### Partial fractions and infinite product

#### by tzy9393

**1. Partial fractions**

Theorem 1 (Mittag-Leffler)Let be a sequence of complex numbers with and let be polynomials without constant term. Then there are functions which are meromorphic in the whole plane with poles at the points and the corresponding singular parts . Moreover, the most general meromorphic function of this kind can be written in the form

where ‘s are suitably chosen polynomials and is analytic in the whole plane.

*Proof:* The idea is to subtract a certain portion of analytic part of to ensure convergence without adding any poles. For each , the singular part is analytic in the ball . So we can obtain Taylor expansion of around the origin up to the degree . Using estimate for the remainder term, we have

where and denotes the maximum of in . In particular, this estimate implies that

For each , we may pick large enough to ensure , which can be done by making . In doing so, we obtain

Given , we shall show that the series converges uniformly in . We may split the series into two categories: the first one with and the second one with . Then represents a meromorphic function with poles at because it is a finite sum. In the latter category, since , we have in the disk.Thus above estimate implies

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

Recall that and so . We then can work out a first few terms in the Laurant expansion at the origin which is

Thus the singular part of at 0 is . Moreover, as has period 1 namely for all , we can see that the singular part at integers are . Clearly converges on all compact subsets which do not contain integers. It follows that we may write

where is entire. Write we see that . From the uniform convergence of the series, we can see that converges to 0 as . Using the idendity

we see that uniformly as . In particular, this implies is bounded in the strip . By periodicity, is bounded in the entire complex plane. Then Liouville’s theorem says the reduces to a constant. In this case, must vanish as the limit is zero. Therefore, we have the identity

Integrate on both sides, we obtain that

In order to make the right-hand side converge, we need to substract extra terms from the Taylor series of . In this case, subtracting all constant term is enough. That is which is comparable to . Thus we have

We may bracket terms of and together, and obtain

Now is necessary because both sides are odd.

Next we exploit this fact to investigate the sum

We may split into odd terms and even terms

Clearly we know that and . Thus in the limit we have

**2. Product representation**

An infinite product of complex numbers

is evaluated by taking limits of the partial products . WLOG, we say that 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 , we see that a necessary condition is that . Thus WLOG, we may write

so that is a necessary condition for convergence. If no factor is zero, it is natural to compare with

We then have the following result:

Lemma 2The infinite product converges simultaneously with the series whose terms represent the values of the principal branch of the logarithm.

*Proof:* Suppose converges. Denote the partial sum . Then clearly and by continuity of we have .

Lemma 3The product converges absolutely if and only if converges.

*Proof:* Recall that . Then if either the product of the series converges, we see that . Then we employ the following inequality

to derive the result. Next we prove Weierstrass products formula.

Theorem 4There exists an entire function with prescribed zeros provided in the infinitely many zeros case that . Every entire function with these zeros and no other zeros can be written in the form

where the product is taken over all , is an entire function.

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

By taking , we see that for , the series as we may take sufficiently large enough to make .

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

*Proof:* Suppose is meromorphic in the whole plane. Let be an entire function with poles of for zeros. Then is an entire function. Then as desired. The product representation becomes more interesting if it is possible to pick a uniform . The previous proof shows that the product

converges to an entire function if the series converges for all . That is to say . Assume that is the smallest integer for which this series converges; the previous product expression is called the **canonical product** associated with the sequence and is the **genus** of the canonical product. If the exponent in the product representation of an entire function with

reduces to a polynomial, then is of finite genus. The genus of is equal to . For instance, an entire function of genus zero is of the form

with . The canonical representation of an entire function of genus 1 is either of the form

with ; or of the form

with .

Next we discuss an example: . It has zeros at . Since adn , we must take so that

In order to determine we form logarithmic derivative on both sides and find

This shows that and reduces to a constant. Since as , we must have identically. Hence we can write

**3. Hadamard product**

Recall that an entire function is said to of order is is the smallest number such that

for all as soon as is sufficiently large. Here denotes the maximum modulus of on . The genus and the order are closely related as seen by the following theorem.

Theorem 6The genus and the order of an entire function satisfies

*Proof:* Suppose that is an entire function of order . First we claim that for any given . This will address the issue of convergence of the canonical product. Given . Recall Jensen’s formula

and the fact

Apply these formulae with and we will get

This implies as . Moreover, note we also have the relation

This implies

Pick so that will show that .