I am trying to factorize $\sin(x)\over x$ which by [Taylor series expansion](http://en.wikipedia.org/wiki/Taylor_theorem) 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](http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem) which is a transcendental generalization of the [Fundamental Theorem of Algebra](http://en.wikipedia.org/wiki/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?



Using Weierstrass’s Factorization Theorem
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.


