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Math Help - Uniform convergence

  1. #1
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    Uniform convergence

    1) Let~M\in\mathbb{R}>0 Show that \sum_{n=1}^{\infty} \frac{z^{2n}}{n!} converges uniformly for |z| < M .


    That part I feel confident on. I use ratio test on \sum_{n=1}^{\infty} \frac{M^{2n}}{n!} and find convergence, then use Weierstrauss M-Test to show the statement as required.


    2) Does \sum_{n=1}^{\infty} \frac{z^{2n}}{n!} converge uniformly on \mathbb{C}?

    Here is where I'm a bit confused. Part 1 shows uniform convergence for |z| < M and doesn't set any requirements on M, but the only example I have says the answer is still no, as z=0 is convergence, but not uniformly. A bit of explanation would be wonderful! Thanks
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  2. #2
    Super Member Rebesques's Avatar
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    Re: Uniform convergence

    For part 2), remember the expansion for e^w=\sum w^n/n!.
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  3. #3
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    Re: Uniform convergence

    Quote Originally Posted by Thomas154321 View Post
    1) Let~M\in\mathbb{R}>0 Show that \sum_{n=1}^{\infty} \frac{z^{2n}}{n!} converges uniformly for |z| < M .


    That part I feel confident on. I use ratio test on \sum_{n=1}^{\infty} \frac{M^{2n}}{n!} and find convergence, then use Weierstrauss M-Test to show the statement as required.


    2) Does \sum_{n=1}^{\infty} \frac{z^{2n}}{n!} converge uniformly on \mathbb{C}?

    Here is where I'm a bit confused. Part 1 shows uniform convergence for |z| < M and doesn't set any requirements on M, but the only example I have says the answer is still no, as z=0 is convergence, but not uniformly. A bit of explanation would be wonderful! Thanks
    Just use the ratio test for part 2.

    The series is convergent wherever $\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \end{align*}$, so solve

    $\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{\frac{z^{2 \left( n + 1 \right) } }{\left( n + 1 \right) !}}{\frac{z^{2n}}{n!}} \right| &< 1 \\ \lim_{n \to \infty} \left| \frac{n!\,z^{2n + 2}}{\left( n + 1 \right) n! \, z^{2n}} \right| &< 1 \\ \lim_{n \to \infty} \left| \frac{z^2}{n + 1} \right| &< 1 \\ \left| z^2 \right| \lim_{n \to \infty} \frac{1}{n + 1} &< 1 \\ \left| z \right| ^2 \cdot 0 &< 1 \\ 0 &< 1 \end{align*}$

    The limit of the ratios is 0, which is ALWAYS less than 1, no matter what value of z. So that means that the series is convergent for all z.
    Last edited by Prove It; December 29th 2014 at 07:05 PM.
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