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Math Help - Another "simple looking" Maclaurin Series

  1. #1
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    Another "simple looking" Maclaurin Series

    Find a Maclaurin series for f(x)=(6)/((1-x)^4)


    looks simple, but I know I am not getting the right answer
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  2. #2
    MHF Contributor chisigma's Avatar
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    Suppose that You have...

    f(x)=\frac{1}{1-x} (1)

    ... and You derive it three times. You obtain...

    \frac{d^{3}}{dx^{3}} f(x)= \frac{6}{(1-x)^4} (2)

    Now if You know the McLaurin expansion of f(x) what is the procedure to obtain the McLaurin expansion of \frac{d^{3}}{dx^{3}} f(x) ?...

    Kind regards

    \chi \sigma
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  3. #3
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    Still confused, on expansion.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Because for |x|<1 is...

    \frac{1}{1-x} = 1+x+x^{2}+ \dots + x^{n} + \dots = \sum_{n=0}^{\infty} x^{n} (1)

    ... deriving 'term by term' we have in sequence...

    \frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^{2}} = 1+2x+3x^{2}+\dots+(n+1)\cdot x^{n} + \dots= \sum_{n=0}^{\infty} (n+1)\cdot x^{n} (2)

    \frac{d^{2}}{dx^{2}} \frac{1}{1-x} = \frac{2}{(1-x)^{3}} = 2+6x+12x^{2}+\dots+(n+1)\cdot(n+2)\cdot x^{n} + \dots= \sum_{n=0}^{\infty} (n+1)\cdot(n+2)\cdot x^{n} (3)

    \frac{d^{3}}{dx^{3}} \frac{1}{1-x} = \frac{6}{(1-x)^{4}} = 6+24x+120 x^{2}+\dots+(n+1)\cdot(n+2)\cdot (n+3)\cdot x^{n} + \dots=

    = \sum_{n=0}^{\infty} (n+1)\cdot(n+2)\cdot (n+3)\cdot x^{n} (4)

    If necessary of course we can proceed further ... so that we obtain in general...

    \frac{1}{(1-x)^{k}} = \frac{1}{(k-1)!}\cdot \sum_{n=0}^{\infty} (n+1)\cdot (n+2)\dots (n+k-1)\cdot x^{n} (5)

    Kind regards

    \chi \sigma
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