1.6 Pade Approximation
1.6 Padé Approximation
NM for a< x< b PN(a)=Po+P1.+P222+.+pN. 1.99 QM()=1+91.C+9204+.+9M2 1.100
f(a)=a0+a1.2+a2. 02+ k and form the difference f(a)QM(a)-PN(a)=z(a) ∑x1(∑g)-∑ ∑ 1.102) j=N+M+1
0 =0 g100+a1-pi q20+q101+a2-P2=0 (1.103) q30+q21+y102+a3-p3=0 gMaN-M+ 9M-1aN-M+1+.+aN-PN ane gMaN-M+1+9M-1aN-M2+.+gIaN +aN+1=0 ④MN-M+2+M=1N-M+3+…+91N+1+aN+2=0 1.104 gMaN +9M-1aN+1+.+glaN+M-1+aN+M=0
Example 1.17. Establish the pade approximation 15,120-69002+313x Cos(x)≈R414(x)= 15,120+660x2+134