# Series Expansion of 1/Polynomial

1. Feb 18, 2014

### Hepth

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. Feb 18, 2014

### Simon Bridge

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.

3. Feb 19, 2014

### Hepth

Thanks!

4. Feb 20, 2014

### Simon Bridge

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.

5. Feb 21, 2014

### Hepth

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.

6. Feb 21, 2014

### Simon Bridge

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.