Is there a way to analytically solve $x^\alpha + y^\alpha = \alpha(x + y)$ for $\alpha$, other than $\alpha = 1$? Suppose $x, y > 0$ are fixed arbitrary positive numbers. Is there a way to solve $x^\alpha + y^\alpha = \alpha(x + y)$ for $\alpha \in \Bbb{R}$ analytically, excluding the trivial solution $\alpha = 1$?
This question cropped up during some research into generalising the inversion in the sphere map to Bregman distances. I have a plausible generalisation, and I'm trying to apply it to a nice Bregman sphere, but out popped this particular equation.
I do understand that such a solution will not involve elementary functions. I have no real skillset with special functions, so all I could do was try solving it with a CAS. I don't have Maple or Mathematica on my computer, but I tried using Wolfram Alpha and SageMath. The latter was useless, returning $\alpha = \frac{x^\alpha + y^\alpha}{x + y}$.
The former gave me nothing, simply telling me that computation time was exceeded. Indeed, even when I substituted a particular value for $y$, Wolfram Alpha still gave me no results. Only once I substituted two particular values for $x$ and $y$ would it give me an approximation with no closed form.
I realise that there's little hope for a nice solution, but I thought I'd check here before I give up completely.
 A: Whilst it is very likely a closed form does not exist (see for instance Two exponential terms equation solution), it is possible to determine a series expansion for $\alpha$.
Let $z=f(\alpha)$ where $f(\alpha)=x^\alpha+y^\alpha-\alpha(x+y)$. Since $f$ is analytic at zero with a non-zero derivative there, we can use the Lagrange inversion theorem to obtain an expansion about zero:$$\alpha=\sum_{n=1}^\infty g_n\frac{(z-f(0))^n}{n!}\bigg\vert_{z=0}=\sum_{n=1}^\infty\frac{(-2)^ng_n}{n!}$$ where \begin{align}g_n=\lim_{\alpha\to0}D_\alpha^{n-1}\left(\frac\alpha{x^\alpha+y^\alpha-\alpha(x+y)-2}\right)^n.\end{align} It is important to note that this only works when $\alpha<1$; that is, the series is valid only if $$y>e^{W(-x(\log x-1)/e)+1},$$ which follows by solving the tangent case $f'(\alpha)=0$ where $1$ is the only root. In addition, the domain of $W$ over the reals is $[-1/e,\infty)$, so when $x>e^{W(1/e)+1}\approx3.6$, there is no restriction on the value of $y$.
In relation to the coefficients, the first term can be evaluated using L'Hopital
\begin{align}g_1&=\lim_{\alpha\to0}\frac\alpha{x^\alpha+y^\alpha-\alpha(x+y)-2}\\&=\lim_{\alpha\to0}\frac1{x^\alpha\log x+y^\alpha\log y-(x+y)}\\&=\frac1{\log x+\log y-(x+y)}\end{align} and the second term requires repeated application \begin{align}g_2&=\lim_{\alpha\to0}D_\alpha\left(\frac\alpha{x^\alpha+y^\alpha-\alpha(x+y)-2}\right)^2\\&=\small\lim_{\alpha\to0}2\left(\frac\alpha{x^\alpha+y^\alpha-\alpha(x+y)-2}\right)\left(\frac{x^\alpha+y^\alpha-\alpha(x+y)-2-\alpha(x^\alpha\log x+y^\alpha\log y-(x+y))}{(x^\alpha+y^\alpha-\alpha(x+y)-2)^2}\right)\\&=\small\frac2{\log x+\log y-(x+y)}\lim_{\alpha\to0}\frac{x^\alpha+y^\alpha-\alpha(x+y)-2-\alpha(x^\alpha\log x+y^\alpha\log y-(x+y))}{(x^\alpha+y^\alpha-\alpha(x+y)-2)^2}\\&=\small\frac2{\log x+\log y-(x+y)}\lim_{\alpha\to0}\frac{-\alpha(x^\alpha\log^2x+y^\alpha\log^2y)}{2(x^\alpha+y^\alpha-\alpha(x+y)-2)(x^\alpha\log x+y^\alpha\log y-(x+y))}\\&=\small\frac{-1}{\log x+\log y-(x+y)}\lim_{\alpha\to0}\frac\alpha{x^\alpha+y^\alpha-\alpha(x+y)-2}\lim_{\alpha\to0}\frac{x^\alpha\log^2x+y^\alpha\log^2y}{x^\alpha\log x+y^\alpha\log y-(x+y)}\\&=-\frac{\log^2x+\log^2y}{(\log x+\log y-(x+y))^3},\end{align} and so on. When $x$ is large (the value of $y$ does not matter), these first two terms $$\alpha\approx-\frac2{\log x+\log y-(x+y)}-\frac{2(\log^2x+\log^2y)}{(\log x+\log y-(x+y))^3}$$ already give a very good approximation of the true value of $\alpha$. Here is a visualisation.
A: A general analytic solution in prescribed classes of functions is possibly not known. But of course that cannot be proven.
$$x^\alpha+y^\alpha=\alpha(x+y)$$
$$e^{\ln(x)\alpha}+e^{\ln(y)\alpha}-\alpha x-\alpha y=0$$
Applying Ritt's theorem on elementary inverses of elementary functions (Ritt 1925) and Schanuel's conjecture yields that the elementary functions $\alpha\to e^{\ln(x)\alpha}+e^{\ln(y)\alpha}-\alpha x-\alpha y$ over non-discrete domains cannot have elementary partial inverses. Therefore the equation cannot be solved by simply rearranging it by applying only finite numbers of elementary functions (elementary operations) readable from the equation. It is not known if all solutions of the equation are elementary numbers (Lin 1983, Chow 1999).
I give some cases where analytic solutions exist.
If $\alpha$ is rational, we get an algebraic equation.
If $x=1$, $y=1$, $x=y$ or $x=\alpha^{\frac{1}{\alpha-1}}$, $\alpha$ can be represented in terms of Lambert W.
A: Here is an explicit solution using Lagrange reversion and more familiar notation:
$$a^x+b^x=cx\implies x=\sum_{n=1}^\infty\frac{\left.\frac{d^{n-1}(a^x+b^x)^n}{dx^{n-1}}\right|_{x=0}}{c^nn!}$$
Using the binomial theorem:
$$a^x+b^x=cx\implies x=\sum_{n=1}^\infty\sum_{k=0}^n\binom nk\frac{\ln^{n-1}\left(a^kb^{n-k}\right)}{c^n n!}= \sum_{n=1}^\infty\sum_{k=0}^n\frac{((n-k)\ln(b)+k\ln(a))^{n-1}}{k!(n-k)!c^n}$$
Where $0\le k\le d\ge n$ shown here. Unfortunately, the convergence is pretty slow. Fortunately, the only inner sum pattern I could find by switching the sums, or not, was with the hypergeometric function $_p\text F_q$:
$$\boxed{e^{ax}+e^{bx}=cx\implies x=\sum_{n=1}^\infty\frac{(bn)^{n-1}}{c^nn!}\,_n\text F_{n-1}\left(-n,\frac{bn}{a-b}+1,\dots, \frac{bn}{a-b}+1;\frac{bn}{a-b},\dots,\frac{bn}{a-b};-1\right)}$$
The hypergeometric function reduces to a polynomial for which we are searching for a form in terms of simpler functions.
