# What is an alternative formulation to a contour integral?

Suppose that a rational generating function is given representing the function $f(x) = \sum_{i=0}^{2n}{c_i x^i}$ where $c_i \in \mathbb{N}$

The goal is to determine if the coefficient in the middle, $c_n$, is nonzero or not.

For example, we could be given $f(x) = \frac{1 - 10x^9 + 9x^{10}}{(x - 1)^2}$, which represents $1 + 2x + 3x^2 + \cdots + 9x^8$. We are interested, in this case, in the middle coefficient, which happens to be 5.

Original Formulation

The actual value can be determined by evaluating a contour integral. This is accomplished by first multiplying $f(x)$ by $x^{-n}$. Then, rewriting the variable $x$ as $x\cdot e^{i t}$ and finally evaluating the integral $\int_C{f(x)dt}$. The result of all of this is the coefficient in question.

The Goal

I seek to find alternative formulations to this problem. Let me give some examples.

Example 1 : "Tiling"

Start with the original formula $f(x)$. Make a copy, for example $f(y)$, by rewriting $x$ as $y$. Then we can subtract the second function from the first, eliminating the first coefficient. We repeat additions and subtractions (possibly infinitely many times) until all coefficients except for the middle one have been eliminated.

Example 2 : "Division"

We multiply $f(x)$ by a function $g(y)$, where $g(y) = \frac{1}{1-x^{2n+1}}$, essentially making the series repeat forever. We then divide this by $x^n\cdot g(y)$, and get the remainder. This remainder is the first half of the original series, and $c_n$ can be easily obtained from this.

What is desired / Motivations

I'd simply like to have alternative formulations for the solution. I'm using this for an algorithm, and I'd like to publish this with anyone who gives a formulation that provides a quick result.

I'm not sure I understand exactly what you want, but if you differentiate the function $n$ times and then evaluate the result at $x=0$ you get $n! c_n$.
• Thanks for your enlightening arguments. I had originally ruled out many methods in my mind, because $n \approx \Theta(2^{q 2^p})$. I hope to be more open minded now, and I've realized there's alot more I can consider. – Matt Groff Oct 21 '10 at 2:30