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Power series and integration >

  1. Mar 31, 2007 #1
    Power series and integration >:(

    Hello, this question is about power series and integration of power series, the question and my working is on the image below.
    I had to write the question and my working plus the correct answer on a piece of paper and scan it, sorry for the hasle

    http://i148.photobucket.com/albums/s38/InsertMe/img015.jpg

    Smilies in the image :P
     
  2. jcsd
  3. Mar 31, 2007 #2

    AKG

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    [tex]-\frac{1}{x} - \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n-1}}{4n^2-1}[/tex]

    [tex]= -\frac{1}{x} + \sum_{n=0}^{\infty}\frac{-(-1)^nx^{2n-1}}{4n^2-1}[/tex]

    [tex]= -\frac{1}{x} + \sum_{n=0}^{\infty}\frac{(-1)^{n+1}x^{2n-1}}{4n^2-1}[/tex]

    [tex]= -\frac{1}{x} + \frac{(-1)^{0+1}x^{2\cdot 0-1}}{4\cdot 0^2-1} + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n-1}}{4n^2-1}[/tex]

    [tex]= -\frac{1}{x} + \frac{x^{-1}}{1} + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n-1}}{4n^2-1}[/tex]

    [tex]=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n-1}}{4n^2-1}[/tex]
     
  4. Apr 1, 2007 #3
    thanks alot AKG

    just one question is it always possible to do that whenever i get a function plus (or minus) a power series?
     
  5. Apr 1, 2007 #4

    HallsofIvy

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    In this particular case you should have noticed that when n= 0, x2n-1 is just x-1, the same as the 1/x you are adding. Strictly speaking, what you have is not a "power series" since the standard definition requires non-negative powers only.
     
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