《数值分析》课程PPT教学课件（英文版）Chapter 01 The Solution of Nonlinear Equations 1.6 Padé Approximation

1.6 Pade Approximation

1.6 Padé Approximation

PN(a) NM()= QM(a) fora<x≤b (1.98) N(a)=Po+P1.+p2 22+.+pN. A PN( 1.99 QM(x)=1+91x+92x2+…+grxn 1.100

f(x)=a0+a1+2x2+…+akx+ (1.101) and form the difference f(a)QM()-PN(a)=2() N ∑1∑9-∑=-∑ (1.102) N+M+1

p=0 g1+a1-p1=0 q200+g101+a2-P2=0 g3+g21+qQ2+a3-3=0 ④MN=M+gM-1(N-M4+1+…+aN-p=0 an MN=M+1+gM-10N-M+2+…+g10N+aN+1=0 ④MON=M+2+91M1-1N-M4+3+…+g1N+1+aN+2=0 qMaN+9M-10N+1+…+9aN+M-1+aN+M=0

Example 1.17. Establish the pade approximation 15,120-6900x2+3134 Cos(x)≈R4(x)= 15,120+660x2+134

R44(X) cos(x) 08 Figure 1.18 Figure 1. 18 The graph of y=cos(a)and its Pade approximation R4.(a)

1.6.1 Continued fraction Form

1.6.1 Continued Fraction Form

15,120-69002+313x Cos(x)≈R14(x) 15,120+660x2+13x4 (1.105) 1753.13609467 R14(x)=2.07692308 354593883+2+620.1992877.30984207+x2

y=ER(X) 0.6-04-0.2 0.8 Figure 1.19(a) Figure 1. 19 (a) Graph of the error ER(a)=cos(a)-R4.(a) for the Pade approximation R4,(a)

Y=E (X) 0.8 .6 0.4 0.2 .2 0.6 08 Figure 1.19(b) Fi gure 1.1 9 (b) Graph of the error Ep()= cos()-P6(r for the Taylor approximation P6 (a)