Consider the function:
$$ f_{q}\left(x\right)=x+x^{q} $$
where $q\in\mathbb{Q},q>0$ and $x \in \mathbb{R}$, $x\geq 0$.
I am wondering what would be a method for inverting this function. It is monotone increasing on the non-negative reals as far as I see and so should be invertible.
Is there a closed form for the inverse function? Otherwise, could we express it as a series or an integral perhaps?
Thank you
Using Lagrange reversion:
$$x^q+x=a\implies x=a+\sum_{n=1}^\infty\frac{(-1)^n}{n!}\frac{d^{n-1}a^{qn}}{da^{n-1}}$$
and evaluating with factorial power $u^{(v)}$:
$$a+\sum_{n=1}^\infty\frac{(-1)^n}{n!}\frac{d^{n-1}a^{qn}}{da^{n-1}} = a+\sum_{n=1}^\infty\frac{(-1)^n}{n!}(qn)^{(n-1)}a^{(q-1)n+1}= a+\sum_{n=1}^\infty\frac{(-1)^n a^{(q-1)n+1}(qn)!}{((q-1)n+1)!n!} $$
Test the formula and look at “Substitution”
For $q\in\Bbb N$ there is a pattern with the hypergeometric function $_u\text F_v(…)$:
$$\boxed{x^q+x=a\implies x=a\,_{q-1}\text F_{q-2}\left(\frac1q,\dots,1-\frac1q;\frac{2}{q-1},\dots,\frac{q+1}{q-1};-\frac{q^qa^{q-1}}{(q-1)^{q-1}}\right),|a|\le(q-1)q^\frac q{1-q}}$$
I wonder if $q\in\Bbb Q$ is expressible in terms of the hypergeometric function?
