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Math Help - Multiplication of series

  1. #1
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    Multiplication of series

    1. Multiply the series for (1-x)^-1 (valid when |x|<1) by itself and so obtain the expansion

    (1-x)^-2 = sigma n=0 to infinity (n+1)x^n, |x|<1.

    2. Derive the expansion

    (1-x)^-3 = sigma n=0 to infinity ((n+1)(n+2)*x^n)/2 (|x|<1)

    by multiplication of series. Use the result of Exercise 1.
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  2. #2
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    Multiplying series is extra work in this case.

    Notice that \displaystyle \frac{d}{dx}\left(\frac{1}{1 - x}\right) = \frac{1}{(1 - x)^2} and \displaystyle \frac{d}{dx}\left[\frac{1}{(1 - x)^2}\right] = \frac{2}{(1-x)^3}...
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    (1-x)^{-1}=\sum_{n=0}^{+\infty}x^n is absolutely convergent for |x|<1 . Use a well known result about Cauchy product of series.


    Fernando Revilla
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