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Math Help - Prove inequality

  1. #1
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    Prove inequality

    Show that sin(x) = O(x) for all x in R.

    From the defn, we must show

    \exists A s.t. |\sin(x)| \le A|x|

    We know that |sin(x)| <= 1.

    But now I'm stuck.

    Similarly, how would I go about showing:

    \exists A>0,c\in \mathbb{R} s.t.

    |e^{-x}|\le A|x^{-m}|, m\in\mathbb{Z}^+,~c<x<\infty

    What's the general method for these things? (I know e^x is always positive, so the mod signs are redundant)

    (The above is derived from e^{-x}=O(x^{-m}), m positive integer.)
    Last edited by scorpion007; July 15th 2010 at 12:44 AM.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    If f(x) is uniformly continuous on [1,inf) then there exist M>0 with |f(x)|<Mx
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  3. #3
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    A little more elementary: By the mean value theorem, for any x, taking f(x)= x- sin(x), we have \frac{f(x)- f(0)}{x- 0}= f'(c) for some c between 0 and x. But f'(x)= 1- cos(x)\ge 0. Therefore, we have f(x)= x- sin(x)\ge 0 so x\ge sin(x).
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  4. #4
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    Thanks -- but you've shown that sin(x) <= x, but that doesn't necessarily imply that |sin(x)|<=|x|, since

    a < b \not\Rightarrow |a|<|b|, right?

    (PS. I don't think I've got enough background for uniform continuity, having not studied metric spaces or almost any functional analysis)
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    You can do something similar(or symmetrical ) to what Mr. HallsofIvy did.
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  6. #6
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    Yes, the first part of my reply was indeed directed at HallsOfIvy. I'm not convinced that it proves what was required, for the reason I mentioned.
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