Is there any deeper meaning to trigonometric identity $${{\sin x}\over x} = \prod_{k=1}^{\infty} \cos\left({x\over{2^k}}\right)$$ beyond it corresponding to characteristic functions of i.i.d. random variables?


I give 2 proofs, that show that there is indeed something more than the pure analytical formula, beyond its intrinsic beauty.

First proof:

Let us consider the formula under the form:

$$\frac{\sin \pi x}{\pi x} = \prod_{k=1}^{\infty}\cos{\frac{\pi x}{2^k}}= \lim_{n \rightarrow \infty}{\prod_{k=1}^{n}\cos{\frac{\pi x}{2^k}}}$$

Its Fourier Transform (using Euler formula $\cos(a)=\frac{1}{2}(e^{ia}+e^{-ia}))$ and the fact that a product is transformed into a convolution, is:

$$\mathbb{1}_{[-\frac{1}{2},\frac{1}{2}]}(u) = \lim_{n \rightarrow \infty}{{\Huge \ast}_{k=1}^{n}\frac{1}{2}\left(\delta(u+\frac{k}{2^{k+1}})+\delta(u-\frac{k}{2^{k+1}})\right)} \ \ \ (*)$$

Now, let us expand the convolution product ${\Huge \ast}_{k=1}^{n}$ on the RHS, in the particular case $n=3$ in order to understand what is going on, i.e.,

(using property $\delta(u-a)*\delta(u-b)=\delta(u-(a+b))$):




The general case being:


Explanation: each term obtained in the development of the product gives rise to a number that can be (up to factorization by $1/2$) written as the base-2 decomposition of a number of the form $\dfrac{N}{2^{n+1}}$, all numbers of this form being represented exactly once. This is why we obtain a regular spacing.

(Nice) interpretation :

  • Intuitively, the uniformly spread "mass" on interval $[-\frac{1}{2},\frac{1}{2}]$ is placed into $2^n$ "Dirac bins".

  • In more probabilistic (or "measure theory") and rigorous terms, the transformed formula expresses a convergence of a regularly spaced discrete uniform distribution to its continuous counterpart.

The figure below illustrates this kind of equivalence between the continuous measure represented by the characteristic function of $[-1/2,1/2]$ and the "Dirac forest" made (in our particular case) of 8 regularly spaced "Dirac trees" of height=mass 1/8 each.

enter image description here

Edit: A second proof using a very different, entirely geometrical, interpretation:

Let us consider the figure below with $A_k(\cos\left(\dfrac{\theta}{2^{k-1}}\right),\sin\left(\dfrac{\theta}{2^{k-1}}\right))$ and $H_k=AA_k \cap OA_k$.

Let us establish relationship

$$AH_{k+1}=\dfrac{AH_k}{2 \cos\left(\dfrac{\theta}{2^{k}}\right)} \ \ \ (1)$$

In order to follow more easily, we will take the particular case $n=2$ WLOG:

Relationship $AH_{3}=\dfrac{AH_2}{2 \cos\left(\dfrac{\theta}{4}\right)}$ is the consequence of two facts:

  • $AA_2=2 AH_3$ ($H_2$ is the foot of the altitude in isosceles triangle $AOA_1$) and

  • $AH_2=AA_2 \cos\left(\dfrac{\theta}{4}\right)$ (inscribed angle theorem: (https://en.wikipedia.org/wiki/Inscribed_angle): the angle under which arc $A_1A_2$ is "seen" from $A$ is half the angle under which it is "seen" from the center $0$).

Thus, from (1),

$$AH_{n+1}=\dfrac{AH_1}{2^{n} \prod_{k=1}^{n}\cos\left(\dfrac{\theta}{2^{k}}\right)} \ \ \ (2)$$ But, if $n$ is large, $$AH_{n+1}=H_{n+1}A_{n}=sin\left(\dfrac{\theta}{2^{n}}\right)\approx \dfrac{\theta}{2^{n}} \ \ \ (3) $$

(classical approximation for small angles). Thus, taking into account that $AH_1=\sin \theta$, grouping (2) and (3) gives:

$$\dfrac{\sin \theta}{2^{n} \prod_{k=1}^{n}\cos\left(\dfrac{\theta}{2^{k}}\right)}\approx\dfrac{\theta}{2^{n}} \ \ \ (4)$$

After cancellation of $2^n$ in both sides, formula (4) is clearly a proof of the infinite product formula.

Note: A rigorous treatment would necessitate to replace the approximation argument by a limit argument, which is rather easy...

enter image description here

  • $\begingroup$ So would it be ok if it was $3^k $ instead of $2^k $ in the denominator? $\endgroup$ – N.S.JOHN Apr 23 '16 at 3:03
  • $\begingroup$ @N.S.JOHN Do you mean changing $2^k$ into $3^k$ into the main formula ? It would be false because $\cos(x/2^k)<\cos (x/3^k)$ for $|x|<\pi $ $\endgroup$ – Jean Marie Apr 23 '16 at 6:28
  • $\begingroup$ Oh I thought that $ x$ tends to 0. $\endgroup$ – N.S.JOHN Apr 23 '16 at 8:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.