How to solve a trinomial with a rational exponent $x^{q}-x-c=0$

Question

I require solutions to equations of the following form: $x^{q}-x-c = 0$ where $q$ is a rational number very close to 1 with $q$ such that: $$q=\frac{p^k+1}{p^k}$$ $$q=\frac{p^k}{p^k-1}$$ $p$ prime, $c$ the log of an integer. I'm hoping to be using a workstation computer to solve several of these equations to a high degree of precision and output some results to file.

Mostly I simply want the high precision solutions, up to 15 significant digits to the right of mantissa if possible. However further understanding about methods and the 'nature and properties' of the solutions is appreciated. For example, in this post: Solving an equation with irrational exponents I believe the author of the accepted answer suggests that if the exponent is rational then it is an algebraic problem with an algebraic solution. If so how can I generally solve these equations algebraically? Other posts have mentioned Lagrange Inversion and Fox-Wright Psi function for 'integral' exponents with the possibility of extrapolating to real values due to continuity. See the 2nd comment in this post: Solving $a\ln\ln(x)=\ln\ln(bx)$ in terms of Lambert W function but my exponent is rational and very close to 1, so I'm not sure that would be the best approach here. I know that use of the word 'integral' may sometimes allow quotients as well - depending on circumstances.

If there's an algebraic solution, I may just implement that in the program. Otherwise which iterative or computational method might work best for this type of equation, as the problem requires confidence in the precision of the results.

edit: I realize this probably should've been two different questions. One concerning numerical methods and the other concerning properties of the solutions. I've since acquired an introductory textbook on algebraic number theory to help understand the solutions better.

Answer 1

The equation $x^q - x - c = 0$ transforms to $y - k y^{1/q} - 1 = 0$ where $k = c^{(1/q)-1}$ and $x = (cy)^{1/q}$. This has a nice series solution in powers of $k$:

$$ y = \sum_{n=0}^\infty \frac{k^n}{n!} \prod_{j=0}^{n-2} \left(\frac{n}{q}-j\right)$$ Note that for $n=0$ and $n=1$ the empty product is $1$, so the first two terms are $1 + k$.

Answer 2

Newton's method is usually a really good way of solving equations like this, especially if you know a good neighbor hood of the solutions.

Start with an initial guess $x_n$. Then subsequent guesses are $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$.

$x_{n+1}=x_n-\frac{x_n^q-(x_n+c)}{qx_n^{q-1}-1}$

$x_{n+1}=\frac{(q-1)x_n^q+c}{qx_n^{q-1}-1}$

Then iterate.

Polynomials are low cost calculations. Rule of thumb is you'll need about half as many steps as you want accurate decimal digits, but you can keep track of that by monitoring the size of $|f(x_n)/f'(x_n)|=|x_{n+1}-x_n|$.

Answer 3

You wrote: "I believe the author of the accepted answer suggests that if the exponent is rational then it is an algebraic problem with an algebraic solution. If so how can I generally solve these equations algebraically?"
"an algebraic problem" means "an algebraic equation over a field" here.
For rational $q$, your equation is related to an algebraic equation (polynomial equation) over the field of $c$.
The algebra of algebraic equations over a field is Galois theory.
Solutions of only some of the algebraic equations over a field can be represented by radicals in dependence of the coefficients of the equation, solutions of the other of such equations cannot be represented in this way.
For algebraic $c$, the equation is an algebraic equation, and the solutions $x$ are algebraic numbers then.

The integral numbers mean the integers.

$$x^q-x-c=0$$

You are considering rational exponents $q$ and real $c$.

For rational $q$, your equation is a rational equation or an algebraic irrational equation - over $\mathbb{R}$.
Those kinds of equations are related to algebraic equations (polynomial equations) over $\mathbb{R}$. So you can use the known solution formulas and methods for algebraic equations over a field ($\mathbb{R}$).
Look e.g. for quintic* and radical*, Bring radical*, hypergeometric function*, Inverse Beta Regularized etc.

Are elementary and generalized hypergeometric functions sufficient to express all algebraic numbers?

Closed form solution for $x^n + x+ C = 0$

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

If you get a solution in closed form, e.g. in terms of Special functions, you get hints for its numeric calculation.

You could use computer algebra system software (CAS). They are quite powerful at solving equations - in closed form or numerically.
see e.g. Wolfram Alpha, Sage Math Cell $\ $

$$x^\frac{a}{b}-x-c=0$$ $x\to t^b$: $$\left(t^b\right)^\frac{a}{b}-t^b-c=0$$ $$t^a-t^b-c=0\tag{1}$$ For complex $x$, you get further solutions. Ask if you need this.

For integer $a,b\neq 0$ with $a\neq b$, equation 1 is a trinomial equation.

For real or complex $a,b\neq 0$ with $a\neq b$, equation 1 is in a form similar to a trinomial equation. A closed-form solution can be obtained using confluent Fox-Wright Function $\ _1\Psi_1$ therefore. So you get hints for the numeric calculations.

Szabó, P. G.: On the roots of the trinomial equation. Centr. Eur. J. Operat. Res. 18 (2010) (1) 97-104

Belkić, D.: All the trinomial roots, their powers and logarithms from the Lambert series, Bell polynomials and Fox–Wright function: illustration for genome multiplicity in survival of irradiated cells. J. Math. Chem. 57 (2019) 59-106

a)

$$x^{\frac{p^k+1}{p^k}}-x-c=0$$ $x\to t^{p^k}$: $$t^{p^k+1}-t^{p^k}-c=0$$ $p^k\to d$: $$t^{d+1}-t^d-c=0$$

b)

$$x^{\frac{p^k}{p^k-1}}-x-c=0$$ $x\to t^{p^k-1}$: $$t^{p^k}-t^{p^k-1}-c=0$$ $p^k\to d$: $$t^d-t^{d-1}-c=0$$

c)

$$x^{\frac{p^k-1}{p^k-p}}-x-c=0$$ $x\to t^{p^k-p}$: $$t^{p^k-1}-t^{p^k-p}-c=0$$ $p^k\to d$: $$t^{d-1}-t^{d-p}-c=0$$