Series solution of the trinomial equation

Question

The roots of trinomial equations $x^p+x-q=0$ ($p\in\mathbb{N}$) can be expressed in terms of the hypergeometric functions. I am wondering if at least one real root, for instance given by the following iterative solution $x_n=\frac{q - (1 - p) x_{n-1}^p}{ 1+p x_{n-1}^{(p - 1)}}$, $x_0=q^{1/p}$ can be expressed in a similar form for $p\in\mathbb{R}$, $p\ge2$.

Answer 1

An analogous formula does hold, although the corresponding functions are not hypergeometric if $p$ is irrational.

For given $p\in\mathbb{R}$, $p>1$, consider the power series $$h(z)=\sum_{k=0}^{\infty} \frac{(-1)^k}{pk-k+1}\binom{pk}{k}\, z^k$$ with radius of convergence $R=(p-1)^{p-1}/p^p.$

Then, for $0\le y\le R^{1/(p-1)}$, the function $g(y):=yh(y^{p-1})$ is the inverse function of $f(x):=x+x^p$. $$*$$ [edit] There is also an analogous inversion formula for three or more terms, to invert e.g. $f(x)=x+ax^p+bx^q$ with real exponents $p>1$ and $q>1$. If $H=H_{p,q}$ is the analytic function $$H(u,v)=\sum_{i\ge0,j\ge0}\frac{(-1)^{i+j}}{ (p-1)i+ (q-1)j+1} {pi+qj \choose i,\, j}u^iv^j,$$ then $g(y):=yH(ay^{p-1},by^{q-1})$ is the local inverse of $f$ at $0$ (the multinomial coefficient in the double series is ${pi+qj \choose i,\, j}:=\frac{(pi+qj)(pi+qj-1)\dots(pi+qj-i-j+1)}{i!j!}$ .)