# Laurent expansion of $\exp(z+\frac 1z)$

This question is similar to this one, but I want to explicitly calculate the Laurent series for the function $$f(z)=\exp\left (z+\frac 1z\right)$$ around $z=0$, i.e. find a closed form for the coefficients $b_n$ of the expression $$f(z)=\sum_{n=-\infty}^{+\infty}b_nz^n$$ Using the usual series for $\exp(\cdot)$, \begin{align} \exp\left(z+\frac 1z\right) &=\sum_{n=0}^\infty\frac{\left(z+\frac 1 z\right)^n}{n!} \\&=\sum_{n=0}^{\infty}\sum_{k=0}^n \frac{\binom n k z^kz^{k-n}}{n!} \\&=\sum_{n=0}^{\infty}\sum_{k=0}^n \frac{1}{k!(n-k)!}z^{2k-n} \\&=\sum_{n=0}^{\infty}z^{-n}\sum_{k=0}^n \frac{1}{k!(n-k)!}z^{2k} \end{align}

Forgetting about coefficients, the $z$ terms in the sum indexed by $k$ are of the kind $$1+z^2+z^4+\ldots +z^{2n}$$ and when multiplied by $z^{-n}$ are symmetric, becoming of the kind $$z^{-n}+z^{-n+2}+z^{-n+4}+\ldots+z^n \tag 1$$ It seems like from this we can reconstruct each coefficient.

Let's try to understand what the coefficient for $z^0$ should be; clearly only even $n$ contribute to this coefficient since $z^{2k-n}$ always has odd exponent for odd $n$. If $n$ is even, the only term which has $0$ exponent in $(1)$ is the central one, i.e. the coefficient of $z^0$ is $$\sum_{n=0: \text{ n is even}}^\infty \frac{1}{\left(\frac n2\right)!\left(\frac n2\right)!}=\sum_{k=0}^\infty\frac{1}{k!k!}$$

Playing a bit with the coefficients, I conjecture that a closed form for $(b_n)_{n\in \mathbb Z}$ is

$$b_n=b_{-n}=\sum_{k=0}^\infty\frac{1}{k!(n+k)!}$$

I have two questions:

1. Can my argument for the calculation of $b_0$ be generalized to yield a closed form for $b_n$?
2. Is the conjectured closed form for $b_n$ correct? How can we find a closed form for $b_n$?

An equivalent but a more straightforward way is to expand each exponential separately. Let us exploit the convention that $1/k! = 0$ for $k = -1, -2, \cdots$. (This convention is simply to forget about the pesky bounds for summation indices.) Then

\begin{align*} e^{z+1/z} = e^z e^{1/z} &= \left( \sum_{j=-\infty}^{\infty} \frac{z^j}{j!} \right)\left( \sum_{k=-\infty}^{\infty} \frac{z^{-k}}{k!} \right) \\ &= \sum_{j,k=-\infty}^{\infty} \frac{z^{j-k}}{j!k!} \\ &= \sum_{n=-\infty}^{\infty} \sum_{k=-\infty}^{\infty} \frac{z^n}{(k+n)!k!}, \end{align*}

where $n = j-k$ in the last equality. Also, rearranging the sums works since they converge absolutely whenever $z \neq 0$. From this we read out that

$$b_n = b_{-n} = \sum_{k=0}^{\infty} \frac{1}{k!(k+n)!}$$

when $n \geq 0$.

• Works like a charm. Thank you very much Sangchul. Apr 4 '17 at 19:12
• @Lonidard, In fact your solution is just one step away from this derivation. Half of your computation is used to establish the identity $e^{z+1/z} = e^z e^{1/z}$ in disguise. Apr 4 '17 at 19:15
• Indeed. I remember going over something similar in my first Calculus course proving $e^{x+y}=e^xe^y$ from the series definition. Apr 4 '17 at 19:16

$$\displaystyle e^{z + 1/z} = \sum_{k =-\infty}^\infty \sum_{m =-\infty}^\infty \frac{z^k1_{k \ge 0}}{k!} \frac{z^{-m}1_{m \ge 0}}{m!} = \sum_{n=-\infty}^\infty \sum_{k =-\infty}^\infty \frac{z^k1_{k \ge 0}}{k!}\frac{z^{n-k}1_{k-n \ge 0}}{(k-n)!}\\ = \sum_{n=-\infty}^\infty z^n \sum_{k =\max(0,n)}^\infty \frac{1}{k!(k-n)!}$$