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