Partial fractions and infinite product

1. Partial fractions

Theorem 1 (Mittag-Leffler) Let ${(a_n)_{n=1}^\infty}$ be a sequence of complex numbers with ${\lim_{n\rightarrow\infty}a_n=\infty}$ and let ${p_n(\cdot)}$ be polynomials without constant term. Then there are functions which are meromorphic in the whole plane with poles at the points ${a_n}$ and the corresponding singular parts ${p_n\left(\frac{1}{z-a_n}\right)}$. Moreover, the most general meromorphic function of this kind can be written in the form

$\displaystyle f(z)=\sum_{n=1}^\infty\left[p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right]+g(z)$

where ${q_n(\cdot)}$‘s are suitably chosen polynomials and ${g(\cdot)}$ is analytic in the whole plane.

Proof: The idea is to subtract a certain portion of analytic part of ${p_n\left(\frac{1}{z-a_n}\right)}$ to ensure convergence without adding any poles. For each ${n\in{\mathbb N}}$, the singular part ${p_n\left(\frac{1}{z-a_n}\right)}$ is analytic in the ball ${B(0,\frac{|a_n|}{2})}$. So we can obtain Taylor expansion ${q_n(\cdot)}$ of ${p_n\left(\frac{1}{z-a_n}\right)}$ around the origin up to the degree ${d_n}$. Using estimate for the remainder term, we have

$\displaystyle \left|p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right|\leq \sum_{d_n+1}^\infty\frac{M_n}{(|a_n|/2)^k}|z|^k=M_n\left(\frac{2|z|}{|a_n|}\right)^{d_n+1}\frac{1}{1-\frac{|z|}{|a_n|/2}}=M_n\frac{\beta_n^{d_n+1}}{1-\beta_n}$

where ${\beta_n=\frac{2|z|}{|a_n|}}$ and ${M_n}$ denotes the maximum of ${|p_n(\cdot)|}$ in ${|z|\leq\frac{|a_n|}{2}}$. In particular, this estimate implies that

$\displaystyle \left|p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right|\leq 2M_n\beta_n^{d_n+1}\text{ whenever }|z|\leq\frac{|a_n|}{4}$

For each ${n}$, we may pick ${d_n}$ large enough to ensure ${2M_n\beta_n^{d_n+1}\leq 2^{-n}}$, which can be done by making ${\frac{M_n}{2^{d_n}}\leq 2^{-n}}$. In doing so, we obtain

$\displaystyle \left|p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right|\leq 2^{-n}\text{ whenever }|z|\leq\frac{|a_n|}{4}$

Given ${R> 0}$, we shall show that the series converges uniformly in ${|z|\leq R}$. We may split the series ${\sum_{n=1}^\infty\left|p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right|}$ into two categories: the first one with ${|a_n|\leq 4R }$ and the second one with ${|a_n|>4R}$. Then ${\sum_{|a_n|\leq 4R}\left[p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right]}$ represents a meromorphic function with poles at ${a_n}$ because it is a finite sum. In the latter category, since ${|a_n|>4R}$, we have ${|z|\leq\frac{|a_n|}{4}}$ in the disk.Thus above estimate implies

$\displaystyle \sum_{|a_n|> 4R}\left|p_n\left(\frac{1}{z-a_n}\right)-q_n(z)\right|\leq \sum 2^{-n}$

This completes the proof. $\Box$ Let us try to investigate several standard examples. Consider the function

$\displaystyle f(z)=\frac{\pi^2}{\sin^2\pi z}$

Recall that ${\sin\pi z=\pi z-\frac{(\pi z)^3}{3!}+\cdots}$ and so ${\sin^2\pi z =(\pi z)^2-\frac{2(\pi z)^4}{3!}+\cdots}$. We then can work out a first few terms in the Laurant expansion at the origin which is

$\displaystyle f(z)=\frac 1 {z^2}-\frac{2\pi^2}{3!}+\text{higher order terms}$

Thus the singular part of ${f}$ at 0 is ${\frac 1z^2}$. Moreover, as ${f}$ has period 1 namely ${f(z+n)=f(z)}$ for all ${n{\mathbb N}}$, we can see that the singular part at integers are ${\frac{1}{(z-n)^2}}$. Clearly ${\sum_{n\in{\mathbb Z}}\frac{1}{(z-n)^2}}$ converges on all compact subsets which do not contain integers. It follows that we may write

$\displaystyle f(z)=\sum_{n\in{\mathbb Z}}\frac{1}{(z-n)^2}+g(z)$

where ${g(z)}$ is entire. Write ${z=x+iy}$ we see that ${\frac{1}{(z-n)^2}=\frac{1}{(x-n)^2-y^2+2iy(x-n)}}$. From the uniform convergence of the series, we can see that ${\sum_{n\in{\mathbb Z}}\frac{1}{(z-n)^2}}$ converges to 0 as ${|y|\rightarrow 0}$. Using the idendity

$\displaystyle |\sin \pi z|^2=\cosh^2\pi y-\cos^2\pi x$

we see that ${f(z)\rightarrow 0}$ uniformly as ${|y|\rightarrow 0}$. In particular, this implies ${g(z)}$ is bounded in the strip ${0\leq x\leq 1}$. By periodicity, ${g(z)}$ is bounded in the entire complex plane. Then Liouville’s theorem says the ${g(z)}$ reduces to a constant. In this case, ${g(z)}$ must vanish as the limit is zero. Therefore, we have the identity

$\displaystyle \frac{\pi^2}{\sin^2\pi z}=\sum_{n=-\infty}^\infty\frac{1}{(z-n)^2}$

Integrate on both sides, we obtain that

$\displaystyle -\pi\cot\pi z=\sum\frac{-1}{z-n}$

In order to make the right-hand side converge, we need to substract extra terms from the Taylor series of ${\frac{1}{z-n}}$. In this case, subtracting all constant term is enough. That is ${\sum_{n=-N}^N\frac{1}{z-n}+\frac{1}{n}=\sum_{n=-N}^N\frac{z}{n(z-n)}}$ which is comparable to ${\frac 1 {n^2}}$. Thus we have

$\displaystyle \pi\cot\pi z=\frac{1}{z}+\sum_{n\neq 0}\left(\frac{1}{z-n}+\frac{1}{n}\right)+c$

We may bracket terms of ${n}$ and ${-n}$ together, and obtain

$\displaystyle \pi\cot\pi z=\frac{1}{z}+\sum_{n\neq 0}\frac{2z}{z^2-n^2}+c$

Now ${c=0}$ is necessary because both sides are odd.

Next we exploit this fact to investigate the sum

$\displaystyle \lim_{m\rightarrow\infty}\sum_{-m}^m\frac{(-1)^n}{z-n}=\frac 1z+\sum_{n=1}^\infty(-1)^n\frac{2z}{z^2-n^2}$

We may split into odd terms and even terms

$\displaystyle \sum_{-2k-1}^{2k+1}\frac{(-1)^n}{z-n}=\sum_{n=-k}^k\frac{1}{z-2n}-\sum_{n=-k-1}^{k}\frac{1}{z-1-2n}$

Clearly we know that ${\frac{\pi }{2}\cot\frac{\pi z}{2}=\lim_{m\rightarrow\infty}\sum_{m}^{-m}\frac{1}{z-2n}}$ and ${\frac{\pi }{2}\cot\frac{\pi (z-1)}{2}=\lim_{m\rightarrow\infty}\sum_{m}^{-m}\frac{1}{z-1-2n}}$. Thus in the limit we have

$\displaystyle \frac{\pi }{2}\cot\frac{\pi z}{2}-\frac{\pi }{2}\cot\frac{\pi (z-1)}{2}=\frac{\pi}{\sin\pi z}$

2. Product representation

An infinite product of complex numbers

$\displaystyle p_1p_2\cdots p_n=\prod_{n=1}^\infty p_n$

is evaluated by taking limits of the partial products ${P_n=p_1\cdots p_n}$. WLOG, we say that ${P_n}$ converges if and only if at most a finite number of the factors are zero and if the partial products formed by the nonvanishing factors tend to a finite nonzero limit. By writing ${p_n=\frac{P_n}{P_{n-1}}}$, we see that a necessary condition is that ${p_n\rightarrow 1}$. Thus WLOG, we may write

$\displaystyle \prod_{n=1}^\infty(1+a_n)$

so that ${a_n\rightarrow 0}$ is a necessary condition for convergence. If no factor is zero, it is natural to compare with

$\displaystyle \sum_{n=1}^\infty\log(1+a_n)$

We then have the following result:

Lemma 2 The infinite product ${\prod_{n=1}^\infty(1+a_n)}$ converges simultaneously with the series ${\sum_{n=1}^\infty\log(1+a_n)}$ whose terms represent the values of the principal branch of the logarithm.

Proof: Suppose ${\sum_{n=1}^\infty\log(1+a_n)}$ converges. Denote the partial sum ${S_n}$. Then clearly ${P_n=e^{S_n}}$ and by continuity of ${\exp}$ we have ${P_n\rightarrow e^S}$. $\Box$

Lemma 3 The product ${\prod_{n=1}^\infty (1+a_n)}$ converges absolutely if and only if ${\sum_{n=1}^\infty |a_n|}$ converges.

Proof: Recall that ${\lim_{z\rightarrow 0}\frac{\lim(1+z)}{z}=1}$. Then if either the product of the series converges, we see that ${a_n\rightarrow 0}$. Then we employ the following inequality

$\displaystyle (1-\epsilon)|a_n|\leq |\log(1+a_n)|\leq (1+\epsilon)|a_n|$

to derive the result. $\Box$ Next we prove Weierstrass products formula.

Theorem 4 There exists an entire function with prescribed zeros ${a_n}$ provided in the infinitely many zeros case that ${a_n\rightarrow\infty}$. Every entire function with these zeros and no other zeros can be written in the form

$\displaystyle f(z)=z^me^{g(z)}\prod_{n=1}^\infty \left(1-\frac{z}{a_n}\right)^{\frac{z}{a_n}+\frac{(z/a_n)^2}{2}+\cdots+\frac{(z/a_n)^{m_n}}{m_n}}$

where the product is taken over all ${a_n\neq 0}$, ${g(z)}$ is an entire function.

Proof: First we shall show that the product converges. This is direct because

$\displaystyle \begin{array}{rcl} \log\left(1-\frac{z}{a_n}\right)^{\frac{z}{a_n}+\frac{(z/a_n)^2}{2}+\cdots+\frac{(z/a_n)^{m_n}}{m_n}}& = \log(1-z/a_n)+\frac{z}{a_n}+\frac{(z/a_n)^2}{2}+\cdots+\frac{(z/a_n)^{m_n}}{m_n}\\ & =-\sum_{k=m_n+1}^\infty\frac{(z/a_n)^{k}}{k} \end{array}$

By taking ${m_n=n}$, we see that for ${|z|\leq R}$, the series ${\sum_{k=n+1}^\infty\frac{(z/a_n)^{k}}{k}\leq \sum_{k=n+1}^\infty\left(\frac{R}{|a_n|}\right)^{k}<\infty}$ as we may take ${|a_n|}$ sufficiently large enough to make ${\frac{R}{|a_n|}<1}$. $\Box$

Corollary 5 Every function which is meromorphic in the whole plane is the quotient of two entire function.

Proof: Suppose ${f(z)}$ is meromorphic in the whole plane. Let ${g(z)}$ be an entire function with poles of ${f(z)}$ for zeros. Then ${h(z)=f(z)g(z)}$ is an entire function. Then ${f(z)=\frac{h}{g}}$ as desired. $\Box$ The product representation becomes more interesting if it is possible to pick a uniform ${m_n=h}$. The previous proof shows that the product

$\displaystyle \prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)^{\frac{z}{a_n}+\frac{z^2}{2a_n^2}+\cdots+\frac{z^h}{ha_n^h}}$

converges to an entire function if the series ${\sum_{n=1}^\infty\frac{1}{h+1}\left(\frac{R}{a_n}\right)^{h+1}}$ converges for all ${R\in{\mathbb R}^+}$. That is to say ${\sum_{n=1}^\infty\frac{1}{|a_n|^{h+1}}<\infty}$. Assume that ${h}$ is the smallest integer for which this series converges; the previous product expression is called the canonical product associated with the sequence ${(a_n)_{n=1}^\infty}$ and ${h}$ is the genus of the canonical product. If the exponent ${g(z)}$ in the product representation of an entire function ${f(z)}$ with

$\displaystyle z^me^{g(z)}\prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)^{\frac{z}{a_n}+\frac{z^2}{2a_n^2}+\cdots+\frac{z^h}{ha_n^h}}$

reduces to a polynomial, then ${f}$ is of finite genus. The genus of ${f}$ is equal to ${\max(h,\deg(g))}$. For instance, an entire function of genus zero is of the form

$\displaystyle Cz^m\prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)$

with ${\sum_{n=1}^\infty 1/|a_n|<\infty}$. The canonical representation of an entire function of genus 1 is either of the form

$\displaystyle Cz^me^{\alpha z}\prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)e^{\frac{z}{a_n}}$

with ${\sum_{n=1}^\infty\frac{1}{|a_n|^2}<\infty, \sum_{n=1}^\infty\frac{1}{|a_n|}=\infty}$; or of the form

$\displaystyle Cz^m e^{\alpha z}\prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)$

with ${\sum_{n=1}^\infty 1/|a_n|<\infty}$.

Next we discuss an example: ${\sin\pi z}$. It has zeros at ${z=\pm n}$. Since ${\sum\frac 1n=\infty }$adn ${\sum\frac 1{n^2}<\infty}$, we must take ${h=1}$ so that

$\displaystyle \sin\pi z=ze^{g(z)}\prod_{n\neq 0}\left(1-\frac{z}{n}\right)e^{\frac{z}{n}}$

In order to determine ${g(z)}$ we form logarithmic derivative on both sides and find

$\displaystyle \pi\cot\pi z=\frac{1}{z}+g'(z)+\sum_{n\neq 0}\left(\frac{1}{z-n}+\frac 1n\right)=g'(z)+\pi\cot\pi z$

This shows that ${g'(z)=0}$ and ${g(z)}$ reduces to a constant. Since ${\frac{\sin\pi z}{z}\rightarrow \pi}$ as ${z\rightarrow 0}$, we must have ${e^{g(z)}=\pi}$ identically. Hence we can write

$\displaystyle \sin\pi z=\pi z\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)$

Recall that an entire function is said to of order ${\lambda}$ is ${\lambda}$ is the smallest number such that

$\displaystyle M(r)\leq e^{r^{\lambda+\epsilon}}$

for all ${\epsilon>0}$ as soon as ${r}$ is sufficiently large. Here ${M(r)}$ denotes the maximum modulus of ${f(z)}$ on ${|z|\leq r}$. The genus and the order are closely related as seen by the following theorem.

Theorem 6 The genus ${h}$ and the order ${\lambda}$ of an entire function satisfies

$\displaystyle h\leq \lambda\leq h+1$

Proof: Suppose that ${f}$ is an entire function of order ${\lambda}$. First we claim that ${\sum_{n=1}^\infty|a_n|^{-(\lambda+\epsilon)}<\infty}$ for any given ${\epsilon>0}$. This will address the issue of convergence of the canonical product. Given ${r>0}$. Recall Jensen’s formula

$\displaystyle \log|f(0)|=\sum_{n=1}^\infty\log\left|\frac{z_n}{r}\right|+\frac{1}{2\pi}\int_0^{2\pi}\log|f(re^{i\theta})|d\theta$

and the fact

$\displaystyle \int_0^{r}\frac{n(x)}{x}dx=-\sum_{n=1}^\infty\log\left|\frac{z_n}{r}\right|$

Apply these formulae with ${2r}$ and we will get

$\displaystyle \begin{array}{rcl} n(r)\log 2&=\int_r^{2r}\frac{n(r)dx}{x}\\ & \leq \int_0^{2r}\frac{n(x)dx}{x}\\ & \leq \frac{1}{2\pi}\int_0^{2\pi}\log|f(2re^{i\theta})|d\theta-\log|f(0)| \end{array}$

This implies ${n(r)r^{-\lambda-\epsilon}\leq \frac{2}{\log 2}r^{\lambda-\lambda-\epsilon}-\frac{\log|f(0)|}{\log 2}r^{-\lambda-\epsilon}\rightarrow 0}$ as ${r\rightarrow\infty}$. Moreover, note we also have the relation

$\displaystyle k\leq n(|z_k|)<|a_n|^{\lambda+\epsilon}$

This implies

$\displaystyle \sum\frac{1}{|a_n|^{h+1}}\leq\sum\frac{1}{n^{\frac{h+1}{\lambda+\epsilon}}}$

Pick ${\epsilon>0}$ so that ${\lambda+\epsilon will show that ${\sum_{n=1}^\infty|a_n|^{-h-1}<\infty}$. $\Box$