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

  • 3
    $\begingroup$ 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. $\endgroup$ – Fabian Jun 12 '12 at 13:43
  • $\begingroup$ I believe you have a typo - the roots of the polynomial should be squared. $\endgroup$ – process91 Jun 12 '12 at 14:17
  • 3
    $\begingroup$ @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. $\endgroup$ – process91 Jun 12 '12 at 16:59
  • 2
    $\begingroup$ @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. $\endgroup$ – process91 Jun 12 '12 at 17:04
  • 1
    $\begingroup$ Very related $\endgroup$ – Pedro Tamaroff Mar 31 '13 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.

| cite | improve this answer | |
  • 1
    $\begingroup$ This is REALLY great $\endgroup$ – Lucas Zanella Jan 12 '14 at 4:20
  • $\begingroup$ 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! $\endgroup$ – ireallydonknow May 18 '14 at 15:25
  • 1
    $\begingroup$ 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 $f(2\pi)=\cos(\pi)=-1\neq 1=f(0)$. $\endgroup$ – PhoemueX Jan 18 '15 at 10:37
  • 1
    $\begingroup$ @man_in_green_shirt, thank you! Fixed the typo. $\endgroup$ – Norbert Jun 24 '17 at 16:25
  • 2
    $\begingroup$ @man_in_green_shirt exponents cancels out $\endgroup$ – Norbert Jun 25 '17 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*}

| cite | improve this answer | |
  • 1
    $\begingroup$ Wait a minute there... What is $sen(z)$? $\endgroup$ – BigbearZzz Sep 11 '16 at 23:55
  • 2
    $\begingroup$ Is sinz, sorry, because in portuguese senx is sinx. $\endgroup$ – Israel Meireles Chrisostomo Sep 12 '16 at 0:00

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.