# Proof of an infinite product formula

1. Oct 11, 2008

### Omri

Hi,

I am looking for a proof of the following formula:

$$\prod_{n=1}^{\infty} \left(1-\frac{q^2}{n^2} \right) = \frac{\sin(\pi q)}{\pi q}$$

Does anyone know where to find such a proof?

Thanks

2. Oct 11, 2008

### Pere Callahan

In any good text book on Complex Analysis. Observe that both the left and the right hand side of your equation is an entire function of q (considered as a complex variable) and that the zeroes of both side coincide, more precisely the zeros are exactly the integers.

The theorem is called http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem" [Broken]

Last edited by a moderator: May 3, 2017
3. Oct 11, 2008

### Omri

First I must confess that I have never heard of the term "entire function" or about the factorization theorem. This is probably because I don't know everything about complex analysis, and have never taken a real university course in the subject (hence I wouldn't have a good textbook...). I only know about it from my tutor, who taught me the basics, and now he asked me to find a proof of the product above.

It is indeed easily seen that the zeros of both sides coincide, and that they are indeed the cases where q is an integer. But is that enough? And also, that factor (pi*q)^-1 on the right-hand side has now effect on the zeros, so in general I could have replaced it with any other numeric factor. Why should it be this specific one?

Thanks a lot for you help.

4. Oct 11, 2008

### gel

This can be proved as a special case of the http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem" [Broken], or more directly by contour integration. However, that requires complex analysis.

A book I have (An introduction to the theory of the Riemann Zeta-Function by S.J. Patterson) contains a couple of simple proofs, given as exercises. They in fact prove
$$\pi\cot\pi z=z^{-1}+\sum_{n>0}2z/(z^2-n^2)$$
which is just the http://en.wikipedia.org/wiki/Logarithmic_derivative" [Broken] of the identity you state.

The first proof only requires knowing that the only bounded entire functions (i.e. differentiable functions on the complex plane) are the constant functions, whereas the second doesn't require any complex analysis). I'll write it out as it appears in the book and let you work through the details of the proofs yourself.

1.12 Show that
$$f(z)=z^{-1}+\sum_{n>0}2z/(z^2-n^2)$$
can be expressed as
$$f(z)=\lim_{N\rightarrow\infty}\sum_{n=-N}^N(z-n)^{-1}.$$
Deduce that
$$f(z+1)=f(z).$$
Show also that if |Im(z)|>4 then |f(z)|<5. Conclude that
$$f(z)-\pi\cot\pi z$$
is a constant. Finally show that this constant is zero.

And, the second proof.

1.13 Show that with f as in Exercise 1.12
\begin{align*} f(z)^2=&z^{-2}+\sum_{n>0}\left\{4z^2(z^2-n^2)^{-2}+4(z^2-n^2)^{-1}\right\}\\ & +4z^2\sum_{\substack{m,n>0\\ m\not=n}}(z^2-n^2)^{-1}(z^2-m^2)^{-1}\\ =& z^{-2}+\sum_{n>0}\left\{4z^2(z^2-n^2)^{-2}+4(z^2-n^2)^{-1}\right\}\\ & - 8z^2\sum_{n>0}(z^2-n^2)^{-1}\sum_{\substack{m>0\\ m\not=n}}(m^2-n^2)^{-1}. \end{align*}
Show that
$$\sum_{\substack{m>0\\ m\not=n}}(m^2-n^2)^{-1}=3/4n^2$$
and deduce that
$$f^\prime(z)=-f(z)^2-6\sum_{n\ge 1}n^{-2}.$$
Finally prove that
$$f(z)=\pi\cot\pi z$$
and
$$\sum_{n>0}n^{-2}=\pi^2/6.$$
(This argument, which avoids the techniques of complex analysis and Fourier theory, is due to Eisenstein).

Last edited by a moderator: May 3, 2017
5. Apr 2, 2011

### futurebird

I found this thread and I just wanted to say that this identity inspired me to find these identities:
• $$- \frac{\pi ^2}{3!} = \displaystyle \sum_{j_1=1}^{\infty} -j_1^{-2}$$
• $$\frac{\pi ^4}{5!} = \displaystyle \sum_{j_1,j_2=1 \atop j_1 \neq j_2}^{\infty} (j_1j_2)^{-2}$$
• $$- \frac{\pi ^6}{7!} = \displaystyle \sum_{j_1,j_2,j_3=1 \atop j_i \neq j_k} - (j_1j_2j_3)^{-2}$$
• $$\vdots$$
• $$\frac{\pi ^{2n}}{(2n+1)!} = \displaystyle \sum_{j_1,...j_n=1 \atop j_i \neq j_k}^{\infty} (j_1j_2...j_n)^{-2}$$
• $$\vdots$$

I explain how at my math blog http://www.futurebird.com/?p=156

It turns out these sums are something called the multi zeta function... Neat, no?