Laurent series $\frac{\exp(z^{-1})}{(z-1)^2}$

Question

Let $g(z)=\frac{e^{1/z}}{(z-1)^2}$.

Then, $$e^{1/z}=\sum_{n=0}^\infty \frac{1}{n!z^n}$$ and $$\frac{1}{(z-1)^2}=\sum_{k=0}^\infty (k+1)z^k.$$ Then, we can write $$g(z)=\sum_{n=0}^\infty \sum_{k=0}^n(k+1)\frac{1}{(n-k)!}z^{2k-n}.$$

I don´t know what to do next, if I want the expression $g(z)=\sum_{n=-\infty}^\infty....$ Any help is greatly appreciated!

Edit: I want to characterize the singularities at $z_0=0, z_1=1$.

Answer 1

We are looking for a Laurent series expansion of \begin{align*} \color{blue}{g(z)=\frac{e^{1/z}}{(z-1)^2}}\tag{1} \end{align*} at the points $0$ and $1$. From theory we know $g(z)$ has two isolated singularities at $z_0=0$ and $z_1=1$ and $g(z)$ is holomorph in $\mathbb{C}\setminus\{0,1\}$. The point $z_0=0$ is an essential singularity of $g(z)$, whereas $z_1=1$ is a pole of order $2$.

Expansion at $z_0=0$:

Since $e^{1/z}$ has an essential singularity at $z_0=0$ we are looking for a representation \begin{align*} g(z)=\frac{e^{1/z}}{(z-1)^2}=\sum_{m=-\infty}^{\infty} c_m x^m \end{align*} with main part containing infinite non-zero terms $c_m x^m, m<0$. We already know from OPs calculation the Laurent series expansion of \begin{align*} e^{1/z}=\sum_{n=0}^\infty\frac{1}{n!}\,\frac{1}{z^n}\qquad\text{and}\qquad\frac{1}{(z-1)^2}=\sum_{k=0}^{\infty}(k+1)z^k \end{align*}

We obtain \begin{align*} \color{blue}{g(z)}&\color{blue}{=\frac{e^{1/z}}{(z-1)^2}} =\sum_{n=0}^\infty \frac{1}{n!}\,\frac{1}{z^n}\sum_{k=0}^\infty(k+1)z^k\\ &=\sum_{m=-\infty}^{\infty}\left(\sum_{{k-n=m}\atop{k,n\geq 0}}\frac{k+1}{n!}\right)z^m\\ &\color{blue}{=\sum_{m=-\infty}^{-1}\sum_{n=-m}^\infty\frac{m+n+1}{n!}z^m +\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{m+n+1}{n!}z^m}\tag{1} \end{align*}

It is convenient to split in (1) the sum since the index regions of the inner sums are somewhat different. Note that (1) is already a perfect Laurent series expansion. But we continue and simplify the inner sums somewhat. We start with the inner sum of the left-hand part of (1). Note that here $m$ is a negative integer. We obtain \begin{align*} \color{blue}{\sum_{n=-m}^\infty}&\color{blue}{\frac{m+n+1}{n!}}\\ &=\sum_{n=0}^\infty\frac{m+n+1}{n!}-\sum_{n=0}^{-m-1}\frac{m+n+1}{n!}\\ &=(m+1)\sum_{n=0}^\infty\frac{1}{n!}+\sum_{n=1}^\infty\frac{1}{(n-1)!}-\sum_{n=0}^{-m-1}\frac{m+n+1}{n!}\\ &=(m+1)e+e-\sum_{n=0}^{-m-1}\frac{m+n+1}{n!}\\ &\,\,\color{blue}{=(m+2)e-\sum_{n=0}^{-m-1}\frac{m+n+1}{n!}}\tag{2.1} \end{align*} and see the coefficient $c_m, m<0$ is an integral multiple of $e$ added by a rational number. The inner sum of the right-hand part of (1) is easier to calculate. We obtain \begin{align*} \color{blue}{\sum_{n=0}^\infty}\color{blue}{\frac{m+n+1}{n!}}&=(m+1)e+\sum_{n=1}^\infty\frac{1}{(n-1)!}\color{blue}{=(m+2)e}\tag{2.2} \end{align*} Putting (2.1) and (2.2) into (1) we finally get \begin{align*} \color{blue}{g(z)}&\color{blue}{=\frac{e^{1/z}}{(z-1)^2}}\\ &\color{blue}{=\sum_{m=-\infty}^{-1}\left((m+2)e-\sum_{n=0}^{-m-1}\frac{m+n+1}{n!}\right)z^m}\\ &\qquad\color{blue}{+\sum_{m=0}^{\infty}(m+2)ez^m}\\ &=\cdots +\left(\frac{11}{2}-2e\right)\frac{1}{z^4}+(3-e)\frac{1}{z^3}+\frac{1}{z^2}+e\frac{1}{z}\\ &\qquad\qquad+2e+3ez+4ez^2+5ez^3+6ez^4+\cdots\\ \end{align*}

Expansion at $z_1=1$:

Since $e^{1/z}$ is holomorphic at $\mathbb{C}\setminus{\{0\}}$ and $\frac{1}{(z-1)^2}$ has a pole of order $2$ at $z_1=1$ we are looking for a representation \begin{align*} g(z)=\frac{e^{1/z}}{(z-1)^2}=\sum_{m=-2}^{\infty} c_m (x-1)^m \end{align*}

Since $\frac{1}{(z-1)^2}$ is already the Laurent series expansion at $z_1=1$ we focus on $e^{1/z}$. We obtain \begin{align*} \color{blue}{e^{1/z}}&=\sum_{n=0}^\infty\frac{1}{n!}\,\frac{1}{z^n}\\ &=\sum_{n=0}^\infty\frac{1}{n!}\,\frac{1}{\left(1+(z-1)\right)^n}\\ &=\sum_{n=0}^\infty\frac{1}{n!}\sum_{m=0}^\infty\binom{-n}{m}(z-1)^m\\ &\,\,\color{blue}{=\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{1}{n!}(-1)^m\binom{n+m-1}{m}(z-1)^m}\tag{3}\\ \end{align*} We observe the coefficients of $z^m$ in (3) are a rational multiple of $e=\sum_{n=0}^\infty\frac{1}{n!}$.

Multiplication of (3) with $\frac{1}{(z-1)^2}$ finally results in \begin{align*} \color{blue}{g(z)}&\color{blue}{=\frac{e^{1/z}}{(z-1)^2}}\\ &=\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{1}{n!}(-1)^m\binom{n+m-1}{m}(z-1)^{m-2}\\ &\,\,\color{blue}{=\sum_{m=-2}^\infty\sum_{n=0}^\infty\frac{1}{n!}(-1)^m\binom{n+m+1}{m}(z-1)^{m}}\\ &=e\frac{1}{(z-1)^2}-e\frac{1}{z-1}+\frac{3}{2}e-\frac{13}{6}e(z-1)+\frac{73}{24}e(z-1)^2+\cdots \end{align*}

The coefficients $1, -1, 3, -13, 73, -501, 4\,051, -37\,633,\ldots$ of the Laurent series expansion of $e^{1/z}$ at $z_1=1$ are archived in OEIS as A293125.