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I just answered this question

distribution of infinite sum of $\sum (2x_n -1)/2^n$

by using the formula in the title which I lifted off a random formula sheet on the internet. My question is, how do we derive this? I have never learnt how to sum infinite products like this.

I believe there is also a formula for $\cosh$ (by Osborn's rule). A justification why this follows from the $\cos$ case would also be nice.

Community
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Lost1
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    Replace $\infty$ by $n$. Multiply by $2^n\sin\left(\frac{x}{2^n}\right)$. Take $n\to \infty$ and use the double angle formula $\sin(2x)=2\sin(x)\cos(x)$ several times. Be happy! – chubakueno Jun 29 '14 at 19:09
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    Hint: Stop the product at a random number n, and multiply it with $\sin\dfrac x{2^n}$ , then compute the limit for the newly obtained formula using l'Hopital. – Lucian Jun 29 '14 at 19:09

3 Answers3

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Hint: $$ \sin(2x) = 2\sin(x)\cos(x) $$ so $$ \cos \left(\frac{x}{2^k}\right) = \frac{\sin\left(\frac{x}{2^k}\right)}{2\sin\left(\frac{x}{2^{k+1}}\right)} $$

the_fox
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Mathmo123
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Use formula $\displaystyle \cos x=\frac{e^{ix}+e^{-ix}}{2}$ and $a^2-b^2=(a-b)(a+b)$ $n$ times for $\displaystyle \frac{(e^{\frac{x}{2^n}}-e^{-\frac{x}{2^n}})}{\frac{x}{2^n}}\Pi_{k=1}^{n} \frac{e^{\frac{x}{2^k}}+e^{-\frac{x}{2^k}}}{2}=\frac{e^{ix}-e^{-ix}}{x}$

Because $\displaystyle \lim_{n \to\infty}\frac{(e^{\frac{x}{2^n}}-e^{-\frac{x}{2^n}})}{\frac{x}{2^n}}=2$ we have:

$\frac{e^{ix}-e^{-ix}}{x}\lim_{n \to \infty}\frac{(e^{\frac{x}{2^n}}-e^{-\frac{x}{2^n}})}{\frac{x}{2^n}}\Pi_{k=1}^{n} \frac{e^{\frac{x}{2^k}}+e^{-\frac{x}{2^k}}}{2}=\lim_{n \to \infty}\frac{(e^{\frac{x}{2^n}}-e^{-\frac{x}{2^n}})}{\frac{x}{2^n}} \Pi_{k=1}^{\infty}\cos \frac{x}{2^k}=2\Pi_{k=1}^{\infty}\cos \frac{x}{2^k}$

Finally $\frac{e^{ix}-e^{-ix}}{2x}=\frac{\sin x}{x}$

agha
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Let

$P_n=\prod\limits_{k=1}^n\cos(\frac{x}{2^{k}})$

Then we have:

$\sin(\frac{x}{2^{n}})P_n=\cos(\frac{x}{2}) \dots \cos(\frac{x}{2^{n}})\sin(\frac{x}{2^{n}})=\frac{1}{2}\cos(\frac{x}{2}) \dots \sin(\frac{x}{2^{n-1}})$

=$\dots$=$\frac{\sin(x)}{2^{n}}$

by applying double angle formula for $\sin(2x)$ repeatedly.

Then

$P_n=\frac{\sin(x)}{2^{n}}\frac{1}{\sin(\frac{x}{2^{n}})}$

Now as $n \rightarrow \infty$ and $\sin(\frac{x}{2^{n}}) \rightarrow \frac{x}{2^{n}}$

and you obtain required result. You can produce rigourous arguments about limit part if you want, but this is essentially how you calculate infinite product.

user160738
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