# Logarithmic Derivative of Infinite Product of Analytic Functions

## Theorem

Let $D\subset\C$ be open.

Let $(f_n)$ be a sequence of analytic functions $f_n:D\to\C$.

Let none of the $f_n$ be identically zero on any open subset of $D$.

Let the product $\displaystyle \prod_{n \mathop = 1}^\infty f_n$ converge locally uniformly to $f$.

Then $\displaystyle \frac{f'}f = \sum_{n \mathop = 1}^\infty \frac{f_n'}{f_n}$

and the series converges locally uniformly in $D\setminus\{z\in D : f(z) = 0\}$.

## Outline of Proof

Using Corollary to Logarithm of Infinite Product of Complex Functions we can write $\log f$ locally as $\displaystyle \sum_{n \mathop = 1}^\infty \log f_n$ up to a constant and some terms.

By Derivative of Uniform Limit of Analytic Functions we can differentiate term-wise.

## Proof

Note that by Infinite Product of Analytic Functions is Analytic, $f$ is analytic.

Let $z_0\in D$ with $f(z_0)\neq0$.

By the Corollary to Logarithm of Infinite Product of Complex Functions, there exist $n_0\in\N$, $k\in\Z$ and a open neighborhood $U$ of $z_0$ such that:

• $f_n(z)\neq0$ for $n\geq n_0$ and $z\in U$
• The series $\displaystyle \sum_{n \mathop = n_0}^\infty \log f_n$ converges uniformly on $U$ to $\log g + 2k\pi i$, where $g = \displaystyle \prod_{n\mathop =n_0}^\infty f_n$.

We have, for $z\in U$:

 $\ds \frac {f'}{f}$ $=$ $\ds \frac {(f_1\cdots f_{n_0-1}\cdot g)'}{f_1\cdots f_{n_0-1}\cdot g}$ $\ds$ $=$ $\ds \frac {f_1'}{f_1} + \cdots + \frac {f_{n_0-1}'}{f_{n_0-1} } + \frac{g'}g$ Logarithmic Derivative of Product of Analytic Functions

and

 $\ds \frac {g'}{g}$ $=$ $\ds (\log g)'$ Logarithmic Derivative is Derivative of Logarithm $\ds$ $=$ $\ds \sum_{n \mathop = n_0}^\infty (\log f_n)' - (2k\pi i)'$ Derivative of Uniform Limit of Analytic Functions $\ds$ $=$ $\ds \sum_{n \mathop = n_0}^\infty \frac {f_n'}{f_n}$ Logarithmic Derivative is Derivative of Logarithm

and the series converges locally uniformly on $U$.

Thus $\frac{f'}f = \displaystyle \sum_{n \mathop = 1}^\infty \frac {f_n'}{f_n}$ converges locally uniformly on $U\setminus\{z\in U : f(z) = 0\}$.

$\blacksquare$