The infinite product of $sin(\pi z)$ is said to be...

$\sin \pi z=\pi z\prod _{{n\neq 0}}\left(1-{\frac {z}{n}}\right)e^{{z/n}}=\pi z\prod _{{n=1}}^{\infty }\left(1-{\frac {z^{2}}{n^{2}}}\right)$

Based on the elementary factors provided by the Weierstrass factorization theorem...

$E_{n}(z)={\begin{cases}(1-z)&{\text{if }}n=0,\\(1-z)\exp \left({\frac {z^{1}}{1}}+{\frac {z^{2}}{2}}+\cdots +{\frac {z^{n}}{n}}\right)&{\text{otherwise}}.\end{cases}}$

I can see why we have...

$\sin \pi z=\pi z\prod _{{n\neq 0}}\left(1-{\frac {z}{n}}\right)e^{{z/n}}$

But I don't understand how the second part was derived. Where did the exponent go? Can someone explain why the following is true?

$\pi z\prod _{{n\neq 0}}\left(1-{\frac {z}{n}}\right)e^{{z/n}}=\pi z\prod _{{n=1}}^{\infty }\left(1-{\frac {z^{2}}{n^{2}}}\right)$


You pair together the $n$ and $-n$ terms on the infinite product.

Note that $(1+\frac{z}{n})(1+\frac{z}{-n})e^{\frac{z}{n}}e^{\frac{z}{-n}} = (1-\frac{z^2}{n^2})$

Thus we can change the index of the product from $n\neq 0$ to $n=1$ to $\infty$ by replacing the $n$th term as above.

  • $\begingroup$ I can see how you got the answer, is it possible to explain briefly why this works? $\endgroup$ – Bolboa Mar 31 at 2:08
  • $\begingroup$ Namely, why does multiplying the $n$ and $-n$ terms together result in $n=1$ to ∞ ? $\endgroup$ – Bolboa Mar 31 at 2:14
  • 1
    $\begingroup$ For each $n = 1$ to $\infty$ in the product, take the $n$th term and multiply it by the $-n$th term. Once we have done this for every $n = 1$ to $\infty$, we will have exhausted all of the terms less than $0$. Thus all that will be left is a product over $n = 1$ to $\infty$. $\endgroup$ – George Dewhirst Mar 31 at 2:18

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