I need to find the Laurent series of the following function around $0$ – $$f(z) = \exp(z+\frac{1}{z})$$
Now by power series expansion, I got $$f(z) = \sum_{m=0}^{\infty} \frac{z^m}{m!} \sum_{k=0}^{\infty}\frac{1}{k! z^k}$$
Now I know I am supposed to use the concept of Cauchy product of power series but I am not much comfortable with that. So, can someone explain to me how to perform the product here ?
Best Answer
The function $f(z)=\exp(z+1/z)$ is analytic in the annular region $0<|z|<\infty$. We can represent the factors $\exp(z)$ and $\exp(1/z)$ as Laurent series over the annulus $0<|z|<\infty$, viz., $\exp(z)=\sum_{k=-\infty}^\infty a_k z^k$ and $\exp(1/z)=\sum_{k=-\infty}^\infty b_k z^k$, where $$a_k=\begin{cases} 1/k!&\text{for $k\ge0$}\\ 0&\text{for $k<0$}\,,\end{cases}$$ and $$b_k=\begin{cases} 0&\text{for $k>0$}\\ 1/(-k)!&\text{for $k\le0$}\,.\end{cases}$$ The Laurent series of the product is $f(z)=\sum_{k=-\infty}^\infty c_k z^k$, where $c_n=\sum_{k=-\infty}^\infty a_k b_{n-k}$. The latter simplifies to $c_n=\sum_{k=0}^\infty\frac1{k!} \frac1{(n+k)!}$. Further, by comparison with the series expression for the modified Bessel function of the first kind, $$I_n(z)=\sum_{k=0}^\infty{(\frac12 z)^{n+2k}\over k!\,\Gamma(n+k+1)}\,,$$ it follows that $c_n=I_n(2)$. Thus we have that $$\exp\left(z+\frac1z\right)= I_0(2)+\sum_{n=1}^\infty I_n(2)\left(z^n+{1\over z^n}\right).$$