Factorization of infinite polynomials

Question

Does the fundamental theorem of algebra apply for infinite polynomials as well? I haven't studied complex analysis so i'm not asking for a proof just an answer.

Answer 1

First off, we need to talk about what kind of "infinite polynomials" (called power series) we want to talk about. Many of them do not converge outside of the point $x=0$, even though algebraically they are distinct looking expressions. For instance $\sum n!x^n$ and $\sum n!^2x^n$ are distinct formal power series but they both define the same function in the complex plane - namely the constant zero function whose domain is only $\{0\}$.

At the other extreme are entire functions. These functions may be defined by a formal power series (in a complex variable) which converges for all values in the plane. Let's discuss these.

Right off the bat we have an issue: the exponential function $\exp(z)$ has no zeros, and we may use it to create an infinite family of distinct entire functions each having no zeros, namely the entire functions of the form $\exp g(z)$ (where $g(z)$ is also entire).

There are all invertible - we can think of them as "units" (multiplicatively invertible elements) in the "ring" of all meromorphic functions defined on the complex plane. (A ring is an algebraic structure where you have addition and multiplication with the usual properties like distributivity.) So we can think of these as like the "leading coefficients" of a polynomial - thinks we can factor out and then put in front of the rest of the factorization.

If the zeros of an entire function $f(z)$ are given by some set $\Lambda=\{\lambda\}$, then we can't use

$$ \prod_{\lambda\in\Lambda}(z-\lambda)$$

because this product never converges. (Its partial products could converge to zero if $z=\lambda$ for some $\lambda\in\Lambda$, but many sources call that a divergent infinite product. And otherwise the partial products do not converge to any limit.) One way to fix this would be to normalize; consider

$$ \prod_{\lambda\in\Lambda}\left(1-\frac{z}{\lambda}\right).$$

If the magnitudes $|\lambda|$ grow fast enough, the factors of this product might tend to $1$ fast enough for this infinite product to converge. This introduces another small glitch though: what if $\lambda=0$ is a zero of $f(z)$? Well, then let's just put $z^m$ in front of the factorization for some multiplicity $m$, which solves that problem.

Unfortunately, the sizes $|\lambda|$ may not grow fast enough. This can be fixed with yet another kind of normalizing factor, but it is not so obvious this time. Define the new factors

$$ E_n(z)=(1-z)\exp\left(z+\frac{z^2}{2}+\cdots+\frac{z^n}{n}\right). $$

Note that $z+z^2/2+\cdots+z^n/n$ is the $n$th partial sum of the Mercator series for $-\ln(1-z)$, so in fact we have $E_n(z)\to 1$ as $n\to\infty$. (Use the convention $E_0(z)=1-z$, the original.)

Then the Weierstrass factorization theorem says we can factor $f(z)$ as

$$ f(z)=z^m e^{\,\large g(z)}\prod_{\lambda\in \Lambda} E_{n(\lambda)}\left(\frac{z}{\lambda}\right) $$

for some multiplicity $m$, entire function $g(z)$, and whole numbers $n(\lambda)$.

The factorization is clearly not unique. For instance, one could replace $g(z)$ with $g(z)+2\pi i$ without affecting $f(z)$. Also, one could replace any instance of $E_n(z/\lambda)$ with $E_{n+1}(n/\lambda)$ as long as one subtracts $(z/\lambda)^{n+1}/(n+1)$ from $g(z)$ and this would not change $f(z)$.