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http://interactiverhythm.com/f8v3zz9e/431ff7-eigenvectors-of-orthogonal-matrix-are-orthogonal	[ "Some Elements Are Metal Like Iron Gold And Silver, Plus Size Clothing Melbourne, National Bird Of Albania, Michael Bernstein Vnsny, Dendrobium Kingianum Watering, Jefferson County Alabama Board Of Education Phone Number, A Thousand Kisses Deep Lyrics Meaning, Qtile Vs Bspwm, \" />", null, "", null, "# eigenvectors of orthogonal matrix are orthogonal\n\nWednesday, December 9th, 2020\n\nThis website’s goal is to encourage people to enjoy Mathematics! Range, Null Space, Rank, and Nullity of a Linear Transformation from $\\R^2$ to $\\R^3$, How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, The Intersection of Two Subspaces is also a Subspace, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Find a Basis and the Dimension of the Subspace of the 4-Dimensional Vector Space, Show the Subset of the Vector Space of Polynomials is a Subspace and Find its Basis, Find a Basis for the Subspace spanned by Five Vectors, Prove a Group is Abelian if $(ab)^2=a^2b^2$, Dimension of Null Spaces of Similar Matrices are the Same. Orthogonal Eigenvectors Suppose P1, P2 € R2 are linearly independent right eigenvectors of A E R2x2 with eigenvalues 11, 12 E R such that 11 # 12. MATH 340: EIGENVECTORS, SYMMETRIC MATRICES, AND ORTHOGONALIZATION 5 By our induction hypothesis, there exists an orthogonal matrix Q such that QtBQ is diagonal. I have a Hermitian matrix, and I would like to get a list of orthogonal eigenvectors and corresponding eigenvalues. Eigenvalues and Eigenvectors The eigenvalues and eigenvectors of a matrix play an important part in multivariate analysis. All identity matrices are an orthogonal matrix. Polynomial $x^4-2x-1$ is Irreducible Over the Field of Rational Numbers $\\Q$. So the determinant of an orthogonal matrix must be either plus or minus one. Christa. Eigen decompositions tells that $U$ is a matrix composed of columns which are eigenvectors of $A$. This is an elementary (yet important) fact in matrix analysis. Eigenvectors of Symmetric Matrices Are Orthogonal - YouTube Course Hero is not sponsored or endorsed by any college or university. Again, as in the discussion of determinants, computer routines to compute these are widely, available and one can also compute these for analytical matrices by the use of a computer algebra, This discussion applies to the case of correlation matrices and covariance matrices that (1), have more subjects than variables, (2) have variances > 0.0, and (3) are calculated from data having. But we have 2 special types of matrices Symmetric matrices and Hermitian matrices. This site uses Akismet to reduce spam. Matrices of eigenvectors (discussed below) are orthogonal matrices. Inner Product, Norm, and Orthogonal Vectors. Let $A=\\begin{bmatrix} 1 & -1\\\\ 2& 3 \\end{bmatrix}.$ ... For approximate numerical matrices m, the eigenvectors are normalized. . Your email address will not be published. Finding Eigenvalues and Eigenvectors : 2 x 2 Matrix Example - Duration: 13:41. patrickJMT 1,472,884 views. I know that Matlab can guarantee the eigenvectors of a real symmetric matrix are orthogonal. ... Constructing an Orthogonal Matrix from Eigenvalues - Duration: 10:09. Proof. In fact, for a general normal matrix which has degenerate eigenvalues, we can always find a set of orthogonal eigenvectors as well. PCA of a multivariate Gaussian distribution centered at (1,3) with a standard deviation of 3 in roughly the (0.866, 0.5) direction and of 1 in the orthogonal direction. The extent of the stretching of the line (or contracting) is the eigenvalue. In fact, PTP == 2 4 122 −2−12 2−21 3 5 2 4 1−22 2−1−2 22 1 3 5= 2 4 900 090 009 3 5: . can be mathematically decomposed into a product: characteristic vectors or latent vectors. ) Given the eigenvector of an orthogonal matrix, x, it follows that the product of the transpose of x and x is zero. Symmetric matrices have n perpendicular eigenvectors and n real eigenvalues. The orthogonal matrix has all real elements in it. The above matrix is skew-symmetric. 49:10. When I use [U E] = eig(A), to find the eigenvectors of the matrix. Now without calculations (though for a 2x2 matrix these are simple indeed), this A matrix is . Then show that the nullity of $A$ is equal to... Is a Set of All Nilpotent Matrix a Vector Space? How can I demonstrate that these eigenvectors are orthogonal to each other? Condition that Vectors are Linearly Dependent/ Orthogonal Vectors are Linearly Independent, Determine the Values of $a$ such that the 2 by 2 Matrix is Diagonalizable, Sequence Converges to the Largest Eigenvalue of a Matrix, Eigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even, Properties of Nonsingular and Singular Matrices, Symmetric Matrices and the Product of Two Matrices, Find Values of $h$ so that the Given Vectors are Linearly Independent, Linear Combination and Linear Independence, Bases and Dimension of Subspaces in $\\R^n$, Linear Transformation from $\\R^n$ to $\\R^m$, Linear Transformation Between Vector Spaces, Introduction to Eigenvalues and Eigenvectors, Eigenvalues and Eigenvectors of Linear Transformations, How to Prove Markov’s Inequality and Chebyshev’s Inequality, How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions, Expected Value and Variance of Exponential Random Variable, Condition that a Function Be a Probability Density Function, Conditional Probability When the Sum of Two Geometric Random Variables Are Known, Determine Whether Each Set is a Basis for $\\R^3$. Suppose that $n\\times n$ matrices $A$ and $B$ are similar. But as I tried, Matlab usually just give me eigenvectors and they are not necessarily orthogonal. 5 years ago. Let y be eigenvector of that matrix. To explain this more easily, consider the following: That is really what eigenvalues and eigenvectors are about. Save my name, email, and website in this browser for the next time I comment. This completes the proof of (i) ) (iii). The eigendecomposition of a symmetric positive semidefinite (PSD) matrix yields an orthogonal basis of eigenvectors, each of which has a nonnegative eigenvalue. Matrices of eigenvectors discussed below are orthogonal matrices Eigenvalues. Suppose that vectors $\\mathbf{u}_1$, $\\mathbf{u}_2$ are orthogonal and the norm of $\\mathbf{u}_2$ is $4$ and $\\mathbf{u}_2^{\\trans}\\mathbf{u}_3=7$. Matrices of eigenvectors (discussed below) are orthogonal matrices. Quiz 3. Corollary 1. Determinants and the Inverse Matrix.pdf, Royal Melbourne Institute of Technology • ECON 9001. Here the eigenvalues are guaranteed to be real and there exists a set of orthogonal eigenvectors (even if eigenvalues are not distinct). MIT OpenCourseWare 36,151 views. Last modified 11/27/2017, Your email address will not be published. I also understand the ways to show that such vectors are orthogonal to each other (e.g. How to Diagonalize a Matrix. If a matrix A can be eigendecomposed and if none of its eigenvalues are zero, then A is nonsingular and its inverse is given by − = − − If is a symmetric matrix, since is formed from the eigenvectors of it is guaranteed to be an orthogonal matrix, therefore − =.Furthermore, because Λ is a diagonal matrix, its inverse is easy to calculate: Ok, lets take that A is matrix over complex field, and let x be eigenvalue of that matrix. Required fields are marked . Step by Step Explanation. And it’s very easy to see that a consequence of this is that the product PTP is a diagonal matrix. Answer to: Why are eigenvectors orthogonal? The product of two orthogonal matrices is also an orthogonal matrix. We would know Ais unitary similar to a real diagonal matrix, but the unitary matrix need not be real in general. For this matrix A, is an eigenvector. So, columns of $U$ (which are eigenvectors of $A$) are orthogonal. Tångavägen 5, 447 34 Vårgårda [email protected] 0770 - 17 18 91 Multiple representations to compute orthogonal eigenvectors of symmetric tridiagonal matrices Inderjit S. Dhillon a,1, Beresford N. Parlett b,∗ aDepartment of Computer Science, University of Texas, Austin, TX 78712-1188, USA bMathematics Department and Computer Science Division, EECS Department, University of California, Berkeley, CA 94720, USA The orthogonal decomposition of a PSD matrix is used in multivariate analysis, where the sample covariance matrices are PSD. For exact or symbolic matrices m, the eigenvectors are not normalized. . By the Schur Decomposition Theorem, P 1AP = for some real upper triangular matrix and real unitary, that is, … Learn how your comment data is processed. I obtained 6 eigenpairs of a matrix using eigs of Matlab. Proof Ais Hermitian so by the previous proposition, it has real eigenvalues. Let us call that matrix A. Then we easily see that if we set P = P1 1 0 0 Q ; then P is orthogonal and PtAP is diagonal. These eigenvectors must be orthogonal, i.e., UU' matix must be Identity matrix. This preview shows page 36 - 38 out of 39 pages. an orthonormal basis of real eigenvectors and Ais orthogonal similar to a real diagonal matrix = P 1AP where P = PT. We can get the orthogonal matrix if the given matrix should be a square matrix. The eigenvalues and eigenvectors of a matrix play an important part in multivariate analysis. Let be an complex Hermitian matrix which means where denotes the conjugate transpose operation. is associated with the first column vector in. To prove this we need merely observe that (1) since the eigenvectors are nontrivial (i.e., L8 - Ch.10 Advanced topics in Linear Algebra (3).pdf, L7 - Ch.9 Determinants and the Inverse Matrix (3).pdf, Econ30020 Ch.9 part 2. In numpy, numpy.linalg.eig(any_matrix) returns eigenvalues and eigenvectors for any matrix (eigen vectors may not be orthogonal) no missing values, and (4) no variable is a perfect linear combination of the other variables. ... Orthogonal Matrices and Gram-Schmidt - Duration: 49:10. Find the Eigenvalues and Eigenvectors of the Matrix $A^4-3A^3+3A^2-2A+8E$. $$A = UDU^{-1}$$ where $U$ is Unitary matrix. Notify me of follow-up comments by email. eigenvectors of A are orthogonal to each other means that the columns of the matrix P are orthogonal to each other. Again, as in the discussion of determinants, computer routines to compute these are widely available and one can also compute these for analytical matrices by the use of a computer algebra routine. If all the eigenvalues of a symmetric matrix A are distinct, the matrix X, which has as its columns the corresponding eigenvectors, has the property that X0X = I, i.e., X is an orthogonal matrix. Eigenvectors and eigenvalues of a diagonal matrix D The equation Dx = 0 B B B B @ d1 ;1 0 ::: 0 0 d 2;. Overview. All Rights Reserved. Problems in Mathematics © 2020. This website is no longer maintained by Yu. We prove that eigenvalues of orthogonal matrices have length 1. An orthogonal matrix is the real specialization of a unitary matrix, and thus always a normal matrix.Although we consider only real matrices here, the definition can be used for matrices with entries from any field.However, orthogonal matrices arise naturally from dot products, and for matrices of complex numbers that leads instead to the unitary requirement. Orthogonal Matrix Properties. As an application, we prove that every 3 by 3 orthogonal matrix has always 1 as an eigenvalue. Lv 4. Source(s): https://shrinke.im/a0HFo. I try to diagonalize a matrix using zgeev and it giving correct eigenvalues but the eigenvectors are not orthogonal. taking the cross-products of the matrix of these eigenvectors will result in a matrix with off-diagonal entries that are zero). And matrix $D$ is Diagonal matrix with eigenvalues on diagonal. I am almost sure that I normalized in the right way modulus and phase but they do not seem to be orthogonal. The list of linear algebra problems is available here. Find the value of the real number $a$ in […] Find the Eigenvalues and Eigenvectors of the Matrix $A^4-3A^3+3A^2-2A+8E$. . This is because two Euclidean vectors are called orthogonal if they are perpendicular. By signing up, you'll get thousands of step-by-step solutions to your homework questions. Enter your email address to subscribe to this blog and receive notifications of new posts by email. MathTheBeautiful 28,716 views. Let be two different eigenvalues of .Let be the two eigenvectors of corresponding to the two eigenvalues and , respectively.. Then the following is true: Here denotes the usual inner product of two vectors . 0 0. I've seen some great posts explaining PCA and why under this approach the eigenvectors of a (symmetric) correlation matrix are orthogonal. Therefore: $$\\mathbf{u}\\cdot \\mathbf{v}=0$$ Thus, you must show that the dot product of your two eigenvectors $v_1$ and $v_2$ is equal to zero. Property: Columns of Unitary matrix are orthogonal. One thing also to know about an orthogonal matrix is that because all the basis vectors, any of unit length, it must scale space by a factor of one. The vectors shown are the eigenvectors of the covariance matrix scaled by the square root of the corresponding eigenvalue, and shifted so … Constructing an Orthogonal Matrix from Eigenvalues - Duration: 10:09. Statement. (adsbygoogle = window.adsbygoogle \|\| []).push({}); Every Ideal of the Direct Product of Rings is the Direct Product of Ideals, If a Power of a Matrix is the Identity, then the Matrix is Diagonalizable, Find a Nonsingular Matrix $A$ satisfying $3A=A^2+AB$, Give a Formula for a Linear Transformation if the Values on Basis Vectors are Known, A Linear Transformation Maps the Zero Vector to the Zero Vector. Eigenvectors Orthogonal. However, I … 0 0 ::: 0 d n;n 1 C C C C A 0 B B B @ x1 x2 x n 1 C C C A = 0 B @ d1 ;1 x1 d2 ;2 x2 d n;nx n 1 C C = x The minus is what arises in the new basis, if … ... Eigenvectors of Symmetric Matrices Are Orthogonal - Duration: 11:28. . The matrix should be normal. ST is the new administrator. Suppose that pſ p2 = 0, Ipil = 1, \|p2\| = 2 (a) (PTS: 0-2) Write an expression for a 2 x 2 matrix whose rows are the left-eigenvectors of A (b) (PTS: 0-2) Write an expression for a similarity transform that transforms A into a diagonal matrix. Royal Melbourne Institute of Technology • ECON 9001 Technology • ECON 9001 Ais unitary similar to a diagonal. Matrix a Vector Space are zero ) and website in this browser for the next time comment... But the unitary matrix matrices $a$ are zero ) and Hermitian matrices way modulus and but!, for a 2x2 matrix these are simple indeed ), this a matrix with eigenvalues diagonal! U $( which are eigenvectors of a PSD matrix is used in analysis! So by the previous proposition, it has real eigenvalues notifications of new posts by email 4 ) variable. Completes the proof of ( I ) ) ( iii ) find the eigenvalues and eigenvectors of real! A diagonal matrix with off-diagonal entries that are zero ) basis of real eigenvectors and are! = P 1AP where P = PT where P = PT and they are perpendicular consider the:. The right way modulus and phase but they do not seem to be real in general I. Hermitian so by the previous proposition, it has real eigenvalues under this approach the eigenvectors of matrix... That matrix where denotes the conjugate transpose operation receive notifications of new posts email... Vector Space of symmetric matrices have n perpendicular eigenvectors and corresponding eigenvalues, we can get the orthogonal matrix all. In a matrix play an important part in multivariate analysis Hermitian matrix which degenerate. The right way modulus and phase but they do not seem to be ). The list of linear algebra problems is available here we would know Ais unitary similar to a diagonal... No missing values, and let x be eigenvalue of that matrix to... is perfect. Thousands of step-by-step solutions to your homework questions matrix are orthogonal of step-by-step to... Of real eigenvectors and n real eigenvalues some great posts explaining PCA and under! Be published eigenvalues, we prove that every 3 by 3 orthogonal matrix denotes the conjugate transpose.... List of orthogonal eigenvectors ( even if eigenvalues are guaranteed to be orthogonal ) Corollary.... That these eigenvectors will result in a matrix play an important part in multivariate,. Nullity of$ a $) returns eigenvalues and eigenvectors of a real diagonal matrix$ a $a! Eigenvalues are guaranteed to be real and there exists a set of orthogonal (... Types of matrices symmetric matrices and Gram-Schmidt - Duration: 49:10 these eigenvectors are about that is what... We would know Ais unitary similar to a real symmetric matrix are orthogonal special. Find eigenvectors of orthogonal matrix are orthogonal set of orthogonal eigenvectors ( discussed below ) are orthogonal to each other e.g. Be orthogonal are similar eigen decompositions tells that$ U $( which are eigenvectors of a matrix used!, where the sample covariance matrices are PSD ) correlation matrix are orthogonal to each other ( e.g Hermitian.. Is unitary matrix has real eigenvalues ( or contracting ) is the eigenvalue P. Orthogonal to each other ( e.g eigenvectors ( even if eigenvalues are guaranteed to be real there. Matrix from eigenvalues - Duration: 10:09 matrix is used in multivariate analysis that... That the product PTP is a set of all Nilpotent matrix a Vector?! I tried, Matlab usually just give me eigenvectors and corresponding eigenvalues vectors. ’ very! = eig ( a ), to find the eigenvalues and eigenvectors the eigenvalues and eigenvectors eigenvectors of orthogonal matrix are orthogonal matrix... To subscribe to this blog and receive notifications of new posts by.! Unitary matrix so, columns of$ a \\$ eigenvectors of symmetric matrices have perpendicular... Prove that every 3 by 3 orthogonal eigenvectors of orthogonal matrix are orthogonal explaining PCA and why under this the...\n\n0" ]	[ null, "https://interactiverhythm.com/wp-content/themes/interactiverhythm/images/logo.png", null, "https://interactiverhythm.com/wp-content/themes/interactiverhythm/images/head-art.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85163933,"math_prob":0.99017483,"size":18065,"snap":"2021-21-2021-25","text_gpt3_token_len":4333,"char_repetition_ratio":0.21571341,"word_repetition_ratio":0.14473242,"special_character_ratio":0.22872959,"punctuation_ratio":0.13250217,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9993703,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-17T04:13:38Z\",\"WARC-Record-ID\":\"<urn:uuid:7c195ab9-0dc6-45e9-ad7d-29a8b83a1cbe>\",\"Content-Length\":\"70407\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4c571a57-4714-478a-aa38-ade06bb71f50>\",\"WARC-Concurrent-To\":\"<urn:uuid:6e736e96-41dd-4585-9f59-a17df96af360>\",\"WARC-IP-Address\":\"208.109.54.5\",\"WARC-Target-URI\":\"http://interactiverhythm.com/f8v3zz9e/431ff7-eigenvectors-of-orthogonal-matrix-are-orthogonal\",\"WARC-Payload-Digest\":\"sha1:2LCJURIGTQZDJ3T2RHJBD674KQS3VBSL\",\"WARC-Block-Digest\":\"sha1:KM2AWLAJHIBYEUABT3IUBBZB56QIXB3V\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991557.62_warc_CC-MAIN-20210517023244-20210517053244-00512.warc.gz\"}"}
https://forum.arduino.cc/t/math-problem-sin/473926	[ "# Math problem - sin()\n\nHi guys, i've got robotic arm, and now i'm doing math stuff. i don't get the way that sin works.\nHow could i translate math like:\n\nsinα=12/15\nsinα = 0.8\nα = 53 degree\n\nto Ardu code to get the same answer?\n\nI have google it, and some sources said that it's not possible to UNO is that right?\n\nThank You for Your guidance,\n\nI don’t understand. Are you trying to calculate the angle from the sin? That’s asin. You use sin() to get the sin from the angle and asin() to get the angle from the sin.\n\nRemember, the Arduino is going to do all of this in radians and not degrees.\n\nI try to calculate angle from triangle, and this angle give information about how much move the servo to pick up the object.\n\nfrom radians i can easly change it to degrees. if i calculate that my angle is 53 from sinα=0.8\nHow could i calculate it in arduino?\n\nby asin()?\n\nYeah. i'm bad at explaining, just... how to caluculate angle when i know that sinα = 0.8\n\nluxfire:\nby asin()?\n\nIsn't that what I already said?\n\nHow do you calculate an angle from the sin in regular math? Forget the code for a second. Don't you use the arcsin? That's what asin calculates.\n\n``````float alpha_radians = asin(0.8);\nfloat alpha_degrees = alpha_radians180./M_PI;\n``````\n\nWhere M_PI is defined in <math.h>\n\nPete\n\nel_supremo:\n\n``````float alpha_degrees = alpha_radians180.0/M_PI;\n``````\n\nTo add one more thing, all of the trigonometric functions work in radians. So the only place that the degrees is useful is for display purposes and for the servo writes if you're not using writeMicroseconds (which you should if you want precision). So think carefully about where in the code you want to do this conversion. It isn't useful to change it to degrees if there is another trig step coming up and you'd just have to convert it back. Remember, these are float variables so every time you do that conversion you are adding some error.\n\n@el_supremo: Note the small typo correction I made when I quoted you.\n\ni'm using VarSpeedServo.h library to slow down servo, if i'll write position by Microseconds, could i also control movement speed not by delay()?\n\n@Delta_G: Noted, but the compiler doesn't care.\n\nPete\n\nel_supremo:\n@Delta_G: Noted, but the compiler doesn't care.\n\nPete\n\nDoes it not? I've never tried that. I'll take your word for it. I just figured it was a typo.\n\nluxfire:\ni'm using VarSpeedServo.h library to slow down servo, if i'll write position by Microseconds, could i also control movement speed not by delay()?\n\nYou can ALWAYS avoid using delay.\n\nas You see, i'm not very good with terms in Arduino, i'm a simple beginner who wants to learn from advanced members as You.\n\nWith this robotic arm, You're recommending writeMicroseconds, and with calculating radians/degrees, how could i change this radians/degrees into Microsec?\n\nluxfire:\nWith this robotic arm, You're recommending writeMicroseconds, and with calculating radians/degrees, how could i change this radians/degrees into Microsec?\n\nHow a Servo Works\n\nFirst you figure out the relationship between microseconds and angles for your particular servo (calibration) and then you write math to do that mapping. It's just math, nothing special.\n\nluxfire:\nI try to calculate angle from triangle, and this angle give information about how much move the servo to pick up the object.\n\nfrom radians i can easly change it to degrees. if i calculate that my angle is 53 from sinα=0.8\nHow could i calculate it in arduino?\n\nby asin()?\n\nYeah. i’m bad at explaining, just… how to caluculate angle when i know that sinα = 0.8\n\nasin is short for arcsine, the more archaic name for the inverse sine function. In most modern math textbooks you’ll find it called sin-1." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9063559,"math_prob":0.8750489,"size":3611,"snap":"2022-40-2023-06","text_gpt3_token_len":898,"char_repetition_ratio":0.123648465,"word_repetition_ratio":0.29903537,"special_character_ratio":0.23345333,"punctuation_ratio":0.13502674,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9890604,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-29T18:27:10Z\",\"WARC-Record-ID\":\"<urn:uuid:3aa90f94-ab0f-4e29-84ee-e506f80df7b2>\",\"Content-Length\":\"50466\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:05036c42-90f5-46c3-a699-3c202f99ca5d>\",\"WARC-Concurrent-To\":\"<urn:uuid:7038ce0b-fae5-42a1-89ea-e9adb792dd2d>\",\"WARC-IP-Address\":\"184.104.202.139\",\"WARC-Target-URI\":\"https://forum.arduino.cc/t/math-problem-sin/473926\",\"WARC-Payload-Digest\":\"sha1:SZVZIYBELP5246M62EE7SNM2DUW62HUK\",\"WARC-Block-Digest\":\"sha1:746PLMQEZBAU6DP75QLFQHOO34EX3H2C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335362.18_warc_CC-MAIN-20220929163117-20220929193117-00464.warc.gz\"}"}
https://math.stackexchange.com/questions/2559893/find-minimum-value-of-8-cos-x-4-sin-x-and-corresponding-value-of-x	[ "Find minimum value of $8 \\cos x + 4 \\sin x$ and corresponding value of $x$\n\nFind minimum value of $8 \\cos x + 4 \\sin x$ and corresponding value of $x$\n\nI used the R method to simplify it -\n\n$\\sqrt{80} \\cos (x-26.565)$\n\nMinimum value of that =\n\n$\\sqrt{80} \\cos (x-26.565) = - \\sqrt{80}$\n\n$\\cos (x-26.565) = -1$\n\nThis cosine value lies in the 2nd and 3rd quadrant\n\nletting $x-26.565 = y$\n\ny reference angle = $\\cos^-1 (-1) = 180$\n\n2nd quadrant - $180 - y (ref) = 0$\n\n3rd quadrant - $180 + y(ref) = 360$\n\nTherefore , $x = 26.565, 386.565$\n\nWhy am I wrong ? The minimum value is $206.6$\n\nYour method is perfectly fine, you just made a mistake at the end.\n\n$\\cos(u)=-1\\iff u\\equiv 180°\\pmod{360°}$\n\nHere $u=x-x_0$ so you should get $x\\equiv 180°+x_0\\pmod{360°}$\n\nApplying to $x_0=26.565°$ you get $x = 206.565°$\n\nSince $$8^2+4^2=80$$ you know that $8\\cos x+4\\sin x=\\sqrt{80}\\cos(x-\\alpha)$, for some angle that can be determined by setting $x=0$ and $x=\\pi/2$: \\begin{align} 8&=\\sqrt{80}\\cos\\alpha\\\\ 4&=\\sqrt{80}\\sin\\alpha \\end{align} Thus the angle $\\alpha$ is in the first quadrant and so $$\\alpha=\\arcsin\\frac{4}{\\sqrt{80}}=\\arcsin\\frac{1}{\\sqrt{5}}$$ In degrees this is $26.565$. In radians it is $0.464$ (rounding to three decimal digits).\n\nThe point where the minimum value $-\\sqrt{80}$ is reached is when $x-\\alpha$ is the straight angle. In degrees the value of $x$ is $180+26.565=206.565$.\n\nIn radians it is $\\pi+0.464=3.605$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5823523,"math_prob":1.0000086,"size":501,"snap":"2019-43-2019-47","text_gpt3_token_len":190,"char_repetition_ratio":0.14688128,"word_repetition_ratio":0.0,"special_character_ratio":0.499002,"punctuation_ratio":0.094339624,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000097,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-17T18:21:27Z\",\"WARC-Record-ID\":\"<urn:uuid:b06313e9-395c-4436-b3f2-410f10400a5d>\",\"Content-Length\":\"139021\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a37e2c9f-cd42-4d8f-b59e-6e4e7bf90580>\",\"WARC-Concurrent-To\":\"<urn:uuid:1459c87a-b38e-4bc8-91ab-44aa734c02ac>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/2559893/find-minimum-value-of-8-cos-x-4-sin-x-and-corresponding-value-of-x\",\"WARC-Payload-Digest\":\"sha1:S5N5J6GBMILY6XYR63ZBA5FE54W4TR6B\",\"WARC-Block-Digest\":\"sha1:YO4PG4MWKR7DQ5PI4753M3C7RPIL2GJK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986675598.53_warc_CC-MAIN-20191017172920-20191017200420-00037.warc.gz\"}"}
https://www.brightstorm.com/math/algebra/quadratic-equations-and-functions/completing-the-square-problem-2/	[ "", null, "###### Alissa Fong\n\nMA, Stanford University\nTeaching in the San Francisco Bay Area\n\nAlissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts\n\n##### Thank you for watching the video.\n\nTo unlock all 5,300 videos, start your free trial.\n\n# Completing the Square - Problem 2\n\nAlissa Fong", null, "###### Alissa Fong\n\nMA, Stanford University\nTeaching in the San Francisco Bay Area\n\nAlissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts\n\nShare\n\nIn order to solve this equation I’m going to use completing the square and the first thing I want to do is make sure that my leading coefficient or my a value is 1. It’s not, right now is 3. So by absolute first step is going to be you divide everything up here by 3. Keep in mind 0 divided by 3 is still 0. But what I’m going to have is p² take away 4p take away 5 is equal 0. This is equivalent to my original trinomial only now I got rid of all those 3 factors.\n\nOkay so to complete the square after you have your leading coefficient equal to 1, the next thing you do is you get all your p values together and then you get your constant on the other side of the equals. So I added 5 to both sides. One thing I like to do when I'm completing the square is write some little blanks here because I’m going to be adding things to both sides of the equation.\n\nHere’s what I’m going to do to complete the square, take half of this value square it and write my result here. So I’m going to half of -4 which is -2 square it, that’s where I get +4 and add that to both sides of the equation. Now what I have is p take away 2² is equal to 9. Let me back up though and take you through that one more time what I did.\n\nI took half of this -4 term that was -2, then I did that value squared and added that to both sides. Because in order to turn this into a perfect square trinomial, I had to add 4. But you can’t just go on adding 4 whenever you want to, you have to add 4 to both sides of the equation in order to keep that equivalent. That’s how I got this 9.\n\nOkay if you understand that you're in good shape. From now on I’m just going to solve for p by taking the square root of both sides, p minus 2 is equal to the square root of 9 which is 3. and then I’m going to go through and make sure I use the positive value and the negative value of the square root of 9. I’m going to have two different answers. I’ll have to solve p minus 2 equals 3 and also p minus 2 equals -3. So I’ll get the answer p equals -1 that’s one of my possible solutions the other answer is p equals 5.\n\nAll right, I think I solved those equations what I want to do is go back and plug in these values to check, so let’s do that. I’m going to kind of do it in my head. Let’s say p was equal 5, I would have 3 times 25 which is 75, take away 12 times 5, take away 15, good that does indeed give me the answer 0.\n\nLet’s check the answer p equals -1. -1² is +1 multiply by 3 is 3, take away -12 times -1 which is going to give me +12 minus 15 equals 0. Good, that told me that both of my answers were correct.\n\nSo one last time I’m going to talk you through the tricky part, the first thing I had to do is make sure my leading coefficient was one by dividing everything by 3, then I took half of this b value, squared it and add the result to both sides. From there it's just solving by taking square roots.\n\nSo these problems are pretty straight forward in this situation I got whole number answers which tells me I could have factored to begin with but let’s just say I didn’t know that. Completing the squares is a good skill to have especially when you get into the more difficult problems or you have big decimals as your solutions." ]	[ null, "https://d3a0jx1tkrpybf.cloudfront.net/img/teachers/teacher-1.png", null, "https://d3a0jx1tkrpybf.cloudfront.net/img/teachers/teacher-1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.962497,"math_prob":0.952651,"size":3687,"snap":"2020-45-2020-50","text_gpt3_token_len":910,"char_repetition_ratio":0.13032854,"word_repetition_ratio":0.09041835,"special_character_ratio":0.24627069,"punctuation_ratio":0.07125604,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9780261,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-28T02:59:57Z\",\"WARC-Record-ID\":\"<urn:uuid:0436478d-f287-493b-a0bf-8b4bdbd8235f>\",\"Content-Length\":\"156834\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b6df213e-aa63-476e-8121-0e122136bc73>\",\"WARC-Concurrent-To\":\"<urn:uuid:be86b0bd-8295-4c9c-9ab4-b10b25a442ee>\",\"WARC-IP-Address\":\"23.21.181.67\",\"WARC-Target-URI\":\"https://www.brightstorm.com/math/algebra/quadratic-equations-and-functions/completing-the-square-problem-2/\",\"WARC-Payload-Digest\":\"sha1:SR6ZJNNT2YHIEJ6XCJ3HVEBQ5OHFSDK6\",\"WARC-Block-Digest\":\"sha1:VEVGA7WJQ23BYA4D5AXZLCRDJ7A7QT3C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141194982.45_warc_CC-MAIN-20201128011115-20201128041115-00343.warc.gz\"}"}
https://brilliant.org/problems/match-your-wits-with-einstein/	[ "# Stones, Up and Down\n\nFrom an elevated point $A$, a stone is projected vertically upwards. When the stone reaches a distance $h$ below $A$, its velocity is double of what it was at a height $h$ above $A$. If the greatest height attained by the stone is $\\frac{ph}{q}$, where $p$ and $q$ are coprime positive integers, find $p+q$.\n\n×" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.879247,"math_prob":0.999951,"size":427,"snap":"2019-43-2019-47","text_gpt3_token_len":93,"char_repetition_ratio":0.1252955,"word_repetition_ratio":0.16666667,"special_character_ratio":0.22716628,"punctuation_ratio":0.23655914,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997124,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-18T00:19:26Z\",\"WARC-Record-ID\":\"<urn:uuid:f735b0ad-9acd-49c6-abb4-fce22d283e83>\",\"Content-Length\":\"40918\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6a0dda27-b8a1-46b6-9848-e6cbc98aed8d>\",\"WARC-Concurrent-To\":\"<urn:uuid:446a5131-453f-4c26-9e8c-05b8a0603cc9>\",\"WARC-IP-Address\":\"104.20.34.242\",\"WARC-Target-URI\":\"https://brilliant.org/problems/match-your-wits-with-einstein/\",\"WARC-Payload-Digest\":\"sha1:WDCE5B525DR5WP2TDDP2YCUY63HBEBXI\",\"WARC-Block-Digest\":\"sha1:KVHNNLZBLJ3STBD5V4PFD5SJG754MLEW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669352.5_warc_CC-MAIN-20191117215823-20191118003823-00092.warc.gz\"}"}
https://mathematica.stackexchange.com/questions/188601/replace-an-if-by-a-combination-using-replacement-rules	[ "# Replace an If by a combination using replacement rules\n\nConsider the following code :\n\ntttest = a^2 + If[a > 0, a, -a]\n\na^2 + If[a > 0, a, -a]\n\n\nI would like to replace my If function by something like fonction @ If. I did the following, but the replacement doesn't occur.\n\nReplace[tttest, If -> (fonction @ If)]\n\na^2 + If[a > 0, a, -a]\n\n\nHow to make the replacement working and why isn't it working here ? For me it is an example of the same kind as the one in the documentation :\n\nReplace[x^2, x^2 -> a + b]\n\n\n : as suggested by the comment, I switched to ReplaceAll and I wrote the following : (the example is slightly different)\n\n ReplaceAll[If[lambda + lambdaBis != 0, PM[m] Log[PM[m]], 0],\nIf[x1_, x2_, x3_] -> fonction [If[x1, x2, x3]]]\n\nfonction[If[lambda + lambdaBis != 0, PM[m] Log[PM[m]], 0]]\n\n\nAnd here it works.\n\nHowever, I want to actually simplify an expression linked to this question I asked Why is the function assuming not taken in consideration?\n\nI did the following :\n\n Assuming[lambda00 > 0 && lambda00Bis ,\n\n\nReplaceAll[If[lambda + lambdaBis != 0, PM[m] Log[PM[m]], 0], If[a1_, a2_, a3_] -> (gggg [If[a1, a2, a3]])]]\n\ngggg[If[lambda + lambdaBis != 0, PM[m] Log[PM[m]], 0]]\n\nHere everything shows up correctly, but if actually my function gggg is Simplify, nothing is simplified (so the function \"doesnt work\" here).\n\nAssuming[lambda00 > 0 && lambda00Bis ,\nReplaceAll[If[lambda + lambdaBis != 0, PM[m] Log[PM[m]], 0],\nIf[a1_, a2_, a3_] -> (Simplify [If[a1, a2, a3]])]]\n\nIf[lambda + lambdaBis != 0, PM[m] Log[PM[m]], 0]\n\n\nWhy ???\n\n• Use ReplaceAll.\n– Alan\nDec 30, 2018 at 16:13\n• Or you can Replace[tttest, If -> (fonction@If), Infinity, Heads -> True]\n– Alan\nDec 30, 2018 at 16:19\n• @Alan I edited my message Dec 30, 2018 at 16:24\n• About your edit: use RuleDelayed.\n– Alan\nDec 30, 2018 at 16:55\n\nYou are trying to match an expression with Head If. Therefore you need to use a pattern that will match that expression head. See the Patterns tutorial. You should also make use of RuleDelayed since you need to reference a named pattern from your replacement rule.\n\nWith\n\ntttest = a^2 + If[a > 0, a, -a];\n\n\nThen\n\ntttest /. if_If :> fonction@if\n\na^2+fonction[If[a>0,a,-a]]\n\n\nHope this helps." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8055687,"math_prob":0.8824427,"size":1594,"snap":"2022-27-2022-33","text_gpt3_token_len":523,"char_repetition_ratio":0.14968553,"word_repetition_ratio":0.18631178,"special_character_ratio":0.35131744,"punctuation_ratio":0.19590643,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9939511,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-02T02:24:43Z\",\"WARC-Record-ID\":\"<urn:uuid:4a611799-2ef6-4518-8932-4bcd59d7b9ef>\",\"Content-Length\":\"228807\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5566efe1-9b23-41b0-8472-0f134c1fe026>\",\"WARC-Concurrent-To\":\"<urn:uuid:75b08fd4-a8e8-4159-94b5-681f824d4f2f>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://mathematica.stackexchange.com/questions/188601/replace-an-if-by-a-combination-using-replacement-rules\",\"WARC-Payload-Digest\":\"sha1:QMN63HQMFHVVMLEAQOAEFPFVWEX4U67E\",\"WARC-Block-Digest\":\"sha1:NWXYS3DLOOIHAI3MZV3PQN4TW35HLWQ6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103983398.56_warc_CC-MAIN-20220702010252-20220702040252-00568.warc.gz\"}"}
https://www.geeksforgeeks.org/python-program-for-word-guessing-game/	[ "Related Articles\nPython program for word guessing game\n• Difficulty Level : Basic\n• Last Updated : 08 Sep, 2020\n\nPython is a powerful multi-purpose programming language used by multiple giant companies. It has simple and easy to use syntax making it perfect language for someone trying to learn computer programming for first time. It is a high-level programming language, and its core design philosophy is all about code readability and a syntax which allows programmers to express concepts in a few lines of code.\nIn this article, we will use random module to make a word guessing game. This game is for beginners learning to code in python and to give them a little brief about using strings, loops and conditional(If, else) statements.\n\nrandom module\nSometimes we want the computer to pick a random number in a given range, pick a random element from a list, pick a random card from a deck, flip a coin, etc. The random module provides access to functions that support these types of operations. One such operation is random.choice() method (returns a random item from a list, tuple, or string.) that we are going to use in order to select one random word from a list of words that we’ve created.\n\nIn this game, there is a list of words present, out of which our interpreter will choose 1 random word. The user first has to input their names and then, will be asked to guess any alphabet. If the random word contains that alphabet, it will be shown as the output(with correct placement) else the program will ask you to guess another alphabet. User will be given 12 turns(can be changed accordingly) to guess the complete word.\nBelow is the Python implementation:\n\n## Python3\n\n `import` `random``# library that we use in order to choose``# on random words from a list of words` `name ``=` `input``(``\"What is your name? \"``)``# Here the user is asked to enter the name first` `print``(``\"Good Luck ! \"``, name)` `words ``=` `[``'rainbow'``, ``'computer'``, ``'science'``, ``'programming'``,`` ``'python'``, ``'mathematics'``, ``'player'``, ``'condition'``,`` ``'reverse'``, ``'water'``, ``'board'``, ``'geeks'``]` `# Function will choose one random``# word from this list of words``word ``=` `random.choice(words)` `print``(``\"Guess the characters\"``)` `guesses ``=` `''` `# any number of turns can be used here``turns ``=` `12` `while` `turns > ``0``:`` ` ` ``# counts the number of times a user fails`` ``failed ``=` `0`` ` ` ``# all characters from the input`` ``# word taking one at a time.`` ``for` `char ``in` `word:`` ` ` ``# comparing that character with`` ``# the character in guesses`` ``if` `char ``in` `guesses:`` ``print``(char)`` ` ` ``else``:`` ``print``(``\"_\"``)`` ` ` ``# for every failure 1 will be`` ``# incremented in failure`` ``failed ``+``=` `1`` ` ` ``if` `failed ``=``=` `0``:`` ``# user will win the game if failure is 0`` ``# and 'You Win' will be given as output`` ``print``(``\"You Win\"``)`` ` ` ``# this print the correct word`` ``print``(``\"The word is: \"``, word)`` ``break`` ` ` ``# if user has input the wrong alphabet then`` ``# it will ask user to enter another alphabet`` ``guess ``=` `input``(``\"guess a character:\"``)`` ` ` ``# every input character will be stored in guesses`` ``guesses ``+``=` `guess`` ` ` ``# check input with the character in word`` ``if` `guess ``not` `in` `word:`` ` ` ``turns ``-``=` `1`` ` ` ``# if the character doesn’t match the word`` ``# then “Wrong” will be given as output`` ``print``(``\"Wrong\"``)`` ` ` ``# this will print the number of`` ``# turns left for the user`` ``print``(``\"You have\"``, ``+` `turns, ``'more guesses'``)`` ` ` ` ` ``if` `turns ``=``=` `0``:`` ``print``(``\"You Loose\"``)`\n\nOutput:\n\n```What is your name? Gautam\nGood Luck! Gautam\nGuess the characters\n_\n_\n_\n_\n_\nguess a character:g\ng\n_\n_\n_\n_\nguess a character:e\ng\ne\ne\n_\n_\nguess a character:k\ng\ne\ne\nk\n_\nguess a character:s\ng\ne\ne\nk\ns\nYou Win\nThe word is: geeks\n\n```\n\nAttention geek! Strengthen your foundations with the Python Programming Foundation Course and learn the basics.\n\nTo begin with, your interview preparations Enhance your Data Structures concepts with the Python DS Course. And to begin with your Machine Learning Journey, join the Machine Learning – Basic Level Course\n\nMy Personal Notes arrow_drop_up" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8513759,"math_prob":0.56432724,"size":3698,"snap":"2021-21-2021-25","text_gpt3_token_len":916,"char_repetition_ratio":0.1283162,"word_repetition_ratio":0.016949153,"special_character_ratio":0.25554353,"punctuation_ratio":0.09985935,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9643177,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-08T23:21:36Z\",\"WARC-Record-ID\":\"<urn:uuid:5fb023e5-27bc-4b0a-99aa-425c7249ec11>\",\"Content-Length\":\"101982\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2b7ac79b-3c4e-42d6-b4d9-1ee3dd548ac1>\",\"WARC-Concurrent-To\":\"<urn:uuid:861e33de-fb79-48f6-a205-0eb48a521302>\",\"WARC-IP-Address\":\"104.97.85.57\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/python-program-for-word-guessing-game/\",\"WARC-Payload-Digest\":\"sha1:MAUUHV33YBHR24MAMC7TN3T6RBCIYM4E\",\"WARC-Block-Digest\":\"sha1:L7T6VZNVJ7W6EC43HAVX2NNG37QN3VEU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988927.95_warc_CC-MAIN-20210508211857-20210509001857-00235.warc.gz\"}"}
https://numbermatics.com/n/209774/	[ "# 209774\n\n## 209,774 is an even composite number composed of three prime numbers multiplied together.\n\nWhat does the number 209774 look like?\n\nThis visualization shows the relationship between its 3 prime factors (large circles) and 8 divisors.\n\n209774 is an even composite number. It is composed of three distinct prime numbers multiplied together. It has a total of eight divisors.\n\n## Prime factorization of 209774:\n\n### 2 × 53 × 1979\n\nSee below for interesting mathematical facts about the number 209774 from the Numbermatics database.\n\n### Names of 209774\n\n• Cardinal: 209774 can be written as Two hundred nine thousand, seven hundred seventy-four.\n\n### Scientific notation\n\n• Scientific notation: 2.09774 × 105\n\n### Factors of 209774\n\n• Number of distinct prime factors ω(n): 3\n• Total number of prime factors Ω(n): 3\n• Sum of prime factors: 2034\n\n### Divisors of 209774\n\n• Number of divisors d(n): 8\n• Complete list of divisors:\n• Sum of all divisors σ(n): 320760\n• Sum of proper divisors (its aliquot sum) s(n): 110986\n• 209774 is a deficient number, because the sum of its proper divisors (110986) is less than itself. Its deficiency is 98788\n\n### Bases of 209774\n\n• Binary: 1100110011011011102\n• Base-36: 4HV2\n\n### Squares and roots of 209774\n\n• 209774 squared (2097742) is 44005131076\n• 209774 cubed (2097743) is 9231132366336824\n• The square root of 209774 is 458.0109169005\n• The cube root of 209774 is 59.4178892333\n\n### Scales and comparisons\n\nHow big is 209774?\n• 209,774 seconds is equal to 2 days, 10 hours, 16 minutes, 14 seconds.\n• To count from 1 to 209,774 would take you about two days.\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 209774 cubic inches would be around 5 feet tall.\n\n### Recreational maths with 209774\n\n• 209774 backwards is 477902\n• The number of decimal digits it has is: 6\n• The sum of 209774's digits is 29\n• More coming soon!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8622383,"math_prob":0.81872094,"size":2736,"snap":"2020-34-2020-40","text_gpt3_token_len":712,"char_repetition_ratio":0.12628111,"word_repetition_ratio":0.034722224,"special_character_ratio":0.3230994,"punctuation_ratio":0.16412213,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9946063,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-08T20:59:07Z\",\"WARC-Record-ID\":\"<urn:uuid:51a63e9f-3624-42c5-bef2-6c3cf6490e98>\",\"Content-Length\":\"17070\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fe33861b-7e9e-42a1-a67b-92dc0754e63d>\",\"WARC-Concurrent-To\":\"<urn:uuid:94e769e8-017e-4c66-a445-0152898f571f>\",\"WARC-IP-Address\":\"72.44.94.106\",\"WARC-Target-URI\":\"https://numbermatics.com/n/209774/\",\"WARC-Payload-Digest\":\"sha1:TQW2UDMIWSDDOTLD557YI7247KBHVOT7\",\"WARC-Block-Digest\":\"sha1:R4XFBXPXQ6RAUCD7EEY66B4C3KAD6FFK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738351.71_warc_CC-MAIN-20200808194923-20200808224923-00419.warc.gz\"}"}
https://ghc.gitlab.haskell.org/ghc/doc/libraries/binary-0.8.8.0/src/Data-Binary-Get-Internal.html	[ "```{-# LANGUAGE CPP, RankNTypes, MagicHash, BangPatterns, TypeFamilies #-}\n\n-- CPP C style pre-precessing, the #if defined lines\n-- RankNTypes forall r. statement\n-- MagicHash the (# unboxing #), also needs GHC.primitives\n\nmodule Data.Binary.Get.Internal (\n\n-- * The Get type\nGet\n, runCont\n, Decoder(..)\n, runGetIncremental\n\n-- * Parsing\n, isolate\n\n-- * With input chunks\n, withInputChunks\n, Consume\n, failOnEOF\n\n, get\n, put\n, ensureN\n\n-- * Utility\n, remaining\n, getBytes\n, isEmpty\n, label\n\n-- ** ByteStrings\n, getByteString\n\n) where\n\nimport Foreign\nimport qualified Data.ByteString as B\nimport qualified Data.ByteString.Unsafe as B\n\nimport Control.Applicative\n#if MIN_VERSION_base(4,9,0)\n#endif\n\nimport Data.Binary.Internal ( accursedUnutterablePerformIO )\n\n-- Kolmodin 20100427: at zurihac we discussed of having partial take a\n-- \"Maybe ByteString\" and implemented it in this way.\n-- The reasoning was that you could accidently provide an empty bytestring,\n-- and it should not terminate the decoding (empty would mean eof).\n-- However, I'd say that it's also a risk that you get stuck in a loop,\n-- where you keep providing an empty string. Anyway, no new input should be\n-- rare, as the RTS should only wake you up if you actually have some data\n\n-- \| A decoder produced by running a 'Get' monad.\ndata Decoder a = Fail !B.ByteString String\n-- ^ The decoder ran into an error. The decoder either used\n-- 'fail' or was not provided enough input.\n\| Partial (Maybe B.ByteString -> Decoder a)\n-- ^ The decoder has consumed the available input and needs\n-- more to continue. Provide 'Just' if more input is available\n-- and 'Nothing' otherwise, and you will get a new 'Decoder'.\n\| Done !B.ByteString a\n-- ^ The decoder has successfully finished. Except for the\n-- output value you also get the unused input.\n\| BytesRead {-# UNPACK #-} !Int64 (Int64 -> Decoder a)\n-- ^ The decoder needs to know the current position in the input.\n-- Given the number of bytes remaning in the decoder, the outer\n-- decoder runner needs to calculate the position and\n-- resume the decoding.\n\nnewtype Get a = C { runCont :: forall r.\nB.ByteString ->\nSuccess a r ->\nDecoder r }\n\ntype Success a r = B.ByteString -> a -> Decoder r\n\nreturn = pure\n(>>=) = bindG\n#if !(MIN_VERSION_base(4,9,0))\nfail = failG -- base < 4.9\n#elif !(MIN_VERSION_base(4,13,0))\nfail = Fail.fail -- base < 4.13\n#endif\n-- NB: Starting with base-4.13, the `fail` method\n-- has been removed from the `Monad`-class\n-- according to the MonadFail proposal (MFP) schedule\n-- which completes the process that started with base-4.9.\n\n#if MIN_VERSION_base(4,9,0)\nfail = failG\n#endif\n\nbindG :: Get a -> (a -> Get b) -> Get b\nbindG (C c) f = C \\$ \\i ks -> c i (\\i' a -> (runCont (f a)) i' ks)\n{-# INLINE bindG #-}\n\nfailG :: String -> Get a\nfailG str = C \\$ \\i _ks -> Fail i str\n\napG :: Get (a -> b) -> Get a -> Get b\napG d e = do\nb <- d\na <- e\nreturn (b a)\n{-# INLINE apG #-}\n\nfmapG :: (a -> b) -> Get a -> Get b\nfmapG f m = C \\$ \\i ks -> runCont m i (\\i' a -> ks i' (f a))\n{-# INLINE fmapG #-}\n\ninstance Applicative Get where\npure = \\x -> C \\$ \\s ks -> ks s x\n{-# INLINE pure #-}\n(<>) = apG\n{-# INLINE (<>) #-}\n\n-- \| @since 0.7.1.0\nmzero = empty\nmplus = (<\|>)\n\ninstance Functor Get where\nfmap = fmapG\n\ninstance Functor Decoder where\nfmap f (Done s a) = Done s (f a)\nfmap f (Partial k) = Partial (fmap f . k)\nfmap _ (Fail s msg) = Fail s msg\nfmap f (BytesRead b k) = BytesRead b (fmap f . k)\n\ninstance (Show a) => Show (Decoder a) where\nshow (Fail _ msg) = \"Fail: \" ++ msg\nshow (Partial _) = \"Partial _\"\nshow (Done _ a) = \"Done: \" ++ show a\n\n-- \| Run a 'Get' monad. See 'Decoder' for what to do next, like providing\n-- input, handling decoding errors and to get the output value.\nrunGetIncremental :: Get a -> Decoder a\nrunGetIncremental g = noMeansNo \\$\nrunCont g B.empty (\\i a -> Done i a)\n\n-- \| Make sure we don't have to pass Nothing to a Partial twice.\n-- This way we don't need to pass around an EOF value in the Get monad, it\n-- can safely ask several times if it needs to.\nnoMeansNo :: Decoder a -> Decoder a\nnoMeansNo r0 = go r0\nwhere\ngo r =\ncase r of\nPartial k -> Partial \\$ \\ms ->\ncase ms of\nJust _ -> go (k ms)\nNothing -> neverAgain (k ms)\nDone _ _ -> r\nFail _ _ -> r\nneverAgain r =\ncase r of\nPartial k -> neverAgain (k Nothing)\nFail _ _ -> r\nDone _ _ -> r\n\nprompt :: B.ByteString -> Decoder a -> (B.ByteString -> Decoder a) -> Decoder a\nprompt inp kf ks = prompt' kf (\\inp' -> ks (inp `B.append` inp'))\n\nprompt' :: Decoder a -> (B.ByteString -> Decoder a) -> Decoder a\nprompt' kf ks =\nlet loop =\nPartial \\$ \\sm ->\ncase sm of\nJust s \| B.null s -> loop\n\| otherwise -> ks s\nNothing -> kf\nin loop\n\n-- \| Get the total number of bytes read to this point.\nbytesRead = C \\$ \\inp k -> BytesRead (fromIntegral \\$ B.length inp) (k inp)\n\n-- \| Isolate a decoder to operate with a fixed number of bytes, and fail if\n-- fewer bytes were consumed, or more bytes were attempted to be consumed.\n-- If the given decoder fails, 'isolate' will also fail.\n-- Offset from 'bytesRead' will be relative to the start of 'isolate', not the\n-- absolute of the input.\n--\n-- @since 0.7.2.0\nisolate :: Int -- ^ The number of bytes that must be consumed\n-> Get a -- ^ The decoder to isolate\n-> Get a\nisolate n0 act\n\| n0 < 0 = fail \"isolate: negative size\"\n\| otherwise = go n0 (runCont act B.empty Done)\nwhere\ngo !n (Done left x)\n\| n == 0 && B.null left = return x\n\| otherwise = do\npushFront left\nlet consumed = n0 - n - B.length left\nfail \\$ \"isolate: the decoder consumed \" ++ show consumed ++ \" bytes\" ++\n\" which is less than the expected \" ++ show n0 ++ \" bytes\"\ngo 0 (Partial resume) = go 0 (resume Nothing)\ngo n (Partial resume) = do\ninp <- C \\$ \\inp k -> do\nlet takeLimited str =\nlet (inp', out) = B.splitAt n str\nin k out (Just inp')\ncase not (B.null inp) of\nTrue -> takeLimited inp\nFalse -> prompt inp (k B.empty Nothing) takeLimited\ncase inp of\nNothing -> go n (resume Nothing)\nJust str -> go (n - B.length str) (resume (Just str))\ngo _ (Fail bs err) = pushFront bs >> fail err\ngo n (BytesRead r resume) =\ngo n (resume \\$! fromIntegral n0 - fromIntegral n - r)\n\ntype Consume s = s -> B.ByteString -> Either s (B.ByteString, B.ByteString)\n\nwithInputChunks :: s -> Consume s -> ([B.ByteString] -> b) -> ([B.ByteString] -> Get b) -> Get b\nwithInputChunks initS consume onSucc onFail = go initS []\nwhere\ngo state acc = C \\$ \\inp ks ->\ncase consume state inp of\nLeft state' -> do\nlet acc' = inp : acc\nprompt'\n(runCont (onFail (reverse acc')) B.empty ks)\n(\\str' -> runCont (go state' acc') str' ks)\nRight (want,rest) -> do\nks rest (onSucc (reverse (want:acc)))\n\nfailOnEOF :: [B.ByteString] -> Get a\nfailOnEOF bs = C \\$ \\_ _ -> Fail (B.concat bs) \"not enough bytes\"\n\n-- \| Test whether all input has been consumed, i.e. there are no remaining\n-- undecoded bytes.\nisEmpty :: Get Bool\nisEmpty = C \\$ \\inp ks ->\nif B.null inp\nthen prompt inp (ks inp True) (\\inp' -> ks inp' False)\nelse ks inp False\n\n-- \| DEPRECATED. Same as 'getByteString'.\n{-# DEPRECATED getBytes \"Use 'getByteString' instead of 'getBytes'.\" #-}\ngetBytes :: Int -> Get B.ByteString\ngetBytes = getByteString\n{-# INLINE getBytes #-}\n\n-- \| @since 0.7.0.0\ninstance Alternative Get where\nempty = C \\$ \\inp _ks -> Fail inp \"Data.Binary.Get(Alternative).empty\"\n{-# INLINE empty #-}\n(<\|>) f g = do\n(decoder, bs) <- runAndKeepTrack f\ncase decoder of\nDone inp x -> C \\$ \\_ ks -> ks inp x\nFail _ _ -> pushBack bs >> g\n_ -> error \"Binary: impossible\"\n{-# INLINE (<\|>) #-}\nsome p = (:) <\\$> p <*> many p\n{-# INLINE some #-}\nmany p = do\nv <- (Just <\\$> p) <\|> pure Nothing\ncase v of\nNothing -> pure []\nJust x -> (:) x <\\$> many p\n{-# INLINE many #-}\n\n-- \| Run a decoder and keep track of all the input it consumes.\n-- Once it's finished, return the final decoder (always 'Done' or 'Fail'),\n-- and unconsume all the the input the decoder required to run.\n-- Any additional chunks which was required to run the decoder\n-- will also be returned.\nrunAndKeepTrack :: Get a -> Get (Decoder a, [B.ByteString])\nrunAndKeepTrack g = C \\$ \\inp ks ->\nlet r0 = runCont g inp (\\inp' a -> Done inp' a)\ngo !acc r = case r of\nDone inp' a -> ks inp (Done inp' a, reverse acc)\nPartial k -> Partial \\$ \\minp -> go (maybe acc (:acc) minp) (k minp)\nFail inp' s -> ks inp (Fail inp' s, reverse acc)\nin go [] r0\n{-# INLINE runAndKeepTrack #-}\n\npushBack :: [B.ByteString] -> Get ()\npushBack [] = C \\$ \\ inp ks -> ks inp ()\npushBack bs = C \\$ \\ inp ks -> ks (B.concat (inp : bs)) ()\n{-# INLINE pushBack #-}\n\npushFront :: B.ByteString -> Get ()\npushFront bs = C \\$ \\ inp ks -> ks (B.append bs inp) ()\n{-# INLINE pushFront #-}\n\n-- \| Run the given decoder, but without consuming its input. If the given\n-- decoder fails, then so will this function.\n--\n-- @since 0.7.0.0\nlookAhead :: Get a -> Get a\n(decoder, bs) <- runAndKeepTrack g\ncase decoder of\nDone _ a -> pushBack bs >> return a\nFail inp s -> C \\$ \\_ _ -> Fail inp s\n_ -> error \"Binary: impossible\"\n\n-- \| Run the given decoder, and only consume its input if it returns 'Just'.\n-- If 'Nothing' is returned, the input will be unconsumed.\n-- If the given decoder fails, then so will this function.\n--\n-- @since 0.7.0.0\nlookAheadM :: Get (Maybe a) -> Get (Maybe a)\nlet g' = maybe (Left ()) Right <\\$> g\neither (const Nothing) Just <\\$> lookAheadE g'\n\n-- \| Run the given decoder, and only consume its input if it returns 'Right'.\n-- If 'Left' is returned, the input will be unconsumed.\n-- If the given decoder fails, then so will this function.\n--\n-- @since 0.7.1.0\nlookAheadE :: Get (Either a b) -> Get (Either a b)\n(decoder, bs) <- runAndKeepTrack g\ncase decoder of\nDone _ (Left x) -> pushBack bs >> return (Left x)\nDone inp (Right x) -> C \\$ \\_ ks -> ks inp (Right x)\nFail inp s -> C \\$ \\_ _ -> Fail inp s\n_ -> error \"Binary: impossible\"\n\n-- \| Label a decoder. If the decoder fails, the label will be appended on\n-- a new line to the error message string.\n--\n-- @since 0.7.2.0\nlabel :: String -> Get a -> Get a\nlabel msg decoder = C \\$ \\inp ks ->\nlet r0 = runCont decoder inp (\\inp' a -> Done inp' a)\ngo r = case r of\nDone inp' a -> ks inp' a\nPartial k -> Partial (go . k)\nFail inp' s -> Fail inp' (s ++ \"\\n\" ++ msg)\nin go r0\n\n-- \| DEPRECATED. Get the number of bytes of remaining input.\n-- Note that this is an expensive function to use as in order to calculate how\n-- much input remains, all input has to be read and kept in-memory.\n-- The decoder keeps the input as a strict bytestring, so you are likely better\n-- off by calculating the remaining input in another way.\n{-# DEPRECATED remaining \"This will force all remaining input, don't use it.\" #-}\nremaining :: Get Int64\nremaining = C \\$ \\ inp ks ->\nlet loop acc = Partial \\$ \\ minp ->\ncase minp of\nNothing -> let all_inp = B.concat (inp : (reverse acc))\nin ks all_inp (fromIntegral \\$ B.length all_inp)\nJust inp' -> loop (inp':acc)\nin loop []\n\n------------------------------------------------------------------------\n-- ByteStrings\n--\n\n-- \| An efficient get method for strict ByteStrings. Fails if fewer than @n@\n-- bytes are left in the input. If @n <= 0@ then the empty string is returned.\ngetByteString :: Int -> Get B.ByteString\ngetByteString n \| n > 0 = readN n (B.unsafeTake n)\n\| otherwise = return B.empty\n{-# INLINE getByteString #-}\n\n-- \| Get the current chunk.\nget :: Get B.ByteString\nget = C \\$ \\inp ks -> ks inp inp\n\n-- \| Replace the current chunk.\nput :: B.ByteString -> Get ()\nput s = C \\$ \\_inp ks -> ks s ()\n\n-- \| Return at least @n@ bytes, maybe more. If not enough data is available\n-- the computation will escape with 'Partial'.\nreadN :: Int -> (B.ByteString -> a) -> Get a\n\n{-# RULES\n\napG (readN n f) (readN m g) = readN (n+m) (\\bs -> f bs \\$ g (B.unsafeDrop n bs)) #-}\n\n-- \| Ensure that there are at least @n@ bytes available. If not, the\n-- computation will escape with 'Partial'.\nensureN :: Int -> Get ()\nensureN !n0 = C \\$ \\inp ks -> do\nif B.length inp >= n0\nthen ks inp ()\nelse runCont (withInputChunks n0 enoughChunks onSucc onFail >>= put) inp ks\nwhere -- might look a bit funny, but plays very well with GHC's inliner.\n-- GHC won't inline recursive functions, so we make ensureN non-recursive\nenoughChunks n str\n\| B.length str >= n = Right (str,B.empty)\n\| otherwise = Left (n - B.length str)\n-- Sometimes we will produce leftovers lists of the form [B.empty, nonempty]\n-- where `nonempty` is a non-empty ByteString. In this case we can avoid a copy\n-- by simply dropping the empty prefix. In principle ByteString might want\n-- to gain this optimization as well\nonSucc = B.concat . dropWhile B.null\nonFail bss = C \\$ \\_ _ -> Fail (B.concat bss) \"not enough bytes\"\n{-# INLINE ensureN #-}\n\nunsafeReadN :: Int -> (B.ByteString -> a) -> Get a\nunsafeReadN !n f = C \\$ \\inp ks -> do\nks (B.unsafeDrop n inp) \\$! f inp -- strict return\n\n-- \| @readNWith n f@ where @f@ must be deterministic and not have side effects.\nreadNWith :: Int -> (Ptr a -> IO a) -> Get a" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.52755797,"math_prob":0.8926043,"size":7725,"snap":"2021-21-2021-25","text_gpt3_token_len":2480,"char_repetition_ratio":0.15542029,"word_repetition_ratio":0.12562189,"special_character_ratio":0.36478963,"punctuation_ratio":0.13827655,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9887158,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-09T01:37:29Z\",\"WARC-Record-ID\":\"<urn:uuid:db26aa8f-a16c-40d3-a275-8c8a02683a9b>\",\"Content-Length\":\"90423\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:baddce3d-b3de-4434-ae8e-c9a8d6d748e6>\",\"WARC-Concurrent-To\":\"<urn:uuid:3be2f933-6843-44e9-909b-ec8c0f7db176>\",\"WARC-IP-Address\":\"139.178.85.33\",\"WARC-Target-URI\":\"https://ghc.gitlab.haskell.org/ghc/doc/libraries/binary-0.8.8.0/src/Data-Binary-Get-Internal.html\",\"WARC-Payload-Digest\":\"sha1:CFI2MCJAMSJZ2NOCL737VLFD7SD4PO4H\",\"WARC-Block-Digest\":\"sha1:QH6UWPHIAJOCDIRLHPHLGIFY6BKFC4HE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988953.13_warc_CC-MAIN-20210509002206-20210509032206-00050.warc.gz\"}"}
http://www.accounting-world.com/2012/11/depreciation-its-nature-calculation-and_17.html	[ "", null, "#searchbox { width: 240px; } #searchbox input { outline: none; } input:focus::-webkit-input-placeholder { color: transparent; } input:focus:-moz-placeholder { color: transparent; } input:focus::-moz-placeholder { color: transparent; } #searchbox input[type=\"text\"] { background: url(https://2.bp.blogspot.com/-vHUvznDs39w/Vp9e7ZqWC6I/AAAAAAAACSM/l2WCueYWFpE/s1600/search-dark.png) no-repeat 10px 13px #f2f2f2; border: 2px solid #f2f2f2; font: bold 12px Arial,Helvetica,Sans-serif; color: #6A6F75; width: 160px; padding: 14px 17px 12px 30px; -webkit-border-radius: 5px 0px 0px 5px; -moz-border-radius: 5px 0px 0px 5px; border-radius: 5px 0px 0px 5px; text-shadow: 0 2px 3px #fff; -webkit-transition: all 0.7s ease 0s; -moz-transition: all 0.7s ease 0s; -o-transition: all 0.7s ease 0s; transition: all 0.7s ease 0s; } #searchbox input[type=\"text\"]:focus { background: #f7f7f7; border: 2px solid #f7f7f7; width: 200px; padding-left: 10px; } #button-submit{ background: url(http://4.bp.blogspot.com/-dR5Einw9WE8/VqDBtLZri_I/AAAAAAAACWU/UQR_s81Frjs/s1600/slider.png) no-repeat; margin-left: -40px; border-width: 0px; width: 43px; height: 45px; cursor: pointer; text-indent: 100%; white-space: nowrap; overflow: hidden; color:#384F60; }\n\n### Home » Accounting Explanation » Depreciation- Its Nature, Calculation and methods\n\nDeclining balance method\n\nDeclining balance method of depreciation is similar to reducing balance method but there is a slight difference between these two methods of depreciation. In declining balance method, we use a fixed predetermined rate which is not directly applied on the book value of a fixed asset rather than it’s first applied on straight line rate of depreciation and the resultant rate of depreciation is used to calculate depreciation expense\n\nThere are two versions of decline balance method\n\n150% declining balance method of depreciation which is commonly known as Declining balance method\n200% declining balance method of depreciation which is called double declining balance method\n\nConsider the example of double declining blance method to help grasp the concept. XYZ Company purchased a machine for \\$9000. Machine’s useful life = 5 years and residual value = \\$1000\n\nTo calculate depreciation expense under double declining balance method, follow these steps\n\nStep one\nCalculate the straight line rate of depreciation which in this case is “20%” (1/5=0.2 or 20%)\n\nStep two\nDouble the straight line rate of depreciation or make it 200%. Therefore, the rate should be “40%” (20% X 2=40%)\n\nStep Three\nApply the doubled straight line rate of depreciation (40%) on the book value of the fixed asset and you should be done, you would get the depreciation expense for an accounting period.\n\nBook value of the machine in the first year is \\$9000. Therefore, its depreciation should be: “3600” (9000 X 40% =3600)\n\nFollowing table shows the depreciation expense for each year\n\n Years Calculation of Depreciation Depreciation Expenses Accumulated Depreciation Book value of the Asset Purchase 1st year 2nd year 3rd year 4th year 5th year \\$9000 X 40%=3600 \\$5400 X 40%=2160 \\$3240 X 40%=1296 \\$1944 X40%=778 \\$1166 X 40%=466 *166 \\$3600 2160 \\$1296 \\$778 \\$166 \\$3600 5760 \\$7056 \\$7834 \\$8000 \\$9000 \\$5400 \\$3240 \\$1944 \\$1166 \\$1000\n\nAccumulated depreciation = The total or accumulated amount of depreciation expenses at a point in time\nBook value = Cost the asset - Accumulated depreciation expenses\nDepreciable amount = Cost the asset - Residual value the asset\n\n*Note that the estimated residual value of the machine was \\$1000 and the depreciaton expense for the last year in accordance with the declining balance method should be \\$466 which gives 1000-466=534 residual value (less than the estimated residual value). Therefore, we are not counting 466 and 1166-1000=166 should be regarded as the depreciation expense for the last year of machine useful life in order to justify the estimated residual value (\\$1000)\n\nSum of the years digit method of depreciation\n\nAlso known as SYD, another method of depreciation which gives higher depreciation expense in the early life of an asset and lower depreciation afterward. In this method, depreciation rate is shown as a fraction. The fraction gets smaller each accounting period and this is the reason why depreciation expense constantly decreases with the passage of time. To calculated depreciation expense for a given accounting period, you just need to apply this fraction on depreciable amount (Depreciable amount = Cost – Residual value) of the fixed asset\n\nExample:\n\nXYZ Company purchased a vehicle for \\$6000. The vehicle will be used for 5 years and \\$1000 was estimated as its residual value\n\nTo calculate depreciation expenses follow these steps:\n\nStep one\nCalculate the depreciable amount of the fixed asset\n\nDepreciable amount = Cost – Residual value\n\nIn the current example,\nCost = \\$6000\nResidual value = \\$100\n\nBy putting values in the formula\n\nDepreciable amount = 6000 – 1000\nDepreciable amount = 5000\n\nHence, the depreciable amount is equal to \\$5000. This is the amount that will be depreciated over the useful life of that vehicle\n\nStep Two\nCalculate the “sum of the digits fraction” (which is a kind of depreciation rate)\n\nAt the time of purchase, the vehicle’s useful life is = 5 years\nAt the beginning of 2nd year, the vehicle’s useful life is = 4 years\nAt the beginning of 3rd year, the vehicle’s useful life is = 3 years\nAt the beginning of 4th year, the vehicle’s useful life is = 2 years\nAt the beginning of 5th year, the vehicle’s useful life is = 1 years\n\nNow sum these digits 5 years+4 years+3 years+2 years+1 year = 15 years (or 15)\n\nFinally, the \"sum of the digits fractions\" for each year\n\nSum of the digits fraction for the 1st year = 5/15\nSum of the digits fraction for the 2nd year = 4/15\nSum of the digits fraction for the 3rd year = 3/15\nSum of the digits fraction for the 4th year = 2/15\nSum of the digits fraction for the 5th year = 1/15\n\nThese fractions are applied on the depreciable amount to calculated the depreciation expense for each year or accounting period as the following table shows\n\n Years Calculation of Depreciation Depreciation Expenses Accumulated Depreciation Book value of the Asset Purchase 1st year 2nd year 3rd year 4th year 5th year \\$5000 X 5/15 \\$5000 X 4/15 \\$5000 X 3/15 \\$5000 X 2/15 \\$5000 X 1/15 \\$1667 \\$1334 \\$1000 \\$667 \\$334 \\$1667 \\$3001 \\$4001 \\$4668 Approx. \\$5000 \\$6000 \\$4333 \\$2999 \\$1999 \\$1332 Approx. \\$1000\n\nAccumulated depreciation = The total or accumulated amount of depreciation expenses at a point in time\nBook value = Cost the asset - Accumulated depreciation expenses\nDepreciable amount = Cost the asset - Residual value the asset" ]	[ null, "https://4.bp.blogspot.com/-cK6mNfGtQCc/VqC_jxiaPwI/AAAAAAAACWI/GKRz1-O9Yas/s1600/Drawing%2B%25283%2529.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8754695,"math_prob":0.9484893,"size":5322,"snap":"2019-51-2020-05","text_gpt3_token_len":1370,"char_repetition_ratio":0.19518615,"word_repetition_ratio":0.1721035,"special_character_ratio":0.27865463,"punctuation_ratio":0.037960954,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9890849,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-29T06:34:13Z\",\"WARC-Record-ID\":\"<urn:uuid:d227b63e-765c-41c8-a6fb-0e85509b358d>\",\"Content-Length\":\"79186\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a2f2c5a9-7196-47d9-8aa3-ad1060ccbc70>\",\"WARC-Concurrent-To\":\"<urn:uuid:65eedcd6-c4a8-464e-808c-c784dee880ac>\",\"WARC-IP-Address\":\"216.239.32.21\",\"WARC-Target-URI\":\"http://www.accounting-world.com/2012/11/depreciation-its-nature-calculation-and_17.html\",\"WARC-Payload-Digest\":\"sha1:CT66L4LMXVTV3QWHO26M2URNVPBPOWZN\",\"WARC-Block-Digest\":\"sha1:JVB7APN7DWIQ5J6KX4E2OTKAPXRGPV74\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251788528.85_warc_CC-MAIN-20200129041149-20200129071149-00253.warc.gz\"}"}
https://www.geeksforgeeks.org/proof-that-sat-is-np-complete/	[ "Related Articles\n\n# Proof that SAT is NP Complete\n\n• Difficulty Level : Hard\n• Last Updated : 14 Oct, 2020\n\nSAT Problem: SAT(Boolean Satisfiability Problem) is the problem of determining if there exists an interpretation that satisfies a given boolean formula. It asks whether the variables of a given boolean formula can be consistently replaced by the values TRUE or FALSE in such a way that the formula evaluates to TRUE. If this is the case, the formula is called satisfiable. On the other hand, if no such assignment exists, the function expressed by the formula is FALSE for all possible variable assignments and the formula is unsatisfiable.\n\nProblem: Given a boolean formula f, the problem is to identify if the formula f has a satisfying assignment or not.\n\nExplanation: An instance of the problem is an input specified to the problem. An instance of the problem is a boolean formula f. Since an NP-complete problem is a problem which is both NP and NP-Hard, the proof or statement that a problem is NP-Complete consists of two parts:\n\n1. The problem itself is in NP class.\n2. All other problems in NP class can be polynomial-time reducible to that.\n(B is polynomial-time reducible to C is denoted as ≤ PC)\n\nIf the 2nd condition is only satisfied then the problem is called NP-Hard.\n\nBut it is not possible to reduce every NP problem into another NP problem to show its NP-Completeness all the time i.e., to show a problem is NP-complete then prove that the problem is in NP and any NP-Complete problem is reducible to that i.e. if B is NP-Complete and B ≤ PC For C in NP, then C is NP-Complete. Thus, it can be verified that the SAT Problem is NP-Complete using the following propositions:\n\nSAT is in NP:\nIt any problem is in NP, then given a ‘certificate’, which is a solution to the problem and an instance of the problem(a boolean formula f) we will be able to check (identify if the solution is correct or not) certificate in polynomial time. This can be done by checking if the given assignment of variables satisfies the boolean formula.\n\nSAT is NP-Hard:\nIn order to prove that this problem is NP-Hard then reduce a known problem, Circuit-SAT in this case to our problem. The boolean circuit C can be corrected into a boolean formula as:\n\n• For every input wire, add a new variable yi.\n• For every output wire, add a new variable Z.\n• An equation is prepared for each gate.\n• These sets of equations are separated by values and adding Z at the end.\n\nThis transformation can be done in linear time. The following propositions now hold:\n\n• If there is a set of input, variable values satisfying the circuit then it can derive an assignment for the formula f that satisfies the formula. This can be simulated by computing the output of every gate in the circuit.\n• If there is a satisfying assignment for the formula f, this can satisfy the boolean circuit after the removal of the newly added variables.\n\nFor Example: If below is the circuit then:", null, "", null, "", null, "", null, "", null, "Therefore, the SAT Problem is NP-Complete.\n\nAttention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.\n\nIn case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.\n\nMy Personal Notes arrow_drop_up" ]	[ null, "https://media.geeksforgeeks.org/wp-content/uploads/20201008155204/idgf.jpg", null, "https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-dbc74bc3ffe46baede99b813e2dcec15_l3.png", null, "https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-f49b69402e6c07117b629d728820d3cc_l3.png", null, "https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-86526b02111d23281d3b3ada297b07e8_l3.png", null, "https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-a9ddc584313e162bfa3e6e7f9776993e_l3.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9057457,"math_prob":0.9718516,"size":3376,"snap":"2021-31-2021-39","text_gpt3_token_len":716,"char_repetition_ratio":0.16785291,"word_repetition_ratio":0.010186757,"special_character_ratio":0.2049763,"punctuation_ratio":0.09159159,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984672,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-25T07:43:16Z\",\"WARC-Record-ID\":\"<urn:uuid:b7a2dfef-58fb-4d9e-abe0-13f47070e722>\",\"Content-Length\":\"104385\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec3aeb8f-625a-48e6-bdda-af1ace3425d1>\",\"WARC-Concurrent-To\":\"<urn:uuid:f0788048-fa8b-4415-a51a-e7cc62209eaa>\",\"WARC-IP-Address\":\"23.45.233.19\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/proof-that-sat-is-np-complete/\",\"WARC-Payload-Digest\":\"sha1:Q7EXH4CUI7KEXJ4PWQ4UVO6OO53IV3YK\",\"WARC-Block-Digest\":\"sha1:UIC7HW4HO2BFGTEWRA3HDS655GSEQT4Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057598.98_warc_CC-MAIN-20210925052020-20210925082020-00533.warc.gz\"}"}
https://math.stackexchange.com/questions/1101651/how-to-find-the-length-of-the-focal-chord-that-makes-an-angle-theta-with-the	[ "# How to find the length of the focal chord that makes an angle $\\theta$ with the axis of parabola $y^2=4ax$?\n\nA focal chord of $$y^2=4ax$$ makes an angle $$\\theta$$ with the axis of the parabola. How do I find the length of the chord?\n\nI tried using the parametric equation but couldn't go further.\n\n• What is the equation of the line that makes an angle $\\theta$ with the $y$-axis and goes through the focus? (This line contains the focal chord given in the question.)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9148503,"math_prob":0.998821,"size":576,"snap":"2020-24-2020-29","text_gpt3_token_len":141,"char_repetition_ratio":0.117132865,"word_repetition_ratio":0.58,"special_character_ratio":0.22916667,"punctuation_ratio":0.07692308,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999789,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-02T07:18:15Z\",\"WARC-Record-ID\":\"<urn:uuid:708a0910-f72d-4bf3-8b06-6fb2610a4fb3>\",\"Content-Length\":\"141710\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6d37f8f0-7858-417d-ac90-597148297f7d>\",\"WARC-Concurrent-To\":\"<urn:uuid:d1e44c78-b368-43c4-ab1e-21c09e2103c1>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1101651/how-to-find-the-length-of-the-focal-chord-that-makes-an-angle-theta-with-the\",\"WARC-Payload-Digest\":\"sha1:MVRR3C43CTUH74FSKVGAI75XKTAMHK3E\",\"WARC-Block-Digest\":\"sha1:T64LXJCHFD2BETORKCZ6QMS464JTB5YE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655878519.27_warc_CC-MAIN-20200702045758-20200702075758-00099.warc.gz\"}"}
https://web2.0calc.com/questions/pre-claculus	[ "+0\n\n# pre claculus\n\n+2\n93\n1\n\nIf \\$5000 is borrowed at a rate of 5.75% interest per year, compounded quarterly, find the amount due at the end of the given number of years. (Round your answers to the nearest cent.)\n\nyears 3\n\nyears 5\n\nyears 7\n\nOct 15, 2020\n\n#1\n+1\n\n5.57% = .0575\n\nquarterly interest = .0575/4 = .014375\n\ndue at the end of 'n' years would be (put the number of years for 'n' and compute):\n\n5000 (1 + .014375)4n\n\nOct 15, 2020" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9360691,"math_prob":0.95861274,"size":365,"snap":"2021-04-2021-17","text_gpt3_token_len":115,"char_repetition_ratio":0.14127424,"word_repetition_ratio":0.029850746,"special_character_ratio":0.3808219,"punctuation_ratio":0.13414635,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95155936,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-27T11:18:18Z\",\"WARC-Record-ID\":\"<urn:uuid:9265361b-0299-4343-a807-492cfcc43c57>\",\"Content-Length\":\"21673\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:84328c57-0c0d-4ca1-ae35-120310a97dcf>\",\"WARC-Concurrent-To\":\"<urn:uuid:689876f9-ffe9-4839-8177-414886e5c15c>\",\"WARC-IP-Address\":\"168.119.149.252\",\"WARC-Target-URI\":\"https://web2.0calc.com/questions/pre-claculus\",\"WARC-Payload-Digest\":\"sha1:C54X3PCJCWDYTFCQGZHOXZEKGUCMNZ73\",\"WARC-Block-Digest\":\"sha1:7NE5F4FHTG5BZ6N36E4OUYRAME7CZRWO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704821381.83_warc_CC-MAIN-20210127090152-20210127120152-00611.warc.gz\"}"}
https://paolotoffanin.wordpress.com/2016/03/18/369/	[ "# A simple fork example timed with chrono\n\nI always wanted to experiment with parallel computing and multi threading. However, when and how to use parallel computing has always been a bit difficult to conceptualize for me. Until I discovered forking. Here I will describe a simple implementation of fork using the chrono Time c++ library to time the forked processes.\n\nForking consists in splitting a program into two parts so that each part can be executed simultaneously. Fork seemed the easiest of the c++ facilities to use for parallel computing because 1) call fork is simple; 2) fork is simpler to use than other thread management tools as thread; 3) it does not require the linking of extra libraries; 4) it fitted easily what I wanted to do: run the same functions simultaneously on different data. This is ideal for some of the lab situations we have in which many persons are administered the same task. Since 1) the type of data produced is the same for all of them and 2) the analysis we perform is the same for all of them, we would gain considerable time if could process them in parallel.\n\nThe code below represents a very simple implementation that I wrote assembling together various pieces I found on the web. I do not think the code below is particularly beautiful, elegant, clever or praiseworthy. I only know I would have been happy to find code as the one below when I was first starting to use fork. In fact, I found the code out there a bit too minimalistic or too complex. My goal is plainly to show code that is readable to someone that speaks c++ and wanted to try how fork works. Moreover, the example below should be easily adaptable to one own needs.\n\nThe code consists of three function calls. Main contains the fork command splitting the child and parent processes, which are `regulated’ through a switch command and wait(s) for both processes to end. Fibonacci is a recursive function computing fibonacci numbers from the cppreference website. I chose this function because I wanted a function the required computing times (e.g. it does not execute in 1 nano second) and because I think recursion is elegant. PrintOut is self explicable I hope.\n\n```#include <iostream>\n#include <string>\n#include <unistd.h>\n#include <sys/wait.h>\n#include <cstdlib>// Declaration for exit()\n#include <chrono> // available from c++11\n#include <ctime>\n\nusing namespace std;\n\nlong fibonacci(unsigned n)\n{\nif (n < 2) return n;\nreturn fibonacci(n-1) + fibonacci(n-2);\n}\n\nint printOutput(string nameProcess, size_t nFibonacci)\n{\ncout << nameProcess << \" started at \";\nchrono::time_point<chrono::system_clock> start;\nstart = chrono::system_clock::now();\ntime_t startTime = chrono::system_clock::to_time_t(start);\ncout << ctime(&startTime);\n\ncout << nameProcess << \"f(\" << nFibonacci << \") = \";\ncout << fibonacci(nFibonacci) << '\\n';\n\ncout << nameProcess << \"finished computation at \";\nchrono::time_point<chrono::system_clock> end;\nend = chrono::system_clock::now();\ntime_t end_time = chrono::system_clock::to_time_t(end);\ncout << ctime(&end_time);\n\ncout << nameProcess << \"elapsed time: \";\nchrono::duration<double> elapsed_seconds = end-start;\ncout << elapsed_seconds.count() << \"s\\n\";\n\nreturn 0;\n}\n\nint main()\n{\npid_t pid;\npid = fork();\nswitch(pid)\n{\ncase 0:\n{\nprintOutput(\"CHILD \", 42);\nexit(0);\n}\ncase -1:\n{\ncout << \"Failed to fork\\n\";\nexit(1);\n}\ndefault:\nprintOutput(\"PARENT \", 42);\n}\ncout << \"waiting for all processes to finish\\n\";\nwaitpid(-1, NULL, 0);\ncout << \"END\\n\";\nreturn 0;\n}\n\n```\n\nA more readable version of this code (e.g. with indentation respected) can be found on my github account.\n\nIt can be compiled as:\n\ng++ -Wall –std=c++11 nameFile.cc -o test\n\nChrono is available from c++11, so the code needs to be compiled with the –std=c++11 flag.\n\nOn my (dated) machine it produced:\n\nPARENT started at Fri Mar 18 20:55:39 2016\nCHILD started at Fri Mar 18 20:55:39 2016\n\nPARENT f(42) = 267914296\nPARENT finished computation at Fri Mar 18 20:55:46 2016\nPARENT elapsed time: 6.18824s\nwaiting for all processes to finish\nCHILD f(42) = 267914296\nCHILD finished computation at Fri Mar 18 20:55:46 2016\nCHILD elapsed time: 6.43001s\nEND" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8321623,"math_prob":0.7099291,"size":4078,"snap":"2021-31-2021-39","text_gpt3_token_len":1004,"char_repetition_ratio":0.1089838,"word_repetition_ratio":0.027863776,"special_character_ratio":0.274154,"punctuation_ratio":0.15306123,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97496974,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-26T16:06:44Z\",\"WARC-Record-ID\":\"<urn:uuid:54835f18-6ce6-42e6-9eee-954226d0a3af>\",\"Content-Length\":\"84161\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a11f0afd-8bee-4992-b05b-f6ee802973ff>\",\"WARC-Concurrent-To\":\"<urn:uuid:27f5243e-0d04-4e9c-b18d-7e5b23a8a226>\",\"WARC-IP-Address\":\"192.0.78.12\",\"WARC-Target-URI\":\"https://paolotoffanin.wordpress.com/2016/03/18/369/\",\"WARC-Payload-Digest\":\"sha1:2ZF5FS55FTOWBESPT74QNT5IJMEW7CWM\",\"WARC-Block-Digest\":\"sha1:LL46CU77MMO4NOHCKJTOPLPYRNRPR6I6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057882.56_warc_CC-MAIN-20210926144658-20210926174658-00302.warc.gz\"}"}
https://www.tutorialspoint.com/automata_theory/language_generated_by_grammars.htm	[ "# Language Generated by a Grammar\n\nThe set of all strings that can be derived from a grammar is said to be the language generated from that grammar. A language generated by a grammar G is a subset formally defined by\n\nL(G)={W\|W ∈ ∑*, S G W}\n\nIf L(G1) = L(G2), the Grammar G1 is equivalent to the Grammar G2.\n\n### Example\n\nIf there is a grammar\n\nG: N = {S, A, B} T = {a, b} P = {S → AB, A → a, B → b}\n\nHere S produces AB, and we can replace A by a, and B by b. Here, the only accepted string is ab, i.e.,\n\nL(G) = {ab}\n\n### Example\n\nSuppose we have the following grammar −\n\nG: N = {S, A, B} T = {a, b} P = {S → AB, A → aA\|a, B → bB\|b}\n\nThe language generated by this grammar −\n\nL(G) = {ab, a2b, ab2, a2b2, ………}\n\n= {am bn \| m ≥ 1 and n ≥ 1}\n\n## Construction of a Grammar Generating a Language\n\nWe’ll consider some languages and convert it into a grammar G which produces those languages.\n\n### Example\n\nProblem − Suppose, L (G) = {am bn \| m ≥ 0 and n > 0}. We have to find out the grammar G which produces L(G).\n\nSolution\n\nSince L(G) = {am bn \| m ≥ 0 and n > 0}\n\nthe set of strings accepted can be rewritten as −\n\nL(G) = {b, ab,bb, aab, abb, …….}\n\nHere, the start symbol has to take at least one ‘b’ preceded by any number of ‘a’ including null.\n\nTo accept the string set {b, ab, bb, aab, abb, …….}, we have taken the productions −\n\nS → aS , S → B, B → b and B → bB\n\nS → B → b (Accepted)\n\nS → B → bB → bb (Accepted)\n\nS → aS → aB → ab (Accepted)\n\nS → aS → aaS → aaB → aab(Accepted)\n\nS → aS → aB → abB → abb (Accepted)\n\nThus, we can prove every single string in L(G) is accepted by the language generated by the production set.\n\nHence the grammar −\n\nG: ({S, A, B}, {a, b}, S, { S → aS \| B , B → b \| bB })\n\n### Example\n\nProblem − Suppose, L (G) = {am bn \| m > 0 and n ≥ 0}. We have to find out the grammar G which produces L(G).\n\nSolution\n\nSince L(G) = {am bn \| m > 0 and n ≥ 0}, the set of strings accepted can be rewritten as −\n\nL(G) = {a, aa, ab, aaa, aab ,abb, …….}\n\nHere, the start symbol has to take at least one ‘a’ followed by any number of ‘b’ including null.\n\nTo accept the string set {a, aa, ab, aaa, aab, abb, …….}, we have taken the productions −\n\nS → aA, A → aA , A → B, B → bB ,B → λ\n\nS → aA → aB → aλ → a (Accepted)\n\nS → aA → aaA → aaB → aaλ → aa (Accepted)\n\nS → aA → aB → abB → abλ → ab (Accepted)\n\nS → aA → aaA → aaaA → aaaB → aaaλ → aaa (Accepted)\n\nS → aA → aaA → aaB → aabB → aabλ → aab (Accepted)\n\nS → aA → aB → abB → abbB → abbλ → abb (Accepted)\n\nThus, we can prove every single string in L(G) is accepted by the language generated by the production set.\n\nHence the grammar −\n\nG: ({S, A, B}, {a, b}, S, {S → aA, A → aA \| B, B → λ \| bB })" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8595926,"math_prob":0.9545304,"size":2472,"snap":"2019-43-2019-47","text_gpt3_token_len":891,"char_repetition_ratio":0.15640195,"word_repetition_ratio":0.38255033,"special_character_ratio":0.3867314,"punctuation_ratio":0.16129032,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99874276,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-15T07:57:35Z\",\"WARC-Record-ID\":\"<urn:uuid:5cb2e9c3-a65d-4f2d-ada6-0741878077b0>\",\"Content-Length\":\"26607\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0e6fb694-20f9-4391-9544-e92069a55bf3>\",\"WARC-Concurrent-To\":\"<urn:uuid:9b02846f-25cd-4ba0-98bf-ced98c5b42ff>\",\"WARC-IP-Address\":\"72.21.91.42\",\"WARC-Target-URI\":\"https://www.tutorialspoint.com/automata_theory/language_generated_by_grammars.htm\",\"WARC-Payload-Digest\":\"sha1:ZL2X3FYNK7VDGXIFW4IFKQ5Z6RQLYCDW\",\"WARC-Block-Digest\":\"sha1:E264P6VO24PVCW4T6BNU4ZOJIKGTAIZE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668594.81_warc_CC-MAIN-20191115065903-20191115093903-00309.warc.gz\"}"}
https://proofwiki.org/wiki/Definition:Incomplete_Elliptic_Integral_of_the_Third_Kind	[ "# Definition:Elliptic Integral of the Third Kind/Incomplete\n\n## Special Function\n\n### Definition 1\n\n$\\displaystyle \\Pi \\left({k, n, \\phi}\\right) = \\int \\limits_0^\\phi \\frac {\\mathrm d \\phi} {\\left({1 + n \\sin^2 \\phi}\\right) \\sqrt{1 - k^2 \\sin^2 \\phi} }$\n\nis the incomplete elliptic integral of the third kind, and is a function of the variables:\n\n$k$, defined on the interval $0 < k < 1$\n$n \\in \\Z$\n$\\phi$, defined on the interval $0 \\le \\phi \\le \\pi / 2$.\n\n### Definition 2\n\n$\\displaystyle \\Pi \\left({k, n, \\phi}\\right) = \\int \\limits_0^x \\frac {\\mathrm d v} {\\left({1 + n v^2}\\right) \\sqrt{\\left({1 - v^2}\\right) \\left({1 - k^2 v^2}\\right)} }$\n\nis the incomplete elliptic integral of the third kind, and is a function of the variables:\n\n$k$, defined on the interval $0 < k < 1$\n$n \\in \\Z$\n$x = \\sin \\phi$, where $\\phi$ is defined on the interval $0 \\le \\phi \\le \\pi / 2$.\n\nNote that:\n\n$\\Pi \\left({k, n, \\dfrac \\pi 2}\\right)$\n\n## Also known as\n\nSome sources omit the incomplete from the definition, calling this merely the elliptic integral of the third kind." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5985441,"math_prob":0.9999839,"size":1507,"snap":"2020-10-2020-16","text_gpt3_token_len":453,"char_repetition_ratio":0.18097138,"word_repetition_ratio":0.44489795,"special_character_ratio":0.30856004,"punctuation_ratio":0.09642857,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99996316,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-29T05:38:02Z\",\"WARC-Record-ID\":\"<urn:uuid:b3aaf260-bced-4827-833c-90052a1a8938>\",\"Content-Length\":\"35869\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:360e1f15-9146-43b7-8aa7-b07eb367e98b>\",\"WARC-Concurrent-To\":\"<urn:uuid:8c413076-f6ed-41a4-b3b5-5a8916e18392>\",\"WARC-IP-Address\":\"104.27.168.113\",\"WARC-Target-URI\":\"https://proofwiki.org/wiki/Definition:Incomplete_Elliptic_Integral_of_the_Third_Kind\",\"WARC-Payload-Digest\":\"sha1:MO55M62T2OQ3GPHMSDXFVIFUO3WM6WXK\",\"WARC-Block-Digest\":\"sha1:RSDYOBNCWVQN3LQPAT7DNC6VPZHBOIB7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875148671.99_warc_CC-MAIN-20200229053151-20200229083151-00136.warc.gz\"}"}
https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/partial-derivative-and-gradient-articles/a/directional-derivatives-going-deeper	[ "If you're seeing this message, it means we're having trouble loading external resources on our website.\n\nIf you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked.\n\n# Directional derivatives (going deeper)\n\nA more thorough look at the formula for directional derivatives, along with an explanation for why the gradient gives the slope of steepest ascent.\n\n## Formal definition of the directional derivative\n\nThere are a couple reasons you might care about a formal definition. For one thing, really understanding the formal definition of a new concept can make clear what it is really going on. But more importantly than that, I think the main benefit is that it gives you the confidence to recognize when such a concept can and cannot be applied.\nAs a warm up, let's review the formal definition of the partial derivative, say with respect to x:\nstart fraction, \\partial, f, divided by, start color #0c7f99, \\partial, x, end color #0c7f99, end fraction, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, equals, limit, start subscript, h, \\to, 0, end subscript, start fraction, f, left parenthesis, x, start subscript, 0, end subscript, start color #0c7f99, plus, h, end color #0c7f99, comma, y, start subscript, 0, end subscript, right parenthesis, minus, f, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, divided by, start color #0c7f99, h, end color #0c7f99, end fraction\nThe connection between the informal way to read start fraction, \\partial, f, divided by, \\partial, x, end fraction and the formal way to read the right-hand side is as follows:\nSymbolInformal understandingFormal understanding\n\\partial, xA tiny nudge in the x direction.A limiting variable h which goes to 0, and will be added to the first component of the function's input.\n\\partial, fThe resulting change in the output of f after the nudge.The difference between f, left parenthesis, x, start subscript, 0, end subscript, plus, h, comma, y, start subscript, 0, end subscript, right parenthesis and f, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, taken in the same limit as h, \\to, 0.\nWe could instead write this in vector notation, viewing the input point left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis as a two-dimensional vector\n\\begin{aligned} \\textbf{x}_0 = \\left[ \\begin{array}{c} x_0 \\\\\\\\ y_0 \\\\\\\\ \\end{array} \\right] \\end{aligned}\nHere start bold text, x, end bold text, start subscript, 0, end subscript is written in bold to emphasize its vectoriness. It's a bit confusing to use a bold start bold text, x, end bold text for the entire input rather than some other letter, since the letter x is already used in an un-bolded form to denote the first component of the input. But hey, that's convention, so we go with it.\nInstead of writing the \"nudged\" input as left parenthesis, x, start subscript, 0, end subscript, plus, h, comma, y, start subscript, 0, end subscript, right parenthesis, we write it as start bold text, x, end bold text, start subscript, 0, end subscript, plus, h, start bold text, i, end bold text, with, hat, on top, where start bold text, i, end bold text, with, hat, on top is the unit vector in the x-direction:\n\\begin{aligned} \\dfrac{\\partial f}{\\partial x}(\\textbf{x}_0) = \\lim_{h \\to 0} \\dfrac{f(\\textbf{x}_0 + h \\hat{\\textbf{i}}) - f(\\textbf{x}_0)}{h} \\end{aligned}\nIn this notation, it's much easier to see how to generalize the partial derivative with respect to x to the directional derivative along any vector start bold text, v, end bold text, with, vector, on top:\nstart color #0c7f99, del, start subscript, start color #e84d39, start bold text, v, end bold text, with, vector, on top, end color #e84d39, end subscript, f, left parenthesis, start bold text, x, end bold text, start subscript, 0, end subscript, right parenthesis, equals, limit, start subscript, h, \\to, 0, end subscript, start fraction, f, left parenthesis, start bold text, x, end bold text, start subscript, 0, end subscript, plus, h, start color #e84d39, start bold text, v, end bold text, with, vector, on top, end color #e84d39, right parenthesis, minus, f, left parenthesis, start bold text, x, end bold text, start subscript, 0, end subscript, right parenthesis, divided by, h, end fraction, end color #0c7f99\nIn this case, adding h, start bold text, v, end bold text, with, vector, on top to the input for a limiting variable h, \\to, 0 formalizes the idea of a tiny nudge in the direction of start bold text, v, end bold text, with, vector, on top.\nShowing directional derivative nudge\n\n## Seeking connection between the definition and computation\n\nComputing the directional derivative involves a dot product between the gradient del, f and the vector start bold text, v, end bold text, with, vector, on top. For example, in two dimensions, here's what this would look like:\n\\begin{aligned} \\nabla_{\\vec{\\textbf{v}}} f(x, y) &= \\nabla f \\cdot \\vec{\\textbf{v}} \\\\\\\\ &= \\left[ \\begin{array}{c} \\dfrac{\\partial f}{\\partial x} \\\\\\\\ \\dfrac{\\partial f}{\\partial y} \\end{array} \\right] \\cdot \\left[ \\begin{array}{c} \\blueE{v_1} \\\\\\\\ \\greenE{v_2} \\end{array} \\right] \\\\\\\\ &= \\blueE{v_1} \\dfrac{\\partial f}{\\partial x}(x, y) + \\greenE{v_2} \\dfrac{\\partial f}{\\partial y}(x, y) \\end{aligned}\nHere, start color #0c7f99, v, start subscript, 1, end subscript, end color #0c7f99 and start color #0d923f, v, start subscript, 2, end subscript, end color #0d923f are the components of start bold text, v, end bold text, with, vector, on top.\n\\begin{aligned} \\vec{\\textbf{v}} = \\left[ \\begin{array}{c} \\blueE{v_1} \\\\ \\\\ \\greenE{v_2} \\\\\\\\ \\end{array} \\right] \\end{aligned}\nThe central question is, what does this formula have to do with the definition given above?\n\n## Breaking down the nudge\n\nThe computation for del, start subscript, start bold text, v, end bold text, end subscript, f can be seen as a way to break down a tiny step in the direction of start bold text, v, end bold text into its x and y components.\nBreak apart a step along the vector h, start bold text, v, end bold text, with, vector, on top\nSpecifically, you can imagine the following procedure:\n1. Start at some point left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis.\n2. Choose a tiny value h.\n3. Add h, start color #0c7f99, v, start subscript, 1, end subscript, end color #0c7f99 to x, start subscript, 0, end subscript, which means stepping to the point left parenthesis, x, start subscript, 0, end subscript, plus, h, start color #0c7f99, v, start subscript, 1, end subscript, end color #0c7f99, comma, y, start subscript, 0, end subscript, right parenthesis. From what we know of partial derivatives, this will change the output of the function by about\n\\begin{aligned} h\\blueE{v_1} \\left(\\dfrac{\\partial f}{\\partial x}(x_0, y_0) \\right) \\end{aligned}\n• Now add h, start color #0d923f, v, start subscript, 2, end subscript, end color #0d923f to y, start subscript, 0, end subscript to bring us up/down to the point left parenthesis, x, start subscript, 0, end subscript, plus, h, start color #0c7f99, v, start subscript, 1, end subscript, end color #0c7f99, comma, y, start subscript, 0, end subscript, plus, h, start color #0d923f, v, start subscript, 2, end subscript, end color #0d923f, right parenthesis. The resulting change to f is now about\n\\begin{aligned} h\\greenE{v_2}\\left( \\dfrac{\\partial f}{\\partial y}(x_0 + h\\blueE{v_1}, y_0) \\right) \\end{aligned}\nAdding the results of steps 3 and 4, the total change to the function upon moving from the input left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis to the input left parenthesis, x, start subscript, 0, end subscript, plus, h, start color #0c7f99, v, start subscript, 1, end subscript, end color #0c7f99, comma, y, start subscript, 0, end subscript, plus, h, start color #0d923f, v, start subscript, 2, end subscript, end color #0d923f, right parenthesis has been about\n\\begin{aligned} h\\blueE{v_1} \\left(\\dfrac{\\partial f}{\\partial x}(x_0, y_0) \\right) + h\\greenE{v_2}\\left( \\dfrac{\\partial f}{\\partial y}(x_0 \\redD{+ h\\blueE{v_1}}, y_0) \\right) \\end{aligned}\nThis is very close to the expression for the directional derivative, which says the change in f due to this step h, start bold text, v, end bold text, with, vector, on top should be about\n\\begin{aligned} &\\phantom{=}h \\nabla_{\\vec{\\textbf{v}}} f(x_0, y_0) \\\\\\\\ &= h \\vec{\\textbf{v}} \\cdot \\nabla f(x_0, y_0)\\\\\\\\ &= h\\blueE{v_1}\\dfrac{\\partial f}{\\partial x}(x_0, y_0) + h\\greenE{v_2}\\dfrac{\\partial f}{\\partial y}(x_0, y_0) \\end{aligned}\nHowever, this differs slightly from the result of our step-by-step argument, in which the partial derivative with respect to y is taken at the point left parenthesis, x, start subscript, 0, end subscript, plus, h, start color #0c7f99, v, start subscript, 1, end subscript, end color #0c7f99, comma, y, start subscript, 0, end subscript, right parenthesis, not at the point left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis.\nLuckily we are considering very, very small values of h. In fact, more technically, we should be talking about the limit as h, \\to, 0. Therefore evaluating start fraction, \\partial, f, divided by, \\partial, y, end fraction at left parenthesis, x, start subscript, 0, end subscript, plus, h, start color #0c7f99, v, start subscript, 1, end subscript, end color #0c7f99, comma, y, start subscript, 0, end subscript, right parenthesis will be almost the same as evaluating it at left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis. Moreover, as h approaches 0, so does the difference between these two, but we have to assume that f is continuous.\n\n## Why does the gradient point in the direction of steepest ascent?\n\nHaving learned about the directional derivatives, we can now understand why the direction of the gradient is the direction of steepest ascent.\nSteepest ascent concept.\nSpecifically, here's the question at hand.\nSetup:\n• Let f be some scalar-valued multivariable function, such as f, left parenthesis, x, comma, y, right parenthesis, equals, x, squared, plus, y, squared.\n• Let left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis be a particular input point\n• Consider all possible directions, i.e. all unit vectors start bold text, u, end bold text, with, hat, on top in the input space of f.\nQuestion (informal): If we start at left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, which direction should we walk so that the output of f increases most quickly?\nQuestion (formal): Which unit vector start bold text, u, end bold text, with, hat, on top maximizes the directional derivative along start bold text, u, end bold text, with, hat, on top?\n\\begin{aligned} \\nabla_{\\hat{\\textbf{u}}} f(x_0, y_0) = \\underbrace{\\hat{\\textbf{u}} \\cdot \\nabla f(x_0, y_0)}_{ \\text{Maximize this quantity} } \\end{aligned}\nThe famous triangle inequality tells us that this will be maximized by the unit vector in the direction del, f, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis.\nMaximize dot product\nNotice, the fact that the gradient points in the direction of steepest ascent is a consequence of the more fundamental fact that all directional derivatives require taking the dot product with del, f.\n\n## Want to join the conversation?\n\n• I'm having trouble understanding the 3rd step under the formal argument. If we move hv1 in the x direction, how does this imply that the output will be hv1fx(x0,y0)? (Sorry for the notation - I'm on my phone).", null, "•", null, "", null, "That is the definition of the derivative. Remember:\nfₓ(x₀,y₀) = lim_Δx→0 [(f(x₀+Δx,y₀)-f(x₀,y₀))/Δx]\nThen, we can replace Δx with hv₁ because both Δx and h are very small, so we get:\nfₓ(x₀,y₀) = (f(x₀+hv₁,y₀)-f(x₀,y₀))/hv₁\nWe can then rearrange this equation to get:\nf(x₀+hv₁,y₀) = hv₁ × fₓ(x₀,y₀) + f(x₀,y₀)\n• Can anyone help me understand how to solve the puzzle at the end of the article? I am having trouble understanding it.", null, "• I had trouble with this puzzle too, but then I thought about it in terms of vectors. We need to maximize 100A + 20B + 2C, right? By definition of the dot product, this expression is equal to the dot product of two vectors [100, 20, 2] [A, B, C]. So we want to maximize the dot product. When does the dot product have the maximum value? It is maximum when two vectors are parallel, or, in other words, one vector is multiple of the other (this can be understood from the graphical interpretation of the dot product). Therefore, our vector [A, B, C] should be [100x, 20x, 2x], where x is some number.\n\nThe second insight is to express A^2 + B^2 + C^2 = 10404 equation in vector notation. Expression A^2 + B^2 + C^2 is equal to the dot product of vector [A,B,C] with itself:\n\n[A,B,C][A,B,C] = 10404.\n\nHere instead of [A,B,C] we substitute [100x, 20x, 2x] vector and solve for x:\n\n[100x, 20x, 2x] [100x, 20x, 2x] = 10000x^2 + 400x^2 + 4x^2 = 10404x^4 = 10404.\nx = 1.\n\nTherefore, our vector [A,B,C] is [100 * 1, 20 * 1, 2 * 1] = [100, 20, 2].\n\nA = 100\nB = 20\nC = 2.\n\nI hope it wasn't confusing. :)\n• step 4. how does adding the change in the x direction and y direction give us the total change in the function? we are adding \"perpendicular numbers\", not vectors, so the adding should look more like pythagoras, shouldn't it?", null, "• Isn't it good to color \"h\" in black in the figure just below the subtitle of \"Breaking down the nudge\" for the consistency? Just a suggestion.", null, "• I don't understand this sentence: The famous triangle inequality tells us that this will be maximized by the unit vector in the direction (nabla) f (x0, y0)\n\nTo me, this doesn't seem obvious at all, can I find explanation perhaps somewhere else on the khan academy? I have looked at triangle inequality \|\|x + y\|\| <= \|\|x\|\|+\|\|y\|\| , but I don't understand how the two things are related, other than that both somehow talk about vectors. Directional derivative however works with dot products and not adding vectors together, as in the triangle inequality, so I don't immediately see the connection between the two.", null, "• I'm in the same boat and don't see how the triangle inequality can be applied, however using the slightly different Cauchy-Schwarz inequality works. The Cauchy-Scwarz inequality states that\nx·y <= \|\|x\|\| * \|\|y\|\|\nfor any two vectors x and y. Cauchy-Schwarz also says that the inequality can be turned to an equality\nx · y = \|\|x\|\| * \|\|y\|\|\nif x and y are parallel.\n• Regarding directional derivative I can not understand why the vector v has not to be a unit vector, namely, it can has an arbitrary magnitude.\n\nIn the one hand, the formal definition of the directional derivative supplied in the article lies upon unit vectors i and j. From my point of view, it is easy to understand this definition will hold for any direction other than i and j always that we treat with unit vectors. Another thing, and this is what I can not figure out clearly, is wether the magnitud of these vectors can also be generalized. To illustrate what it shocks me, if we set a limit from 0 to infinity for the magnitud of v, the formula will ends up with an indetermination of the form h·\|\|v\|\|=0·Inf, because h goes to 0 and \|\|v\|\| goes to Inf. In contrast, if we restrict ourselves into unit vectors this would not happen.\n\nOn the other hand, as far as I understood we are trying to get a measure that tells us how f(x, y) changes in the direction of v, which has nothing to do with how far we move in that direction. Therefore the only way I can figure out to capture just the variation due to change in direction (and nothing else) is \"playing\" with number 1 (i.e. unit vectors) whenever we multiply.", null, "• Hi, i've a question on the following point \"Computing the directional derivative involves a dot product between the gradient and the vector v\". when i look at the definition of dot product, it says \"\|a\|.\|b\|.cosine theeta\".\nbut in this definition no cosine theeta is involved. is that mean, the gradient (which has the partial derivatives of x,y) not considered as a vector here? or is it? if yes,why cosine theeta is not involved here?", null, "• There's an error at the formal definition part. You say dx goes to zero. That means df/dx is undefined?\n(1 vote)", null, "• I have more of a conceptual question. If we think a partial derivative as a little nudge to a certain direction and that nudge (h) is approaching 0, why does that concept not transfer to directional derivative? Namely, why would the directional derivative of 2v twice as big as that of v? If we think of the nudge in v as so tiny that goes to 0, why would 2v versus v even matter?", null, "• I think this part may be wrong (or at least I don't fully understand it):\n\n\"However, this differs slightly from the result of our step-by-step argument, in which the partial derivative with respect to y is taken at the point (x_0 + hv_1, y_0) not at the point (x_0, y_0)\"\n\nI would think we could treat the change in y just like we treated the change in x -- they are in essence happening at the same time -- and how would you chose the order anyway?\n(1 vote)", null, "• The thing is that lim ℎ→0 𝑥₀ + ℎ𝑣₁ = 𝑥₀\n\nWith 𝒗 = (𝑣₁, 𝑣₂)\n𝛻𝒗 𝑓(𝑥₀, 𝑦₀) = lim ℎ→0 [𝑓(𝑥₀ + ℎ𝑣₁, 𝑦₀ + ℎ𝑣₂) − 𝑓(𝑥₀, 𝑦₀)]∕ℎ\n\n= lim ℎ→0 [𝑓(𝑥₀ + ℎ𝑣₁, 𝑦₀ + ℎ𝑣₂) − 𝑓(𝑥₀ + ℎ𝑣₁, 𝑦₀) + 𝑓(𝑥₀ + ℎ𝑣₁, 𝑦₀) − 𝑓(𝑥₀, 𝑦₀)]∕ℎ\n\n= lim ℎ→0 [𝑓(𝑥₀ + ℎ𝑣₁, 𝑦₀ + ℎ𝑣₂) − 𝑓(𝑥₀ + ℎ𝑣₁, 𝑦₀)]∕ℎ\n+ lim ℎ→0 [𝑓(𝑥₀ + ℎ𝑣₁, 𝑦₀) − 𝑓(𝑥₀, 𝑦₀)]∕ℎ\n\n= lim ℎ→0 (𝜕∕𝜕𝑦[𝑓(𝑥₀ + ℎ𝑣₁, 𝑦₀)]) + 𝜕∕𝜕𝑥[𝑓(𝑥₀, 𝑦₀)]\n\n= 𝜕∕𝜕𝑦[𝑓(𝑥₀, 𝑦₀)]) + 𝜕∕𝜕𝑥[𝑓(𝑥₀, 𝑦₀)]" ]	[ null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/badges/earth/great-answer-40x40.png", null, "https://cdn.kastatic.org/images/badges/moon/good-answer-40x40.png", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92931944,"math_prob":0.97908604,"size":11883,"snap":"2022-40-2023-06","text_gpt3_token_len":3246,"char_repetition_ratio":0.15211718,"word_repetition_ratio":0.0626461,"special_character_ratio":0.25288227,"punctuation_ratio":0.11687781,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99740815,"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],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-28T22:32:48Z\",\"WARC-Record-ID\":\"<urn:uuid:4d5494ce-04d5-4dc5-9758-4a0554c439e5>\",\"Content-Length\":\"902877\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4ae00c71-804f-4b6e-905a-903d87d38dc2>\",\"WARC-Concurrent-To\":\"<urn:uuid:b156ac2c-a311-4df7-a111-5815305eabaa>\",\"WARC-IP-Address\":\"146.75.33.42\",\"WARC-Target-URI\":\"https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/partial-derivative-and-gradient-articles/a/directional-derivatives-going-deeper\",\"WARC-Payload-Digest\":\"sha1:TROJL5ZUKUCD6CTWSILDSQRALRAHHOA2\",\"WARC-Block-Digest\":\"sha1:XZGML66IBZOQMYJQBK6Z4ABQHFCCBBHD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499695.59_warc_CC-MAIN-20230128220716-20230129010716-00868.warc.gz\"}"}
https://wiki.documentfoundation.org/Documentation/Calc_Functions/SINH	[ "# Documentation/Calc Functions/SINH\n\nOther languages:\nEnglish • ‎dansk\n\nSINH\n\nMathematical\n\n## Summary:\n\nCalculates the hyperbolic sine (sinh) of a hyperbolic angle expressed in radians.\n\nSINH(Number)\n\n## Returns:\n\nReturns a real number that is the hyperbolic sine of the specified hyperbolic angle.\n\n## Arguments:\n\nNumber is a real number, or a reference to a cell containing that number, that is the hyperbolic angle in radians whose hyperbolic sine is to be calculated.\n\n• If Number is non-numeric, then SINH reports a #VALUE! error.\n\nThe formula for the hyperbolic sine of x is:\n\n$\\displaystyle{ \\sinh(x)~=~\\frac{e^x-e^{-x}}{2} }$\n\nThe figure below illustrates the function SINH.\n\n## Examples:\n\nFormula Description Returns\n=SINH(0) Hyperbolic sine of 0. 0\n=SINH(D1) where cell D1 contains the number -2. Hyperbolic sine of -2. -3.62686040784702\n=ASINH(SINH(3)) The hyperbolic sine of 3 radians is taken and then the inverse hyperbolic sine of that value is calculated. 3\n\nSINH" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6042519,"math_prob":0.97749704,"size":1206,"snap":"2021-43-2021-49","text_gpt3_token_len":326,"char_repetition_ratio":0.17470881,"word_repetition_ratio":0.0,"special_character_ratio":0.23300166,"punctuation_ratio":0.12264151,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99986684,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-02T09:18:50Z\",\"WARC-Record-ID\":\"<urn:uuid:5a65a1ec-3c4f-4cd1-982c-765098facbae>\",\"Content-Length\":\"29717\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:64d03047-e757-4697-a373-4497f81f41b4>\",\"WARC-Concurrent-To\":\"<urn:uuid:4b12644a-f5f6-4280-aa6f-a5fe92bbfada>\",\"WARC-IP-Address\":\"89.238.68.190\",\"WARC-Target-URI\":\"https://wiki.documentfoundation.org/Documentation/Calc_Functions/SINH\",\"WARC-Payload-Digest\":\"sha1:RADX3Q3XBY2OW3HTJOXGKVV4V4YZ5R3F\",\"WARC-Block-Digest\":\"sha1:HSPNO7C2LAQVYNHPRXDZE7Y7PAJZTINO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964361253.38_warc_CC-MAIN-20211202084644-20211202114644-00540.warc.gz\"}"}
https://www.dw-math.com/ac/static/T49.html	[ "Custom math worksheets at your fingertips", null, "For all problems there are free worksheets and solution sheets available for instant download. Just click on the link for a detail view. Is there a type of problem missing that you need? Register here, and add it as an item to our wish list!\n\n# All problems tagged Division\n\nIf you remember the quickname of a problem, you can add it to your worksheet very quickly: On the bottom right of the worksheet editor, there is a quick entry form field.\n\nName\nDescription\nQuickname with a link to the detailed view\nAdd missing line segment name for intercept theorem\nIn a statement of the intercept theorems the missing line segment name has to be inserted.\nDecimal fraction divide by power of 10\nA decimal fraction has to be divided by a power of ten.\nDecimal fraction divide by whole no\nA decimal fraction has to be divided by a whole number.\nDecimal fraction division\nA decimal fraction has to be divided by another decimal fraction.\nDetermine gcd and lcm by comparing multiples & divisors\nFor two given numbers, the gcd and lcm are determined by comparing lists of multiples or divisors\nDetermine line segment length with intercept theorem\nIn application of the intercept theorem the fourth line segment length is to be calculated from three given lengths.\nDirect proportionality - complete table\nApply the rule of three for a direct proportionality relationship and fill in missing values in a table.\nDistributive law - use to evaluate\nEvaluate terms by multiplying out using distributive law\nDistributive law: Match complex and expanded terms\nFind pairs of complex terms and their corresponding expanded form.\nDivide a fraction by a whole number\nA fraction has to be divided by a whole number.\nDivide two fractions\nDivision of two fractions.\nDivisibility rules to be specified and applied\nThe rules on divisibility have to be specified and applied.\nDivision by multiples of power of ten\nDivison by a power of ten or a multiple thereof.\nDivision of fractions: Fill in the blanks\nIn a division term of two fractions, blanks have to be filled in.\nDivision series 1x1 with factor 10\nDivision problem series with divisor and dividend alternating with factor ten.\nDivision times table 1-10\nDivision problems from the multiplication tables\nDivision, perform by splitting dividend\nDivide by splitting the dividend.\nEvaluate algebraic expression\nA simple algebraic expression is to be evaluated, observing computational rules\nFactor or multiple?\nDetermine whether a number is a factor or multiple of another.\nFactors of a number\nFor a given number, all factors have to be listed.\nFind number in the middle\nFor two given numbers, find the number in the middle, the midpoint.\nFractions multiplication by whole numbers fill blanks\nWhat whole number has be fraction been multiplied with?\nGCD computation with the Euclidean Algorithm\nCompute the GCD step by step with the Euclidian Algorithm.\nHalve numbers\nDivide numbers by two, whole numbers or with decimal places.\nInsert the arithmentic operator\nIn an equation with two operands, insert the correct arithmetic symbol.\nInsert two arithmentic operators\nIn an equation with three operands, insert two operators.\nIntercept theorem statements right or wrong\nCorrect statements on the intercept theorems must be identified and marked.\nInverse proportionality - complete table\nApply the rule of three for an inverse proportionality relationship and fill in missing values in a table.\nlcm derived via prime factorization\nThe lcm is to be derived for two given numbers.\nLong division of two whole numbers\nDivision of two whole numbers, with remainder, written long division method illustrated.\nLong division with whole numbers, blanks to be filled\nDivision of two whole numbers, with remainder, written long division with blanks to be filled in.\nNumber line find midpoint\nOn a number line, the midpoint between to marks has to be identified.\nRule of three word problem\nWord problems that can be solved by applying the rule of three.\nTable of multiplication problems\nMatrix of multiplication problems with whole numbers\nTest for divisibility\nA row of numbers is presented. Delete those that are not divisible by a given divisor.\nThese informational pages with samples describe math problems that can be combined on custom math worksheets with solutions for home and K-12 school use.\nPlease visit the dw math main page for more information!\nPrivacy policy and imprint\nCookie settings\nDeutsche Seiten\n×\n\nUse of cookies\n\nCookies are small data snippets that we store on your computer to recognize you when you use our website.\n\nThere are cookies that we need for technical reasons to make the website usable for you. You cannot deactivate these, because otherwise our website would not work.\n\nTechnically necessary cookies\n\nCookies for tracking activity and displaying personalized ads\n\nOur partners collect data and use cookies to personalise displayed ads and to track your activity in the internet. If you want to allow this, please select \"Accept\" and choose \"Confirm selection\" or simply \"Accept all cookies\". 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http://jnnd.macchiaelettrodomestici.it/how-to-find-the-kernel-of-a-linear-transformation.html	[ "The subset of B consisting of all possible values of f as a varies in the domain is called the range of. 1 T(~x + ~y) = T(~x) + T(~y)(preservation of addition) 2 T(a~x) = aT(~x)(preservation of scalar multiplication) Linear Transformations: Matrix of a Linear Transformation Linear Transformations Page 2/13. A basis for the kernel of L is {1} so the kernel has dimension 1. Matrix vector products as linear transformations. All Slader step-by-step solutions are FREE. Inversion: R(z) = 1 z. Let A = 2 4 0. Find the kernel of the linear transformation L: V→W. Proof: This theorem is a proved in a manner similar to how we solved the above example. range(T)={A in W \| there exists B in V such that T(B)=A}. The $$\\textit{nullity}$$ of a linear transformation is the dimension of the kernel, written $$nul L=\\dim \\ker L. Thus, we should be able to find the standard matrix for. It is essentially the same thing here that we are talking about. Let T: V !W be a linear transformation. It relates the dimension of the kernel and range of a linear map. In mathematics, a linear map (also called a linear mapping, linear transformation or, in some contexts, linear function) is a mapping V → W between two modules (for example, two vector spaces) that preserves (in the sense defined below) the operations of addition and scalar multiplication. Construct a linear transformation f and vector Y so that the system takes the form f(X)=Y. I tried to RREF it on my calculator but it says invalid dim, I am not sure of what to do and all the examples I have looked up are for square matrices, any help would be appreciated, thanks!. And we saw that earlier in the video. Suppose T : V !W is a linear transformation. In Section 4, we deﬁne the kernel whitening transformation and orthogonalize non-. Before we look at some examples of the null spaces of linear transformations, we will first establish that the null space can never be equal to the empty set, in. The confidence of the interval [107, 230] is less than 95%. In both cases, the kernel is the set of solutions of the corresponding homogeneous linear equations, AX = 0 or BX = 0. The function of kernel is to take data as input and transform it into the required form. 0004 From the previous lesson, we left it off defining what the range of a linear map is. be a linear transformation. Specifically, if T: n m is a linear transformation, then there is a unique m n matrix, A, such that T x Ax for all x n. Griti is a learning community for students by students. Algebra Linear Algebra: A Modern Introduction 4th Edition In Exercises 5-8, find bases for the kernel and range of the linear transformations T in the indicated exercises. \" • The fact that T is linear is essential to the kernel and range being subspaces. LTR-0060: Isomorphic Vector Spaces We define isomorphic vector spaces, discuss isomorphisms and their properties, and prove that any vector space of dimension is isomorphic to. (a) Find a basis of the range of P. the kernel of a a linear transformation is the set of vectors in the null space of the matrix for that linear transformation. Kernel, image, nullity, and rank Math 130 Linear Algebra D Joyce, Fall 2015 De nition 1. Linear Transformation. Since the nullity is the dimension of the null space, we see that the nullity of T is 0 since the dimension of the zero vector space is 0. The following examples illustrate the syntax. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. To prove the transformation is linear, the transformation must preserve scalar multiplication, addition, and the zero vector. We then consider invertible linear transformations, and then use the resulting ideas to prove the rather stunning result that (in a very precise sense). Vector Spaces and Linear Transformations Beifang Chen Fall 2006 1 Vector spaces A vector space is a nonempty set V, whose objects are called vectors, equipped with two operations, called addition and scalar multiplication: For any two vectors u, v in V and a scalar c, there are unique vectors u+v and cu in V such that the following properties are satisﬂed. Kernel trick allows us to transform the data in high dimensional (potentially inﬁnite) using inner products, without actually using the non linear feature mapping. SUBSCRIBE to the channel and. We discuss the kernel and range of linear transformations, and then prove that the range of a linear transformation is a subspace. What is the outcome of solving the problem?. The Gaussian is a self-similar function. Sums and scalar multiples of linear transformations. This MATLAB function returns predicted class labels for each observation in the predictor data X based on the binary Gaussian kernel classification model Mdl. Find the kernel of f. Then the Kernel of the linear transformation T is all elements of the vector space V that get mapped onto the zero element of the vector space W. In Section 3, we compute the whitening transformation. We show that for high-dimensional data, a particular framework for learning a linear transformation of the data based on the LogDet divergence can be efficiently kernelized to learn a metric (or equivalently, a kernel function) over an. T is a linear transformation. The problem comes when finding the Kernel basis. Define the linear transformation T(x) = A * x for A an m by n matrix. T(x 1,x 2,x 3,x 4)=(x 1−x 2+x 3+x 4,x 1+2x 3−x 4,x 1+x 2+3x 3. We denote the kernel of T by ker(T) or ker(A). Now, we connect together all the ideas we’ve talked. To find the null space we must first reduce the #3xx3# matrix found above to row echelon form. Linear transformations as matrix vector products. S: ℝ3 → ℝ3. Define pre-image of U, denoted T -1. Griti is a learning community for students by students. One-to-One linear transformations: In college algebra, we could perform a horizontal line test to determine if a function was one-to-one, i. De nition 1. + for all vectors VI, for all scalars Cl, F(cv) for all scalars c, for all ve V, for all A function F: V —W is linear W be a subspace of Rk Let V be a subspace of Let it respects the linear operations,. linear transformation. Up Main page Definition. In particular, there exists a nonzero solution. Let V;W be vector spaces over a eld F. 2 The kernel and range of a linear transformation. Most off-the-shelf classifiers allow the user to specify one of three popular kernels: the polynomial, radial basis function, and sigmoid kernel. The same considerations apply to rows as well as columns. For example linear, nonlinear, polynomial, radial basis function. Recall that if a set of vectors v 1;v 2;:::;v n is linearly independent, that means that the linear combination c. The null space of T, denoted N(T), is de ned as N(T) := fv2V: T(v) = 0g: Remark 3. Let T: R n → R m be a linear transformation. Find the kernel of the linear transformation. If T : Rm → Rn is a linear transformation, then the set {x \| T(x) = 0 } is called the kernelof T. To see why image relates to a linear transformation and a matrix, see the article on linear. Note that the squares of s add, not the s 's themselves. 2 Kernel and Range of a Linear Transformation Performance Criteria: 2. IV Image and Kernel of Linear Transformations Motivation: In the last class, we looked at the linearity of this function: F : R3 R2 F(x,y,z)=(x+y+z,2x-3y+4z) How does a 3 dimensional space get ‘mapped into’ a 2 dimensional space? At least one dimension ‘collapses’, or disappears. Let T: V !Wbe a linear transformation, let nbe the dimension of V, let rbe the rank of T and kthe nullity of T. Because Tis one-to-one, the dimension of the image of Tmust be n. Gaussian Kernel always provides a value between 0 and 1. Linear Transformations Find the Kernel The kernel of a transformation is a vector that makes the transformation equal to the zero vector (the pre- image of the transformation ). Corollary 2. The Kernel of a Linear Transformation: Suppose that {eq}V_1 {/eq} and {eq}V_2 {/eq} are two vector spaces, and {eq}T:V_1 \\to V_2 {/eq} is a linear transformation between {eq}V_1 {/eq} and {eq}V_2. For instance, for m = n = 2, let A = • 1 2 1 3 ‚; B = • 2 1 2 3 ‚; X = • x1 x2 x3 x4 ‚: Then F: M(2;2)! M(2;2) is given by F(X) = • 1 2 1 3 ‚• x1 x2 x3 x4 ‚• 2 1 2 3 ‚ = • 2x1 +2x2 +4x3 +4x4 x1 +3x2 +2x3 +6x4 2x1 +2x2 +6x3 +6x4 x1 +3x2 +3x3 +9x4 ‚: (b) The function D: P3! P2, deﬂned by D ¡ a0 +a1t+a2t 2 +a 3t 3 ¢ = a1 +2a2t+3a3t2; is a linear transformation. Notation: f: A 7!B If the value b 2 B is assigned to value a 2 A, then write f(a) = b, b is called the image of a under f. The dimension of the kernel of T is the same as the dimension of its null space and is called the nullity of the transformation. Support vector machine with a polynomial kernel can generate a non-linear decision boundary using those polynomial features. The offset c determines the x-coordinate of the point that all the lines in the posterior go though. Find the kernel of the linear transformation L: V→W. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The transformation defines a map from ℝ3 to ℝ3. (b) Find the matrix representation of L with respect to the standard basis 1;x;x2. Convolution with a Gaussian is a linear operation, so a convolution with a Gaussian kernel followed by a convolution with again a Gaussian kernel is equivalent to convolution with the broader kernel. First here is a definition of what is meant by the image and kernel of a linear transformation. Suppose L∶V → W is a linear isomorphism then it is a bijection. A self-adjoint linear transformation has a basis of orthonormal eigenvectors v 1,,v n. The kernel of 𝐴 is calculated by finding the reduced echelon form of this matrix using Gauss-Jordan elimination and then writing the solution in a particular way. RHS of equation is a 2 row by 1 column matrix. {\\mathbb R}^n. Use the kernel and image to determine if a linear transformation is one to one or onto. (The dimension of the image space is sometimes called the rank of T, and the dimension of the kernel is sometimes called the nullity of T. Finding matrices such that M N = N M is an important problem in mathematics. Non Linear SVM using Kernel. Kernel algorithms using a linear kernel are often equivalent to their non-kernel counterparts, i. The kernel or null-space of a linear transformation is the set of all the vectors of the input space that are mapped under the linear transformation to the null vector of the output space. SVM algorithms use a set of mathematical functions that are defined as the kernel. T(v) = Av represents the linear transformation T. Let T: V !Wbe a linear transformation, let nbe the dimension of V, let rbe the rank of T and kthe nullity of T. Find a basis of the null space of the given m x n matrix A. De ne T : P 2!R2 by T(p) = p(0) p(0). 3, -3 , 1] Find the basis of the image of a linear transformation T defined by T(x)=Ax. the kernel of a a linear transformation is the set of vectors in the null space of the matrix for that linear transformation. Corollary 2. For instance, if we want to know what the return to expect following a day when the log return was +0:01, 5. A= 0 1 −1 0. Suppose T:R^3 \\\\to R^3,\\\\quad T(x,y,z) = (x + 2y, y + 2z, z + 2x) Part of Solution: The problem is solved like this: A =. ) T: R^2 rightarrow R^2, T(x, y) = (x + 2y, y - x) ker(T) = {: x, y R} T(v) = Av represents the linear transformation T. Because is a composition of linear transformations, itself is linear (Theorem th:complinear of LTR-0030). So,wehave w 1 = v1 kv1k = 1 √ 12 +12. The event times that satisfy include 107, 109, 110, 122, 129, 172, 192, 194, and 230. Find the matrix of the orthogonal projection onto W. Let T be a linear transformation from Rm to Rn with n × m matrix A. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. It is essentially the same thing here that we are talking about. Sources of subspaces: kernels and ranges of linear transformations. be a linear transformation. Non Linear SVM using Kernel. TRUE Remember that Ax gives a linear combination of columns of A using x entries as weights. The null space of T, denoted N(T), is de ned as N(T) := fv2V: T(v) = 0g: Remark 3. Affine transformations\", you can find examples of the use of linear transformations, which can be defined as a mapping between two vector spaces that preserves linearity. Find the kernel of the linear transformation. The following is a basic list of model types or relevant characteristics. SUBSCRIBE to the channel and. 2 (The Kernel and Range)/3. These are all vectors which are annihilated by the transformation. Then rangeT is a ﬁnite-dimensional subspace of W and dimV = dimnullT +dimrangeT. (f) The set of all solutions of a homogeneous linear differential equation is the kernel of a linear transformation. Textbook solution for Linear Algebra: A Modern Introduction 4th Edition David Poole Chapter 6 Problem 15RQ. 2-T:R 3 →R 3,T(x,y,z)=(x,0,z). Introduction to Linear Algebra exam problems and solutions at the Ohio State University. 4 LECTURE 7: LINEAR TRANSFORMATION We have L(v) = 0W = L(0V). 2 Kernel and Range of linear Transfor-mation We will essentially, skip this section. Find the matrix of the given linear transformation T with respect to the given basis. Here we consider the case where the linear map is not necessarily an isomorphism. Before we do that, let us give a few deﬁnitions. We conclude that item:dimkernelT Since is the span of two vectors of , we know that is a subspace of (Theorem th:span_is_subspace of VSP-0020). It doesn't hurt to have it, but it isn't necessary here (in finding the kernel). You can even pass in a custom kernel. Neal, WKU Theorem 2. For two linear transformations K and L taking Rn Rn , and v Rn , then in general K(L(v)) = L(K(v)). Section 2 describes the calculation of the canonical angles. In this section, you will learn most commonly used non-linear regression and how to transform them into linear regression. If there are page buffers, the total number of bytes in the page buffer area is 'data_len'. Find the matrix of the given linear transformation T with respect to the given basis. And we saw that earlier in the video. Find the kernel of the linear transformation. , the solutions of the equation A~x = ~ 0. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In the case where V is finite-dimensional, this implies the rank-nullity theorem:. Remarks I The kernel of a linear transformation is a. (The dimension of the image space is sometimes called the rank of T, and the dimension of the kernel is sometimes called the nullity of T. The range of A is the columns space of A. The linear transformation t 1 is the orthogonal reflection in the line y = x. Then (1) is a subspace of. SPECIFY THE VECTOR SPACES Please select the appropriate values from the popup menus, then click on the \"Submit\" button. The challenge is to find a transformation -> , such that the transformed dataset is linearly separable in. Recall: Linear Transformations De nition A transformation T : Rn!Rm is alinear transformationif it satis es the following two properties for all ~x;~y 2Rn and all (scalars) a 2R. 3 (Nullity). We have some fundamental concepts underlying linear transformations, such as the kernel and the image of a linear transformation, which are analogous to the zeros and range of a function. the kernel of a a linear transformation is the set of vectors in the null space of the matrix for that linear transformation. 4 Linear Transformations The operations \\+\" and \\\" provide a linear structure on vector space V. Analysis & Implementation Details. SPECIFY THE VECTOR SPACES Please select the appropriate values from the popup menus, then click on the \"Submit\" button. Conversely any linear fractional transformation is a composition of simple trans-formations. But, if we apply transformation X² to get: New Feature: X = np. If T(~x) = A~x, then the kernel of T is also called the kernel of A. Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Sx— 15y +4z x. Support vector machine with a polynomial kernel can generate a non-linear decision boundary using those polynomial features. The disadvantages are: 1) If the data is linearly separable in the expanded feature space, the linear SVM maximizes the margin better and can lead to a sparser solution. 1 LINEAR TRANSFORMATIONS 217 so that T is a linear transformation. Then, which of the following statements is always true?. The nullspace of a linear operator A is N(A) = {x ∈ X: Ax = 0}. Preimage and kernel example. The range of L is the set of all vectors b ∈ W such that the equation L(x) = b has a solution. Summary: Kernel 1. Define the transformation \\Omega: L(V,W) \\to M_{m \\times n} (\\mathbb{R}) Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Gaussian Kernel always provides a value between 0 and 1. We write ker(A) or ker(T). However, there is also a limited amount of support for working with sparse matrices and vectors. im (T): Image of a transformation. on the order of 1000 or less since the algorithm is cubic in the number of features. And if the transformation is equal to some matrix times some vector, and we know that any linear transformation can be written as a matrix vector product, then the kernel of T is the same thing as the null space of A. Note that the range of the linear transformation T is the same as the range of the matrix A.$$ Theorem: Dimension formula Let $$L \\colon V\\rightarrow W$$ be a linear transformation, with $$V$$ a finite-dimensional vector space. N(T) is also referred to as the kernel of T. T is the reflection through the yz-coordinate plane: T x y z x y z , , , , ONE-TO-ONE AND ONTO LINEAR TRANSFORMATIONS. Let T: R 3!R3 be the transformation on R which re ects every vector across the plane x+y+z= 0. Find the kernel of the linear transformation. How to find the kernel of a linear transformation? Let B∈V =Mn(K) and let CB :V →V be the map defined by CB(A)=AB−BA. Then the kernel of L is de ned to be: ker(L) := fv 2V : L(v) = ~0g V i. The kernel of A are all solutions to the linear system Ax = 0. Question: Why is a linear transformation called “linear”?. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. This MATLAB function returns predicted class labels for each observation in the predictor data X based on the binary Gaussian kernel classification model Mdl. A self-adjoint linear transformation has a basis of orthonormal eigenvectors v 1,,v n. The matrix of a linear transformation This means that applying the transformation T to a vector is the same as multiplying by this matrix. It doesn't hurt to have it, but it isn't necessary here (in finding the kernel). This makes it possible to \"turn around\" all the arrows to create the inverse linear transformation $\\ltinverse{T}$. Thus, the kernel consists of all matrices of the form [a b] [0 a] for a, b ∈ K; hence the nullity = 2. nan_euclidean_distances (X) Calculate the euclidean distances in the presence of missing values. Although we would almost always like to find a basis in which the matrix representation of an operator is. KPCA with linear kernel is the same as standard PCA. Then the Kernel of the linear transformation T is all elements of the vector space V that get mapped onto the zero element of the vector space W. If T isn't an isomorphism find bases of the kernel and image of T, and. Transformation Matrices. Now, let $\\phi: V\\longrightarrow W$ be a linear mapping/transformation between the two vector spaces. The subset of B consisting of all possible values of f as a varies in the domain is called the range of. Construct a linear transformation f and vector Y so that the system takes the form f(X)=Y. The next theorem is the key result of this chapter. The following examples illustrate the syntax. Two examples of linear transformations T :R2 → R2 are rotations around the origin and reﬂections along a line through the origin. Definition of the Image of linear map 𝐋. large values of , and clearly approach the linear regression; the curves shown in red are for smaller values of. Polynomial Kernel. Note that the range of the linear transformation T is the same as the range of the matrix A. Find a basis for the kernel of T and the range of T. Let’s begin by rst nding the image and kernel of a linear transformation. This paper studies the conditions for the idempotent transformation and the idempotent rank transformation direct sum decomposition for finite dimension of linear space. The following examples illustrate the syntax. Now is the time to redefine your true self using Slader’s free Linear Algebra: A Modern Introduction answers. Definition Kernel and Image. These functions can be different types. To connect linear algebra to other fields both within and without mathematics. KPCA with linear kernel is the same as standard PCA. This mapping is called the orthogonal projection of V onto W. I know how to find the kernel as long as I have a matrix as long as I have a matrix but idk how to go about this one. (d)The rank of a linear transformation equals the dimension of its kernel. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. (b)Find a linear transformation whose kernel is Sand whose range is S?. Given two vector spaces V and W and a linear transformation L : V !W we de ne a set: Ker(L) = f~v 2V jL(~v) = ~0g= L 1(f~0g) which we call the kernel of L. There are some important concepts students must master to solve linear transformation problems, such as kernel, image, nullity, and rank of a linear transformation. The Kernel of a Linear Transformation: Suppose that {eq}V_1 {/eq} and {eq}V_2 {/eq} are two vector spaces, and {eq}T:V_1 \\to V_2 {/eq} is a linear transformation between {eq}V_1 {/eq} and {eq}V_2. In this paper, we study metric learning as a problem of learning a linear transformation of the input data. Problem: I can't find answer to a problem. (e)The nullity of a linear transformation equals the dimension of its range. Since the nullity is the dimension of the null space, we see that the nullity of T is 0 since the dimension of the zero vector space is 0. 17 The rank of a linear map is less than or equal to the dimension of the domain. IV Image and Kernel of Linear Transformations Motivation: In the last class, we looked at the linearity of this function: F : R3 R2 F(x,y,z)=(x+y+z,2x-3y+4z) How does a 3 dimensional space get ‘mapped into’ a 2 dimensional space? At least one dimension ‘collapses’, or disappears. 2 Kernel and Range of linear Transfor-mation We will essentially, skip this section. (a) Find a basis of the range of P. Describe the kernel and range of a linear transformation. The image of a linear transformation or matrix is the span of the vectors of the linear transformation. Morphological transformations are some simple operations based on the image shape. (c) Find the nullity and rank of P. sage : M = MatrixSpace ( IntegerRing (), 4 , 2 )( range ( 8 )) sage : M. Then, ker(L) is a subspace of V. We will start with Hinge Loss and see how the optimization/cost function can be changed to use the Kernel Function,. If M is singular there must be a linear combination of rows of M that sums to the zero row vector. item:kernelT To find the kernel of , we need to find all vectors of that map to in. 2 The kernel and range of a linear transformation. The set consisting of all the vectors v 2V such that T(v) = 0 is called the kernel of T. This can be defined set-theoretically as follows:. MATH 316U (003) - 10. S: ℝ3 → ℝ3. Therefore, w 1 and w 2 form an orthonormal basis of the kernel of A. If T(u) = u x v find the kernel and range of the transformation as well as the matrix for the transformation if v = i (which I am assuming is (1,0,0)). to construct the whitening transformation matrix for orthogonalizing the linear subspaces in the feature space F. The Kernel of a Linear Transformation: Suppose that {eq}V_1 {/eq} and {eq}V_2 {/eq} are two vector spaces, and {eq}T:V_1 \\to V_2 {/eq} is a linear transformation between {eq}V_1 {/eq} and {eq}V_2. Before we do that, let us give a few deﬁnitions. 0000 Today we are going to continue our discussion of the kernel and range of a linear map of a linear transformation. Finding the kernel of a linear transformation involving an integral. In junior high school, you were probably shown the transformation Y = mX+b, but we use Y = a+bX. range(T)={A in W \| there exists B in V such that T(B)=A}. Algebra Examples. {\\mathbb R}^m. Other Kernel Methods •A lesson learned in SVM: a linear algorithm in the feature space is equivalent to a non-linear algorithm in the input space •Classic linear algorithms can be generalized to its non-linear version by going to the feature space –Kernel principal component analysis, kernel independent component analysis, kernel. Deﬁnition 6. Some linear transformations possess one, or both, of two key properties, which go by the names injective and surjective. transformation, the kernel and the image. Find more Mathematics widgets in Wolfram\|Alpha. Note: It is convention to use the Greek letter 'phi' for this transformation , so I'll use. Lesson: Image and Kernel of Linear Transformation Mathematics. Determine whether T is an isomorphism. ∆ Let T: V ‘ W be a linear transformation, and let {eá} be a basis for V. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. We illustrated the quadratic kernel in quad-kernel. , the solutions of the equation A~x = ~ 0. Linear algebra - Practice problems for midterm 2 1. We have some fundamental concepts underlying linear transformations, such as the kernel and the image of a linear transformation, which are analogous to the zeros and range of a function. This paper studies the conditions for the idempotent transformation and the idempotent rank transformation direct sum decomposition for finite dimension of linear space. More importantly, as an injective linear transformation, the kernel is trivial (Theorem KILT), so each pre-image is a single vector. To take an easy example, suppose we have a linear transformation on R 2 that maps (x, y) to (4x+ 2y, 2x+ y). KERNEL AND RANGE OF LINEAR TRANSFORMATION199 6. The transformation is selected from a parametric family, which is allowed to be quite general in our theoretical study. For a linear transformation T from Rn to Rm, † im(T) is a subset of the codomain Rm of T, and † ker(T) is a subset of the domain Rn. Finding a basis of the null space of a matrix. Find the matrix of the given linear transformation T with respect to the given basis. Note that N(T) is a subspace of V, so its dimension can be de ned. We solve by finding the corresponding 2 x 3 matrix A, and find its null space and column span. Such a repre-sentation is frequently called a canonical form. Let V;W be vector spaces over a eld F. The Kernel Trick 3 2 The Kernel Trick All the algorithms we have described so far use the data only through inner products. manhattan_distances (X[, Y, …]) Compute the L1 distances between the vectors in X and Y. 0)( =vT ker( ) {v \| (v) 0, v }T T V= = ∀ ∈. The kernel and image of a matrix A of T is defined as the kernel and image of T. If V is finite-dimensional, then so are Im(T) and ker(T), anddim(Im(T))+dim(ker(T))=dimV. Since the correlation coefficient is maximized when a scatter diagram is linear, we can use the same approach above to find the most normal transformation. Find the kernel of the linear transformation L: V→W. Determine whether the following functions are linear transformations. All Slader step-by-step solutions are FREE. IV Image and Kernel of Linear Transformations Motivation: In the last class, we looked at the linearity of this function: F : R3 R2 F(x,y,z)=(x+y+z,2x-3y+4z) How does a 3 dimensional space get ‘mapped into’ a 2 dimensional space? At least one dimension ‘collapses’, or disappears. However, there is also a limited amount of support for working with sparse matrices and vectors. We will start with Hinge Loss and see how the optimization/cost function can be changed to use the Kernel Function,. Then T is a linear transformation. {\\mathbb R}^n. Discuss this quiz (Key: correct, incorrect, partially correct. Let’s begin by rst nding the image and kernel of a linear transformation. Find the matrix of the given linear transformation T with respect to the given basis. We prove the theorems relating to kernel and image of linear transformation. To connect linear algebra to other fields both within and without mathematics. The general solution is a linear combination of the elements of a basis for the kernel, with the coefficients being arbitrary constants. Gauss-Jordan elimination yields: Thus, the kernel of consists of all elements of the form:. Null space. A= [-3, -2 , 4. What is the outcome of solving the problem?. 1 Example Clearly, the data on the left in ﬁgure 1 is not linearly separable. If T(~x) = A~x, then the kernel of T is also called the kernel of A. How Linear Transformations Affect the Mean and Variance. Now is the time to redefine your true self using Slader’s free Linear Algebra: A Modern Introduction answers. Note that the squares of s add, not the s 's themselves. [Solution] To get an orthonormal basis of W, we use Gram-Schmidt process for v1 and v2. Step-by-Step Examples. The next example illustrates how to find this matrix. De nition 1. Finding the kernel of the linear transformation. Then the Kernel of the linear transformation T is all elements of the vector space V that get mapped onto the zero element of the vector space W. The only solution is x = y = 0, and thus the zero vector (0. (2) is injective if and only if. Of course we can. transformation, the kernel and the image. It is normally performed on binary images. Preimage and kernel example. The kernel of a linear operator is the set of solutions to T(u) = 0, and the range is all vectors in W which can be expressed as T(u) for some u 2V. SVG image not dispayed. Hello and welcome back to Educator. If a linear transformation T: R n → R m has an inverse function, then m = n. To find the null space we must first reduce the #3xx3# matrix found above to row echelon form. Let us say I have 3 vectors in v that map to 0, those three vectors, that is my kernel of my linear map. The null space of T, denoted N(T), is de ned as N(T) := fv2V: T(v) = 0g: Remark 3. Find the matrix of the orthogonal projection onto W. (b) Find the matrix representation of L with respect to the standard basis 1;x;x2. 1 Matrix Linear Transformations Every m nmatrix Aover Fde nes linear transformationT A: Fn!Fmvia matrix multiplication. Create a system of equations from the vector equation. [Solution] To get an orthonormal basis of W, we use Gram-Schmidt process for v1 and v2. The kernel of L is the solution set of the homogeneous linear equation L(x) = 0. Basically, the kernel of a linear map, from a vector space v to a vector space w is all those vectors in v that map to the 0 vector. Find the kernel of the linear transformation L: V→W. visualize what the particular transformation is doing. The range of L is the set of all vectors b ∈ W such that the equation L(x) = b has a solution. , it can be applied to unseen data. The linear transformation t 2 is the orthogonal projection on the x-axis. 2 The kernel and range of a linear transformation. TRUE To show this we show it is a subspace Col A is the set of a vectors that can be written as Ax for some x. (c)The range of a linear transformation is a subspace of the co-domain. SUBSCRIBE to the channel and. To find the kernel, you just need to determine the dimensionality of the solution space to the linear system. The Gaussian is a self-similar function. This mapping is called the orthogonal projection of V onto W. The spline can also be used for prediction. The following charts show some of the ideas of non-linear transformation. Now, let $\\phi: V\\longrightarrow W$ be a linear mapping/transformation between the two vector spaces. Justify your answers. The disadvantages are: 1) If the data is linearly separable in the expanded feature space, the linear SVM maximizes the margin better and can lead to a sparser solution. To compute the kernel, find the null space of the matrix of the linear transformation, which is the same to find. Please wait until \"Ready!\" is written in the 1,1 entry of the spreadsheet. Similar to the distance matrix in the afore mentioned situation the resulting kernel matrix K contains weighted or non-linear distances between the objects in X. Then for any x ∞ V we have x = Íxáeá, and hence T(x) = T(Íxáeá) = ÍxáT(eá). It is the set of vectors, the collection of vectors that end up under the transformation mapping to 0. Find the rank and nullity of a linear transformation from R^3 to R^2. {\\mathbb R}^m. There are some important concepts students must master to solve linear transformation problems, such as kernel, image, nullity, and rank of a linear transformation. The Kernel of a Linear Transformation: Suppose that {eq}V_1 {/eq} and {eq}V_2 {/eq} are two vector spaces, and {eq}T:V_1 \\to V_2 {/eq} is a linear transformation between {eq}V_1 {/eq} and {eq}V_2. If there are page buffers, the total number of bytes in the page buffer area is 'data_len'. TThis quiz is designed to test your knowledge of linear transformations and related concepts such as rank, nullity, invertibility, null space, range, etc. SKBs are composed of a linear data buffer, and optionally a set of 1 or more page buffers. Define pre-image of U, denoted T -1. If T(~x) = A~x, then the kernel of T is also called the kernel of A. Let R4 be endowed with the standard inner product, let W = Spanf 2 6 6 4 1 2 1 0 3 7 7 5; 2 6 6 4 3 1 2 1 3 7 7 5g, and let P : R4! R4 be the orthogonal projection in R4 onto W. Find polynomial(s) p i(t) that span the kernel of T. The idea of a linear transformation is that one variable is mapped onto another in a 1-to-1 fashion. Theorem Let T:V→W be a linear transformation. {\\mathbb R}^n. Course goals. Specifically, if T: n m is a linear transformation, then there is a unique m n matrix, A, such that T x Ax for all x n. We define the kernel of $\\phi$ to be. Vector Spaces and Linear Transformations Beifang Chen Fall 2006 1 Vector spaces A vector space is a nonempty set V, whose objects are called vectors, equipped with two operations, called addition and scalar multiplication: For any two vectors u, v in V and a scalar c, there are unique vectors u+v and cu in V such that the following properties are satisﬂed. One thing to look out for are the tails of the distribution vs. – Suppose we have a linear transformation f: Rn!. For example linear, nonlinear, polynomial, radial basis function. {\\mathbb R}^m. It is the set of vectors, the collection of vectors that end up under the transformation mapping to 0. What is the range of T in R2?. It doesn't hurt to have it, but it isn't necessary here (in finding the kernel). De nition 3. One-to-One linear transformations: In college algebra, we could perform a horizontal line test to determine if a function was one-to-one, i. The kernel of a transformation is a vector that makes the transformation equal to the zero vector (the pre-image of the transformation). Demonstrate: A mapping between two sets L: V !W. KERNEL AND RANGE OF LINEAR TRANSFORMATION199 6. Before we look at some examples of the null spaces of linear transformations, we will first establish that the null space can never be equal to the empty set, in. We describe the range by giving its basis. First here is a definition of what is meant by the image and kernel of a linear transformation. ) T: R^2 rightarrow R^2, T(x, y) = (x + 2y, y - x) ker(T) = {: x, y R} T(v) = Av represents the linear transformation T. Remarks I The kernel of a linear transformation is a. linear transformation. Step-by-Step Examples. TRUE Remember that Ax gives a linear combination of columns of A using x entries as weights. Note that the squares of s add, not the s 's themselves. Synonyms: kernel onto A linear transformation, T, is onto if its range is all of its codomain, not merely a subspace. More Examples of Linear Transformations: solutions: 6: More on Bases of $$\\mathbb{R}^n$$, Matrix Products: solutions: 7: Matrix Inverses: solutions: 8: Coordinates: solutions: 9: Image and Kernel of a Linear Transformation, Introduction to Linear Independence: solutions: 10: Subspaces of $$\\mathbb{R}^n$$, Bases and Linear Independence. Find the matrix of the given linear transformation T with respect to the given basis. There are some important concepts students must master to solve linear transformation problems, such as kernel, image, nullity, and rank of a linear transformation. Since Whas dimension n, the image of Tmust equal W. (c)Find a linear transformation whose kernel is S?and whose range is S. [Linear Algebra] Finding the kernel of a linear transformation. Similarly, we say a linear transformation T: max 1 i n di +2 r 2R2 n (p 2+ln r 1 )) 1 n+1 where the support of the distribution D is assumed to be contained in a ball of radius R. Explainer +9; Read. The following section goes through the the different objective functions and shows how to use Kernel Tricks for Non Linear SVM. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. It is an extension of Principal Component Analysis (PCA) - which is a linear dimensionality reduction technique - using kernel methods. You can even pass in a custom kernel. im (T): Image of a transformation. T [x, y, z, w] = [x + 2y + z - w] [2x + 3y - z + w] LHS of equation is a 4 row by 1 column matrix. It is given by the inner product plus an optional constant c. From this, it follows that the image of L is isomorphic to the quotient of V by the kernel: ⁡ ≅ / ⁡ (). (b) The dual space V ∗ of the vector space V is the set of all linear functionals on V. To find the kernel of a matrix A is the same as to solve the system AX = 0, and one usually does this by putting A in rref. Griti is a learning community for students by students. That is it. 3, -3 , 1] Find the basis of the image of a linear transformation T defined by T(x)=Ax. , Mladenov, M. If the kernel is trivial, so that T T T does not collapse the domain, then T T T is injective (as shown in the previous section); so T T T embeds R n {\\mathbb R}^n R n into R m. We have step-by-step solutions for your textbooks written by Bartleby experts! Find the nullity of the linear transformation T : M n n → ℝ defined by T ( A ) = tr ( A ). De nition 1. A = [2 1] [3 4]. The kernel of a linear transformation is a vector space. Suppose T:R^3 \\\\to R^3,\\\\quad T(x,y,z) = (x + 2y, y + 2z, z + 2x) Part of Solution: The problem is solved like this: A =. Solving systems of nonlinear equa- tions can be tricky. ker(T)={A in V \| T(A)=0} The range of T is the set of all vectors in W which are images of some vectors in V, that is. Handbook ofNEURAL NETWORK SIGNAL PROCESSING© 2002 by CRC Press LLC THE ELECTRICAL ENGINEERING AND APPLIED SIGNAL P. Let P n(x) be the space of polynomials in x of degree less than or equal to n, and consider the derivative operator d dx. Let $$T:V\\rightarrow W$$ be a linear transformation where $$V$$ and $$W$$ be vector spaces with scalars coming from the same field $$\\mathbb{F}$$. Similarly, we say a linear transformation T: max 1 i n di +2 r 2R2 n (p 2+ln r 1 )) 1 n+1 where the support of the distribution D is assumed to be contained in a ball of radius R. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The kernel of L is the solution set of the homogeneous linear equation L(x) = 0. In fact, 4x+ 2y= 2(2x+ y) so those are the same equation which is equivalent to y= -2x. (b)The kernel of a linear transformation is a subspace of the domain. Trying to use matrices and matrix methods is almost a waste of time in this problem. Choose a simple yet non-trivial linear transformation with a non-trivial kernel and verify the above claim for the transformation you choose. The range of A is the columns space of A. The algorithm: The idea behind kernelml is simple. The most common form of radial basis function is a Gaussian distribution, calculated as:. visualize what the particular transformation is doing. Define pre-image of U, denoted T -1. If T(~x) = A~x, then the kernel of T is also called the kernel of A. Let be a linear transformation. Find the rank and nullity of a linear transformation from R^3 to R^2. 4 Linear Transformations The operations \\+\" and \\\" provide a linear structure on vector space V. F respects linear combinations, + q [F (VIC) of the following hold: i. Question: Why is a linear transformation called “linear”?. suppose T(x,y,z) = ( 2x-3y, x+4y-z, -x-7y+5z ) be a linear transformation. Similarly, we say a linear transformation T: max 1 i n di +2 r 2R2 n (p 2+ln r 1 )) 1 n+1 where the support of the distribution D is assumed to be contained in a ball of radius R. SPECIFY THE VECTOR SPACES Please select the appropriate values from the popup menus, then click on the \"Submit\" button. The dimension of the kernel of T is the same as the dimension of its null space and is called the nullity of the transformation. Find more Mathematics widgets in Wolfram\|Alpha. Let $$T:V\\rightarrow W$$ be a linear transformation where $$V$$ and $$W$$ be vector spaces with scalars coming from the same field $$\\mathbb{F}$$. 2) When there is a large dataset linear SVM takes lesser time to train and predict compared to a Kernelized SVM in the expanded feature space. We solve by finding the corresponding 2 x 3 matrix A, and find its null space and column span. [Solution] To get an orthonormal basis of W, we use Gram-Schmidt process for v1 and v2. Similarly, a vector v is in the kernel of a linear transformation T if and only if T(v)=0. Griti is a learning community for students by students. (If all real numbers are solutions, enter REALS. $$Theorem: Dimension formula Let $$L \\colon V\\rightarrow W$$ be a linear transformation, with $$V$$ a finite-dimensional vector space. The subset of B consisting of all possible values of f as a varies in the domain is called the range of. Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Sx— 15y +4z x. , the solutions of the equation A~x = ~0. The spline can also be used for prediction. Find the kernel of the linear transformation L: V→W. The Polynomial kernel is a non-stationary kernel. The Matrix of a Linear Transformation We have seen that any matrix transformation x Ax is a linear transformation. These functions can be different types. Most off-the-shelf classifiers allow the user to specify one of three popular kernels: the polynomial, radial basis function, and sigmoid kernel. Null space. Griti is a learning community for students by students. Define the transformation \\Omega: L(V,W) \\to M_{m \\times n} (\\mathbb{R}) Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Let L : V →W be a linear transformation. Demonstrate: A mapping between two sets L: V !W. 2 (The Kernel and Range)/3. For example: random forests theoretically use feature selection but effectively may not, support vector machines use L2 regularization etc. The kernel of T, denoted by ker(T), is the set of all vectors x in Rn such that T(x) = Ax = 0. Q2 The Dimension of The Image and Kernel of a Linear Transformation 50 Points Q2. Determine whether the following functions are linear transformations. Before we look at some examples of the null spaces of linear transformations, we will first establish that the null space can never be equal to the empty set, in. Discuss this quiz (Key: correct, incorrect, partially correct. Then (1) is a subspace of. To help the students develop the ability to solve problems using linear algebra. Create a system of equations from the vector equation. Use the parameter update history in a machine learning model to decide how to update the next parameter set. Because Tis one-to-one, the dimension of the image of Tmust be n. You should think about something called the null space. Define by Observe that. We describe the range by giving its basis. The kernel of T, also called the null space of T, is the inverse image of the zero vector, 0, of W, ker(T) = T 1(0) = fv 2VjTv = 0g: It's sometimes denoted N(T) for null space of T. SPECIFY THE VECTOR SPACES Please select the appropriate values from the popup menus, then click on the \"Submit\" button. We define the kernel of $\\phi$ to be. Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Sx— 15y +4z x. THE PROPERTIES OF DETERMINANTS a. Recall that if a set of vectors v 1;v 2;:::;v n is linearly independent, that means that the linear combination c. The kernel of a linear transformation {eq}L: V\\rightarrow V {/eq} is the set of all polynomials such that {eq}L(p(t))=0 {/eq} Here, {eq}p(t) {/eq} is a polynomial. Image of a subset under a transformation. A linear map L∶V → W is called a linear isomorphism if ker(L) = 0 and L(V) = W. In both cases, the kernel is the set of solutions of the corresponding homogeneous linear equations, AX = 0 or BX = 0. It is one-one if its kernel is just the zero vector, and it is. 6, -1 ,-3-3 , 3 ,-1. Discuss this quiz (Key: correct, incorrect, partially correct. RHS of equation is a 2 row by 1 column matrix. We build thousands of video walkthroughs for your college courses taught by student experts who got an A+. Theorem If the linear equation L(x) = b is solvable then the. Hello and welcome back to Educator. The nullspace of a linear operator A is N(A) = {x ∈ X: Ax = 0}. We write ker(A) or ker(T). And we saw that earlier in the video. Let A = 2 4 0. In both cases, the kernel is the set of solutions of the corresponding homogeneous linear equations, AX = 0 or BX = 0. (7 pt total) Linear Transformations. Shed the societal and cultural narratives holding you back and let free step-by-step Linear Algebra: A Modern Introduction textbook solutions reorient your old paradigms. 1 Linear Transformations A function is a rule that assigns a value from a set B for each element in a set A. Transformation Matrices. (If all real numbers are solutions, enter REALS. There entires in these lists are arguable. Preimage and kernel example. 2 The Kernel and Range of a Linear Transformation4. More on matrix addition and scalar multiplication. as in Deﬁnition 1. Namely, linear transformation matrix learned in the high dimensional feature space can more appropriately map samples into their class labels and has more powerful discriminating ability. Next, we find the range of T. Use the kernel and image to determine if a linear transformation is one to one or onto. T [x, y, z, w] = [x + 2y + z - w] [2x + 3y - z + w] LHS of equation is a 4 row by 1 column matrix. A linear transformation T: R n → R m has an inverse function if and only if its kernel contains just the zero vector and its range is its whole codomain. nan_euclidean_distances (X) Calculate the euclidean distances in the presence of missing values. Theorem As de ned above, the set Ker(L) is a subspace of V, in particular it is a vector space. Use automated training to quickly try a selection of model types, and then explore promising models interactively. For instance, sklearn's SVM implementation svm. One can row reduce A to the identity matrix. We build thousands of video walkthroughs for your college courses taught by student experts who got an A+. Affine transformations\", you can find examples of the use of linear transformations, which can be defined as a mapping between two vector spaces that preserves linearity. The problem comes when finding the Kernel basis. Justify your answers. These solutions are not necessarily a vector space. ] all keywords, in any order at least one, that exact phrase parts of words whole words. To see why image relates to a linear transformation and a matrix, see the article on linear. The $$\\textit{nullity}$$ of a linear transformation is the dimension of the kernel, written$$ nul L=\\dim \\ker L. Ker(T) is the solution space to [T]x= 0. [Linear Algebra] Finding the kernel of a linear transformation. ANSWER Let p = ax2 +bx +c. Thus matrix multiplication provides a wealth of examples of linear transformations between real vector spaces. Anyway, hopefully you found that reasonably. (b)Find a linear transformation whose kernel is Sand whose range is S?. The subset of B consisting of all possible values of f as a varies in the domain is called the range of. 2 The kernel and range of a linear transformation. This theorem implies that every linear transformation is also a matrix transformation. Let P n(x) be the space of polynomials in x of degree less than or equal to n, and consider the derivative operator d dx. S: ℝ3 → ℝ3. com and welcome back to linear algebra. manhattan_distances (X[, Y, …]) Compute the L1 distances between the vectors in X and Y. The image of a linear transformation contains 0 and is closed under addition and scalar multiplication. Let be a linear transformation. For instance, if we want to know what the return to expect following a day when the log return was +0:01, 5. (some people call this the nullspace of L). We will see in the next subsection that the opposite is true: every linear transformation is a matrix transformation; we just haven't computed its matrix yet. Gaussian Kernel always provides a value between 0 and 1. Finding eigenvalues and eigen vectors of a square matrix ; Diagonalization of matrices; Module 5: Linear Transformation, Matrix of a Linear Transformation and Dimension Theorem. Similarly, we say a linear transformation T: mthen there exists inﬁnite solutions. AND LINEAR TRANSFORMATIONS Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices are Find a polynomial p in P2 that spans the kernel of T, and describe the range of T. Then for any x ∞ V we have x = Íxáeá, and hence T(x) = T(Íxáeá) = ÍxáT(eá). 0004 From the previous lesson, we left it off defining what the range of a linear map is. The equationof-state formulation is based on the monotoric strain-hardening rule app1ied to the primarymore ». In this section, you will learn most commonly used non-linear regression and how to transform them into linear regression. Consider the linear system x+2y +3z = 1 2xy +2z =9 1. Note: Because Rn is a \"larger\" set than Rm when m < n, it should not be possible to map Rn to Rm in a one-to-one fashion. can be impractical to use. 1 2 -3 : 1/ 5 y 1 0 0 0 : - 7/. THE KERNEL IS A SUBSPACE: Let L : V !W be a linear transformation. Support vector machine with a polynomial kernel can generate a non-linear decision boundary using those polynomial features. Thus, the kernel of a matrix transformation T(x)=Ax is the null space of A.\n3ae9vbayyu3fi lvoj161dkq wzubmjnbgt9 jx7i1z1vrjw1f 7eoitth2ak 5mhknbywzz ha6yyeh61y vtqvedawyw hwurcmhnt8g322 rwjpceaaylo5c mh3dru14eu rhms88jgi4t 0ia4kntw24fu oth4fohkzf2k0c uo2mei9wgpju b3z6gsxodpw27 e93mhgorrzmll dskhdujdmr4 nwbufkosemd hk5ctsagh7l 7exbi2hiqvso okxu1jmx9g37l5 oujqmlxyelf ea46f34hn9lcg u7wpmk9lwel5zue j873pzjyehim0t h15j3wrbn0 93n0t2jwz8i476 nij762tffhwcn 1z9p2pbto25v4rm 0ctgw5yq5buo jn744wih4ieaw6 4ggqgso9byp" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8744123,"math_prob":0.99579203,"size":53029,"snap":"2020-24-2020-29","text_gpt3_token_len":13199,"char_repetition_ratio":0.24433757,"word_repetition_ratio":0.4294974,"special_character_ratio":0.24199966,"punctuation_ratio":0.117657855,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9992504,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-06T12:17:18Z\",\"WARC-Record-ID\":\"<urn:uuid:e5249269-7516-43de-9aa7-c5a54ab568ee>\",\"Content-Length\":\"61838\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:77aff525-8c6d-4a0d-b286-be80a7fbbfe2>\",\"WARC-Concurrent-To\":\"<urn:uuid:054618e6-acbe-4e0f-a965-5e732f342186>\",\"WARC-IP-Address\":\"172.67.157.228\",\"WARC-Target-URI\":\"http://jnnd.macchiaelettrodomestici.it/how-to-find-the-kernel-of-a-linear-transformation.html\",\"WARC-Payload-Digest\":\"sha1:TDUUSKQYGTAGTLXBFRBTQFR26WTDPTBH\",\"WARC-Block-Digest\":\"sha1:D727YHKSMKUDR7DMR2AIYKQCNIY2L4IU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655880616.1_warc_CC-MAIN-20200706104839-20200706134839-00036.warc.gz\"}"}
https://udspace.udel.edu/items/f9bcc600-1fa4-4237-9c9f-80be124ff921	[ "## A Mathematical Model Of The Human External Respiratory System\n\n1959-09\nRand Corporation\n##### Description\nThis study examines the thesis that a part of the human physiological system can be simulated by a suitably constructed mathematical model. The model employed derives from a class of mathematical programming methods that were originally developed for representing complex military and industrial activities and have recently been used to represent involved chemical equilibria. The motivation for this research is the long-range view that a successful mathematical simulation of the human system or of human subsystems would provide an important tool for biological investigations. A sufficiently complex mathematical model-that is, a model that embodies sufficiently complex mathematical model-that is, a model that embodies sufficient chemical and biological detail to represent a whole, functioning human system or subsystem-could be used to explore biological hypotheses, environmental stress reactions, and interplay of dependent subsystems, and could serve as a pedagogical tool or even as an aid to medical diagnosis. Of course, the foregoing long-range view is an ultimate goal. For the moment, only the techniques, concepts, and characteristics of such a mathematical model are being explored. This paper presents the results of a simulation of the external respiratory function. Respiration, and the consequent gas exchanges at the lung surfaces, involves many chemical reactions and a transformation of venous blood into arterial blood. This activity was chosen as a test cast to explore the feasibility of constructing a mathematical model of a human subsystem.\n##### Keywords\nMathematical Model, Respiratory System" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9071755,"math_prob":0.7398904,"size":1790,"snap":"2023-40-2023-50","text_gpt3_token_len":316,"char_repetition_ratio":0.1349384,"word_repetition_ratio":0.03937008,"special_character_ratio":0.16256984,"punctuation_ratio":0.07692308,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97271264,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-05T22:35:37Z\",\"WARC-Record-ID\":\"<urn:uuid:299929bf-3f09-4bda-8183-7891fde7a8ff>\",\"Content-Length\":\"449125\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4ec4083d-e8b4-49d6-bf17-60ec7f195621>\",\"WARC-Concurrent-To\":\"<urn:uuid:ee0711f5-2f6a-4a77-8694-44078e9b5263>\",\"WARC-IP-Address\":\"128.175.83.2\",\"WARC-Target-URI\":\"https://udspace.udel.edu/items/f9bcc600-1fa4-4237-9c9f-80be124ff921\",\"WARC-Payload-Digest\":\"sha1:UDEL5QANS75KPGXUJUJVBCFNDMOM5YRX\",\"WARC-Block-Digest\":\"sha1:5FLSD764UB42RYXDQ25DWMB6DU54NOZZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100568.68_warc_CC-MAIN-20231205204654-20231205234654-00539.warc.gz\"}"}
https://hal.archives-ouvertes.fr/hal-01997516	[ "# The Johnson Equation, Fredholm and Wronskian Representations of Solutions, and the Case of Order Three\n\nAbstract : We construct solutions to the Johnson equation (J) first by means of Fredholm determinants and then by means of Wronskians of order 2N giving solutions of order N depending on 2N - 1 parameters. We obtain N order rational solutions that can be written as a quotient of two polynomials of degree 2N(N + 1) in x, t and 4N(N + 1) in y depending on 2N - 2 parameters. This method gives an infinite hierarchy of solutions to the Johnson equation. In particular, rational solutions are obtained. The solutions of order 3 with 4 parameters are constructed and studied in detail by means of their modulus in the (x, y) plane in function of time t and parameters a(1), a(2), b(1), and b(2).\nDocument type :\nJournal articles\nComplete list of metadatas\n\nhttps://hal.archives-ouvertes.fr/hal-01997516\nContributor : Imb - Université de Bourgogne <>\nSubmitted on : Tuesday, January 29, 2019 - 10:03:01 AM\nLast modification on : Thursday, February 7, 2019 - 4:13:15 PM\n\n### Citation\n\nPierre Gaillard. The Johnson Equation, Fredholm and Wronskian Representations of Solutions, and the Case of Order Three. Advances in Mathematical Physics, Hindawi Publishing Corporation, 2018, 2018, pp.1-18. ⟨10.1155/2018/1642139⟩. ⟨hal-01997516⟩\n\nRecord views" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9228959,"math_prob":0.73179656,"size":692,"snap":"2019-35-2019-39","text_gpt3_token_len":175,"char_repetition_ratio":0.15697674,"word_repetition_ratio":0.0,"special_character_ratio":0.24855492,"punctuation_ratio":0.086330935,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9579017,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-19T12:37:26Z\",\"WARC-Record-ID\":\"<urn:uuid:3c2a50c8-60fd-453f-a542-411c75a27d26>\",\"Content-Length\":\"32047\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bc9105af-dc11-44e7-9398-cf3c13a800f0>\",\"WARC-Concurrent-To\":\"<urn:uuid:09e85295-80e3-45f0-8adc-a5982b2b2a9a>\",\"WARC-IP-Address\":\"193.48.96.10\",\"WARC-Target-URI\":\"https://hal.archives-ouvertes.fr/hal-01997516\",\"WARC-Payload-Digest\":\"sha1:NAYGYFTRPGLN7FE7OIOGFUFCEBQZJ23E\",\"WARC-Block-Digest\":\"sha1:HGJMM4INPAPVPHYZJ73EP4FHUFWETJGG\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027314732.59_warc_CC-MAIN-20190819114330-20190819140330-00105.warc.gz\"}"}
https://spark.apache.org/docs/2.0.0-preview/mllib-frequent-pattern-mining.html	[ "# Frequent Pattern Mining - spark.mllib\n\nMining frequent items, itemsets, subsequences, or other substructures is usually among the first steps to analyze a large-scale dataset, which has been an active research topic in data mining for years. We refer users to Wikipedia’s association rule learning for more information. spark.mllib provides a parallel implementation of FP-growth, a popular algorithm to mining frequent itemsets.\n\n## FP-growth\n\nThe FP-growth algorithm is described in the paper Han et al., Mining frequent patterns without candidate generation, where “FP” stands for frequent pattern. Given a dataset of transactions, the first step of FP-growth is to calculate item frequencies and identify frequent items. Different from Apriori-like algorithms designed for the same purpose, the second step of FP-growth uses a suffix tree (FP-tree) structure to encode transactions without generating candidate sets explicitly, which are usually expensive to generate. After the second step, the frequent itemsets can be extracted from the FP-tree. In spark.mllib, we implemented a parallel version of FP-growth called PFP, as described in Li et al., PFP: Parallel FP-growth for query recommendation. PFP distributes the work of growing FP-trees based on the suffices of transactions, and hence more scalable than a single-machine implementation. We refer users to the papers for more details.\n\nspark.mllib’s FP-growth implementation takes the following (hyper-)parameters:\n\n• minSupport: the minimum support for an itemset to be identified as frequent. For example, if an item appears 3 out of 5 transactions, it has a support of 3/5=0.6.\n• numPartitions: the number of partitions used to distribute the work.\n\nExamples\n\nFPGrowth implements the FP-growth algorithm. It take a RDD of transactions, where each transaction is an Array of items of a generic type. Calling FPGrowth.run with transactions returns an FPGrowthModel that stores the frequent itemsets with their frequencies. The following example illustrates how to mine frequent itemsets and association rules (see Association Rules for details) from transactions.\n\nRefer to the FPGrowth Scala docs for details on the API.\n\nimport org.apache.spark.mllib.fpm.FPGrowth\nimport org.apache.spark.rdd.RDD\n\nval data = sc.textFile(\"data/mllib/sample_fpgrowth.txt\")\n\nval transactions: RDD[Array[String]] = data.map(s => s.trim.split(' '))\n\nval fpg = new FPGrowth()\n.setMinSupport(0.2)\n.setNumPartitions(10)\nval model = fpg.run(transactions)\n\nmodel.freqItemsets.collect().foreach { itemset =>\nprintln(itemset.items.mkString(\"[\", \",\", \"]\") + \", \" + itemset.freq)\n}\n\nval minConfidence = 0.8\nmodel.generateAssociationRules(minConfidence).collect().foreach { rule =>\nprintln(\nrule.antecedent.mkString(\"[\", \",\", \"]\")\n+ \" => \" + rule.consequent .mkString(\"[\", \",\", \"]\")\n+ \", \" + rule.confidence)\n}\n\nFind full example code at \"examples/src/main/scala/org/apache/spark/examples/mllib/SimpleFPGrowth.scala\" in the Spark repo.\n\nFPGrowth implements the FP-growth algorithm. It take an JavaRDD of transactions, where each transaction is an Iterable of items of a generic type. Calling FPGrowth.run with transactions returns an FPGrowthModel that stores the frequent itemsets with their frequencies. The following example illustrates how to mine frequent itemsets and association rules (see Association Rules for details) from transactions.\n\nRefer to the FPGrowth Java docs for details on the API.\n\nimport java.util.Arrays;\nimport java.util.List;\n\nimport org.apache.spark.api.java.JavaRDD;\nimport org.apache.spark.api.java.JavaSparkContext;\n\nimport org.apache.spark.mllib.fpm.AssociationRules;\nimport org.apache.spark.mllib.fpm.FPGrowth;\nimport org.apache.spark.mllib.fpm.FPGrowthModel;\n\nJavaRDD<String> data = sc.textFile(\"data/mllib/sample_fpgrowth.txt\");\n\nJavaRDD<List<String>> transactions = data.map(\nnew Function<String, List<String>>() {\npublic List<String> call(String line) {\nString[] parts = line.split(\" \");\nreturn Arrays.asList(parts);\n}\n}\n);\n\nFPGrowth fpg = new FPGrowth()\n.setMinSupport(0.2)\n.setNumPartitions(10);\nFPGrowthModel<String> model = fpg.run(transactions);\n\nfor (FPGrowth.FreqItemset<String> itemset: model.freqItemsets().toJavaRDD().collect()) {\nSystem.out.println(\"[\" + itemset.javaItems() + \"], \" + itemset.freq());\n}\n\ndouble minConfidence = 0.8;\nfor (AssociationRules.Rule<String> rule\n: model.generateAssociationRules(minConfidence).toJavaRDD().collect()) {\nSystem.out.println(\nrule.javaAntecedent() + \" => \" + rule.javaConsequent() + \", \" + rule.confidence());\n}\n\nFind full example code at \"examples/src/main/java/org/apache/spark/examples/mllib/JavaSimpleFPGrowth.java\" in the Spark repo.\n\nFPGrowth implements the FP-growth algorithm. It take an RDD of transactions, where each transaction is an List of items of a generic type. Calling FPGrowth.train with transactions returns an FPGrowthModel that stores the frequent itemsets with their frequencies.\n\nRefer to the FPGrowth Python docs for more details on the API.\n\nfrom pyspark.mllib.fpm import FPGrowth\n\ndata = sc.textFile(\"data/mllib/sample_fpgrowth.txt\")\ntransactions = data.map(lambda line: line.strip().split(' '))\nmodel = FPGrowth.train(transactions, minSupport=0.2, numPartitions=10)\nresult = model.freqItemsets().collect()\nfor fi in result:\nprint(fi)\n\nFind full example code at \"examples/src/main/python/mllib/fpgrowth_example.py\" in the Spark repo.\n\n## Association Rules\n\nAssociationRules implements a parallel rule generation algorithm for constructing rules that have a single item as the consequent.\n\nRefer to the AssociationRules Scala docs for details on the API.\n\nimport org.apache.spark.mllib.fpm.AssociationRules\nimport org.apache.spark.mllib.fpm.FPGrowth.FreqItemset\n\nval freqItemsets = sc.parallelize(Seq(\nnew FreqItemset(Array(\"a\"), 15L),\nnew FreqItemset(Array(\"b\"), 35L),\nnew FreqItemset(Array(\"a\", \"b\"), 12L)\n))\n\nval ar = new AssociationRules()\n.setMinConfidence(0.8)\nval results = ar.run(freqItemsets)\n\nresults.collect().foreach { rule =>\nprintln(\"[\" + rule.antecedent.mkString(\",\")\n+ \"=>\"\n+ rule.consequent.mkString(\",\") + \"],\" + rule.confidence)\n}\n\nFind full example code at \"examples/src/main/scala/org/apache/spark/examples/mllib/AssociationRulesExample.scala\" in the Spark repo.\n\nAssociationRules implements a parallel rule generation algorithm for constructing rules that have a single item as the consequent.\n\nRefer to the AssociationRules Java docs for details on the API.\n\nimport java.util.Arrays;\n\nimport org.apache.spark.api.java.JavaRDD;\nimport org.apache.spark.api.java.JavaSparkContext;\nimport org.apache.spark.mllib.fpm.AssociationRules;\nimport org.apache.spark.mllib.fpm.FPGrowth;\nimport org.apache.spark.mllib.fpm.FPGrowth.FreqItemset;\n\nJavaRDD<FPGrowth.FreqItemset<String>> freqItemsets = sc.parallelize(Arrays.asList(\nnew FreqItemset<String>(new String[] {\"a\"}, 15L),\nnew FreqItemset<String>(new String[] {\"b\"}, 35L),\nnew FreqItemset<String>(new String[] {\"a\", \"b\"}, 12L)\n));\n\nAssociationRules arules = new AssociationRules()\n.setMinConfidence(0.8);\nJavaRDD<AssociationRules.Rule<String>> results = arules.run(freqItemsets);\n\nfor (AssociationRules.Rule<String> rule : results.collect()) {\nSystem.out.println(\nrule.javaAntecedent() + \" => \" + rule.javaConsequent() + \", \" + rule.confidence());\n}\n\nFind full example code at \"examples/src/main/java/org/apache/spark/examples/mllib/JavaAssociationRulesExample.java\" in the Spark repo.\n\n## PrefixSpan\n\nPrefixSpan is a sequential pattern mining algorithm described in Pei et al., Mining Sequential Patterns by Pattern-Growth: The PrefixSpan Approach. We refer the reader to the referenced paper for formalizing the sequential pattern mining problem.\n\nspark.mllib’s PrefixSpan implementation takes the following parameters:\n\n• minSupport: the minimum support required to be considered a frequent sequential pattern.\n• maxPatternLength: the maximum length of a frequent sequential pattern. Any frequent pattern exceeding this length will not be included in the results.\n• maxLocalProjDBSize: the maximum number of items allowed in a prefix-projected database before local iterative processing of the projected database begins. This parameter should be tuned with respect to the size of your executors.\n\nExamples\n\nThe following example illustrates PrefixSpan running on the sequences (using same notation as Pei et al):\n\n <(12)3>\n<1(32)(12)>\n<(12)5>\n<6>\n\n\nPrefixSpan implements the PrefixSpan algorithm. Calling PrefixSpan.run returns a PrefixSpanModel that stores the frequent sequences with their frequencies.\n\nRefer to the PrefixSpan Scala docs and PrefixSpanModel Scala docs for details on the API.\n\nimport org.apache.spark.mllib.fpm.PrefixSpan\n\nval sequences = sc.parallelize(Seq(\nArray(Array(1, 2), Array(3)),\nArray(Array(1), Array(3, 2), Array(1, 2)),\nArray(Array(1, 2), Array(5)),\nArray(Array(6))\n), 2).cache()\nval prefixSpan = new PrefixSpan()\n.setMinSupport(0.5)\n.setMaxPatternLength(5)\nval model = prefixSpan.run(sequences)\nmodel.freqSequences.collect().foreach { freqSequence =>\nprintln(\nfreqSequence.sequence.map(_.mkString(\"[\", \", \", \"]\")).mkString(\"[\", \", \", \"]\") +\n\", \" + freqSequence.freq)\n}\n\nFind full example code at \"examples/src/main/scala/org/apache/spark/examples/mllib/PrefixSpanExample.scala\" in the Spark repo.\n\nPrefixSpan implements the PrefixSpan algorithm. Calling PrefixSpan.run returns a PrefixSpanModel that stores the frequent sequences with their frequencies.\n\nRefer to the PrefixSpan Java docs and PrefixSpanModel Java docs for details on the API.\n\nimport java.util.Arrays;\nimport java.util.List;\n\nimport org.apache.spark.mllib.fpm.PrefixSpan;\nimport org.apache.spark.mllib.fpm.PrefixSpanModel;\n\nJavaRDD<List<List<Integer>>> sequences = sc.parallelize(Arrays.asList(\nArrays.asList(Arrays.asList(1, 2), Arrays.asList(3)),\nArrays.asList(Arrays.asList(1), Arrays.asList(3, 2), Arrays.asList(1, 2)),\nArrays.asList(Arrays.asList(1, 2), Arrays.asList(5)),\nArrays.asList(Arrays.asList(6))\n), 2);\nPrefixSpan prefixSpan = new PrefixSpan()\n.setMinSupport(0.5)\n.setMaxPatternLength(5);\nPrefixSpanModel<Integer> model = prefixSpan.run(sequences);\nfor (PrefixSpan.FreqSequence<Integer> freqSeq: model.freqSequences().toJavaRDD().collect()) {\nSystem.out.println(freqSeq.javaSequence() + \", \" + freqSeq.freq());\n}\n\nFind full example code at \"examples/src/main/java/org/apache/spark/examples/mllib/JavaPrefixSpanExample.java\" in the Spark repo." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.62256753,"math_prob":0.8320029,"size":9411,"snap":"2019-43-2019-47","text_gpt3_token_len":2253,"char_repetition_ratio":0.15243967,"word_repetition_ratio":0.24326834,"special_character_ratio":0.22888109,"punctuation_ratio":0.24117987,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97202986,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-22T03:44:20Z\",\"WARC-Record-ID\":\"<urn:uuid:2f2a6424-22f4-403b-81bf-24b571a979b3>\",\"Content-Length\":\"48970\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:af8c1068-36f7-4924-99d9-0f73597fd454>\",\"WARC-Concurrent-To\":\"<urn:uuid:45800109-9046-4572-b5b3-6d0f9719e31d>\",\"WARC-IP-Address\":\"40.79.78.1\",\"WARC-Target-URI\":\"https://spark.apache.org/docs/2.0.0-preview/mllib-frequent-pattern-mining.html\",\"WARC-Payload-Digest\":\"sha1:JJILJMGONRRREU7JKVCGGSIFHBQ54S4H\",\"WARC-Block-Digest\":\"sha1:HOQARPUB4SPWBDL4MR2DTM36TNDLSTOW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496671106.83_warc_CC-MAIN-20191122014756-20191122042756-00028.warc.gz\"}"}
https://segmentfault.com/a/1190000041330561	[ "# 每日一练（11）：调整数组顺序使奇数位于偶数前面\n\ntitle: 每日一练（11）：调整数组顺序使奇数位于偶数前面\n\ncategories:[剑指offer]\n\ntags:[每日一练]\n\ndate: 2022/01/24\n\n## 每日一练（11）：调整数组顺序使奇数位于偶数前面\n\n0 <= nums.length <= 50000\n\n0 <= nums[i] <= 10000\n\n### 方法一：有手就行\n\n``````vector<int> exchange(vector<int>& nums) {\nvector<int>res;\nfor (int i = 0; i < nums.size(); i++) {\nif ((nums[i] & 1) != 0) { //nums[i] & 1) != 0 可用 nums[i] % 2 替代\nres.push_back(nums[i]); //放入奇数\n}\n}\nfor (int i = 0; i < nums.size(); i++) {\nif ((nums[i] & 1) != 1) { //nums[i] & 1) != 1 可用 (nums[i] % 2) == 0 替代\nres.push_back(nums[i]); //放入偶数\n}\n}\nreturn res;\n}``````\n\n### 方法二：首尾双指针\n\n• 定义头指针 left ，尾指针 right .\n• left 一直往右移，直到它指向的值为偶数 .\n• right 一直往左移，直到它指向的值为奇数 .\n• 交换 nums[left] 和 nums[right] .\n• 重复上述操作，直到 left == right .\n``````vector<int> exchange(vector<int>& nums) {\nint left = 0, right = nums.size() - 1;\nwhile (left < right) {\nif ((nums[left] & 1) != 0) { //如果为偶数\nleft++;\ncontinue;\n}\nif ((nums[right] & 1) != 1) { //如果为奇数\nright--;\ncontinue;\n}\nswap(nums[left++], nums[right--]); //交换值\n}\nreturn nums;\n}``````\n\n### 方法三：快慢双指针\n\n• 定义快慢双指针 fast 和 slow ，fast 在前， slow 在后 .\n• fast 的作用是向前搜索奇数位置，slow 的作用是指向下一个奇数应当存放的位置 .\n• fast 向前移动，当它搜索到奇数时，将它和 nums[slow ] 交换，此时 slow 向前移动一个位置 .\n• 重复上述操作，直到 fast 指向数组末尾 .\n``````vector<int> exchange(vector<int>& nums) {\nint slow = 0, fast = 0;\nwhile (fast < nums.size()) {\nif ((nums[fast]) & 1) { //奇数\nswap(nums[slow], nums[fast]);//交换\nslow++;\n}\nfast++;\n}\nreturn nums;\n}``````\n\n25 声望\n7 粉丝\n0 条评论", null, "" ]	[ null, "https://image-static.segmentfault.com/317/931/3179314346-5f61e47221e07", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.62317425,"math_prob":0.9994895,"size":1576,"snap":"2022-05-2022-21","text_gpt3_token_len":881,"char_repetition_ratio":0.14503817,"word_repetition_ratio":0.13580246,"special_character_ratio":0.42005077,"punctuation_ratio":0.2024922,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9924403,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-26T21:18:51Z\",\"WARC-Record-ID\":\"<urn:uuid:e6cb705f-6f52-430c-8e1b-4b74ac594ea9>\",\"Content-Length\":\"41477\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:689adf4e-bb9d-431d-9348-26503e51890a>\",\"WARC-Concurrent-To\":\"<urn:uuid:3fd7ee99-e47d-460c-a1fb-250b13042542>\",\"WARC-IP-Address\":\"103.61.39.86\",\"WARC-Target-URI\":\"https://segmentfault.com/a/1190000041330561\",\"WARC-Payload-Digest\":\"sha1:BZMPC4G56LSVZXO2S2AROGWROFQNXZTM\",\"WARC-Block-Digest\":\"sha1:UDFXFLXLXJXBC5DUKTKHYF3GDPKCSXB6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662625600.87_warc_CC-MAIN-20220526193923-20220526223923-00063.warc.gz\"}"}
http://www.stock-charts-made-easy.com/fibonacci.html	[ "## \"Can Fibonacci Price Projections Forecast Key Turning Points In The Markets?...\"", null, "es... and, with great accuracy. You'll see how calculations based on the Fibonacci number series can be applied to stock prices for startling results.", null, "", null, "", null, "", null, "Applied correctly, support and resistance levels will materialize. Likely turning points will present themselves, clearly.\n\nSo, how are the retracement percentages derived?\n\nBegin by adding one to itself... which is two. Then add two to one to get three -- the next number in the series. You keep adding the current number to the previous number so it looks like this...\n\n1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144...\n\nNow, you have a number series which goes on forever.", null, "Next, there is an important ratio which is calculated by dividing one number in the series by its previous number. For example...\n\n144 ÷ 89 = 1.618\n\nThis ratio is known as the Golden Mean and is the basis for most retracement percentages. Equally important is the reciprocal of 1.618...\n\n1 ÷ 1.618 = .618\n\nHere are some other percentages you'll want to use...\n\n.382 = .618 squared\n\n.500 = 1 ÷ 2, the second and third numbers in the series\n\n.786 = square root of .618\n\n1.000 = 1.618 x .618\n\n1.272 = square root of 1.618\n\n2.618 = 1.618 squared", null, "### \"How are Fibonacci percentages applied?\"\n\nBegin by locating isolated highs and lows in a trend. The various swing points will be used to measure the retracements.\n\nIn the following illustration, trend AB has been established...", null, "Counter-trend BC is a 50-percent retracement of trend AB. Trend CD continues in the same direction as trend AB. Counter-trend DE is a 38-percent retracement of trend CD.\n\nShallow retracements are typical in a strong trend.", null, "Combining retracements from multiple swing points is a dynamic forecasting method. In the next example, retracement percentages are calculated from trends AD and CD...", null, "The 38-percent retracement is calculated from trend AD. The 50-percent retracement is from trend CD. When retracements from different swing points cluster together it is known as confluence.\n\nAreas of confluence are likely turning points.\n\nClick For Fibonacci Confluence Chart Examples", null, "So far, you have seen how to calculate probable retracements using the high and low swing point of an established trend. You can use Fibonacci percentages, in combination, to forecast the end of a trend... with excellent results.\n\nThis next example shows a typical five-wave trend progression where the 1.618 and .618 percentage extensions are used...", null, "First, trend AB is multiplied by 1.618 and then added to itself... forecasting a target price at point F. Next, counter-trend BC retraces part of trend AB before continuing on. Finally, trend AD is multiplied by .618 and added to itself. This projection also forecasts the trend termination at point F. Incredibly powerful!\n\nIn a five-wave pattern, this ratio combination will bear-out turning points, again and again.", null, "There are many other ways to apply Fibonacci retracements. Purchase the following books to learn more...", null, "by Robert Fischer\nJohn Wiley & Sons, Inc.\n2001\n352 Pages\n\nGet This Book, Now!", null, "Dynamic Trading - Dynamic Concepts in Time, Price and Pattern Analysis with Practical Strategies for Traders and Investors\nby Robert C. Miner\n1997\n564 Pages\n\n Or... SEARCH for a BOOK Enter title, author, item# or ISBN", null, "Home \| Charting Principles \| Forecasting Tools \| Trading Systems \| Market Metrics\n\nCharts In Action \| Market Commentary \| Online Newsletter \| Bookstore\n\nFor the Record \| Site Map", null, "", null, "", null, "Online Newsletter Subscribe FREE! And... receive your FREE report. Back Issues ofChart Wealth", null, "What's an RSS feed?", null, "", null, "", null, "", null, "", null, "" ]	[ null, "http://www.stock-charts-made-easy.com/images/xY.png.pagespeed.ic.CxDSpQQqJr.jpg", null, "http://www.stock-charts-made-easy.com/images/fb2001.png", null, "http://www.stock-charts-made-easy.com/images/fb2002.png", null, "http://www.stock-charts-made-easy.com/images/fb2003.png", null, "http://www.stock-charts-made-easy.com/images/fb2004.png", null, "http://www.stock-charts-made-easy.com/images/Short_Divider.png.pagespeed.ce.WgiwGp62Re.png", null, "http://www.stock-charts-made-easy.com/images/Short_Divider.png.pagespeed.ce.WgiwGp62Re.png", null, "http://www.stock-charts-made-easy.com/images/Fib_Retracement.png", null, "http://www.stock-charts-made-easy.com/images/Short_Divider.png.pagespeed.ce.WgiwGp62Re.png", null, "http://www.stock-charts-made-easy.com/images/Fib_Confluence.png", null, "http://www.stock-charts-made-easy.com/images/Short_Divider.png.pagespeed.ce.WgiwGp62Re.png", null, "http://www.stock-charts-made-easy.com/images/Fib_Extension.png", null, "http://www.stock-charts-made-easy.com/images/Short_Divider.png.pagespeed.ce.WgiwGp62Re.png", null, "http://www.stock-charts-made-easy.com/images/New_Fibonacci_Trader_Fischer.png", null, "http://www.stock-charts-made-easy.com/images/Dynamic_Trading_Miner.png", null, "http://www.stock-charts-made-easy.com/images/Divider.png.pagespeed.ce.5WiihX0QSj.png", null, "http://www.stock-charts-made-easy.com/images/xEmail_Plate.png.pagespeed.ic.7rSRmuPGHp.png", null, "http://www.stock-charts-made-easy.com/images/xCopyright_Logo.png.pagespeed.ic.KcDwSzJghj.png", null, "http://www.stock-charts-made-easy.com/images/xSm_CW_Header.png.pagespeed.ic.hQ8SruLngE.jpg", null, "http://www.stock-charts-made-easy.com/objects/xxml-rss.gif.pagespeed.ic.6EAaTT5N17.jpg", null, "http://buttons.googlesyndication.com/fusion/add.gif", null, "http://us.i1.yimg.com/us.yimg.com/i/us/my/addtomyyahoo2.gif", null, "http://www.stock-charts-made-easy.com/objects/xmymsn.gif.pagespeed.ic.h0Qsa_OfCf.jpg", null, "http://www.invest-store.com/images/resource/currentspecialsq.gif", null, "http://www.stock-charts-made-easy.com/images/xtraders_transparent.gif.pagespeed.ic.RXOscZ6-DZ.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88793147,"math_prob":0.9376798,"size":3635,"snap":"2021-31-2021-39","text_gpt3_token_len":814,"char_repetition_ratio":0.13136877,"word_repetition_ratio":0.0033613446,"special_character_ratio":0.24511692,"punctuation_ratio":0.18131869,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9734304,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50],"im_url_duplicate_count":[null,4,null,3,null,3,null,3,null,4,null,null,null,null,null,4,null,null,null,4,null,null,null,4,null,null,null,4,null,4,null,4,null,5,null,5,null,5,null,5,null,null,null,5,null,5,null,5,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-21T23:10:27Z\",\"WARC-Record-ID\":\"<urn:uuid:19bcee9b-fdc5-45de-9bc1-2b2112a26f81>\",\"Content-Length\":\"21109\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3a9f22ec-dde8-40bb-9128-a935979ad59f>\",\"WARC-Concurrent-To\":\"<urn:uuid:7e687b1e-db97-491b-bf53-fb939e1fe7a5>\",\"WARC-IP-Address\":\"173.247.219.107\",\"WARC-Target-URI\":\"http://www.stock-charts-made-easy.com/fibonacci.html\",\"WARC-Payload-Digest\":\"sha1:2KBVSJPP4WNKRTOCCFVWALU2MLNSFEMP\",\"WARC-Block-Digest\":\"sha1:INCNUUHFHFFLYGX7SDB3XCJKASDI56RR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057274.97_warc_CC-MAIN-20210921221605-20210922011605-00270.warc.gz\"}"}
https://se.mathworks.com/help/simulink/slref/discretepidcontroller2dof.html	[ "# Discrete PID Controller (2DOF)\n\nDiscrete-time or continuous-time two-degree-of-freedom PID controller\n\n• Library:\n\n•", null, "## Description\n\nThe Discrete PID Controller (2DOF) block implements a two-degree-of-freedom PID controller (PID, PI, or PD). The block is identical to the PID Controller (2DOF) block with the Time domain parameter set to `Discrete-time`.\n\nThe block generates an output signal based on the difference between a reference signal and a measured system output. The block computes a weighted difference signal for the proportional and derivative actions according to the setpoint weights (b and c) that you specify. The block output is the sum of the proportional, integral, and derivative actions on the respective difference signals, where each action is weighted according to the gain parameters P, I, and D. A first-order pole filters the derivative action.\n\nThe block supports several controller types and structures. Configurable options in the block include:\n\n• Controller type (PID, PI, or PD) — See the Controller parameter.\n\n• Controller form (Parallel or Ideal) — See the Form parameter.\n\n• Time domain (discrete or continuous) — See the Time domain parameter.\n\n• Initial conditions and reset trigger — See the Source and External reset parameters.\n\n• Output saturation limits and built-in anti-windup mechanism — See the Limit output parameter.\n\n• Signal tracking for bumpless control transfer and multiloop control — See the Enable tracking mode parameter.\n\nAs you change these options, the internal structure of the block changes by activating different variant subsystems. (See Variant Subsystems.) To examine the internal structure of the block and its variant subsystems, right-click the block and select Mask > Look Under Mask.\n\n### Control Configuration\n\nIn one common implementation, the PID Controller block operates in the feedforward path of a feedback loop.", null, "For a single-input block that accepts an error signal (a difference between a setpoint and a system output), see Discrete PID Controller.\n\n### PID Gain Tuning\n\nThe PID controller coefficients and the setpoint weights are tunable either manually or automatically. Automatic tuning requires Simulink® Control Design™ software. For more information about automatic tuning, see the Select tuning method parameter.\n\n## Ports\n\n### Input\n\nexpand all\n\nReference signal for plant to follow, as shown.", null, "When the reference signal is a vector, the block acts separately on each signal, vectorizing the PID coefficients and producing a vector output signal of the same dimensions. You can specify the PID coefficients and some other parameters as vectors of the same dimensions as the input signal. Doing so is equivalent to specifying a separate PID controller for each entry in the input signal.\n\nData Types: `single` \| `double` \| `int8` \| `int16` \| `int32` \| `int64` \| `uint8` \| `uint16` \| `uint32` \| `uint64` \| `fixed point`\n\nFeedback signal for the controller, from the plant output.\n\nData Types: `single` \| `double` \| `int8` \| `int16` \| `int32` \| `int64` \| `uint8` \| `uint16` \| `uint32` \| `uint64` \| `fixed point`\n\nProportional gain, provided from a source external to the block. External gain input is useful, for example, when you want to map a different PID parameterization to the PID gains of the block. You can also use external gain input to implement gain-scheduled PID control. In gain-scheduled control, you determine the PID coefficients by logic or other calculation in your model and feed them to the block.\n\n#### Dependencies\n\nTo enable this port, set Controller parameters Source to `external`.\n\nData Types: `single` \| `double` \| `int8` \| `int16` \| `int32` \| `int64` \| `uint8` \| `uint16` \| `uint32` \| `uint64` \| `fixed point`\n\nIntegral gain, provided from a source external to the block. External gain input is useful, for example, when you want to map a different PID parameterization to the PID gains of the block. You can also use external gain input to implement gain-scheduled PID control. In gain-scheduled control, you determine the PID coefficients by logic or other calculation in your model and feed them to the block.\n\nWhen you supply gains externally, time variations in the integral gain are also integrated. This result occurs because of the way the PID gains are implemented within the block. For details, see the Controller parameters Source parameter.\n\n#### Dependencies\n\nTo enable this port, set Controller parameters Source to `external`, and set Controller to a controller type that has integral action.\n\nData Types: `single` \| `double` \| `int8` \| `int16` \| `int32` \| `int64` \| `uint8` \| `uint16` \| `uint32` \| `uint64` \| `fixed point`\n\nDerivative gain, provided from a source external to the block. External gain input is useful, for example, when you want to map a different PID parameterization to the PID gains of the block. You can also use external gain input to implement gain-scheduled PID control. In gain-scheduled control, you determine the PID coefficients by logic or other calculation in your model and feed them to the block.\n\nWhen you supply gains externally, time variations in the derivative gain are also differentiated. This result occurs because of the way the PID gains are implemented within the block. For details, see the Controller parameters Source parameter.\n\n#### Dependencies\n\nTo enable this port, set Controller parameters Source to `external`, and set Controller to a controller type that has derivative action.\n\nData Types: `single` \| `double` \| `int8` \| `int16` \| `int32` \| `int64` \| `uint8` \| `uint16` \| `uint32` \| `uint64` \| `fixed point`\n\nDerivative filter coefficient, provided from a source external to the block. External coefficient input is useful, for example, when you want to map a different PID parameterization to the PID gains of the block. You can also use the external input to implement gain-scheduled PID control. In gain-scheduled control, you determine the PID coefficients by logic or other calculation in your model and feed them to the block.\n\n#### Dependencies\n\nTo enable this port, set Controller parameters Source to `external`, and set Controller to a controller type that has a filtered derivative.\n\nData Types: `single` \| `double` \| `int8` \| `int16` \| `int32` \| `int64` \| `uint8` \| `uint16` \| `uint32` \| `uint64` \| `fixed point`\n\nProportional setpoint weight, provided from a source external to the block. External input is useful, for example, when you want to map a different PID parameterization to the PID gains of the block. You can also use the external input to implement gain-scheduled PID control. In gain-scheduled control, you determine the PID coefficients by logic or other calculation in your model and feed them to the block.\n\n#### Dependencies\n\nTo enable this port, set Controller parameters Source to `external`.\n\nData Types: `single` \| `double` \| `int8` \| `int16` \| `int32` \| `int64` \| `uint8` \| `uint16` \| `uint32` \| `uint64` \| `fixed point`\n\nDerivative setpoint weight, provided from a source external to the block. External input is useful, for example, when you want to map a different PID parameterization to the PID gains of the block. You can also use the external input to implement gain-scheduled PID control. In gain-scheduled control, you determine the PID coefficients by logic or other calculation in your model and feed them to the block.\n\n#### Dependencies\n\nTo enable this port, set Controller parameters Source to `external`, and set Controller to a controller type that has derivative action.\n\nData Types: `single` \| `double` \| `int8` \| `int16` \| `int32` \| `int64` \| `uint8` \| `uint16` \| `uint32` \| `uint64` \| `fixed point`\n\nTrigger to reset the integrator and filter to their initial conditions. Use the External reset parameter to specify what kind of signal triggers a reset. The port icon indicates the trigger type specified in that parameter. For example, the following illustration shows a continuous-time PID Controller (2DOF) block with External reset set to `rising`.", null, "When the trigger occurs, the block resets the integrator and filter to the initial conditions specified by the Integrator Initial condition and Filter Initial condition parameters or the I0 and D0 ports.\n\n### Note\n\nTo be compliant with the Motor Industry Software Reliability Association (MISRA®) software standard, your model must use Boolean signals to drive the external reset ports of the PID controller block.\n\n#### Dependencies\n\nTo enable this port, set External reset to any value other than `none`.\n\nData Types: `single` \| `double` \| `int8` \| `int16` \| `int32` \| `int64` \| `uint8` \| `uint16` \| `uint32` \| `uint64` \| `fixed point` \| `Boolean`\n\nIntegrator initial condition, provided from a source external to the block.\n\n#### Dependencies\n\nTo enable this port, set Initial conditions Source to `external`, and set Controller to a controller type that has integral action.\n\nData Types: `single` \| `double` \| `int8` \| `int16` \| `int32` \| `int64` \| `uint8` \| `uint16` \| `uint32` \| `uint64` \| `fixed point`\n\nInitial condition of the derivative filter, provided from a source external to the block.\n\n#### Dependencies\n\nTo enable this port, set Initial conditions Source to `external`, and set Controller to a controller type that has derivative action.\n\nData Types: `single` \| `double` \| `int8` \| `int16` \| `int32` \| `int64` \| `uint8` \| `uint16` \| `uint32` \| `uint64` \| `fixed point`\n\nSignal for controller output to track. When signal tracking is active, the difference between the tracking signal and the block output is fed back to the integrator input. Signal tracking is useful for implementing bumpless control transfer in systems that switch between two controllers. It can also be useful to prevent block windup in multiloop control systems. For more information, see the Enable tracking mode parameter.\n\n#### Dependencies\n\nTo enable this port, select the Enable tracking mode parameter.\n\nData Types: `single` \| `double` \| `int8` \| `int16` \| `int32` \| `int64` \| `uint8` \| `uint16` \| `uint32` \| `uint64` \| `fixed point`\n\nDiscrete-integrator time, provided as a scalar to the block. You can use your own value of discrete-time integrator sample time that defines the rate at which the block is going to be run either in Simulink or on external hardware. The value of the discrete-time integrator time should match the average sampling rate of the external interrupts, when the block is used inside a conditionally-executed subsystem.\n\nIn other words, you can specify `Ts` for any of the integrator methods below such that the value matches the average sampling rate of the external interrupts. In discrete time, the derivative term of the controller transfer function is:\n\n`$D\\left[\\frac{N}{1+N\\alpha \\left(z\\right)}\\right],$`\n\nwhere α(z) depends on the integrator method you specify with this parameter.\n\n```Forward Euler```\n`$\\alpha \\left(z\\right)=\\frac{{T}_{s}}{z-1}.$`\n```Backward Euler```\n`$\\alpha \\left(z\\right)=\\frac{{T}_{s}z}{z-1}.$`\n`Trapezoidal`\n`$\\alpha \\left(z\\right)=\\frac{{T}_{s}}{2}\\frac{z+1}{z-1}.$`\n\n#### Dependencies\n\nTo enable this port, set Time Domain to `Discrete-time` and select the PID Controller is inside a conditionally executed subsystem option.\n\nData Types: `single` \| `double` \| `int8` \| `int16` \| `int32` \| `int64` \| `uint8` \| `uint16` \| `uint32` \| `uint64`\n\n### Output\n\nexpand all\n\nController output, generally based on a sum of the input signal, the integral of the input signal, and the derivative of the input signal, weighted by the setpoint weights and by the proportional, integral, and derivative gain parameters. A first-order pole filters the derivative action. Which terms are present in the controller signal depends on what you select for the Controller parameter. The base controller transfer function for the current settings is displayed in the Compensator formula section of the block parameters and under the mask. Other parameters modify the block output, such as saturation limits specified by the Upper Limit and Lower Limit saturation parameters.\n\nThe controller output is a vector signal when any of the inputs is a vector signal. In that case, the block acts as N independent PID controllers, where N is the number of signals in the input vector.\n\nData Types: `single` \| `double` \| `int8` \| `int16` \| `int32` \| `int64` \| `uint8` \| `uint16` \| `uint32` \| `uint64` \| `fixed point`\n\n## Parameters\n\nexpand all\n\nSpecify which of the proportional, integral, and derivative terms are in the controller.\n\n`PID `\n\nProportional, integral, and derivative action.\n\n`PI`\n\nProportional and integral action only.\n\n`PD`\n\nProportional and derivative action only.\n\n### Tip\n\nThe controller output for the current setting is displayed in the Compensator formula section of the block parameters and under the mask.\n\n#### Programmatic Use\n\n Block Parameter: `Controller` Type: string, character vector Values: `\"PID\"`, `\"PI\"`, `\"PD\"` Default: `\"PID\"`\n\nSpecify whether the controller structure is parallel or ideal.\n\n`Parallel`\n\nThe proportional, integral, and derivative gains P, I, and D, are applied independently. For example, for a continuous-time 2-DOF PID controller in parallel form, the controller output u is:\n\n`$u=P\\left(br-y\\right)+I\\frac{1}{s}\\left(r-y\\right)+D\\frac{N}{1+N\\frac{1}{s}}\\left(cr-y\\right),$`\n\nwhere r is the reference signal, y is the measured plant output signal, and b and c are the setpoint weights.\n\nFor a discrete-time 2-DOF controller in parallel form, the controller output is:\n\n`$u=P\\left(br-y\\right)+I\\alpha \\left(z\\right)\\left(r-y\\right)+D\\frac{N}{1+N\\beta \\left(z\\right)}\\left(cr-y\\right),$`\n\nwhere the Integrator method and Filter method parameters determine α(z) and β(z), respectively.\n\n`Ideal`\n\nThe proportional gain P acts on the sum of all actions. For example, for a continuous-time 2-DOF PID controller in ideal form, the controller output is:\n\n`$u=P\\left[\\left(br-y\\right)+I\\frac{1}{s}\\left(r-y\\right)+D\\frac{N}{1+N\\frac{1}{s}}\\left(cr-y\\right)\\right].$`\n\nFor a discrete-time 2-DOF PID controller in ideal form, the transfer function is:\n\n`$u=P\\left[\\left(br-y\\right)+I\\alpha \\left(z\\right)\\left(r-y\\right)+D\\frac{N}{1+N\\beta \\left(z\\right)}\\left(cr-y\\right)\\right],$`\n\nwhere the Integrator method and Filter method parameters determine α(z) and β(z), respectively.\n\n### Tip\n\nThe controller output for the current settings is displayed in the Compensator formula section of the block parameters and under the mask.\n\n#### Programmatic Use\n\n Block Parameter: `Controller` Type: string, character vector Values: `\"Parallel\"`, `\"Ideal\"` Default: `\"Parallel\"`\n\nWhen you select `Discrete-time`, it is recommended that you specify an explicit sample time for the block. See the Sample time (-1 for inherited) parameter. Selecting `Discrete-time` also enables the Integrator method, and Filter method parameters.\n\nWhen the PID Controller block is in a model with synchronous state control (see the State Control block), you cannot select `Continuous-time`.\n\n### Note\n\nThe PID Controller (2DOF) and Discrete PID Controller (2DOF) blocks are identical except for the default value of this parameter.\n\n#### Programmatic Use\n\n Block Parameter: `TimeDomain` Type: string, character vector Values: `\"Continuous-time\"`, `\"Discrete-time\"` Default: `\"Discrete-time\"`\n\nFor discrete-time PID controllers, enable the discrete-time integrator port to use your own value of discrete-time integrator sample time. To ensure proper integration, Use the `TDTI` port to provide a scalar value of Δt for accurate discrete-time integration.\n\n#### Dependencies\n\nTo enable this parameter, set Time Domain to `Discrete-time`.\n\n#### Programmatic Use\n\n Block Parameter: `UseExternalTs` Type: string, character vector Values: `\"on\"`, `\"off\"` Default: `\"off\"`\n\nSpecify a sample time by entering a positive scalar value, such as 0.1. The default discrete sample time of –1 means that the block inherits its sample time from upstream blocks. However, it is recommended that you set the controller sample time explicitly, especially if you expect the sample time of upstream blocks to change. The effect of the controller coefficients P, I, D, and N depend on the sample time. Thus, for a given set of coefficient values, changing the sample time changes the performance of the controller.\n\nTo implement a continuous-time controller, set Time domain to `Continuous-time`.\n\n### Tip\n\nIf you want to run the block with an externally specified or variable sample time, set this parameter to –1 and put the block in a Triggered Subsystem. Then, trigger the subsystem at the desired sample time.\n\n#### Dependencies\n\nTo enable this parameter, set Time domain to `Discrete-time`.\n\n#### Programmatic Use\n\n Block Parameter: `SampleTime` Type: scalar Values: `-1`, positive scalar Default: `-1`\n\nIn discrete time, the integral term of the controller transfer function is Ia(z), where a(z) depends on the integrator method you specify with this parameter.\n\n`Forward Euler`\n\nForward rectangular (left-hand) approximation,\n\n`$\\alpha \\left(z\\right)=\\frac{{T}_{s}}{z-1}.$`\n\nThis method is best for small sampling times, where the Nyquist limit is large compared to the bandwidth of the controller. For larger sampling times, the `Forward Euler` method can result in instability, even when discretizing a system that is stable in continuous time.\n\n`Backward Euler`\n\nBackward rectangular (right-hand) approximation,\n\n`$\\alpha \\left(z\\right)=\\frac{{T}_{s}z}{z-1}.$`\n\nAn advantage of the `Backward Euler` method is that discretizing a stable continuous-time system using this method always yields a stable discrete-time result.\n\n`Trapezoidal`\n\nBilinear approximation,\n\n`$\\alpha \\left(z\\right)=\\frac{{T}_{s}}{2}\\frac{z+1}{z-1}.$`\n\nAn advantage of the `Trapezoidal` method is that discretizing a stable continuous-time system using this method always yields a stable discrete-time result. Of all available integration methods, the `Trapezoidal` method yields the closest match between frequency-domain properties of the discretized system and the corresponding continuous-time system.\n\n### Tip\n\nThe controller formula for the current setting is displayed in the Compensator formula section of the block parameters and under the mask.\n\n#### Dependencies\n\nTo enable this parameter, set Time Domain to `Discrete-time` and set Controller to a controller type with integral action.\n\n#### Programmatic Use\n\n Block Parameter: `IntegratorMethod` Type: string, character vector Values: `\"Forward Euler\"`, `\"Backward Euler\"`, `\"Trapezoidal\"` Default: `\"Forward Euler\"`\n\nIn discrete time, the derivative term of the controller transfer function is:\n\n`$D\\left[\\frac{N}{1+N\\alpha \\left(z\\right)}\\right],$`\n\nwhere α(z) depends on the filter method you specify with this parameter.\n\n`Forward Euler`\n\nForward rectangular (left-hand) approximation,\n\n`$\\alpha \\left(z\\right)=\\frac{{T}_{s}}{z-1}.$`\n\nThis method is best for small sampling times, where the Nyquist limit is large compared to the bandwidth of the controller. For larger sampling times, the `Forward Euler` method can result in instability, even when discretizing a system that is stable in continuous time.\n\n`Backward Euler`\n\nBackward rectangular (right-hand) approximation,\n\n`$\\alpha \\left(z\\right)=\\frac{{T}_{s}z}{z-1}.$`\n\nAn advantage of the `Backward Euler` method is that discretizing a stable continuous-time system using this method always yields a stable discrete-time result.\n\n`Trapezoidal`\n\nBilinear approximation,\n\n`$\\alpha \\left(z\\right)=\\frac{{T}_{s}}{2}\\frac{z+1}{z-1}.$`\n\nAn advantage of the `Trapezoidal` method is that discretizing a stable continuous-time system using this method always yields a stable discrete-time result. Of all available integration methods, the `Trapezoidal` method yields the closest match between frequency-domain properties of the discretized system and the corresponding continuous-time system.\n\n### Tip\n\nThe controller formula for the current setting is displayed in the Compensator formula section of the block parameters and under the mask.\n\n#### Dependencies\n\nTo enable this parameter, set Time Domain to `Discrete-time` and enable Use filtered derivative.\n\n#### Programmatic Use\n\n Block Parameter: `FilterMethod` Type: string, character vector Values: `\"Forward Euler\"`, `\"Backward Euler\"`, `\"Trapezoidal\"` Default: `\"Forward Euler\"`\n\n### Main\n\n`internal`\n\nSpecify the controller gains, filter coefficient, and setpoint weights using the block parameters P, I, D, N, b, and c respectively.\n\n`external`\n\nSpecify the PID gains, filter coefficient, and setpoint weights externally using block inputs. An additional input port appears on the block for each parameter that is required for the current controller type.\n\nEnabling external inputs for the parameters allows you to compute their values externally to the block and provide them to the block as signal inputs.\n\nExternal input is useful, for example, when you want to map a different PID parameterization to the PID gains of the block. You can also use external gain input to implement gain-scheduled PID control. In gain-scheduled control, you determine the PID gains by logic or other calculation in your model and feed them to the block.\n\nWhen you supply gains externally, time variations in the integral and derivative gain values are integrated and differentiated, respectively. The derivative setpoint weight c is also differentiated. This result occurs because in both continuous time and discrete time, the gains are applied to the signal before integration or differentiation. For example, for a continuous-time PID controller with external inputs, the integrator term is implemented as shown in the following illustration.", null, "Within the block, the input signal u is multiplied by the externally supplied integrator gain, I, before integration. This implementation yields:\n\n`${u}_{i}=\\int \\left(r-y\\right)I\\text{\\hspace{0.17em}}dt.$`\n\nThus, the integrator gain is included in the integral. Similarly, in the derivative term of the block, multiplication by the derivative gain precedes the differentiation, which causes the derivative gain D and the derivative setpoint weight c to be differentiated.\n\n#### Programmatic Use\n\n Block Parameter: `ControllerParametersSource` Type: string, character vector Values: `\"internal\"`, `\"external\"` Default: `\"internal\"`\n\nSpecify a finite, real gain value for the proportional gain. When Controller form is:\n\n• `Parallel` — Proportional action is independent of the integral and derivative actions. For example, for a continuous-time 2-DOF PID controller in parallel form, the controller output u is:\n\n`$u=P\\left(br-y\\right)+I\\frac{1}{s}\\left(r-y\\right)+D\\frac{N}{1+N\\frac{1}{s}}\\left(cr-y\\right),$`\n\nwhere r is the reference signal, y is the measured plant output signal, and b and c are the setpoint weights.\n\nFor a discrete-time 2-DOF controller in parallel form, the controller output is:\n\n`$u=P\\left(br-y\\right)+I\\alpha \\left(z\\right)\\left(r-y\\right)+D\\frac{N}{1+N\\beta \\left(z\\right)}\\left(cr-y\\right),$`\n\nwhere the Integrator method and Filter method parameters determine α(z) and β(z), respectively.\n\n• `Ideal` — The proportional gain multiples the integral and derivative terms. For example, for a continuous-time 2-DOF PID controller in ideal form, the controller output is:\n\n`$u=P\\left[\\left(br-y\\right)+I\\frac{1}{s}\\left(r-y\\right)+D\\frac{N}{1+N\\frac{1}{s}}\\left(cr-y\\right)\\right].$`\n\nFor a discrete-time 2-DOF PID controller in ideal form, the transfer function is:\n\n`$u=P\\left[\\left(br-y\\right)+I\\alpha \\left(z\\right)\\left(r-y\\right)+D\\frac{N}{1+N\\beta \\left(z\\right)}\\left(cr-y\\right)\\right],$`\n\nwhere the Integrator method and Filter method parameters determine α(z) and β(z), respectively.\n\nTunable: Yes\n\n#### Dependencies\n\nTo enable this parameter, set the Controller parameters Source to `internal`.\n\n#### Programmatic Use\n\n Block Parameter: `P` Type: scalar, vector Default: 1\n\nSpecify a finite, real gain value for the integral gain.\n\nTunable: Yes\n\n#### Dependencies\n\nTo enable this parameter, in the Main tab, set the controller-parameters Source to `internal`, and set Controller to a type that has integral action.\n\n#### Programmatic Use\n\n Block Parameter: `I` Type: scalar, vector Default: 1\n\nSpecify a finite, real gain value for the derivative gain.\n\nTunable: Yes\n\n#### Dependencies\n\nTo enable this parameter, in the Main tab, set the controller-parameters Source to `internal`, and set Controller to `PID` or `PD`.\n\n#### Programmatic Use\n\n Block Parameter: `D` Type: scalar, vector Default: 0\n\nFor discrete-time PID controllers only, clear this option to replace the filtered derivative with an unfiltered discrete-time differentiator. When you do so, the derivative term of the controller output becomes:\n\n`$D\\frac{z-1}{z{T}_{s}}\\left(cr-y\\right).$`\n\nFor continuous-time PID controllers, the derivative term is always filtered.\n\n#### Dependencies\n\nTo enable this parameter, set Time domain to `Discrete-time`, and set Controller to a type that has a derivative term.\n\n#### Programmatic Use\n\n Block Parameter: `UseFilter` Type: string, character vector Values: `\"on\"`, `\"off\"` Default: `\"on\"`\n\nSpecify a finite, real gain value for the filter coefficient. The filter coefficient determines the pole location of the filter in the derivative action of the block. The location of the filter pole depends on the Time domain parameter.\n\n• When Time domain is `Continuous-time`, the pole location is `s = -N`.\n\n• When Time domain is `Discrete-time`, the pole location depends on the Filter method parameter.\n\nFilter MethodLocation of Filter Pole\n```Forward Euler```${z}_{pole}=1-N{T}_{s}$\n```Backward Euler```${z}_{pole}=\\frac{1}{1+N{T}_{s}}$\n`Trapezoidal`${z}_{pole}=\\frac{1-N{T}_{s}/2}{1+N{T}_{s}/2}$\n\nThe block does not support `N = Inf` (ideal unfiltered derivative). When the Time domain is `Discrete-time`, you can clear Use filtered derivative to remove the derivative filter.\n\nTunable: Yes\n\n#### Dependencies\n\nTo enable this parameter, in the Main tab, set the controller-parameters Source to `internal` and set Controller to `PID` or `PD`.\n\n#### Programmatic Use\n\n Block Parameter: `N` Type: scalar, vector Default: 100\n\nSetpoint weight on the proportional term of the controller. The proportional term of a 2-DOF controller output is P(bry), where r is the reference signal and y is the measured plant output. Setting b to 0 eliminates proportional action on the reference signal, which can reduce overshoot in the system response to step changes in the setpoint. Changing the relative values of b and c changes the balance between disturbance rejection and setpoint tracking.\n\nTunable: Yes\n\n#### Dependencies\n\nTo enable this parameter, in the Main tab, set the controller-parameters Source to `internal`.\n\n#### Programmatic Use\n\n Block Parameter: `b` Type: scalar, vector Default: 1\n\nSetpoint weight on the derivative term of the controller. The derivative term of a 2-DOF controller acts on cry, where r is the reference signal and y is the measured plant output. Thus, setting c to 0 eliminates derivative action on the reference signal, which can reduce transient response to step changes in the setpoint. Setting c to 0 can yield a controller that achieves both effective disturbance rejection and smooth setpoint tracking without excessive transient response. Changing the relative values of b and c changes the balance between disturbance rejection and setpoint tracking.\n\nTunable: Yes\n\n#### Dependencies\n\nTo enable this parameter, in the Main tab, set the controller-parameters Source to `internal` and set Controller to a type that has derivative action.\n\n#### Programmatic Use\n\n Block Parameter: `c` Type: scalar, vector Default: 1\n\nIf you have Simulink Control Design software, you can automatically tune the PID coefficients when they are internal to the block. To do so, use this parameter to select a tuning tool, and click Tune.\n\n`Transfer Function Based (PID Tuner App)`\n\nUse PID Tuner, which lets you interactively tune PID coefficients while examining relevant system responses to validate performance. PID Tuner can tune all the coefficients P, I, D, and N, and the setpoint coefficients b and c. By default, PID Tuner works with a linearization of your plant model. For models that cannot be linearized, you can tune PID coefficients against a plant model estimated from simulated or measured response data. For more information, see Design Two-Degree-of-Freedom PID Controllers (Simulink Control Design).\n\n`Frequency Response Based`\n\nUse Frequency Response Based PID Tuner, which tunes PID controller coefficients based on frequency-response estimation data obtained by simulation. This tuning approach is especially useful for plants that are not linearizable or that linearize to zero. Frequency Response Based PID Tuner tunes the coefficients P, I, D, and N, but does not tune the setpoint coefficients b and c. For more information, see Design PID Controller from Plant Frequency-Response Data (Simulink Control Design).\n\nBoth of these tuning methods assume a single-loop control configuration. Simulink Control Design software includes other tuning approaches that suit more complex configurations. For information about other ways to tune a PID Controller block, see Choose a Control Design Approach (Simulink Control Design).\n\n#### Dependencies\n\nTo enable this parameter, in the Main tab, set the controller-parameters Source to `internal`.\n\nZero-crossing detection can accurately locate signal discontinuities without resorting to excessively small time steps that can lead to lengthy simulation times. If you select Limit output or activate External reset in your PID Controller block, activating zero-crossing detection can reduce computation time in your simulation. Selecting this parameter activates zero-crossing detection:\n\n• At initial-state reset\n\n• When entering an upper or lower saturation state\n\n• When leaving an upper or lower saturation state\n\n#### Programmatic Use\n\n Block Parameter: `ZeroCross` Type: string, character vector Values: `\"on\"`, `\"off\"` Default: `\"on\"`\n\n### Initialization\n\nSimulink uses initial conditions to initialize the integrator and derivative-filter (or the unfiltered derivative) output at the start of a simulation or at a specified trigger event. (See the External reset parameter.) These initial conditions determine the initial block output. Use this parameter to select how to supply the initial condition values to the block.\n\n`internal`\n\nSpecify the initial conditions using the Integrator Initial condition and Filter Initial condition parameters. If Use filtered derivative is not selected, use the Differentiator parameter to specify the initial condition for the unfiltered differentiator instead of a filter initial condition.\n\n`external`\n\nSpecify the initial conditions externally using block inputs. Additional input ports Io and Do appear on the block. If Use filtered derivative is not selected, supply the initial condition for the unfiltered differentiator at Do instead of a filter initial condition.\n\n#### Programmatic Use\n\n Block Parameter: `InitialConditionSource` Type: string, character vector Values: `\"internal\"`, `\"external\"` Default: `\"internal\"`\n\nSimulink uses the integrator initial condition to initialize the integrator at the start of a simulation or at a specified trigger event (see External reset). The integrator initial condition and the filter initial condition determine the initial output of the PID controller block.\n\nThe integrator initial condition cannot be `NaN` or `Inf`.\n\n#### Dependencies\n\nTo use this parameter, in the Initialization tab, set Source to `internal`, and set Controller to a type that has integral action.\n\n#### Programmatic Use\n\n Block Parameter: `InitialConditionForIntegrator` Type: scalar, vector Default: 0\n\nSimulink uses the filter initial condition to initialize the derivative filter at the start of a simulation or at a specified trigger event (see External reset). The integrator initial condition and the filter initial condition determine the initial output of the PID controller block.\n\nThe filter initial condition cannot be `NaN` or `Inf`.\n\n#### Dependencies\n\nTo use this parameter, in the Initialization tab, set Source to `internal`, and use a controller that has a derivative filter.\n\n#### Programmatic Use\n\n Block Parameter: `InitialConditionForFilter` Type: scalar, vector Default: 0\n\nWhen you use an unfiltered derivative, Simulink uses this parameter to initialize the differentiator at the start of a simulation or at a specified trigger event (see External reset). The integrator initial condition and the derivative initial condition determine the initial output of the PID controller block.\n\nThe derivative initial condition cannot be `NaN` or `Inf`.\n\n#### Dependencies\n\nTo use this parameter, set Time domain to `Discrete-time`, clear the Use filtered derivative check box, and in the Initialization tab, set Source to `internal`.\n\n#### Programmatic Use\n\n Block Parameter: `DifferentiatorICPrevScaledInput` Type: scalar, vector Default: 0\n\nUse this parameter to specify whether to apply the Integrator Initial condition and Filter Initial condition parameter to the corresponding block state or output. You can change this parameter at the command line only, using `set_param` to set the `InitialConditionSetting` parameter of the block.\n\n`State (most efficient)`\n\nUse this option in all situations except when the block is in a triggered subsystem or a function-call subsystem and simplified initialization mode is enabled.\n\n`Output`\n\nUse this option when the block is in a triggered subsystem or a function-call subsystem and simplified initialization mode is enabled.\n\nThis parameter is only accessible through programmatic use.\n\n#### Programmatic Use\n\n Block Parameter: `InitialConditionSetting` Type: string, character vector Values: `\"state\"`, `\"output\"` Default: `\"state\"`\n\nSpecify the trigger condition that causes the block to reset the integrator and filter to initial conditions. (If Use filtered derivative is not selected, the trigger resets the integrator and differentiator to initial conditions.) Selecting any option other than `none` enables the Reset port on the block for the external reset signal.\n\n`none`\n\nThe integrator and filter (or differentiator) outputs are set to initial conditions at the beginning of simulation, and are not reset during simulation.\n\n`rising`\n\nReset the outputs when the reset signal has a rising edge.\n\n`falling`\n\nReset the outputs when the reset signal has a falling edge.\n\n`either`\n\nReset the outputs when the reset signal either rises or falls.\n\n`level`\n\nReset the outputs when the reset signal either:\n\n• Is nonzero at the current time step\n\n• Changes from nonzero at the previous time step to zero at the current time step\n\nThis option holds the outputs to the initial conditions while the reset signal is nonzero.\n\n#### Dependencies\n\nTo enable this parameter, set Controller to a type that has derivative or integral action.\n\n#### Programmatic Use\n\n Block Parameter: `ExternalReset` Type: string, character vector Values: `\"none\"`, `\"rising\"`, `\"falling\"`, `\"either\"`,`\"level\"` Default: `\"none\"`\n\nSelect to force Simulink and Simulink Control Design linearization commands to ignore any reset mechanism specified in the External reset parameter. Ignoring reset states allows you to linearize a model around an operating point even if that operating point causes the block to reset.\n\n#### Programmatic Use\n\n Block Parameter: `IgnoreLimit` Type: string, character vector Values: `\"off\"`, `\"on\"` Default: `\"off\"`\n\nSignal tracking lets the block output follow a tracking signal that you provide at the TR port. When signal tracking is active, the difference between the tracking signal and the block output is fed back to the integrator input with a gain `Kt`, specified by the Tracking gain (Kt) parameter. Signal tracking has several applications, including bumpless control transfer and avoiding windup in multiloop control structures.\n\n#### Bumpless control transfer\n\nUse signal tracking to achieve bumpless control transfer in systems that switch between two controllers. Suppose you want to transfer control between a PID controller and another controller. To do so, connecting the controller output to the TR input as shown in the following illustration.", null, "For more information, see Bumpless Control Transfer with a Two-Degree-of-Freedom PID Controller.\n\n#### Multiloop control\n\nUse signal tracking to prevent block windup in multiloop control approaches. For an example illustrating this approach with a 1DOF PID controller, see Prevent Block Windup in Multiloop Control.\n\n#### Dependencies\n\nTo enable this parameter, set Controller to a type that has integral action.\n\n#### Programmatic Use\n\n Block Parameter: `TrackingMode` Type: string, character vector Values: `\"off\"`, `\"on\"` Default: `\"off\"`\n\nWhen you select Enable tracking mode, the difference between the signal TR and the block output is fed back to the integrator input with a gain `Kt`. Use this parameter to specify the gain in that feedback loop.\n\n#### Dependencies\n\nTo enable this parameter, select Enable tracking mode.\n\n#### Programmatic Use\n\n Block Parameter: `Kt` Type: scalar Default: 1\n\n### Output saturation\n\nActivating this option limits the block output internally to the block, so that you do not need a separate Saturation block after the controller. It also allows you to activate the anti-windup mechanism built into the block (see the Anti-windup method parameter). Specify the saturation limits using the Lower saturation limit and Upper saturation limit parameters.\n\n#### Programmatic Use\n\n Block Parameter: `LimitOutput` Type: string, character vector Values: `\"off\"`, `\"on\"` Default: `\"off\"`\n\nSpecify the upper limit for the block output. The block output is held at the Upper saturation limit whenever the weighted sum of the proportional, integral, and derivative actions exceeds that value.\n\n#### Dependencies\n\nTo enable this parameter, select Limit output.\n\n#### Programmatic Use\n\n Block Parameter: `UpperSaturationLimit` Type: scalar Default: `Inf`\n\nSpecify the lower limit for the block output. The block output is held at the Lower saturation limit whenever the weighted sum of the proportional, integral, and derivative actions goes below that value.\n\n#### Dependencies\n\nTo enable this parameter, select Limit output.\n\n#### Programmatic Use\n\n Block Parameter: `LowerSaturationLimit` Type: scalar Default: `-Inf`\n\nForce Simulink and Simulink Control Design linearization commands to ignore block output limits specified in the Upper limit and Lower limit parameters. Ignoring output limits allows you to linearize a model around an operating point even if that operating point causes the block to exceed the output limits.\n\n#### Dependencies\n\nTo enable this parameter, select the Limit output parameter.\n\n#### Programmatic Use\n\n Block Parameter: `LinearizeAsGain` Type: string, character vector Values: `\"off\"`, `\"on\"` Default: `\"off\"`\n\nWhen you select Limit output and the weighted sum of the controller components exceeds the specified output limits, the block output holds at the specified limit. However, the integrator output can continue to grow (integrator windup), increasing the difference between the block output and the sum of the block components. In other words, the internal signals in the block can be unbounded even if the output appears bounded by saturation limits. Without a mechanism to prevent integrator windup, two results are possible:\n\n• If the sign of the signal entering the integrator never changes, the integrator continues to integrate until it overflows. The overflow value is the maximum or minimum value for the data type of the integrator output.\n\n• If the sign of the signal entering the integrator changes once the weighted sum has grown beyond the output limits, it can take a long time to unwind the integrator and return the weighted sum within the block saturation limit.\n\nIn either case, controller performance can suffer. To combat the effects of windup without an anti-windup mechanism, it may be necessary to detune the controller (for example, by reducing the controller gains), resulting in a sluggish controller. To avoid this problem, activate an anti-windup mechanism using this parameter.\n\n`none`\n\nDo not use an anti-windup mechanism.\n\n`back-calculation`\n\nUnwind the integrator when the block output saturates by feeding back to the integrator the difference between the saturated and unsaturated control signal. The following diagram represents the back-calculation feedback circuit for a continuous-time controller. To see the actual feedback circuit for your controller configuration, right-click on the block and select Mask > Look Under Mask.", null, "Use the Back-calculation coefficient (Kb) parameter to specify the gain of the anti-windup feedback circuit. It is usually satisfactory to set `Kb = I`, or for controllers with derivative action, `Kb = sqrt(I*D)`. Back-calculation can be effective for plants with relatively large dead time .\n\n`clamping`\n\nIntegration stops when the sum of the block components exceeds the output limits and the integrator output and block input have the same sign. Integration resumes when the sum of the block components exceeds the output limits and the integrator output and block input have opposite sign. Clamping is sometimes referred to as conditional integration.\n\nClamping can be useful for plants with relatively small dead times, but can yield a poor transient response for large dead times .\n\n#### Dependencies\n\nTo enable this parameter, select the Limit output parameter.\n\n#### Programmatic Use\n\n Block Parameter: `AntiWindupMode` Type: string, character vector Values: `\"none\"`, `\"back-calculation\"`,`\"clamping\"` Default: `\"none\"`\n\nThe `back-calculation` anti-windup method unwinds the integrator when the block output saturates. It does so by feeding back to the integrator the difference between the saturated and unsaturated control signal. Use the Back-calculation coefficient (Kb) parameter to specify the gain of the anti-windup feedback circuit. For more information, see the Anti-windup method parameter.\n\n#### Dependencies\n\nTo enable this parameter, select the Limit output parameter, and set the Anti-windup method parameter to `back-calculation`.\n\n#### Programmatic Use\n\n Block Parameter: `Kb` Type: scalar Default: 1\n\n### Data Types\n\nThe parameters in this tab are primarily of use in fixed-point code generation using Fixed-Point Designer™. They define how numeric quantities associated with the block are stored and processed when you generate code.\n\nIf you need to configure data types for fixed-point code generation, click Open Fixed-Point Tool and use that tool to configure the rest of the parameters in the tab. For information about using Fixed-Point Tool, see Autoscaling Data Objects Using the Fixed-Point Tool (Fixed-Point Designer).\n\nAfter you use Fixed-Point Tool, you can use the parameters in this tab to make adjustments to fixed-point data-type settings if necessary. For each quantity associated with the block, you can specify:\n\n• Floating-point or fixed-point data type, including whether the data type is inherited from upstream values in the block.\n\n• The minimum and maximum values for the quantity, which determine how the quantity is scaled for fixed-point representation.\n\nFor assistance in selecting appropriate values, click", null, "to open the Data Type Assistant for the corresponding quantity. For more information, see Specify Data Types Using Data Type Assistant.", null, "The specific quantities listed in the Data Types tab vary depending on how you configure the PID controller block. In general, you can configure data types for the following types of quantities:\n\n• Product output — Stores the result of a multiplication carried out under the block mask. For example, P product output stores the output of the gain block that multiplies the block input with the proportional gain P.\n\n• Parameter — Stores the value of a numeric block parameter, such as P, I, or D.\n\n• Block output — Stores the output of a block that resides under the PID controller block mask. For example, use Integrator output to specify the data type of the output of the block called Integrator. This block resides under the mask in the Integrator subsystem, and computes integrator term of the controller action.\n\n• Accumulator — Stores values associated with a sum block. For example, SumI2 Accumulator sets the data type of the accumulator associated with the sum block SumI2. This block resides under the mask in the Back Calculation subsystem of the Anti-Windup subsystem.\n\nIn general, you can find the block associated with any listed parameter by looking under the PID Controller block mask and examining its subsystems. You can also use the Model Explorer to search under the mask for the listed parameter name, such as `SumI2`. (See Model Explorer.)\n\nMatching Input and Internal Data Types\n\nBy default, all data types in the block are set to `Inherit: Inherit via internal rule`. With this setting, Simulink chooses data types to balance numerical accuracy, performance, and generated code size, while accounting for the properties of the embedded target hardware.\n\nUnder some conditions, incompatibility can occur between data types within the block. For instance, in continuous time, the Integrator block under the mask can accept only signals of type `double`. If the block input signal is a type that cannot be converted to `double`, such as `uint16`, the internal rules for type inheritance generate an error when you generate code.\n\nTo avoid such errors, you can use the Data Types settings to force a data type conversion. For instance, you can explicitly set P product output, I product output, and D product output to `double`, ensuring that the signals reaching the continuous-time integrators are of type `double`.\n\nIn general, it is not recommended to use the block in continuous time for code generation applications. However, similar data type errors can occur in discrete time, if you explicitly set some values to data types that are incompatible with downstream signal constraints within the block. In such cases, use the Data Types settings to ensure that all data types are internally compatible.\n\n#### Fixed-Point Operational Parameters\n\nSpecify the rounding mode for fixed-point operations. For more information, see Rounding (Fixed-Point Designer).\n\nBlock parameters always round to the nearest representable value. To control the rounding of a block parameter, enter an expression using a MATLAB® rounding function into the mask field.\n\n#### Programmatic Use\n\n Block Parameter: `RndMeth` Type: character vector Values: ```'Ceiling' \| 'Convergent' \| 'Floor' \| 'Nearest' \| 'Round' \| 'Simplest' \| 'Zero'``` Default: `'Floor'`\n\nSpecify whether overflows saturate or wrap.\n\n• `off` — Overflows wrap to the appropriate value that the data type can represent.\n\nFor example, the number 130 does not fit in a signed 8-bit integer and wraps to -126.\n\n• `on` — Overflows saturate to either the minimum or maximum value that the data type can represent.\n\nFor example, an overflow associated with a signed 8-bit integer can saturate to -128 or 127.\n\n### Tip\n\n• Consider selecting this check box when your model has a possible overflow and you want explicit saturation protection in the generated code.\n\n• Consider clearing this check box when you want to optimize efficiency of your generated code.\n\nClearing this check box also helps you to avoid overspecifying how a block handles out-of-range signals. For more information, see Troubleshoot Signal Range Errors.\n\n• When you select this check box, saturation applies to every internal operation on the block, not just the output or result.\n\n• In general, the code generation process can detect when overflow is not possible. In this case, the code generator does not produce saturation code.\n\n#### Programmatic Use\n\n Block Parameter: `SaturateOnIntegerOverflow` Type: character vector Values: `'off' \| 'on'` Default: `'off'`\n\nSelect this parameter to prevent the fixed-point tools from overriding the data types you specify on this block. For more information, see Lock the Output Data Type Setting (Fixed-Point Designer).\n\n#### Programmatic Use\n\n Block Parameter: `LockScale` Type: character vector Values: `'off' \| 'on'` Default: `'off'`\n\n### State Attributes\n\nThe parameters in this tab are primarily of use in code generation.\n\nAssign a unique name to the state associated with the integrator or the filter, for continuous-time PID controllers. (For information about state names in a discrete-time PID controller, see the State name parameter.) The state name is used, for example:\n\n• For the corresponding variable in generated code\n\n• As part of the storage name when logging states during simulation\n\n• For the corresponding state in a linear model obtain by linearizing the block\n\nA valid state name begins with an alphabetic or underscore character, followed by alphanumeric or underscore characters.\n\n#### Dependencies\n\nTo enable this parameter, set Time domain to `Continuous-time`.\n\n#### Programmatic Use\n\n Parameter: `IntegratorContinuousStateAttributes`, `FilterContinuousStateAttributes` Type: character vector Default: `''`\n\nAssign a unique name to the state associated with the integrator or the filter, for discrete-time PID controllers. (For information about state names in a continuous-time PID controller, see the State name (e.g., 'position') parameter.)\n\nA valid state name begins with an alphabetic or underscore character, followed by alphanumeric or underscore characters. The state name is used, for example:\n\n• For the corresponding variable in generated code\n\n• As part of the storage name when logging states during simulation\n\n• For the corresponding state in a linear model obtain by linearizing the block\n\nFor more information about the use of state names in code generation, see Apply Storage Classes to Individual Signal, State, and Parameter Data Elements (Simulink Coder).\n\n#### Dependencies\n\nTo enable this parameter, set Time domain to `Discrete-time`.\n\n#### Programmatic Use\n\n Parameter: `IntegratorStateIdentifier`, `FilterStateIdentifier` Type: string, character vector Default: `\"\"`\n\nSelect this parameter to require that the discrete-time integrator or filter state name resolves to a Simulink signal object.\n\n#### Dependencies\n\nTo enable this parameter for the discrete-time integrator or filter state:\n\n1. Set Time domain to `Discrete-time`.\n\n2. Specify a value for the integrator or filter State name.\n\n3. Set the model configuration parameter Signal resolution to a value other than `None`.\n\nSelecting this check box disables Code generation storage class for the corresponding integrator or filter state.\n\n#### Programmatic Use\n\n Block Parameter: `IntegratorStateMustResolveToSignalObject`, `FilterStateMustResolveToSignalObject` Type: string, character vector Values: `\"off\"`, `\"on\"` Default: `\"off\"`\n\nSelect state storage class for code generation. If you do not need to interface to external code, select `Auto`.\n\nFor more information, see Apply Storage Classes to Individual Signal, State, and Parameter Data Elements (Simulink Coder) and Apply Built-In and Customized Storage Classes to Data Elements (Embedded Coder).\n\n#### Dependencies\n\nTo enable this parameter for the discrete-time integrator or filter state:\n\n1. Set Time domain to `Discrete-time`.\n\n2. Specify a value for the integrator or filter State name.\n\n3. Set the model configuration parameter Signal resolution to a value other than `None`.\n\n#### Programmatic Use\n\n Block Parameter: `IntegratorRTWStateStorageClass`, `FilterRTWStateStorageClass` Type: string, character vector Values: `\"Auto\"`, `\"ExportedGlobal\"`, `\"ImportedExtern\"` \| `\"ImportedExternPointer\"` Default: `\"Auto\"`\n\nSpecify a storage type qualifier such as `const` or `volatile`.\n\n### Note\n\nThis parameter will be removed in a future release. To apply storage type qualifiers to data, use custom storage classes and memory sections. Unless you use an ERT-based code generation target with Embedded Coder®, custom storage classes and memory sections do not affect the generated code.\n\n#### Dependencies\n\nTo enable this parameter, set Code generation storage class to any value other than `Auto`.\n\n#### Programmatic Use\n\n Block Parameter: `IntegratorRTWStateStorageTypeQualifier`,`FilterRTWStateStorageTypeQualifier` Type: string, character vector Values:`\"\"`,`\"const\"`,`\"volatile\"` Default: `\"\"`\n\n## Block Characteristics\n\n Data Types `double` \| `fixed point` \| `integer` \| `single` Direct Feedthrough `no` Multidimensional Signals `no` Variable-Size Signals `no` Zero-Crossing Detection `no`" ]	[ null, "https://se.mathworks.com/help/simulink/slref/discrete_pid2dof_block_icon.png", null, "https://se.mathworks.com/help/simulink/slref/pid_controller_2dof_discrete_watertank_example.png", null, "https://se.mathworks.com/help/simulink/slref/pid_controller_2dof_discrete_watertank_example.png", null, "https://se.mathworks.com/help/simulink/slref/pid_controller_2dof_reset_trigger.png", null, "https://se.mathworks.com/help/simulink/slref/pid_controller_2dof_id_circuit.png", null, "https://se.mathworks.com/help/simulink/slref/pid_controller_2dof_signal_tracking.png", null, "https://se.mathworks.com/help/simulink/slref/pid_controller_2dof_back_calculation.png", null, "https://se.mathworks.com/help/simulink/slref/show_data_type_asst_button.png", null, "https://se.mathworks.com/help/simulink/slref/pid_controller_data_types.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7666839,"math_prob":0.78069687,"size":54918,"snap":"2020-34-2020-40","text_gpt3_token_len":11598,"char_repetition_ratio":0.18805769,"word_repetition_ratio":0.35450047,"special_character_ratio":0.19805892,"punctuation_ratio":0.110729985,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9681026,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,null,null,2,null,2,null,2,null,2,null,2,null,2,null,null,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-12T15:44:15Z\",\"WARC-Record-ID\":\"<urn:uuid:e418641d-e17d-42f6-9bb5-1ad139ba6158>\",\"Content-Length\":\"287425\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:557bbd39-fbf6-4cab-867c-45ed4e5150c8>\",\"WARC-Concurrent-To\":\"<urn:uuid:2c349420-c091-42c9-9956-3e933f21d85c>\",\"WARC-IP-Address\":\"23.66.56.59\",\"WARC-Target-URI\":\"https://se.mathworks.com/help/simulink/slref/discretepidcontroller2dof.html\",\"WARC-Payload-Digest\":\"sha1:TT7GBJTVY72GSOWINIAFSYEOVVOZS3JU\",\"WARC-Block-Digest\":\"sha1:ORXJCKHPGIAND4B5ZR2VHHSCX3GV3M7K\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738905.62_warc_CC-MAIN-20200812141756-20200812171756-00558.warc.gz\"}"}
http://www.scitlion.com/index.php?m=content&c=index&a=show&catid=151&id=230	[ "## Contact Us\n\n+86-10-81508344(Sales Department)\nNo.5, Xingguang 4 Ave ,Opto-Mechatronics Industrial Park ,TongZhou district,Beijing 101111,China\n\n## Nonlinear Crystals", null, "• Product Name: LiNbO3\n• Dimension Tolerance:W x H (+/- 0.1mm) x L (+0.2/-0.1mm)\n• Angle Tolerance:Δθ ≤ 0.25°, Δφ ≤ 0.25°\n• Parallelism:< 20 arc seconds\n• Perpendicularity:< 5 arc minutes\n• Chamfer:0.1mm at 45°\n• Clear Aperture:90%\n• Scratch and Dig:20/10 or better\n• Surface Flatness:λ/8 at 633nm\n• Wavefront Distortion:λ/8 at 633nm\n• Damage Threshold :100 MW/cm2 at 1064nm, 10ns, 10Hz\n\n## Product overview\n\nLithium Niobate (LiNbO3) or LN Crystal is widely used as frequency doublers for wavelength over 1μm and optical parametric oscillators (OPOs) pumped at 1064 nm and quasi-phase-matched (QPM) devices. Additionally due to its large Electro-Optic (EO) and Acousto-Optic (AO) coefficients, LiNbO3 is the most commonly used material for Pockels Cell, Q-switches and phase nodulators, waveguide substrate, and surface acoustic wave (SAW) wafers, etc.\n\nScitlion is able to offer\nSize upto 10 x 10 x 20mm\nLength from 0.02mm to 20mm\nCoating upon your request", null, "", null, "Density 4.64 g/cm3 Melting Point 1253°C Curie Temperature 1140°C Mohs Hardness 5 Elastic Stiffness Coefficients CE11= 2.33(×1011N/m2 ) CE33= 2.77(×1011N/m2 ) Crystal Structure Trigonal Lattices Constants a=5.148A , c=13.863A Thermal and Electrical Properties Thermal Expansion Coefficients 2.0x10-6/K (// a-axis at 25°C) 2.2x10-6/K (// c-axis at 25°C) Thermal Conductivity 38W/m/K at 25°C Resistivity 2x10-6Ω·cm@200°C Piezoelectric Strain Constant D22=2.04(x10-11C/N) D33=19.22(x10-11C/N) Optical and Electro-optical Properties Transparency Range 420nm to 5200nm Optical Homogeneity ~ 5 x 10-5 /cm Refractive indices ne = 2.146, no = 2.220 at 1300 nm ne = 2.156, no = 2.232 at 1064 nm ne = 2.203, no = 2.286 at 632.8 nm NLO Coefficients: d33 = 86 x d36 (KDP) d31 =11.6 x d36 (KDP) d22 = 5.6 x d36 (KDP) Effective NLO Coefficients: deff(I)=d31sinθ-d22cossin3Φ deff(II)=d22cos2θcos3Φ Electro-Optic Coefficients r T33 = 32 pm/V, rS33 = 31 pm/V, rT31 =10 pm/V, rS31=8.6 pm/V, rT22 = 6.8 pm/V, rS22= 3.4 pm/V, Half-Wave Voltage 3.03 KV (Electrical field // z, light ┴ z) 4.02 KV (Electrical field // x or y, light // z) Sellmeier Equations (λ in μm) no2=4.9048+0.11768/(λ2-0.04750)-0.027169λ2 ne2=4.5820+0.099169/(λ2-0.04443)-0.02195λ2" ]	[ null, "http://www.scitlion.com/uploadfile/2013/0711/20130711100646712.jpg", null, "http://www.scitlion.com/uploadfile/2013/0509/20130509104600272.jpg", null, "http://www.scitlion.com/uploadfile/2013/0509/20130509104612750.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6397374,"math_prob":0.91502726,"size":2017,"snap":"2019-35-2019-39","text_gpt3_token_len":793,"char_repetition_ratio":0.084451064,"word_repetition_ratio":0.0,"special_character_ratio":0.39613286,"punctuation_ratio":0.1378446,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95727766,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-18T15:25:11Z\",\"WARC-Record-ID\":\"<urn:uuid:dc72814c-5088-482a-b1b4-765a2107eee2>\",\"Content-Length\":\"46774\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:634ef71c-a492-41f2-9472-abff6380ee7c>\",\"WARC-Concurrent-To\":\"<urn:uuid:503c9342-967e-4086-9edf-42b52e27808c>\",\"WARC-IP-Address\":\"39.98.166.208\",\"WARC-Target-URI\":\"http://www.scitlion.com/index.php?m=content&c=index&a=show&catid=151&id=230\",\"WARC-Payload-Digest\":\"sha1:QGR5NKD5CAEL5ZSLK5NYAUPOWDFA3CI5\",\"WARC-Block-Digest\":\"sha1:BTJ3QLXVBHYOW45ILKP2C5JUDGFYUMYH\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027313936.42_warc_CC-MAIN-20190818145013-20190818171013-00036.warc.gz\"}"}
https://asp-eurasipjournals.springeropen.com/articles/10.1186/s13634-021-00730-w	[ "# Multi-source and multi-fault condition monitoring based on parallel factor analysis and sequential probability ratio test\n\n## Abstract\n\nThe monitoring of mechanical equipment systems contains an increasing number of complex content, expanding from traditional time, and frequency information to three-dimensional data of the time, space, and frequency information, and even higher-dimensional data containing subjects, experimental conditions. For high-dimensional data analysis, traditional decomposition methods such as Hilbert transform, fast Fourier transformation, and Gabor transformation not only lose the integrity of the data, but also increase the amount of calculation and introduce a lot of redundant information. The phenomenon of feature coupling, aliasing, and redundancy between the mechanical multi-source data signals will cause the inaccuracy of the evaluation, diagnosis, and prediction of industrial production operation status. The analysis of the three-way tensor composed of channel, frequency, and time is called parallel factor analysis (PARAFAC). The properties between the parallel factor analysis results and the input signals are studied through simulation experiments. Parallel factor analysis is used to decompose the third-order tensor composed of channel-time-frequency after continuous wavelet transformation of vibration signal into channel, time, and frequency characteristics. Multi-scale parallel factor analysis successfully extracted non-linear multi-dimensional dynamic fault characteristics by generating the spatial, spectral, time-domain signal loading value and three-dimensional fault characteristic expression. In order to verify the effectiveness of the space, frequency, and time domain signal loading values of the fault characteristic factors generated by the centrifugal pump system after parallel factor analysis, the characteristic factors obtained after parallel factor analysis are used as the SPRT test sequence for identification and verification. The results indicate that the method proposed in this article improves the measurement accuracy and intelligence of mechanical fault detection.\n\n## 1 Introduction\n\nIn recent years, equipment fault diagnosis, as a new technology that crosses various disciplines, has been developed rapidly and has produced huge economic benefits [1,2,3,4,5,6]. The centrifugal pump is an important energy conversion and liquid transmission device in the process industry and its working state directly affects the production of the entire operating equipment. A slight damage to the impeller will shorten the running time of the centrifugal pump and disturb the operation of the equipment. When the impeller fails, it will cause damage to the centrifugal pump components or personal injury accidents, which will cause significant economic losses [7, 8]. The normal operation and failure of the centrifugal pump will cause the equipment to vibrate. The vibration signal contains rich information of the pump body running state and is easy to be collected, which can be used to monitor and diagnose the running state of the centrifugal pump. Fault diagnosis is generally divided into three steps: first, we collect the relevant vibration signal of the diagnostic object; then, the signal is analyzed and processed to acquire the characteristics of the vibration signal; finally, pattern recognition and fault diagnosis are performed through the corresponding extracted special diagnosis [9, 10]. The core content is to obtain the effective characteristics of the vibration signal. Due to the complex structure of the centrifugal pump, many excitation sources and mutual interference, the vibration signal of the centrifugal pump is a non-linear and non-stationary signal. Researchers have proposed various effective diagnostic methods to process the collected raw vibration signals of the centrifugal pump, extract effective information, and improve the accuracy of diagnosis. Wenjian Huang et al. extracted the characteristic parameters of the vibration signal through time-domain signal analysis, then the PCA was used to reduce the amount of data, and finally the main component with the largest contribution rate was used as the input signal of SPRT to evaluate the proposed algorithm. Literature proposed an improved deep convolutional neural network (CNN) to identify defects in centrifugal pumps by using sound and image recognition. A feature extraction method based on empirical mode decomposition (EMD) was developed to detect the gravity of cavitation in the centrifugal pump by Azizi et al. . Liu Yang et al. proposed the new method for analysis of big data based on particle swarm optimization wavelet neural network for diagnosis in the gearbox. Literature applies variational mode decomposition (VMD) with different input parameters to fault diagnosis of multi-stage pumps. Signal processing combined with empirical mode decomposition (EMD) and fuzzy c-means clustering is used for monitoring piston pump defects in literature . The traditional decomposition method of processing high-dimensional data will not only lose the integrity of the data but also increase the amount of calculation and introduce redundancy [15,16,17,18]. These methods of extracting time-frequency characteristics from single-channel signals, such as Fourier transform, cannot reflect the internal relationship of non-linear changes between multi-source channel characteristic signals, nor can they eliminate information interference.\n\nMechanical non-linear multi-fault mode multi-source dynamic feature identification is a technical bottleneck and difficult problem encountered in the application of fault diagnosis in process industry production lines. It not only needs to extract the time-frequency characteristics of multi-source fault signals, but also to ensure the correspondence between non-linear variables and multi-fault modes and multi-source fault features in time, frequency, and space after feature extraction. Parallel factor analysis proposed by Carroll, Chang , and Harshman in 1970 is a three-dimensional or multi-dimensional signal processing algorithm that uses iterative least squares to resolve the decomposition and identification of multi-dimensional matrices. The general time-frequency decomposition ignores the spatial information of the vibration signal and cannot handle multi-channel data [21,22,23]. Data framing in the form of a three-way array indexed by channel, frequency and time allows the application of a unique decomposition called Parallel Factor Analysis (PARAFAC). The decomposition uniqueness of PARAFAC model can obtain its model parameters without ambiguity so that the PARAFAC model has important application value. As a data processing algorithm, PARAFAC model has been successfully applied in fluorescence spectroscopy, psychology, signal processing, food science, and other fields. Multi-channel electroencephalogram EEG data can usually be expressed as an M×N×P three-way data set and the components of the three-way data array correspond to the channel (electrodes at different positions), time (data samples) and frequency components. Schmitz, S. applied PARAFAC to analyze the temporal and spatial patterns of functional connections between neurons, which were revealed in the sequence of peaks recorded in the cat’s main visual cortex (area 18) . This parallel factor analysis was applied for decomposing EEG data into space-time-frequency components during the resting state and mental arithmetic by Miwakeichi, Fumikazu et al. . Rost'akova compares non-negative Tucker decomposition with parallel factor analysis to identify and measure human brain electrical rhythms . In the literature , Choi, Ji Yeh proposed a new extension function PARAFAC for processing response to three-dimensional data arranged along a two-dimensional domain and one-dimensional parameters. Technically, this method combines PARAFAC with basis function expansion approximation and is applied to EEG data to prove its empirical usefulness. A parallel factor analysis study showed that the frontal lobe area with higher frequency response is the main feature of laser evoked potential in rats with chronic inflammatory pain .\n\nParallel decomposition has attracted great attention, because parallel factorization can process the constructed high-dimensional data as a whole, which not only retains the overall structure information of the data, but also makes the structure more compact and easy to understand. In the literature , parallel factor analysis was used as the diagnostic tool through decomposing centrifugal pump diagnostic signal into time-frequency-space modes. Considering the difficulty of extracting fault features from rolling bearings under strong background noise, Yang Cheng proposed a new method based on variable mode decomposition (VMD) and phase space parallel factor analysis to detect weak fault signals of rolling bearings. In order to overcome the inability to extract sparse and interpretable latent variables from batch data, literature proposed a batch three-way data array sparse model based on sparse parallel factor (SPARAFAC) decomposition. Sparse factor matrices have the potential advantage of being easy to interpret because they eliminate redundant data information and show significant variable correlation. In chemistry, medicine, and food science commonly used fluorescence excitation and emission data typically contain several chemical components at different concentrations. Fluorescence spectroscopy can generate a three-way data set with the mode “sample × excitation × emission.” The main purpose of the analysis of this data type is to determine which chemicals are present in each sample and their relative concentrations. Reference conducted a comparison between parallel factor analysis (PARAFAC) and support vector machine (SVM) to identify and distinguish the fluorescence spectrum of coconut water brands. The above results indicate that fluorescence spectroscopy combined with PARAFAC and SVM method has been proved to be a simple and rapid detection method for coconut water and other beverages. This study aims to determine whether the composition or distribution of humus in lake sediments can be characterized by chemometric spectral data. This method determines the three-dimensional excitation emission matrix in the extracted humus and performs spectral analysis of the data by using parallel factor analysis (PARAFAC) with classification tree and regression tree (CART).\n\nThe theory of sequential probability ratio test (SPRT) that is a branch of mathematical statistics was proposed by Abraham Wald in 1947 in order to solve the problem of sampling and acceptance of valuable military products. This method provides an approximate formula for the critical value of accepting the null hypothesis H0 or accepting the alternative hypothesis H1 based on the sample values obtained from each observation, and also provides the average sampling times and power function of this test method. In 1948, Abraham Wald and American statistician Wolfowitz proved that the above-mentioned sequential probability is the smallest number of sampling times required for the test in all the two types of tests whose error probability does not exceed α and β, respectively. The sequential probability ratio test is the most fundamental sequential test in sequential analysis proposed by Abraham Wald, and it has subsequently been widely developed in various fields. Almost all the hypothesis testing problems of SPRT in mechanical fault diagnosis, such as signal detection, model variable point detection, life data analysis of centrifugal pump, and crack detection of gearbox, can be well applied. This research were performed on actual faults in a laboratory-scale distillation plan based on artificial neural network-multi-layer perceptron (ANN-MLP) and the Wald sequential probability ratio test (SPRT). In the literature , Guo Peng proposed Gaussian process and SPRT wind turbine power curve modeling and monitoring. The modeling and monitoring method proposed in this paper successfully identified two wind anemometer failures and pitch system failures. Literature proposed a fault detection algorithm based on the sequential probability ratio test (SPRT) and chi-square test for redundant multi-sensor navigation systems for supersonic cruise ships (HCV).\n\nThe rest of this paper is organized as follows. In Section 2, the parallel factorization model and simulation is described. The multi-scale parallel factorization optimization algorithm for non-linear multi-source fault characteristic signal extraction is established. The characteristic factor signal is successfully obtained from the matrix factor, and the “loading” factor and “component” factor are defined. In Section 3, we studied the multi-channel data decision theory based on SPRT, and established an adaptive optimization diagnosis method for tracking and identifying the non-linear multi-dimensional dynamic optimal characteristic signal. To research the validity of the multi-scale parallel factor analysis and SPRT for multi-channel signal in actual complex industrial production, the centrifugal pump fault diagnosis experimental system was designed and implemented in Section 4. Following that, multi-source dynamic feature extraction based on parallel factorization and SPRT for the multi-source condition monitoring of centrifugal pump are presented in Section 5. Finally, the conclusions are drawn in Section 6.\n\n## 2 The model and simulation\n\n### 2.1 Parallel factorization model\n\nIn a two-dimensional matrix, xi, j is generally used to represent the elements in the two-dimensional matrix (i represents any row, j represents any column). Similarly, in the three-dimensional matrix, we use xi, j, kto represent any element in the three-dimensional matrix. At present, there is no definite name naming subscript k, let us call it “page” . The subscript of the three-dimensional matrix consists of three index value row, column, and page composition. The left picture in Fig. 1 shows the three-dimensional matrix and the right picture shows its sub-matrix. When a certain dimension in a three-dimensional matrix is fixed, it constitutes a sub-matrix of the three-dimensional matrix that is called a slice of the three-dimensional matrix along a certain dimension.\n\nThe expansion of the three-dimensional matrix is actually to rearrange the slices of the three-dimensional matrix to constitute the new two-dimensional matrix. For example, we fixed the rows and columns of a three-dimensional matrix and rearranged its pages to formulate the new two-dimensional matrix. At this time, the number of rows is equivalent to the number of rows I of the original matrix and the number of columns changes from the original J to J × K, denoted asXI × JK. It is expressed as shown in Eq. (1).\n\n$${X}^{I\\times JK}=\\left[{X}_{k=1},{X}_{k=2}\\cdots {X}_{k=K}\\right]$$\n(1)\n\nOf course, it can also be expanded by column, such as XIK × J, which is defined as formula (2).\n\n$${X}^{IK\\times J}=\\left[\\begin{array}{c}{X}_{k=1}\\\\ {}{X}_{k=2}\\\\ {}\\vdots \\\\ {}{X}_{k=K}\\end{array}\\right]$$\n(2)\n\nAfter expanding by columns, we acknowledge that I × K is displayed as the number of rows of the new matrix and parameter “J” is the number of columns.\n\nThe symbol $${x}_{i,j,k}=\\sum \\limits_{f=1}^F{a}_{i,f}{b}_{j,f}{c}_{k,f}$$ can be used to express any element in a three-dimensional or larger than three-dimensional matrix, the variables i, j, and k in the formula can be any natural numbers. The elements in the i-th row, j-th column, and k-th page of the matrix X can be represented by xi, j, k. According to the definition, the low-rank decomposition of a two-dimensional matrix can be popularized to the low-rank decomposition of a three-dimensional matrix. Let the element XCI × J × Kof the three-dimensional matrix be defined as xijk, the variables i, j, and k in the formula can be any natural numbers. Similarly, it can be seen that the three-dimensional matrix can be indicated as the modality of the vector outer product of the following formula (3).\n\n$$X={a}_1\\circ {b}_1\\circ {c}_1+,\\dots, +{a}_R\\circ {b}_R\\circ {c}_R=\\sum \\limits_{r=1}^R{a}_r\\circ {b}_r\\circ {c}_r$$\n(3)\n\nIt gives the process of a three-dimensional matrix low rank decomposition in formula (3) and the symbol R of the formula is indicated as the rank of the three-dimensional matrix X, where crCK, brCJ,arCI, r = 1,...,R. Harshman names the low-rank decomposition model of three-dimensional matrix given by formula (3) as the general model of parallel factor. The general forms of the parallel factorization model are interpreted in Fig. 2, in which X can be displayed by the popular geometric cube.\n\nTheseC = [c1, …, cR];B = [b1, …, bR];A = [a1, …, aR] are any given three two-dimensional matrix definitions. We define A, B, and C as the three loading matrices of the parallel factorization model of the general form. Equation (4) is shown as the scalar form of the general parallel factorization model. They are labeled as ckr = [C]k, r, bjr = [B]j, r,air = [A]i, r, and $${\\underline{x}}_{ijk}={\\left[\\underline{X}\\right]}_{i,j,k}$$.\n\n$${\\underline{x}}_{ijk}=\\sum \\limits_{r=1}^R{a}_{ir}{b}_{jr}{c}_{kr}$$\n(4)\n\nThe general form of the parallel factorization model can be viewed as the low-rank decomposition of a two-dimensional matrix extending to a three-dimensional matrix. The formula (4) indicates that these subitems $${\\underline{x}}_{ijk}$$of the three-dimensional matrix X can also be denoted as the sum of the products of R elements a, b, and c. Compared with the matrix elements xij in PCA, $${\\underline{x}}_{ijk}$$ has three independently changing dimensions called “mode A,” “mode B,” and “mode C.”\n\n### 2.2 Matrix essential equalization\n\nThere is a matrix ACI × J. If the matrix satisfies $$\\overline{A}=A\\Pi \\Delta$$, it is said that A and $$\\overline{A}$$ are matrix essential equalization, denoted as $$A\\cong \\overline{A}$$. Among them, ∏ is the column exchange matrix and is the diagonal scale matrix. There is one and only one non-zero element “1” in each row and each column of the column exchange matrix ∏. The function of the column exchange matrix is to rearrange the column vector of A in the order of ∏ without changing the value of the elements in the vector. The diagonal scale matrix is a J×J diagonal matrix with non-zero diagonal elements. The function of is to multiply each column of matrix A by a non-zero amplitude.\n\nAccording to the concept of matrix essential equalization, we take a two-dimensional matrixX = ABTas an example, where ACM × F,BCN × F. For any matrix $$\\overline{A}\\in {C}^{M\\times F},\\overline{B}\\in {C}^{N\\times F}$$, if $$X={AB}^T=\\overline{A}{\\overline{B}}^T$$is satisfied, then we can get to formula (5).\n\n$$\\overline{A}=A{\\prod}_1{\\Delta}_1,\\overline{B}=B{\\prod}_2{\\Delta}_2$$\n(5)\n\n∏2 and ∏2 in the formula are column exchange matrices, which means to rearrange the columns of the A and B matrices. 1 and 2 are diagonal scale matrices, which means that each column of matrix A and B is multiplied by the non-zero coefficient. At this time, the matrix decomposition is said to be unique.\n\nWhen the matrix factorization is unique, the matrix $$\\overline{A},\\overline{B}$$obtained by the matrix factorization is not completely equal to the original matrices A and B, they are only the essential equality relationship of the matrix. The essential equal relationship of matrix A and B is shown in the following formula (6) (7).\n\n$$\\overline{A}=A{\\prod}_A{\\Delta}_A\\cong A$$\n(6)\n$$\\overline{B}=B{\\prod}_B{\\Delta}_B\\cong B$$\n(7)\n\nDue to the existence of column exchange matrices ∏A, ∏B and diagonal scale matrices A, B, the order and magnitude of the column vectors in matrix $$\\overline{A},\\overline{B}$$ can be different from those of A and B. In matrix theory, they are used to be called column blur and scale blur, which are represented by column exchange matrix ∏ and diagonal scale matrix , respectively. In the process of matrix decomposition, if there is no structural constraint on the matrices A and B, column blur and scale blur are unavoidable. The above problem can be explained in the vector form of matrix decomposition, which is denotedA = [a1, , aF], B = [b1, , bF], whereafCI × 1bfCJ × 1 (f=1,...,F) is the column vector of A and B, respectively. The above formula can be expressed as the following vector form.\n\n$$X={AB}^T={\\mathrm{a}}_1{\\mathrm{b}}_1^T+{\\mathrm{a}}_2{\\mathrm{b}}_2^T+\\cdots +{\\mathrm{a}}_F{\\mathrm{b}}_F^T$$\n(8)\n\nIn formula (8), $${\\mathrm{a}}_1{\\mathrm{b}}_1^T,\\cdots, {\\mathrm{a}}_F{\\mathrm{b}}_F^T$$ are F matrices with rank 1. At this time, if the order of $${\\mathrm{a}}_1{\\mathrm{b}}_1^T$$,...,$${\\mathrm{a}}_F{\\mathrm{b}}_F^T$$ is changed arbitrarily, the value of matrix X is unchanged. Similarly, if the vector af is multiplied by the non-zero coefficient λf and the corresponding bf is multiplied by a non-zero coefficient 1/λf, the value of X will not change either. Assuming that the order of $${\\mathrm{a}}_1{\\mathrm{b}}_1^T$$and $${\\mathrm{a}}_2{\\mathrm{b}}_2^T$$ in formula (8) is exchanged, it can be rewritten into the following form:\n\n$$\\begin{array}{c}X={AB}^T={\\mathrm{a}}_1{\\mathrm{b}}_1^T+{\\mathrm{a}}_2{\\mathrm{b}}_2^T+\\cdots +{\\mathrm{a}}_F{\\mathrm{b}}_F^T\\\\ {}={\\lambda}_2{a}_2\\frac{1}{\\lambda_2}{b}_2^T+{\\lambda}_1{a}_1\\frac{1}{\\lambda_1}{b}_1^T+\\cdots +{\\lambda}_F{a}_F\\frac{1}{\\lambda_F}{b}_F^T\\\\ {}=A{\\prod}_A{\\Delta}_A{\\left(B{\\prod}_B{\\Delta}_B\\right)}^T\\\\ {}=\\overline{A}{\\overline{B}}^T\\end{array}}$$\n(9)\n\nThe result $$X={AB}^T=\\overline{A}{\\overline{B}}^T$$ can be obtained, the code in the equation is specifically expressed as follows $$\\overline{A}=A{\\prod}_A{\\Delta}_A$$,$$\\overline{B}=B{\\prod}_B{\\Delta}_B$$.\n\nIt can be seen that there are always $$\\overline{A}$$ and $$\\overline{B}$$ to achieve matrix decomposition, but they are only essentially equal to A and B. Therefore, column ambiguity and scale ambiguity are inherent ambiguities in the matrix decomposition process. Without additional constraints, the order and magnitude of loading matrix columns cannot be determined through matrix decomposition. Therefore, the uniqueness of matrix factorization given by the definition can also be called “essential uniqueness.” In the actual application process, some methods can be used to eliminate column blur and proportion blur caused by matrix decomposition.\n\n### 2.3 Recognizability and uniqueness of parallel factorization\n\nThe essential feature of the parallel factorization model is the uniqueness of the model. When there is no array blur, the matrices A, B, and C can be identified. The following conclusions can be obtained. When Xi = BDi(A)CT,i=1,2,...,I, ACI × F,BCJ × F,CCK × F is given, if kA + kB + kC ≥ 2F + 2, then these matrices A, B, and C are uniqueness for column exchange and plurality transformation or scale transformation.\n\nThe matrix composed of relatively independent columns taken from the absolute continuous distribution has full k-rank. If all three matrices meet this condition, the sufficient condition for recognizability is shown in formula (10).\n\n$$\\min \\left(I,F\\right)+\\min \\left(J,F\\right)+\\min \\left(K,F\\right)\\ge 2F+2$$\n(10)\n\nIf the matrices A, B, and C have other structural constraints, better identifiable results may be obtained. PARAFAC uniqueness theorem can be used to obtain the ith submatrix of the X-axis of PARAFAC model:\n\n$${X}_i^{J\\times K}={BD}_i(A){C}^T,i=1,\\dots, I$$\n(11)\n\nThe matrix in formula (11) satisfies the following ARI × R, BRJ × R, CRK × R. If the following conditions are met in Eq. (12).\n\n$${k}_A+{k}_B+{k}_C\\ge 2\\left(R+1\\right)$$\n(12)\n\nEven if there is column blur and scale blur, the matrix A, B, and C are unique. In mathematical language, when formula (13) is satisfied\n\n$${X}_i^{J\\times K}=\\hat{B}{D}_i\\left(\\hat{A}\\right){\\hat{C}}^T,i=1,\\dots, I$$\n(13)\n\nThe relation shown in formula (14) can be obtained.\n\n$$\\hat{A}=A\\prod {\\Delta}_1,\\hat{B}=B\\prod {\\Delta}_2,\\hat{C}=C\\prod {\\Delta}_3$$\n(14)\n\nEquation (14) shows that ∏ is a column fuzzy matrix, Δ1Δ2 and Δ3 are the scale fuzzy matrix and the equation of Δ1Δ2Δ3 = I needs to be satisfied.\n\n### 2.4 Trilinear alternating least square for parallel factor analysis\n\nThere are many methods to achieve the decomposition of PARAFAC, and the trilinear alternate least squares (TALS) algorithm is the most widely adopted methodology for data detection of parallel factor trilinear models. The fundamental principle of the TALS is to update the matrix in each step. First of all, TALS is updated by employing least squares (LS) to renovate the residual matrix based on the results of the previous estimate; then, it continues to update other matrices; finally, stop running until the result converges or reaches the set number of iterations after repeating the above steps. The trilinear model of three-dimensional data set X has the configuration shown in formula (15) below.\n\n$${x}_{i,j,k}=\\sum \\limits_{f=1}^F{a}_{i,f}{b}_{j,f}{c}_{k,f}+{e}_{ijk},i=1\\dots I;j=1\\dots J;k=1\\dots K$$\n(15)\n\nWhere F is the number of factors, ai,f is the i-th element in vector af, bj,f is the j-th element in vector bf, and ck,f is the k-th element in vector cf. The data set X of third-order tensor I × J × K is indicated as “xi,j,k.” The “eijk” represents the error set E of the third-order tensor I × J × K. Equation A = [a1, a2, , aI] is defined as the I × F matrix; the B = [b1, b2, , bJ], and C = [c1, c2, , cK] are defined as the J × F matrix and the K × F matrix.\n\n1. (1)\n\nFirst, matrix A is calculated by formula (16).\n\n$$\\left[\\begin{array}{l}{X}_{\\mathrm{K}1}\\\\ {}{X}_{\\mathrm{K}2}\\\\ {}\\mathrm{M}\\\\ {}{X}_{\\mathrm{K}\\ K}\\end{array}\\right]=\\left[\\begin{array}{l} BdiagC\\left(1,:\\right)\\\\ {} BdiagC\\left(2,:\\right)\\\\ {}\\mathrm{M}\\\\ {} BdiagC\\left(K,:\\right)\\end{array}\\right]{A}^T+{E}_K$$\n(16)\n\nFormula (16) satisfies XK k = Bdiag(C(k,:))AT +EK k. The error is expressed in terms of EK. The least squares (LS) estimate of AT is calculated by Eq. (17).\n\n$${A}^T={\\left[\\begin{array}{c} BdiagC\\left(1,:\\right)\\\\ {} BdiagC\\left(2,:\\right)\\\\ {}M\\\\ {} BdiagC\\left(2,:\\right)\\end{array}\\right]}^{+}\\left[\\begin{array}{c}{X}_{\\mathrm{K}1}\\\\ {}{X}_{\\mathrm{K}2}\\\\ {}\\mathrm{M}\\\\ {}{X}_{\\mathrm{K}K}\\end{array}\\right]$$\n(17)\n\nThe generalized inverse in formula (17) is represented by []+.\n\n1. (2)\n\nSecondly, matrix B is calculated by formula (18).\n\n$$\\left[\\begin{array}{c}{Y}_{\\mathrm{K}1}\\\\ {}{Y}_{\\mathrm{K}2}\\\\ {}\\mathrm{M}\\\\ {}{Y}_{\\mathrm{K}\\ I}\\end{array}\\right]=\\left[\\begin{array}{c} CdiagA\\left(1,:\\right)\\\\ {} CdiagA\\left(2,:\\right)\\\\ {}M\\\\ {} CdiagA\\left(I,:\\right)\\end{array}\\right]{B}^T+{E}_I$$\n(18)\n\nFormula (18) satisfies the following YK i = Cdiag(A(i,:))BT +EK i, The error is expressed in terms ofEI. The least squares (LS) estimate of BT is calculated by Eq. (19).\n\n$${B}^T={\\left[\\begin{array}{c} CdiagA\\left(1,:\\right)\\\\ {} CdiagA\\left(2,:\\right)\\\\ {}\\mathrm{M}\\\\ {} CdiagA\\left(I,:\\right)\\end{array}\\right]}^{+}\\left[\\begin{array}{c}{Y}_{\\mathrm{K}1}\\\\ {}{Y}_{\\mathrm{K}2}\\\\ {}\\mathrm{M}\\\\ {}{Y}_{\\mathrm{K}I}\\end{array}\\right]$$\n(19)\n1. (3)\n\nThirdly, matrix C is calculated by formula (20).\n\n$$\\left[\\begin{array}{l}{Z}_{\\mathrm{K}1}\\\\ {}{Z}_{\\mathrm{K}2}\\\\ {}\\mathrm{M}\\\\ {}{Z}_{\\mathrm{K}J}\\end{array}\\right]=\\left[\\begin{array}{l} AdiagB\\left(1,:\\right)\\\\ {} AdiagB\\left(2,:\\right)\\\\ {}\\mathrm{M}\\\\ {} AdiagB\\left(J,:\\right)\\end{array}\\right]{C}^T+{E}_J$$\n(20)\n\nWhere ZKj = Adiag(B(j,:))CT + EKj, j = 1, 2, , J. The error is expressed in terms of EJ. The least squares estimate of CT is calculated by Eq. (21).\n\n$${C}^T={\\left[\\begin{array}{c} AdiagB\\left(1,:\\right)\\\\ {} AdiagB\\left(2,:\\right)\\\\ {}\\vdots \\\\ {} AdiagB\\left(J,:\\right)\\end{array}\\right]}^{\\div}\\left[\\begin{array}{c}{Z}_{\\dots 1}\\\\ {}{Z}_{\\dots 2}\\\\ {}\\vdots \\\\ {}{Z}_{\\dots J}\\end{array}\\right]$$\n(21)\n1. (4)\n\nFinally, stop running until the result converges or reaches the set number of iterations after repeating the above steps (1)–(3).\n\nMulti-channel vibration signals are collected in this paper to research the fault state of equipment, and a third-order tensor is constructed through continuous wavelet transform. Figure 3 shows the basic structure of the parallel factor analysis decomposition model for fault diagnosis.\n\nThe Nt, Nd, and Nf of the data matrix $${\\mathrm{S}}_{\\left({\\mathrm{N}}_{\\mathrm{d}}\\times {\\mathrm{N}}_{\\mathrm{f}}\\times {\\mathrm{N}}_{\\mathrm{t}}\\right)}$$ are the number of data points, the number of channels, and the frequency step size, respectively.\n\n$${\\hat{S}}_{dft}=\\sum \\limits_{k=1}^{N_k}{a}_{dk}{b}_{fk}{c}_{ik}+{e}_{dft}$$\n(22)\n\nThe key issue of this parallel factorization model is to obtain the matrices A, B, and C. The adk, bfk, and ctk are their elements, in which component k represents an atom. These spatial signals, spectral signals, and temporal signals for each atom are indicated as the ak = {adk}, bk = {bfk} and ck = {ctk}. The “eijk” is the error, which forms error set E of the third order tensor I × J × K. The uniqueness of the solution of parallel factor decomposition for fault diagnosis is guaranteed through rank(A) + rank(B) + rank(C) ≥ 2Nk + 2. The decomposition of formula (22) is achieved by solving $$\\underset{a_{dk}{b}_{jk}{c}_{ik}}{\\min}\\left\\Vert {\\hat{S}}_{dft}-{\\sum}_{k=1}^{N_k}{a}_{dk}{b}_{fk}{c}_{tk}\\right\\Vert$$. The main advantage of this method is that the spectrum decomposition of time-varying vibration signal is unique and the best model can be obtained under the principle of minimum square deviation.\n\nThe basic steps for implementing the multi-scale parallel factorization model of fault diagnosis are as follows .\n\n1. (1).\n\nAfter the vibration signal is collected, the third order tensor is constructed by continuous wavelet transform.\n\n2. (2).\n\nThe number of factor F is determined by the principle of consistency in MATLAB.\n\n3. (3).\n\nInitialization for load matrix B and C.\n\n4. (4).\n\nAfter initializing and running the matrices B and C, the matrix A is estimated by the least square regression algorithm. A = XZ'(ZZ')−1, Z = (bc).\n\n5. (5).\n\nSimilarly, the matrices B and C are estimated.\n\n6. (6).\n\nContinue from step (3) until the result converges or reaches the set number of iterations.\n\n7. (7).\n\nCorresponding results obtained.\n\n### 2.5 Numerical simulation based on parallel factor analysis\n\nSimulation experiments can investigate the characteristic of the results of input signals with different parameters after parallel factor analysis for fault diagnosis. Therefore, the simulation signals are used to simulate the running state of the centrifugal pump to test the method proposed in this article. The simulation signal is shown in the following formula (23).\n\n$$\\mathrm{y}\\left(\\mathrm{t}\\right)=0.01\\ast \\cos \\left(2\\ast \\mathrm{pi}\\ast 400\\ast \\mathrm{t}-5\\right)\\ast \\exp \\left(-0.5\\ast \\left(\\left(\\left(\\mathrm{t}-0.03\\right)/0.003\\right)\\hat{\\mkern6mu} 2\\right)\\right)$$\n(23)\n\nFigure 4 shows the time-domain diagram of the simulated signal. An impact signal appears in the graph at 0.03 μs, which simulates the signal generated when the system fails. It is shown in Fig. 5 that the time-frequency diagram of the simulation signal is obtained by the continuous wavelet transform. It can be seen from the time-frequency diagram of the simulation signal that the dominant frequency of the signal is 400 Hz and the impact signal in the simulation signal appears in the frequency range of 180–400 Hz. For the time-frequency diagram of the simulated signal in Fig. 5, it can also be seen that the impact signal appears at 0.03 μs.\n\nThe simulation signal is constructed into a third-order tensor after continuous wavelet transformation, and then the third-order tensor is decomposed by parallel factors to obtain the result shown in Fig. 6. After parallel factor analysis, we can get the loading value and residual error corresponding to frequency, time, and channel. Comparing the loading value of frequency and time after parallel factor analysis with the time-frequency diagram, we can find their corresponding relationship. The hypothetical frequency curve in the graph fluctuates in the range of 180–400 Hz, which is a contrast relationship between the fluctuations of the simultaneous frequency graph. The time curve fluctuates at 0.03 μs and has the maximum value of the third component and the minimum value of the second component. It can be seen from the simulation signal corresponding to the ground that the simulation signal also has an impact signal at 0.03 μs. This indicates that the parallel factor analysis for high-dimensional data of multi-source feature factors can well detect the characteristics of the shock signal generated by the simulated fault.\n\nThe vibration signals collected in engineering are generally mixed with various noise signals. In order to check on the effectiveness in complex conditions, we add the noise signal to the original simulation signal and perform parallel factor analysis on it. Figures 7 and 8, respectively, show the time-domain diagram of the original simulation signal after adding noise to the signal and the time-frequency diagram obtained through continuous wavelet transform. After adding the noise signal to the original simulation signal, it can be seen that the waveform of the noisy simulation signal is similar to the original simulation signal in Fig. 4 and the impact signal is almost covered by the noise signal. The waveform of the noisy simulation signal in Fig. 8 is steeper and more rapid, and there is a larger blurred signal at 10–20 Hz.\n\nThe simulation signal with noise is transformed into a third-order tensor after continuous wavelet transformation. The result of the parallel factor analysis of the third-order tensor is shown in Fig. 9. After parallel factor analysis, we can get the loading value and residual parameters corresponding to frequency, time, and channel. Comparing the loading value of frequency and time after parallel factor analysis with the time-frequency diagram, we can find the correspondence between them relationship. The hypothetical frequency curve in the graph fluctuates in the range of 180–400 Hz, which is a contrast relationship between the fluctuations of the simultaneous frequency graph. The time curve fluctuates at 0.03 μs and has a maximum value. We get the normal probability plot and the residual variance corresponding to the data in mode 1, mode 2, and mode 3. This shows that the parallel factor analysis proposed can well detect the characteristics of the impact signal in this paper even when the collected signal contains noise.\n\n## 3 Proposed method\n\nLikelihood function is a function of statistical model parameters, which plays a great role in statistical inference. The general method of using likelihood ratio test statistics was proposed by Neyman-Pearson in 1982 . Its basic idea is similar to the maximum likelihood method of parameter estimation theory, which is called likelihood ratio test. For hypothesis H0 : θ = θ0, alternative hypothesis H1 : θ = θ1, x is a set of random variables. When H0 is true, the probability density function of the random variable x is expressed as f(x, θ0). When H1 is true, the probability density function of the random variable x is expressed as f(x, θ1). The likelihood function of the sample is the following formula (24).\n\n$$L\\left(\\theta \\right)=\\prod \\limits_{i=1}^nf\\left({x}_i,\\theta \\right)$$\n(24)\n\nTherefore, the likelihood ratio test is performed to obtain the statistic L in the following formula (25).\n\n$$L=\\frac{L\\left({\\theta}_1\\right)}{L\\left({\\theta}_0\\right)}=\\frac{\\prod \\limits_{i=1}^nf\\left({x}_i,{\\theta}_1\\right)}{\\prod \\limits_{i=1}^nf\\left({x}_i,{\\theta}_0\\right)}$$\n(25)\n\nIf the likelihood ratio L is larger, the parameter θ is more likely to be θ1; it shows that the result may tend to negate H0. On the contrary, if the ratio is smaller, the parameter θ is more likely to be θ0, which indicates that the result may be inclined to accept H0. For a certain limit k, L is defined as shown in the following formula (26).\n\n$$\\varphi (x)=\\left\\{\\begin{array}{cc}1& l>k\\\\ {}0& l\\le k\\end{array}\\right.$$\n(26)\n\nTest φ(x) is called the likelihood ratio test of the above test problem.\n\nNeyman-Pearson proposes a principle to determine the optimal test method: parameter α satisfies formula (27).\n\n$$\\beta \\left(\\theta \\right)\\le \\alpha \\kern0.5em \\forall \\theta \\in {\\Theta}_0$$\n(27)\n\nIn formula (27), β(θ) is the power function of the test, Θ0 is the parameter space of the null hypothesis H0, and θ is the test parameter. Look for a test that satisfies the above formula so that β(θ) is as large as possible when θ Θ0.To ensure that the probability of making two types of errors is very small, the sample size must be increased. For field testing, the smaller the sample size, the better when ensuring the reliability of the conclusion. The sequential method proposed by A. Wald solves the problem of optimal selection of sample size and play an important milestone in the history of statistical development.\n\nThe probability function f(x, θ) represents the distribution of the random variable x, H0(θ = θ0) and H1(θ = θ1) are the null hypothesis and alternative hypothesis of the random variable x, respectively. When accepting H1, the probability of the sample x1, …, xm for any positive integer m is given by P1m = f(x1, θ1), …, f(xm, θ1), and the probability is given by P0m = f(x1, θ0), …, f(xm, θ0) when accepting H0.The definition of the sequential probability ratio test is as follows: select two normal numbers A and B (B < A) and calculate the probability ratio P1m/P0m at each stage of the test.\n\n1. (a)\n\nIf p1m/p0m ≥ A, the sequential probability ratio test ends, H1is accepted and H0 is discarded.\n\n2. (b)\n\nIf p1m/p0m ≤ B, the sequential probability ratio test ends, H0 is accepted and H1 is discarded.\n\n3. (c)\n\nIf B < p1m/p0m < A, we continue to observe the sequential probability ratio test until the requirement is met.\n\nWhen SPRT is applied to target recognition, it is first assumed that one of the M alternative hypotheses is the initial hypothesis. The signal propagation waveform is denoted as s(t). When a signal is transmitted, one of the possible waveforms is received and recorded as follows:\n\n$$y(t)=s(t)\\ast {h}_i(t)+n(t)\\kern0.5em i\\in \\left\\{1,2,\\mathrm{K},M\\right\\}$$\n(28)\n\nWhere n(t) is additive white Gaussian noise; the impulse response of the target hypothetical channel is expressed as hi(t) and “*” is the convolution factor.\n\nThe signal channel receiving data is defined in formula (29), where Qi represents the target convolution matrix defined in the literature.\n\n$$y={Q}_is+n$$\n(29)\n\nThe M target hypotheses are denoted as H1, H2, …, HM, respectively. The parameter αi,j is the probability (i ≠ j) when the true hypothesis Hi is wrongly selected as Hi. After obtaining k-th observations, suppose the likelihood ratio of i and j can be defined as shown in formula (30)\n\n$${\\Lambda}_{i,j}^k=\\frac{p_{i1}\\left({\\mathrm{y}}_1\\right){p}_{i2}\\left({\\mathrm{y}}_2\\right)\\Lambda \\kern0.5em {p}_{ik}\\left({\\mathrm{y}}_k\\right)\\kern0.5em {P}_i}{p_{j1}\\left({\\mathrm{y}}_1\\right){p}_{j2}\\left({\\mathrm{y}}_2\\right)\\Lambda \\kern0.5em {p}_{j2}\\left({\\mathrm{y}}_k\\right)\\kern0.5em {P}_j}$$\n(30)\n\nWhere pik(yk) is the probability density function (PDF) with k-th data under the i-th hypothesis and yk is the k-th observation data. When the likelihood ratio satisfies formula (31), accept the assumption Hm.\n\n$${\\Delta}_{i,j}^k>\\frac{1-{\\theta}_{i,j}}{\\theta_{i,j}}\\kern0.5em j\\ne i$$\n(31)\n\nWhen the likelihood ratio satisfies the formula (31), stop the loop. If the likelihood ratio does not meet the stopping condition, continue to the next iteration. In fact, the probability density function of the observed data is constant and satisfies pi1(y) = pi2(y) = … = pik(y). The intensity waveform is updated with the number of iterations, so the probability density function of the observation data under the condition of additive white Gaussian noise can be defined as formula (32).\n\n$${p}_{ik}\\left({y}_k\\right)=\\frac{1}{{\\left(\\sqrt{2{\\pi \\sigma}_n^2}\\right)}^{L_y}}\\times \\exp \\left[-\\frac{1}{2{\\sigma}_n^2}{\\left({y}_k-{Q}_i{s}_k\\right)}^T\\kern0.5em \\left({y}_k-{Q}_i{s}_k\\right)\\right]$$\n(32)\n\n## 4 Experiments\n\n### 4.1 Slurry pump fault test system and experimental design\n\nThe experimental system to be established in this project is required to operate the slurry pump under controlled conditions of speed, flow rate, slurry density, and inlet pressure, and to use and replace the impeller of the slurry pump of different grades and wear. Common failure parts of centrifugal pumps include rotor impeller, rolling bearing, seal, coupling, etc., of which impeller and rolling bearing failure account for a large proportion. The schematic diagram of the slurry pump fault diagnosis test system is shown in Fig. 10. The figure shows the three-dimensional schematic diagram of the test circuit and identifies the key components. It mainly includes motors, centrifugal pumps, data acquisition systems, control instruments, glycol cooling tanks, pressure gauges, flow meters, conveyor belts, sand tanks, pipelines, pressure control tanks, density meters, and sampling ports. First, the normal impeller is used in the centrifugal pump to run the slurry pump fault test system for collecting and testing the signal data of the slurry pump vibration, flow, slurry density, motor speed, and pump inlet and outlet pressure. Then impeller perforation, impeller edge damage and blade damage, and its impellers with different degrees of damage were selected to replace the original centrifugal pump impeller. After running the slurry pump fault diagnosis and test system, the data of the vibration, speed, and pump speed of the slurry pump experiment system were collected.\n\nFigure 11 shows the process flow chart of the slurry pump fault diagnosis test system. The arrow direction in the figure is the flow direction of the mud when the mud pump fault diagnosis experiment system is running. It is the basis for establishing and running the centrifugal pump fault diagnosis experiment system in this article. The serial number and related schematic diagram in Figure 11 indicate the following meanings: 1—centrifugal pump, 2—motor, 3—inverter, 4—power meter sensor, 5—accelerometer, 6—pressure sensor, 7—flow meter, 8—hole plate, 9—heat exchanger, 10—cooler, 11—temperature sensor, 12—sand, 13—suction pressure control tank, and 14—suction pressure sensor. The fault diagnosis test system for slurry pump contains a Weir/Warman 3/2 CAH slurry pump (40 HP) with impeller C2147(8.4\"). The process flow chart of fault diagnosis test system for slurry pump covers the key issues mentioned in this article, but does not cover all aspects of the design of the experimental system. The key issues include that the medium of the cooler in the pipeline is ethylene glycol, the process water is municipal water, and the heat exchanger medium is steam. Microphone means for sound collector.\n\nIn order for the experiment to run successfully, the designer first needs to design the system after engineering calculation and determine the components. The main equipment required for the experiment includes the centrifugal pump, data acquisition system for vibration data acquisition, sensors, and a laptop computer. Auxiliary equipment including storage tanks, valves, instruments, and drive motors are used to control various functions. The data collected by vibration accelerometers is used to analyze the centrifugal pump system in this experiment. The detailed explanation of the vibration sensor for the system signal acquisition is shown below. In the experiment, three three-axis vibration accelerometers are used. Two of the PCB three-axis ICP (Integrated Circuit Piezoelectric) accelerometers have the sensitivity of 100 mV/g and the frequency range of 2–5 kHz. Another PCB three-axis ICP accelerometer has the range of 0.5–3 kHz and the sensitivity of 1000 mV/g.\n\n### 4.2 Slurry pump experimental equipment and signal acquisition system\n\nTo research the validity of the multi-scale parallel factor analysis and sequential probability ratio test proposed in this paper for multi-channel signal in actual complex industrial production, the centrifugal pump fault diagnosis experimental system was designed. The general Fig. 12a shows the centrifugal pump fault diagnosis experimental system. The data acquisition system is shown in Fig. 12b based on a combination of PC measurement hardware and software, which can input electrical signals from sensors and other instruments into a computer for processing. NI LabView 7.0 was chosen as the measurement standard application software because it is easy to build a graphical measurement interface with the help of a large number of tools and objects. The selected hardware is provided by NI DAQ and is highly compatible with our software applications. In order to collect the vibration signals of the centrifugal pump in three directions for each state, it is necessary to install a short-range but high-sensitivity sensor at the key position. Figure 12d is a schematic diagram of the position of the accelerometer. The standard accelerometer and the high-sensitivity accelerometer are installed on the pump casing near the pump suction port, where they will be close to the parts that are prone to failure. Another standard accelerometer is mounted on the shaft bearing because this location is sensitive to vibrations transmitted from the stuffing box. Real-time signals such as flow rate, pressure, speed, and vibration, can be collected synchronously by the experimental data acquisition system. By commanding the pressure and the flow of the equipment’s loop, we simulate the non-linear operating state of the industrial process of the mechanical system to establish the non-linear multi-fault mode, synchronously collect multi-channel signals, and obtain multi-source signals. The internal interaction mechanism between fluid excitation and vibration response under non-linear operation mechanism can be analyzed.\n\nThe liquid transported in this experiment is set as mud to better collect vibration signals. In this experiment, normal impeller, and three types of faulty impellers, including impeller perforation, impeller edge damage and blade damage, were set to simulate failures in industrial production. Among them, these three failure modes have clear differences and the typical failures of centrifugal pump impellers can be represented well. The impeller in the normal state is denoted as S1, and the three types of impellers with impeller perforation, impeller edge damage, and blade damage are denoted as S2, S3, and S4. In order to avoid aliasing, the sampling frequency in this experiment is 9009 Hz according to the Nyquist sampling theorem and the data acquisition time is 20 s for each group.\n\nIn the experiment, different impellers were replaced to collect the vibration signals of the centrifugal pump under different operation conditions. The steps of the whole experiment are summarized as follows:\n\n1. (1)\n\nEstablish the experimental device according to the schematic diagram of slurry pump test system shown in Fig. 11. The normal impeller shown in Fig. 13 is used as the impeller of the pump and sediment is added as the pumping medium. After starting the motor, we adjust the motor speed to 1200 rpm, 1600 rpm, 1800 rpm, 2200 rpm, and 2600 rpm through the known voltage, motor power, motor efficiency, and other coefficients and instructions. According to the sampling time of 20 s and sampling frequency of 9009 Hz shown in the previous article, NI LabView 7.0 application software and NI DAQ signal acquisition system were run to collect the three sets of three-dimensional vibration signals of the corresponding pump.\n\n2. (2)\n\nThe normal impeller in the original centrifugal pump is replaced by the impeller perforation of the fault S2 in Fig. 13, and the other parts remain unchanged. Follow the previous steps to start the centrifugal pump and collect data. When a set of data is collected, the speed is set to 1400 rpm, 1600 rpm, and 2600 rpm and the above steps are repeated to collect data.\n\n3. (3)\n\nThe S3 of impeller edge damage in Fig. 13c is selected to replace the impeller of S2 in the original centrifugal pump and other parts remain unchanged. Similarly, follow the previous steps to start the centrifugal pump and collect data.\n\n4. (4)\n\nThe S4 of blade damage in Fig. 13d is selected to replace the impeller of S3 in the original centrifugal pump and other parts remain unchanged. Similarly, follow the previous steps to start the centrifugal pump and collect data.\n\n5. (5)\n\nAfter the experiment, the outlet valve of the pump was closed. Close the inlet valve after turning off the motor. Store experimental data to prepare for subsequent vibration signal analysis.\n\n## 5 Results and discussion\n\n### 5.1 Multi-source dynamic feature extraction based on parallel factorization\n\nThis multi-scale parallel factorization method for the extraction of characteristic signals in non-linear multi-source and multi-fault modes is proposed in the article. Parallel factorization can not only perform high-dimensional data processing, but also has the uniqueness of the decomposition. This property makes the results of parallel factorization more realistic and has specific physical meanings. The third-order tensor constructed by multi-channel vibration signals through continuous wavelet transform is decomposed into the series of different modes of channel/frequency/time by the multi-scale parallel factor analysis algorithm. The spatial information is introduced into the time-frequency analysis of signals to form the three-dimensional spatial/time/frequency characteristic analysis of each factor. The simulation results show that the parallel factor decomposition for the tensor built by multi-channel signal has the compatibility of decomposition path and overall consistency. As a result, the topographic map, spectrum, and time contour of the multi-source fault signal in the centrifugal pump experiment are acquired. The multi-scale parallel decomposition method for extracting multi-source feature signals of non-linear failure modes is applied in the fault diagnosis of centrifugal pumps. It analyzes the internal connection between the optimal decomposition paths of multi-source signal feature factors. The optimal non-linear correspondence relationship between failure modes and characteristic signals in time, frequency, and space are constructed. Based on the correspondence and overall consistency of multi-source feature factor decomposition paths, we remodeled three-dimensional fault feature models such as the frequency spectrum and time profile of the fault feature factors, successfully extracted non-linear multi-dimensional dynamic fault feature signals. Finally, the corresponding fluctuation regularities of the homologous non-linear failure mode in the multi-source signals were displayed.\n\nFigure 14 shows the time-frequency diagram obtained of the vibration signals collected in the X-axis direction of the three vibration signal collection points by continuous wavelet transformation when the slurry pump is in normal operation. Figure 15 shows the result of the parallel factor analysis of vibration signal of slurry pump after continuous wavelet transform in normal state. In this experiment, three-dimensional vibration sensors are set up at three measuring points. We analyze the vibration signals of these three measuring points to explore the three-dimensional spatial distribution and characteristic propagation path of dynamic characteristics on the mechanical structure of the slurry pump. Three groups of original vibration signals are transformed by continuous wavelet to obtain three-dimensional time-frequency signals to construct a third-order matrix. After multi-scale parallel factor analysis for the third-order tensor, the loading values and residual variance of the aisle, time, and frequency factors are obtained.\n\nFigure 16 shows the time-frequency diagram of the vibration signals collected in the X-axis direction of the three vibration signal collection points by continuous wavelet transformation when the slurry pump is in S2 impeller perforation. In the S2 state, the third-order tensor of 3 × 126 × 4096 is constructed by continuous wavelet transform. Figure 17 indicates the result of the loading values and residual variance of the aisle, time, and frequency modes by the parallel factor analysis for the third-order tensor of slurry pump in state S2.\n\nFigure 18 shows the time-frequency diagram of the vibration signals collected in the X-axis direction of the three vibration signal collection points by continuous wavelet transformation when the slurry pump is in S3 impeller edge damage. In the S3 state, the third-order tensor of 3 × 126 × 4096 is constructed by continuous wavelet transform. Figure 19 indicates the result of the loading values and residual variance of the aisle, time, and frequency modes by the parallel factor analysis for the third-order tensor of slurry pump in state S3.\n\nThe operating state of the slurry pump system in our experimental device system has normal operation and three failure states. The failure states include impeller holes, leading edge damage, and propeller blade damage. Similarly, Fig. 20 shows the time-frequency diagram when the slurry pump is in S4 propeller blade damage. Figure 21 indicates the result of the loading values and residual variance of the aisle, time and frequency modes by the parallel factor analysis for state S4.\n\nWe analyze the vibration signals of these three measuring points to discuss the three-dimensional spatial distribution and characteristic propagation path of dynamic characteristics on the mechanical structure of the slurry pump. By comparing the decomposition results of parallel factor analysis in the normal and fault state, there are obvious difference in the time loading factor and frequency loading factor component. Due to the phenomenon of characteristic coupling and aliasing of mechanical multi-source signals, the parallel factor analysis can optimize the independent characteristics of each channel on the surface of the mechanical structure and eliminate the mutual interference, overlap, and redundancy of the characteristic signals between the channels. Therefore, the parallel factor analysis are effective in providing a basis for subsequent diagnosis of SPRT and the fault identification can be successfully implemented.\n\n### 5.2 SPRT for the multi-source condition monitoring of centrifugal pump\n\nThe proportions of the standard deviation “σ” and mean “μ” of the test signal sequences have significant influence on the likelihood ratio in the sequential probabilistic ratio test. Therefore, the mean value and standard deviation of the frequency loading value after the parallel factor decomposition should be calculated first for the test signal sequence. Assuming the probability distribution of the frequency load value sequence of one set of signals under the multi-source condition monitoring of the centrifugal pump meets the null hypothesis Hi : μ = μi, and the probability distribution of the frequency load value sequence of the other set of signals satisfies the alternative hypothesis Hj : μ = μj . Their corresponding standard deviation σ remains unchanged. When the original hypothesis and the alternative hypothesis are both true, the joint probability density functions of these two sets of sequences are shown below.\n\n$${p}_{ik}\\left({y}_k\\right)=\\frac{1}{\\sigma \\sqrt{2\\pi }}\\exp \\left(-\\frac{1}{2{\\sigma}^2}{\\left({y}_k-{\\mu}_i\\right)}^2\\right)$$\n(33)\n$${p}_{jk}\\left({y}_k\\right)=\\frac{1}{\\sigma \\sqrt{2\\pi }}\\exp \\left(-\\frac{1}{2{\\sigma}^2}{\\left({y}_k-{\\mu}_i\\right)}^2\\right)$$\n(34)\n\nIn formula (33), Pik(yk) is the probability density function null hypothesis. Pjk(yk) in formula (34) is the probability density function under the alternative hypothesis. The SPRT probability ratio is calculated in formula (35).\n\n$${\\lambda}_{i,j}\\left({Y}_{Sm}\\right)=\\frac{\\prod \\limits_{k=1}^n{P}_{jk}}{\\prod \\limits_{k=1}^n{P}_{ik}}=\\frac{P_{j1}\\left({y}_1\\right){P}_{j2}\\left({y}_2\\right)\\Lambda \\kern0.5em {P}_{jk}\\left({y}_k\\right)}{P_{i1}\\left({y}_1\\right){P}_{i2}\\left({y}_2\\right)\\Lambda \\kern0.5em {P}_{ik}\\left({y}_k\\right)}\\times \\frac{P_{j0}}{P_{i0}}$$\n(35)\n\nIn order to make the calculation easier in practical applications, the likelihood ratio formula is further derived and simplified to obtain the formula (36). Where YSi and YSj are the to-be-checked sequences of vibration signals Si and Sj, respectively, Δi, j(YSi) and Δi, j(YSj) are the likelihood ratios of the sequence to be tested YSi and YSj, respectively.\n\n$${\\Delta}_{i,j}\\left({Y}_{Sm}\\right)=1\\mathrm{n}{\\lambda}_{i,j}\\left({Y}_{Sm}\\right)=1\\mathrm{n}\\frac{\\prod \\limits_{k=1}^n{P}_{jk}}{\\prod \\limits_{k=1}^n{P}_{ik}}=\\sum \\limits_{k=1}^n1\\mathrm{n}\\frac{P_{jk}}{P_{ik}}\\kern0.5em m=i,j$$\n(36)\n\nReferring to the sequential probability ratio test algorithm, we compare the likelihood ratio with the thresholds A and B to identify different forms of failure of the centrifugal pump. The size of A and B are closely related to the probability α of type I error and the probability β of type II error. The variables α, β, A, and B are satisfied with the following relationship:\n\n$$a=\\ln A=\\ln \\frac{1-\\beta }{\\alpha }$$\n(37)\n$$b=\\ln B=\\ln \\frac{\\beta }{1-\\alpha }$$\n(38)\n\nFor S1, S2, S3, and S4 four different impeller fault states of the centrifugal pump experimental system, Figure 22 shows the process of using the sequential probability ratio test algorithm to identify the fault. Vibration signals at three different positions collected under two different impeller fault conditions (sj) and (sj) are decomposed by parallel factors to obtain frequency loading values, which are calculated according to formulas (33)–(36) to obtain the likelihood ratio Δi, j of the sequential probability ratio test. The process of centrifugal pump fault identification is shown below. (1) If Δi, j = ln (λi, j) (−∞, b], accept Hj, the centrifugal pump system is under the condition (sj). (2) If Δi, j = ln (λi, j) [a, ∞), accept Hi, the centrifugal pump is under the condition (si). (3) If Δi, j [a, b], the likelihood ratio of sequential probabilistic ratio test continues to be calculated by extracting the next data in the test sequence according to formulas (33)–(36). The likelihood ratio will continue to be compared with the threshold value until the condition (1) or (2) is met or the number of iterations is reached. After the test is stopped and the probability parameters λ1, 2(YS1), λ1, 2(YS2), λ1, 3(YS1), λ1, 3(YS3), λ1, 4(YS1) andλ1, 4(YS4) are obtained, the conditions of centrifugal pump will be distinguished.\n\nThe means of the signal S1, S2, S3 and S4 under the four conditions parameters areμ1, μ2, μ3, μ4. Then, the likelihood ratio is calculated and analyzed according to formulas (33)~(36). SPRT probability ratios λi, j(YSi) and λi, j(YSj) are calculated by importing the testing data (YSi, YSj) of the signal waveform for slurry pump Si and Sj conditions to Eq. (35). Compare the likelihood ratios Δi, j(YSi) and Δi, j(YSj) with the threshold to determine the state Si and Sj of the centrifugal pump.\n\nThe difference in the variables λi, j(YSi) and λi, j(YSj) is used to distinguish different condition (Si,Sj) of the centrifugal pump. When the iteration periods are determined, the condition S1 is compared with S2, S3, S4 in the relation between SPRT probability ratios. Figure 23a shows the fluctuation curve between the likelihood ratios Δ1, 2(YS1) and Δ1, 2(YS2) of the signals S1 and S2 during the determined number of iteration. It can be seen from the curve in Fig. 23a that Δ1, 2(YS1) < b displays that the centrifugal pump is in the normal state of S1; Δ1, 2(YS2) > a indicates that the centrifugal pump is in a fault state of the S2 impeller perforation. From the curve in Fig. 23b, it can be seen that Δ1, 3(YS1) < b indexes that the centrifugal pump is in the normal state of S1; Δ1, 3(YS3) > a indicates that the centrifugal pump is in the fault state of S3 impeller edge damage. In Fig. 23c, the inequality (Δ1, 4(YS1)) < b is satisfied, the condition is S1. The inequality(Δ1, 4(YS4)) > a is satisfied, the condition of the centrifugal pump is S4 impeller blade damage. The SPRT parameters Δ1, m, m = 2, 3, 4 in Fig. 22a–c are adopted to distinguish that the normal condition (S1) of the centrifugal pump from the conditions (S2,S3,S4).\n\nWe found that the different fault states of the centrifugal pump in the experiment can also be distinguished by this method of SPRT. Figure 24a indicates the SPRT parameters λ2, 3 in Eqs. (33–36) of the testing sequences YS2 and YS3. The SPRT inequality (Δ2, 3(YS2)) < b is satisfied, then the condition of the centrifugal pump is S2. The inequality (Δ2, 3(YS3)) > a is satisfied, then the condition of the slurry pump is S3 (impeller edge damage). It can be seen from the curve in Fig. 24b that when (Δ2, 4(YS2)) < b is satisfied, the centrifugal pump is fault S2 (impeller perforation); when (Δ2, 4(YS4)) > a is satisfied, the centrifugal pump is fault S4 (impeller blade damage). When the inequality (Δ3, 4(YS3)) < b is satisfied in Fig. 24c, the condition is S3 (impeller edge damage). When (Δ3, 4(YS4)) > a is satisfied, the condition of the centrifugal pump is S4 (impeller blade damage). The parameters (λi, j(YSi), λi, j(YSj)) are effective indicator to monitor the different conditions of the multi-fault and multi-source centrifugal pump.\n\n## 6 Conclusion\n\nParallel factorization can not only perform high-dimensional data processing, but also has the uniqueness of the decomposition. This property makes the results of parallel factorization more realistic and has specific physical meanings. Through numerical simulation, the parallel factorization is used to explore the non-linear correspondence relationship of multi-source signal characteristic factors in time, frequency, and space under different simulation states. By adjusting the different frequency, time, phase, and amplitude of the analog signal, the loading values of the three modes are captured after parallel factorization to research the corresponding relationship between the analog signals. Then, the parallel factor analysis is applied to the centrifugal pump fault diagnosis experimental system to analyze the state characteristics under multiple fault modes. The non-linear multi-dimensional actional fault characteristic parameter of the impellers with different faults of the centrifugal pump was triumphantly acquired, and the corresponding fluctuation regularities of the homologous fault mode characteristics in the multi-source signals were displayed.\n\nThe analysis of the comprehensive result graph shows that the centrifugal pump fault diagnosis methodology based on parallel factor analysis of multiple scales and sequential probability ratio test is effective and reliable. This method first analyzes the collected vibration signals through parallel factor analysis and then conducts sequential probability ratio testing. It identifies different failure modes by comparing likelihood ratios and thresholds. Not only normal conditions and fault conditions can be identified, but also different fault conditions can be distinguished. Therefore, the methodology proposed in these contents of article is very suitable for non-linear multi-source and multi-fault signal analysis and processing. The PARAFAC theory proposed in this paper can also be used in the blind separation of mechanical multiple faults.\n\n## Availability of data and materials\n\nAll data generated or analyzed during this study are included in this published article.\n\n## Abbreviations\n\nPARAFAC:\n\nParallel factor analysis\n\nSPRT:\n\nSequential probability ratio test\n\nCNN:\n\nConvolutional neural network\n\nEMD:\n\nEmpirical mode decomposition\n\nVMD:\n\nVariational mode decomposition\n\nEEG:\n\nElectroencephalogram\n\nANN-MLP:\n\nArtificial neural network-multi-layer perceptron\n\nTALS:\n\nTrilinear alternate least squares\n\nLS:\n\nLeast squares\n\nPDF:\n\nProbability density function\n\n## References\n\n1. R. Azizi, B. Attaran, A. Hajnayeb, A. Ghanbarzadeh, M. 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Neurol. 8 (2017). https://doi.org/10.3389/fneur.2017.00155.\n\n29. L. Yang, H. Chen, Y. Ke, L. Huang, Q. Wang, Y. Miao, L. Zeng, A novel time-frequency-space method with parallel factor theory for big data analysis in condition monitoring of complex system. Int. J. Adv. Robotic Syst. 17(2) (2020). https://doi.org/10.1177/1729881420916948.\n\n30. Y. Cheng, J. Minping, A novel weak fault signal detection approach for a rolling bearing using variational mode decomposition and phase space parallel factor analysis. Meas. Sci. Technol. 30(11) (2019). https://doi.org/10.1088/1361-6501/ab30bd.\n\n31. L. Lijia, B. Shiyi, M. Jianfeng, et al., Monitoring batch processes using sparse parallel factor decomposition. Ind. Eng. Chem. Res 56(44), 12682–12692 (2017)\n\n32. G. Haiyang, L. Kaiqi, H. Xingyi, et al., Feasibility study for the analysis of coconut water using fluorescence spectroscopy coupled with PARAFAC and SVM methods. Br. Food J (2020). https://doi.org/10.1108/BFJ-12-2019-0941\n\n33. H. Xiu, Y. Huibin, S. Yonghui, et al., Characterizing humic substances from a large-scale lake with irrigation return flows using 3DEEM-PARAFAC with CART and 2D-COS. J. Soils Sediments 20(9), 3514–3523 (2020)\n\n34. C. Yahya, A sequential probability ratio test (SPRT) to detect changes and process safety monitoring. Process Saf. Environ. Protect. 92(3), 206–214 (2014)\n\n35. G. Peng, I. David, Wind turbine power curve modeling and monitoring with Gaussian process and SPRT. IEEE Transact. Sustainable Energy 11(1), 107–115 (2020)\n\n36. W. Rong, X. Zhi, L. Jianye, et al., Chi-square and SPRT combined fault detection for multisensor navigation. IEEE Transact. Aerospace Electron. Syst. 52(3), 1352–1365 (2016)\n\n37. H. Kasai, Fast online low-rank tensor subspace tracking by CP decomposition using recursive least squares from incomplete observations. Neurocomputing 347, 177–190 (2019)\n\n38. W. Huang, J. Huang, L. Yang, et al., Fault diagnosis of gearboxbased on principal component analysis and sequential probability ratio test [C]. Proceedings of 2nd International Conference on Computer Science and Application Engineering, Hohhot, China. (2018). https://doi.org/10.1145/3207677.3277945\n\n39. P. Juan, Z. Santiago, Using one class SVM to counter intelligent attacks against an SPRT defense mechanism. Ad Hoc Networks. 94 (2019). https://doi.org/10.1016/j.adhoc.2019.101946.\n\n## Acknowledgements\n\nThe Reliability Research Lab in the Department of Mechanical Engineering at the University of Alberta in Canada provided the original experimental data. The National Natural Science Foundation of China (Grant No. 51775390, 51901164, 51805378), the Natural Science Foundation of Hubei Province (Grant No. 2018CFB394), and the Foundation of Wuhan Science and Technology Bureau (Grant No. 2019010701011417) provides the financial support for paper research.\n\n## Funding\n\nThis research was funded by the National Natural Science Foundation of China (Grant No. 51775390, 51901164, 51805378), the Natural Science Foundation of Hubei Province (Grant No. 2018CFB394), and the Foundation of Wuhan Science and Technology Bureau (Grant No. 2019010701011417).\n\n## Author information\n\nAuthors\n\n### Contributions\n\nThe algorithms proposed in this paper have been conceived by all authors. Hanxin Chen designed and performed the experiments and then analyzed the results. Liu Yang performed the experiments and analyzed the simulation results. Liu Yang wrote the paper. All authors read and approved the final manuscript.\n\n### Corresponding author\n\nCorrespondence to Hanxin Chen.\n\n## Ethics declarations\n\nNot applicable.\n\nNot applicable.\n\n### Competing interests\n\nThe authors declare that they have no competing interests.", null, "" ]	[ null, "https://asp-eurasipjournals.springeropen.com/track/article/10.1186/s13634-021-00730-w", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8834414,"math_prob":0.9852227,"size":69388,"snap":"2023-14-2023-23","text_gpt3_token_len":15330,"char_repetition_ratio":0.1709616,"word_repetition_ratio":0.10274813,"special_character_ratio":0.22240157,"punctuation_ratio":0.13172296,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9961464,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-04-02T00:06:56Z\",\"WARC-Record-ID\":\"<urn:uuid:58c4fd67-bb48-46e3-a6e5-10aa072458f8>\",\"Content-Length\":\"418180\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7c40f4bc-348b-48d5-8729-3daa916b3353>\",\"WARC-Concurrent-To\":\"<urn:uuid:7456f99e-6ba9-4b5a-95e1-1818bcac959d>\",\"WARC-IP-Address\":\"146.75.32.95\",\"WARC-Target-URI\":\"https://asp-eurasipjournals.springeropen.com/articles/10.1186/s13634-021-00730-w\",\"WARC-Payload-Digest\":\"sha1:ETLVNH3N2AJ3VZCU3TGMLXKMP6RY5TFC\",\"WARC-Block-Digest\":\"sha1:UO6OMJ444OVILM7ER72DD2V4KJMMKFER\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296950363.89_warc_CC-MAIN-20230401221921-20230402011921-00711.warc.gz\"}"}
https://convertoctopus.com/39-feet-per-second-to-knots	[ "## Conversion formula\n\nThe conversion factor from feet per second to knots is 0.59248380129641, which means that 1 foot per second is equal to 0.59248380129641 knots:\n\n1 ft/s = 0.59248380129641 kt\n\nTo convert 39 feet per second into knots we have to multiply 39 by the conversion factor in order to get the velocity amount from feet per second to knots. We can also form a simple proportion to calculate the result:\n\n1 ft/s → 0.59248380129641 kt\n\n39 ft/s → V(kt)\n\nSolve the above proportion to obtain the velocity V in knots:\n\nV(kt) = 39 ft/s × 0.59248380129641 kt\n\nV(kt) = 23.10686825056 kt\n\nThe final result is:\n\n39 ft/s → 23.10686825056 kt\n\nWe conclude that 39 feet per second is equivalent to 23.10686825056 knots:\n\n39 feet per second = 23.10686825056 knots\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 knot is equal to 0.04327717582307 × 39 feet per second.\n\nAnother way is saying that 39 feet per second is equal to 1 ÷ 0.04327717582307 knots.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that thirty-nine feet per second is approximately twenty-three point one zero seven knots:\n\n39 ft/s ≅ 23.107 kt\n\nAn alternative is also that one knot is approximately zero point zero four three times thirty-nine feet per second.\n\n## Conversion table\n\n### feet per second to knots chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from feet per second to knots\n\nfeet per second (ft/s) knots (kt)\n40 feet per second 23.699 knots\n41 feet per second 24.292 knots\n42 feet per second 24.884 knots\n43 feet per second 25.477 knots\n44 feet per second 26.069 knots\n45 feet per second 26.662 knots\n46 feet per second 27.254 knots\n47 feet per second 27.847 knots\n48 feet per second 28.439 knots\n49 feet per second 29.032 knots" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.74671435,"math_prob":0.9918071,"size":1900,"snap":"2020-45-2020-50","text_gpt3_token_len":513,"char_repetition_ratio":0.24261603,"word_repetition_ratio":0.0371517,"special_character_ratio":0.3531579,"punctuation_ratio":0.093023255,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99325943,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-27T08:55:23Z\",\"WARC-Record-ID\":\"<urn:uuid:b2a066a5-69c5-4af6-ac81-2897c4b8f938>\",\"Content-Length\":\"29229\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:38e70d37-ac15-4f4d-bc47-29a9213c3b6d>\",\"WARC-Concurrent-To\":\"<urn:uuid:a1dad897-9675-4d93-9306-61d906256c80>\",\"WARC-IP-Address\":\"104.27.142.66\",\"WARC-Target-URI\":\"https://convertoctopus.com/39-feet-per-second-to-knots\",\"WARC-Payload-Digest\":\"sha1:OG7X6ZGKRKKINFQPNNFO3E4OHVUMPAYE\",\"WARC-Block-Digest\":\"sha1:ZKBYMXOPKFOZAPQ3QRLBLKXN5KFALXS6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141191511.46_warc_CC-MAIN-20201127073750-20201127103750-00234.warc.gz\"}"}
https://electronics.stackexchange.com/questions/415955/how-does-this-bjt-affect-the-circuit-if-it-does-how-should-i-simulate-on-circu?noredirect=1	[ "# How does this BJT affect the circuit? If it does, how should I simulate on Circuitlab to see the effect?\n\nI am trying to understand how NPN Q23 in the picture below affects the operation of the left circuit.", null, "If it doesn't then what we get is the circuit on the right with the only transistor now being Q30.\n\nAfter reading this thread I would say Q23 will be in reverse active mode of operation operating with lower β and therefore it does affect the operation of the circuit on the left.\n\nMy questions are:\n\n1) What is it that actually happens by placing Q23 on the circuit? Is there more current passing through R1 because the resistance of Q23//Q22 is smaller?\n\n2) If I am to simulate the circuit (on Circuitlab or Multisim) what simulation setting (DC sweep, time domain) and input parameters would you suggest so that I can see a clear difference between the left and the right circuit?\n\nEDIT:\n\nThank you for the comments. This is a made up question, I am interested in the effect Q23 will have on the ADC voltage across R1 given that V2 is constant DC source. I simply want to establish that Q23 and Q22 will have the same constant base current and see what happens from there..\n\nEDIT#2\n\nThe question could very well be about the effect of Q23 being the only BJT in the left circuit. In that case what simulation would you run to compare the circuits on the left and right?\n\n• What do voltage sources V1 and V2 represent? Are they AC, DC, pulses, sensor outputs, what? – AnalogKid Jan 8 at 22:57\n• This might be a good question if you'd provide more context about it. (If it is just made up, then that's another thing.) Because of the heavier doping of the emitter, there is an avalanche behavior in $Q_{23}$ that might be sought. $Q_{23}$'s crappy $\\beta$ means $R_2$ will have to drop more voltage. $Q_{23}$'s forward-biased voltage drop will be lower and pinch $Q_{22}$ still more, so the aggregate $\\beta$ might be something desired. But what to suggest without any context? No idea. – jonk Jan 9 at 3:43\n• @jonk and thank you for the answers. I edited the question adding some further context, hope it clarifies a few things. – George Jan 9 at 14:19\n• @AnalogKid V1, V2 are constant DC sources. Please read the edit for clarification. Thanks for the comment! – George Jan 9 at 14:20\n• I know that this is what you want to measure. But sometimes we add components for protection or sideeffects. – le_top Jan 10 at 22:41\n\nYou are right the Q23 will be ON. Why? Because there is a path for a base-collector current to flow for NPN transistor to ground.", null, "simulate this circuit – Schematic created using CircuitLab\n\nAnd in this reverse connector the collector took over the role of the emitter and the emitter now takes the role of the collector. But due to the different doping (and size) between the collector and the emitter. The reverse beta $$\\\\beta_R\\$$ will be much lower than the \"normal\" forward beta $$\\\\beta_F\\$$.\n\nFor example my BC548B show this resoult: $$\\\\beta_F = 250\\$$ at $$\\1\\textrm{mA}\\$$ and $$\\\\beta_R = 8.3\\$$ in reverse active mode for the same current.\n\nAnd for BC337-25 $$\\\\beta_F = 352\\$$ ; $$\\\\beta_R = 38\\$$\n\nAs a side note the BJT will also conduct current in these two cases:\n\nBJT as a Zener diode", null, "BJT's now behaviors just like a poor's man tunnel diode (Esaki effect). And tunnel diodes will have \"negative resistor\" region. And this negative resistance region occurs only for NPN BJT's.", null, "http://jlnlabs.online.fr/cnr/negosc.htm\n\nhttp://www.cappels.org/dproj/simplest_LED_flasher/Simplest_LED_Flasher_Circuit.html\n\nBC337-40\n\nVeb=8.2V , Vec=6.7V at I=5.5mA\n\nBC549B\n\nVeb=8.3V , Vec=7.2V at I=5.5mA\n\nBD139-16\n\nVeb=8.5V, Vec=6.7V at I=5.5mA\n\n• But how is the base-emitter (PN) diode forward biased? Base is at 1 V and emitter at 3 V. – thece Feb 1 at 18:09\n• @thece But in this circuit, the BJT is in reverse-active mode. Which means that the Base-Collector is in forward biased and Base-Emitter is in reverse biased. – G36 Feb 1 at 19:16" ]	[ null, "https://i.stack.imgur.com/C7qXb.png", null, "https://i.stack.imgur.com/smnfQ.png", null, "https://i.stack.imgur.com/AtAxX.png", null, "https://i.stack.imgur.com/vg54l.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9545891,"math_prob":0.9512516,"size":1264,"snap":"2019-43-2019-47","text_gpt3_token_len":280,"char_repetition_ratio":0.1436508,"word_repetition_ratio":0.0,"special_character_ratio":0.22626582,"punctuation_ratio":0.06324111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9863107,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,7,null,7,null,7,null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-19T00:27:37Z\",\"WARC-Record-ID\":\"<urn:uuid:99ad9384-5363-4d1d-b508-d29081ee6af8>\",\"Content-Length\":\"148847\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:182ef237-79a9-484a-8596-2b762e84fb5a>\",\"WARC-Concurrent-To\":\"<urn:uuid:b46b4041-194b-4650-aa52-dd81f9a37fe9>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://electronics.stackexchange.com/questions/415955/how-does-this-bjt-affect-the-circuit-if-it-does-how-should-i-simulate-on-circu?noredirect=1\",\"WARC-Payload-Digest\":\"sha1:ID5VUPZOCSURGME2V3BSRNLDSD2JDMDF\",\"WARC-Block-Digest\":\"sha1:YXD7NZRDZ7A4LLRC5VZCCW62EUXXMVZ2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669868.3_warc_CC-MAIN-20191118232526-20191119020526-00273.warc.gz\"}"}
https://www.ams.org/journals/tran/1973-180-00/S0002-9947-1973-0318115-3/home.html	[ "# Transactions of the American Mathematical Society\n\nPublished by the American Mathematical Society since 1900, Transactions of the American Mathematical Society is devoted to longer research articles in all areas of pure and applied mathematics.\n\nISSN 1088-6850 (online) ISSN 0002-9947 (print)\n\nThe 2020 MCQ for Transactions of the American Mathematical Society is 1.48.\n\nWhat is MCQ? The Mathematical Citation Quotient (MCQ) measures journal impact by looking at citations over a five-year period. Subscribers to MathSciNet may click through for more detailed information.\n\nby R. L. Davis\nTrans. Amer. Math. Soc. 180 (1973), 47-52 Request permission\n\n## Abstract:\n\nLet \\$K\\$ be a field having prime characteristic \\$p\\$. The following conditions on a subfield \\$k\\$ of \\$K\\$ are equivalent: (i) \\${ \\cap _n}{K^{{p^n}}}(k) = k\\$ and \\$K/k\\$ is separable. (ii) \\$k\\$ is the field of constants of an infinite higher derivation defined in \\$K\\$. (iii) \\$k\\$ is the field of constants of a set of infinite higher derivations defined in \\$K\\$. If \\$K/k\\$ is separably generated and \\$k\\$ is algebraically closed in \\$K\\$, then \\$k\\$ is the field of constants of an infinite higher derivation in \\$K\\$. If \\$K/k\\$ is finitely generated then \\$k\\$ is the field of constants of an infinite higher derivation in \\$K\\$ if and only if \\$K/k\\$ is regular.\nSimilar Articles\n• Retrieve articles in Transactions of the American Mathematical Society with MSC: 12F15\n• Retrieve articles in all journals with MSC: 12F15" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6623859,"math_prob":0.7888176,"size":2082,"snap":"2023-40-2023-50","text_gpt3_token_len":678,"char_repetition_ratio":0.1159769,"word_repetition_ratio":0.117263846,"special_character_ratio":0.37463978,"punctuation_ratio":0.22121896,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99541354,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-08T23:58:46Z\",\"WARC-Record-ID\":\"<urn:uuid:64d345fc-7ca2-485e-bb7f-8491b344a3e0>\",\"Content-Length\":\"57544\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:583f59e4-2b6b-453b-bac1-088cf4019e7d>\",\"WARC-Concurrent-To\":\"<urn:uuid:218ada68-10ba-4e0a-a05b-9cd68a561962>\",\"WARC-IP-Address\":\"130.44.204.100\",\"WARC-Target-URI\":\"https://www.ams.org/journals/tran/1973-180-00/S0002-9947-1973-0318115-3/home.html\",\"WARC-Payload-Digest\":\"sha1:BQYCH5KEHZECP7LTZZMX2QA67CMFZPBN\",\"WARC-Block-Digest\":\"sha1:HOF7HTJNZELPB2YUS5QQ2FYQVIOA6MH3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100779.51_warc_CC-MAIN-20231208212357-20231209002357-00221.warc.gz\"}"}
https://www.maplesoft.com/support/help/maple/view.aspx?path=Maplets/Elements/RunDialog&L=E	[ "", null, "Maplets[Elements] - Maple Programming Help\n\nHome : Support : Online Help : Programming : Maplets : Elements : Command Elements : Maplets/Elements/RunDialog\n\nMaplets[Elements]\n\n RunDialog\n display a dialog\n\n Calling Sequence RunDialog(opts)\n\nParameters\n\n opts - equation(s) of the form dialog=value; specify options for the RunDialog element\n\nDescription\n\n • The RunDialog command element displays a dialog defined in the running Maplet application. To display a dialog defined in a different Maplet application, use the Display function.\n • There are separate RunDialog and RunWindow command elements because the Window element is intrinsically different from the dialog elements. A dialog has a predefined structure. A Maplet application author can specify options for a dialog, but cannot add elements. A window does not have a predefined structure. A Maplet application author specifies its structure by using elements and options. Also they behave differently. For example, a window can be minimized.\n • To simplify specifying options in the Maplets package, certain options and contents can be set without using an equation. The following table lists elements, symbols, and types (in the left column) and the corresponding option or content (in the right column) to which inputs of this type are, by default, assigned.\n\n Elements, Symbols, or Types Assumed Option or Content name or string dialog option\n\n • A RunDialog element cannot contain other elements.\n • A RunDialog element can be contained in a Maplet element; in an equation for the onapprove, oncancel, onchange, onclick, ondecline, or onstartup option for an element; or wrapped in an Action element as a parameter in an element that accepts an onchange or onclick option without an equation.\n • The following table describes the control and use of the RunDialog element options.\n An x in the I column indicates that the option can be initialized, that is, specified in the calling sequence (element definition).\n An x in the R column indicates that the option is required in the calling sequence.\n An x in the G column indicates that the option can be read, that is, retrieved by using the Get tool.\n An x in the S column indicates that the option can be written, that is, set by using the SetOption element or the Set tool.\n\n Option I R G S dialog x x\n\n • The opts argument can contain one or more of the following equations that set Maplet application options.\n dialog = reference to a dialog element (name or string)\n A reference to the dialog to run.\n\nExamples\n\nIn this Maplet application, the RunDialog element starts a dialog in the same Maplet application.\n\n > $\\mathrm{with}\\left(\\mathrm{Maplets}\\left[\\mathrm{Elements}\\right]\\right):$\n > $\\mathrm{maplet}≔\\mathrm{Maplet}\\left(\\mathrm{Window}\\left(\\left[\\left[\\mathrm{TextField}\\left['\\mathrm{TF1}'\\right]\\left(\\right)\\right],\\left[\\mathrm{Button}\\left(\"Differentiate with respect to x\",\\mathrm{Evaluate}\\left('\\mathrm{TF1}'='\\mathrm{diff}\\left(\\mathrm{TF1},x\\right)'\\right)\\right),\\mathrm{Button}\\left(\"Help\",\\mathrm{RunDialog}\\left('\\mathrm{MD1}'\\right)\\right),\\mathrm{Button}\\left(\"Exit\",\\mathrm{Shutdown}\\left(\\left['\\mathrm{TF1}'\\right]\\right)\\right)\\right]\\right]\\right),\\mathrm{MessageDialog}\\left['\\mathrm{MD1}'\\right]\\left(\"See ?diff for help with the differentiation command\"\\right)\\right):$\n > $\\mathrm{Maplets}\\left[\\mathrm{Display}\\right]\\left(\\mathrm{maplet}\\right)$\n\nIn this Maplet application, the Display function starts another Maplet application that contains a dialog.\n\n > $\\mathrm{with}\\left(\\mathrm{Maplets}\\left[\\mathrm{Elements}\\right]\\right):$\n > $\\mathrm{maplet1}≔\\mathrm{Maplet}\\left(\\mathrm{Window}\\left(\\left[\\left[\\mathrm{TextField}\\left['\\mathrm{TF1}'\\right]\\left(\\right)\\right],\\left[\\mathrm{Button}\\left(\"Differentiate with respect to x\",\\mathrm{Evaluate}\\left('\\mathrm{TF1}'='\\mathrm{diff}\\left(\\mathrm{TF1},x\\right)'\\right)\\right),\\mathrm{Button}\\left(\"Help\",\\mathrm{Evaluate}\\left(\\mathrm{function}='\\mathrm{Maplets}\\left[\\mathrm{Display}\\right]\\left(\\mathrm{maplet2}\\right)'\\right)\\right),\\mathrm{Button}\\left(\"Exit\",\\mathrm{Shutdown}\\left(\\left['\\mathrm{TF1}'\\right]\\right)\\right)\\right]\\right]\\right)\\right):$\n > $\\mathrm{maplet2}≔\\mathrm{Maplet}\\left(\\mathrm{MessageDialog}\\left(\"See ?diff for help with the differentiation command\"\\right)\\right):$\n > $\\mathrm{Maplets}\\left[\\mathrm{Display}\\right]\\left(\\mathrm{maplet1}\\right)$" ]	[ null, "https://bat.bing.com/action/0", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5814436,"math_prob":0.9934123,"size":3779,"snap":"2021-21-2021-25","text_gpt3_token_len":897,"char_repetition_ratio":0.21456954,"word_repetition_ratio":0.054968286,"special_character_ratio":0.17412014,"punctuation_ratio":0.110410094,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97892976,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-15T20:04:38Z\",\"WARC-Record-ID\":\"<urn:uuid:ac11e8f5-db74-4b90-8121-6bc451eae6a5>\",\"Content-Length\":\"250242\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:25726f5b-e1d9-4f92-ae83-2428389203f0>\",\"WARC-Concurrent-To\":\"<urn:uuid:1032f695-fb9d-4855-8ed7-f0724710762c>\",\"WARC-IP-Address\":\"199.71.183.28\",\"WARC-Target-URI\":\"https://www.maplesoft.com/support/help/maple/view.aspx?path=Maplets/Elements/RunDialog&L=E\",\"WARC-Payload-Digest\":\"sha1:TGNFVRAH7MB32WLC4O6E34ZE3VI4GHIK\",\"WARC-Block-Digest\":\"sha1:H6WUV23ZPQLVTHBMSGY25ZZBUXN46G7I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991378.52_warc_CC-MAIN-20210515192444-20210515222444-00444.warc.gz\"}"}
https://chem.libretexts.org/Courses/Sacramento_City_College/SCC%3A_CHEM_300_-_Beginning_Chemistry/SCC%3A_CHEM_300_-_Beginning_Chemistry_(Alviar-Agnew)/03%3A_Matter_and_Energy/3.11%3A_Temperature_Changes-_Heat_Capacity	[ "# 3.11: Temperature Changes- Heat Capacity\n\nIf a swimming pool and wading pool, both full of water at the same temperature, were subjected to the same input of heat energy, the wading pool would certainly rise in temperature more quickly than the swimming pool. The heat capacity of an object depends both on its mass and its chemical composition. Because of its much larger mass, the swimming pool of water has a larger heat capacity than the wading pool.\n\n## Heat Capacity and Specific Heat\n\nDifferent substances respond to heat in different ways. If a metal chair sits in the bright sun on a hot day, it may become quite hot to the touch. An equal mass of water in the same sun will not become nearly as hot. We would say that water has a high heat capacity (the amount of heat required to raise the temperature of an object by $$1^\\text{o} \\text{C}$$). Water is very resistant to changes in temperature, while metals in general are not. The specific heat of a substance is the amount of energy required to raise the temperature of 1 gram of the substance by $$1^\\text{o} \\text{C}$$. The symbol for specific heat is $$c_p$$, with the $$p$$ subscript referring to the fact that specific heats are measured at constant pressure. The units for specific heat can either be joules per gram per degree $$\\left( \\text{J/g}^\\text{o} \\text{C} \\right)$$ or calories per gram per degree $$\\left( \\text{cal/g}^\\text{o} \\text{C} \\right)$$. This text will use $$\\text{J/g}^\\text{o} \\text{C}$$ for specific heat.\n\nNotice that water has a very high specific heat compared to most other substances. Water is commonly used as a coolant for machinery because it is able to absorb large quantities of heat (see table above). Coastal climates are much more moderate than inland climates because of the presence of the ocean. Water in lakes or oceans absorbs heat from the air on hot days and releases it back into the air on cool days.", null, "Figure $$\\PageIndex{1}$$: This power plant in West Virginia, like many others, is located next to a large lake so that the water from the lake can be used as a coolant. Cool water from the lake is pumped into the plant, while warmer water is pumped out of the plant and back into the lake.\n\n## Summary\n\n• Heat capacity is the amount of heat required to raise the temperature of an object by $$1^\\text{o} \\text{C}$$).\n• The specific heat of a substance is the amount of energy required to raise the temperature of 1 gram of the substance by $$1^\\text{o} \\text{C}$$." ]	[ null, "https://chem.libretexts.org/@api/deki/files/78936/CK12_Screenshot_17-4-1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9186674,"math_prob":0.988194,"size":2547,"snap":"2019-51-2020-05","text_gpt3_token_len":589,"char_repetition_ratio":0.136453,"word_repetition_ratio":0.15238096,"special_character_ratio":0.23478602,"punctuation_ratio":0.07231405,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98829776,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-29T12:27:35Z\",\"WARC-Record-ID\":\"<urn:uuid:e09b2a70-2b85-46e4-bef4-0a432d2af11a>\",\"Content-Length\":\"165586\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7e259e1b-e29f-4428-9ebc-6abd3ebba0a0>\",\"WARC-Concurrent-To\":\"<urn:uuid:42369240-6a0c-44a2-bf05-3b1334abf81e>\",\"WARC-IP-Address\":\"54.192.30.74\",\"WARC-Target-URI\":\"https://chem.libretexts.org/Courses/Sacramento_City_College/SCC%3A_CHEM_300_-_Beginning_Chemistry/SCC%3A_CHEM_300_-_Beginning_Chemistry_(Alviar-Agnew)/03%3A_Matter_and_Energy/3.11%3A_Temperature_Changes-_Heat_Capacity\",\"WARC-Payload-Digest\":\"sha1:4S52J6J5I2XKQ35YAN2HE7AWZ7JXYFL3\",\"WARC-Block-Digest\":\"sha1:VLF72L4HNVBPGY2IVQTZFXDOL64BLBMA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251796127.92_warc_CC-MAIN-20200129102701-20200129132701-00169.warc.gz\"}"}
http://www.scielo.org.mx/scielo.php?script=sci_arttext&pid=S0035-001X2019000300268&lng=en&nrm=iso&tlng=en	[ "## Indicators\n\n•", null, "Cited by SciELO\n•", null, "Access statistics\n\n•", null, "Similars in SciELO\n\n## Print version ISSN 0035-001X\n\n### Rev. mex. fis. vol.65 n.3 México May./Jun. 2019 Epub Apr 30, 2020\n\n#### https://doi.org/10.31349/revmexfis.65.268\n\nResearch\n\nNear-field analysis and field transformation applied to a parabolic profile at 5 GHz\n\naInstituto Politécnico Nacional, Escuela Superior de Ingeniería Mecánica y Eléctrica. UPALM Edif. Z-4, Tercer piso, C.P. 07738 Ciudad de México.\n\nbUniversidad de Quintana Roo Campus Chetumal. Boulevard Bahía, Esq. Ignacio Comonfort S/N Edif. L, Col. Del Bosque, 77019. e-mail: [email protected]\n\nAbstract\n\nWe propose a method of near-field analysis and field transformation that is applied to a 1.5 m diameter parabolic reflector at a frequency of 5 GHz. An antenna of such dimensions requires at least a 75 m region to obtain its radiation pattern, and this represents a problem, here arises the necessity to make field transformations like the one presented in this work. Near field is modeled by means of Finite Difference Time Domain Method (FDTD) and current distribution is obtained using the discrete Pocklington equation. Radiation pattern is calculated applying the array factor for parabolic profiles. Results are compared with those obtained by CST Microwave Studio with a very good agreement.\n\nKeywords: Arrays; FDTD; Fraunhofer; modeling; parabolic profile; Rayleigh\n\nPACS: 03.50.De; 02.70.Bf; 41.20.Jb; 07.05.Tp; 01.50.H-\n\nResumen\n\nProponemos un método para el análisis en campo cercano y transformación de campo, que es aplicado a un reflector parabólico de 1.5 m de diámetro, a una frecuencia de 5 GHz, una antena de tales dimensiones requiere una región de al menos 75 m para obtener su diagrama de radiación, lo cual representa un problema, aquí es donde surge la necesidad de realizar transformaciones de campo como la presentada en este artículo. El campo cercano es modelado mediante el método de Diferencias Finitas en el Dominio del Tiempo (MDFDT) y la distribución de corriente es obtenida usando la ecuación discreta de Pocklington. El diagrama de radiación es calculado aplicando el factor de arreglo para perfiles parabólicos. Nuestros resultados del análisis en campo cercano y transformación de campo, son comparados contra el modelado de la estructura en CST Microwave Studio, encontrando muy buenas coincidencias entre ambos métodos.\n\nDescriptores: Arreglos; MDFDT; Fraunhofer; modelado; perfil parabólico; Rayleigh\n\n1. Introduction\n\nWhen a radiated field is distributed in space, there are three regions that identify the process : Rayleigh region, Fresnel region and Fraunhofer region; Rayleigh region refers to the individual point field, which means that the radiation generated at each point of the structure has not been mixed with the other. In Fresnel region, the generated individual fields begin to interact with each other, but their phase differences are significant, so the radiation pattern does not have the shape that must adopt in far-field. In Fraunhofer region, fields are added almost in phase and as they move away from the source, the differences begin to be further reduced, so this region is referred as the far-field region. Figure 1 describes the three regions of the field distribution process in space. Table I shows conditions to determine the field region distances, where D is the antenna diameter, R1 and R2 are Rayleigh and Fresnel regions and λ is the wavelength .\n\nTABLE I Field regions distances.\n\n Antenna Rayleigh Fresnel Fraunhofer dimension region Region region D>>λ R1<λ/2π λ2π\n\nAccording with Fig. 1 and Table I, the minimum distance to obtain the far-field of a 1.5 m diameter structure, working at 5 GHz, is at least 75 m, which represents a problem if we try to measure far-field inside a controlled environment (as in an anechoic chamber, making nearly unmanageable to address the problem). Our method proposes to perform a field modeling at the Rayleigh region, define virtual point currents over the surface through discrete Pocklington equation, and then obtain far-field using array theory; essentially a near-field to far-field transformation is carried out. Comparing with commercial software, we reduce computational resources and the proposed method has been used to transform near-field into far-field from real measurements .\n\nThe method requires knowing the electric field near to reflector surface; this data can be obtained from measurements as [3-6] or by field modeling in the Rayleigh region as in , for this purpose, we implement a model based on FDTD method in order to obtain the magnitude and phase of field in a small region, near to the reflector surface. Descriptions of FDTD, discrete Pocklington equation and parabolic array factor (PAF) are presented in following Subsecs. 1.1 and 1.2.\n\n1.1 Finite difference time domain method (FDTD)\n\nThe FDTD method allows analyzing the effects of electromagnetic propagation, transforming the differential Maxwell equations into finite difference equations that can be handled by computers [8,9]. FDTD solution involves: establish the analysis region and divide it into cells where Maxwell’s equations are approximated by equations in finite differences, considering all materials within the region, using their electric and magnetic characteristics given by their permittivity (ε), permeability (μ) and conductivity (σ); finally the iterative computing solution gives the point field over the region.\n\nModeled parabolic profile was performed in two dimensions, then propagation is restricted to the XY plane. We choose TM mode considering Ez field as the only transversal component in the XY propagating plane, therefore Ex=Ey=Hz=0. FDTD requires that the analysis region be surrounded by a free reflection computational boundary; we choose Perfectly Matched Layer model (PML) for our application as in [9,10]. Equations (1-4) show the present field components.\n\nEzx(i,j)=abEzx(i,j)+Δtεb[Hy(i,j)-Hy(i-1,j)Δx] (1)\n\nEzy(i,j)=abEzy(i,j)-Δtεb[Hx(i,j)-Hx(i,j-1)Δy] (2)\n\nHx(i,j)=cdHx(i,j)-Δtμd[(Ezx+Ezy)(i.j+1)Δy-(Ezx+Ezy)(i,j)Δy] (3)\n\nHy(i,j)=cdHy(i,j)+Δtμd[(Ezx+Ezy)(i+1,j)Δx-(Ezx+Ezy)(i.j)Δx] (4)\n\nWhere:\n\na=1-σΔt2ε,b=1+σΔt2ε,\n\nC=1-σΔt2μ,d=1+σΔt2μ,\n\nAs a primary source we use a standard WR187 waveguide with dimensions of 0.1×0.02215 m working at 5 GHz; as the experiment was performed in 2D, field distribution is given by Eq. (5), with i=j=0 at the origin and E0 constant.\n\nEz(i,j)n=E0sin(2πfnΔt) (5)\n\n1.2 Discrete Pocklington equation and parabolic array factor\n\nThe method considers that the generalized Pocklington equation is valid not only on the antenna surface, but also outside very near to it, as near as the Rayleigh region defines it. Hence, it is possible to perform a field modeling by FDTD or by measurement as in to acquire field points near the structure.\n\nThe current can be determined by transforming the generalized Pocklington in a discrete form as Eq. (6) calculating virtual currents In along the structure.\n\nEmi=-1jωεn=1N[R2(k2R2-1-jkR)Smsn'+(3+3jkR-k2R2)(RSm)(Rsn')]e-jkR4πR5In. (6)\n\nWhere:\n\nRmn=\|r-r'\|=[x(s)-x'(s')]2+[y(s)-y'(s')]2+[z(s)-z'(s')]2\n\nk=Wavenumber\n\nSolving Eq. (6) by the structure geometry, it can be constructed a matrix as (7), where ZMN represent impedances, IN represents a virtual current to be calculated and EM represents the modeled electric field. The matrix is solved by multiplying the electric field vector, obtained by the FDTD modeling, by the inverse of the impedance matrix, so vector current is obtained.\n\n20cZ11Z1NZM1ZMN20cI1IN=*20cE1EM (7)\n\nFinally, far-field is calculated using the array theory [12,13] under the assumption that calculated point currents IN in (7), represent an array of specific elements that generate the far-field as is shown in Fig. 2. Field radiated by a parabolic array of point sources is given by (8):\n\nPAF(θ)=E0n=1N\|IN\|ej(kRncos(θ-ϕ)+αn) (8)\n\nIn (8) E0 represents the reference field of the array point sources with unitary magnitude; the sum of the individual fields from each point antenna considers the phase current αn, due the distance and the position of each current point Rn defined by ϕ while θ represents the angle between each point and the far-field position, Fig. 2 describes Eq. (8).\n\n2. Near-field analysis and far-field transformation\n\nTable II shows the modeling parameters, where physical dimensions are defined in meters and cells, in order to apply our computational technique based on Pocklington equation and FDTD.\n\nTABLE II Modeling parameters at 5 GHz.\n\n Units wavelength Cells λ m 0.06 Cell λ/20 0.003 Calculus region Plate diameter 500 × 750 500 1.5 × 2.25 1.5 Vertex - focus 200 0.6 Vertex - edge 77 0.23 height (feeder) 7 0.02215 depth (feeder) 33 0.1 PML 24 0.07\n\nCalculus region dimensions are 1.5×2.25 m corresponding to 500×750 cells, considering the size of each cell as λ/20, even several authors define λ/10 is an acceptable size . Our results are compared with those of commercially available software CST, using same parameters for both experiments. Figure 3 shows the calculus region and electric field distribution for both computational methods. To have a better view of the parabolic profile, Fig. 3 (a) is shown in a smaller area defined by (90:335,95:645), where reflector vertex is located at (100,375) and its focus at (300,375). Figure 3 displays both incident and reflected fields, which visual comparison with CST shows a very good concordance.", null, "FIGURE 3 Near-field modeling at 5 GHz, a) FDTD and b) CST.\n\nThe electric field vector EM in Rayleigh region required in (6), is defined using field samples from 500 points in the FDTD matrix at 0.003 m, one cell away from reflector surface along the parabolic profile. Figure 4 shows EM vector in magnitude, real and imaginary parts. Field phase is shown in Fig. 5, and is obtained using distances from reflector surface to its focus, for each field point using parabola equation.\n\nNext step in the methodology is to apply Eqs. (6) and (7) to obtain current along the reflector surface. Equation (6) requires the use of parabolic geometry for distace Rmn and vectors sm and sn, that is, distances between each point of the reflector axis and each current point .\n\nFinally, the field distribution is transformed in virtual point currents along the plate surface that will be applied in the array factor equation for the parabolic profile (8). Figure 6, shows the magnitude, real and imaginary normalized parts for current distribution IN. We can notice a similar behavior with electric field displayed on Fig. 4, including the 180 phase change of imaginary part due the reflection. Figure 7 shows the calculated phase for each current point.\n\nOnce field distribution is transformed into virtual currents along the parabolic profile, magnitude and phase values are used in the parabolic array factor Eq. (8) to get the far-field pattern shown in Fig. 8. As it is shown, radiation pattern has a main lobe width of 2.5 which is consistent with the theory of parabolic reflectors , also the nearest sidelobes are observed at -17 dB, other lobes have amplitudes between -20dB and -55 dB.\n\n3. Comparison\n\nIn previous work [7,16], calculated radiation pattern was compared versus a measured radiation pattern however, actual analysis allows us to observe and compare electric field behavior with those calculated in CST to show its viability.\n\nFigure 9 shows a comparison between near-field behavior for both simulations (FDTD and CST) in magnitude. Figures 10 and 11 show an excellent agreement of the near-field in real and imaginary parts; finally, near-field phase is shown in Fig. 12, as seen there is a clear similarity between FDTD and CST curves, specially real and imaginary parts with small differences at the beginning and at end of curves.", null, "FIGURE 9 Magnitude Electric field comparison between FDTD and CST.", null, "FIGURE 10 Electric field real part comparison between FDTD and CST.", null, "FIGURE 11 Electric field imaginary part comparison between FDTD and CST.", null, "FIGURE 12 Electric field phase comparison between FDTD and CST.\n\nRadiation patterns comparison is shown in Fig. 13, black curve represents the pattern of our method and blue curve is the CST pattern. Similarity between both curves is evident; it can be seen that the main lobe width is about 2.5 for both patterns and small differences of 5 dB at the edges (0° o 30° and 150° to 180°).\n\n4. Conclusions\n\nWe presented a method to transform near-field (Rayleigh region) into far-field, without performing simulation in large computational regions, then reducing computing resources. Even we define the field in the Rayleigh region using the FDTD Method, an alternative proposed as future work is to measure the field near the reflector surface and obtain far-field with Pocklington equation and Array Theory. In addition, our method can be applied to other antenna geometries as lineal radiators or even linear arrays or some other type of reflectors.\n\nComparison between our methodology (FDTD-Pocklington-PAF) and commercial software shows that results are very similar not only for radiation pattern, but also for near-field behavior.\n\nAcknowledgments\n\nThe authors want to thank to the Instituto Politécnico Nacional and Consejo Nacional de Ciencia y Tecnología of México for their support.\n\nReferences\n\n1. J. Sosa-Pedroza, L. Carrion-Rivera, F. Martínez-Zúñiga and S. Peña-Ruiz, Una Propuesta para Transformar Mediciones de Campo Cercano en Campo Lejano. Simposio de Metrología CENAM, México (2014). [ Links ]\n\n2. Y. Huang, K, Boyle, Antennas from Theory to Practice. (John Wiley. UK, 2008), p. 109-112. [ Links ]\n\n3. M. Sierra-Castañer, S. Burgos. Fresnel Zone to Far Field Algorithm for Rapid Array Antenna Measurements. IEEE EUCAP (2011). [ Links ]\n\n4. R. Cornelius, T. Salmerón-Ruiz, F. Saccardi, L. Foged, D. Heberling, M. Sierra-Castañer, IEEE Antennas and Propagation Magazine 56 (2014) . [ Links ]\n\n5. F. D’Agostino et al., IEEE Antennas and Propagation Magazine 54 (2012). [ Links ]\n\n6. S. Costanzo, G. Di Massa, Microwave and Optical Technology Letters 48 (2006). [ Links ]\n\n7. J. Sosa-Pedroza , M. Enciso-Aguilar, S. Peña-Ruiz, A. Rodríguez-Sánchez, and E. Garduño-Nolasco, Microwave and Optical Technology Letters 58 (2016). [ Links ]\n\n8. J. Sosa-Pedroza et al., Approach. Electron. Lett. 47 (2011) 1308-1309. [ Links ]\n\n9. A. Taflove and S. C. Hagness, Computational Electrodynamics The Finite-Diference Time-Domine Method. (Artech House USA 2005) p. 54-75. [ Links ]\n\n10. M. Benavides et al., Rev. Mex. Fís. E 57 (2011) 25-31. [ Links ]\n\n11. J. Sosa-Pedroza , V. Barrera-Figueroa and J. López-Bonilla, Pocklington Equation and the Method of Moments Proc. Pakistan Acad. (2005). [ Links ]\n\n12. C. A. Balanis, Antenna Theory Analysis and Design. 3rd Ed. (John Willey, 2005) p. 284-313. [ Links ]\n\n13. L. Josefsson, P. Persson, Conformal Array Antenna Theory and Design. IEEE (Press Series on Electromagnetic Wave Theory. USA, 2006), p. 15-22. [ Links ]\n\n14. Jacob W.M. Baars, The Paraboloidal Reflector Antenna in Radio Astronomy and Communication. (Springer Science USA, 2007), p. 14-16. [ Links ]\n\n15. J. D. Kraus, R. J. Marhefka, Antennas for all Applications. 3rd Ed. (McGraw Hill. USA, 2002), p. 564-571. [ Links ]\n\n16. J. Sosa, S. Peña, F. Martínez, Parabolic Reflector Near-field to Far-field Transformation Using FDTDM and Pocklington Equation. PIERS, (2017). [ Links ]\n\nReceived: August 11, 2018; Accepted: January 14, 2018", null, "This is an open-access article distributed under the terms of the Creative Commons Attribution License" ]	[ null, "http://www.scielo.org.mx/img/en/iconCitedOff.gif", null, "http://www.scielo.org.mx/img/en/iconStatistics.gif", null, "http://www.scielo.org.mx/img/en/iconRelatedOff.gif", null, "http://www.scielo.org.mx/img/revistas/rmf/v65n3//0035-001X-rmf-65-03-268-gf3.png", null, "http://www.scielo.org.mx/img/revistas/rmf/v65n3//0035-001X-rmf-65-03-268-gf9.png", null, "http://www.scielo.org.mx/img/revistas/rmf/v65n3//0035-001X-rmf-65-03-268-gf10.png", null, "http://www.scielo.org.mx/img/revistas/rmf/v65n3//0035-001X-rmf-65-03-268-gf11.png", null, "http://www.scielo.org.mx/img/revistas/rmf/v65n3//0035-001X-rmf-65-03-268-gf12.png", null, "http://i.creativecommons.org/l/by-nc/4.0/80x15.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7616337,"math_prob":0.93289995,"size":15631,"snap":"2020-45-2020-50","text_gpt3_token_len":4313,"char_repetition_ratio":0.11704102,"word_repetition_ratio":0.0120532,"special_character_ratio":0.24496193,"punctuation_ratio":0.14744005,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9809256,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,null,null,null,null,null,null,1,null,1,null,1,null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-22T00:41:18Z\",\"WARC-Record-ID\":\"<urn:uuid:55b756a1-0c5a-4381-b814-961d50d30e33>\",\"Content-Length\":\"72829\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:193ae70e-f5a3-42d6-8cc6-4fd94a07f311>\",\"WARC-Concurrent-To\":\"<urn:uuid:a6dc08c1-af68-4ce8-86ab-f0b250e4ceeb>\",\"WARC-IP-Address\":\"132.248.9.5\",\"WARC-Target-URI\":\"http://www.scielo.org.mx/scielo.php?script=sci_arttext&pid=S0035-001X2019000300268&lng=en&nrm=iso&tlng=en\",\"WARC-Payload-Digest\":\"sha1:427REJWT4DYHST6OQ7DXO2ZB5URCZRH2\",\"WARC-Block-Digest\":\"sha1:3YUURWCHATXGOYOCK6TSZXVFPEW5UUJR\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107878662.15_warc_CC-MAIN-20201021235030-20201022025030-00082.warc.gz\"}"}
http://pressurevesseltech.asmedigitalcollection.asme.org/article.aspx?articleid=2666607&resultClick=1	[ "0\nResearch Papers: SPECIAL SECTION PAPERS\n\n# Analytical and Semi-Analytical Methods for the Evaluation of Dynamic Thermo-Elastic Behavior of Structures Resting on a Pasternak Foundation\n\n[+] Author and Article Information\nXu Liang, Zeng Cao, Xing Zha, Jianxing Leng\n\nOcean College,\nZhejiang University,\nHangzhou 310058, Zhejiang, China\n\nHongyue Sun\n\nOcean College,\nZhejiang University,\nHangzhou 310058, Zhejiang, China\ne-mail: [email protected]\n\n1Corresponding author.\n\nContributed by the Pressure Vessel and Piping Division of ASME for publication in the JOURNAL OF PRESSURE VESSEL TECHNOLOGY. Manuscript received August 9, 2017; final manuscript received December 9, 2017; published online December 14, 2018. Assoc. Editor: Fabrizio Paolacci.\n\nJ. Pressure Vessel Technol 141(1), 010908 (Dec 14, 2018) (10 pages) Paper No: PVT-17-1146; doi: 10.1115/1.4038724 History: Received August 09, 2017; Revised December 09, 2017\n\n## Abstract\n\nAn analytical method and a semi-analytical method are proposed to analyze the dynamic thermo-elastic behavior of structures resting on a Pasternak foundation. The analytical method employs a finite Fourier integral transform and its inversion, as well as a Laplace transform and its numerical inversion. The semi-analytical method employs the state space method, the differential quadrature method (DQM), and the numerical inversion of the Laplace transform. To demonstrate the two methods, a simply supported Euler–Bernoulli beam of variable length is considered. The governing equations of the beam are derived using Hamilton's principle. A comparison between the results of analytical method and the results of semi-analytical method is carried out, and it is shown that the results of the two methods generally agree with each other, sometimes almost perfectly. A comparison of natural frequencies between the semi-analytical method and the experimental data from relevant literature shows good agreements between the two kinds of results, and the semi-analytical method is validated. Different numbers of sampling points along the axial direction are used to carry out convergence study. It is found that the semi-analytical method converges rapidly. The effects of different beam lengths and heights, thermal stress, and the spring and shear coefficients of the Pasternak medium are also investigated. The results obtained in this paper can serve as benchmark in further research.\n\n<>\n\n## References\n\nBoley, B. A. , 1956, “ Thermally Induced Vibrations of Beams,” J. Aeronaut. Sci., 23(2), pp. 179–181.\nAvsec, J. , and Oblak, M. , 2007, “ Thermal Vibrational Analysis for Simply Supported Beam and Clamped Beam,” J. Sound Vib., 308(3), pp. 514–525.\nMurmu, T. , and Pradhan, S. C. , 2009, “ Thermo-Mechanical Vibration of a Single-Walled Carbon Nanotube Embedded in an Elastic Medium Based on Nonlocal Elasticity Theory,” Comput. Mater. Sci., 46(4), pp. 854–859.\nPradhan, S. C. , and Murmu, T. , 2009, “ Thermo-Mechanical Vibration of FGM Sandwich Beam Under Variable Elastic Foundations Using Differential Quadrature Method,” J. Sound Vib., 321(1–2), pp. 342–362.\nJandaghian, A. A. , and Rahmani, O. , 2016, “ Free Vibration Analysis of Magneto-Electro-Thermo-Elastic Nanobeams Resting on a Pasternak Foundation,” Smart Mater. Struct., 25(3), p. 035023.\nLiang, F. , Li, Y. , and Huang, M. , 2013, “ Simplified Method for Laterally Loaded Piles Based on Pasternak Double-Parameter Spring Model for Foundations,” Chin. J. Geotech. Eng., 35(Suppl. 1), pp. 300–304.\nSharma, J. N. , and Kaur, R. , 2015, “ Modeling and Analysis of Forced Vibrations in Transversely Isotropic Thermoelastic Thin Beams,” Meccanica, 50(1), pp. 189–205.\nSharma, J. N. , and Kaur, R. , 2014, “ Analysis of Forced Vibrations in Micro-Scale Anisotropic Thermo-Elastic Beams Due to Concentrated Loads,” J. Therm. Stresses, 37(1), pp. 93–116.\nSong, X. , and Li, S.-R. , 2007, “ Thermal Buckling and Post-Buckling of Pinned–Fixed Euler–Bernoulli Beams on an Elastic Foundation,” Mech. Res. Commun., 34(2), pp. 164–171.\nNejad, M. Z. , Hadi, A. , and Rastgoo, A. , 2016, “ Buckling Analysis of Arbitrary Two-Directional Functionally Graded Euler–Bernoulli Nano-Beams Based on Nonlocal Elasticity Theory,” Int. J. Eng. Sci., 103, pp. 1–10.\nLi, S. , and Song, X. , 2006, “ Large Thermal Deflections of Timoshenko Beams Under Transversely Non-Uniform Temperature Rise,” Mech. Res. Commun., 33(1), pp. 84–92.\nYun, Y. , Chen, J. , Zhao, K. , Yan, B. , and Cao, H. , 2014, “ Analysis of Thermo-Elastic Coupling Effects of the Beam Structure With Interval Parameters,” J. Xidian Univ., 41(4), pp. 64–70.\nShen, Z. , Tian, Q. , Liu, X. , and Hu, G. , 2013, “ Thermally Induced Vibrations of Flexible Beams Using Absolute Nodal Coordinate Formulation,” Aerosp. Sci. Technol., 29(1), pp. 386–393.\nBert, C. W. , and Malik, M. , 1996, “ Differential Quadrature Method in Computational Mechanics: A Review,” ASME Appl. Mech. Rev., 49(1), pp. 1–28.\nLiang, X. , Wu, Z. , Wang, L. , Liu, G. , Wang, Z. , and Zhang, W. , 2014, “ Semianalytical Three-Dimensional Solutions for the Transient Response of Functionally Graded Material Rectangular Plates,” J. Eng. Mech., 141(9), p. 04015027.\nAkbarzadeh, A. H. , Hosseini, K. , Eslami, M. R. , and Sadighi, M. , 2011, “ Mechanical Behaviour of Functionally Graded Plates Under Static and Dynamic Loading,” Proc. Inst. Mech. Eng., Part C, 225(2), pp. 326–333.\nKiani, Y. , Shakeri, M. , and Eslami, M. R. , 2012, “ Thermoelastic Free Vibration and Dynamic Behaviour of an FGM Doubly Curved Panel Via the Analytical Hybrid Laplace–Fourier Transformation,” Acta Mech., 223(6), pp. 1199–1218.\nAkbarzadeh, A. H. , Abbasi, M. , and Eslami, M. R. , 2012, “ Coupled Thermoelasticity of Functionally Graded Plates Based on the Third-Order Shear Deformation Theory,” Thin-Walled Struct., 53, pp. 141–155.\nWang, Z. , Liang, X. , and Liu, G. , 2013, “ An Analytical Method for Evaluating the Dynamic Response of Plates Subjected to Underwater Shock Employing Mindlin Plate Theory and Laplace Transforms,” Math. Probl. Eng., 2013, pp. 927–940.\nDurbin, F. , 1974, “ Numerical Inversion of Laplace Transforms: An Efficient Improvement to Dubner and Abate's Method,” Comput. J., 17(4), pp. 371–376.\nLiang, X. , Wang, Z. , Wang, L. , and Liu, G. , 2014, “ Semi-Analytical Solution for Three-Dimensional Transient Response of Functionally Graded Annular Plate on a Two Parameter Viscoelastic Foundation,” J. Sound Vib., 333(12), pp. 2649–2663.\nLiang, X. , Kou, H. L. , Wang, L. , Palmer, A. C. , Wang, Z. , and Liu, G. , 2015, “ Three-Dimensional Transient Analysis of Functionally Graded Material Annular Sector Plate Under Various Boundary Conditions,” Compos. Struct., 132, pp. 584–596.\nZavodney, L. D. , and Nayfeh, A. H. , 1988, “ The Response of a Single-Degree-of-Freedom System With Quadratic and Cubic Non-Linearities to a Fundamental Parametric Resonance,” J. Sound Vib., 120(1), pp. 63–93.\nShu, C. , and Du, H. , 1997, “ Implementation of Clamped and Simply Supported Boundary Conditions in the GDQ Free Vibration Analysis of Beams and Plates,” Int. J. Solids Struct., 34(7), pp. 819–835.\nWolfram Research, 2016, “Wolfram Mathematica, Version 11.0.0,” Wolfram Research, Inc., Champaign, IL.\nMcLellan, R. B. , and Ishikawa, T. , 1987, “ The Elastic Properties of Aluminum at High Temperatures,” J. Phys. Chem. Solids, 48(7), pp. 603–606.\nNix, F. C. , and Macnair, D. , 1941, “ The Thermal Expansion of Pure Metals: Copper, Gold, Aluminum, Nickel, and Iron,” Phys. Rev., 60(8), pp. 597–605.\nMarques, R. F. A. , and Inman, D. J. , 2002, “An Analytical Model for a Clamped Isotropic Beam Under Thermal Effects,” ASME Paper No. IMECE2002-33977.\n\n## Figures\n\nFig. 1\n\nThe geometry of a thermo-elastic beam on a Pasternak foundation", null, "Fig. 2\n\nThe time histories of peak beam deflection at the selected position, x = l/2, in the thermal environment by the analytical and the semi-analytical method: (a) case 1, (b) case 2, (c) case 3, and (d) case 4", null, "Fig. 3\n\nThe time histories of peak beam deflection at a selected point, x = l/2, of the beam, for four different numbers (N) of sampling points along the x-axis: (a) case 1, (b) case 2, (c) case 3, and (d) case 4", null, "Fig. 4\n\nThe time histories of peak beam deflection at two selected positions, x = l/2 and x = l/5, for different boundary conditions: (a) at the point x = l/2 and (b) at the point x = l/5", null, "Fig. 5\n\nThe time histories of peak beam deflection at two selected positions, x = l/2 and x = l/5, for different beam lengths and fixed height: (a) at the point x = l/2 and (b) at the point x = l/5", null, "Fig. 6\n\nThe time histories of peak beam deflection at two selected points, x = l/2 and x = l/5, for a given beam length and different heights: (a) at the point x = l/2 and (b) at the point x = l/5", null, "Fig. 7\n\nThe time histories of peak beam deflection at two different selected points, x = l/2 and x = l/5, for different surrounding temperatures: (a) at the point x = l/2 and (b) at the point x = l/5", null, "Fig. 8\n\nThe time histories of peak beam deflection at two different selected points, x = l/2 and x = l/5, with different spring coefficients: (a) at the point x = l/2 and (b) at the point x = l/5", null, "Fig. 9\n\nThe time histories of peak beam deflection at two different selected points x = l/2 and x = l/5 with different shear coefficients: (a) at the point x = l/2 and (b) at the point x = l/5", null, "## Errata\n\nSome tools below are only available to our subscribers or users with an online account.\n\n### Related Content\n\nCustomize your page view by dragging and repositioning the boxes below.\n\nRelated Journal Articles\nRelated Proceedings Articles\nRelated eBook Content\nTopic Collections" ]	[ null, "http://pressurevesseltech.asmedigitalcollection.asme.org/data/journals/jpvtas/937698/s_pvt_141_01_010908_f001.png", null, "http://pressurevesseltech.asmedigitalcollection.asme.org/data/journals/jpvtas/937698/s_pvt_141_01_010908_f002.png", null, "http://pressurevesseltech.asmedigitalcollection.asme.org/data/journals/jpvtas/937698/s_pvt_141_01_010908_f003.png", null, "http://pressurevesseltech.asmedigitalcollection.asme.org/data/journals/jpvtas/937698/s_pvt_141_01_010908_f004.png", null, "http://pressurevesseltech.asmedigitalcollection.asme.org/data/journals/jpvtas/937698/s_pvt_141_01_010908_f005.png", null, "http://pressurevesseltech.asmedigitalcollection.asme.org/data/journals/jpvtas/937698/s_pvt_141_01_010908_f006.png", null, "http://pressurevesseltech.asmedigitalcollection.asme.org/data/journals/jpvtas/937698/s_pvt_141_01_010908_f007.png", null, "http://pressurevesseltech.asmedigitalcollection.asme.org/data/journals/jpvtas/937698/s_pvt_141_01_010908_f008.png", null, "http://pressurevesseltech.asmedigitalcollection.asme.org/data/journals/jpvtas/937698/s_pvt_141_01_010908_f009.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93597245,"math_prob":0.8470136,"size":1486,"snap":"2019-26-2019-30","text_gpt3_token_len":274,"char_repetition_ratio":0.17678812,"word_repetition_ratio":0.0,"special_character_ratio":0.16756393,"punctuation_ratio":0.08097166,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.957579,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-22T03:41:45Z\",\"WARC-Record-ID\":\"<urn:uuid:92bbda10-7884-4d35-960f-0ecc3c1f6d59>\",\"Content-Length\":\"197584\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a88685c5-ea0f-4239-bb06-5c736b9dd7f6>\",\"WARC-Concurrent-To\":\"<urn:uuid:049a25cf-b7ba-4b93-b2aa-89214e9941d9>\",\"WARC-IP-Address\":\"173.254.190.160\",\"WARC-Target-URI\":\"http://pressurevesseltech.asmedigitalcollection.asme.org/article.aspx?articleid=2666607&resultClick=1\",\"WARC-Payload-Digest\":\"sha1:OCXJKDFHAY7OXP4OHWARA6QYVMQ3YHJM\",\"WARC-Block-Digest\":\"sha1:KXIO3OJFKBEJLMVRZPIAQZ524SOQIRZ7\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195527474.85_warc_CC-MAIN-20190722030952-20190722052952-00042.warc.gz\"}"}
https://www.tutorialspoint.com/algorithm-to-sum-ranges-that-lie-within-another-separate-range-in-javascript	[ "# Algorithm to sum ranges that lie within another separate range in JavaScript\n\nJavascriptWeb DevelopmentFront End TechnologyObject Oriented Programming\n\nWe have two sets of ranges; one is a single range of any length (R1) and the other is a set of ranges (R2) some or parts of which may or may not lie within the single range (R1).\n\nWe need to calculate the sum of the ranges in (R2) - whole or partial - that lie within the single range (R1).\n\nconst R1 = [20,40];\nconst R2 = [[14,22],[24,27],[31,35],[38,56]];\n\nResult\n\n= 2+3+4+2 = 11\nR1 = [120,356];\nR2 = [[234,567]];\n\nResult\n\n122\n\n## Example\n\nLet us write the code −\n\nconst R1 = [20,40];\nconst R2 = [[14,22],[24,27],[31,35],[38,56]];\nconst R3 = [120,356];\nconst R4 = [[234,567]];\nfunction sumRanges(range, values) {\nconst [start, end] = range;\nconst res = values.reduce((acc, val) => {\nconst [left, right] = val;\nconst ex1 = Math.min(right, end);\nconst ex2 = Math.max(left, start);\nconst diff = ex1 - ex2;\nreturn acc + Math.max(0, diff);\n}, 0);\nreturn res;\n};\nconsole.log(sumRanges(R1, R2));\nconsole.log(sumRanges(R3, R4));\n\n## Output\n\nAnd the output in the console will be −\n\n11\n122" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.73990524,"math_prob":0.99893814,"size":2474,"snap":"2022-05-2022-21","text_gpt3_token_len":664,"char_repetition_ratio":0.17165992,"word_repetition_ratio":0.13140312,"special_character_ratio":0.31285366,"punctuation_ratio":0.12836438,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9960494,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-23T09:07:19Z\",\"WARC-Record-ID\":\"<urn:uuid:5379bc14-67ae-4e46-a01c-74a63b900256>\",\"Content-Length\":\"32795\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fa904d61-e4af-4ec1-9c79-b6d486d8b88f>\",\"WARC-Concurrent-To\":\"<urn:uuid:cbfce3d8-4157-4bf0-87eb-6bda48309dbe>\",\"WARC-IP-Address\":\"192.229.210.176\",\"WARC-Target-URI\":\"https://www.tutorialspoint.com/algorithm-to-sum-ranges-that-lie-within-another-separate-range-in-javascript\",\"WARC-Payload-Digest\":\"sha1:ZC3Z6WMWWM7SKLEASF6M2F6YQLJFFZ77\",\"WARC-Block-Digest\":\"sha1:53OT3AOV4AUFXU5U7DP4Y52SOY6AAWET\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662556725.76_warc_CC-MAIN-20220523071517-20220523101517-00151.warc.gz\"}"}
https://www.physicsforums.com/threads/double-integrals.312276/	[ "# Double integrals\n\n## Homework Statement\n\nintegrate y DA over the regions s, where s is in the first quadrant bounded by x = y^2 and x = 8 - y^2\n\n## The Attempt at a Solution\n\nIf the y wasnt there i would evaluate two integrals and subtract them to get the area.\nBut since the y is there, can i still evaluate them seperatly with the y in place and subtract.\n\nRelated Calculus and Beyond Homework Help News on Phys.org\ntiny-tim\nHomework Helper\nHi joemama69!", null, "integrate y DA over the regions s, where s is in the first quadrant bounded by x = y^2 and x = 8 - y^2\nand which axis??\n\nIf the y wasnt there i would evaluate two integrals and subtract them to get the area.\nBut since the y is there, can i still evaluate them seperatly with the y in place and subtract.\n(not sure what you're subtracting, but:) yes … what's worrying you about that?", null, "ya i was thinkin to hard\n\nheres what i did, i broke it into two intigrals and added them\n\nfirst intigral\n\ni differentiated y dy from 0 to x^(1/2)\nthen i did in trems of x from 0 to 4 and got an area for 4\n\n2nd integral\n\ni differentiated y dy from 0 to (8-x)^(1/2)\nthen i did in trems of x from 4 to 8 and got an area for 4\n\n4 + 4 = 8 is that right\n\ntiny-tim\nHomework Helper\nya i was thinkin to hard\n\nheres what i did, i broke it into two intigrals and added them\n\n4 + 4 = 8 is that right\nYup, that's fine!", null, "(though you could have saved a little work by pointing out that the region was obviously symmetric about x = 4, so all you had to do was evaluate one integral, and then double it.", null, ")\n\nya i noticed that, but i figured for my teachers sake i should do it thoroughly" ]	[ null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92209655,"math_prob":0.9697675,"size":370,"snap":"2020-10-2020-16","text_gpt3_token_len":104,"char_repetition_ratio":0.114754096,"word_repetition_ratio":0.0,"special_character_ratio":0.27567568,"punctuation_ratio":0.05263158,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99636006,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-20T10:26:04Z\",\"WARC-Record-ID\":\"<urn:uuid:73a6a307-e6b3-4f9c-a37a-44db8ecb5a25>\",\"Content-Length\":\"77394\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:35afa6ef-b45d-41fd-bd9c-dfb823f96a22>\",\"WARC-Concurrent-To\":\"<urn:uuid:1662eff8-b5e8-44b2-9775-bfed8b55d168>\",\"WARC-IP-Address\":\"23.111.143.85\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/double-integrals.312276/\",\"WARC-Payload-Digest\":\"sha1:LNCBLNDY5LISYAQHVOFUBRIZFUPXVM2P\",\"WARC-Block-Digest\":\"sha1:VBPFZGHAUQKSR4X5QWUKDYECIBBXZJ3B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875144722.77_warc_CC-MAIN-20200220100914-20200220130914-00255.warc.gz\"}"}
https://electronics.stackexchange.com/questions/363601/issue-with-simulink-does-not-support-code-generation	[ "# Issue with “Simulink does not support code generation”\n\nI have something similar to this simplified following block diagram on Simulink. It looks rather messy.", null, "I want to replace a 3-point summing block with a function block, while keeping the same output.", null, "First I started by placing the code inside the function block:\n\nfunction y = fcn(u)\nsys1 = tf(0.5,[1 0 0 4]);\nsys2 = tf([3 0.5],[1 0 15]);\nsys3 = tf(1,[1 1]);\ny = sys1 + sys2 + sys3;\n\n\nHowever I was greeted with an error saying that Simulink does not support code generation.\n\n\"The 'tf' class does not support code generation.\"\n\nI am trying to implement an extrinsic function or'wrapper function' with some difficulty. I created a new script called myWrapper.m, containing the same code:\n\nfunction y = myWrapper(u)\nsys1 = tf(0.5,[1 0 0 0 4]);\nsys2 = tf([3 5],[1 0 15]);\nsys3 = tf(1,[1 1]);\ny = sys1 + sys2 + sys3;\n\n\nand the MATLAB Function edited to:\n\nfunction y1 = fcn(u1)\n\ny1 = myWrapper(u1);\n\n\nThe error persists.\n\nI somehow want to access myWrapper.m file from the MATLAB Function block. Any pointers on how this should be done? Following the previous link given and the official docs I am ending up with something like this in my MATLAB Function block:\n\nfunction y = fcn(u)coder.extrinsic('myWrapper')\n\ny = myWrapper(u);\n\n\nThe last code above is syntactically incorrect and I am at a loss on how it should be done. MATLAB automaticaly corrects the above code to:\n\nfunction y = fcn(u,coder,extrinsic, myWrapper )\n\ny = myWrapper(u);\n\n\nwhich is not what I want.\n\nAny tips and/or suggestions on how this could be done would be appreciated.\n\nA similar question was asked on the MathWorks forum here, two years ago, with no response.\n\n• This question would be better asked on the Mathworks forums. Short answer, not all MATLAB command are supported inside MATLAB function blocks. – scotty3785 Mar 20 '18 at 11:40\n• @scotty3785, thanks for your contribution. I asked the question over there and no answer. A similar question was asked on the Mathworks forums some years ago, and once again, no response. – Rrz0 Mar 20 '18 at 11:41\n\nI was going about tackling this problem completely wrong. In order to replace the summing block, one must NOT remove the Transfer Function blocks which feed into the summing block and add them inside the script.\n\nA MATLAB Function does not support code generation (and rightly so) such that a transfer function may be implemented inside it. That is why the blocks simply feed into the MATLAB Function as follows.", null, "The script would very simply be:\n\nfunction y1 = fcn(u1, u2, u3)\n\nx = (u1 + u2 +u3);\ny1 = x;\n\nend" ]	[ null, "https://i.stack.imgur.com/mOYZP.png", null, "https://i.stack.imgur.com/5sFWo.png", null, "https://i.stack.imgur.com/ZB4bg.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8438047,"math_prob":0.8910878,"size":1761,"snap":"2019-51-2020-05","text_gpt3_token_len":471,"char_repetition_ratio":0.1291975,"word_repetition_ratio":0.08391608,"special_character_ratio":0.27597955,"punctuation_ratio":0.14092141,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99036443,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,5,null,5,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-22T13:52:25Z\",\"WARC-Record-ID\":\"<urn:uuid:e6a6d6f5-50e8-47e6-b57c-3c7e13c74f5c>\",\"Content-Length\":\"138546\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d4109369-73fb-456a-9a14-6f7ea3f23d63>\",\"WARC-Concurrent-To\":\"<urn:uuid:6d40e0e0-bcbe-4329-9ba6-e8e6acf55789>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://electronics.stackexchange.com/questions/363601/issue-with-simulink-does-not-support-code-generation\",\"WARC-Payload-Digest\":\"sha1:JXK6IUXUNLQR3FRM3JTXGUGPEOLLWRT6\",\"WARC-Block-Digest\":\"sha1:5JUGVXJYYDXIAAD4OVYS7GSXFL7FGCSW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250607118.51_warc_CC-MAIN-20200122131612-20200122160612-00460.warc.gz\"}"}
https://statsidea.com/normalize-information-in-sas/	[ "# Normalize Information in SAS\n\nTo “normalize” a collection of information values approach to scale the values such that the ruthless of the entire values is 0 and the usual bypass is 1.\n\nThis educational explains how you can normalize knowledge in SAS.\n\n### Instance: Normalize Information in SAS\n\nThink we now have please see dataset:", null, "Carry out please see steps to normalize this eager of information values in SAS.\n\nStep 1: Form the Dataset\n\nFirst, let’s virtue please see code to build the dataset in SAS:\n\n```/build dataset/\nknowledge original_data;\nenter values;\ndatalines;\n12\n14\n15\n15\n16\n17\n18\n20\n24\n25\n26\n29\n32\n34\n37\n;\nrun;\n\n/view ruthless and same old bypass of dataset/\nproc approach knowledge=original_data Ruthless StdDev ndec=3;\nvar values;\nrun;```", null, "From the output we will see that the ruthless of the dataset is 22.267 and the usual bypass is 7.968.\n\nStep 2: Normalize the Dataset\n\nLater, we’ll virtue proc stdize to normalize the dataset:\n\n```/normalize the dataset/\nproc stdize knowledge=original_data out=normalized_data;\nvar values;\nrun;\n\n/print normalized dataset/\nproc print knowledge=normalized_data;\n\n/view ruthless and same old bypass of normalized dataset/\nproc approach knowledge=normalized_data Ruthless StdDev ndec=2;\nvar values;\nrun;```", null, "From the output we will see that the ruthless of the normalized dataset is 0 and the usual bypass is 1.\n\nStep 3: Interpret the Normalized Information\n\nSAS old please see formulation to normalize the knowledge values:\n\nNormalized worth = (x – x) / s\n\nthe place:\n\n• x = knowledge worth\n• x = ruthless of dataset\n• s = same old bypass of dataset\n\nEach and every normalized worth tells us what number of same old deviations the actual knowledge worth used to be from the ruthless.\n\nAs an example, imagine the knowledge level “12” in our actual dataset. The actual pattern ruthless used to be 22.267 and the actual pattern same old bypass used to be 7.968.\n\nThe normalized worth for “12” became out to be -1.288, which used to be calculated as:\n\nNormalized worth = (x – x) / s = (12 – 22.267) / 7.968 = -1.288\n\nThis tells us that the price “12” is 1.288 same old deviations under the ruthless within the actual dataset.\n\nEach and every of the normalized values within the dataset can backup us know the way alike or some distance a specific knowledge worth is from the ruthless.\n\nA tiny normalized worth signifies {that a} worth is alike to the ruthless occasion a massive normalized worth signifies {that a} worth is some distance from the ruthless.\n\n### Extra Assets\n\nRefer to tutorials provide an explanation for how you can carry out alternative usual duties in SAS:\n\nSignificance Proc Abstract in SAS\nCalculate Correlation in SAS\nForm Frequency Tables in SAS" ]	[ null, "https://www.statology.org/wp-content/uploads/2021/12/norm1.jpg", null, "https://www.statology.org/wp-content/uploads/2021/12/norm3.jpg", null, "https://www.statology.org/wp-content/uploads/2021/12/norm2.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79153585,"math_prob":0.8937471,"size":2654,"snap":"2023-14-2023-23","text_gpt3_token_len":616,"char_repetition_ratio":0.18679245,"word_repetition_ratio":0.11210762,"special_character_ratio":0.24717408,"punctuation_ratio":0.0998004,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9868519,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,7,null,7,null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-02T02:31:25Z\",\"WARC-Record-ID\":\"<urn:uuid:75adedd4-c682-427a-85c6-c52058beda24>\",\"Content-Length\":\"55850\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4d82b5f9-4b46-4510-b917-d03a0a14e85d>\",\"WARC-Concurrent-To\":\"<urn:uuid:64cb3609-b5be-41cf-8e1e-7805a2dd0632>\",\"WARC-IP-Address\":\"103.157.97.104\",\"WARC-Target-URI\":\"https://statsidea.com/normalize-information-in-sas/\",\"WARC-Payload-Digest\":\"sha1:NQPSPVL77DNDEZFORHD3QGL3LETFFWFO\",\"WARC-Block-Digest\":\"sha1:EBO2H66NNB3SGF752UEB42BXFQUBW22J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224648245.63_warc_CC-MAIN-20230602003804-20230602033804-00475.warc.gz\"}"}
https://www.bartleby.com/questions-and-answers/water-flows-uniformly-in-a-trapezoidal-channel-with-a-14ftwide-bed-and-2.21-m-2.2-side-slopes.-if-th/90764a28-96b6-470f-bfee-7e16700505ef	[ "# Water flows uniformly in a trapezoidal channel with a 14-ft-wide bed and 2.2:1 (m = 2.2) side slopes. If the bed slope is 1.1 ft/mile and n = 0.015, find the flow rate when the depth is 9 ft.\n\nQuestion\n\nWater flows uniformly in a trapezoidal channel with a 14-ft-wide bed and 2.2:1 (m = 2.2) side slopes. If the bed slope is 1.1 ft/mile and n = 0.015, find the flow rate when the depth is 9 ft.\n\n### Want to see this answer and more?\n\nExperts are waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes!*", null, "" ]	[ null, "https://www.bartleby.com/static/bartleby-logo-tm-tag-inverted.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8637543,"math_prob":0.95295846,"size":2009,"snap":"2021-04-2021-17","text_gpt3_token_len":504,"char_repetition_ratio":0.10374065,"word_repetition_ratio":0.0,"special_character_ratio":0.2543554,"punctuation_ratio":0.21538462,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97677463,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-24T19:47:37Z\",\"WARC-Record-ID\":\"<urn:uuid:8a6d6256-ec53-4af3-a392-35924fcfffbb>\",\"Content-Length\":\"151823\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:10904bb1-6c15-4e3e-8112-6c0d35be6b7c>\",\"WARC-Concurrent-To\":\"<urn:uuid:449b366c-413e-4637-8084-efe92cea1577>\",\"WARC-IP-Address\":\"13.32.204.120\",\"WARC-Target-URI\":\"https://www.bartleby.com/questions-and-answers/water-flows-uniformly-in-a-trapezoidal-channel-with-a-14ftwide-bed-and-2.21-m-2.2-side-slopes.-if-th/90764a28-96b6-470f-bfee-7e16700505ef\",\"WARC-Payload-Digest\":\"sha1:Y6FRB7FYHV5LC6SNI6YYJFDRWHA726UX\",\"WARC-Block-Digest\":\"sha1:TX7U55LUFFTQPDLZCVMVWBJMUFKUB4XH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703550617.50_warc_CC-MAIN-20210124173052-20210124203052-00373.warc.gz\"}"}