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How to compute this product ? :

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

where $\cosh$ is the hyperbolic cosine .

  • 1
    Essentially a duplicate of [How to evaluate $\lim\limits_{n\to \infty}\prod\limits_{r=2}^{n}\cos\left(\frac{\pi}{2^{r}}\right)$](http://math.stackexchange.com/questions/1057003/how-to-evaluate-lim-limits-n-to-infty-prod-limits-r-2n-cos-left-frac) because $\cosh(x) = \cos(ix)$ – 2017-01-27

1 Answers 1

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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 :

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

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    I didn't understand this part : $\prod\limits_{k=0}^{n}\frac{\sinh(\frac{x}{2^{k-1}})}{\sinh\left(\frac{x}{2^k}\right)}=\frac{\sinh(2x)}{\sinh\left(\frac{x}{2^n}\right)}$ – 2017-01-27
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    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 : $$\prod_{k=1}^n\frac{a_{k-1}}{a_k}=\frac{a_0}{a_n}$$ – 2017-01-27
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    ok thank you very much it's clearer now :-) . – 2017-01-27