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An equation like

$$a^x+b^x=1$$

can be turned to the form

$$t^\alpha+t=1$$ by a suitable change of variable.

When $\alpha$ is a rational we can put that in a polynomial form

$$u^p+u^q=1$$ and use Galois theory to refute the existence of a solution expressible by radicals.

But what about $\alpha$ being irrational, and what about the (non-)existence of closed-form formulas ?

Update:

From the comment by @mercio, we know that for every $\alpha$ of the form $\log_c(1-c)$ where $c$ has a closed form expression, we have the closed-form solution $t=c$. This settles the first part of the question.

So is there a way to characterize the $\alpha$'s for which a closed-form solution exists ?

  • the equation describes a bijection from $t \in (0;1)$ to $\alpha \in (0 ; + \infty)$, and you can express $\alpha$ in terms of $\log (1-t)$ and $\log t$ so this gives you plenty of cases where $t$ and $\alpha$ have a "closed form". – mercio Jan 14 '16 at 11:59
  • @mercio: can you give an example ? –  Jan 14 '16 at 12:22
  • pick the nice closed form $t=1/3$. Then the corresponding $\alpha$ is $\log (1-t)/\log t = 1-\log 2/\log 3$, so if you start with this particular irrational $\alpha$, you get a closed form for $t$. – mercio Jan 14 '16 at 12:35
  • @mercio. Indeed. Then a related question is if there is an algorithmic process to go from the closed expression of $\alpha$ to that of $t$, or exclude its existence. –  Jan 14 '16 at 13:10
  • Wait a second - how do you get from $a^x+b^x=1$ (an equation with unknown in exponential) to $t^\alpha+t=1$ (where, unless I misunderstand, $\alpha$ is fixed and the unknown is in the base now)? – Wojowu Jan 14 '16 at 17:00
  • @Wojowu: $t=b^x$, $t^\alpha=(b^\alpha)^x$. –  Jan 14 '16 at 17:22
  • Non-rational $\alpha$ means: The equation cannot be transformed into an equation solved for $t$ by applying only elementary functions according to Liouville. – IV_ Dec 10 '16 at 20:41
  • 1
    Would you accept the Fox-Wright function as a closed form? I just found this article online which describes a solution for some values of $\alpha$ in Section 4. – pregunton Mar 04 '18 at 11:06
  • @pregunton: no, why should I ? –  Mar 04 '18 at 15:51

1 Answers1

1

My answer is for solutions in terms of elementary functions.

Let $\alpha,a,b,t,x\in\mathbb{C}$.

$$t^\alpha+t=1\tag{1}$$

You already treated the case of rational $\alpha$.
$\ $

1.) Solutions in the algebraic numbers

Let $\alpha$ be an irrational algebraic number and $t$ algebraic.
a) $0$ and $1$ aren't solutions of equation 1. Therefore $t\notin\{0,1\}$.
b) Assume that $t\notin\{0,1\}$. Then Gelfond–Schneider theorem implies that $t^\alpha$ is transcendental. Because equation 1 is an algebraic equation of $t^\alpha$ and $t^\alpha$ is transcendental, equation 1 is a contradiction. $t\notin\{0,1\}$ cannot be algebraic therefore.
Because of a) and b), $t$ cannot be algebraic.

2.) Solutions by elementary partial inverse functions

The elementary functions are generated from their complex function variable by applying finite numbers of $\exp$, $\ln$ and/or unary or multiary algebraic functions.

Solving equation 1 by rearranging it by operations we can read from the equation means applying appropriate partial inverse functions of the elementary function $f\colon t\mapsto t^\alpha+t$.

The incomprehensibly unfortunately hardly noticed theorem in [Ritt 1925], that is proved also in [Risch 1979], implies what kinds of elementary functions have elementary partial inverse functions: The elementary functions having elementary partial inverse functions are generated from their complex function variable by applying finite numbers of $\exp$, $\ln$ and/or unary algebraic functions.

We have $f(t)=A(t^\alpha,t)$, wherein $A$ is an algebraic function. For irrational $\alpha$, $\text{trdeg}_{\overline{\mathbb{Q}}}(t^\alpha,t)>0$, and with help of Ritt's theorem we conclude that $f$ doesn't have elementary partial inverse functions. Therefore equation 1 is not solvable by rearranging it by operations we can read from the equation.

[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90

[Risch 1979] Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math. 101 (1979) (4) 743-759

3.) Solutions in the elementary numbers

The elementary numbers are the numbers that are generated from the rational numbers by applying only elementary functions.

For irrational $\alpha$, the problem of solving equation 1 by elementary numbers is unsolved because the only general statement we have for this kinds of problems is Lin's theorem in [Lin 1983]. See e.g. Example equation which does not have a closed-form solution. But Lin's theorem is applicable only to irreducible algebraic equations of simultaneously $z\in\mathbb{C}$ and $e^z$, and equation 1 possibly cannot be transformed into this form.

[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50

[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448

IV_
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