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Thread: 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

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


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

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

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

    So $\displaystyle |-x^2| < 1$

    $\displaystyle |x|^2 <1$

    $\displaystyle |x| < 1$.



    So we have

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

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

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

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

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

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


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

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

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

    So $\displaystyle |-x^2| < 1$

    $\displaystyle |x|^2 <1$

    $\displaystyle |x| < 1$.



    So we have

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

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

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

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

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