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
06 y=R44(X) cos(X) Figure 1.18 Figure 1. 18 The graph of y= cos()and its Pade approximation R4.4(r)
1.6.1 Continued fraction For
1.6.1 Continued Fraction Form
cos(r)≈R4()= 15,120-69002+3132 15,120+660x2+13 1105) 1753.13609467 R14(x)=24.07692308 35.45938873+x2+620.19922877(5.30984207+x2
y=E。(X) 08 .2 Figure 1.19(a) Figure 1.19(a) Graph of the error ER(a)=cos(a)-R4, (a) for the Pade approximation R4. 4 (a)
X10 y=E(x) 0.8 0.2 .8 Figure 1.19 (b) Figure 1.19(b) Graph of the error Ep(a)= cos (a)-P6(a) for the Taylor approximation P6(a)
