Tài liệu Integral Equations and Inverse Theory part 1 pptx

... variable t. 790 Chapter 18 . Integral Equations and Inverse Theory Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92 by Cambridge ... kernel. Equation (18 .0 .1) is analogous to the matrix equation K · f = g (18 .0.2) whose solution is f = K 1 · g,whereK 1 is the matrix inverse. Like equation (18 .0.2...

Ngày tải lên: 21/01/2014, 18:20

... (18 .1. 2) to equation (18 .1. 1), we get f(t)=λ N  j =1 w j K(t, s j )f(s j )+g(t) (18 .1. 3) Evaluate equation (18 .1. 3) at the quadrature points: f(t i )=λ N  j =1 w j K(t i ,s j )f(s j )+g(t i ) (18 .1. 4) Let ... [4] Smithies, F. 19 58, Integral Equations (Cambridge, U.K.: Cambridge University Press). Kanwal, R.P. 19 71, Linear Integral Equations (New York: Academic Press...

Ngày tải lên: 21/01/2014, 18:20

... diag(w j ) and its square root, D 1/ 2 = diag( √ w j ). Then equation (18 .1. 7) becomes K · D · f = σf Multiplying by D 1/ 2 ,weget  D 1/ 2 ·K·D 1/ 2  ·h=σh (18 .1. 8) where h = D 1/ 2 ·f. Equation (18 .1. 8) ... equation (4 .1. 11) :  t i a K(t i ,s)f(s)ds = h   1 2 K i0 f 0 + i 1  j =1 K ij f j + 1 2 K ii f i   (18 .2.3) Thus the trapezoidal method for equation (18 .2...

Ngày tải lên: 21/01/2014, 18:20

... result  b a w(x)f (x)dx = 1 6 f (kh)  (k +1) (k+2)(k+3)W 0 −(3k 2 +12 k+ 11 )W 1 +3(k+2)W 2 −W 3  + 1 2 f([k +1] h)  −k(k+2)(k+3)W 0 +(3k 2 +10 k+6)W 1 −(3k +5)W 2 +W 3  + 1 2 f([k +2]h)  k(k +1) (k+3)W 0 −(3k 2 +8k+3)W 1 +(3k+4)W 2 −W 3  + 1 6 f([k ... formula is  b a w(x)f(x)dx = 1 2 f (kh)  (k +1) (k+2)W 0 −(2k +3)W 1 +W 2  +f([k +1] h)  −k(k+2)W 0 +2(k +1)...

Ngày tải lên: 26/01/2014, 15:20

... r i (x µ )(x µ +1 − x µ 1 )/2 (18 .4.8) 806 Chapter 18 . Integral Equations and Inverse Theory Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92 ... data, χ 2 = N  i =1 N  j =1  c i − M  µ =1 R iµ u(x µ )  S 1 ij  c j − M  µ =1 R jµ u(x µ )  ≈ N  i =1  c i −  M µ =1 R iµ u(x µ ) σ i...

Ngày tải lên: 26/01/2014, 15:20

... =                   1 −33 10 0000··· 0 − 310 12 6 10 000··· 0 3 12 19 15 6 10 00··· 0 16 15 20 15 6 10 0··· 0 0 16 15 20 15 6 10 ··· 0 . . . . . . . . . 0 ··· 0 16 15 20 15 6 10 0··· 00 16 15 20 15 6 1 0 ... =               1 − 210 000··· 0 −25− 410 00··· 0 1 −46− 410 0··· 0 01 46− 410 ··· 0 . . . . . . . . . 0 ··· 01 46− 410 0··· 0 01 46− 41 0···...

Ngày tải lên: 26/01/2014, 15:20

... [u( x )] (18 .6 .1) 816 Chapter 18 . Integral Equations and Inverse Theory Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92 by ... q(x) ·  W(x)+λS  ·q(x) (18 .6 .11 ) subject to the constraint (18 .6.7) that q(x) · R =1. The solution of equation (18 .6 .11 ) is q(x)= [W(x)+λS] 1 ·R R·[W(x)+λS] 1...

Ngày tải lên: 17/10/2013, 22:15

... some y the linear system (W(x)+λS)·y =R,andthen substitute y into both the numerators and denominators of 18 .6 .12 or 18 .6 .13 .) Equations (18 .6 .12 ) and (18 .6 .13 ) have a completely different character ... exp  − 1 2 χ 2  exp[−H(u)] (18 .7 .10 ) or, equivalently, minimize: − ln [ Prob(u|c)]= 1 2 χ 2 [u]+H(u)= 1 2 χ 2 [u]+ M  µ =1 u µ ln(u µ /U) (18 .7 .11 ) This oug...

Ngày tải lên: 24/10/2013, 15:15

Ngày tải lên: 14/12/2013, 22:15

... IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92 by Cambridge University Press.Programs Copyright (C) 19 88 -19 92 by Numerical Recipes Software. Permission is ... IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92 by Cambridge University Press.Programs Copyright (C) 19 88 -19 92 by Numerical Recipes Softwa...

Ngày tải lên: 15/12/2013, 04:15