I am trying to formulate $\sinh^{-1} x$.
I know that $$\sinh x = \dfrac{\exp(x)-\exp(-x)}{2}$$
but how do we translate $\sinh^{-1} x$ ?
Please excuse my lack of knowledge in math.
$\endgroup$1 Answer
$\begingroup$If $y=\operatorname{arsinh}x$ then $x=\sinh y=\frac{e^y-e^{-y}}{2}$, which rearranges to $e^{2y}-2xe^y-1=0$, a quadratic in $e^y=x\pm\sqrt{x^2+1}$. We need the $\pm=+$ branch because, when $y$ is large, $x\sim\tfrac12e^y$ (or even easier, note $e^y>0$). So $$\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right).$$You can show as a similar exercise that$$\operatorname{arcosh}x=\ln\left(x+\sqrt{x^2-1}\right),\,\operatorname{artanh}x=\tfrac12\ln\tfrac{1+x}{1-x}.$$
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