Is there analytical solution to $x^{a+1} + x^{a} = c$ with $0<a<1$?

Question

How to solve the nonlinear equation of type $x^{a+1} + x^{a} = c$ with $0<a<1$ and $c>0$? Sorry, I don't know which tag is appropriate.

Answer 1

In the most general case, there is no analytical solution for the zeros of $$f(x)=x^{a+1} + x^{a} - c$$ and I guess that only numerical methods could be used.

Consider the simplest, that is to say Newton method : starting from a reasonable guess $x_0$, it will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ so, for this case, $$x_{n+1}=x_n-\frac{x_n^{a+1}+x_n^a-c}{a x_n^{a-1}+(a+1) x_n^a}$$

For illustration purposes, let us take $a=0.2345$ and $c=10$; being lazy, I shall start with $x_0=10$. Then Newton iterates will be $5.88834$, $5.66079$, $5.65988$ which is the solution for six significant figures.

Being less lazy, instead of starting at $x_0=c$, we could start at $x_0=c^{\frac{1}{a+1}}$ (which is a upper bound of the solution) and convergence will be slightly faster.

You could also look at the problem as the intersection of $y=\frac{c}{x+1}$ and $y=x^a$

Answer 2

Rewrite as $$t^{\alpha}=c-t,$$ where $x^a=t$ and $\alpha=1+1/a>2$. This is the intersection of a "parabola" with a straight line inclined by 45°.

For $c<0$, there is no solution ($x,t\ge 0$ assumed).

For $c>0$, there is a single solution. A good starting value for numerical resolution is given by assuming $\alpha=2$:

$$t_0=\frac{\sqrt{4c+1}-1}2.$$

The formula for $\alpha=3$ (given by Wolfram) is also manageable:

$$t_0=\sqrt[3]{\frac s2}-\frac13\sqrt[3]{\frac2{s}},$$ where $s=\sqrt{c^2+\frac4{81}}+c$.

For large values of $\alpha$, it may be better to use $t^\alpha\approx0$, hence $t_0=c$ for $c<1$ and $t_0=1$ for $c>1$.

The question can also be approached as $t^\beta=c-t$, where $x^{a+1}=t$ and $0<\beta=a/(a+1)<\frac12$.