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Geometric power series

  1. Nov 27, 2006 #1

    fsm

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    I was wondering if someone could check my work:

    Find the geometric power series representation of
    f(x)=ln(1+2x), c=0

    I get [tex]\\sum_{n=0}^ \\infty[/tex]2(-2x)^n+1 on -1/2<x<1/2
     
  2. jcsd
  3. Nov 27, 2006 #2
    [tex] f(x) = \ln(1+2x) [/tex]

    [tex] \ln(1+2x) = \frac{1}{2} \int \frac{1}{1+2x} [/tex]

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


    [tex] \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-2x)^{n+1}}{n+1} = \sum_{n=0}^{\infty} \frac{(-2x)^{n+1}}{2n+2} [/tex]
     
  4. Nov 27, 2006 #3

    fsm

    User Avatar

    Thanks!! I see what I did.
     
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