Weierstrass Factorization Theorem and Its Applications (ongoing)

April 20, 2024

Sometimes, we need to be able to express a transcendental function in a form of a product of infinitely many linear polynomials which become 0 at each zero.

For examples, when we derive eigenstates of 1-dim quantum harmonic oscillator using Feynmann path integral (Das. Lectures on quantum mechanics. p.515), or deduce a partition function of a system consists of $N$ ideal spin-0 bosons through imaginary time path integral technique (Brown. Quantum field theory. p.96), expressing some trigonometric functions in forms of products of infinite linear polynomials is demanded.

0. Motivation

From the "Fundamental Theorem of Algebra", we know the fact that
"Any polynomial \[ P(z) \;=\; a_0 + a_1 z + a_2 z^2 + \cdots + a_n z^n \] of degree $n$ where each $a_n$ is nonzere has at least one zero."
(Can be deriven from Liouville’s theorem)

Using the mathematical induction, we may generalize our previous statement as
"Any polynomial $P(z)$ of degree $n$ can be expressed as a product of $n$ linear factors : \[ P(z) \;=\; c\, (z-z_1)(z-z_2)\cdots (z-z_n) \] where $c$ and $z_k$ are complex constants."
(Actually, we can directly derive this statement from Rouche’s theorem.)

Then one may ask a plausible question :
"Can any entire function, say $f(z)$, be factorized as \[ f(z) \;=\; c\,\prod_{i=0}^{\infty} \, (z-z_i) \] where $c$ is a complex constant and $\{z_i\}$ is a set of all zeros of $f(z)$?"
In order to answer to this question, we need to check whether the infinite product above converges to $f(z)$ or not.

This notion is the first step toward the Weierstrass Factorization Theorem !

1. Tour De Weierstrass Factorization Theorem

Definition. Weierstrass elementary factor
Let $n$ be a positive integer. An elementary factor $E_n$ (which is a function) is defined as equations below. \[ \begin{aligned} &E_0 (z) \;=\; 1-z \\ &E_n (z) \;=\; (1-z) \, \mathrm{exp}\left(z + \frac{z^2}{2} + \cdots + \frac{z^n}{n} \right) \qquad , n \,\ge\, 1 \end{aligned}\]

Observe that the function $E_n (z/a)$ has an unique zero of order 1 in which $z=a$.

Theorem 1
Let $|z| \, \le \, 1$ and $p \,\ge\, 0$. Then $| E_n (z) - 1 | \; \le \; |z|^{n+1}$.

proof. It is trivial for $n=0$ since $|E_n(z) - 1 | = | (1-z)-1 | = |z|$. Hence, let us focus our attention on positive $n$.

Consider a Laurent series expansion of $E_n (z)$ about $z=0$. Since $E_n (z)$ is non-singular at $z=0$ (bcs it is entire), \[ E_n (z) \;=\; 1 + \sum_{k=1}^{\infty} a_k z^k \] By differentiating $E_n(z)$ with respect to $z$ we obtain \[ E'_n (z) \;=\; \sum_{k=1}^{\infty} k\, a_k\, z^{k-1} \;=\; - z^n \,\mathrm{exp}\left( z + \cdots + \frac{z^n}{n} \right) \] The second equality comes from the definition of $E_n (z)$. By comparing the two expressions of $E'_n(z)$ we obtain $a_1 = a_2 = \cdots = a_n = 0$ since the lowest order of $E'_n (z)$ is 1. Furthermore, since the coefficients of the expantions of $\mathrm{exp}(z+\cdots+z^p /p)$, $a_k \,\le\, 0$ for $k \,\ge\, n+1$.

Observe that \[ E_n (1) \;=\; 0 \;=\; 1 + \sum_{k = n+1}^{\infty} a_k \;=\; - \sum_{k = n+1}^{\infty} |a_k| \qquad \longrightarrow \qquad \sum_{k = n+1}^{\infty} |a_k| \;=\; 1 \]

Hence, for the region in which $|z|\,\le\, 1$, \[ | E_n (z) - 1 | \;=\; \left| \sum_{k=n+1}^{\infty} a_k \, z^k \right| \;=\; |z|^{n+1} \, \left| \sum_{k = n+1}^{\infty} a_k \, z^{n-p-1} \right| \;\le\; |z|^{n+1} \, \sum_{k=n+1}^{\infty} | a_n | \;=\; |z|^{n+1} \]

and this is what we wanted. $\quad \blacksquare $

Theorem 2
Let $\{ a_n \}$ be a complex number sequence such that $ \lim_{n \to \infty} | a_n | = \infty\ $ and anyone is nonzero.
Let $\{ \xi_n \}$ be any sequence of non-negative integers for which \[ \sum_{n=1}^{\infty} \left( \frac{r}{|a_n|} \right)^{\xi_n + 1} \qquad (2) \] converges for all $r>0$.
Then \[ f(z) \;=\; \prod_{n=1}^{\infty} E_{\xi_n} \left( \frac{z}{a_n} \right) \] converges uniformly absolutely and entire with zeros only at points $a_n$.

proof. Suppose that there are integers $\xi_n$ such that eqn(1) is satisfied. Then, according to the previous theorem,

\[ \left| 1 - E_{\xi_n} \left(\frac{z}{a_n}\right) \right| \; \le \; \left| \frac{z}{a_n} \right|^{\xi_n + 1} \; \le \; \left( \frac{r}{|a_n|} \right)^{\xi_n + 1} \qquad (3) \]

whenever $|z| \,\le\, r$ and $r/|a_n| \,\le\, 1$.

Consider an infinite sum which is defined by

\[ \sum_{n=1}^{\infty} \left[ {1 - E_{\xi_n} \left( \frac{z}{a_n} \right) } \right] \; . \]

From the just previous inequality(2) and our hypothesis(3), the infinite sum above is convergent uniformly according to the Weierstrass M-test.

What we need to show is, however, the convergence of $f(z)$. Hence, we need to set up a strategy so as to proceed the present proof.

Weierstrass Factorization Theorem
Let $f$ be an entire function.
Let $\{ a_n \}$ be the set of all non-zero zeros of $f$ repeated according to multiplicity.
Let $f$ has a zero at $z=0$ of order $m \ge 0$.
Then, there is an entire function $f$ and a sequance of integers $\xi_n\} such that \[f(z) \;=\; z^m e^{g(z)} \, \prod_{n=1}^{\infty} E_{\xi_n} \left( \frac{z}{a_n} \right) \; .\]
proof. ALSO WILL BE FILLED SOON...... $\quad \blacksquare$

2. Applications

Trigonometric Functions

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Gamma Function