solhw6

solhw6 - Solutions for Homework 6 Foundations of...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solutions for Homework 6 Foundations of Computational Math 2 Spring 2011 Problem 6.1 Consider a minimax approximation to a function f ( x ) on [ a, b ]. Assume that f ( x ) is contin- uous with continuous first and second order derivatives. Also, assume that f ′′ ( x ) < 0 on for a ≤ x ≤ b , i.e., f is concave on the interval. 6.1.a . Derive the equations you would solve to determine the linear minimax approx- imation, p 1 ( x ) = αx + β , to f ( x ) on [ a, b ] and describe their use to solve the problem. 6.1.b . Apply your approach to determine p 1 ( x ) = αx + β for f ( x ) = − x 2 on [ − 1 , 1]. 6.1.c . How does p 1 ( x ) relate to the quadratic monic Chebyshev polynomial t 2 ( x )? 6.1.d . Apply your approach to determine ˜ p 1 ( x ) = ˜ αx + ˜ β for f ( x ) = − x 2 on [0 , 1]. 6.1.e . How could the quadratic monic Chebyshev polynomial t 2 ( y ) on − 1 ≤ y ≤ 1 be used to provide and alternative derivation of ˜ p 1 ( x ) on 0 ≤ x ≤ 1? 6.1.f . Suppose you adapt your approach to derive a constant approximation, p ( x ). What points will you use as the extrema of the error? Solution: We have f ( x ) ∈ C 2 [ a, b ] with f ′′ ( x ) < 0 on [ a, b ]. Define the error function e ( x ) = f ( x ) − αx − β e ′ ( x ) = f ′ ( x ) − α e ′′ ( x ) = f ′′ ( x ) < So e ( x ) is also concave on [ a, b ]. It therefore has at most one point, x = c , in the interval where e ′ ( c ) = 0. Determining p 1 ( x ) requires three points where e ( x ) is maximal and of alternating sign. Since e ( x ) is concave the potential extrema are a , b , and c (if it exists). Assume that c exists and that a < c < b . We then have the equations e ( a ) = − e ( c ) e ( b ) = − e ( c ) or e ( a ) = e ( b ) f ′ ( c ) = α to determine the unknowns α , β , and c . The discussion of f ( x ) = e x on [0 , 1] in the notes is an example of this case. 1 Consider f ( x ) = − x 2 on [ a, b ]. f ′ ( c ) = α → c = − α 2 e ( a ) = e ( b ) → − a 2 − αa = − b 2 − αb → α = ( a 2 − b 2 ) ( b − a ) → c = 1 2 ( a + b ) − e ( a ) = e ( c ) → a 2 + αa + β = − c 2 − αc − β → β = α 2 8 − a α 2 − a 2 2 Taking a = − 1 and b = 1 yields: α = 0 c = 0 β = − 1 2 p 1 ( x ) = p ( x ) = − 1 2 It is easy to verify that e ( − 1) = e (1) = − 1 / 2 and e ( c ) = e (0) = 1 / 2 satisfying the necessary and sufficient minimax conditions....
View Full Document

This note was uploaded on 11/10/2011 for the course MAD 5404 taught by Professor Gallivan during the Spring '11 term at FSU.

Page1 / 7

solhw6 - Solutions for Homework 6 Foundations of...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online