Show that for every and ,

thanks

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- August 28th 2008, 08:08 PM #1

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- August 28th 2008, 08:32 PM #2

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Define and show . Now . If there was such that then since satisfies Rolle's theorem on it would mean there is a point such that and this is a contradiction. Thus, for . If there was such that then since satisfies IVT it means there would be such that . But this is a contradiction because is increasing on . Thus, .

- August 29th 2008, 01:48 PM #3

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well, if we prove that for , then we're done because that means is increasing and thus:

the inequality is clearly true for x = 0. so we assume that x > 0. so the claim is:

suppose for now that a > 1. since we have thus: hence which obviously holds

for a = 1 as well. therefore: which proves (1). Q.E.D.

- August 29th 2008, 05:48 PM #4

- August 29th 2008, 07:16 PM #5

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- September 8th 2008, 10:04 PM #6
A simple proof for the case .

By the Taylor's expansion, we have since and by the binomial expansion, we have

Using the fact that and the inequalities above, we get the resired result since is increasing in when