I am trying to find the inverse of $f$, knowing that \begin{equation} f \left( x \right) = \sum_{i = 1}^{n} \mu_i \left( 1 - x \right)^{i - 1} \end{equation} where $\sum_{i = 1}^n \mu_i = 1$, $x \in \left[ 0 , 1 \right]$.
It seems that $f$ is a bijection of $\left[ 0, 1 \right]$, so $f$ has an inverse. For example, it is easy to check that $f \left( x \right) = \frac{1 - \left( 1 - x \right)^n}{nx}$ is stricly monotonous, where $\mu_i = \frac{1}{n}$. However, even in this case, I am unsure of the existence of a closed-form solution for the inverse.
And I am just wondering if a closed-form exists in general.
Thank you!
I do not think that you could find a closed form for the general case.
However, letting $t=1-x$, you could consider that $$y= \sum_{i = 1}^{n} \mu_i\,t^{i - 1}$$ is a Taylor expansion built around $t=0$ to $O(t^{n+1})$ and use series reversion for an approximate solution. This would give quite nasty things such as $$t=z-\frac{\mu_3 }{\mu_2}z^2+\frac{2 \mu_3^2-\mu_2 \mu_4 }{\mu_2^2} z^3+\frac{-5 \mu_3^3+5 \mu_2 \mu_4 \mu_3-\mu_2^2 \mu_5 }{\mu_2^3}z^4+ \frac{14 \mu_3^4-21 \mu_2 \mu_4 \mu_3^2+6 \mu_2^2 \mu_5 \mu_3+3 \mu_2^2 \mu_4^2 }{\mu_2^4}z^5+\cdots \qquad\text{where} \qquad z=\frac{y-\mu_1}{\mu_2}$$
The case where $\mu_i = \frac{1}{n}\,\,\forall i$, it seems that $$t=z \sum_{i=0}^n (-1)^i z^i=\frac{z}{z+1}\left(1+(-1)^n z^{n+1} \right) \qquad\text{where} \qquad z=n y-1$$