divB
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Hi,
I found the following formula:
<br /> \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)}}}<br />
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
I found the following formula:
<br /> \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)}}}<br />
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