# Logarithmic Derivative of Infinite Product of Analytic Functions

## Theorem

Let $D \subseteq \C$ be open.

Let $\sequence {f_n}$ 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 $\ds \prod_{n \mathop = 1}^\infty f_n$ converge locally uniformly to $f$.

Then:

$\ds \dfrac {f'} f = \sum_{n \mathop = 1}^\infty \frac {f_n'} {f_n}$

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

## Outline of Proof

Using Corollary to Logarithm of Infinite Product of Complex Functions we can write $\log f$ locally as $\ds \sum_{n \mathop = 1}^\infty \log f_n$ 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 $\map f {z_0} \ne 0$.

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:

$\map {f_n} z \ne 0$ for $n \ge n_0$ and $z \in U$
The series $\ds \sum_{n \mathop = n_0}^\infty \log f_n$ converges uniformly on $U$ to $\log g + 2 k \pi i$, where $g = \ds \prod_{n \mathop = n_0}^\infty f_n$.

We have, for $z \in U$:

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

and

 $\ds \frac {g'} g$ $=$ $\ds \paren {\log g}'$ Logarithmic Derivative is Derivative of Logarithm $\ds$ $=$ $\ds \sum_{n \mathop = n_0}^\infty \paren {\log f_n}' - \paren {2 k \pi i}'$ 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 $\dfrac {f'} f = \ds \sum_{n \mathop = 1}^\infty \frac {f_n'} {f_n}$ converges locally uniformly on $U \setminus \set {z \in U : \map f z = 0}$.

$\blacksquare$