0
$\begingroup$

I need to show that $$ \cos(\pi x) = \prod_{n=0}^{\infty} \left( 1-\frac{x^2}{(n+\frac{1}{2})^2}\right)$$.

For $x \notin \mathbb Z$ one can use $\cos(\pi x ) = \frac{\sin(2 \pi x )}{2 \sin(\pi x )}$ and the fact that $$ \sin(\pi x) = \pi x \prod_{n=0}^{\infty} \left( 1-\frac{x^2}{n^2}\right)$$ I am not sure how to prove the equality in the case $x \in \mathbb Z$.

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Hint: Use the Identity theorem from complex analysis. If they are equal on an open set (in fact on any set with limit points), then they are equal everywhere.

Think of them as complex functions. $$\prod_{n=0} ^{\infty} \left(1 - \frac{z^2}{(n+1/2)^2}\right )$$ gives you an entire function since $\sum_{n=0}^{\infty} \frac{1}{ (n+1/2)^2}$ coverges.

You have shown that for all $z \not \in \mathbb N$, the product equals to an entire function $\cos z$. Therefore, they are equal for all $z\in \mathbb C$.

Edit: To avoid complex analysis, note that $\prod _{n=0} ^{\infty}\left( 1 - \frac{x^2} {(n+1/2)^2} \right)$ is a continuous function, since $\sum_{n=0}^{\infty} \frac{1}{ (n+1/2)^2}$ converges. If you have proven the result for all $x \not \in \mathbb N$, then the result simply extends by continuity.

$\endgroup$
8
  • $\begingroup$ thanks for the answer, but in the course of my studies I am not "supposed" to be familiar with complex analysis. This problem is supposed to be solved using only tools of the real analysis. $\endgroup$
    – zesy
    Commented May 8, 2016 at 12:29
  • 1
    $\begingroup$ See if the edit helps you $\endgroup$
    – Shubhodip Mondal
    Commented May 8, 2016 at 12:39
  • $\begingroup$ But the product is not convergent for all $x=\pm (2p+1), \, p \in \mathbb{N}$, since in this case a certain factor is zero. $\endgroup$
    – Olivier Oloa
    Commented May 8, 2016 at 14:45
  • $\begingroup$ @OlivierOloa: you mean p+1/2 right? But if a certain factor is zero, does it obviously not converge to zero? Don't confuse the sum log definition for product, that is valid only for a non-zero sequence. $\endgroup$
    – Shubhodip Mondal
    Commented May 8, 2016 at 15:50
  • $\begingroup$ Sorry, I meant $x=\pm (p+1/2), \, p \in \mathbb{N}$ then one factor is equal to $0$ in this case does the infinite product converge? No confusion. Please, have a look at this: en.wikipedia.org/wiki/Infinite_product $\endgroup$
    – Olivier Oloa
    Commented May 8, 2016 at 15:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .