# Simple partial (?) fraction?

1. Jul 15, 2010

### divB

Hi,

I found the following formula:

$$\prod_{m=1}^{N_d} \frac{1}{(1-e^{\alpha_{(m)}}z^{-1})^{n_{(m)}}} = \sum_{m=1}^{N_d} \sum_{n=1}^{n_{(m)}} \frac{c_{m,n} }{(1-e^{\alpha_{(m)}}z^{-1})^{n_{(m)}}}$$

What I want is finding the coefficients $$c_{m,n}$$. This looks like a simple partial fraction method.

In fact I am able to find the coefficients in the following form:

$$\dots = \sum_{m=1}^{N_d} \sum_{n=1}^{n_{(m)}} \frac{c_{m,n} }{(z-e^{\alpha_{(m)}})^{n_{(m)}}}$$

using the standard partial fraction method. This is also what I get using Apart in Mathematica.

But I need the coefficients for the form described above: Only the denominator should contain $$z^{-1}$$ and nothing more.

Can anybody tell me how to get this form? Is it possible at all? (it should be because this formula is used in a paper...)

Is there any Mathematica command which produces the desired form?

Thank you very much.

Regards,
divB

2. Jul 15, 2010

### trambolin

use a change of variables for the exponential and the 1/z, you would be back in the standard form again. Find the partial fractional form and replace the variables.

3. Jul 15, 2010

### divB

Oh, I am so stupid .... sure, this works!

Thank you very much!!