analysis (PCA) is a well-known tool for dimension re...\n01/15/2020\n\n### Profile least squares estimators in the monotone single index model\n\nWe consider least squares estimators of the finite dimensional regressio...\n07/16/2019\n\n### On the L_p-error of the Grenander-type estimator in the Cox model\n\nWe consider the Cox regression model and study the asymptotic global beh...\n10/06/2019\n\n### Distributed filtered hyperinterpolation for noisy data on the sphere\n\nProblems in astrophysics, space weather research and geophysics usually ...\n09/04/2018\n\n### More is Less: Perfectly Secure Oblivious Algorithms in the Multi-Server Setting\n\nThe problem of Oblivious RAM (ORAM) has traditionally been studied in a ...\n10/20/2020\n\n### Distributed Learning of Finite Gaussian Mixtures\n\nAdvances in information technology have led to extremely large datasets ..." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8525289,"math_prob":0.70651585,"size":1564,"snap":"2023-14-2023-23","text_gpt3_token_len":278,"char_repetition_ratio":0.15641026,"word_repetition_ratio":0.0,"special_character_ratio":0.16432226,"punctuation_ratio":0.070866145,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96066713,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-26T18:44:23Z\",\"WARC-Record-ID\":\"<urn:uuid:93ba6e03-4fb8-4e73-befd-68f2c355e422>\",\"Content-Length\":\"125939\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a9d37d7e-f454-4335-af9b-9b81747b3a49>\",\"WARC-Concurrent-To\":\"<urn:uuid:8ae44cfc-37b3-4d04-b2fb-e6fc29d04fe0>\",\"WARC-IP-Address\":\"18.67.76.81\",\"WARC-Target-URI\":\"https://deepai.org/publication/removing-the-curse-of-superefficiency-an-effective-strategy-for-distributed-computing-in-isotonic-regression\",\"WARC-Payload-Digest\":\"sha1:2DPJEUENNGRXBJ2AYHHAGJ2INZDA5GOA\",\"WARC-Block-Digest\":\"sha1:F5DHRLYC76P5OVDBBSQNJYA6PNDBHJZW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296946445.46_warc_CC-MAIN-20230326173112-20230326203112-00141.warc.gz\"}"}
https://www.bionicturtle.com/forum/tags/garch/	[ "What's new\n\n# garch\n\n1. ### YouTube T2-26: Maximum likelihood estimation of GARCH parameters\n\nGARCH(1,1) is the popular approach to estimating volatility, but its disadvantage (compared to STDDEV or EWMA) is that you need to fit three parameters. Maximum likelihood estimation, MLE, is an immensely useful statistical approach that can be used to find \"best fit\" parameters. In this video...\n2. ### YouTube T2-25: Comparing volatility approaches: MA versus EWMA versus GARCH\n\nThe general form for all three is: σ^2(n) = γV(L) + αu^2(n-1) + σ^2(n-1).\n3. ### YouTube T2-24: Forecast volatility with GARCH(1,1)\n\nThe GARCH(1,1) volatility forecast is largely a function of the first term omega, ω = γ*V(L), which itself is the product of a rate of reversion, γ, and a reversion level, V(L); aka, long-run or unconditional variance David's XLS is here: https://trtl.bz/2yGdnjv\n4. ### YouTube T2-23: Volatility: GARCH 1,1\n\nThe GARCH(1,1) volatility estimate shares a similarity to EWMA volatility: both assign greater (lesser) weight to recent (distant) returns. But the GARCH(1,1) has an additional feature: it models a long-run (aka, unconditional) variance toward which the volatility series is pulled. David's XLS...\n5. ### P1.T2.706. Bivariate normal distribution (Hull)\n\nLearning objectives: Calculate covariance using the EWMA and GARCH(1,1) models. Apply the consistency condition to covariance. Describe the procedure of generating samples from a bivariate normal distribution. Describe properties of correlations between normally distributed variables when using...\n6. ### P1.T2.704. Forecasting volatility with GARCH (Hull)\n\nLearning objectives: Explain mean reversion and how it is captured in the GARCH(1,1) model. Explain the weights in the EWMA and GARCH(1,1) models. Explain how GARCH models perform in volatility forecasting. Describe the volatility term structure and the impact of volatility changes. Questions...\n7. ### P1.T2.703. EWMA versus GARCH volatility (Hull)\n\nLearning objectives: Apply the exponentially weighted moving average (EWMA) model to estimate volatility. Describe the generalized autoregressive conditional heteroskedasticity (GARCH(p,q)) model for estimating volatility and its properties. Calculate volatility using the GARCH(1,1) model...\n8. ### R16.P1.T2. Hull - expected value of u(n+t-1)^2\n\nIn Hull - Risk Management and Financial Institutions, it is stated, in page 222 (10.10 using GARCH(1,1) to forecase future volatility), that: \"the expected value of u(n+t−1)^2 is σ(n+t−1)^2\". Is this something obvious? Can anybody explain why this should be the case? Thanks!\n9. ### GARCH(1,1) vs EWMA for Forecasting Volatility\n\nSo I link this video which explains GARCH(1,1) as a measure to forecast future volatility. Now we know EWMA is a special case of GARCH which sums alpha and beta equal to 1 and therefore ignores any impact on long run variance, implying that variance is not mean reverting.. Again when we...\n10. ### P1.T2.502. Covariance updates with EWMA and GARCH(1,1) models\n\nLearning outcomes: Define correlation and covariance, differentiate between correlation and dependence. Calculate covariance using the EWMA and GARCH (1,1) models. Apply the consistency condition to covariance. Questions: 502.1. About the consistency condition, each of the following is true...\n11. ### P1.T2.409 Volatility, GARCH(1,1) and EWMA\n\nConcept: These on-line quiz questions are not specifically linked to AIMs, but are instead based on recent sample questions. The difficulty level is a notch, or two notches, easier than bionicturtle.com's typical AIM-by-AIM question such that the intended difficulty level is nearer to an actual..." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8650544,"math_prob":0.97174466,"size":2908,"snap":"2021-43-2021-49","text_gpt3_token_len":646,"char_repetition_ratio":0.13464187,"word_repetition_ratio":0.02877698,"special_character_ratio":0.21492435,"punctuation_ratio":0.17142858,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9934743,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-28T18:27:21Z\",\"WARC-Record-ID\":\"<urn:uuid:3dc6fb01-ec9f-4531-b33a-bc7423e5b3fb>\",\"Content-Length\":\"64742\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6ee8fa4b-d037-4fca-9f0a-e73a23dce49f>\",\"WARC-Concurrent-To\":\"<urn:uuid:74924898-9e05-4125-ae5e-a3e837015f67>\",\"WARC-IP-Address\":\"172.66.43.16\",\"WARC-Target-URI\":\"https://www.bionicturtle.com/forum/tags/garch/\",\"WARC-Payload-Digest\":\"sha1:FMJFPVKMOTXMGGO6FZMHNSUPL2Y2EFLI\",\"WARC-Block-Digest\":\"sha1:YIWDYFT3HUK2AWM4J3GD4Q2YQLO3MV5X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358570.48_warc_CC-MAIN-20211128164634-20211128194634-00417.warc.gz\"}"}
https://zh-min-nan.m.wikipedia.org/wiki/Pang-b%C3%B4%CD%98:Infobox_road/doc	[ "```{{Infobox road\n\|country=\n\|type=\n\|route=\n\|map=\n\|length_mi=\n\|length_km=\n\|length_round=\n\|length_ref=\n\|established=\n\|direction_a=\n\|terminus_a=\n\|junction=\n\|direction_b=\n\|terminus_b=\n\|previous_type=\n\|previous_route=\n\|next_type=\n\|next_route=\n}}\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6980704,"math_prob":0.4765237,"size":263,"snap":"2023-14-2023-23","text_gpt3_token_len":86,"char_repetition_ratio":0.17374517,"word_repetition_ratio":0.0,"special_character_ratio":0.29277566,"punctuation_ratio":0.03846154,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95608014,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-04-02T04:51:18Z\",\"WARC-Record-ID\":\"<urn:uuid:d94b394b-0e30-464f-9574-0a48af9d9eb0>\",\"Content-Length\":\"21524\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:62a53edf-f64a-4f94-ab23-bbf12795a4e9>\",\"WARC-Concurrent-To\":\"<urn:uuid:4081ef03-48cd-4ea0-8d9c-50faa8fae347>\",\"WARC-IP-Address\":\"208.80.154.224\",\"WARC-Target-URI\":\"https://zh-min-nan.m.wikipedia.org/wiki/Pang-b%C3%B4%CD%98:Infobox_road/doc\",\"WARC-Payload-Digest\":\"sha1:5ZTQ3TEWXGXVMYF6XY2MPSCOJPJIOY3N\",\"WARC-Block-Digest\":\"sha1:LEGVKWYF4NNJSTZBK7ZM53KUCEWCJPVY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296950383.8_warc_CC-MAIN-20230402043600-20230402073600-00295.warc.gz\"}"}
https://www.mdpi.com/1099-4300/22/10/1143	[ "", null, "Next Article in Journal\nThe Topp-Leone Generalized Inverted Exponential Distribution with Real Data Applications\nPrevious Article in Journal\nApproximate Learning of High Dimensional Bayesian Network Structures via Pruning of Candidate Parent Sets\n\nFont Type:\nArial Georgia Verdana\nFont Size:\nAa Aa Aa\nLine Spacing:\nColumn Width:\nBackground:\nArticle\n\n# Partial Classifier Chains with Feature Selection by Exploiting Label Correlation in Multi-Label Classification\n\n1\nDepartment of Computer Science and Technology, China University of Mining and Technology, Beijing 100083, China\n2\nSchool of Software and IoT Engineering, Jiangxi University of Finance & Economics, Nanchang 330013, China\n3\nSchool of Technology and Business Studies, Dalarna University, 79188 Falun, Sweden\n\nAuthors to whom correspondence should be addressed.\nEntropy 2020, 22(10), 1143; https://doi.org/10.3390/e22101143\nReceived: 17 August 2020 / Revised: 4 October 2020 / Accepted: 6 October 2020 / Published: 10 October 2020\n\n## Abstract\n\n:\nMulti-label classification (MLC) is a supervised learning problem where an object is naturally associated with multiple concepts because it can be described from various dimensions. How to exploit the resulting label correlations is the key issue in MLC problems. The classifier chain (CC) is a well-known MLC approach that can learn complex coupling relationships between labels. CC suffers from two obvious drawbacks: (1) label ordering is decided at random although it usually has a strong effect on predictive performance; (2) all the labels are inserted into the chain, although some of them may carry irrelevant information that discriminates against the others. In this work, we propose a partial classifier chain method with feature selection (PCC-FS) that exploits the label correlation between label and feature spaces and thus solves the two previously mentioned problems simultaneously. In the PCC-FS algorithm, feature selection is performed by learning the covariance between feature set and label set, thus eliminating the irrelevant features that can diminish classification performance. Couplings in the label set are extracted, and the coupled labels of each label are inserted simultaneously into the chain structure to execute the training and prediction activities. The experimental results from five metrics demonstrate that, in comparison to eight state-of-the-art MLC algorithms, the proposed method is a significant improvement on existing multi-label classification.\n\n## 1. Introduction\n\nIn machine learning applications, the traditional single label classification (SLC) problem has been explored substantially. However, more recently, the multi-label classification (MLC) problem has attracted increasing research interest because of its wide range of applications, such as text classification [1,2], social network analysis , gene function classification , and image/video annotation . With SLC, one instance only belongs to one category, whereas with MLC, it can be allocated to multiple categories simultaneously. MLC is a generalization of SLC, which makes it a more difficult and general problem in the machine learning community. Due to multiple labels and the possible links between them, multi-label correlations become very complex . On the one hand, for example, it is more likely for a piece of news tagged with “war” to have another tag “army” than “entertainment”. On the other hand, in a classification of nature scenes with the set of picture labels (“beach”, “building”, “desert”, “sailboat”, “camel”, “city”), it is less likely that a picture of scenery is labelled by both “desert” and “beach”. Thus, exploring these complex couplings is an important challenge in MLC since label correlations can improve classification performance. Based on the order of correlations, the exploitation of label correlations can be divided into roughly three categories :\n(1)\nFirst-order strategy: This divides the MLC problem into a number of independent binary classification problems. The prominent advantage is their conceptual simplicity and high efficiency even though the related classifiers might not acquire the optimal results because of ignoring the label couplings.\n(2)\nSecond-order strategy: This considers pairwise relationships between labels. The resulting classifiers can achieve good levels of generalization performance since label couplings are exploited to some extent. However, they are only able to exploit label-coupling relationships to a limited extent. Many real-world applications go beyond these second-order assumptions.\n(3)\nHigh-order strategy: This tackles the MLC problem by considering high-order relationships between labels. This strategy is said to have stronger correlation-modeling capabilities than the other two strategies, and its corresponding classifiers to have a higher degree of time and space complexity.\nBroadly speaking, there are many relevant works discussing coupling learning of complex interactions . Using a data-driven approach, Wang et al. [8,9] showed how complex coupling relationships could demonstrate learning on categorical and continuous data, respectively, including intra-coupling within objects and inter-coupling between objects. Complex coupling relationships have also been discussed in different applications, such as clustering , outlier detection and behavior analysis . Aiming to address the MLC problem, the classifier chain (CC) is a well-known method that adopts a high-order strategy to extract label couplings. Its chaining mechanism allows each individual classifier to incorporate the predictions of the previous one as additional information. However, CC suffers from two obvious drawbacks: (1) the label order is randomly inserted into the chain, which usually has a strong effect on classification performance ; (2) all of the labels are inserted into the chain under the assumption that they each have coupling relationships when, in fact, this assumption is too idealistic. Irrelevant labels presenting in the chain actually reduce the predictive results of the CC approach. In this work, we will address the two problems identified here simultaneously. We propose a partial classifier chain method with feature selection (PCC-FS) that exploits the coupling relationships in the MLC problem. The main contributions of this paper include:\n(1)\nA new construction method of chain mechanism that only considers the coupled labels (partial labels) and inserts them into the chain simultaneously, and thus improves the prediction performance;\n(2)\nA novel feature selection function that is integrated into the PCC-FS method by exploiting the coupling relationships between features and labels, thus reducing the number of redundant features and enhancing the classification performance;\n(3)\nLabel couplings extracted from the MLC problem based on the theory of coupling learning, including intra-couplings within labels and inter-couplings between features and labels, which makes the exploration of label correlation more comprehensive.\nThe rest of this paper is organized as follows. The background and reviews of the related work about CC and feature selection in the MLC problem are discussed in Section 2. Based on our reviewed previous research, we outline our PCC-FS approach in Section 3. This consists of three components: feature selection with inter-coupling exploration, intra-coupling exploration in label set, and label set prediction. In Section 4, we discuss the experiment’s environment, the datasets, the evaluation criteria, and the analysis based on the experimental results. Conflicting criteria for algorithm comparison and a series of statistical tests have been carried out for validating the experimental results in Section 5. We conclude the paper in Section 6 by identifying our contribution to this particular research area and our planned future work in this direction.\n\n## 2. Preliminaries\n\nIn this section, we begin by introducing the concepts of MLC, CC, and feature selection. This will establish the theoretical foundation of our proposed approach. Then, we will review the CC-based MLC algorithms.\n\n#### 2.1. MLC Problem and CC Approach\n\nMLC is a supervised learning problem where an object is naturally associated with multiple concepts. It is important to explore the couplings between labels because they can improve the prediction performance of MLC methods. In order to describe our algorithm, some basic aspects of MLC and CC are outlined first. Suppose $( x , y )$ represents a multi-label sample where $x$ is an instance and $y ⊆ L$ is its corresponding label set. $L$ is the total label set, which is defined as follows:\nWe assume that $x = ( x 1 , x 2 , ⋯ , x D ) ∈ X$ is the D-dimensional feature vector corresponding to $x$, where $X ⊆ R D$ is the feature vector space and denotes a specific feature. $y = ( y 1 , y 2 , ⋯ , y Q ) , ∈ { 0 , 1 } Q$ is the Q-dimensional label vector corresponding to $y$, and $y q$ is described as:\nThus, the multi-label classifier $h$ can be defined as:\nWe further assume that there are $m + n$ samples, in which $m$ samples form the training set $X t r a i n$ and $n$ samples form the test set $X t e s t$. They are defined as follows:\nAmong the MLC algorithms, CC may be the most famous MLC method concerning label correlations. It involves $Q$ binary classifiers as in the binary relevance (BR) method. The BR method transforms MLC into a binary classification problem for each label and it trains $Q$ binary classifiers . In the CC algorithm, classifiers are linked along a chain where each classifier handles the BR problem associated with . The feature space of each link in the chain is extended with the 0 or 1 label associations of all previous links. The training and prediction phases of CC are described in Algorithm 1 :\n Algorithm 1. The training phase of the classifier chain (CC) algorithm Input: $X t r a i n$Output: Stepsfor ;do /the jth binary transformation and training/;$X t r a i n ′ → { }$;for , ;do ;$C j : X t r a i n ′ → l j ∈ { 0 , 1 }$ /train $C j$ to predict binary relevance of $l j$ /.\nAfter the training step, a chain $C 1 , … , C Q$ of binary classifiers is generated. As shown in Algorithm 2, each classifier $C j$ in the chain learns and predicts the binary association of label $l j$, $j = 1 , 2 , … , Q$, augmented by all prior binary relevance predictions in the chain $l 1 , … , l j − 1$.\n Algorithm 2. The prediction phase of the CC algorithm Input: $X t e s t$, Output: predicted label sets of each instance in $X t e s t$Stepsfor , $i = m + 1 , … , m + n ;$do; $y i ′ ← { }$;for do ;return $( x i , y i ′ )$·$i = m + 1 , … , m + n$/the classified test samples */.\nThe chaining mechanism of the CC algorithm transmits label information among binary classifiers, which considers label couplings and thus overcomes the label independence problem of the BR method.\n\n#### 2.2. Feature Selection in the MLC Problem\n\nSome comprehensive literature reviews and research articles [15,16,17,18] have discussed the problem of feature selection (FS) in the MLC problems. There are different methods used to select relevant features from the MLC datasets. They can be divided into three main types: filters, wrappers, and embedded methods. Filter methods rank all the features with respect to their relevance and cut off the irrelevant features according to some evaluation function. Generally, filter methods adopt an evaluation function that only depends on the properties of the dataset and hence are independent of any particular learning algorithm. Wrappers use an MLC algorithm to search for and evaluate relevant subsets of features. Wrappers usually integrate a search strategy (for example, genetic algorithm or forward selection method) to reduce the high computational burden. For embedded methods, FS is an integral element of the classification process. In other words, the classification process itself performs feature selection as part of the learning process. Therefore, the proposed FS method can be included in the group of filter methods. It adopts covariance to express the coupling relationships between feature set and label set and adaptively cuts off irrelevant features according to the standard deviation of Gaussian-like distribution.\n\n#### 2.3. Related Work of CC-Based Approaches\n\nThe CC algorithm uses a high-order strategy to tackle the MLC problem; however, its performance is sensitive to the choice of label order. Much of the existing research has focused on solving this problem. Read et al. proposed using the ensemble of classifier chains (ECC) method, where the CC procedure is repeated several times with randomly generated orders and all the classification results are fused to produce the final decision by the vote method. Chen et al. adopted kernel alignment to calculate the consistency between the label and kernel function and then assigned a label order according to the consistency result. Read et al. presented a novel double-Monte Carlo scheme to find a good label sequence. The scheme explicitly searches the space for possible label sequences during the training stage and makes a trade-off between predictive performance and scalability. Genetic algorithm (GA) was used to optimize the label ordering since GA has a global search capability to explore the extremely large space of label permutation [22,23]. Their difference is that the one of the works adopts the method of multiple objective optimization to balance the classifier performance through considering the predictive accuracy and model simplicity. Li et al. applied the community division technology to divide label set and acquire the relationships among labels. All the labels are ranked by their importance.\nSome of the existing literature [25,26,27,28,29,30,31,32,33] adopted Graph Representation to express label couplings and rank labels simultaneously. Sucar et al. introduced a method of chaining Bayesian classifiers that integrates the advantages of CC and Bayesian networks (BN) to address the MLC problem. Specifically, they adopted the tree augmented naïve (TAN) Bayesian network to represent the probabilistic dependency relationships among labels and only inserted the parent nodes of each label into the chain according to the specific selection strategy of the tree root node. Zhang et al. used mutual information (MI) to describe the label correlations and constructed a corresponding TAN Bayesian network. The authors then applied a stacking ensemble method to build the final learning model. Fu et al. adopted MI to present label dependencies and then built a related directed acyclic graph (DAG). The Prim algorithm was then used to generate the maximum spanning tree (MST). For each label, this algorithm found its parent labels from MST and added them into the chain. Lee et al. built a DAG of labels where the correlations between parents and child nodes were maximized. Specifically, they quantified the correlations with the conditional entropy (CE) method and found a DAG that maximized the sum of CE between all parent and child nodes. They discovered that highly correlated labels can be sequentially ordered in chains obtained from the DAG. Varando et al. studied the decision boundary of the CC method when Bayesian network-augmented naïve Bayes classifiers were used as base models. It found polynomial expressions for the multi-valued decision functions and proved that the CC algorithm provided a more expressive model than the binary relevance (BR) method. Chen et al. firstly used the Affinity Propagation (AP) clustering approach to partition the training label set into several subsets. For each label subset, it adopted the MI method to capture label correlations and constructed a complete graph. Then the Prim algorithm was applied to learn the tree-structure constraints (in MST style) among different labels. In the end, the ancestor nodes were found from MST and inserted into the chain for each label. Huang et al. firstly used a k-means algorithm to cluster the training dataset into different groups. The label dependencies of each group were then expressed by the co-occurrence of the label pairwise and corresponding labels were then modeled by a DAG. Finally, the parent labels of each label were inserted into the chain. Sun et al. used the CE method to model label couplings and constructed a polytree structure in the label space. For each label, its parent labels were inserted into the chain for further prediction. Targeting the two drawbacks of the CC algorithm mentioned in Section 1, Kumar et al. adopted the beam search algorithm to prune the label tree and found the optimal label sequence from the root to one of the leaf nodes.\nIn addition to the aforementioned graph-based CC algorithms and considering conditional label dependence, Dembczyński et al. introduced probability theory into the CC approach and outlined their probabilistic classifier chains (PCC) method. Read et al. extended the CC approach to the classifier trellises (CT) method for large datasets, where the labels were placed in an ordered procedure according to the MI measure. Wang et al. proposed the classifier circle (CCE) method, where each label was traversed several times (just once in CC) to adjust the classification result of each label. This method is insensitive to label order and avoids the problems caused by improper label sequences. Jun et al. found that the label with higher entropy should be placed after those with lower entropy when determining label order. Motivated by this idea, they went on to propose four ordering methods based on CE and, after considering each, suggested that the proposed methods did not need to train more classifiers than the CC approach. In addition, Teisseyre and Teisseyre, Zufferey and Słomka proposed two methods that combine the CC approach and elastic-net. The first integrated feature selection into the proposed CCnet model and the second combined the CC method and regularized logistic regression with modified elastic-net penalty in order to handle cost-sensitive features in some specific applications (for example, medical diagnosis).\nIn summary, in order to address the label ordering problem, almost all of the published CC-based algorithms adopted different ranking methods to determine a specific label order (by including all of the labels or just a part of them). All of these methods are reasonable, but it is hard to judge which label order (or label ordering method) is the best one for a specific application. Furthermore, some of these studies adopted different methods (for example, MI, CE, conditional probability, co-occurrence, and so on) to explore label correlations, but they only focused on label space; the coupling relationships were insufficiently exploited. In addition, the CC-based algorithms used in these published studies added previous labels into feature space to predict the current label, which resulted in an excessively large feature space, especially for large label sets. Thus, feature selection is a necessary stage in the CC-based algorithms. In this work, we propose a novel MLC algorithm based on the CC method and feature selection which avoids the label ranking problem and exploits the coupling relationships both in label and feature spaces. Section 3 provides a detailed description of the proposed method.\n\n## 3. The Principle of the PCC-FS Algorithm\n\nInspired by the CCE method , the research presented here organizes labels as a circular structure that can overcome the label ordering problem. However, there are two obvious differences between the PCC-FS algorithm and the CCE algorithm. First, the CCE algorithm included all of the labels in the training and prediction tasks while the PCC-FS algorithm only uses coupled labels to perform these tasks. Second, CCE does not take advantage of label correlations while the PCC-FS algorithm not only exploits the intra-couplings within labels but also explores the inter-couplings between features and labels.\n\n#### 3.1. Overall Description of the PCC-FS Algorithm\n\nThe PCC-FS algorithm aims to solve the MLC problem by exploring coupling relationships in feature and label spaces. The workflow is described in Figure 1a and includes the following three steps:\n(1)\nIn the feature selection stage, we explore the inter-couplings between each feature and label set. The features with low levels of inter-couplings would be cut off on the assumption that all the inter-couplings follow Gaussian-like distribution;\n(2)\nIntra-couplings among labels are extracted to provide the measurement used to select the relevant labels (partial labels) of each label. Irrelevant labels are not then able to hinder the classification performance;\n(3)\nAfter feature selection, as described in Figure 1b, the coupled labels of each label are inserted simultaneously into the chain, and label prediction is executed in an iterative process, thus avoiding the label ordering problem.\nIn order to elaborate on the PCC-FS algorithm in greater detail, we will discuss the above three steps in the following sections. Feature selection, as the data preprocessed step, is discussed in Section 3.2. Intra-coupling exploration is introduced in Section 3.3 and Section 3.4 gives a detailed description of the training and prediction steps in the PCC-FS algorithm.\n\n#### 3.2. Feature Selection with Inter-Coupling Exploration\n\nIn order to eliminate the differences of the various features, the values of each feature are normalized, which is denoted as $x d : n o r m$ for . The PCC-FS algorithm adopts the absolute value of covariance, denoted as $C o v ( x d : n o r m , l j )$, to represent the coupling relationships between normalized feature $x d : n o r m$ and label $l j$. The reason for adopting absolute covariance is that both positive and negative covariance can indicate the correlation between features and labels. In order to calculate $C o v ( x d : n o r m , l j )$, $l j$ is encoded by binary value (1 for containing label $l j$ and 0 otherwise) (We give a concrete example to illustrate the numerical coding. Suppose that we classify bird species by their acoustic features. One of the methods is to convert the audio signal to a spectrogram, which is further represented by an image. Four sample images are digitalized and vectorized to generate the feature matrix $x d : n o r m$. The first label of all images—e.g., —has the same dimension to the first feature—e.g., . Thus, the unbiased sample covariance will be $C o v ( x 1 : n o r m , l 1 ) = \| E ( x 1 : n o r m ∗ l 1 ) − E ( x 1 : n o r m ) E ( l 1 ) \| = 0.22$.). It should be noted that there are many other encoding methods to be developed and our algorithm works only for numerical encoding. The inter-coupling between $x d : n o r m$ and label set is defined in Equation (5):\nFor every feature, we suppose that all the covariance values between feature set and label set follow the Gaussian-like distribution with a concave density function. Since too small inter-coupling has to be discarded, what we need to figure out is the quantile where we truncate the sample irrespective of the distribution. Thus, we do not restrict them to be exactly in Gaussian distribution. Under this relaxation, we can apply the same criteria described in Equation (6) for the truncation, where $μ$ is estimated by sample mean and $σ$ by sample standard deviation:\nWe further suppose that there are $m$ training instances and $n$ test instances. The feature-selected training dataset $Ƶ t r a i n$ is defined as follows in Equation (7):\n$X t r a i n$ represents the matrix that contains all the feature values of $m$ training instances, and $Y t r a i n$ is the matrix of labels for all the training instances. $D ′$ is the number of dimensions after feature selection. Similarly, the test dataset $X t e s t$ is described in Equation (8):\n$X t e s t = [ x 1 1 … x 1 D ′ … x i d … x n 1 … x n D ′ ] n × D ′ .$\n\n#### 3.3. Intra-Coupling Exploration in Label Set\n\nIn the CC algorithm, it is unreasonable that all of the previous labels participate in the learning activity of the current label because it is a highly idealistic assumption that all of the previous labels couple with the current label. In this study, we use the absolute covariance to measure the intra-couplings among labels, named as . The covariance matrix of labels is described in Equation (9):\n$ϕ = [ I a C ( l 1 , l 1 ) … I a C ( l 1 , l Q ) … I a C ( l j , l k ) … . I a C ( l Q , l 1 ) … I a C ( l Q , l Q ) ] Q × Q .$\nThe threshold method is used to judge the intra-coupling relationships, where the threshold $I a C t h r e$ is determined by experimental analysis instead of subjective empirical constants. Specifically, the alternative threshold values are sampled $K$ (for example, $K = 100$) groups with equal space in $[ M a x I a C , M i n I a C ]$, where $M a x I a C$ and $M i n I a C$ are the maximum and minimum values in $ϕ$, excluding the values on the main diagonal. The threshold $I a C t h r e$ is determined by calculating the comprehensive average ranking across five criteria. According to $I a C t h r e$, $ϕ$ is transformed into a matrix with only 0 and 1 values, where one indicates that the corresponding value is no less than the threshold and zero denotes the other cases. For example, if the first line of transformed $ϕ$ is $[ 1 1 0 ⋯ 0 1 1 ] 1 × Q$, it indicates that $l 1$ only has coupling relationships with $l 2$, $l Q − 1 ,$ and $l Q$. For each label $l j$, , we defined its coupled label set in Equation (10):\n\n#### 3.4. Label Prediction of the PCC-FS Algorithm\n\nIn the PCC-FS algorithm, the label set $L$ is organized as a circular structure with random order as described in Figure 1b. For each label , the proposed algorithm constructs binary classifiers by the order of ${ l 1 , l 2 , ⋯ , l Q } ,$ and this process iterates $T$ times. In each iteration, $Ɣ j$ is regarded as an additional feature set for the binary classifier related to label $l j$. We suppose that the binary classifier is defined as follows:\nThe PCC-FS algorithm generates $T ∗ Q$ binary classifiers, $ƕ r , j$, and $Ɓ$ represents the binary learning method. In this work, logistic regression is used as the base binary classifier. In addition, other binary learning methods can also be applied to our algorithm. $Ƈ$ contains the latest predicted values of all of the labels on m training instances, as described in Equation (12):\n$Ƈ = [ l 11 … l 1 Q … l j k … . l m 1 … l m Q ] m × Q .$\nFor the first iteration of the training process, $Ƈ$ is initialized by zero matrix $[ 0 ] m × Q$. $Ƈ ( j )$ represents the vector that contains the latest predicted values of $l j$ on all the training instances. $Ƈ ( j )$ is derived from $Ƈ$ and iteratively updated by $ƕ r , j$. Similarly, $Ƈ ( Ɣ j )$ is the matrix that contains the latest predicted values of $Ɣ j$ on all the training instances. $χ t r a i n r , j$ is the binary dataset related to $l j$ in the rth iteration, which is defined as follows:\nand\n$x r , j = [ X t r a i n , l p r e r , j ] = [ X t r a i n , Ƈ ( Ɣ j ) ] ,$\nwhere . $Y t r a i n ( j )$ is the label vector related to label $l j$ and $x r , j$ is the extended feature matrix. $l p r e r , j$ represents the latest predicted results of $Ɣ j$, which can be acquired by $Ƈ ( Ɣ j )$. The training steps of the PCC-FS algorithm is, therefore, described in Algorithm 3.\n Algorithm 3. The training phase of the PCC-FS algorithm Input: original training data, iterative times T Output: $T ∗ Q$ binary classifiers $ƕ r , j ,$ Steps generate $X t r a i n$ through feature selection process;learn label couplings and generate $Ɣ j$ for each label by Equation (14), ;initialize $χ t r a i n 1 , j$ with zero matrix $[ 0 ] m × Q$;for ;for ;$l p r e r , j = Ƈ ( Ɣ j )$;$x r , j = [ X t r a i n , l p r e r , j ]$;$χ t r a i n r , j = [ x r , j , Y t r a i n ( j ) ]$;$ƕ r , j ← B ( χ t r a i n r , j$);$Ƈ ( j ) = ƕ r , j ( x r , j )$;end for;end for.\nAs described in Algorithm 3, the PCC-FS algorithm learns $ƕ r , j$ and updates $χ t r a i n r , j$ iteratively. The latest prediction results of each label are then integrated into the classifiers $ƕ r , j$ to perform the current prediction activities. After iterating $T$ times, the PCC-FS algorithm should acquire $T ∗ Q$ binary classifiers $ƕ r , j$ for . The prediction steps of the PCC-FS algorithm are described in Algorithm 4 where the trained $T ∗ Q$ binary classifiers $ƕ r , j$ predict the label set for each test sample in an iterative process by using $X t e s t$ and $Ɣ j$.\n Algorithm 4. The prediction phase of the PCC-FS algorithm Input: original test data, $T ∗ Q$ binary classifiers $ƕ r , j ,$ ; Output: predicted label sets for $X t e s t$; Steps generate $X t e s t$ through feature selection process;initialize $χ t e s t 1 , j$ with zero matrix $[ 0 ] n × Q$;for ;for ;$l p r e r , j = Ƈ ( Ɣ j )$;$χ t e s t r , j = [ X t e s t , l p r e r , j ]$;$Ƈ ( j ) = ƕ r , j ( χ t e s t r , j )$;end for;end for;return the predicted results $Ƈ$.\n\n## 4. Experimental Results and Analysis\n\nUsing the introduction of an experimental environment and datasets, this section provides an experimental analysis and comparison of the proposed PCC-FS method and eight other state-of-the-art MLC algorithms.\n\n#### 4.1. Experiment Environment and Datasets\n\nWe included seven datasets in the experiments conducted for this article, all of which are extensively used to evaluate MLC algorithms. The themes of the data include emotions, CAL500, yeast, flags, scene, birds, and enron (for detailed information about these public datasets please see: http://mulan.sourceforge.net/datasets-mlc.html). They cover text, image, music, audio, and biology fields, and the number of labels varies from 6 to 174, as described in Table 1.\n\n#### 4.2. Evaluation Criteria\n\nIn this work, five popular MLC criteria were adopted to validate our method including Hamming Loss (HL), Ranking Loss (RL), One Error (OE), Coverage (Cove), and Average Precision (AP). We used $f$ to denote the function of predicted probability. The predicted probabilities that test instance belonging to each label were sorted in descending order, and $r a n k f ( x i f s , l )$ presents the corresponding rank of label $l$. The symbol $\| · \|$ in the following criteria indicates the number of the element number in a set. Hamming Loss computes the average number of times that labels are misclassified. $Δ$ is the symmetric difference between two sets:\n$H L ( h ) = 1 n ∑ i = m + 1 m + n 1 Q \| h ( x i f s ) Δ y i \| .$\nRanking Loss computes the average number of times when irrelevant labels are ranked before the relevant labels. $y i ¯$ is the complement of $y i$ in $L$:\n$R L ( h ) = 1 n ∑ i = m + 1 m + n 1 \| y i \| \| y i ¯ \| \| { ( l , l ′ ) ∈ y i × y i ¯ \| f ( x i f s , l ) ≤ f ( x i f s , l ′ ) } \| .$\nOne Error calculates the average number of times that the top-ranked label is irrelevant to the test instance:\n$O E ( h ) = 1 n ∑ i = m + 1 m + n \| { l ∉ y i \| l = arg max l ∈ L r a n k f ( x i f s , l ) } \| .$\nCoverage calculates the average number of steps that are in the ranked list to find all the relevant labels of the test instance:\nAverage Precision evaluates the degree for the labels that are prior to the relevant labels and that are still relevant labels:\n\n#### 4.3. Experimental Results Analysis and Comparison\n\nEight state-of-the-art MLC algorithms were chosen for a comparison study in order to act as a contrast to the proposed PCC-FS algorithm. These are HOMER , LP , RAkEL , Rank-SVM , BP_MLL , CC , CCE , and LLSF-DL . HOMER, LP, RAkEL, Rank-SVM, and BP_MLL are classic benchmark algorithms. CC and CCE are the CC-based algorithms, and LLSF_DL is a recently developed algorithm whose learning of label-specific data representation for each class label and class-dependent labels has performed outstandingly. For comparative objectivity, 10-fold cross validation was adopted, and the average values of 10 experimental repetitions were regarded as the final values for every evaluation criterion. The base classifier of CC, CCE, and PCC-FS is linear logistic regression, which was implemented by the Liblinear Toolkit. The full names of the compared algorithms in Table 2, Table 3, Table 4, Table 5, Table 6 and Table 7 are as follows: HOMER: hierarchy of multi-label classifiers; LP: label powerset; RAkEL: RAndom k-labELsets; Rank-SVM: rank support vector machine; BP_MLL: back-propagation for multi-label learning; CC: classifier chains; CCE: classifier circle; PCC-FS: partial classifier chains with feature selection; LLSF-DL: Learning Label-Specific Features and Class-Dependent Labels.\nGenerally speaking, as observed from Table 2, Table 3, Table 4, Table 5, Table 6, Table 7 and Table 8, the LP algorithm demonstrated poor performance across all seven datasets. Five algorithms (Rank-SVM, BP_MLL, HOMER, RAkEL, and LLSF-DL) showed poor performance on some datasets. More specifically, the Rank-SVM algorithm achieved the worst comprehensive performance on the datasets of emotions, flags, birds, and enron. The BP_MLL algorithm did not perform well on the datasets of emotions and scene, while it achieved good results on dataset enron. HOMER’s results were not good on the datasets of yeast, scene, and enron. The RAkEL algorithm did not achieve good performance on dataset yeast, but its performance on dataset flags showed obvious advantages over other algorithms except PCC-FS. The results of LLSF-DL algorithm were not good on the datasets of flags or yeast, besides showing no obvious advantages among the nine algorithms. For dataset scene, the CC algorithm achieved the best results among the nine algorithms on Ranking Loss, Coverage, and Average Precision. The RAkEL algorithm attained the best values for the Hamming Loss criterion but only on the datasets of birds and flags. RAkEL also obtained the best value for the Ranking Loss criterion on dataset birds. For five of the tested evaluation criteria, the proposed PCC-FS algorithm outperformed all of the eight other algorithms with the best comprehensive performance on the datasets of emotions, CAL500, yeast, flags, scene, and birds, and it achieved above-average performance on dataset enron. In order to evaluate the performance of all nine algorithms across these five criteria, their average ranks are presented in Figure 2.\nThe Rank-SVM algorithm achieved the worst average rank on the datasets of emotions, flags, birds, and enron; the LP algorithm obtained the worst average rank on CAL500 and yeast; and the HOMER algorithm the worst average rank on scene. For our proposed PCC-FS algorithm, its average rank on seven datasets (emotions, yeast, CAL500, flags, scene, birds, and enron) was 1, 1.2, 1.4, 1.4, 1.6, 2.6, and 3, respectively. The comparison results demonstrate that the proposed PCC-FS algorithm achieved stable results and significant classification effects on the seven commonly-used datasets in contrast to the eight other most-cited algorithms.\n\n## 5. Conflicting Criteria for Algorithm Comparison and Statistical Test\n\n#### 5.1. Conflicting Criteria\n\nConflicting criteria may exist when comparing algorithms under multiple evaluation methods, because the five methods used in this work did not give consistent ranking results. To conduct a fair comparison between algorithms, we presented the outcome of the sum of ranking differences (SRDs) that is a multi-criteria decision-making tool before making statistical test [47,48]. The absolute values of differences between a reference vector and actual ranking were summed up for each algorithm. Since we have five evaluation methods and seven data sets, the reference was a vector of 35 elements with each element being the best score across each algorithm. The theoretical distribution for SRD was approximately normal after scaling it onto an interval of $[ 0 , 100 ]$. Thus, the normal quantile for each algorithm represented an empirical SRD compared with the reference vector. A detailed implementation is also available in a recent work . The comparison was shown in Figure 3.\nFrom Figure 3, we can see that PCC-FS was located to the left of the curve indicating that PCC-FS is the ideal algorithm and the closest one to the reference. Meanwhile, CC and CCE were in the vicinity of PCC-FS, which means that PCC-FS, CC, and CCE are comparable to each other. Moreover, except for HOMER and LP, the remaining seven algorithms were significantly ($p = 0.05$) different from a random ranking by chance. In this sense, we obtained an overview of the group of ideal algorithms and their significance level. Statistical tests and confidence intervals will further quantify the differences.\n\n#### 5.2. F-Test for All Algorithms\n\nIn this work, we also conducted a Friedman test to analyze performance among the compared algorithms. Table 9 provided the Friedman statistics $F F$ and the corresponding critical value in terms of each evaluation criterion. As shown in Table 9, the null hypothesis (that all of the compared algorithms will perform equivalently) was clearly rejected for each evaluation criterion at a significance level of $α = 0.05$. Consequently, we then proceeded to conduct a post-hoc test in order to analyze the relative performance among the compared algorithms.\nThe Nemenyi test was used to test whether each of the algorithms performed competitively against the other compared algorithms, where PCC-FS was included. Within the test, the performance between two classifiers was considered to be significantly different if the corresponding average ranks differed by at least the critical difference $CD = q α k ( k + 1 ) 6 N$. For the test, $q α$ is equal to 3.102 at the significance level $α = 0.05$, and thus $CD$ takes the value of $4.5409$ (k = 9, N = 7). Figure 4 shows the CD diagrams for each of the five evaluation criteria, with any compared algorithm whose average rank was within one CD to that of PCC-FS connected to it with a red line. Algorithms that were unconnected to PCC-FS were otherwise considered to have a significantly different performance between them. In Hamming Loss, for example, the average rank for PCC-FS was 2.14, and the critical value would be 6.68 by adding CD. Since LP and Rank-SVM got 6.71 and 7.57 for their respect average rankings, they were classified as worse algorithms. However, we could not distinguish the performance of the remaining algorithms from PCC-FS. Since the F-test is based on all algorithms, we will further consider PCC_FS as a control algorithm and make a pairwise comparison in Section 5.3.\n\n#### 5.3. PCC-FS as Control Algorithm\n\nIn order to increase the power of the test, we also considered PCC-FS as a control algorithm and compared it against all other algorithms. For this, we used Bonferroni correction for controlling the family-wise error or the probability of making at least one Type 1 error in multiple hypothesis tests. The comparison was made by examining the critical difference, $C D B o n$, while considering the Bonferroni correction that is conservative in that the critical value for $q α$ becomes 2.724 when $α = 0.05$. The results for that the average ranking difference, $Δ ξ = ξ ¯ o t h e r − ξ ¯ P C C − F S$, is larger $C D B o n$ are marked by “√” in Table 10. It has already been seen in Section 5.2 that $Δ ξ > 0$ for all cases. Empty cells indicate that $Δ ξ$ was within $C D B o n$. From Table 10, we can see that PCC-FS outperforms five algorithms (HOMER, LP, Rank-SVM, BP_MLL, and LLSF-DL) under at least one evaluation criterion.\n\n#### 5.4. Confidence Intervals\n\nConfidence interval can further imply how much better it performs when PCC-FS is compared with other algorithms. To quantify the difference, we constructed the intervals for all eight comparisons. The normality assumption was made on the ranking differences:\n$Δ ξ k ( k + 1 ) / 6 N ~ N ( 0 , 1 ) .$\nUnder the 95% confidence level, we show the intervals for each algorithm in Figure 5. Furthermore, five criteria were grouped. Among the five worse algorithms, all intervals for Rank-SVM seemed to be greater than 0, which indicated a significant difference compared to PCC-FS. The extreme upper bound was close to 10 for One Error. For the remaining four, the majority of lower bounds was greater than or close to 0, while the overall upper bounds were slightly less than that of Rank-SVM. For the other three seemingly indifferent algorithms, four out of five intervals for CC and CCE presented positive values and three out of five for RAkEL. Even though some of the criteria indicated a negative lower bound for CC and CCE, the average values for lower bounds were positive. However, RAkEL seemed to be a comparable algorithm to PCC-FS.\n\n#### 5.5. Summaries\n\nBased on these experimental results, the following observations can be made:\n(1)\nThe proposed PCC-FS algorithm achieves the top average rank among nine algorithms across all five criteria, and the CC-based high-order algorithms (CC, CCE, PCC-FS) in general achieve better performance than the other algorithms. This is because these types of algorithms exploit the label couplings thoroughly.\n(2)\nFour out of five ranking differences for CC and CCE have shown positive intervals meaning that the probability of obtaining a higher rank for PCC-FS compared with CC or CCE is 80% for a given dataset even though they are comparable.\n(3)\nPCC-FS outperforms LP, HOMER, Rank-SVM, BP_MLL, and LLSF-DL because the ranking differences for them are significantly larger than the critical value across the five tested criteria. Corresponding confidence interval gives an overview of the quantified amount in ranking difference.\n(4)\nLP, HOMER, and Rank-SVM perform the worst on all of the five criteria because LP and HOMER transform MLC into one or more single label subproblems. Rank-SVM divides MLC into a series of pairwise classification problems and cannot be seen to describe label couplings very well.\n(5)\nRAkEL performs neutrally among the test algorithms.\n\n## 6. Conclusions\n\nThe MLC problem is an important research issue in the field of data mining, which has a wide range of applications in the real world. Exploring label couplings can improve the classification performance of the MLC problem. The CC algorithm is a well-known way to do this. It adopts a high-order strategy in order to explore label correlations, but it does have two obvious drawbacks. Aiming to address both problems at the same time, we proposed the PCC-FS algorithm, which extracts intra-couplings within label sets and inter-couplings between features and labels. In doing so, our new algorithm makes three major contributions to the MLC problem. First, it uses a new chain mechanism which only considers the coupled labels of each label and organizes them to train and predict data and thus improving prediction performance. Second, by integrating a novel feature selection method into the algorithm to exploit the coupling relationships between features and labels, PCC-FS is able to reduce the number of redundant features and improve classification performance. Third, extracting label couplings in the MLC problem based on the theory of coupling learning, including intra-couplings within labels and inter-couplings between features and labels, makes the exploration of label couplings more sufficient and comprehensive. Compared with other testing algorithms, PCC-FS has the best average ranking and CC-based algorithms are comparable. The analytical results given by multi-criteria decision-making and statistical test are consistent. Using confidence intervals for ranking differences further implies how much better PCC-FS has performed.\nIn the future, some effort will be required to improve and extend the proposed PCC-FS algorithm. First, in our tests, we only used the logistic regression method as the binary classifier. Any further work on the algorithm should investigate the performance of different basic binary classifiers more thoroughly. Second, more adaptive methods of threshold selection should be studied in order to enhance the accuracy and automation of the PCC-FS algorithm. Third, more normalization methods, for example, rank transformation, will be applied to normalize the feature values.\n\n## Author Contributions\n\nConceptualization, Z.W. and T.W.; methodology, Z.W. and M.H.; software, T.W.; formal analysis, Z.W. and M.H.; investigation, Z.W., B.W. and M.H.; data curation, B.W. and T.W.; writing—original draft preparation, Z.W.; writing—review and editing, M.H.; visualization, Z.W. and M.H.; supervision, B.W.; project administration, Z.W.; funding acquisition, Z.W. and B.W. All authors have read and agreed to the published version of the manuscript.\n\n## Funding\n\nThis research was funded by National Science Foundation of China, grant number 61302157 and 61363075; National High Technology Research and Development Program, grant number 2012AA12A308; Yue Qi Young Scholars Program of China University of Mining and Technology, Beijing, grant number 800015Z1117. The APC was funded by Yue Qi Young Scholars Program of China University of Mining and Technology, Beijing.\n\n## Conflicts of Interest\n\nThe authors declare no conflict of interest.\n\n## References\n\n1. Schapire, R.E.; Singer, Y. BoosTexter: A Boosting-based System for Text Categorization. Mach. Learn. 2000, 39, 135–168. [Google Scholar] [CrossRef][Green Version]\n2. Ñanculef, R.; Flaounas, I.; Cristianini, N. Efficient classification of multi-labeled text streams by clashing. Expert Syst. Appl. 2014, 41, 5431–5450. [Google Scholar] [CrossRef][Green Version]\n3. 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Parameters of the Gaussian fit are m = 66.81 s = 7.28. Probability levels 5% (XX1), Median (Med), and 95% (XX19) are also given.\nFigure 3. Evaluation of algorithms using sum of ranking differences. Scaled sum of ranking differences (SRD) values are plotted on x-axis and left y-axis, and right y-axis shows the relative frequencies (black curve). Parameters of the Gaussian fit are m = 66.81 s = 7.28. Probability levels 5% (XX1), Median (Med), and 95% (XX19) are also given.\nFigure 4. Comparison of PCC-FS (control algorithm) against other compared algorithms using the Nemenyi test.\nFigure 4. Comparison of PCC-FS (control algorithm) against other compared algorithms using the Nemenyi test.\nFigure 5. Confidence interval for ranking difference.\nFigure 5. Confidence interval for ranking difference.\nTable 1. Description of multi-label datasets.\nTable 1. Description of multi-label datasets.\nDatasetInstance NumberLabel NumberContinuous Feature NumberDiscrete Feature NumberDensityField\nemotions59367200.311Music\nCAL5005021746800.150Music\nyeast24171410300.303biology\nflags19471090.485Image\nscene2407629400.1790Image\nbirds6451925820.053Audio\nenron170253010010.064Text\nTable 2. Performance comparison of nine algorithms on the emotions dataset.\nTable 2. Performance comparison of nine algorithms on the emotions dataset.\nAlgorithmHamming LossRanking LossOne ErrorCoverageAverage Precision\nHOMER0.2525(5)0.3040(6)0.4032(5)2.5863(6)0.6897(6)\nLP0.2808(6)0.3393(7)0.4605(7)2.6669(7)0.6643(7)\nRAkEL0.2153(3)0.1891(4)0.3120(3)1.9464(4)0.7758(4)\nRank-SVM0.3713(9)0.4273(9)0.6154(9)3.0264(8)0.5714(9)\nBP_MLL0.3519(8)0.4143(8)0.5868(8)3.0759(9)0.5732(8)\nCC0.2171(4)0.1729(3)0.3124(4)1.8134(3)0.7852(3)\nCCE0.2064(2)0.1646(2)0.2719(2)1.7699(2)0.8001(2)\nLLSF-DL0.2893(7)0.2867(5)0.4273(6)2.3407(5)0.6936(5)\nPCC-FS0.2008(1)0.1503(1)0.2550(1)1.7174(1)0.8130(1)\nTable 3. Performance comparisons of nine algorithms on the CAL500 dataset.\nTable 3. Performance comparisons of nine algorithms on the CAL500 dataset.\nAlgorithmHamming LossRanking LossOne ErrorCoverageAverage Precision\nHOMER0.2104(8)0.4071(8)0.8565(8)169.3291(8)0.2609(8)\nLP0.1993(7)0.6559(9)0.9880(9)171.1590(9)0.1164(9)\nRAkEL0.1686(6)0.2870(7)0.3529(7)165.3036(7)0.4008(7)\nRank-SVM0.1376(2)0.1824(5)0.1156(2)129.2589(2)0.4986(4)\nBP_MLL0.1480(5)0.1815(4)0.1186(3)130.084(6)0.4967(5)\nCC0.1386(4)0.1814(3)0.1216(4)129.698(4)0.5025(3)\nCCE0.1377(3)0.1781(2)0.1255(5)129.851(5)0.5094(2)\nLLSF-DL0.2330(9)0.1985(6)0.3500(6)126.967(1)0.4463(6)\nPCC-FS0.1370(1)0.1766(1)0.1155(1)129.616(3)0.5125(1)\nTable 4. Performance comparisons of nine algorithms on the yeast dataset.\nTable 4. Performance comparisons of nine algorithms on the yeast dataset.\nAlgorithmHamming LossRanking LossOne ErrorCoverageAverage Precision\nHOMER0.2619(8)0.3287(8)0.2871(7)9.2457(8)0.6259(8)\nLP0.2768(9)0.3977(9)0.5143(9)9.3607(9)0.5733(9)\nRAkEL0.2270(6)0.2143(6)0.2946(8)7.5086(6)0.7144(6)\nRank-SVM0.2450(7)0.1928(5)0.2521(5)6.3554(3)0.7217(5)\nBP_MLL0.2112(5)0.1761(4)0.2429(4)6.5101(5)0.7473(4)\nCC0.1993(1)0.1699(3)0.2238(2)6.4200(4)0.7596(3)\nCCE0.2015(3)0.1679(2)0.2284(3)6.3475(2)0.7645(2)\nLLSF-DL0.2019(4)0.2585(7)0.2790(6)8.5881(7)0.7004(7)\nPCC-FS0.2013(2)0.1665(1)0.2156(1)6.3335(1)0.7666(1)\nTable 5. Performance comparisons of nine algorithms on the flags dataset.\nTable 5. Performance comparisons of nine algorithms on the flags dataset.\nAlgorithmHamming LossRanking LossOne ErrorCoverageAverage Precision\nHOMER0.2683(3)0.2895(7)0.3703(7)4.1221(7)0.7630(7)\nLP0.2962(6)0.5011(8)0.5587(8)4.9495(8)0.6407(8)\nRAkEL0.2428(1)0.2332(5)0.2255(3)3.7824(3)0.8118(2)\nRank-SVM0.5931(9)0.7052(9) 0.7618(9) 5.5303(9)0.4987(9)\nBP_MLL0.3225(8)0.2226(3)0.2288(4)3.8734(5)0.8028(5)\nCC0.2785(5)0.2136(2)0.2344(5)3.7400(1)0.8117(3)\nCCE0.2749(4)0.2252(4)0.2283(2)3.8513(4)0.8032(4)\nLLSF-DL0.2987(7)0.2786(6)0.2838(6)4.0921(6)0.7683(6)\nPCC-FS0.2619(2)0.2068(1)0.2135(1)3.7492(2)0.8214(1)\nTable 6. Performance comparisons of nine algorithms on the scene dataset.\nTable 6. Performance comparisons of nine algorithms on the scene dataset.\nAlgorithmHamming LossRanking LossOne ErrorCoverageAverage Precision\nHOMER0.1488(7)0.2345(9)0.4595(9)1.2685(9)0.6946(9)\nLP0.1439(6)0.2121(8)0.3984(7)1.1550(8)0.7308(8)\nRAkEL0.1014(4)0.0998(4)0.2672(4)0.5854(4)0.8378(4)\nRank-SVM0.1501(8)0.1039(5) 0.2863(5)0.6266(5)0.8275(5)\nBP_MLL0.1859(9)0.1383(6)0.4550(8)0.7725(6)0.7411(7)\nCC0.1137(5)0.0801(1)0.2439(2)0.4844(1)0.8556(1)\nCCE0.0949(2)0.0922(3)0.2447(3)0.5439(3)0.8495(3)\nLLSF-DL0.0998(3)0.1898(7)0.3482(6)1.0536(7)0.7579(6)\nPCC-FS0.0940(1)0.0872(2)0.2401(1)0.5193(2)0.8540(2)\nTable 7. Performance comparisons of nine algorithms on the birds dataset.\nTable 7. Performance comparisons of nine algorithms on the birds dataset.\nAlgorithmHamming LossRanking LossOne ErrorCoverageAverage Precision\nHOMER0.0641(4)0.2076(3)0.8231(6)5.2117(6)0.3806(5)\nLP0.0731(5)0.2733(7)0.9009(9)6.0743(8)0.2673(6)\nRAkEL0.0502(1)0.1523 (1)0.6957(4)4.0188(4)0.5271(3)\nRank-SVM0.1080(9)0.5359(9)0.8273(7)6.2380(9)0.2421(7)\nBP_MLL0.0587(3)0.4486(8)0.8637(8)5.2591(7)0.2023(9)\nCC0.0923(8)0.2487(6)0.4959(2)3.3094(2)0.5311(2)\nCCE0.0878(7)0.2521(5)0.5162(3)3.3771(3)0.5138(4)\nLLSF-DL0.0536(2)0.1809(2)0.8229(5)4.2128(5)0.2239(8)\nPCC-FS0.0827(6)0.2265(4)0.4457(1)3.1203(1)0.5646(1)\nTable 8. Performance comparisons of nine algorithms on the enron dataset.\nTable 8. Performance comparisons of nine algorithms on the enron dataset.\nAlgorithmHamming LossRanking LossOne ErrorCoverageAverage Precision\nHOMER0.0606(7)0.2471(7)0.4918(7)28.0953(7)0.5067(7)\nLP0.0707(8)0.5480(8)0.8144(8)39.5441(8)0.2246(8)\nRAkEL0.0484(5)0.2011(6)0.2774(2)25.2163(6)0.6156(6)\nRank-SVM0.0737(9)0.6256(9)0.8854(9)45.4454(9)0.1471(9)\nBP_MLL0.0553(6)0.0716(1)0.2679(1)11.2064(1)0.6847(1)\nCC0.0481(4)0.0781(2)0.3099(4)11.8147(2)0.6817(3)\nCCE0.0480(3)0.0809(4)0.3264(6)12.0097(4)0.6720(4)\nLLSF-DL0.0454(1)0.1545(5)0.2954(3)20.9599(5)0.6408(5)\nPCC-FS0.0474(2)0.0784(3)0.3152(5)11.8254(3)0.6832(2)\nTable 9. Summary of the Friedman Statistics $F F$ $( k = 9 , N = 7 )$ and the critical value in terms of each evaluation criterion (k: #comparing algorithms; N: #datasets).\nTable 9. Summary of the Friedman Statistics $F F$ $( k = 9 , N = 7 )$ and the critical value in terms of each evaluation criterion (k: #comparing algorithms; N: #datasets).\nEvaluation Criteria$F F$\nHamming Loss4.25582.1382\nRanking Loss8.9239\nOne-Error7.8679\nCoverage9.6106\nAverage Precision13.6875\nTable 10. Comparisons between PCC-FS and other algorithms.\nTable 10. Comparisons between PCC-FS and other algorithms.\nHamming LossRanking LossOne ErrorCoverageAverage Precision\nHOMER\nLP\nRAkEL\nRank-SVM\nBP_MLL\nCC\nCCE\nLLSF-DL\n\n## Share and Cite\n\nMDPI and ACS Style\n\nWang, Z.; Wang, T.; Wan, B.; Han, M. Partial Classifier Chains with Feature Selection by Exploiting Label Correlation in Multi-Label Classification. Entropy 2020, 22, 1143. https://doi.org/10.3390/e22101143\n\nAMA Style\n\nWang Z, Wang T, Wan B, Han M. Partial Classifier Chains with Feature Selection by Exploiting Label Correlation in Multi-Label Classification. Entropy. 2020; 22(10):1143. https://doi.org/10.3390/e22101143\n\nChicago/Turabian Style\n\nWang, Zhenwu, Tielin Wang, Benting Wan, and Mengjie Han. 2020. \"Partial Classifier Chains with Feature Selection by Exploiting Label Correlation in Multi-Label Classification\" Entropy 22, no. 10: 1143. https://doi.org/10.3390/e22101143\n\nNote that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here." ]	[ null, "https://px.ads.linkedin.com/collect/", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87290865,"math_prob":0.9415253,"size":60728,"snap":"2023-14-2023-23","text_gpt3_token_len":14930,"char_repetition_ratio":0.16061194,"word_repetition_ratio":0.097427905,"special_character_ratio":0.25767356,"punctuation_ratio":0.16842192,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97634774,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-07T09:15:26Z\",\"WARC-Record-ID\":\"<urn:uuid:447c528e-c533-4d91-bb22-60359fb83c73>\",\"Content-Length\":\"429592\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f8d887f1-9ace-403e-90d0-706f7d673b4f>\",\"WARC-Concurrent-To\":\"<urn:uuid:00901f06-8047-4821-b796-cbb9d0be00de>\",\"WARC-IP-Address\":\"104.18.24.151\",\"WARC-Target-URI\":\"https://www.mdpi.com/1099-4300/22/10/1143\",\"WARC-Payload-Digest\":\"sha1:XBC2ELWYZYROSUQ5UT4XBLM2FJEQQBCE\",\"WARC-Block-Digest\":\"sha1:GKFQJOJZ3JILBVIMOSXUTR2ZOVKPAOGC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224653631.71_warc_CC-MAIN-20230607074914-20230607104914-00007.warc.gz\"}"}
https://www.education.com/resources/division-facts/?page=3	[ "# Division Facts\n\n137 filtered results\n137 filtered results\nDivision Facts\nSort by\nMultiplication & Division: Picnicking Signs #2\nWorksheet\nMultiplication & Division: Picnicking Signs #2\nKids complete each equation on this third grade math worksheet by determining whether an equation is multiplication or division and writing in the correct sign.\nMath\nWorksheet\nTransportation Fact Families\nWorksheet\nTransportation Fact Families\nThird graders are learning all about about multiplication and division fact families, and this transporation-themed worksheet gives them lots of practice.\nMath\nWorksheet\nSolve for the Unknown\nWorksheet\nSolve for the Unknown\nChildren learn the steps for solving division problems with multiplication using a variety of strategies in this third grade math worksheet.\nMath\nWorksheet\nDivision: Finding the Quotient!\nWorksheet\nDivision: Finding the Quotient!\nThis third grade math worksheet explains division and lets kids try their hand at a few division problems. It also defines division vocabulary words.\nMath\nWorksheet\nFun With Division\nWorksheet\nFun With Division\nMake math practice fun with this division sheet! As she solves the problems, your child will uncover a hidden phrase at the bottom.\nMath\nWorksheet\nMultiplication & Division: Mushroom Math\nWorksheet\nMultiplication & Division: Mushroom Math\nKids multiply or divide to complete each equation on this third grade math worksheet.\nMath\nWorksheet\nInverse Equations: Multiplication\nWorksheet\nInverse Equations: Multiplication\nDid you know some math equations are related? Your student can review her times tables while learning about inverse relationships.\nMath\nWorksheet\nWorksheet\nMastering division facts is essential for every fourth grade student. Review the basics with your fourth grader on this fun worksheet.\nMath\nWorksheet\nSea Creatures Word Problems: Multi-step Addition and Division\nWorksheet\nSea Creatures Word Problems: Multi-step Addition and Division\nThis underwater-themed worksheet is a playful way for learners to practice solving multi-step word problems.\nMath\nWorksheet\nMath Tiles Mash-up\nWorksheet\nMath Tiles Mash-up\nChildren solve a series of problems to trace a path from start to finish in this fun math puzzle worksheet.\nMath\nWorksheet\nSt. Patrick's Day Division #5\nWorksheet\nSt. Patrick's Day Division #5\nCan your little leprechaun collect all of these lucky clovers? Give him a fun challenge this St. Patrick's Day, with some division problems to solve.\nMath\nWorksheet\nMultiplication and Division: Which One Doesn’t Belong?\nWorksheet\nMultiplication and Division: Which One Doesn’t Belong?\nUse this worksheet to get your students thinking about the relationship between numbers in regards to multiplication and division. They’ll look at four numbers and determine which does not belong.\nMath\nWorksheet\nInverse Equations: Division\nWorksheet\nInverse Equations: Division\nGuide your third grader in learning the inverse relationship between division and multiplication with a few practice problems.\nMath\nWorksheet\nMultiplication and Division: How Do the Numbers Relate?\nWorksheet\nMultiplication and Division: How Do the Numbers Relate?\nIn this worksheet, students will be asked to identify multiplication and division facts.\nMath\nWorksheet\nDivision: Skip-Count to the Quotient\nWorksheet\nDivision: Skip-Count to the Quotient\nPractice using number lines to solve division problems with this matching activity.\nMath\nWorksheet\nGuess the Operation\nLesson Plan\nGuess the Operation\nWant to strengthen your student’s multiplication and division skills? This lesson uses fact families to help students practice those skills.\nMath\nLesson Plan\nDivision Practice Sheet\nWorksheet\nDivision Practice Sheet\nDoes your child need help with division? He'll practice both one and two digit division with both vertical and linear equations.\nMath\nWorksheet\nDo You Know Your Math Facts?\nLesson Plan\nDo You Know Your Math Facts?\nIn this lesson, your students will create their own math facts and practice them! They will understand that multiplication and division are opposites.\nMath\nLesson Plan\nSt. Patrick's Day Division #1\nWorksheet\nSt. Patrick's Day Division #1\nCollect all of these lucky clovers by solving each simple division problem! Your little math whiz can get a great review of his times tables.\nMath\nWorksheet\nDonut Division Problems!\nWorksheet\nDonut Division Problems!\nDo some donut division with this math worksheet. Kids will solve division problems alongside a batch of tasty donuts.\nMath\nWorksheet\nBaseball Division\nWorksheet\nBaseball Division\nTake a swing at math practice with some baseball division! Here are 12 simple division problems that will help your child review his times tables.\nMath\nWorksheet\nDivision Practice Sheet #2\nWorksheet\nDivision Practice Sheet #2\nDivision daredevils, try your hand at these simple division problems! Your child will get to review his times tables with some inverse operations.\nMath\nWorksheet\nMusical Chair Multiplication\nLesson Plan\nMusical Chair Multiplication\nGet your students moving while practicing their multiplication facts! Bring an old favorite to the classroom as students move about and work problems out. This is a fun way to incorporate movement and learning with a twist.\nMath\nLesson Plan\nDivision Word Problem Cards\nWorksheet\nDivision Word Problem Cards\nUse these word problem cards to give your students practice with division!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8061023,"math_prob":0.6250151,"size":8201,"snap":"2021-21-2021-25","text_gpt3_token_len":1906,"char_repetition_ratio":0.23057216,"word_repetition_ratio":0.17045455,"special_character_ratio":0.17827094,"punctuation_ratio":0.062547095,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9959197,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-11T04:23:38Z\",\"WARC-Record-ID\":\"<urn:uuid:08a6d997-eee4-4d1f-b2d8-5937c56cefe1>\",\"Content-Length\":\"145399\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f0b14431-8612-416b-9289-d242094e2d0a>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c8b45d4-1b95-4edc-91ff-0f495481b651>\",\"WARC-IP-Address\":\"151.101.201.185\",\"WARC-Target-URI\":\"https://www.education.com/resources/division-facts/?page=3\",\"WARC-Payload-Digest\":\"sha1:OFU7SIYV2KITECL7QTKYZHFL42SP63VJ\",\"WARC-Block-Digest\":\"sha1:ZGC2XW2XYT6QLVQC3H3YUOGNPVSFMGEI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991641.5_warc_CC-MAIN-20210511025739-20210511055739-00193.warc.gz\"}"}
https://www.icp.az/2021/1-06.php	[ "A.S. Shukurov.\nAsymptotic results for a semi-Markov process describing the behavior of somestochastic systems\n\nA semi-Markov process with discrete interference of events with one screen, which describes the behavior of a stochastic system with partial recovery of the resource, is investigated. A semi-Markov process is constructed and asymptotic formulas are found for the mathematical expectation and variance of the first moment when the semi-Markov process enters the zero state.\n\nKeywords: Semi-Markov process, Partial recovery, Recovery rate, Asymptotic methods, Stochastic model" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87297213,"math_prob":0.7791004,"size":1152,"snap":"2022-05-2022-21","text_gpt3_token_len":235,"char_repetition_ratio":0.16550523,"word_repetition_ratio":0.96103895,"special_character_ratio":0.16319445,"punctuation_ratio":0.110526316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9784634,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-18T04:34:50Z\",\"WARC-Record-ID\":\"<urn:uuid:ce5913ab-0fc9-4113-bd45-7d031201e208>\",\"Content-Length\":\"15139\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9d3d2a13-99e1-44e6-ac96-7f80d2bcb604>\",\"WARC-Concurrent-To\":\"<urn:uuid:e4151bab-a3d4-40fc-9c0c-3a93a8f3ec86>\",\"WARC-IP-Address\":\"104.21.81.215\",\"WARC-Target-URI\":\"https://www.icp.az/2021/1-06.php\",\"WARC-Payload-Digest\":\"sha1:SLSWQR2XQRXB3I7G253UQ3TNCDOU4MBD\",\"WARC-Block-Digest\":\"sha1:5NN5NMUNCMUPLEFE4A3AS3RBFOJEXFHI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320300722.91_warc_CC-MAIN-20220118032342-20220118062342-00103.warc.gz\"}"}
https://www.transtutors.com/questions/reaction-rate-constant-and-ca-is-the-concentration-of-a-in-the-reaction-tank-at-t-0--1201915.htm	[ "# reaction rate constant, and CA is the concentration of A in the reaction tank. At t = 0, a v L/s aqu", null, "reaction rate constant, and CA is the concentration of A in the reaction tank. At t = 0, a v L/s aqueous stream of CA0 mol A/L solution is pumped into a V-L tank, initially filled with water. The solution leaves the tank at v L/s and at a concentration of CA mol A/L. The density of the solution in the tank is the same as water, and does not change upon reaction.\n\nShow transcribed image text A plant is starting up the production of compound B by utilizing the reaction A right arrow 28 The reaction rate is given by rA = kCA where rA is the reaction rate (rate of consumption of A), k is the reaction rate constant, and CA is the concentration of A in the reaction tank. At t = 0, a v L/s aqueous stream of CA0 mol A/L solution is pumped into a V-L tank, initially filled with water. The solution leaves the tank at v L/s and at a concentration of CA mol A/L. The density of the solution in the tank is the same as water, and does not change upon reaction. Write a differential balance for the total mass within the tank and solve the balance for the change in volume of solution in the tank with time. After simplifying, what is dV/dt? NOTE: For this question, if you need to write an equation using the number e, type the expression exp in the equation editor. Click on the hint panel for more help. Write a differential balance for compound A in terms of v, CA0, k, CA, and V and solve for dCA/dt. Integrate the above expression to obtain an expression for CA as a function of time. (Your answer should be in terms of v, CAO, V, k, and t.) What is the steady-state concentration of A leaving the tank, if CAO = 1.20 mol A/L, V 2200.0 L, k = 0.300 s^-1, and v = 2.40 L/s? What is the steady-state concentration of B leaving the tank under these same conditions?", null, "## Plagiarism Checker\n\nSubmit your documents and get free Plagiarism report\n\nFree Plagiarism Checker" ]	[ null, "https://files.transtutors.com/questions/transtutors004/images/transtutors004_fe338e16-191f-41e0-961f-d213bc2a9bb4.png", null, "https://files.transtutors.com/resources/images/answer-blur.webp", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.918779,"math_prob":0.98686373,"size":659,"snap":"2021-21-2021-25","text_gpt3_token_len":160,"char_repetition_ratio":0.12519084,"word_repetition_ratio":0.0,"special_character_ratio":0.2215478,"punctuation_ratio":0.092857145,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9913753,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-11T17:33:09Z\",\"WARC-Record-ID\":\"<urn:uuid:4aea7787-4dca-4e57-bce8-2f888e9e3ccc>\",\"Content-Length\":\"80028\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:de89e4c3-7b32-4861-b341-945b71e41938>\",\"WARC-Concurrent-To\":\"<urn:uuid:9f770737-efe0-45f0-9059-93b029313925>\",\"WARC-IP-Address\":\"34.224.249.39\",\"WARC-Target-URI\":\"https://www.transtutors.com/questions/reaction-rate-constant-and-ca-is-the-concentration-of-a-in-the-reaction-tank-at-t-0--1201915.htm\",\"WARC-Payload-Digest\":\"sha1:ASKTY2PJDRTEIFSWO3NAA3XWR4DIAZFP\",\"WARC-Block-Digest\":\"sha1:X2GQMVL376INXJL3WPLZFVSSCGZLY5BT\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991648.10_warc_CC-MAIN-20210511153555-20210511183555-00010.warc.gz\"}"}
https://www.education.com/lesson-plan/skip-counting-by-5s/	[ "Guided Lessons\npremium\n\n# Roll the Dice: Skip Counting by Fives\n\nFive, ten, fifteen... Help your students practice their multiplication skills by teaching them to skip count by fives. In this lesson, they will use dice to practice math!\nNeed extra help for EL students? Try the Skip Counting with Visuals pre-lesson.\nEL Adjustments\nGrade Subject View aligned standards\n\nNo standards associated with this content.\n\nNo standards associated with this content.\n\nWhich set of standards are you looking for?\n\nNeed extra help for EL students? Try the Skip Counting with Visuals pre-lesson.\n\nStudents will be able to multiply single digit numbers by five. Students will be able to skip count.\n\nThe adjustment to the whole group lesson is a modification to differentiate for children who are English learners.\nEL adjustments\n(5 minutes)\n• Begin the lesson by asking your students if anyone has a birthday that ends in a 0 or 5.\n• Explain to your class that they will learn about multiplying by 5 through skip counting.\n(15 minutes)\n• Display a blank Hundreds Chart, and remind your students that multiplication is repeated addition.\n• Display a Hundreds Chart with the 5s and 10s column highlighted, and show them the visual pattern made by the numbers that end in fives and zeros.\n• Point to the first number on the highlighted hundreds chart.\n• Model how to skip count by 5s to the number fifty while pointing to the hundreds chart.\n• Point out that the digit in the ones place alternates between 5 and 0.\n• Explain to students that the first number, five, represents how many times the number has been multiplied, which is 1.\n• Explain to students that the equation for this problem would be 5 x 1 = 5.\n• Ask students to create another equation based on the next highlighted square.\n• Ask one volunteer to skip count by five on the chart and produce the number equation.\n(5 minutes)\n• Show the students the blank Hundreds Chart again and instruct them to take out their whiteboard and whiteboard marker.\n• Give your students each one dice, and instruct them to roll the dice once. Instruct them to multiply the number displayed on the dice by five.\n• Have them say the equations quietly, draw the equation by showing the numbers in groups, and write the equation.\n• Once they’ve finished four examples, have them switch with a peer.\n(5 minutes)\n• Distribute a copy of the Skip Counting: Roll, Skip, Draw, and Write worksheet to each learner.\n• Instruct them to continue rolling the dice and practicing independently. Encourage them to do this without the Hundreds Chart in front of them.\n\nEnrichment:\n\n• Give your students two dice, and instruct them to add together the numbers on the dice to multiply by five\n\nSupport:\n\n• Instruct them to work on the Count by Fives or the Dot to Dot Zoo worksheet with the help of the grid.\n(5 minutes)\n• Gauge how your students are doing as you walk around the room, jotting notes about areas of struggle and success. Observe whether or not they are using the chart to multiply.\n• Instruct students to take out a whiteboard and whiteboard marker. Roll the dice and show students the number. Have them write the equation on their whiteboard and find the answer.\n• Ask students to display their answers, and note which students show proficiency and which require reteaching.\n• Display an example solution with the equation and a drawing on the board. Put a mistake in the drawing and ask students to identify the error and how to fix it.\n(5 minutes)\n• Gather the students together, and call out single digit numbers,\n• Have a volunteer come up to the board to write the equation out.\n• Instruct another student to solve and write the answer next to the equation.\n\n### Add to collection\n\nCreate new collection\n\n0\n\n0 items" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9198382,"math_prob":0.8707196,"size":3369,"snap":"2021-04-2021-17","text_gpt3_token_len":730,"char_repetition_ratio":0.15364042,"word_repetition_ratio":0.01369863,"special_character_ratio":0.20926091,"punctuation_ratio":0.08945687,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97547376,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-22T03:52:49Z\",\"WARC-Record-ID\":\"<urn:uuid:25b2b3b9-5fe3-4293-92ef-93ebbe397102>\",\"Content-Length\":\"200040\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b7688a1e-f3d8-4e92-bc4e-f0da29038350>\",\"WARC-Concurrent-To\":\"<urn:uuid:7961670c-3217-453a-aae9-c703aab0f577>\",\"WARC-IP-Address\":\"151.101.201.185\",\"WARC-Target-URI\":\"https://www.education.com/lesson-plan/skip-counting-by-5s/\",\"WARC-Payload-Digest\":\"sha1:EYL6JJPG7ZRBAUM4YXP4HNJ7W7O372EB\",\"WARC-Block-Digest\":\"sha1:2BAVN6NI4D5FJCSPJ6EBQQX46UTQHP5D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039560245.87_warc_CC-MAIN-20210422013104-20210422043104-00019.warc.gz\"}"}
https://spoonless.livejournal.com/191878.html	[ "No account? Create an account\n\n## wandering sets, part 6: quantum mechanics and the role of the observer", null, "First, a quick comment about something pretty basic that I completely forgot to mention in part 4 on the \"Boltzmann's brain\" paradox: why do they call it Boltzmann's brain? Seems obvious I should have explained this, but it's because the type of dilemna raised by it is similar to the \"brain in a vat\" thought experiment that philosophers of metaphysics like to argue about. The basic question is depicted explicitly in the movie The Matrix: how do you know that you aren't just a brain in a vat somewhere, with wires plugged into you, feeding this brain electrical impulses that it interprets as sensory experiences? I think for the most part, the answer is: you don't. I mentioned that the Boltzmann's brain paradox could involve the vacuum fluxuation of a spontaneously generated galaxy, or solar system, or even just a single room. But the ultimate limit of this would be, just a single brain in a vat, with no actual world around it at all. Anyway, just wanted to add that to make part 4 more clear, because you may have been wondering why it was called \"Boltzmann's brain\". If I ever decided to make these notes into a book, I guess this would be one of the chapters. Now on to the matter at hand...\n\nOne of the most puzzling things about quantum mechanics, especially when you first learn it, is why there appear to be two completely different types of rules for how physics works. One is the microscopic set of rules which physicists usually refer to as \"unitary time evolution of the wavefunction\". In quantum mechanics, what's called the wavefunction is similar to a probability distribution either in regular space or momentum space (not phase space, the combination of the two) for which state a particle (or field, or string) could be in. (Actually, it's more like the square root of a probability, but no matter.) While it is similar to a probability distribution in regular space or momentum space, it can be much more simply represented as a single vector in a much larger space called a Hilbert space. The Hilbert space in quantum mechanics is usually infinite dimensional, so much much bigger than even the 6N dimensional phase space we talked about in part 5. As the system progresses further into the future, this state vector traces out a single path through the Hilbert space, and that path is always exactly reversible according to these microscopic rules. The key word here is \"unitary\". The vector is moved around in the Hilbert space mathematically by applying a unitary matrix to it. (Heisenberg's formulation of quantum mechanics was orginally called \"matrix mechanics\" because of this.) Unitary matrices are matrices whose transpose is equal to their complex conjugate (replacing all imaginary components with real components and vice versa, where by imaginary and real I'm talking about the mathematical notion of i, the square root of -1). If this doesn't make any sense, don't worry about it, the only important thing to understand is that this property of unitarity guarantees many nice things about time evolution in quantum mechanics. It makes things nice and smooth, so that a single state always moves to a single state, and if the total probability of the particle being anywhere is initially 100% then it will stay 100% in the future. But most importantly, it guarantees reversibility. Because the time evolution matrix in quantum mechanics is unitary, it means Louiville's theorem holds and you can always reverse time and go backwards exactly to where the system came from.\n\nBut wait--if this is the case, then that also means that all quantum mechanical systems are conservative, ie non-dissipative. Does dissipation not happen in quantum mechanics at all? This brings us to the second type of rule in quantum mechanics: the process of measurement. Originally, this rule was called the \"collapse of the wavefunction\". Because when looked at through Schrodinger's wave mechanics, it appears that when a macroscopic observer makes a measurement in the lab, of a property of a microscopic system (for instance if he asks the question \"where is this particle actually located?\") the rule is that what used to be a probability distribution over many different states--suddenly that distribution is reduced to, or collapses to, a single state. Before the measurement, we mathematically describe the particle as being in a \"superposition\" of many different positions at once, but after the measurement it is only found at one of those positions. Mathematically, this is achieved with a projection matrix. A projection matrix is very different from a unitary matrix. Its action in the Hilbert space is that it maps many vectors onto a single vector, instead of mapping one vector onto a single vector. Because of this, the action is irreversible, and does not satisfy Liouville's theorem. The measurement process is therefore dissipative rather than conservative. In other words, the rule for how an observer of a microscopic system makes measurements in quantum mechanics seems completely opposite to the rule for how microscopic systems evolve in time when they are not being observed. One is nice and smooth and reversible, the other is a sudden reduction, an irreversible collapse. Dissipation only seems to happen during observation.\n\nBut the way I've explained this highlights the paradox, called the \"measurement problem in quantum mechanics\". This paradox is what has spawned many different interpretations of quantum mechanics, which philosophers still argue fiercely about today. But the real question is how to reconcile reversibility with irreversibility, and the answer lies in thermodynamics. And once you understand it, you realize that it doesn't really make a whole lot of sense to talk about measurement in this way, as a sudden \"collapse\" of the wavefunction. And it leads to deeper questions about whether the wave function is real or just a mathematical abstraction--and if there is anything in quantum mechanics which can be said to be real at all. To be continued...\n\n#### Tags:", null, "" ]	[ null, "https://l-userpic.livejournal.com/109317127/845823", null, "https://xc3.services.livejournal.com/ljcounter/", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9527329,"math_prob":0.8279332,"size":4823,"snap":"2019-13-2019-22","text_gpt3_token_len":949,"char_repetition_ratio":0.11869682,"word_repetition_ratio":0.007751938,"special_character_ratio":0.1895086,"punctuation_ratio":0.09175377,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9602862,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-23T18:50:02Z\",\"WARC-Record-ID\":\"<urn:uuid:96a8c8fc-573e-4b9c-ac65-901932020307>\",\"Content-Length\":\"204748\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:74b016f5-b599-4a43-9658-dbebb3e64cbf>\",\"WARC-Concurrent-To\":\"<urn:uuid:ff15cfbd-ed4d-4614-817a-2921e00d7087>\",\"WARC-IP-Address\":\"13.88.179.33\",\"WARC-Target-URI\":\"https://spoonless.livejournal.com/191878.html\",\"WARC-Payload-Digest\":\"sha1:7YNSWZY4B3EVMJIIMKPNUODLDG2DA6N3\",\"WARC-Block-Digest\":\"sha1:KU3VQGDEJJGEKQNFEITRICTH6UIIUHNW\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202924.93_warc_CC-MAIN-20190323181713-20190323203713-00196.warc.gz\"}"}