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Math Help - sequence of functions

  1. #1
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    sequence of functions

    **Problem:**

    prove that the sequence of functions \left \{ f_{n} \right \} defined by:
    f_{n}\left ( x \right )= n\sin\sqrt{4\pi ^{2}n^{2}+x^{2}} converges uniformly on \left [ 0,\alpha  \right ] where \alpha > 0.
    Does \left \{ f_{n} \right \} converges uniformly on \mathbb{R}

    Here is what I did: I proved that the pointwise limit of \left \{ f_{n} \right \} is the function f\left ( x \right )=\frac{x^{2}}{4\pi }. Then, in order to prove the uniform convergence, I need to prove that \sup_{x\in\left [ 0, \alpha  \right ] } \left \{ \left | f_{n}\left ( x \right )-f\left ( x \right ) \right |  \right \}n \to 0 as n \to \infty  and that's where I am stuck. In the book, there is a hint saying that for x in the mentioned interval x\in\left [ 0, \alpha  \right ], and using the inequality \sin\left ( x \right )\geq x-\frac{x^{3}}{3!},
    we get:
    \left| n \sin\sqrt{4\pi ^{2}n^{2}+x^{2}} \right| -\frac{x^{2}}{4\pi }|\leqslant \frac{a^{2}}{4\pi }\left ( 1-\frac{2}{\sqrt{1+\frac{a^{2}}{4\pi ^{2}n^{2}}}+1} \right )+\frac{n}{3!}\frac{\alpha ^{6}}{8n^{3}\pi ^{3}}

    I don't understand how the book got this inequality based on \sin\left ( x \right )\geq x-\frac{x^{3}}{3!}. Can anyone give me a detailed proof how to get that inequality given by the book? From that point, I can easily prove the uniform convergence.

    For the uniform convergence on R, there is hint saying that I should use \left | \sin\left ( x \right ) \right | \leqslant \left | x \right | to get:
    \left| n \sin\sqrt{4\pi ^{2}n^{2}+x^{2}} \ -\frac{x^{2}}{4\pi }\right|\geq  \frac{x^{2}}{4\pi }\left ( 1-\frac{2}{\sqrt{1+\frac{x^{2}}{4\pi ^{2}n^{2}}}+1} \right)

    Any help please how can we derive the last inequality too? because from this inequality, I can easily prove the non uniform convergence on R.
    Last edited by Opalg; November 16th 2011 at 12:15 AM. Reason: fixed TEX tags
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  2. #2
    MHF Contributor Amer's Avatar
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    Re: sequence of functions

    How this f_n(x) = n \sin \left(\sqrt{4 \pi ^2 n^2+x^2}\right)

    converges to this in  \left[0 , \alpha ] pointwise
     f(x) = \frac{x^2}{4\pi} ??
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