Inverse of $f(x)=x^n+x$ on $[0,\infty)$ Fix integer $n > 1$. The function $f_n(x) = x^n + x$ is monotone increasing on $[0,\infty)$, and so has an inverse $f_n^{-1}(x)$ that is also monotone increasing on $[0,\infty)$.  
I'm interested in properties of $f_n^{-1}(x)$ (in particular, in its Taylor series). I'm sure that it's a well-studied function (or rather, family of functions), but I can't find any literature on it, mostly because I don't know what keywords to use.
Does anyone know a name for this function?   
 A: Suppose a Taylor expansion around the origin exists with positive radius of convergence, say $$f_n^{-1}(y)=:g(y)=\sum_{k=0}^\infty a_ky^k.$$ Since $g(0)=0$ it follows that $a_0=0$. Consider the relation $$y=g(y)^n+g(y)=\left(\sum_{k\ge 1}a_ky^k\right)^n+\sum_{k\ge 1}a_ky^k $$ which "simplifies" to $$ y= \sum_{k\ge 1}a_ky^k+\sum_{m\ge n}\left(\sum_{k_1+\ldots k_n=m}a_{k_1} \ldots a_{k_n}\right)y^m.$$ Since the latter sum contains monomials of degree $\ge n$ we see right away that $a_1=1$ and $a_2=\ldots a_{n-1}=0$.Let $$c_m= \sum_{k_1+\ldots k_n=m}a_{k_1} \ldots a_{k_n}.$$ The only way $n$ indices $\ge 1$ sum up to $n$ is that $k_1=\ldots k_n=1$, whence $c_n=1$, yielding the coefficient $a_n+1$ for $y^n$ on the RHS. Since this must vanish we have $a_n=-1$. 
Continuing like this we see that for $n<m<2n-1$ all the terms in the sum defining $c_m$ involve only products of $a_j$ for $1\le j <n$, therefore $a_m=0$ for $n<m<2n-1$. On the other hand $c_{2n-1}=n\cdot (-1)$ and since $a_{2n-1}+c_{2n-1}=0$ we get $a_{2n-1}=n$. 
After this things get more complicated but in principle one can go on and get expressions for further coefficients. All in all, assuming $f_n^{-1}$ has a Taylor expansion around zero with positive radius of convergence, it must look like $$g(x)=x-x^n+nx^{2n-1}+\ldots $$ (Actually, $a_{2n}$ must also vanish..)
A: $$f_1^{-1}(x) = \frac{x}{2}$$ 
$$f_2^{-1}(x) = \frac{1}{2}(-1 \pm \sqrt{4x+1})$$
$$f_3^{-1}(x) = \frac{\sqrt[3]{\frac{2}{3}}}{\sqrt[3]{\sqrt{3}\sqrt{27x^2+4}-9x}} - \frac{\sqrt[3]{\sqrt{3}\sqrt{27x^2+4}-9x}}{\sqrt[3]{2} \cdot 3^{2/3}}$$
You can check this answers with WolframAlpha.
In general $f_n^{-1}(x)$ is the root of polynomial $y^n+y-x$ for $y$. 
I don't know if there is a name for this problem.
