How to prove the following product? $$\frac{\sin(x)}{x}= \left(1+\frac{x}{\pi}\right) \left(1-\frac{x}{\pi}\right) \left(1+\frac{x}{2\pi}\right) \left(1-\frac{x}{2\pi}\right) \left(1+\frac{x}{3\pi}\right) \left(1-\frac{x}{3\pi}\right)\cdots$$
Proving $\frac{\sin x}{x} =\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right) \left(1-\frac{x^2}{3^2\pi^2}\right)\cdots$
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3It 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. – 2012-06-12
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0I believe you have a typo - the roots of the polynomial should be squared. – 2012-06-12
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0Thanks. 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? – 2012-06-12
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3@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. – 2012-06-12
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2@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. – 2012-06-12
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1[Very related](http://math.stackexchange.com/questions/215963/how-to-prove-sin-x-1-fracx3-pi1-fracx2-pi1-fracx-pix/215966#215966) – 2013-03-31
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0@process91 Wrong. If polynomials $f$ and $g$ have the same roots, it doesn't mean one is a constant multiple of another - consider $x$ and $x^2$. The statement becomes true when equal multiplicities of each root are assumed (obviously we're speaking about the complex domain, otherwise it's very untrue, like $1$ and $x^2+1$). Also, $e^x+1$ has infinitely many roots $x_k = \pi i + 2 k \pi i$ for $k \in \mathbb{Z}$ and similarly $e^x-1$ has infinitely many roots $2 k \pi i$ for $k \in \mathbb{Z}$. – 2017-08-02
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0However, the following is true: if $f, g : \mathbb{C} \to \mathbb{C}$ are analytic and have the same roots along with their multiplicities, then they differ by a quotient $e^{h(z)}$ for some analytic function $h : \mathbb{C} \to \mathbb{C}$. The converse is obviously true, i.e. $e^{h(z)} \cdot f(z)$ always has the same roots and their multiplicities as $f(z)$. – 2017-08-02
2 Answers
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.
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1This is REALLY great – 2014-01-12
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0One 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! – 2014-05-18
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0@ireallydonknow A you sure this approaches will give the formula valid for complex values, not only for reals ones? – 2014-05-18
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0@Norbert Yes, you can check out Konrad Knopp's Theory of Functions II. – 2014-05-18
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0@ireallydonknow Thank you! – 2014-05-18
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0Very lucrative stuff! – 2015-01-14
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1The function $f(x)=\cos(\alpha x)$ you define is not actually $2\pi$ periodic (as you claim). For example if $\alpha=1/2$, then $f(2\pi)=\cos(\pi)=-1\neq 1=f(0)$. – 2015-01-18
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0@PhoemueX this was a sloppy definition, now fixed – 2015-07-31
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0Thanks, very nice! Two points which might further improve the answer: 1. It is not true in genera that the Fourier series of continuous functios converges pointwise. Here, we even have uniform convergence of the series for $\cos(\alpha x))$ by the M-test. (Equality is then a consequence of the uniqueness of Fourier coefficients). 2. I did not know directly how to compute the Fourier series (ie the coefficients). This is simple once one recalls $ 2\cos(\alpha x)= e^{i\alpha x}+e^{-i\alpha x}$. – 2015-08-01
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0I think there's a typo: in the complex analysis approach, it should read $$ \ln\sin z=h(z)+\ln(z)+\sum\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(\ln\left(1\boldsymbol{-}\frac{z}{\pi n}\right)+\frac{z}{\pi n}\right) $$ (boldness of minus sign added for emphasis) – 2017-06-24
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1@man_in_green_shirt, thank you! Fixed the typo. – 2017-06-24
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0No problem! This part is unclear to me - any chance you could shed some light on why this identity holds? $$\lim\limits_{N\to\infty}\prod\limits_{n=-N,n\neq 0}^N\left(1-\frac{z}{\pi n}\right)\exp\left(\frac{z}{\pi n}\right)= \lim\limits_{N\to\infty}\prod\limits_{n=1}^N\left(1-\frac{z^2}{\pi^2 n^2}\right)$$ – 2017-06-25
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1@man_in_green_shirt exponents cancels out – 2017-06-25
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*}
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0Wait a minute there... What is $sen(z)$? – 2016-09-11
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2Is sinz, sorry, because in portuguese senx is sinx. – 2016-09-12