Does anyone knows a good way to approximate $\operatorname{arcosh}$ between $1.0$ and $1.1$ precisely? Me and some others are using the standard series $$\ln(2x)-\sum_{n=1}^\infty\left(\frac{(2n)!}{2^{2n}(n!)^2}\right)\frac{x^{-2n}}{2n}$$ but but between $1.0$ and $1.1$ it's only precise to about $2$ decimal places, which isn't enough.
Any help will be greatly appreciated.
You have a very accurate approximation $$\cosh ^{-1}(x)=\sum_{n=0}^\infty (-1) ^n\, a_n\,t^{2n+1}\qquad \text{where} \qquad t=\sqrt{2 (x-1)}$$ where the first $a_n$ form the sequence $$\left\{1,\frac{1}{24},\frac{3}{640},\frac{5}{7168},\frac{35}{294912},\frac{63}{288358 4},\frac{231}{54525952},\frac{143}{167772160},\cdots\right\}$$
We can transform the series in $t$ into a simple Padé approximant which, back to $x$ would give $$\cosh ^{-1}(x)=\sqrt{2(x-1)}\, \frac{ 17 x+103}{27 x+93}$$ whose error is $\frac{61 (x-1)^{7/2}}{22400 \sqrt{2}}$ that is to say $6.09\times 10^{-7}$.
Edit
A much better one would be $$\cosh ^{-1}(x)=\frac{\sqrt{2(x-1)}}{15}\, \frac{69049 x^2+1297462 x+2322769}{9675 x^2+96850 x+139427}$$ which, for an absolute maximum error of $0.01$ can be used up to $x=6$.