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Let $$f(z)=\frac{\sinh(z)}{(z-1)^4}.$$

First I do the following:

$$\sinh(z)=\sinh((z-1)+1)=\sinh(z-1)\cosh(1)+\cosh(z-1)\sinh(1)$$

The expansions of $\sinh(z)$ and $\cosh(z)$ are $$\begin{align*} \sinh(z) &= z+\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots \\ \cosh(z) &= 1+\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots \end{align*}$$

Then $$\begin{align*} \sinh(z-1) &= (z-1)+\frac{(z-1)^3}{3!}+\frac{(z-1)^5}{5!}+\cdots \\ \cosh(z-1) &= 1+\frac{(z-1)^2}{2!}+\frac{(z-1)^4}{4!}+\cdots \end{align*}$$

Now let's see what $$\begin{align*} \frac{\cosh(1)\sinh(z-1)}{(z-1)^4} &= \cosh(1) \biggl( \frac{1}{(z-1)^3}+\frac{1}{3!(z-1)}+\frac{(z-1)}{5!}+\cdots \biggr) \\ \frac{\sinh(1)\cosh(z-1)}{(z-1)^4} &= \sinh(1) \biggl( \frac{1}{(z-1)^4}+\frac{1}{2!(z-1)^2}+\frac{1}{4!}+\cdots \biggr) \end{align*}$$

Rewriting these last two equalities we have $$\begin{align*} \frac{\cosh(1)\sinh(z-1)}{(z-1)^4} &= \sum_{n=1}^{\infty} \frac{\cosh(1)}{(2n-1)!}(z-1)^{2n-5} \\ \frac{\sinh(1)\cosh(z-1)}{(z-1)^4} &= \sum_{n=0}^{\infty} \frac{\sinh(1)}{(2n)!}(z-1)^{2n-4} \end{align*}$$

Hence the Taylor and Laurent series of $f(z)$ is $$ f(z) = \sum_{n=1}^{\infty} \frac{\cosh(1)}{(2n-1)!}(z-1)^{2n-5} + \sum_{n=0}^{\infty} \frac{\sinh(1)}{(2n)!}(z-1)^{2n-4} $$

Where we have $$ a_{n} = \frac{\cosh(1)}{(2n-1)!} \quad \text{and} \quad b_{n} = \frac{\sinh(1)}{(2n)!}. $$

But I don't know if it's okay. I feel a bit confused regarding these series. Could you tell me if I'm okay? or in another case, could you give me any suggestions?

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  • $\begingroup$ Do you mean $\sinh z$? $\endgroup$
    – user403337
    Feb 18, 2021 at 4:00
  • $\begingroup$ @ChrisCuster Yes $\endgroup$
    – Kale_1729
    Feb 18, 2021 at 4:00
  • $\begingroup$ Ok. Well, it may work. $\endgroup$
    – user403337
    Feb 18, 2021 at 4:03
  • $\begingroup$ You can probably write it as one series. $\endgroup$
    – user403337
    Feb 18, 2021 at 4:47
  • $\begingroup$ I edited your post to make the $\LaTeX$ more readable. (Click Edit to view the code.) $\endgroup$ Feb 18, 2021 at 5:43

1 Answer 1

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Your work is good but you can make life a bit easier starting with $z=t+1$ (this is what you implicitly did) $$\frac{\sinh (z)}{(z-1)^4}=\frac{\sinh (t+1)}{t^4}$$ Expanding as you did $$\sinh (t+1)=\sinh (1) \cosh (t)+\cosh (1) \sinh (t)$$ Now, using the expansions of $\cosh (t)$ and $\sinh (t)$ around $t=0$ you then have $$\frac{\sinh (t+1)}{t^4}=\sum_{n=0}^\infty a_n\, t^{n-4}$$ where $$a_{2n+1}=\frac {\cosh(1)}{(2n+1)!}\qquad \text{and} \qquad a_{2n}=\frac {\sinh(1)}{(2n)!}$$ which is your result (just make $t=z-1$).

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