## Product representation

- Sun, 10 Jan, 2010

In dealing with complex analytic functions, it is often quite handy to represent them as their Taylor expansions, i.e. a summation representation. We can generalise a little bit and expand even around non-essential, isolated singularities with a Laurent series. However, quite often we really care about the zeros of a function (e.g. Yang-Lee circle theorem on zeros of the grand partition function for Ising-esque models), and extracting those out of summations is unwieldy. Therefore, it would be much nicer to have a product representation instead.

For polynomials, such a representation is obvious, and unique — the fundamental theorem of algebra guarantees the factorisation into linear factors: $$p(z) = p(0) \prod_{\mathrm{finite~}n} \left(1 - \frac{z}{z_n}\right). $$ The various $z_n$ are then the location of the zeros. We would like to extend this to more general functions.

However, in general this is difficult, and non-unique (see Weierstrass factorization theorem for an existence statement). For one particular subset however, we can create an effective procedure for manufacturing these: functions with only simple isolated zeros and no poles.

So let $f(z)$ be such a function. If $f$ has a zero of order $m$ at $z=0$, then we can divide out by $z^m$ and get something without a zero at the origin, and so without loss of generality that's what we'll assume. Let $z_n$ be the location of remaining (infinite number of) zeros. Then $g(z) = [\ln f(z)]^\prime = f^\prime(z)/f(z)$ has only simple isolated poles with unit residues at $z_n$; thus if we find a summation representation of $g$ we could integrate and exponentiate to find a product representation for $f$.

Now consider: $$\frac{1}{2\pi i} \oint_{\Gamma_n} \frac{g(z^\prime)}{z^\prime-z} dz^\prime = g(z) + \sum^n_j \frac{1}{z_j - z}$$ where $\Gamma_n$ is a contour which encloses the closest $n$ poles to the origin. Then $$g(z) - g(0) = \frac{z}{2\pi i} \oint_{\Gamma_n} \frac{g(z^\prime)}{z^\prime(z-z^\prime)} dz^\prime + \sum_j^n \left(\frac{1}{z-z_j} + \frac{1}{z_j}\right).$$ Thus if we can find a sequence of contours $\Gamma_n$ such that $g$ remains bounded on them, the integral will converge to zero as $n \rightarrow 0$. In that case, we find $$g(z) = g(0) + \sum_n \left(\frac{1}{z-z_n} + \frac{1}{z_n}\right).$$

So now we can return to factorising $f$. Integrating $g$ gives $$\ln f(z) - \ln f(0) = cz + \sum_n \left[ \ln\left(1-\frac{z}{z_n}\right) + \frac{z}{z_n} \right]$$ where $c = g(0)$; re-exponentiating gives $$f(z) = f(0) e^{cz} \prod_n \left(1 - \frac{z}{z_n}\right) e^{z/z_n}.$$

As an example, consider $f(z) = \sin(z)/z$. Then $g(z) = -1/z + \cot(z)$; we can pick contours $\Gamma_n$ which sit between the poles in $\cot z$, and our procedure will converge. A bit of limit work shows that $g(0) = 0$ and $f(0) = 1$. The zeros sit at $z_n = n \pi$, with $n$ being any non-zero integer. Thus we find \begin{align*} \frac{\sin(z)}{z} &= \prod_{n \neq 0} \left(1 - \frac{z}{\pi n}\right) e^{z/n\pi} \\ &= \prod_{n=1}^{\infty} \left(1 - \frac{z^2}{\pi^2 n^2}\right). \end{align*}