# Proving $\frac{\sin x}{x} =\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right) \left(1-\frac{x^2}{3^2\pi^2}\right)\cdots$

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

• 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. Jun 12, 2012 at 13:43
• I believe you have a typo - the roots of the polynomial should be squared. Jun 12, 2012 at 14:17
• @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. Jun 12, 2012 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. Jun 12, 2012 at 17:04
• Very related
– Pedro
Mar 31, 2013 at 22:46

Real analysis approach.

Let $\alpha\in(0,1)$, then define on the interval $[-\pi,\pi]$ the function $f(x)=\cos(\alpha x)$ and $2\pi$-periodically extended it the real line. It is straightforward to compute its Fourier series. Since $f$ is $2\pi$-periodic and continuous 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 $$\ln\sin z=h(z)+\ln(z)+\sum\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(\ln\left(1-\frac{z}{\pi n}\right)+\frac{z}{\pi n}\right)$$ $$\cot z=\frac{d}{dz}\ln\sin z=h'(z)+\frac{1}{z}+\sum\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(\frac{1}{z-\pi n}+\frac{1}{\pi n}\right)$$ 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.

• This is REALLY great
– PPP
Jan 12, 2014 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! May 18, 2014 at 15:25
• 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 $f(2\pi)=\cos(\pi)=-1\neq 1=f(0)$. Jan 18, 2015 at 10:37
• @man_in_green_shirt, thank you! Fixed the typo. Jun 24, 2017 at 16:25
• @man_in_green_shirt exponents cancels out Jun 25, 2017 at 15:09

It is easy to prove , using complex , worth equality below: \begin{equation} \sin((2n+1)z)=\sum_{k=0}^{n}{2n+1 \choose 2k+1}(-1)^k(\cos z)^{2n-2k} \tag{1}(\sin z)^{2k+1} \end{equation}

Dividing above equality by $\displaystyle \sin(z)\cos^{2n}z$, we get: \begin{equation} \frac{\sin((2n+1)z)}{\sin(z)\cos^{2n}z}=\sum_{k=0}^{n}{2n+1 \choose 2k+1}(-1)^k\tan^{2k}(z) \tag{2} \end{equation}

Dividing (1) by $\displaystyle sin^{2n+1}z$ , we get: \begin{equation} \frac{\sin((2n+1)z)}{\sin^{2n+1}z}=\sum_{k=0}^{n}{2n+1 \choose 2k+1}(-1)^k\cot^{2(n-k)}(z) \tag{3} \end{equation}

Making $\displaystyle \cot^2z=\zeta$, we obtain a polynomial in the variable $\displaystyle \zeta$:

\begin{equation} \frac{\sin((2n+1)z)}{\sin^{2n+1}z}=\sum_{k=0}^{n}{2n+1 \choose 2k+1}(-1)^k\zeta^{n-k} \tag{4} \end{equation}

Which is a polynomial of degree n in $\displaystyle \zeta$.As $\displaystyle \zeta$ It is a function of z, we can find the roots of this polynomial by sine the left side, see that: $\\ \displaystyle \sin((2n+1)z)=0\Rightarrow (2n+1)z=k\pi \Rightarrow z=\frac{k\pi}{2n+1}, k \in \mathbb{N}^{*}|k\leq n \\ \\$

The roots of polynomial is:

\begin{equation} \zeta_k=\cot^2\left(\frac{k\pi}{2n+1}\right), k \in \mathbb{N}^{*}|k\leq n \end{equation}

The fundamental theorem of algebra can factor a polynomial by its roots , then we can rewrite ( 4 ) as:

\begin{equation} \frac{\sin((2n+1)z)}{\sin^{2n+1}z}={2n+1 \choose 1}\prod_{k=1}^{n}(\zeta-\zeta_k) \end{equation} ut we have $\displaystyle \cot^2z=\zeta$ and $\displaystyle \zeta_k=\cot^2\left(\frac{k\pi}{2n+1}\right)$, and we get:

\begin{equation} \frac{\sin((2n+1)z)}{\sin^{2n+1}z}=(2n+1)\prod_{k=1}^{n}\left(\cot^2(z)-\cot^2\left(\frac{k\pi}{2n+1}\right)\right) \end{equation}

Multiplying both sides of the above equality by $\displaystyle \tan^{2n}z$, we get:

\begin{equation} \frac{\sin((2n+1)z)}{\sin z\cos^{2n}z}=(2n+1)\prod_{k=1}^{n}\left(1-\tan^2(z) \cot^2\left(\frac{k\pi}{2n+1}\right)\right) \tag{5} \end{equation}

Comparing ( 2 ) and ( 5) , we have:

\begin{equation} \sum_{k=0}^{n}{2n+1 \choose 2k+1}(-1)^k\tan^{2k}(z)=(2n+1)\prod_{k=1}^{n}\left(1-\tan^2(z) \cot^2\left(\frac{k\pi}{2n+1}\right)\right) \end{equation}

Replacing z by $\displaystyle \arctan \frac{z}{2n+1}$, we get:

\begin{equation} \sum_{k=0}^{n}{2n+1 \choose 2k+1}(-1)^k\left( \frac{z}{2n+1}\right)^{2k}=(2n+1)\prod_{k=1}^{n}\left(1-\left( \frac{z}{2n+1}\right)^2 \cot^2\left(\frac{k\pi}{2n+1}\right)\right) \end{equation} Multiplying both sides of the above equality by $\displaystyle \frac{z}{2n+1}$, we have:

\begin{equation} \sum_{k=0}^{n}{2n+1 \choose 2k+1}(-1)^k\left( \frac{z}{2n+1}\right)^{2k+1}=z\prod_{k=1}^{n}\left(1-\left( \frac{z}{2n+1}\right)^2 \cot^2\left(\frac{k\pi}{2n+1}\right)\right) \tag{6} \end{equation}

Notice that: \begin{equation} \sum_{k=0}^{n}{2n+1 \choose 2k+1}(-1)^k\left( \frac{z}{2n+1}\right)^{2k+1}=\frac{1}{2i}\left[\left(1+\frac{zi}{2n+1}\right)^{2n+1}-\left(1-\frac{zi}{2n+1}\right)^{2n+1}\right] \tag{7} \end{equation}

Substituting ( 7) into ( 6) and taking the limit to infinity , it follows that : \begin{equation} \lim_{n \rightarrow \infty} \frac{1}{2i}\left(\left(1+\frac{zi}{2n+1}\right)^{2n+1}-\left(1-\frac{zi}{2n+1}\right)^{2n+1}\right)=\lim_{n \rightarrow \infty} z\prod_{k=1}^{n}\left(1-\left( \frac{z}{2n+1}\right)^2 \cot^2\left(\frac{k\pi}{2n+1}\right)\right) \end{equation}

And this implies:

\begin{equation} \sin z=\lim_{n \rightarrow \infty} z\prod_{k=1}^{n}\left(1-\left( \frac{z}{2n+1}\right)^2 \cot^2\left(\frac{k\pi}{2n+1}\right)\right) \end{equation} Applying "Tannery's Theorem" we get:

\begin{equation*} \sin z=z\prod_{k=1}^{ \infty}\left(1- \frac{z^2}{k^2\pi^2}\right) \end{equation*}

• Wait a minute there... What is $sen(z)$? Sep 11, 2016 at 23:55
• Is sinz, sorry, because in portuguese senx is sinx. Sep 12, 2016 at 0:00

We will use Hardamard Factoriztation theorem to prove it (see https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem#Hadamard_factorization_theorem).

Observe $$|\sin(\pi z)|\le e^{\pi |z|}$$, hence it has order of growth less than equal to one. Further it has simple zero at every integer $$n \in Z$$. Hence by Hardamard Factorization theorem we get $$\sin (\pi z)= z e^{az+b}\displaystyle \prod_{0 \ne n \in Z}\left(1-\frac zn\right)e^{\frac z n}, \text { for some } a, b \in C.$$ Pairing $$n$$ and $$-n$$ together in the product we get $$\begin{equation} \sin (\pi z)= z e^{az+b}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right). \label{on} \end{equation}$$ From above equation we will get $$\frac {\sin (\pi z)}{\pi z}= \frac{e^{az+b}}{\pi}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$ Taking $$z \to 0$$, in both side we get $$\frac{e^{b}}{\pi}=1$$. From above equation we will get $$\frac {\sin (\pi z)}{\pi z}= {e^{az}}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$ Again using the fact that $$\frac {\sin (\pi z)}{\pi z}$$ is an even function, it can be easily seen that $$a=0$$. Hence we get $$\frac {\sin (\pi z)}{\pi z}= \displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$ Thus $${\sin (\pi z)}= \pi z\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$

$$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}{\operatorname{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$$ \begin{align} {\cal P}_{N}\pars{x} & \equiv \prod_{n = 1}^{N}\pars{1 + {x \over n\pi}} = \prod_{n = 1}^{N}\pars{n + x/\pi \over n} = {1 \over N!}{\Gamma\pars{1 + x/\pi + N} \over \Gamma\pars{1 + x/\pi}} \\[5mm] & = {1 \over N!}{\pars{N + x/\pi}! \over \Gamma\pars{1 + x/\pi}} \sr{{\rm as}\ N\ \to\ \infty}{\sim} {1 \over \Gamma\pars{1 + x/\pi}} {\root{2\pi}\pars{N + x/\pi}^{N + x/\pi + 1/2}\,\,\expo{-N - x/\pi} \over \root{2\pi}N^{N + 1/2}\,\,\expo{-N}} \\[5mm] & \sr{{\rm as}\ N\ \to\ \infty}{\sim} {1 \over \Gamma\pars{1 + x/\pi}} {N^{N + x/\pi +1/2}\,\,\,\,\bracks{1 + \pars{x/\pi}/N}^{N}\,\,\expo{- x/\pi} \over N^{N + 1/2}} \\[5mm] & \sr{{\rm as}\ N\ \to\ \infty}{\sim} {N^{x/\pi} \over \Gamma\pars{1 + x/\pi}} \\[1cm] & \mbox{Therefore,} \\ &\color{#44f}{\lim_{N \to \infty}\bracks{{\cal P}_{N}\pars{x}{\cal P}_{N}\pars{-x}}} = {1 \over \Gamma\pars{1 + x/\pi}\Gamma\pars{1 - x/\pi}} = {1 \over \pars{x/\pi}\Gamma\pars{x/\pi}\Gamma\pars{1 - x/\pi}} \\[5mm] & = {\pi \over x}{1 \over \pi/\sin\pars{\pi\bracks{x/\pi}}} = \bbx{\color{#44f}{\sin\pars{x} \over x}} \\ & \end{align}