Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am trying to factorize $\sin(x)\over x$ which by Taylor series expansion and using the roots is $$a \cdot \left(1 - \frac{x}{\pi} \right) \left(1 + \frac{x}{\pi} \right) \left(1 - \frac{x}{2\pi} \right) \left(1 + \frac{x}{2\pi} \right) \left(1 - \frac{x}{3\pi} \right) \left(1 + \frac{x}{3\pi} \right) \cdots$$

Now I was told that this nasty factor $a$ conveniently becomes $1$ due to Weierstrass’s Factorization Theorem which is a transcendental generalization of the Fundamental Theorem of Algebra.

My question
Could you please show me how $a$ is being neutralized using this theorem? Or don't you even need this theorem to do so?

share|improve this question
4  
Look at the limit of both sides as x tends to 0? –  Kevin Buzzard Feb 26 '10 at 11:22
    
Ah, now I see - thank you! –  vonjd Feb 26 '10 at 11:51
    
By the way, you might be interested to know that this is how Euler cracked the Basel problem –  castal Feb 26 '10 at 16:52
    
Thank you, yes - ironically this was the place I was coming from - have a look at the discussion page below ;-) But now everything is clear and I clarified the Wikipedia article with some additional comments. –  vonjd Feb 26 '10 at 16:55
    
But note that you should consider products of $\exp\left(\frac{x}{\pi n}\right) \left(1-\frac{x}{\pi n}\right)$ over all $n \ne 0$, which is necessary for an absolutely convergent infinite product allowing you rearrange the order of the terms. But then the $n$ and $-n$ terms, for $n>0$, combine together and the exponentials cancel! –  Zen Harper Jul 23 '10 at 0:02

2 Answers 2

up vote 4 down vote accepted

The value of this product for small x's is the product of $(1-x^2/(n \pi)^2)$ which, when you take logs (and due to the second power in x), behaves like the sum over n of $-x^2/(n\pi)^2$, which approaches 0 as x approaches 0.

share|improve this answer
    
This is already very helpful - thank you. Just for clarification: shouldn't it be "1 - sum over n of $-x^2/(n\pi)^2$"? And why are you taking logs? And even if you do the difference between $log(1-x^2)$ and $log(-x^2)$ is not neglectable (esp. not for small x because of $x^2$). Or what am I confusing here? Thank you again! –  vonjd Feb 26 '10 at 10:51
    
And another thing: the value of this product is the product $(1-x^2/(n \pi)^2$ - this holds true not only for small x's but for all x, doesn't it? And what I still don't understand is how the $a$ becomes $1$... Thank you again! –  vonjd Feb 26 '10 at 11:42
1  
The point is that both $(\sin x)/x$ and the product approach 1 as $x\to0$, and this serves to determine the value of $a$. And the point of the logs is that the theory of infinite products (en.wikipedia.org/wiki/Infinite_product) is mostly equivalent to that of infinite series, the correspondence coming from taking logs. Furthermore, $\log(1+u)\sim u$ for small $u$; apply to $u=-x^2/(n\pi)^2$. –  Harald Hanche-Olsen Feb 26 '10 at 12:42
    
@vonjd: I hope Herald clarified it –  David Lehavi Feb 26 '10 at 14:16
    
...although the proper setting is to consider complex x, for which logarithms are considerably more subtle than for real x. I regard taking logs in this case as more of a heuristic method than a rigorous proof (it takes a lot of effort to make it rigorous); the "proper" proof is to develop the theory of infinite products directly, similarly to the theory of infinite sums. –  Zen Harper Jul 22 '10 at 23:58

The Weierstrass factorization theorem as usually stated tells you only that $a=e^{g(x)}$ for some entire function $g(x)$. Hadamard's refinement says a little more, based on the growth rate of the function. In your case, since $\left| \frac{\sin x}{x} \right| < \exp\left(|x|^{1+o(1)} \right)$ as the complex number $x$ grows, Hadamard tells you that $g(x)$ is a polynomial of degree at most $1$. Since $\frac{\sin x}{x}$ and $\prod_{n=1}^{\infty} \left(1 - \frac{x^2}{n^2 \pi^2} \right)$ are both even functions, so is $e^{g(x)}$. Thus $g(-x)-g(x)$ is a (constant) integer multiple of $2\pi i$. Hence $g(x)$ is constant, and so is $a$. Finally, as everyone else has pointed out, taking the limit as $x$ goes to $0$ shows that $a=1$.

See Ahlfors, Complex analysis for more about Hadamard's refinement, which relates the "order" and "genus" of an entire function.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.