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This is a homework problem from section 5.2.3 in Ahlfors that I have been struggling on for a long time. We want to show the canonical product representation $$\sin \pi(z+\alpha)=e^{\pi z \cot(\pi \alpha)}\prod_{n=-\infty}^\infty \left(1+\frac{z}{n+\alpha}\right)e^{-z/(n+\alpha)}$$ whenever $\alpha$ is not an integer. A hint is to let the factor at the front of the product be $g(z)$ and find $g'(z)/g(z)$.

I see that $\sin \pi(z+\alpha)$ has simple zeroes at $z=n-\alpha$ for $n\in \mathbb{Z}$, and we can use the Weierstraß factorization theorem, but I do not really know how to use the hint, as well as why we have $+\frac{z}{n+\alpha}$ instead of $-\frac{z}{n-\alpha}$. Maybe I'm just missing something here-- would anyone have any pointers?

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1 Answer 1

Actually, this follows straightforwardly by taking log derivatives of both sides. Here's a more detailed write-up, which I think is correct:

We would like to show that $$\sin \pi(z+\alpha)=e^{\pi z \cot(\pi \alpha)}\prod_{n=-\infty}^\infty \left(1+\frac{z}{n+\alpha}\right)e^{-z/(n+\alpha)}$$ whenever $\alpha$ is not an integer. Note that the zeroes are at $z=n-\alpha$ for $n$ an integer. We use the Weierstrass theorem to get a representation $$\sin \pi(z+\alpha)=e^{g(z)}\prod_{n\neq -\alpha} \left(1-\frac{z}{n-\alpha}\right)e^{\frac{z}{n-\alpha}}=e^{g(z)}\prod_{n\neq \alpha}\left(1+\frac{z}{n+\alpha}\right)e^{-\frac{z}{n+\alpha}}$$ $$=e^{g(z)}\prod_{n=-\infty}^\infty\left(1+\frac{z}{n+\alpha}\right)e^{-\frac{z}{n+\alpha}},$$ where the penultimate equality comes from reversing the product and the last equality follows because $\alpha$ is never an integer. It remains to find $g'(z)$, which we do so by taking the logarithmic derivative on both sides so that $$\pi \cot \pi (z+\alpha)=g'(z)+\sum_{n\neq \alpha}\left(\frac{1}{z-n+\alpha}+\frac{1}{n-\alpha}\right)=g'(z)+\left(\frac{1}{z+\alpha}-\frac{1}{\alpha} \right)+\sum_{n\neq 0}\left(\frac{1}{z-n+\alpha}+\frac{1}{n-\alpha}\right)$$ Now from Ahlfors we also know a series expression for $\pi \cot \pi z$, and using it for $\pi \cot \pi (z+\alpha)$ gives $\pi \cot \pi (z+\alpha)=\frac{1}{z+\alpha}+\sum_{n\neq 0}\left(\frac{1}{z+\alpha-n}+\frac{1}{n}\right)$. Equating terms, we see that $g'(z)=\frac{1}{\alpha}+\sum_{n\neq 0}\left(\frac{1}{n}-\frac{1}{n-\alpha}\right)=\frac{1}{\alpha}+\sum_{n\neq 0}\left(\frac{1}{n}+\frac{1}{\alpha-n}\right)$. But from the $\cot$ formula again, we see that $g'(z)=\frac{1}{\alpha}+\sum_{n\neq 0}\left(\frac{1}{n}+\frac{1}{\alpha-n}\right)=\pi \cot (\pi \alpha)$, which implies that $g(z)=\pi z \cot (\pi \alpha)$ and we have $$\sin \pi(z+\alpha)=e^{\pi z \cot(\pi \alpha)}\prod_{n=-\infty}^\infty \left(1+\frac{z}{n+\alpha}\right)e^{-z/(n+\alpha)}$$ whenever $\alpha$ is not an integer, as desired.

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How is that? ${}$ –  joriki Oct 11 '12 at 19:03
    
@joriki I've updated the answer to show more details. –  ff90 Apr 8 '13 at 5:41

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