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.
Best Answer
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.