Using Weierstrass’s Factorization Theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T20:08:36Z http://mathoverflow.net/feeds/question/16487 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16487/using-weierstrasss-factorization-theorem Using Weierstrass’s Factorization Theorem vonjd 2010-02-26T08:39:00Z 2012-04-18T04:04:43Z <p>I am trying to factorize $\sin(x)\over x$ which by <a href="http://en.wikipedia.org/wiki/Taylor_theorem" rel="nofollow">Taylor series expansion</a> and using the roots is $$a \cdot \left(1 - \frac{x}{\pi} \right) \left(1 + \frac{x}{\pi} \right) \left(1 - \frac{x}{2\pi} \right) \left(1 + \frac{x}{2\pi} \right) \left(1 - \frac{x}{3\pi} \right) \left(1 + \frac{x}{3\pi} \right) \cdots$$</p> <p>Now I was told that this nasty factor $a$ conveniently becomes $1$ due to <a href="http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem" rel="nofollow">Weierstrass’s Factorization Theorem</a> which is a transcendental generalization of the <a href="http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra" rel="nofollow">Fundamental Theorem of Algebra</a>.</p> <p><strong>My question</strong><br> Could you please show me how $a$ is being neutralized using this theorem? Or don't you even need this theorem to do so?</p> http://mathoverflow.net/questions/16487/using-weierstrasss-factorization-theorem/16493#16493 Answer by David Lehavi for Using Weierstrass’s Factorization Theorem David Lehavi 2010-02-26T09:38:42Z 2010-02-26T09:38:42Z <p>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.</p> http://mathoverflow.net/questions/16487/using-weierstrasss-factorization-theorem/16572#16572 Answer by Bjorn Poonen for Using Weierstrass’s Factorization Theorem Bjorn Poonen 2010-02-27T01:37:23Z 2010-02-27T01:37:23Z <p>The Weierstrass factorization theorem as usually stated tells you only that $a=e^{g(x)}$ for some entire function $g(x)$. <a href="http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem#Hadamard_Factorization_Theorem" rel="nofollow">Hadamard's refinement</a> says a little more, based on the growth rate of the function. In your case, since $\left| \frac{\sin x}{x} \right| &lt; \exp\left(|x|^{1+o(1)} \right)$ 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 $a=1$.</p> <p>See Ahlfors, <em>Complex analysis</em> for more about Hadamard's refinement, which relates the "order" and "genus" of an entire function.</p>