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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?

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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
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.

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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
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 ( 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.

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