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Coefficients of Series

  1. Apr 13, 2010 #1
    [PLAIN]http://img243.imageshack.us/img243/156/55139700.png [Broken]

    Well I know that
    [tex]\sum \frac{1}{1-x}[/tex] = 1+[tex]x^{2}[/tex]+[tex]x^{3}[/tex]....

    I know the integral is

    [PLAIN]http://www4a.wolframalpha.com/Calculate/MSP/MSP32919a6dgiib73geda500000hd72bah4i062h1b?MSPStoreType=image/gif&s=36&w=241&h=38 [Broken]

    I've first started with the known sum, then i replaced [x] with [PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP15219a6e2bf832h2gb6000039if18068de4ia05?MSPStoreType=image/gif&s=55&w=20&h=39 [Broken]

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

    Then I tried to integrate the new sum [too long and latek is messing up for me] to get arctan. Then I multiplied it by 16.

    I feel that i'm messing up somewhere, I just don't know where.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 13, 2010 #2
    For the integral: In the denominator factor the 4, that is


    and use the hint.
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