Partial fractions and infinite product

by tzy9393

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)

3. Hadamard product

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<h+1} will show that {\sum_{n=1}^\infty|a_n|^{-h-1}<\infty}. \Box

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