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Math Help - Taylor Expansion

  1. #1
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    Taylor Expansion

    Write the taylor expansion of x/(1-x)^2
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  2. #2
    Super Member Random Variable's Avatar
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    I'm assuming you mean the Taylor series about x=0.

    We know from the sum of a geometric series that  1+x+x^{2}+x^{3}+ ... = \frac {1}{1-x}

    then  \frac{d}{dx} (1+x+x^{2}+x^{3}+ ...) = 1+2x+3x^{2}+4x^{3} +... = \frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^{2}}

    so  \frac{x}{(1-x)^{2}} = x (1+2x+3x^{2}+4x^{3}+...) = (x+2x^{2}+3x^{3}+4x^{4} + ... ) = \sum_{n=1}^{\infty}nx^{n}
    Last edited by Random Variable; July 28th 2009 at 01:04 AM.
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  3. #3
    MHF Contributor chisigma's Avatar
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    The binomial series...

    (a+x)^{n} = a^{n} + \binom{n}{1} a^{n-1}x + \binom{n}{2} a^{n-2}x^{2} + \binom{n}{3} a^{n-3} x^{3} + \dots (1)

    ... in the case a=1 ,  n=-2 and changing x into -x becomes...

     (1-x)^{-2} = 1 + 2x + 3 x^{2} + 4 x^{3} + \dots (2)

    If you multiply each term of (2) by x you obtain what you have requested. That isn't the only way to arrive to the result...

    Kind regards

    \chi \sigma
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