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

Question

I have to compute from the following, its principal part of the Laurent series around $z=0$.

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

I have trouble with computing the series of $(\exp(z)-1)^2$. I wanted to take out the $\frac{1}{z}$ and then multiply it by the series of the exp term.

Thanks already.

Answer 1

It is convenient to represent $\exp(z)-1$ using the big-O notation. \begin{align*} \exp(z)-1&=\sum_{n=1}^\infty\frac{z^n}{n!}\\ &=z+\frac{1}{2}z^2+\frac{1}{6}z^3+\frac{1}{24}z^4+O(z^5)\tag{1} \end{align*}

We obtain \begin{align*} \frac{1}{z\left(\exp(z)-1\right)^2}&=\frac{1}{z\left(z+\frac{1}{2}z^2+\frac{1}{6}z^3+\frac{1}{24}z^4+O(z^5)\right)^2}\tag{2}\\ &=\frac{1}{z\left(z^2+z^3+\frac{7}{12}z^4+\frac{1}{4}z^5+O(z^6)\right)}\tag{3}\\ &=\frac{1}{z^3}\cdot\frac{1}{1+z+\frac{7}{12}z^2+\frac{1}{4}z^3+O(z^4)}\tag{4}\\ &=\frac{1}{z^3}\sum_{n=0}^\infty(-1)^n\left(z+\frac{7}{12}z^2+\frac{1}{4}z^3+O(z^4)\right)^n\tag{5}\\ &=\frac{1}{z^3}\left(1-\left(z+\frac{7}{12}z^2+\frac{1}{4}z^3+O(z^4)\right)\right.\\ &\qquad\qquad\left.+\left(z^2+\frac{7}{6}z^3+O(z^4)\right)-\left(z^3+O(z^4)\right)+O(z^4)\right)\tag{6}\\ &=\frac{1}{z^3}\left(1-z+\frac{5}{12}z^2-\frac{1}{12}z^3+O(z^4)\right)\tag{7}\\ &=\color{blue}{\frac{1}{z^3}-\frac{1}{z^2}+\frac{5}{12z}}-\frac{1}{12}+O(z)\tag{8} \end{align*}

with the blue marked part in (8) being the principal part of $$\frac{1}{z\left(\exp(z)-1\right)^2}$$ at $z=0$.

Comment:

• In (2) we use the representation of $\exp(z)-1$ from (1).

• In (3) we multiply out and collect all terms with power greater or equal $6$ in $O(z^6)$.

• In (4) we factor out $\frac{1}{z^3}$.

• In (5) we apply the geometric series expansion.

• In (6) we multiply out, write the terms for $n=0,1,2$ and $n=3$ and collect all other summands in $O(z^4)$.

• In (7) we collect the terms accordingly.

Answer 2

Hint: The calculation may be simplified if one first notices that the symmetrized generator of Bernoulli numbers, the so-called $A$-roof function $$\widehat{A}(z)~:=~\frac{z/2}{\sinh\frac{z}{2}}~=~1-\frac{z^2}{24}+ O(z^4)\tag{1}$$ is even and hence cannot have a third order term.

Then OP's function $f(z)$ satisfies

$$ z^3f(z)~=~\frac{z^2}{(e^z-1)^2}~=~ \widehat{A}(z)^2e^{-z}$$ $$~=~\left(1-\frac{z^2}{12}+ O(z^4) \right)\left(1-z+\frac{z^2}{2}-\frac{z^3}{6} + O(z^4) \right) $$ $$~=~\left(1-z+\frac{5z^2}{12}-\frac{z^3}{12} + O(z^4) \right),$$

from which the principal part of the Laurent series can be directly read off.