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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.

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

3 Answers 3

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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 $v = 0$.

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Similar to José Carlos Santos'answer

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 $(z+1)$.

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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*}

References

  1. https://math.stackexchange.com/a/4700985
  2. https://math.stackexchange.com/a/4253870
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