# How to derive the identity $\sin x/x=\prod_{n=1}^\infty \cos(x/2^n)$ without using telescoping?

I am wondering how to derive the following equality $$\frac{\sin x}{x}=\prod_{n=1}^\infty \cos(x/2^n)\tag{1}$$ without using the method of telescoping. I know that there is already a question on deriving this infinite product representation of $$\sin x/x$$, but all the answers in the link telescope the product. Here is the method, for completeness.

First, we can use the trigonometric identity $$\sin x=2\cos (x/2)\sin(x/2)$$ to yield $$\cos(x/2)=\frac{\sin x}{2 \sin(x/2)}$$. More generally, this implies that $$\cos(x/2^{n})=\frac{\sin (x/2^{n-1})}{2 \sin(x/2^{n})}$$

Our infinite product is thus $$\prod_{n=1}^\infty\cos(x/2^n)=\frac{\sin (x)}{2 \sin(x/2)}\cdot \frac{\sin (x/2)}{2 \sin(x/4)} \cdot \frac{\sin (x/4)}{2 \sin(x/8)} \cdots$$ Treating the product as a limit of a finite product $$f_k(x)=\prod_{n=1}^k \cos(x/2^n)$$, we notice that $$f_k(x)=\frac{\sin(x)}{2^k\sin(x/2^k)},$$ with $$\lim_{k\to\infty} f_k(x)=\sin x/x$$. Thus, $$\frac{\sin x}{x}=\prod_{n=1}^\infty \cos(x/2^n).$$

Question:

How to show that $$(1)$$ is true without using telescoping?

• Why do you want to avoid telescoping? Commented Jan 2, 2020 at 23:33
• Note the Weierstrass factorization theorem:\begin{align*} \frac{\sin x}{x}&=\prod_{n=1}^\infty\left(1-\frac{x^2}{n^2\pi^2}\right)\\ \cos\frac{x}{2^n}&=\prod_{m=0}^\infty\left(1-\frac{x^2}{(2m+1)^2 2^{2(n-1)}\pi^2}\right) \end{align*}
– user632577
Commented Jan 2, 2020 at 23:44
• You could use the fact that $\sum_{n\ge1}\pm 2^{-n}$ (where the signs are chosen independently and uniformly) is uniformly distributed on $[-1,1]$; the desired equality relates two formulas for the characteristic function of such a random variable. Commented Jan 2, 2020 at 23:50
• @kimchilover very nice Commented Jan 3, 2020 at 0:10
• @Lucas Henrique I am curious to know if there are other ways to evaluate this product. Commented Jan 3, 2020 at 2:13

You can do this by a trick that is essentially the same as looking at this product in frequency domain.

To avoid any analytical difficulty, let's examine finite sums; we have $$\prod_{n=1}^{k}\cos(x/2^n)=\prod_{n=1}^k\left(\frac{e^{ix/2^n}+e^{-ix/2^n}}{2}\right)$$ We can expand the sum on the right as $$\frac{1}{2^k}\sum_{\sigma\in\{-1,1\}^k}\exp\left(ix\cdot \left(\sigma_1\cdot \frac{1}2+\sigma_2\cdot \frac{1}{2^2}+\ldots+\sigma_k\cdot \frac{1}{2^k}\right)\right)$$ where $$\sigma$$ is a string of $$k$$ terms in $$\{-1,1\}$$ representing which side of the sum within the former product was followed.

One can see that for $$n=1$$, the angular frequencies encountered (i.e. the coefficient of $$ix$$) are $$1/2$$ and $$-1/2$$ . For $$n=2$$, the frequencies are $$-3/4,\,-1/4,\,1/4,\,3/4$$. We can prove via induction that the possible values of that coefficient are just the set of numbers of the form $$a/2^k$$ for odd integers $$a$$ between $$-2^k$$ and $$2^k$$. Thus, the partial sum works out to: $$\frac{1}{2^k}\cdot \sum_{\substack{a\text{ odd}\\ -2^k < a < 2^k}}\exp\left(ix \cdot \frac{-a}{2^k}\right)$$ We could bail out at this step and recognize that the sum is actually a geometric series (with ratio $$\exp\left(\frac{ix}{2^{k-1}}\right)$$), which would lead us back to the expression you derived for the partial sums. However, we could also recognize this an average of equally spaced evaluations of the function $$z\mapsto \exp(ix\cdot z)$$ over the interval $$[-1,1]$$ with more evaluations as $$k$$ increases; thus, in the limit, this product becomes an integral giving the average value of $$\exp(ixt)$$ over the interval $$[-1,1]$$: $$\lim_{k\rightarrow\infty}\prod_{n=1}^k\cos(x/2^n) = \frac{1}2\int_{-1}^1\exp(ixt)\,dt$$ Of course, this is just integrating an exponential function, which can be done easily, and works out to $$\frac{\sin(x)}x$$.

Here is a different packaging of the same basic argument presented in Milo Brandt's answer.

Let $$X_k=\sum_{j=1}^k \sigma_j 2^{-j}$$, where the $$\sigma_i$$ are iid $$\pm1$$ random variables. This has uniform distribution on the $$2^k$$ points uniformly spaced $$2^{1-k}$$ apart in the range from $$-1+2^{-k}$$ to $$1-2^{-k}$$, as can be seen from the binary expansions of the integers from $$0$$ to $$2^k$$. One can verify directly that $$X_k$$ converges in distribution to the continuous uniform distribution on $$[-1,1]$$.

The characteristic function of $$X_k$$, namely the function $$\phi_k(t)=E[\exp(itX_k)]$$ is given by $$\prod_{j=1}^k E[\exp(i \sigma_j 2^{-j})] = \prod_{j=1}^k \cos(t2^{-j})$$.

By Lévy's continuity theorem, for each $$t$$, one has $$\lim_{k\to\infty}\varphi_k(t)=\varphi(t)$$, where $$\varphi(t)$$ is the characteristic function of the uniform distribution on $$[-1,1]$$, which is $$\varphi(t)=\frac 12\int_{-1}^1 \exp(itx)\,dx = \frac{\sin(t)}t.$$