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Simple partial (?) fraction?

  1. Jul 15, 2010 #1

    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 [tex]c_{m,n}[/tex]. This looks like a simple partial fraction method.

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

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

    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 [tex]z^{-1}[/tex] 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.

  2. jcsd
  3. Jul 15, 2010 #2
    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.
  4. Jul 15, 2010 #3
    Oh, I am so stupid .... sure, this works!

    Thank you very much!!
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