solving systems of linear equations using substitution worksheet 8 2
Tài liệu Đề tài " Isomonodromy transformations of linear systems of difference equations" pptx
... example, for n = 2 we havep2s1p2s= p12sp12s+1,t(p12s)=t(p2s),t(p12s+1)=t(p2s−1),s∈ Z,p2sp2s+1= p 2 2s+1p 2 2s +2 ,t(p 2 2s+1)=t(p2s+1),t(p 2 2s +2 )=t(p2s),s∈ Z.It ... maps and integrable dynamics, Phys. Lett. A 314 (20 03), 21 4 22 1.(Received October 4, 20 02) ISOMONODROMY TRANSFORMATIONS1149Proof. As in the proof of Proposition 1.1, we may assume that n =0.Comparing ... [k1,k 2 ] ···[k2s−1,k2s]. As a function in the endpoints k1, ,k2s∈{0, 1, ,N}; this Annals of Mathematics Isomonodromy transformations of linear systems of difference equations ...
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Tài liệu Solution of Linear Algebraic Equations part 1 docx
... this:a11x1+ a 12 x 2 + a13x3+ ···+a1NxN=b1a 21 x1+ a 22 x 2 + a 23 x3+ ···+a2NxN=b 2 a31x1+ a 32 x 2 + a33x3+ ···+a3NxN=b3··· ···aM1x1+aM2x 2 +aM3x3+···+aMNxN= ... 1 -80 0 -87 2- 7 423 (North America only),or send email to trade@cup.cam.ac.uk (outside North America).Chapter 2. Solution of Linear Algebraic Equations 2. 0 IntroductionA set of linear algebraic equations ... RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by Cambridge University Press.Programs Copyright (C) 1 988 -19 92 by Numerical Recipes Software. Permission...
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Tài liệu Solution of Linear Algebraic Equations part 2 ppt
... equationa11a 12 a13a14a 21 a 22 a 23 a 24 a31a 32 a33a34a41a 42 a43a44·x11x 21 x31x41x 12 x 22 x 32 x 42 x13x 23 x33x43y11y 12 y13y14y 21 y 22 y 23 y 24 y31y 32 y33y34y41y 42 y43y44=b11b 21 b31b41b 12 b 22 b 32 b 42 b13b 23 b33b431000010000100001 (2. 1.1)Here ... equationa11a 12 a13a14a 21 a 22 a 23 a 24 a31a 32 a33a34a41a 42 a43a44·x11x 21 x31x41x 12 x 22 x 32 x 42 x13x 23 x33x43y11y 12 y13y14y 21 y 22 y 23 y 24 y31y 32 y33y34y41y 42 y43y44=b11b 21 b31b41b 12 b 22 b 32 b 42 b13b 23 b33b431000010000100001 (2. 1.1)Here ... RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by Cambridge University Press.Programs Copyright (C) 1 988 -19 92 by Numerical Recipes Software. Permission...
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Tài liệu Solution of Linear Algebraic Equations part 11 ppt
... matrices,a11a 12 a 21 a 22 Ãb11b 12 b 21 b 22 =c11c 12 c 21 c 22 (2. 11.1)Eight, right? Here they are written explicitly:c11= a11ì b11+ a 12 ì b 21 c 12 = a11ì b 12 + a 12 ì b 22 c 21 = a 21 ì ... set of formulas was, in fact, discovered by Strassen[1]. The formulas are:Q1 (a11+ a 22 ) ì (b11+ b 22 )Q 2 (a 21 + a 22 ) ì b11Q3 a11ì (b 12 b 22 )Q4 a 22 ì (b11+ b 21 )Q5 ... b11+ a 12 ì b 21 c 12 = a11ì b 12 + a 12 ì b 22 c 21 = a 21 ì b11+ a 22 ì b 21 c 22 = a 21 ì b 12 + a 22 ì b 22 (2. 11 .2) Do you think that one can write formulas for the c’s that involve only...
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Tài liệu Solution of Linear Algebraic Equations part 3 pdf
... 42 Chapter 2. Solution of Linear Algebraic Equations Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by Cambridge ... isxi=1aiibi−Nj=i+1aijxj (2. 2.4)The procedure defined by equation (2. 2.4) is called backsubstitution.Thecom-bination of Gaussian elimination and backsubstitution yields a solution to the set of equations. The ... ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by Cambridge University Press.Programs Copyright (C) 1 988 -19 92 by Numerical Recipes Software. Permission is granted for...
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Tài liệu Solution of Linear Algebraic Equations part 4 docx
... j : αi1β1j+ αi2β2j+ ···+αiiβjj= aij (2. 3.9)i>j: αi1β1j+αi2β2j+···+αijβjj= aij (2. 3.10) Equations (2. 3 .8) – (2. 3.10) total N 2 equations for the N 2 + N unknown α’s ... β 22 β 23 β 24 00β33β34000β44=a11a 12 a13a14a 21 a 22 a 23 a 24 a31a 32 a33a34a41a 42 a43a44 (2. 3 .2) We can use a decomposition such as (2. 3.1) to solve the linear setA · x =(L·U)·x=L·(U·x)=b (2. 3.3)by ... case of a 4 ì 4 matrix A, for example, equation (2. 3.1) would look like this:α11000α 21 α 22 00α31α 32 α330α41α 42 α43α44·β11β 12 β13β140 β 22 β 23 β 24 00β33β34000β44=a11a 12 a13a14a 21 a 22 a 23 a 24 a31a 32 a33a34a41a 42 a43a44 (2. 3 .2) We...
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Tài liệu Solution of Linear Algebraic Equations part 5 docx
... aNbN·u1u 2 ···uN−1uN=r1r 2 ···rN−1rN (2. 4.1) 2. 4 Tridiagonal and Band Diagonal Systems of Equations 51Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by ... 2. 4 Tridiagonal and Band Diagonal Systems of Equations 53Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by Cambridge ... ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by Cambridge University Press.Programs Copyright (C) 1 988 -19 92 by Numerical Recipes Software. Permission is granted for...
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Tài liệu Solution of Linear Algebraic Equations part 12 pdf
... Inverse(a11)R 2 = a 21 ì R1R3= R1ì a 12 R4= a 21 ì R3R5= R4 a 22 R6= Inverse(R5)c 12 = R3ì R6c 21 = R6ì R 2 R7= R3ì c 21 c11= R1 R7c 22 = R6 (2. 11.6) ... call 1 -80 0 -87 2- 7 423 (North America only),or send email to trade@cup.cam.ac.uk (outside North America).in terms of whichc11= Q1+ Q4− Q5+ Q7c 21 = Q 2 + Q4c 12 = Q3+ Q5c 22 = Q1+ ... matricesa11a 12 a 21 a 22 andc11c 12 c 21 c 22 (2. 11.5)are inverses of each other. Then the c’s can be obtained from the a’s by the followingoperations (compare equations 2. 7 .22 and 2. 7 .25 ):R1= Inverse(a11)R 2 = a 21 ì R1R3=...
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Tài liệu Solution of Linear Algebraic Equations part 6 pptx
... reassuring. 58 Chapter 2. Solution of Linear Algebraic Equations Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by Cambridge ... 56Chapter 2. Solution of Linear Algebraic Equations Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by Cambridge University ... Machinery). [1] 2. 5 Iterative Improvement of a Solution to Linear Equations 57Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by Cambridge...
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Tài liệu Solution of Linear Algebraic Equations part 7 docx
... (scale) { 62 Chapter 2. Solution of Linear Algebraic Equations Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by Cambridge ... n ≤ N (2. 6.3) 66Chapter 2. Solution of Linear Algebraic Equations Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by Cambridge ... RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by Cambridge University Press.Programs Copyright (C) 1 988 -19 92 by Numerical Recipes Software. Permission...
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Tài liệu Solution of Linear Algebraic Equations part 8 docx
... 1 2 3 4 5 6 7 8 9 10 11ija[k] 7 8 8 10 11 12 3 2 4 5 4sa[k] 3. 4. 5. 0. 5. x 1. 7. 9. 2. 6. (2. 7. 28 ) Here x is an arbitrary value. Notice that, according to the storage rules, the value of ... inverseA−1→ A−1−z ⊗ w1+λ (2. 7.5) 2. 7 Sparse Linear Systems 83 Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by Cambridge University ... multiplication of its transpose and a vector. As 2. 7 Sparse Linear Systems 81 Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by Cambridge...
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Enclosing solutions of systems of equations
... attractorNDSolve4.1540 680 321 2944451.6 484 28 5 51076 28 2 25 .9 721 0749 686 6006(13)IDSolve[ −57.0 724 33 985 95995, 79 . 82 067 089 488 7 58 ][ 25 3.4101 326 6 12 387 2, 24 2.97953 722 353 984 ][ 21 0.347760367 329 9, 29 3.06501763 924 2 ... number):Mathematica1 .81 26 53 486 19 726 650.97 089 484 8346366 2. 8. 8910653497975930.906 326 7430 986 333 (22 )Algorithm 2 1 .81 26 53 486 4 526 2[63, 76]0.97 089 484 8 28 5 20 50[0, 6]1.999999999999999[3, ... are:Mathematica7 .25 351756 183 085 0 . 82 507349 120 27 126 4.6 486 983 64 72 580 56−5 .86 54314 28 8 10479 2. 0767 326 60 789 20 9 (24 )Algorithm 2 7 .25 3517570611[5 42, 623 ]0 . 82 50734961640 [26 9, 723 ]4.6 486 983 6 781 35[045, 440]−5 .86 5431440461[656, 85 6] 2. 0767 326 619734[114,...
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Tài liệu Solution of Linear Algebraic Equations part 9 docx
... xj.Inotherwords,Pj(xi)=δij=Nk=1Ajkxk−1i (2 .8. 4)But (2 .8. 4) saysthat Ajkis exactly the inverse of the matrix of componentsxk−1i,whichappears in (2 .8. 2) , with the subscript as the column index. Therefore the solution of (2 .8. 2) is ... +1−j (2 .8. 25 )Now, starting with the initial valuesx(1)1= y1/R0G(1)1= R−1/R0H(1)1= R1/R0 (2 .8. 26 )we can recurse away. At each stage M we use equations (2 .8. 23 ) and (2 .8. 24 ) ... 92 Chapter 2. Solution of Linear Algebraic Equations Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by Cambridge...
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Tài liệu Solution of Linear Algebraic Equations part 10 docx
... +1,i +2, ,N (2. 9.5) 92 Chapter 2. Solution of Linear Algebraic Equations Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0- 521 -431 08- 5)Copyright (C) 1 988 -19 92 by ... +1−j (2 .8. 25 )Now, starting with the initial valuesx(1)1= y1/R0G(1)1= R−1/R0H(1)1= R1/R0 (2 .8. 26 )we can recurse away. At each stage M we use equations (2 .8. 23 ) and (2 .8. 24 ) ... xj.Inotherwords,Pj(xi)=δij=Nk=1Ajkxk−1i (2 .8. 4)But (2 .8. 4) says that Ajkis exactly the inverse of the matrix of componentsxk−1i,whichappears in (2 .8. 2) , with the subscript as the column index. Therefore the solution of (2 .8. 2) is...
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