# Laurent series for $\sin z \sin(1/z)$?

How can I find a laurent series for $\sin z \sin(1/z)$ for $z \neq 0$?

Is it possible to multiply two laurent series? I saw that in wikipedia, it's not generally possible.

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But $z \mapsto \sin z$ is analytic, and this is a very special case of multiplication. – Siminore Sep 14 '12 at 9:33
Isn't the singularity at $z=0$ an essential one? – Hagen von Eitzen Sep 14 '12 at 9:33
I don't know if you know Gauchy-product, if not have a look: en.wikipedia.org/wiki/Cauchy_product – Lucien Sep 14 '12 at 9:38
@HagenvonEitzen Yes, it is. For a pole, the number of terms of the Laurent series with negative exponent is finite. Here is will be infinite. – Sasha Sep 14 '12 at 12:15

Operate formally. Fix finite $z\not=0$. Then each expand each sine into Taylor series: $$\begin{eqnarray} \sin(z) \sin\left(\frac{1}{z}\right) &=& \sum_{n=0}^\infty (-1)^n \frac{z^{2n+1}}{(2n+1)!} \sum_{m=0}^\infty (-1)^n \frac{z^{-2m-1}}{(2m+1)!} \\ &=& \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-z^2)^{n-m}}{(2n+1)!(2m+1)!} = \Big| \text{insert identity}\Big| \\ &=& \sum_{k=-\infty}^\infty \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-z^2)^{n-m}}{(2n+1)!(2m+1)!} \delta_{k,n-m} \\ &=& \sum_{k=-\infty}^\infty (-z^2)^k \sum_{m=\max(0,-k)}^\infty \frac{1}{(2(k+m)+1)!(2m+1)!} \\ &=& \sum_{k=-\infty}^\infty (-z^2)^k \sum_{m=0}^\infty \frac{1}{(2m+2 |k| +1)!(2m+1)!} \end{eqnarray}$$ The latter sum can be evaluated in terms of Bessel functions, giving: $$\sin(z) \sin\left(\frac{1}{z}\right) = \frac{1}{2} \sum_{k=-\infty}^\infty (-z^2)^k \left(I_{2|k|}(2)- J_{2|k|}(2)\right)$$ Indeed: $$\begin{eqnarray} \sum_{m=0}^\infty \frac{1}{(2m+2 |k| +1)!(2m+1)!} &=& \sum_{m=0}^\infty \frac{1+(-1)^m}{2} \frac{1}{(m+2 |k| +1)!(m+1)!} \\ &=& \sum_{m=0}^\infty \frac{1-(-1)^m}{2} \frac{1}{(m+2 |k|)!(m)!} \\ &=& \frac{1}{2} I_{2|k|}(2) - \frac{1}{2} J_{2|k|}(2) \end{eqnarray}$$