# Infinite product of sine function

How to prove the following product? $$\frac{\sin(x)}{x}= \left(1+\frac{x}{\pi}\right) \left(1-\frac{x}{\pi}\right) \left(1+\frac{x}{2\pi}\right) \left(1-\frac{x}{2\pi}\right) \left(1+\frac{x}{3\pi}\right) \left(1-\frac{x}{3\pi}\right)\cdot\ldots$$

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It is rather easy to see that the roots of the left and right hand side are equal (what are the roots of $\sin x$?). However, I do not believe that this proves equality, as e.g., $x$ and $5x$ are polynomials with the same root but they aren't equal. What I want to say is that one has to additional fix the overall multiplicative constant. –  Fabian Jun 12 '12 at 13:43
I believe you have a typo - the roots of the polynomial should be squared. –  process91 Jun 12 '12 at 14:17
Thanks. Obviously once a polynomial is multiplied by a constant, it has the same roots as the original one. Conversely, does it exist a pair of polynomials f and g, such that we cannot multiply f by a constant and reach g, has the same roots? –  Michael Li Jun 12 '12 at 15:20
@Michael For finite polynomials the answer is no. If $f$ and $g$ have the same roots, then for some constant $\alpha$, $f=\alpha g$. This is an immediate consequence of the fundamental theorem of algebra. For infinite polynomials, however, the polynomial may not even have a root (think of the exponential function), and clearly we can exploit this to make different polynomials with the same roots - consider the power series for $e^x$ and $e^x +1$. Here neither function has any roots, but they are not a constant multiple of each other. –  process91 Jun 12 '12 at 16:59
@Michael We could even consider situations like this: $f = e^x-1$ and $g = x$. Then the power series for $f$ and the polynomial $g$ share the same finite number of roots, but neither is a scalar multiple of the other. –  process91 Jun 12 '12 at 17:04

Real analysis approach.

Let $\alpha\in(0,1)$, then define on the interval $[-\pi,\pi]$ the following function $f(x)=\cos(\alpha x)$. It is straightforward to compute its Fourier series. Since $f$ is $2\pi$-periodic and continuously differentiable on $[-\pi,\pi]$, then its Fourier series converges pointwise to $f$ on $[-\pi,\pi]$: $$f(x)=\frac{2\alpha\sin\pi\alpha}{\pi}\left(\frac{1}{2\alpha^2}+\sum\limits_{n=1}^\infty\frac{(-1)^n}{\alpha^2-n^2}\cos nx\right), \quad x\in[-\pi,\pi]\tag{1}$$ Now take $x=\pi$, then we get $$\cot\pi\alpha-\frac{1}{\pi\alpha}=\frac{2\alpha}{\pi}\sum\limits_{n=1}^\infty\frac{1}{\alpha^2-n^2}, \quad\alpha\in(-1,1)\tag{2}$$ Fix $t\in(0,1)$. Note that for each $\alpha\in(0,t)$ we have $|(\alpha^2-n^2)^{-1}|\leq(n^2-t^2)^{-1}$ and the series $\sum_{n=1}^\infty(n^2-t^2)^{-1}$ is convergent. By Weierstrass $M$-test the series in the right hand side of $(2)$ is uniformly convergent for $\alpha\in(0,t)$. Hence we can integrate $(2)$ over the interval $[0,t]$. And we get $$\ln\frac{\sin \pi t}{\pi t}=\sum\limits_{n=1}^\infty\ln\left(1-\frac{t^2}{n^2}\right), \quad t\in(0,1)$$ Finally, substitute $x=\pi t$, to obtain $$\frac{\sin x}{x}=\prod\limits_{n=1}^\infty\left(1-\frac{x^2}{\pi^2 n^2}\right), \quad x\in(0,\pi)$$

Complex analysis approach

We will need the following theorem (due to Weierstrass).

Let $f$ be an entire function with infinite number of zeros $\{a_n:n\in\mathbb{N}\}$. Assume that $a_0=0$ is zero of order $r$ and $\lim\limits_{n\to\infty}a_n=\infty$, then $$f(z)= z^r\exp(h(z))\prod\limits_{n=1}^\infty\left(1-\frac{z}{a_n}\right) \exp\left(\sum\limits_{k=1}^{p_n}\frac{1}{k}\left(\frac{z}{a_n}\right)^{k}\right)$$ for some entire function $h$ and sequence of positive integers $\{p_n:n\in\mathbb{N}\}$. The sequence $\{p_n:n\in\mathbb{N}\}$ can be chosen arbitrary with only one requirement $-$ the series $$\sum\limits_{n=1}^\infty\left(\frac{z}{a_n}\right)^{p_n+1}$$ is uniformly convergent on each compact $K\subset\mathbb{C}$.

Now we apply this theorem to the entire function $\sin z$. In this case we have $a_n=\pi n$ and $r=1$. Since the series $$\sum\limits_{n=1}^\infty\left(\frac{z}{\pi n}\right)^2$$ is uniformly convergent on each compact $K\subset \mathbb{C}$, then we may choose $p_n=1$. In this case we have $$\sin z=z\exp(h(z))\prod\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(1-\frac{z}{\pi n}\right)\exp\left(\frac{z}{\pi n}\right)$$ Let $K\subset\mathbb{C}$ be a compact which doesn't contain zeros of $\sin z$. For all $z\in K$ we have $$\ln\sin z=h(z)+\frac{1}{z}+\sum\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(\ln\left(1+\frac{z}{\pi n}\right)+\frac{z}{\pi n}\right)$$ $$\cot z=\frac{d}{dz}\ln\sin z=h'(z)+1+\sum\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(\frac{1}{z-\pi n}+\frac{1}{\pi n}\right)$$ It is known that (here you can find the proof) $$\cot z=\frac{1}{z}+\sum\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(\frac{1}{z-\pi n}+\frac{1}{\pi n}\right).$$ hence $h'(z)=0$ for all $z\in K$. Since $K$ is arbitrary then $h(z)=\mathrm{const}$. This means that $$\sin z=Cz\prod\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(1-\frac{z}{\pi n}\right)\exp\left(\frac{z}{\pi n}\right)$$ Since $\lim\limits_{z\to 0}z^{-1}\sin z=1$, then $C=1$. Finally, $$\frac{\sin z}{z}=\prod\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(1-\frac{z}{\pi n}\right)\exp\left(\frac{z}{\pi n}\right)= \lim\limits_{N\to\infty}\prod\limits_{n=-N,n\neq 0}^N\left(1-\frac{z}{\pi n}\right)\exp\left(\frac{z}{\pi n}\right)=$$ $$\lim\limits_{N\to\infty}\prod\limits_{n=1}^N\left(1-\frac{z^2}{\pi^2 n^2}\right)= \prod\limits_{n=1}^\infty\left(1-\frac{z^2}{\pi^2 n^2}\right)$$ This result is much more stronger because it holds for all complex numbers. But in this proof I cheated because series representation for $\cot z$ given above require additional efforts and use of Mittag-Leffler's theorem.

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Thanks. Nice and useful proofs. –  Michael Li Jun 13 '12 at 3:09
This is REALLY great –  Lucas Zanella Jan 12 '14 at 4:20
One doesn't need the Mittag-Leffler's theorem for the expression of $\cot z$; there are alternative ways of getting it. Nice answer there though! –  ireallydonknow May 18 '14 at 15:25
@ireallydonknow A you sure this approaches will give the formula valid for complex values, not only for reals ones? –  Norbert May 18 '14 at 15:49
The function $f(x)=\cos(\alpha x)$ you define is not actually $2\pi$ periodic (as you claim). For example if $\alpha=1/2$, then $f(2\pi)=\cos(\pi)=-1\neq 1=f(0)$. –  PhoemueX Jan 18 at 10:37