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Series Representation for Function

  1. Dec 10, 2011 #1
    1. The problem statement, all variables and given/known data
    Fin the Taylor series about x = 0 for:
    f(x) = 1 / (1-x2)2


    2. Relevant equations
    g(x) = 1 / (1-x2)

    3. The attempt at a solution
    Differentiating g(x), the series representation of g'(x) is Ʃn(x2)n-1
    Since f(x) = g'(x)/2x
    f(x) = Ʃ(n/2)x2n-1
    I'm pretty sure that's right, but what I don't get is that the solution for this problem is also written as Ʃn(x2)n-1, which is just g'(x).

    What am I misunderstanding here?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 10, 2011 #2

    SammyS

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    The Taylor series about x = 0 for g(x) = 1 / (1-x2) is: [itex]\displaystyle g(x)=\sum_{n=0}^{\infty}(x^2)^n=\sum_{n=0}^{\infty}(x^{2n})\,,[/itex] where x0=1 .

    Differentiating term by term gives: [itex]\displaystyle g'(x)=\sum_{n=1}^{\infty}2n(x^{2n-1})=\sum_{n=0}^{\infty}2(n+1)x(x^{2n})\,,\text{ or equivalently }\sum_{n=1}^{\infty}\frac{2n}{x}(x^2)^n[/itex]

    Dividing by 2x gives: [itex]\displaystyle f(x)=\sum_{n=1}^{\infty}n(x^2)^{n-1}=\sum_{n=0}^{\infty}(n+1)(x^{2})^n\,.[/itex]

    They're all somewhat different.
     
  4. Dec 10, 2011 #3
    [itex]\displaystyle g'(x)=\sum_{n=1}^{\infty}2n(x^{2n-1})=\sum_{n=0}^{\infty}2(n+1)x(x^{2n})\,,\text{ or equivalently }\sum_{n=1}^{\infty}\frac{2n}{x}(x^2)^n[/itex]



    Where did the x in front of the x^2n come from? I know you just shifted n=1 to n = 0, but I can't see where the x comes from?
     
  5. Dec 10, 2011 #4

    SammyS

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    [itex]\displaystyle g'(x)=\sum_{n=1}^{\infty}2n(x^{2n-1}) =\sum_{n=0}^{\infty}2(n+1)(x^{2n+1})=\sum_{n=0}^{ \infty}2(n+1)(x\cdot x^{2n})[/itex]

    Change n to n+1 so 2n - 1 becomes 2(n+1)-1 = 2n + 2 - 1 = 2n+1 .
     
  6. Dec 10, 2011 #5

    vela

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    If you were wondering why your attempt didn't work out, it's because you forgot to use the chain rule when you differentiated (x2)n. That's where the factor of 2x went missing in your calculation.
     
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