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

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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 Taylors formula to := x and = x + 2 h and obtain f ( x + 2 h ) = f ( x ) + 2 hf ( x ) + 2 h 2 f 00 ( ) for some ( x, x + 2 h ) . This yields f ( x ) = ( f ( x + 2 h )- f ( x )) / 2 h- hf 00 ( ) . (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 > 0 and find > 0 so that | ( x )- ( x ) | < / 2 whenever | x- x | , x [ a, b ] . Pick a partition P = { , .. . , n } of [ a, b ] such that each subinterval has length at most and no i equals...
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Copy of s11 - Solutions to homework # 11. 1. Since f is...

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