Is there an analytic solution for $x^r - x + a = 0$?

Question

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.

Answer 1

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 $4.9415299$

Answer 2

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.

Answer 3

Background

There is a good accepted answer by @Claude Leibovici, but we use the Inverse of the Regularized Incomplete Beta function $\text I^{-1}_x(a,b)$ defined as the median of the Beta distribution which uses the Regularized Beta function $\text I_x(a,b)$ with the following definitions: $$\text I_x(a,b)=\frac{\Gamma(a+b)}{\Gamma(b)\Gamma(a)}\int_0^x t^{a-1} (1-t)^{b-1}dt,\frac{\int_0^{\text I^{-1}_x(a,b)}t^{a-1} (1-t)^{b-1}dt }{\text B(a,b)}=\text I_{\text I^{-1}_x(a,b)}(a,b)\mathop=^{0\le x\le1}x$$ Now note that when: $$\text I_{1-x}(2,r)=rx^{r+1}-(r+1)x^r+1$$ for example let’s solve for $x$: $$\text I_{1-x}(2,e)= ex^{e+1}-(e+1)x^e+1=\frac12\implies 2ex-2(e+1)+x^{-e}=0\implies x=0.587686… =1-\text I^{-1}_\frac12(2,e)$$ with this verification.

If a special case of the Regularized Beta function can be solved in closed form for its subscripted $x$ value, then you can derive many special cases for the Inverse of the Regularized Beta function in terms of other functions. Unfortunately there is a small domain, little research on the inverse function, and a limited amount of polynomials that we can use it for, but let’s try the following to get a general, but slightly complicated, closed form for $x$ in $x^r-x+a$ in terms of $\text I^{-1}_x(a,b)$ which may simplify into simpler functions if the $a,b,x$ are not too complicated.

Solution Derivation:

Assume the equation is in the form of:

$$\text I_{1-ax}(2,-r)+b=(ax)^{-r}((ax)^r-arx+r-1)+b=0\implies (ax)^r-arx+r-1+(ax)^rb=(b+1)(ax)^r-arx+r-1=0\implies x^r-\frac{ar}{(b+1)a^r}x+\frac{r-1}{(b+1)a^r}=0$$

The goal is to have the coefficient of $x$ to be $-1$, so let:

$$\frac{ar}{(b+1)a^r} =1\implies b=ra^{1-r}-1; a^r,ar\ne 0$$

Now that we have gotten rid of one variable:

$$x^r-\frac{ar}{(b+1)a^r}x+\frac{r-1}{(b+1)a^r}= x^r-\frac{ar}{(ra^{1-r}-1 +1)a^r}x+\frac{r-1}{(ra^{1-r}-1 +1)a^r}=x^r-x+\frac{r-1}{ar}=0 $$

Now we can let $\frac{r-1}{ar}=A$ to solve for $x^r-x+A=0$. Let’s see what happened to the other side of the equation:

$$(ax)^r \text I_{1-ax}(2,-r)+b(ax)^r = (b+1)(ax)^r-arx+r-1=0\implies \frac x{b+1} \text I_{1-ax}(2,-r)+\frac{bx}{b+1} = x^r-\frac{ar}{(b+1)a^r}x+\frac{r-1}{(b+1)a^r}=0 \implies \frac x{ra^{1-r}-1 +1} \text I_{1-ax}(2,-r)+\frac{ra^{1-r}-1 x}{ra^{1-r}-1 +1} = \frac{xa^{r-1}}{r}\text I_{1-ax}(2,-r)+\left(1-\frac{a^{r-1}}r\right)x=x^r-x+\frac{r-1}{ar}=0 $$

Finally:

$$\frac{xa^{r-1}}{r}\text I_{1-ax}(2,-r)+\left(1-\frac{a^{r-1}}r\right)x=x^r-x+\frac{r-1}{ar}=0 $$

Notice that both sides will always equal zero, so we can split them up and do each case separately which works:

$$ x^r-x+\frac{r-1}{ar} =0, \frac{a^{r-1}}{r}\text I_{1-ax}(2,-r)+1-\frac{a^{r-1}}r=0 \implies \text I_{1-ax}(2,-r)+ \frac r{a^{r-1}} - 1 =0$$

Old Series Approximation Method:

$$\boxed{x^r-x+\frac{r-1}{ar} =0\implies x=\frac{1-\text I^{-1}_{1-r a^{1-r}}(2,-r)}a} ,0\le 1-r a^{1-r} \le 1,r<0$$

using the “Inversebetaregularized$(z,a,b)$” command in Wolfram Alpha or Mathematica for $\text I^{-1}_z(a,b)$.

It looks like you may need complex $a,r$, but let me see later.

The formula itself looks useless, but here is proof of how it could be used with multiple domain extensions.

Here is an example using the series expansion Now use $$x^r-x+\frac{r-1}{ar}=\sqrt[e]x-x+\frac1e-\frac1{e^2}=0, a=e^2,r=\frac1e\implies x=0.592109…, 0.0250461…$$

Using our derived formula:

$$\sqrt[e]x-x+\frac1e-\frac1{e^2}=0 \implies x=\frac{1-\text I^{-1}_{1-e^{1-\frac2e}}\left(2,-\frac1e\right)}{e^2}$$

Now plug into our series from Wolfram Functions:

First term+second term

Third term

Fourth term

Final result after adding terms, coefficients, and constants

After 4 terms in the Inverse Regularized Beta series, we get that $x≈0.02547$ when the root is actually $x=0.02504…$. Please correct me and give me feedback.

Therefore when we use the series expansion for $\text I^{-1}_z(2,b)$, we have the following final solution:

$$\boxed{x^r-x+a=0\implies x=\frac{ar\left(1-\text I^{-1}_{1-r\left(\frac{r-1}{ar}\right)^{1-r}}(2,-r)\right)}{r-1},\text I^{-1}_z(2,-r)= -\frac{2(r+1)z}{3r(r-1)}+\sqrt{\frac{2z}{r(r-1)}}+\frac{(r+1)(11r+2)\left(\frac z{r(r-1)}\right)^\frac32}{18\sqrt 2} -\frac{(r+1)(43r^2+5r-2)}{135r^2(r-1)^2}+…\in\Bbb R}$$

”Closed form:”

Transforming $$x^{-r}-x+a=0\implies x=\frac{ar\left(1-\text I^{-1}_{1+r\left(\frac{r+1}{ar}\right)^{-1-r}}(2,r)\right)}{r+1},-1\le r\left(\frac{r+1}{ar}\right)^{-1-r}\le0 $$

does not need any slow series for:

$$\boxed{x^r-x+a=0\implies x=\frac{br}{(1-r)\left(\text I^{-1}_{\frac{1-r}b\left(\frac{br}{r-1}\right)^r+1}(2,r-1)-1\right)},-1\le \frac{1-r}b\left(\frac{br}{r-1}\right)^r \le 0}$$

Test here or see formula $1$ here