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Math Help - A Power Series

  1. #1
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    A Power Series

    Hi everyone ! I need a help with this stuff

    Show that a power series c(sub n)x^n has the same radius of convergence as c(sub n+m)x^n, for any positive integer m.

    Thanks.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by drpawel View Post
    Hi everyone ! I need a help with this stuff

    Show that a power series c(sub n)x^n has the same radius of convergence as c(sub n+m)x^n, for any positive integer m.

    Thanks.
    Note that making this change will only make a constant change in the series. See if you can extrapolate a proof from that. If not check back and I will present my solution.
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  3. #3
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    Actually,I figured out that it is constant change,but still I can't find the way to show it.
    Can you help me?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by drpawel View Post
    Hi everyone ! I need a help with this stuff

    Show that a power series c(sub n)x^n has the same radius of convergence as c(sub n+m)x^n, for any positive integer m.

    Thanks.
    Suppose that sum series \sum{c_nx^n} converges \forall{x}\in(-\delta,\delta). This implies that \lim_{n\to\infty}\sqrt[n]{|c_nx^n|}<1~\forall{x}\in(-\delta,\delta)

    Now we are looking at \sum{c_{n+m\in\mathbb{N}}x^n}. So using the ratio test we need to find the value of x such that \lim_{n\to\infty}\sqrt[n]{|c_{n+m}x^n|}<1. Now letting n+m=z we have \lim_{z\to\infty}\sqrt[z-m]{c_zx^{z-m}}=|x|\lim_{z\to\infty}\sqrt[z-m]{c_z}\sim|x|\lim_{x\to\infty}\sqrt[z]{c_z}<1~\forall{x}\in(-\delta,\delta)\quad\blacksquare
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  5. #5
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    Quote Originally Posted by drpawel View Post
    Hi everyone ! I need a help with this stuff

    Show that a power series c(sub n)x^n has the same radius of convergence as c(sub n+m)x^n, for any positive integer m.

    Thanks.
    both series are convergent at x = 0. for x \neq 0 let p(x)=c_0 + c_1x + \cdots + c_{m-1}x^{m-1}. now the claim trvially follows from this identity: \sum_{n=0}^{\infty}c_nx^n = x^m\sum_{n=0}^{\infty}c_{n+m}x^n +p(x). \ \ \Box
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