Logarithmic Derivative of Infinite Product of Analytic Functions
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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
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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 $U \setminus \set {z \in U : \map f z = 0}$.
$\blacksquare$