Power series and integration >

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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 [Broken]

Smilies in the image :P
 
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  • #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]
 
  • #3
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thanks alot AKG

just one question is it always possible to do that whenever i get a function plus (or minus) a power series?
 
  • #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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