# Laurent series calculation(they seem to calculate it without Laurent series?)

Laurent sreies expansion of the function $f(z)=z^{-1}\sinh(z^{-1})$ about the point $0$.

I thought I was meant to use this:

$$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n + \sum_{n=1}^\infty \frac{b_n}{(z-z_0)^n}$$

$r_1\lt |z-z_0|\lt r_2$

Where $$a_n = \frac{1}{2\pi i} \int_c \frac{f(\xi)}{(\xi - z_0)^{n+1}} d\xi, b_n = \frac{1}{2\pi i} \int_c \frac{f(\xi)}{(\xi - z_0)^{-(n+1)}} d\xi$$

But the answer didn't do any of this? why??

For $f(z)=\sinh z$, we have $$f^{(n)}(z) = \left\{ \begin{array}{cc}\sinh z&z\text{ even}\\ \cosh z& z\text{ odd} \end{array}\right.$$

$$\implies f^{(n)} = 1, n \text{ odd}. =0, n \text{ even}$$

Gives Maclaurin series for $\sinh z$ is $$z + \frac{z^3}{3!} + \frac{z^5}{5!}+\cdots$$ Laurent series for $\sinh z^{-1}$ is $$\frac{1}{z} + \frac{1}{3!z^3} + \frac{1}{5!z^5}+\cdots$$

Laurent series for $z^{-1}\sinh(z^{-1})$ is $$\frac{1}{z^2}+\frac{1}{3!z^4}+\frac{1}{5!z^6}+\cdots$$

Note that $z^{-1}\sinh(z^{-1})$ is analytic for $z\ne 0$, so the origin is an isolated singularity from the Laurent series, there are arbitrary large negative powers of $z$, so this is essential.

• Ahhh perhaps I have some idea. The $a_n$ and $b_n$ look like they could be run with the generalised cauchy integral formula perhaps – Powerfulseriesee Jun 13 '15 at 4:22
• If you know the taylor series of $f(z)$ you know the laurent series for $f(z^{-1})$. Why would you want to compute those crazy integrals when you can just use your knowledge of simple taylor series instead? – whacka Jun 13 '15 at 5:33
• The Perl's motto applies here: There's more than one way to do it. And using the Taylor series is the best way in this case. – Martín-Blas Pérez Pinilla May 26 '16 at 7:10

The solution simply calculated the taylor series of $\sinh(z)$ and went from there.