# how to evaluate $\prod \limits_{k=0}^{n} \cosh\left(\frac{x}{2^{k}}\right)$?

How to compute this product ? :

$\prod \limits_{k=0}^{n} \cosh(\frac{x}{2^{k}})$

where $\cosh$ is the hyperbolic cosine .

It is well known that $\sinh(x+y)=\sinh(x)\cosh(y)+\cosh(x)\sinh(y)$.

Taking $x=y$ leads to the "duplication formula" :

$\sinh(2x)=2\sinh(x)\cosh(x)$

So, if $x\neq0$ then :

$$\cosh(x)=\frac{\sinh(2x)}{2\sinh(x)}$$

so that, for all $(x,n)\in\mathbb{R}^\times\times\mathbb{N}$ :

$$\prod_{k=0}^n\cosh\left(\frac{x}{2^k}\right)=\frac{1}{2^{n+1}}\prod_{k=0}^n\frac{\sinh(\frac{x}{2^{k-1}})}{\sinh\left(\frac{x}{2^k}\right)}=\boxed{\frac{\sinh(2x)}{2^{n+1}\sinh\left(\frac{x}{2^n}\right)}}$$

We can now compute the limit, as $n\to\infty$, of this product and get :

$$\forall x\in\mathbb{R}^\times,\quad\prod_{k=0}^\infty\cosh\left(\frac{x}{2^k}\right)=\frac{\sinh(2x)}{2x}$$

• I didn't understand this part : $\prod\limits_{k=0}^{n}\frac{\sinh(\frac{x}{2^{k-1}})}{\sinh\left(\frac{x}{2^k}\right)}=\frac{\sinh(2x)}{\sinh\left(\frac{x}{2^n}\right)}$ – Hilbert Jan 27 '17 at 20:13
• This product looks like : $$\prod_{k=1}^n\frac{a_{k-1}}{a_k}=\frac{a_0}{a_1}\,\frac{a_1}{a_2}\cdots\frac{a_{n-1}}{a_n}$$ if you look carefully at it, you will notice that almost all factors cancel ! And after that remains : $$\prod_{k=1}^n\frac{a_{k-1}}{a_k}=\frac{a_0}{a_n}$$ – Adren Jan 27 '17 at 20:20
• ok thank you very much it's clearer now :-) . – Hilbert Jan 27 '17 at 20:26