# generating functions

• Feb 28th 2010, 11:52 PM
jmedsy
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.
• Mar 1st 2010, 12:44 AM
NonCommAlg
Quote:

Originally Posted by jmedsy
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.

$\displaystyle (x^2+x^3+x^4+x^5+x^6)^5=x^{10}(1-x^5)^5(1-x)^{-5}.$ so you only need to find the coefficient of $\displaystyle x^{10}$ in $\displaystyle (1-x^5)^5(1-x)^{-5}.$ now we have:

$\displaystyle (1-x^5)^5(1-x)^{-5}=(1-5x^5+10x^{10}-10x^{15} + 5x^{20}-x^{25})\sum_{n=0}^{\infty} \binom{n+4}{n}x^n.$ so the coefficient you're looking for is: $\displaystyle \binom{14}{10}-5\binom{9}{5}+10.$
• Mar 1st 2010, 01:15 AM
jmedsy
Quote:

Originally Posted by NonCommAlg
$\displaystyle (x^2+x^3+x^4+x^5+x^6)^5=x^{10}(1-x^5)^5(1-x)^{-5}.$ so you only need to find the coefficient of $\displaystyle x^{10}$ in $\displaystyle (1-x^5)^5(1-x)^{-5}.$ now we have:

$\displaystyle (1-x^5)^5(1-x)^{-5}=(1-5x^5+10x^{10}-10x^{15} + 5x^{20}-x^{25})\sum_{n=0}^{\infty} \binom{n+4}{n}x^n.$ so the coefficient you're looking for is: $\displaystyle \binom{14}{10}-5\binom{9}{5}+10.$

thanks, i spent a lot of time on that