Solution to generalized-polynomial equation?

Question

Is it possible to obtain the solution to this generalized-polynomial equation? $$b x^a - x +c =0$$ with $-1<a<1$, $b>0$, $c>0$ and $x>0$.

Answer 1

Hint: Define $$f(x)=bx^a-x+c$$ and use calculus.

$$f'(x)=bax^{a-1}-1$$ we have $$f(0)=c>0$$ if $a<0$ then we get $$f'(x)<0$$ If $0<a<1$ then we get $$f'(x)=0$$ if $$x=\left(\frac{1}{ab}\right)^{1/{a-1}}$$

Answer 2

In general, only by numerical methods or series. The following series solution in powers of $b c^{a-1}$ converges if $b c^{a-1}$ is small :

$$ \eqalign{x &= c + b c^a + \frac{ca}{b} (b c^{a-1})^2 + \frac{ca(3a-1)}{2} (b c^{a-1})^3 + \ldots\cr &= c + b c^a + c \sum_{k=2}^\infty \frac{(b c^{a-1})^k}{k!} \prod_{j=0}^{k-2} (ka-j)}$$

Answer 3

For convenience, rewrite as

$$x^a=mx+p$$ which describes the intersection(s) of a power law and a straight line.

If $m<0$, the equation has a single root when $p>0$ and none otherwise.

If $m>0$, then if $p>0$, the equation has a single solution. If $p<0$, consider the tangent of slope $m$, i.e. the $\bar x$ such that

$$a\bar x^{a-1}=m.$$

Then if $\bar x^a<m\bar x+p$, the equation has two solutions, on either side of $\bar x$, else none.