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Math Help - Another Taylor series question

  1. #1
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    Another Taylor series question

    I don't know, maybe it's late and I just need to knock off, but I can't quite nail the power series for the function \frac{4}{x-1} with the center at zero. \frac{1}{x-1} is (-1)^nx^n, however, so could I simply put the 4 out in front as a coefficient and use that? 4(-1)^nx^n perhaps?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Here

    Yes the power series would be \sum_{n=0}^{\infty}4(-1)^{n}x^{n}\...since all you do is say that \frac{4}{1-x}= 4\frac{1}{1-x}= 4\sum_{n=0}^{\infty}(-1)^{n}x^{n} and distribute the 4 to get \sum_{n=0}^{\infty}4(-1)^{n}x^{n}\
    Last edited by Mathstud28; April 4th 2008 at 06:14 AM.
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  3. #3
    GAMMA Mathematics
    colby2152's Avatar
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    Just multiply it by 4.

    Think... \sum x = 1 + 2 + 3 + 4 + 5 \ldots

    4 \sum x \Rightarrow \sum 4x = 4(1) + 4(2) + 4(3) + 4(4) + 4(5) + \ldots \Rightarrow 4(1 + 2 + 3 + 4 + 5) + \ldots
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  4. #4
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    Ok, so that is all I do...thanks guys!
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