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

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