I am wondering, if I can find a good approximant for this function
$$f(z)=\log \left[ \frac{1-z^2}{z \left(3-z^2\right)}\sinh \left\{\frac{z \left(3-z^2\right)}{1-z^2}\right\}\right]$$
assuming $z \in [0,1)$. A first naive approach would be to take the Taylor expension. But as this function has a pole at one I would have to take more and more terms into account as z takes values closer to one. Is there a suitable method to approximate this function over the whole interval?
EDIT: I add this plot to show the proposed assymptotics down below
But the basic question remains. Can I find a single function, that is a good approximant of $f$?
EDIT: Changed the plot to account for the small mistake.
For $z \rightarrow 0$ a taylor approximation is fine. I leave this part to you.
For $z\rightarrow 1$ we may observe that the relevant contributions arise from the terms containing a $1-z^2$. The other contributions will only give finite corrections in this limit and can be approximated by their values at $z=1$ . Furthermore we observe that $\sinh(x)\sim \frac{e^x}{\bf2} $ as $x\rightarrow \infty$. Taken all this into account we may write
$$ \lim_{z\rightarrow 1_-}f(z)\sim \log\left(\frac{1-z^2}{2}\frac{e^{\frac{2}{1-z^2}}}{{\bf2}}\right)=\frac{2}{1-z^2}+\log(1-z^2)-\log(4) $$
Edit: A forgotten $\frac{1}{\bf2}$ was inserted. Now the graph looks even better!