Rational Function Approximation - Lecture Slides | MATH 375, Study Guides, Projects, Research of Mathematical Methods for Numerical Analysis and Optimization

Download Rational Function Approximation - Lecture Slides | MATH 375 and more Study Guides, Projects, Research Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity!

Rational Function Approximation

MATH 375

J. Robert Buchanan

Department of Mathematics

Fall 2006

Introduction

So far we have used polynomials to approximate arbitrary continuous functions. Pros: easy to evaluate easy to integrate and differentiate Cons: tend to oscillate error estimates can be much larger than the actual error

Assumptions

Assume the interval over which we are approximating f ( x ) contains x = 0.

r ( 0 ) = p ( 0 ) q ( 0 )

p 0 q 0

=⇒ q 0 6 = 0

Assume q 0 = 1

r ( x ) = p 0 + p 1 x + · · · + pnxn q 0 + q 1 x + · · · + qnxm = ( p 0 + p 1 x + · · · + pnxn )/ q 0 ( q 0 + q 1 x + · · · + qnxm )/ q 0

=

ˆ p ( x ) ˆ q ( x ) where^ q ˆ 0 = 1

Padé Technique

If r ( x ) is of degree N , then r ( x ) contains N + 2 coefficients. q 0 = 1 leaving N + 1 coefficients undetermined.

Theorem (Padé Technique) Given f ( x ) and a rational function r ( x ) with q 0 = 1 , choose the remaining N + 1 coefficients of r ( x ) such that f ( n )( 0 ) = r ( n )( 0 ) for n = 0 , 1 ,... , N.

Multiple Root

Definition A solution p of f ( x ) = 0 is a root of multiplicity m of f if for x 6 = p we can write f ( x ) = ( x − p ) mq ( x ), where lim x → p q ( x ) 6 = 0.

Theorem A function f ∈ C m [ a , b ] has a root of multiplicity m at p ∈ ( a , b ) iff

0 = f ( p ) = f ′( p ) = · · · = f ( m −^1 )( p ) but f ( m )( p ) 6 = 0.

Remark: The design goal is to select the coefficients of r ( x ) so that f ( x ) − r ( x ) has a root of multiplicity N + 1 at x = 0. Equivalently ( (^) ∞ ∑

k = 0

ak xk

) ( (^) m ∑

k = 0

qk xk

∑^ n

k = 0

pk xk

has no terms of degree less than N + 1.

Coefficient List

Consider ( (^) ∞ ∑

k = 0

ak xk

) ( (^) m ∑

k = 0

qk xk

∑^ n

k = 0

pk xk^ =

( a 0 + a 1 x + a 2 x^2 · · · )( 1 + q 1 x + · · · + qmxm ) − ( p 0 + p 1 x + · · · + pnxn )

n Coefficient of xn 0 a 0 − p 0 1 a 0 q 1 + a 1 − p 1 2 a 0 q 2 + a 1 q 1 + a 2 − p 2 .. .

k − pk +

∑^ k

i = 0

ai qk − i

for k = 0 , 1 ,... , N.

Example

Consider f ( x ) = x ln( x + 1 ). Compare the Maclaurin polynomial approximation of degree 6 to the Padé rational approximation of degree 6. Maclaurin Polynomial:

P 6 ( x ) = x^2 −

x^3 +

x^4 −

x^5 +

x^6

Let m = n = 3.

( x^2 −

x^3 +

x^4 −

x^5 +

x^6 )( 1 + q 1 x + q 2 x^2 + q 3 x^3 )

− ( p 0 + p 1 x + p 2 x^2 + p 3 x^3 )

Example Linear System

− p 0 = 0 − p 1 = 0 1 − p 2 = 0 −

− p 3 + q 1 = 0 1 3

q 1 + q 2 = 0

−

q 1 −

q 2 + q 3 = 0 1 5

q 1 +

q 2 −

− q 3 = 0

Solution:

p 0 = 0 , p 1 = 0 , p 2 = 1 , p 3 =

, q 1 =

, q 2 =

, q 3 = −

Continued Fraction

r ( x ) =

x^2 + 1930 x^3 1 + 1715 x + 307 x^2 − 901 x^3

=

90 x^2 + 57 x^3 90 + 102 x + 21 x^2 − x^3

= − 57 + 1287 x^2 + 5814 x + 5130 90 + 102 x + 21 x^2 − x^3 = − 57 +

90 + 102 x + 21 x^2 − x^3 1287 x^2 + 5814 x + 5130 .. .

Continued Fraction (continued)

r ( x ) = − 57 −

x − 3649143 +

(^172671859) x + 80407222469181 − 46898883159756517691410 x + (^239520189937) This form requires only 3 MUL/DIV and 6 ADD/SUB.

Error Comparison (continued)

x | P 6 ( x ) − f ( x )| | r ( x ) − f ( x )|

1. 0 0. 0 0. 0
2. 2 1. 82197 × 10 −^6 1. 54536 × 10 −^8
3. 4 2. 03639 × 10 −^4 1. 24973 × 10 −^6
4. 6 3. 08902 × 10 −^3 1. 41712 × 10 −^5
5. 8 0. 02081 7. 32780 × 10 −^5
6. 0 0. 09019 2. 49046 × 10 −^4

Project Example

Design a rational function approximation to ex^ that is valid for x ∈ R and determine the maximum error. We begin by constructing a rational function approximation to ex^ where m = n = 3 and determining the maximum error on the interval [− ln 2, ln 2]. Maclaurin polynomial:

P 6 ( x ) = 1 + x +

x^2 +

x^3 +

x^4 +

x^5 +

x^6

Approximation

r ( x ) =

1 + x 2 + x 2 10 +^

x^3 120 1 − x 2 + x 2 10 −^

x^3 120

-0.6 -0.4 -0.2 0.2 0.4 0. x

2.5· 10 -

5 · 10 -

7.5· 10 -

error

Error Comparison

• − 0 693147 1 40263 × 10 −^5 3 88490 × 10 − x | P 6 ( x ) − f ( x )| | r ( x ) − f ( x )|
• − 0 593147 4 76879 × 10 −^6 1 43551 × 10 −
• − 0 493147 1 32514 × 10 −^6 4 33816 × 10 −
• − 0 393147 2 74481 × 10 −^7 9 77895 × 10 −
• − 0 293147 3 56024 × 10 −^8 1 38127 × 10 −
• − 0 193147 1 94268 × 10 −^9 8 21306 × 10 −
• − 0 093147 1 19323 × 10 −^11 5 50337 × 10 −
• 0 0068528 2 22045 × 10 −^16 2 22045 × 10 −
• 0 1068528 3 19818 × 10 −^11 1 75637 × 10 −
• 0 2068528 3 30022 × 10 −^9 1 98028 × 10 −
• 0 3068528 5 28434 × 10 −^8 3 46660 × 10 −
• 0 4068528 3 85623 × 10 −^7 2 76750 × 10 −
• 0 5068528 1 81951 × 10 −^6 1 42945 × 10 −
• 0 6068528 6 50278 × 10 −^6 5 59607 × 10 −