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Math Help - Expressing functions as power series

  1. #1
    Senior Member Pinkk's Avatar
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    Expressing functions as power series

    Professor really hasn't explained any of this and gave us a practice exam on it.

    f(x)=\frac{x}{4x+1}

    Would this simply be \sum_{k=0}^{\infty}x(-1)^{k}(4x)^{k}=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{4}(4x)^{k+1}


    I reasoned this since:

    \frac{1}{1+4x}=\sum_{k=0}^{\infty}(-1)^{k}(4x)^{k}
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  2. #2
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    Quote Originally Posted by Pinkk View Post
    Professor really hasn't explained any of this and gave us a practice exam on it.

    f(x)=\frac{x}{4x+1}

    Would this simply be \sum_{k=0}^{\infty}x(-1)^{k}(4x)^{k}=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{4}(4x)^{k+1}


    I reasoned this since:

    \frac{1}{1+4x}=\sum_{k=0}^{\infty}(-1)^{k}(4x)^{k}
    Looks good to me.

    Just remember that the series

    \sum_{k=0}^\infty{(-1)^k(4x)^k} converges only when

    |-4x|< 1

    |4x| < 1

    4|x| < 1

    |x| < \frac{1}{4}

     -\frac{1}{4} < x < \frac{1}{4}.
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  3. #3
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    Correct.

    However it looks better as
     {(-1)^{n}}{x^{n+1}4^{n}}

    And the interval of convergence by prove it is correct no need to check endpoints.
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  4. #4
    Senior Member Pinkk's Avatar
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    For these problems I do not need to find the interval of convergence or radius of convergence but finding those are probably the only things of infinite series I can do without any problem.
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