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??

Given answer method:

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.

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    $\begingroup$ 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 $\endgroup$ – Powerfulseriesee Jun 13 '15 at 4:22
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    $\begingroup$ 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? $\endgroup$ – whacka Jun 13 '15 at 5:33
  • $\begingroup$ 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. $\endgroup$ – Martín-Blas Pérez Pinilla May 26 '16 at 7:10

For any given function, the Laurent series about a given point is unique. So you can pretty much use any method you want to find it. Using Cauchy integrals in the fashion you have described may work theoretically, but I'd imagine it would be really tedious to calculate.

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


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