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Topic: Finding a closed form for a generating function
Replies: 2   Last Post: Oct 10, 2012 3:40 PM

 Messages: [ Previous | Next ]
 Inverse 18 Mathematics Posts: 175 Registered: 7/23/10
Re: Finding a closed form for a generating function
Posted: Oct 10, 2012 2:58 PM

On October 10, 2012, 8:38AM, (unknown) wrote:
> I have the generating function:
>
> C(z) = [1/(1 - z^10)] * [1/(1 - z^25)]
>
>
>
> And I'm trying to find a closed form for this.
>
> I can't use a substitution for z^10 that would simplify
>
> my denominator because it doesn't divide 25.
>
>
>
> Any ideas on how I could simplify C(z) and further
>
> reduce it to a closed form for finding the coefficient
>
> of any z^i term?

First, let's set u = z^5 in your equation, and express it using geometric series expansion. It becomes:

(1 - u^2)^(-1) * (1 - u^5)^(-1) = Sum[u^(2 * i + 5 * j), {i, 0, Infinity}, {j, 0, Infinity}]

If I expand this sum, it leads to:

(1 - u^2)^(-1) * (1 - u^5)^(-1) = Sum[s(n) * u^n, {n, 0, Infinity}]

where s(n) is the number of (i, j) pairs, i , j >= 0 and integer, satisfying the diophantine equation 2 * i + 5 * j = n. For example, s(0) = 1, s(1) = 0, and s(12)=2, because only (6,0) and (1,2) satisfy the equation.

The next question : is there a formula that gives this number of pairs?

Date Subject Author
10/10/12 herakleides@gmail.com
10/10/12 Inverse 18 Mathematics
10/10/12 Rancid Moth