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Series Expansion of 1/Polynomial

  1. Feb 18, 2014 #1

    Hepth

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    Is there a simple way to series expand a function of the form
    $$
    \frac{1}{\sum_{n=0}^{\infty} a_n x^n}
    $$
    about the point ##x=0##, such that it can be expressed as another sum ##\sum_n c_n x^n##?

    I tried doing it by taylor expansion but I end up with a sum of sums of products of sums :) and its been too long for me to remember a lot of the more advanced simplifications.

    Thanks.
     
  2. jcsd
  3. Feb 18, 2014 #2

    Simon Bridge

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    Well, I'd try... $$\sum_{n=0}^\infty b_nx^n = \frac{1}{\sum_{n=0}^\infty a_nx^n}\\
    \implies \left(\sum_{n=0}^\infty b_nx^n\right)\left(\sum_{n=0}^\infty a_nx^n\right)=1\\
    \implies \sum_{n=0}^\infty \left( \sum_{k=0}^n b_ka_{n-k}\right)x^n = 1\\
    \implies \sum_{n=0}^\infty c_n x^n = 1$$... you know what the convolution values are: ##c_0=1, c_{n>0}=0##

    If you also know all the ##a_n## you can find the corresponding ##b_n##
    If you are lucky, there's a pattern.
     
  4. Feb 19, 2014 #3

    Hepth

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    Thanks!
     
  5. Feb 20, 2014 #4

    Simon Bridge

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    Glad you like it - I figured you'd kinda got as far as line 3 or I'd have forced you to work it out yourself :)

    Note: I didn;t use your notation - so where you wanted ##\small{c_n}##'s I gave you ##\small{b_n}##'s

    If the series converges, then it gets easier...
    i.e. In the special case where ##a_n=a## is a constant with ##n##, and ##|x|<1##, then the RHS of line #1 converges to ##(1-x)/a## making ##b_0=1/a,\; b_1=-1/a,\; b_{n>1}=0##

    In general - the series will have a radius of convergence depending on the value of x (and the series). In physics it is common practice to use the first 2-3 terms only, as an approximation.

    Enjoy.
     
  6. Feb 21, 2014 #5

    Hepth

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    I just needed a general approach other than taking a taylor expansion. It IS physics actually, and what I really have are double summations in both the numerator and denominator over two variables that go to arbitrary order.

    So

    $$ \sum_{ij} c_{ij} x^i y^j = \frac{\sum_{nm} a_{nm} x^n y^m}{\sum_{kl} b_{kl} x^k y^l} $$

    And I wanted to find the c's. I know the anm and bkl to any order I really want.
     
  7. Feb 21, 2014 #6

    Simon Bridge

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    If you keep getting double summations like that you may want to consider if you can get where you want more easily through matrix algebra.

    i.e. if ##\vec x## and ##\vec y## are column vectors where the nth element is ##x^{n-1}## and ##y^{n-1}## respectively, then we can make a matrix ##\text{A}## of the coefficients so that: $$\vec y^t A x = \sum_{j=1}^N y^{j-1} \sum_{i=1}^N a_{ij}x^{i-1}$$... you want the powers to start at 0 right?

    Then the problem looks like -
    ... given the following:$$\vec y^t \text{C}\vec x = \frac{\vec y^t\text A \vec x}{\vec y^t \text{B} \vec x}$$... express C in terms of A and B.

    ... but I don't know the problem you are trying to solve.
     
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