Copy of s11 - Solutions to homework 11 1 Since f is twice...

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Unformatted text preview: Solutions to homework # 11. 1. Since f is twice differentiable, by Theorem 5.2 f and f are continuous on ( a, ∞ ) and f 00 exists on ( a, ∞ ). So, for any x ∈ ( a, ∞ ) and h > 0, we apply Taylor’s formula to α := x and β = x + 2 h and obtain f ( x + 2 h ) = f ( x ) + 2 hf ( x ) + 2 h 2 f 00 ( ξ ) for some ξ ∈ ( x, x + 2 h ) . This yields f ( x ) = ( f ( x + 2 h )- f ( x )) / 2 h- hf 00 ( ξ ) . (1) If M = sup | f ( x ) | , M 2 = sup | f 00 ( x ) | , then (1) implies | f ( x ) | ≤ ( M + M ) / 2 h + hM 2 for all x ∈ ( a, ∞ ). Taking the supremum over x therefore gives us M 1 ≤ M /h + hM 2 , with h > being arbitrary. Taking h to be q M /M 2 , we get M 1 ≤ 2 q M M 2 , hence M 2 1 ≤ 4 M M 2 since all involved quantities are positive. 2. Pick ε > 0 and find δ > 0 so that | α ( x )- α ( x ) | < ε/ 2 whenever | x- x | ≤ δ, x ∈ [ a, b ] . Pick a partition P = { ξ , .. . ,ξ n } of [ a, b ] such that each subinterval has length at most δ and no ξ i equals...
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This document was uploaded on 11/17/2010.

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Copy of s11 - Solutions to homework 11 1 Since f is twice...

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