# Find the Taylor and Laurent series with center at $z_{o}= 1$ of the function $\frac{\sinh(z)}{(z-1)^4}$

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?

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

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 $$t=z-1$$).