Computing the Laurent series of $e^{z+\frac{1}{z}}$

Question

I'm really stuck on this, and I have no idea how to start. Writing it at $f(z)=e^z e^{\frac{1}{z}}$ and their expansions didn't really give any insight. I am aware it is possible to multiply the series of these functions, but this was not covered in this course. I am allowed to assume $f$ is holomorphic on $\mathbb{C}\setminus\{0\}$.

Answer 1

Best to expand the series as follows:

$$f(z) = \sum_{k=0}^{\infty} \frac1{k!} \left (z+\frac1{z} \right )^k $$

Look at the first few terms:

$$1+\frac1{1!} \left (z+\frac1{z}\right)+\frac1{2!} \left (z^2+2+\frac1{z^2}\right)+ \frac1{3!} \left (z^3+3 z + \frac{3}{z}+\frac1{z^3}\right)+\cdots$$

First note that we may express the Laurent series as

$$f(z) = \sum_{k=-\infty}^{\infty} c_k z^k$$

where $c_k = c_{-k}$. So let's consider nonnegative $k$.

$$c_0 = 1+ \frac1{2!} 2 + \frac1{4!} 6 + \cdots = \sum_{\ell=0}^{\infty} \frac1{(2 \ell)!} \binom{2 \ell}{\ell} $$

which may be simplified to

$$c_0 = \sum_{\ell=0}^{\infty} \frac1{(\ell!)^2} = I_0(2)$$

where $I_0$ is the modified Bessel function of the first kind, of zero order. In fact, you may find, in general, that

$$c_k= \sum_{\ell=0}^{\infty} \frac1{(k+2 \ell)!} \binom{k+2 \ell}{\ell} = \sum_{\ell=0}^{\infty} \frac1{(k+\ell)! \ell!} = I_k(2)$$

where $I_k$ is the modified Bessel function of the first kind, of $k$th order. Thus,

$$f(z) = I_0(2) + \sum_{k=1}^{\infty} I_k(2) \left (z^k + \frac1{z^k} \right ) $$

Answer 2

Let $C_{n}$ ($n \in \mathbb{Z}$) be a coefficient of the Laurent expansion of $e^{z+\frac{1}{z}}$ at the origin.

By definition,

$$ C_{n} = \frac{1}{2 \pi i} \int_{|z|=1} \frac{e^{z+\frac{1}{z}}}{z^{n+1}} \ dz$$

since the unit circle lies in an annulus in which $e^{z+\frac{1}{z}}$ is analytic.

Let $z= e^{i \theta}$.

Then $$C_{n} = \frac{1}{2 \pi i} \int_{-\pi}^{\pi} \frac{e^{e^{i \theta}+e^{- i \theta}}}{e^{i(n+1) \theta}} i e^{i \theta} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{2 \cos \theta} e^{-in \theta}$$

$$ = \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{2 \cos \theta} \cos(n \theta) \ d \theta = \frac{1}{\pi} \int_{0}^{\pi} e^{2 \cos \theta} \cos (n \theta) \ d \theta$$

But for an integer $n$, the modified Bessel function of the first kind has the integral representation

$$I_{n}(x) = \frac{1}{\pi} \int_{0}^{\pi} e^{x \cos \theta} \cos(n \theta) d \theta$$

http://dlmf.nist.gov/10.32

Therefore,

$$ C_{n} = I_{n}(2)$$

Also notice from the integral representation that $I_{n}(x) = I_{-n}(x)$.