I would like to show that $\cosh(z) – \cos(z) = z^2\prod_{n = 1}^{\infty}\left(1 + \frac{z^4}{4\pi^4n^4} \right)$.
Firstly, I have found that solutions to the equation $\cosh(z) – \cos(z) = 0$ are of the form $z = (1 \pm i)n\pi$ ; $n \in \mathbb{Z}$.
I then would like to appeal to the Weierstrass factorization theorem.
By theorem, it will follow that:
$$\cosh(z) – \cos(z) = ze^{g(z)}\prod_{n = 1}^{\infty}\left(\frac{z}{(1 \pm i)n\pi}\right)E_{p_n} $$
where the $E_{p_n}$ are the Weierstrass elementary factors and $g$ is some entire function.
At this point, I am pretty stuck as to how to transform this into the desired infinite product shown above. Any tips?
Best Answer
Let's start on the RHS. It is $$\left[z\prod_{n=1}^\infty\left(1+\frac{iz^2}{2\pi^2 n^2}\right)\right] \left[z\prod_{n=1}^\infty\left(1-\frac{iz^2}{2\pi^2 n^2}\right)\right].$$ The first bracket here is $$z\prod_{n=1}^\infty\left(1-\frac{((1-i)z)^2}{4\pi^2 n^2}\right) =(1+i)\sin\frac{(1-i)z}2$$ and the second is $$(1-i)\sin\frac{(1+i)z}2.$$ The product is $$2\sin\frac{(1-i)z}2\sin\frac{(1+i)z}2 =\cos iz-\cos z=\cosh z-\cos z.$$