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, $$\frac{\sin x}{x}=\prod_{n=1}^\infty \cos(x/2^n).$$


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

  • 1
    $\begingroup$ Why do you want to avoid telescoping? $\endgroup$ Commented Jan 2, 2020 at 23:33
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    $\begingroup$ 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*} $\endgroup$
    – user632577
    Commented Jan 2, 2020 at 23:44
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    $\begingroup$ 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. $\endgroup$ Commented Jan 2, 2020 at 23:50
  • $\begingroup$ @kimchilover very nice $\endgroup$ Commented Jan 3, 2020 at 0:10
  • $\begingroup$ @Lucas Henrique I am curious to know if there are other ways to evaluate this product. $\endgroup$
    – Diffusion
    Commented Jan 3, 2020 at 2:13

2 Answers 2


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 $$\varphi(t)=\frac 12\int_{-1}^1 \exp(itx)\,dx = \frac{\sin(t)}t.$$


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