<s>[INST] An inverse hyperbolic cosine approximation Given $-1<a<b<0<c<d<1$, $\gamma=\frac{a+b}{a-b}$, how do we approximate $\left\lceil \frac{\cosh^{-1}\gamma}{\log\gamma}\right\rceil$ as
$$
\left\lceil \sqrt{\frac{a}{b}}\left(1+\frac{1}{3}\frac{b}{a}+O(b^2/a^2)\right)\right\rceil?
$$
In general how to approximate $\left\lceil \frac{\cosh^{-1}\delta}{\log\gamma}\right\rceil$ where $\delta=\frac{c+d}{c-d}$?
 [/INST] Presumably you're taking $b/a \to 0$ here.  If $t = b/a$, then $\gamma = (1+t)/(1-t) = 1 + 2 t + 2 t^2 + O(t^3)$.  Now if $\gamma = 1 + u$, so $u = 2 t + 2 t^2 + O(t^3)$, 
$$\eqalign{\cosh^{-1}(\gamma) &= \log\left(\gamma + \sqrt{\gamma^2-1}\right) = \log\left(1+u + \sqrt{2u + u^2}\right)\cr
&= \log\left(1 + \sqrt{2} u^{1/2} + u + \dfrac{\sqrt{2}}{4} u^{3/2} + O(u^{5/2})\right)\cr
&= \sqrt{2} u^{1/2} - \dfrac{\sqrt{2}}{12} u^{3/2} + O(u^{5/2}) \cr
\log(\gamma) &= \log(1+u) = u - \dfrac{u^2}{2} + O(u^3)\cr
\dfrac{\cosh^{-1}(\gamma)}{\log(\gamma)} & = \sqrt{2} u^{-1/2} + \dfrac{5 \sqrt{2}}{12} u^{1/2} + O(u^{3/2})\cr
&= t^{-1/2} + \dfrac{1}{3} t^{1/2} + O(t^{3/2})\cr
}$$
as desired.
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