Infinite products and the Weierstrass factorization theorem

Weierstrass_factorization.tex

% This is the tex file used in the video explaining the Weierstrass factorization theorem:
% https://youtu.be/SEHPs8nrqLY
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% In this video we're going to explain the Weierstrass factorization theorem, giving rise to infinite product representations of functions. Classical examples are that of the Gamma function or the sine function.
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% https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem
% https://en.wikipedia.org/wiki/Infinite_product

\begin{abstract}
   We give a quick and dirty motivation for the Weierstrass factorization theorem. 
   The motivation is gaining conceptual familiarity with the form of the objects involved.
   The tradeoff for conciseness is not elaborating on convergence discussions much.
\end{abstract}

$L_S(x) = \sum_{n\in{S}} \frac{1}{n}x^n,$

which has

$L_S(0)=0.$

For $S={\mathbb N}_+$, we have

$L_{{\mathbb N}_+}(x) = \sum_{n=1}^\infty \frac{1}{n} x^n = -\log( 1-x ) = \log( \frac{1}{1-x} ), \ \ \ \ |x|<1$

$\vspace{.03cm}$

Consider a finite set $F\subset {\mathbb N}_+$ and the infinite $I = {\mathbb N}_+ \setminus F$, such that

${\rm e}^{L_{ I}(x) } \cdot {\rm e}^{L_F(x) }  = {\rm e}^{L_{ {\mathbb N}_+ }(x) }  = \frac{1}{1-x} ,$

and so

${\rm e}^{L_{ I}(x) } \cdot (1-x) \,{\rm e}^{L_F(x) }  = 1 .$

As $0\notin A$, at $x=0$ both factors are independently $1$ and also have rather flat derivatives. 
For $\lim_{x\to 1}$, the right factor goes to $0$.
We're not going to discuss the nice convergence results stemming from this behaviour in these notes.

For a (finite or infinite) sequence $(z_k)_k$ with $z_k\neq{0}$, wherever it converges, the product

$\prod_k (1-\frac{z}{z_k})\, {\rm e}^{L_{F_k}(\frac{z}{z_k}) }, \ \ \ \ |z|<{\rm min}(\{z_k\})$

is zero at all $z_k$. 
Inductively using ${\rm e}^a{\rm e}^b={\rm e}^{a+b}$, the right hand term, whenever it actually converges, then reads
${\rm e}^{\sum_{k} L_{I_k}(\frac{z}{z_k}) } $.

Consider now a sequence with $z_k$ ascending, so that the factors have to come back to $0$ infinitely many times as you go outwards on the complex plane.
Add zero of order $m$ are $z=0$ by multiplying $z^m$. 
Moreover, modify other values by multiplication with a generic non-zero function ${\rm e}^{g(z)}$. 

$f(z) = {\rm e}^{g(z)} z^m \cdot \prod_k (1-\frac{z}{z_k})\, {\rm e}^{L_{F_k}(\frac{z}{z_k}) }.$

$\vspace{.03cm}$

Logarithmic derivative and its derivative:

$\dfrac{f'(z)}{f(z)} = \log(f(z))' = g'(z) + \dfrac{m}{z} + \sum_k \left( \dfrac{1}{z-z_k} + \dfrac{1}{z} \sum_{n\in{F}} \left(\frac{z}{z_k}\right)^n \right),$

$\left(\dfrac{f'(z)}{f(z)}\right)' = \log(f(z))'' = g''(z) - \dfrac{m}{z^2} + \sum_k \left( \dfrac{1}{(z-z_k)^2} + \dfrac{1}{z^2} \sum_{n\in{F}} \left(\frac{z}{z_k}\right)^n (n-1) \right)$