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.

Solution 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 $4.9415299$

Solution 2:

I know there is a good accepted answer by @Claude Leibovici, and I did mention that I would add more, but I still had a few ideas, like a my recursive definition, but let’s do something new and get a closed form. We will 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.

I will list some later. Note the restrictions when using the zero product property. 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:

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

I actually cannot find any cases as $-r>0$, but you could assume an analytic continuation of the $\text I^{-1}_z(a,b)$ function until we define one, but maybe someone could derive one that works. In conclusion the solution is:

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

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 which is valid when the series, for the values in this answer is real. The root given will be the smallest one for the $x^r-x+a=0$. Define an extended function using the link as:

$$\text I^{-1}_z(a,b)\mathop=^\text{def}(az\text B(a,b))^\frac1a+\frac{b-1}{a+1}(az\text B(a,b))^\frac2a+\frac{(b-1)(a^2+3ab-a+5b-4)}{2(a+1)^2(a+2)} (az\text B(a,b))^\frac3a + \frac{(b-1)(a^4+(6b-1)a^3+(b+2)(8b-5)a^2+(33b^2-30b+4)a+b(31b-47)+18)}{3(a+1)^3(a+2)(a+3)}(az\text B(a,b))^\frac4a+… \implies \boxed{\text I^{-1}_z(2,b)=\frac{2(b-1)z}{3b(b+1)}+\sqrt{\frac{2z}{b(b+1)}}+\frac{(b-1)(11b-2)\left(\frac z{b(b+1)}\right)^\frac32}{18\sqrt 2} +\frac{(b-1)(43b^2-5b-2)}{135b^2(b+1)^2}+…}$$

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 $x=0.02504…$ which is ok convergence, but the point is to illustrate that this method works and gives a closed form when defining $\text I^{-1}_z(a,b)$ as a series. I would give the series expansion with the substituted values in this problem, but the method above works just as well. Here is a graph of the inverse function where $\text I_{1-e^2 x}\left(2,-\frac1e\right)+e^{1-\frac2e}-1=0\iff x^\frac1e-x-e^{-1}+e^{-2}=0$

We can now get rid of the $0\le 1-r a^{1-r} \le 1 ,r<0$ restriction. 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:

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

If only we had a way to represent the nth series coefficient in closed form.

If you want a more traditional closed form, then please try using the Fox-Wright function to maybe get a closed form of the Inverse Lagrange expression like in @Somos’s answer:

$$\,_pΨ_q\left(^{(a_1,A_1),…,(a_p,A_p)}_{(b_1,B_1),…,(b_q,B_q)}\ z\right)\mathop =^\text{def}\sum_{n=0}^\infty \frac{\Gamma(a_1+A_1 n)\cdots\Gamma(a_p+A_p n)z^n}{\Gamma(b_1+B_1 n)\cdots\Gamma(b_p+B_p n)}$$

This function is also already in Wolfram Functions and in Wikipedia because it is a special case of the more general Fox-H function, as seen in the article.

If you have Mathematica, you can potentially simplify the function further using Mathematica commands which I do not have, but it may be in Wolfram Alpha. Therefore our closed form is:

$$x^r-x+a=0\implies x=\sum_{n=0}^\infty \binom{rn}n \frac{a^{(r-1)n+1}}{(r-1)n+1}=a \sum_{n=0}^\infty \frac{\Gamma(rn+1)(a^{r-1})^n}{\Gamma(n+1)\Gamma(rn-n+1)((r-1)n+1)} = a \sum_{n=0}^\infty \frac{\Gamma(rn+1)(a^{r-1})^n\Gamma((r-1)n+1)}{n!\Gamma((r-1)n+1)\Gamma((r-1)n+2)}=\\ a\,_2Ψ_2\left(^{\ \ (1,r),(1,r-1)}_{(1,r-1),(2,r-1)}\:\ a^{r-1}\right),|a|<(r-1)r^\frac r{1-r}$$

Maybe we can equate our $2$ solutions?