Laurent series of $e^{z+1/z}$

Question

What is the Laurent series of $e^{z+1/z}$?

I had used $$a_k= \frac{1}{2\pi i}\int_c \frac{f(z)}{z^{k+1}}\,dz $$ for a curve $c$ in which we can use $e^z$ as an analytic func. and expanded the $e^{1/z}$ series expansion.

Answer 1

Assuming you are looking for the Laurent series around $z=0$: Since $$ e^z = 1 + z + \frac{z^2}{2!} + \cdots = \sum_{k=0}^\infty \frac{z^k}{k!} $$ for all $z \in \mathbb{C}$, we also have $$ e^{1/z} = 1 + \frac{1}{z} + \frac{1}{2!z^2} + \cdots = \sum_{m=0}^\infty \frac{1}{m!z^m} $$ for all $z\neq 0$. Multiplying the two series together we get the Laurent series we are looking for: \begin{align} e^{z+1/z} &= \sum_{n=-\infty}^\infty \left(\sum_{k-m=n} \frac{1}{k!m!} \right) z^n \\ &= \sum_{n=-\infty}^\infty \left(\sum_{m=0}^\infty \frac{1}{(m+n)!m!} \right) z^n \end{align}

Answer 2

$$ \begin{align} e^ze^{\frac1z} &=\sum_{k=0}^\infty\frac{z^k}{k!}\sum_{j=0}^\infty\frac1{j!z^j}\\ &=\sum_{k=-\infty}^\infty z^k\color{#0000F0}{\sum_{j=0}^\infty\frac1{(k+j)!j!}}\\ &=\sum_{k=-\infty}^\infty\color{#0000F0}{I_{|k|}(2)}\,z^k \end{align} $$ where it is convention that $\frac1{k!}=0$ when $k<0$.

$I_n$ is the Modified Bessel Function of the Second Kind.

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Why is the Sum in $\boldsymbol{j}$ Even in $\boldsymbol{k}$?

In the preceding, it is stated that $$ e^{z+\frac1z}=\sum_{k=-\infty}^\infty I_{|k|}(2)z^k $$ which implies that $\sum\limits_{j=0}^\infty\frac1{(k+j)!j!}$ is even in $k$; and in fact it is, because of the convention, mentioned above, that $\frac1{k!}=0$ when $k<0$.

For $k\ge0$, we have $$ \begin{align} \sum_{j=0}^\infty\frac1{(-k+j)!j!} &=\sum_{j=k}^\infty\frac1{(-k+j)!j!}\tag{1a}\\ &=\sum_{j=0}^\infty\frac1{j!(j+k)!}\tag{1b} \end{align} $$ Explanation:
$\text{(1a)}$: the terms with $j\lt k$ are $0$ due to the convention
$\text{(1b)}$: substitute $j\mapsto j+k$

Answer 3

This is an alternative approach yielding an integral formula.

If $$f(z)=e^{z+1/z}=\sum_{n=-\infty}^{\infty} a_nz^n$$ and we take $z=e^{i\theta},$ then you get $$f(e^{i\theta})=e^{2\cos\theta}=\sum_{n=-\infty}^{\infty} a_ne^{in\theta}.$$

So the $a_n$ are the Fourier coefficients of the function $f(\theta)=e^{2\cos(\theta)}.$

So $$a_n=\frac1{2\pi}\int_{-\pi}^{\pi} e^{2\cos(\theta)-in\theta}d\theta$$

You can use $a_n=a_{-n}$ to also see:

$$a_n=\frac1{2\pi}\int_{-\pi}^{\pi} e^{2\cos(\theta)}\cos(n\theta)\,d\theta$$

That might be enough of a closed form for some.

Of course, there are a lot easier ways to get that.

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This can also be written in terms of a complex integral, using the residue theorem, with $\gamma(t)=e^{it},$

$$a_n=\frac1{2\pi i}\int_{\gamma} \frac{e^{z+z^{-1}}}{z^{n+1}}\,dz=\frac1{2\pi i}\int_0^{2\pi} e^{2\cos t}e^{-i (n+1)t}\cdot ie^{it}\,dt$$

Yielding the same integral.

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This approach assumes the Laurent series converges on $|z|=1.$ That is not hard to prove here.

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We can get back the other answers by expanding $e^{2\cos\theta}:$ $$a_n=\frac1{2\pi}\sum_{k=0}^{\infty}\frac{2^k}{k!}\int_{-\pi}^{\pi}\cos^k\theta e^{-in\theta}\,d\theta$$

But: using $\cos(\theta)=\frac{1}2\left(e^{i\theta}+e^{-i\theta}\right)$ and expanding with the binomial theorem, we get: $$\int_{-\pi}^{\pi}\cos^k\theta e^{-in\theta}\,d\theta=\begin{cases}0&|n|>k\\0&n-k\text{ odd}\\\frac{2\pi}{2^k}\binom{k}{\frac{n+k}2}&\text{otherwise} \end{cases}$$

We will assume $n$ is non-negative, since $a_n=a_{-n}.$

Thus, letting $i=\frac{k-n}{2},$ this becomes:

$$a_n=\sum_{i=0}^{\infty}\frac{1}{(n+2i)!}\binom{n+2i}{n+i}=\sum_{i=0}^{\infty}\frac1{i!(n+i)!}$$