I am interested in the following function
$$ f(x) = \sqrt{3x \left(1-\frac{x}{\theta} \right) \ln{\left[a \left(1-\frac{\theta}{x} \right) \right]}}, $$
with $ 0<a \leq 1$ and $\theta<x$. Now to make things simple, assume $a= \theta = 1$.
The function seems complicated to be able to find its inverse directly, however is there a way to get maybe an expansion of its inverse? Some sort of infinite series maybe? I used desmos to sketch both the function (red) and its inverse (green), just to get an idea of its shape.
Thanks.
Let $x=\theta t$ and consider $$\frac{f^2(x)}{3\theta}=(1-t)\,t\, \log \left(a\frac{ t-1}{t}\right)=y$$
Expand for large values of $t$ to get $$y=- \log (a)t^2+ (1+\log (a))t-\frac{1}{2}-\frac{1}{6 t}-\frac{1}{12 t^2}-\frac{1}{20 t^3}+O\left(\frac{1}{t^4}\right)$$ Using power series reversion $$t=z+\frac{\log (a)+1}{2 \log (a)}+\frac{\log ^2(a)+1}{8 z \log ^2(a)}-\frac{1}{12 z^2 \log (a)}+O\left(\frac{1}{z^3}\right)$$ where $z=\sqrt{-\frac{y}{\log (a)}}$
Trying for $y=100$ and $a=\frac 1e$ gives $t=\frac{12031}{1200}=10.0258$ while the "exact" solution is $\cdots$ the same.
Edit
If $a=1$, the result is $$t=y+\frac 12+\frac 1{6y}\left(1-\frac{7}{60 y^2}+\frac{139}{5040 y^4}+O\left(\frac{1}{y^6}\right) \right)$$ For $y=100$, this gives $t=100.502$ while the "exact" solution is $\cdots$ the same.
Using this expression and expanding again for large values of $y$ $$\text{lhs - rhs}\sim \frac 1{734 \, y^7}$$