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Power series representation, I really need help :-]

  1. Dec 1, 2007 #1
    [SOLVED] Power series representation, I really need help!!! :-]

    please let me know if i did this correctly

    [tex]f(x)=\arctan{(\frac{x}{3})}[/tex]

    [tex]f'(x)=\frac{\frac{1}{3}}{1-(-\frac{x^2}{3^2})}[/tex]

    [tex]\frac{1}{3}\int[\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{3^{2n}}]dx[/tex]

    [tex]\frac{1}{3}\int[1-(\frac{x}{3})^{2}+(\frac{x}{3})^{4}-(\frac{x}{3})^{6}+(\frac{x}{3})^{8}+...]dx[/tex]

    [tex]\frac{1}{3}[C+x-\frac{(\frac{x}{3})^{3}}{3}+\frac{(\frac{x}{3})^{5}}{5}-\frac{(\frac{x}{3})^{7}}{7}+\frac{(\frac{x}{3})^{9}}{9}][/tex]

    [tex]C+\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{3^{2n+1}(2n+1)}[/tex]
     
    Last edited: Dec 1, 2007
  2. jcsd
  3. Dec 1, 2007 #2

    Avodyne

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    Science Advisor

    Yes, it's correct! And of course, C=0.

    To make Sigma bigger, use \sum instead of \Sigma:

    [tex]C+\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{3^{2n+1}(2n+1)}[/tex]
     
  4. Dec 2, 2007 #3
    NO WAY!!! woohoo :-] Thanks, I really appreciate it. This concept is really weird to me, but I'm trying my best.
     
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