4InterpolationandPolynomialApproximation17-4.6PadeApproximationNumericalMiethods
4 Interpolation and Polynomial Approximation 17-4.6 PadeApproximation
Pade ApproximationsThe notion of rational approximations for f(x) that will be approximatedoverasmallportionofitsdomainArational approximationtof(x)on[a,bl isthequotientoftwopolynomialPn(x) and Qm(x)P(x)Rn,m(x)=fora≤x<bOm(x)GoalistomakethemaximumerrorassmallaspossibleThe method of pade requires that f(x) and its derivative be continuous at x =0.Pr(x)= Po + pix+ pax? +..-+ pnxOm(x)=1+qix+q2x2 +...+qmxmThe rational function Rn,m(x) has N+M+1 unknown coefficients. Assume f(x)isanalyticandhastheMaclaurinexpansiongf(x)=ao+ax+azxL...Formthedifferencef(x)Qm(x)-Pn(x)=Z(x)9/22/20265:57AM
9/22/2026 5:57 AM 2 ⚫ The notion of rational approximations for 𝒇(𝒙) that will be approximated over a small portion of its domain. ⚫ A rational approximation to 𝒇(𝒙) on [𝒂,𝒃] is the quotient of two polynomial 𝑷𝑵(𝒙) and 𝑸𝑴 𝒙 Goal is to make the maximum error as small as possible. The method of pade requires that 𝒇(𝒙) and its derivative be continuous at 𝒙 = 𝟎. The rational function 𝑹𝑵,𝑴(𝒙) has N+M+1 unknown coefficients. Assume 𝒇(𝒙) is analytic and has the Maclaurin expansiong Form the difference 𝒇 𝒙 𝑸𝑴 𝒙 − 𝑷𝑵 𝒙 = 𝒁(𝒙) 2 0 1 2 2 1 2 1 ( ) ( ) N N N M M M P x p p x p x p x Q x q x q x q x = + + + + = + + + + , ( ) ( ) ( ) N N M M P x R x for a x b Q x = 2 0 1 2 ( ) . k k f x a a x a x a x = + + + + +
PadeApproximationsZapx=.cThelower indexj=M+N+1inthe summation on the right sideof Eq.becausethe first N+M derivatives of f(x) and Rn.M(x) are to agree at x = 0. Set equaltozerofork=0,1,2,...,N+M to obtainN+M+1 linearequations.ao-Po=0q% +a -P, =0+aN+I=0qMaN-M+I+qM-iaN-M+2+...+q,aN2a + qia, +az- p2 = 0IMaN-M+2+qM-1N-M+3+...+q,aN+1+aN+2=0+q2a+qa+as-P,=0·+qm-Ian++ +.+qan+M-I+an+m =0IMaN-M+I +...+an -PN =0AMaN-M+AM-9/22/20265:57AM
9/22/2026 5:57 AM 3 The lower index j=M+N+1 in the summation on the right side of Eq. because the first N+M derivatives of 𝒇(𝒙) and 𝑹𝑵,𝑴(𝒙) are to agree at 𝒙 = 𝟎. Set equal to zero for k=0,1,2,.,N+M to obtain N+M+1 linear equations. 0 0 0 1 M N j j j j j j j j j j j j N M a x q x p x c x = = = = + + − = 0 0 1 0 1 1 2 0 1 1 2 2 3 0 2 1 1 2 3 3 0 0 0 0 a p q a a p q a q a a p q a q a q a a p − = + − = + + − = + + + − = 1 1 0 M N M M N M N N q a q a a p − − − + + + + − = 1 1 2 1 1 2 1 3 1 1 2 1 1 1 1 0 0 0 M N M M N M N N M N M M N M N N M N M N N M N M q a q a q a a q a q a q a a q a q a q a a − + − − + + − + − − + + + − + + − + + + + + = + + + + = + + + + =
PadeApproximationsEx.4.17EstablishthePadeapproximationR4.4(x)over[-5,5)15120-6900x2+313x4cos(x) ~ R(x) =15120+660x2+13x4Simplify the computations if we start with f(x) = cos(x1/2)f(x)=3Po-px-pax1+qix+q2x240320=0+0x+0x2+0x+0x4+c.x+cx+Po =11- p。 = 0=11/252P,=-115/252-1/2+q - p, = 092=13/15120P2=313/151201/24-1/2q, +q2- p, = 01/720+1/24q-1/2q2=01-115x/252+313x2/15120cos(x)= f(x)f(x)1+11x/252+13x2/151201/40320-1/720q,+1/24q2=09/22/20265:57AM
9/22/2026 5:57 AM 4 Ex. 4.17 Establish the Pade approximation 𝑹𝟒,𝟒(𝒙) over [−𝟓, 𝟓] Simplify the computations if we start with 𝒇 𝒙 = 𝒄𝒐𝒔(𝒙 𝟏/𝟐 ) 2 4 4 4 2 4 15120 6900 313 15120 660 13 , cos( ) ( ) x x x R x x x − + = + + 1 1 1 1 2 3 4 1 2 24 720 40320 f x x x x x ( ) = − + − + − ( ) 2 3 4 2 2 1 2 0 1 2 2 3 4 5 6 5 6 1 1 1 1 1 1 2 24 720 40320 0 0 0 0 0 x x x x q x q x p p x p x x x x x c x c x − + − + − + + − − − = + + + + + + + 0 1 1 1 2 2 1 2 1 2 1 0 1 2 0 1 24 1 2 0 1 720 1 24 1 2 0 1 40320 1 720 1 24 / / - / / / / / / / p q p q q p q q q q − = − + − = + − = − + − = − + = 0 1 2 11 252 13 15120 / / q q = = 0 1 2 1 115 252 313 15120 / / p p p = = − = 2 2 1 115 252 313 15120 1 11 252 13 15120 / / ( ) / / x x f x x x − + + + 2 cos( ) ( ) x f x =
Continued Fraction Form1.00.5X1-2-12T4-3345y=cos(x)0.5-1.0FFigure 4.18The graph of y = cos(x) and its PadéapproximationR4.4(x)9/22/20265:57AM
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