If the generating function is (1-x^200)(1-x^240)/((1-x^10)(1-x^20)(1-x^25)). Then expand all the terms through x^100. How can I make it in the form of a recurrence relation hn. The initial conditions are given.

Printable View

- Dec 10th 2008, 11:15 PMkkkkkkgenerating function problem
If the generating function is (1-x^200)(1-x^240)/((1-x^10)(1-x^20)(1-x^25)). Then expand all the terms through x^100. How can I make it in the form of a recurrence relation hn. The initial conditions are given.

- Dec 13th 2008, 01:34 PMawkward
The generating function is

$\displaystyle (1-x^{200})(1-x^{240})(1-x^{10})^{-1}(1-x^{20})^{-1}(1-x^{25})^{-1}$

$\displaystyle =(1-x^{200})(1-x^{240})\sum_{i=0}^\infty x^{10i} \sum_{j=0}^\infty x^{20j} \sum_{k=0}^\infty x^{25k}$

The first two factors, $\displaystyle =(1-x^{200})(1-x^{240})$, don't make any contribution to $\displaystyle x^n$ with $\displaystyle n < 200$, so all you have to do is expand enough of

$\displaystyle \sum_{i=0}^\infty x^{10i} \sum_{j=0}^\infty x^{20j} \sum_{k=0}^\infty x^{25k}$

to get the powers of $\displaystyle x$ up to $\displaystyle x^{100}$.

I would start by expanding

$\displaystyle \sum_{j=0}^5 x^{20j} \sum_{k=0}^4 x^{25k}$

up to $\displaystyle x^{100}$.

I don't know enough about what you want in the way of a recurrence to help you with that part of the problem.