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Math Help - Convergence of a series

  1. #1
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    Convergence of a series

    Hi, i need help in the following question:

    I need to prove that there are a_0, a_1,... in R and r>0 such that the series ∑(a_n)x^n converge when |x|< r, and such that ∑(a_n)x^n = arctanx for every |x|< r

    What should i do here? Use the Taylor series of arctanx or something?

    Thanks for any help, and sorry for the mess
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  2. #2
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    I think you need to prove that Arctan(x) has a convergent power series on the given domain. For example when does a function have a Taylor series?
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  3. #3
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    Well the way I would think about it is to note that

    \frac{d}{dx}(\arctan{x}) = \frac{1}{1 + x^2}.


    You should know that for a geometric series \sum_{n = 0}^{\infty}ar^n, its infinite sum is \frac{a}{1 - r} provided |r| < 1.

    I think you can see that if you write \frac{1}{1 + x^2} = \frac{1}{1 - (-x^2)} = \frac{a}{1 - r} then you can see a = 1, r = -x^2.

    So \frac{1}{1 + x^2} is the closed form of \sum_{n = 0}^{\infty}1(-x^2)^n = \sum_{n = 0}^{\infty}(-1)^nx^{2n}. This is provided of course that |r| < 1.

    So |-x^2| < 1

    |x|^2 <1

    |x| < 1.



    So we have

    \frac{1}{1 + x^2} = \sum_{n = 0}^{\infty}(-1)^nx^{2n} for |x| < 1

    \frac{d}{dx}(\arctan{x}) = \sum_{n = 0}^{\infty}(-1)^nx^{2n} for |x| < 1

    \arctan{x} = \int{\sum_{n = 0}^{\infty}(-1)^nx^{2n}\,dx} for |x| < 1

    \arctan{x} = \sum_{n = 0}^{\infty}\int{(-1)^nx^{2n}\,dx} for |x|<1

    \arctan{x} = \sum_{n = 0}^{\infty}\frac{(-1)^nx^{2n + 1}}{2n + 1} for |x| < 1.
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Great solution!
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Prove It View Post
    Well the way I would think about it is to note that

    \frac{d}{dx}(\arctan{x}) = \frac{1}{1 + x^2}.


    You should know that for a geometric series \sum_{n = 0}^{\infty}ar^n, its infinite sum is \frac{a}{1 - r} provided |r| < 1.

    I think you can see that if you write \frac{1}{1 + x^2} = \frac{1}{1 - (-x^2)} = \frac{a}{1 - r} then you can see a = 1, r = -x^2.

    So \frac{1}{1 + x^2} is the closed form of \sum_{n = 0}^{\infty}1(-x^2)^n = \sum_{n = 0}^{\infty}(-1)^nx^{2n}. This is provided of course that |r| < 1.

    So |-x^2| < 1

    |x|^2 <1

    |x| < 1.



    So we have

    \frac{1}{1 + x^2} = \sum_{n = 0}^{\infty}(-1)^nx^{2n} for |x| < 1

    \frac{d}{dx}(\arctan{x}) = \sum_{n = 0}^{\infty}(-1)^nx^{2n} for |x| < 1

    \arctan{x} +C= \int{\sum_{n = 0}^{\infty}(-1)^nx^{2n}\,dx} for |x| < 1 (*)

    \arctan{x} = \sum_{n = 0}^{\infty}\int{(-1)^nx^{2n}\,dx} for |x|<1

    \arctan{x} = \sum_{n = 0}^{\infty}\frac{(-1)^nx^{2n + 1}}{2n + 1} for |x| < 1.
    Maybe... IN(*) you should prove that C=0 by substituting x=0 ?
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