asymptotic behaviour of inverse hyperbolic sine function

Question

Good evening everyone,

I'm solving a physics problem and I would like to approximate the funcion arcsinh(ax)/x to a easier to work with function in the x>> region.

I did a Taylor series to approximate the x<< region, but now I don't know what to do

Answer 1

As you inverse hyperbolic functions are logarithmic functions.

for example, $$ \sinh^{-1} ax = \ln|ax+\sqrt {a^2x^2 +1}|$$

so you may use that function instead.

Answer 2

Ais said in comments and answer, assuming $a>0$, let $y=ax$ to make $$\frac{\sinh ^{-1}(a x)}{x}=\frac a y \frac{\sinh ^{-1}(y)}{y}=\frac a y\log\left(y+\sqrt{1+y^2}\right)$$

Now $$\log\left(y+\sqrt{1+y^2}\right)=\log(y)+\log \left(1+\sqrt{1+\frac{1}{y^2}}\right)$$ and use Taylor $$\sqrt{1+t}=1+\frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}-\frac{5 t^4}{128}+O\left(t^5\right)$$ $$1+\sqrt{1+t}=2+\frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}-\frac{5 t^4}{128}+O\left(t^5\right)$$ $$\log \left(1+\sqrt{1+t}\right)=\log (2)+\frac{t}{4}-\frac{3 t^2}{32}+\frac{5 t^3}{96}-\frac{35 t^4}{1024}+O\left(t^5\right)$$ Make $t=\frac 1{y^2}=\frac 1{a^2x^2}$ to get the desired result.