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Math Help - Maclaurin series of a function

  1. #1
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    Maclaurin series of a function

    Obtain Maclaurin series for the given function

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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by racewithferrari View Post
    Obtain Maclaurin series for the given function

    Easy, if you know the MacLaurin series of \tan^{-1}(x^3):

    x^2\tan^{-1}(x^3)=x^2\cdot\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}\left(x^3\right)^{2k+1}=\sum_{k=0}^\inf  ty\frac{(-1)^k}{2k+1}x^{6k+5}=:\sum_{n=0}^\infty a_n x^n
    where

    a_n :=\begin{cases}\frac{(-1)^{(n-5)/6}}{2\cdot\frac{n-5}{6}+1}, &\text{if } \frac{n-5}{6}\in\mathbb{Z}\\0, &\text{otherwise}\end{cases}
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  3. #3
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    What have you tried?

    Can you do this one? atan(x)? If you can, you are almost done.

    This might help.

    \frac{d}{dx}atan(x)\;=\;\frac{1}{1+x^{2}}\;=\;\fra  c{1}{1-(-x^{2})}\;=\;1 - x^{2} + x^{4} - x^{6} + ...

    Note: Is there a Uniform Convergence consideration in here, somewhere?
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  4. #4
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    Quote Originally Posted by Failure View Post
    Easy, if you know the MacLaurin series of \tan^{-1}(x^3):

    x^2\tan^{-1}(x^3)=x^2\cdot\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}\left(x^3\right)^{2k+1}=\sum_{k=0}^\inf  ty\frac{(-1)^k}{2k+1}x^{6k+5}=:\sum_{n=0}^\infty a_n x^n
    where

    a_n :=\begin{cases}\frac{(-1)^{(n-5)/6}}{2\cdot\frac{n-5}{6}+1}, &\text{if } \frac{n-5}{6}\in\mathbb{Z}\\0, &\text{otherwise}\end{cases}
    where is there 5 when you multiply 3 with 2k+1.
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  5. #5
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    It comes from the product of x^2 and the series.

    x^2\cdot\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}\left(x^3\right)^{2k+1}=\sum_{k=0}^\inf  ty\frac{(-1)^k}{2k+1}x^{6k+3}\cdot x^2=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}x^{6k+5}
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