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Math Help - Use Differentiation to Find a Power Series of Rep. of the Following Function

  1. #1
    Junior Member Coop's Avatar
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    Question Use Differentiation to Find a Power Series of Rep. of the Following Function

    Hi,

    I am not too good with this stuff yet, I am still learning, so I have some questions about why I am doing the things I am.

    I am supposed to find a power series representation of f(x) = 1/(4+x)^2 and find its radius of convergence

    So I realize that the integral of f(x) = -1/(4+x) = (-1/4)*[1/(1+(x/4))] = (-1/4)*[1/(1-(-x/4))]. This is a geometric series with a = 1 and r = -x/4, which means that it has a radius of convergence = |x|/4 < 1 => radius of convergence = 4

    (-1/4)*[1/(1-(-x/4)) can be rewritten as (-1/4)[SERIES(n=0,inf) (-x/4)^n]

    and when I distribute and multiple I get

    SERIES(n=0,inf) [-1*(-1)^n(x^n)]/4^(n+1)

    So since I found the series of the integral of f(x) I have to take its derivative, right?

    The derivative of the above series is

    SERIES(n=1,inf) [-1*(-1)^n(n*x^(n-1))]/4^n+1

    Why didn't I have to do the power rule and differentiate the (-1)^n and 4^(n+1), is it because they don't vary by x?
    Is it because I want the "a," the first term in the geometric series to = 1? And it would equal 0 if n was 0.
    Why did I have to change the bounds of the series to n=1 instead of starting at n=0?


    So to get back to having a bound starting at n=0 I add +1 to all the n values.

    SERIES(n=0,inf) [-1*(-1)^(n+1)((n+1)*x^n)]/4^n+2

    This is equal to SERIES(n=0,inf) [(-1)^n*(n+1)*x^n)]/4^n+2

    And that's what I am told is the right answer. So could someone just make sure I am doing the steps right and help with the questions I had? I really appreciate it, bear with me, I am pretty confused by this right now, because my professor hasn't really gone over this and I am having trouble finding resources :s.

    Thank you I really appreciate any help.
    Last edited by Coop; April 12th 2013 at 11:01 PM.
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  2. #2
    MHF Contributor
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    Re: Use Differentiation to Find a Power Series of Rep. of the Following Function

    Yes it looks correct. And you are right about the reason you don't differentiate those terms, they are just constants.
    Thanks from Coop
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  3. #3
    Junior Member Coop's Avatar
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    Re: Use Differentiation to Find a Power Series of Rep. of the Following Function

    Quote Originally Posted by Prove It View Post
    Yes it looks correct. And you are right about the reason you don't differentiate those terms, they are just constants.
    Thanks So I was right about why the starting bound changes from n=0 to n=1?
    Last edited by Coop; April 13th 2013 at 10:30 AM.
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