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Math Help - generating functions

  1. #1
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    generating functions

    find the coefficient x^20 in (x^2 + x^3 + x^4 + x^5 + x^6)^5

    i transform this into the product of x^10 and two summations. I'm not sure where to go from there. my professor mentioned something about convolution but i havent been able to apply what the book does.
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  2. #2
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    Quote Originally Posted by jmedsy View Post
    find the coefficient x^20 in (x^2 + x^3 + x^4 + x^5 + x^6)^5

    i transform this into the product of x^10 and two summations. I'm not sure where to go from there. my professor mentioned something about convolution but i havent been able to apply what the book does.
    (x^2+x^3+x^4+x^5+x^6)^5=x^{10}(1-x^5)^5(1-x)^{-5}. so you only need to find the coefficient of x^{10} in (1-x^5)^5(1-x)^{-5}. now we have:

    (1-x^5)^5(1-x)^{-5}=(1-5x^5+10x^{10}-10x^{15} + 5x^{20}-x^{25})\sum_{n=0}^{\infty} \binom{n+4}{n}x^n. so the coefficient you're looking for is: \binom{14}{10}-5\binom{9}{5}+10.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    (x^2+x^3+x^4+x^5+x^6)^5=x^{10}(1-x^5)^5(1-x)^{-5}. so you only need to find the coefficient of x^{10} in (1-x^5)^5(1-x)^{-5}. now we have:

    (1-x^5)^5(1-x)^{-5}=(1-5x^5+10x^{10}-10x^{15} + 5x^{20}-x^{25})\sum_{n=0}^{\infty} \binom{n+4}{n}x^n. so the coefficient you're looking for is: \binom{14}{10}-5\binom{9}{5}+10.
    thanks, i spent a lot of time on that
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