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Math Help - Radius of convergence

  1. #1
    MHF Contributor alexmahone's Avatar
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    Radius of convergence

    Find the radius of convergence of:

    c_0\left(1-\frac{x^2}{2}+\frac{x^4}{2^2}-...\right)+c_1\left(x-\frac{x^3}{2}+\frac{x^5}{2^2}-...\right)
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  2. #2
    Super Member girdav's Avatar
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    Re: Radius of convergence

    We have a sum of two power series. What is the radius of convergence of each one?
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    Re: Radius of convergence

    "Radius of convergence" only applies to a power series, not a sum of two power series. The function itself will exist on the intersection of the two intervals of convergence of the two power series.

    You can find the radius of convergence of these power series by using the "ratio test" or "root test". Here, because the nth term in the first series is (-1)^n(x/2)^{2n}, the root test should work nicely. The nth root of (x/2)^{2n} is (x/2)^2. For what x is that less than 1? In order to avoid the "nth root of x", I would use the ratio test for the second: [x(x/2)^{2(n+1)}]/[x(x/2)^{2n}]= x/2. For what x is that less than 1? The "ratio of converence" for the entire sum (although I don't think that's the correct term- the interval on which the function exists) is the smaller of those two.
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Re: Radius of convergence

    Quote Originally Posted by HallsofIvy View Post
    "Radius of convergence" only applies to a power series, not a sum of two power series. The function itself will exist on the intersection of the two intervals of convergence of the two power series.

    You can find the radius of convergence of these power series by using the "ratio test" or "root test". Here, because the nth term in the first series is (-1)^n(x/2)^{2n}, the root test should work nicely. The nth root of (x/2)^{2n} is (x/2)^2. For what x is that less than 1? In order to avoid the "nth root of x", I would use the ratio test for the second: [x(x/2)^{2(n+1)}]/[x(x/2)^{2n}]= x/2. For what x is that less than 1? The "ratio of converence" for the entire sum (although I don't think that's the correct term- the interval on which the function exists) is the smaller of those two.
    Some issues with your post:

    1) The nth term of the first series is (-1)^n(x^2/2)^n and not (-1)^n(x/2)^{2n}.

    2) [x(x^2/2)^{n+1}]/[x(x^2/2)^n]=x^2/2

    So, the radius of convergence is \sqrt{2}.
    Last edited by alexmahone; November 6th 2011 at 08:36 AM.
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