Is there an analytic solution for $x^r - x + a = 0$? I am trying to find the solutions to
$$x^r - x + a = 0$$
for
$$r \in \mathbb{R}, \quad 0 \leq r < 1\\
a \in \mathbb{R}, \quad a \geq 0$$
$a$ and $r$ are both fixed and I am trying to solve for $x$.
Is there a way of rewriting this as a closed-form expression? If there isn't, is there an easy way of approximating the solution?
From what I've seen graphing the equation, there is guaranteed to be exactly one root with the constraints given. This matches the context that the equation came from so I am confident that this is true.
For context, this is not homework. I am trying to rewrite an equation and am stuck at this part.
 A: Consider that you lookk for the zero's of function
$$f(x)=x^r-x+a \qquad \text{with}\qquad 0<r<1 \qquad \text{and}\qquad a>0$$ I shall assume that $r$ is non rational, which means that $x>0$.
The solution is larger than $a$ since $f(a)>0$ and $f(x)$tends to $-\infty$. Using one single iteration of Householder method with $x_0=a$, we have, as an approximation,
$$x_1=a-\frac{3 a^{r+1} \left(r (r+1) a^{2 r}-4 r a^{r+1}+2 a^2\right)}{r \left(r^2+3
   r+2\right) a^{3 r}+18 r a^{r+2}-6 r (2 r+1) a^{2 r+1}-6 a^3}$$ Trying with $a=\pi$ and $r=\frac 1e$, this would give $x_1=4.94840$ while the solution is $x=4.94153$.
Edit
What we can also do is to make two iterations of Newton method and have
$$x=a+\frac{a^{r+1}}{a-r a^r}-\frac{\frac{a^r}{r a^{r-1}-1}+\left(\frac{a^{r+1}}{a-r
   a^r}+a\right)^r}{r \left(\frac{a^{r+1}}{a-r a^r}+a\right)^{r-1}-1}$$ For the worked example, this would give $x=4.94156$.
Update
In fact, the solution is larger than $x_0=a+a^r > a$ (then better).
The first iteration of Halley method will give
$$x_1=x_0-\frac{2 f(x_0) f'(x_0)}{2 f'(x_0)^2-f(x_0) f''(x_0)}$$ with
$$f'(x)=r x^{r-1}-1 \qquad \text{and} \qquad f''(x)=(r-1) r x^{r-2}$$ For the worked example, this would give $x_0=4.66526$ and $x_1=4.94150$
We could even improve the value of $x_0$ drawing the straight line joining points $[a,f(a)]$ and $[a+a^r,f(a+a^r)]$ and have
$$x_0=a+\frac{a^{2 r}}{2 a^r-\left(a+a^r\right)^r}$$ For the worked example, this would give $x_0=4.94813$
Making one iteration of Newton method $\big[$notice that I decreased the order of the method from $4$ (Householder) to $3$ (Halley) and now to $2$ (Newton)$\big]$
$$x_1=a+\frac{a^{2 r}}{2 a^r-\left(a^r+a\right)^r}+\frac{\frac{a^{2 r}}{2 a^r-\left(a^r+a\right)^r}-\left(\frac{a^{2 r}}{2
   a^r-\left(a^r+a\right)^r}+a\right)^r}{r \left(\frac{a^{2 r}}{2
   a^r-\left(a^r+a\right)^r}+a\right)^{r-1}-1}$$ which, for the worked example, gives $x_1=4.9415303$ while the "exact" solution is
$4.9415299$
A: The question asks

I am trying to find the solutions to $$x^r - x + a = 0$$ for
$$r \in \mathbb{R}, \quad 0 \leq r < 1\\
a \in \mathbb{R}, \quad a \geq 0$$
$a$ and $r$ are both fixed and I am trying to solve for $x$.
Is there a way of rewriting this as a closed-form expression? If there isn't, is there an easy way of approximating the solution?

The Wikipedia article
Lambert W function
section on History states:

Lambert first considered the related Lambert's Transcendental Equation in $1758,^{[3]}$ which led to an article by Leonhard Euler in $1783^{[4]}$ that discussed the special case of $we^w$.
The function Lambert considered was
$$    x=x^{m}+q. $$
Euler transformed this equation into the form
$$  x^{a}-x^{b}=(a-b)cx^{a+b}. $$
Both authors derived a series solution for their equations.

Concerning the Lambert trinomial equation the Wikipedia article
Lagrange inversion theorem
Example section states

For instance, the algebraic equation of degree $p$
$$ x^{p}-x+z=0 $$
can be solved for $x$ by means of the Lagrange inversion formula for the function $f(x) = x − x^p$, resulting in a formal series solution
$$ x=\sum _{k=0}^{\infty }{\binom {pk}{k}}{\frac {z^{(p-1)k+1}}{(p-1)k+1}}.$$
By convergence tests, this series is in fact convergent for
$ |z|\leq (p-1)p^{-p/(p-1)}, $
which is also the largest disk in which a local inverse to $f$ can be defined.

Thus, the solution can be expressed as a power series in terms of
$\,z^{p-1},\,$ hence analytic. Note that here the degree $p$ does
not need to be a positive integer. The binomial coefficient is a
rational function of $p$ and thus the power series solution makes
sense even if it is a real number. In your case, the power series
solution is
$$ x = a + a^r + r a^{2r-1} + \frac{r(3r-1)}2 a^{3r-2} +
 \frac{r(4r-2)(4r-1)}6 a^{4r-3} + \dots. $$
I doubt if it has a closed form.
