# Laurent Expansion Coefficients of $\exp\left(\frac{z}{1-z}\right)$

Given the function $$f(z)=\exp\left(\frac{z}{1-z}\right)$$, I want to find the coefficients $$a_0$$, $$a_{-1}$$, and $$a_{-2}$$ of the Laurent expansion $$f(z)=\sum_{n=-\infty}^{\infty}a_n(z+1)^n$$ about $$z=-1$$, on the annulus $$\{z\in\mathbb{C}:|z+1|>2\}$$.

We know that the power series for $$\exp(z)$$ centered about $$z=-1$$ is $$\sum_{n=0}^{\infty}\frac{(z+1)^n}{e\cdot n!}$$, so \begin{align*} \exp\left(\frac{z}{1-z}\right) &= \sum_{n=0}^{\infty}\frac{1}{e\cdot n!}\left(\frac{z}{1-z}+1\right)^n\\ &= \sum_{n=0}^{\infty}\frac{1}{e\cdot n!}\left(\frac{1}{1-z}\right)^n\\ &= \sum_{n=-\infty}^0\frac{(-1)^n}{e\cdot (-n)!}(z-1)^n \end{align*}

But this gives us the Laurent series of f(z) centered at 1 which does not seem to be helpful.

• I would not call that an annulus. Commented Apr 9, 2021 at 13:16

Letting $$z = w - 1$$, you want the Laurent expansion of $$e^{w-1 \over 2 - w}$$ on $$|w| > 2$$. Noting that $${w -1 \over 2 - w} = -1 - {1 \over w - 2}$$, you are looking for the Laurent expansion of $$e^{-1}e^{-{1 \over w - 2}}$$ on $$|w| > 2$$.

Changing variables once again, this time to $$v = {1 \over w}$$, you are seeking the Taylor expansion of $$e^{-1}e^{v \over 2v - 1} = e^{-{1 \over 2}}e^{1 \over 4v - 2}$$ on $$|v| < {1 \over 2}$$. So $$a_0 = e^{-1}$$ and you can find $$a_1$$ and $$a_2$$ by taking the first two derivatives of $$e^{-{1 \over 2}}e^{1 \over 4v - 2}$$ at $$v = 0$$.

Let $$z=x-1\implies \frac z {1-z}=\frac{1-x}{x-2}=-\frac 12 + \sum_{n=1}^\infty 2^{-(n+1)} x^n$$ which make $$e^{\frac{1-x}{x-2}}=\frac{1}{\sqrt{e}}\left(1+\frac{x}{4 }+\frac{5 x^2}{32 }+\frac{37 x^3}{384 }+\frac{361 x^4}{6144 }\right)+O\left(x^5\right)$$ The numerator of the coefficients correspond to sequence $$A025168$$ in $$OEIS$$ and the denominator of the coefficients is just $$4^n n!$$.
Replave $$x$$ by $$(z+1)$$.
It is easy to see that $$\exp\biggl(\frac{z}{1-z}\biggr) =\exp\biggl(\frac{1}{2-(z+1)}-1\biggr) =\frac{1}{e}\exp\biggl(\frac{1}{2-u}\biggr), \quad u=z+1.$$ By the Faa di Bruno formula and some properties for the Bell polynomials of the second kind, we obtain \begin{align*} \biggl[\exp\biggl(\frac{1}{2-z}\biggr)\biggr]^{(n)} &=\sum_{k=0}^{n}\exp^{(k)}\biggl(\frac{1}{2-z}\biggr)B_{n,k}\biggl(\frac{1!}{(2-z)^2},\frac{2!}{(2-z)^3},\dotsc, \frac{(n-k+1)!}{(2-z)^{n-k+2}}\biggr)\\ &=\sum_{k=0}^{n}\exp\biggl(\frac{1}{2-z}\biggr) \frac{1}{(2-z)^{n+k}} B_{n,k}(1!,2!,\dotsc,(n-k+1)!)\\ &=\sum_{k=0}^{n}\exp\biggl(\frac{1}{2-z}\biggr) \frac{1}{(2-z)^{n+k}} \binom{n-1}{k-1}\frac{n!}{k!}\\ &\to n!\frac{\operatorname{e}^{1/2}}{2^n}\sum_{k=0}^{n}\binom{n-1}{k-1}\frac{1}{(2k)!!} \end{align*} as $$z\to0$$, where $$B_{n,k}(x_1,x_2,\dotsc, x_{n-k+1})$$ denotes the Bell polynomials of the second kind. Consequently, we acquire $$\begin{equation*} \exp\biggl(\frac{1}{2-z}\biggr)=\operatorname{e}^{1/2}\sum_{n=0}^{\infty}\Biggl[\frac{1}{2^n}\sum_{k=0}^{n} \binom{n-1}{k-1}\frac{1}{(2k)!!}\Biggr]z^n, \quad z\in[-1,1). \end{equation*}$$ Consequently, we obtain $$\begin{equation*} \exp\biggl(\frac{z}{1-z}\biggr) =\frac{1}{\operatorname{e}^{1/2}}\sum_{n=0}^{\infty}\Biggl[\frac{1}{2^n}\sum_{k=0}^{n} \binom{n-1}{k-1}\frac{1}{(2k)!!}\Biggr](z+1)^n, \quad z\in[-2,0). \end{equation*}$$