Is there any theory (analogous to Galois theory) for solving equations with irrational exponents like:
$ x^{\sqrt{2}}+x^{\sqrt{3}}=1$
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Well, for one thing, this is now a transcendental equation instead of an algebraic one, and the theory of seeing whether a transcendental equation has (unique) solutions is a lot less developed than in the algebraic case (i.e., one usually proceeds on a case-to-case basis for determining if a transcendental equation has solutions). – J. M. ain't a mathematician Aug 30 '10 at 00:54
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In my physics classes, we would just graph these things and read off the answer. I don't know if there is a better way, but I think this is an interesting question. There must be some transcendental equations with an analytic solution, but I suspect the situation could be like Diophantine equations, where there is just no hope for a general method. – Matt Calhoun Aug 30 '10 at 13:25
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I am by no means an expert on Galois theory, far from it, but I didn't think it gave a method to solve polynomials. My understanding is that this theory provides an explanation as to why 5th degree and higher polynomials are not generally solvable by radicals. As for analytic methods for transcendental equations which have nothing to do with Galois theory, check out Newtons method (this approximation will under certain conditions actually converge to the correct answer) http://en.wikipedia.org/wiki/Newton%27s_method – Matt Calhoun Aug 30 '10 at 13:48
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The key point is "some"; the difficulty is in finding general analytic solutions for transcendental functions. – J. M. ain't a mathematician Aug 30 '10 at 13:48
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Matt: the thing to understand with Galois theory is that one has to keep adding "tools" to your repertoire to (analytically) solve polynomial equations of increasing degree. For linears, you only need division; for quadratics to quartics, you need radicals; for quintics you need to use hypergeometric or theta functions, and even higher degree polynomial equation need even more complicated functions. Newton is an approximation method; it does not give analytic solutions. – J. M. ain't a mathematician Aug 30 '10 at 13:52
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@J.Mangaldan: From the wiki article I linked, it says "The (Newton) method will usually converge, provided this initial guess is close enough to the unknown zero". This implies to me that this method will (under the right conditions) generate a sequence of real numbers which converges to the desired solution. I would call this an "analytic solution" since I think of "analysis" as being primarily about converging sequences. Perhaps there is something I am misunderstanding here. I call the finite iterations "approximations", not the limit (if it exists). Am I wrong to use this terminology? – Matt Calhoun Aug 30 '10 at 15:40
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The study of such equations is not "abstract algebra" as it is usually understood. The reason is that to even define the function $x^{\sqrt{2}}$, for example, requires analysis; one has to prove certain properties of $\mathbb{R}$ to ensure that such a function exists. This is in marked contrast to the case of integer or rational powers, where one has a purely algebraic definition and the background theory is equational. To define the function $x^{\sqrt{2}}$ one has to either define $e^x$ and the logarithm or consider a limit of functions $x^{p_n}$ where $p_n$ form a sequence of rational approximations to $\sqrt{2}$, and this is irreducibly non-algebraic stuff.
In particular, while polynomials can be studied in an absurdly general setting, transcendental equations like those you describe are more or less restricted to $\mathbb{R}$ (or $\mathbb{C}$ if you really want to pick a branch of the logarithm). The LHS is an increasing function of $x$, so there is at most one root, which probably one can really only compute numerically if it exists. (Its nonexistence can be ruled out by computing local minima in $(0, 1)$.)
This is another question which touches on a theme which has come up several times on math.SE, which is that exponentiation should really not be thought of as one operation. Instead, it is a collection of related operations with various degrees of generality and applicability which happen to share the same algebraic properties, and one should not infer too much about how similar these operations are.
Qiaochu Yuan
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9One should be careful about making strong broad statements such as so-and-so "cannot be called abstract algebra". A large part of elementary calculus can be algebraicized. That's how computer algebra systems work. One has differential algebra and Galois theory for computing integrals and solving ODEs, Hardy-Rosenlicht fields and transseries for computing limits and asymptotics, etc. – Bill Dubuque Aug 30 '10 at 01:30
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@QiaochuYuan "This is in marked contrast to the case of integer or rational powers, where one has a purely algebraic definition and the background theory is equational." I realize this is an old post but I need to solve equations of the form $x^{p/q}-x-c=0$. Is that considered algebraic, and how does one go about solving it? – Anaxagoras Mar 13 '23 at 02:29
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Your equations are algebraic equations in dependence of at least two algebraically independent monomials. With help of the main theorem in [Ritt 1925], that is also proved in [Risch 1979], we can conclude that the elementary function on the left-hand side of the equations doesn't have a partial inverse with non-discrete domain that is an elementary function. Therefore, the equations cannot be rearranged for $x$ by applying only elementary functions/operations we can read from the equation.
2.)
Your equations are transcendental equations, namely exponential equations, namely exponential sum equations. There are some theorems for the solutions of exponential sum equations in some number fields.
3.)
Topological Galois Theory of Askold Khovanskii meets your question.
[Khovanskii 2014]:
"Vladimir Igorevich Arnold discovered that many classical questions in mathematics are unsolvable for topological reasons. In particular, he showed that a generic algebraic equation of degree 5 or higher is unsolvable by radicals precisely for topological reasons. Developing Arnold’s approach, I constructed in the early 1970s a one-dimensional version of topological Galois theory. According to this theory, the way the Riemann surface of an analytic function covers the plane of complex numbers can obstruct the representability of this function by explicit formulas. The strongest known results on the unexpressibility of functions by explicit formulas have been obtained in this way.
...
The monodromy group of an algebraic function is isomorphic to the Galois group of the associated extension of the field of rational functions. Therefore, the monodromy group is responsible for the representability of an algebraic function by radicals. However, not only algebraic functions have a monodromy group. It is defined for the logarithm, arctangent, and many other functions for which the Galois group does not make sense. It is thus natural to try using the monodromy group for these functions instead of the Galois group to prove that they do not belong to a certain Liouville class. This particular approach is implemented in one-dimensional topological Galois theory ..."
[Khovanskii 2014] Khovanskii, A.: Topological Galois Theory - Solvability and Unsolvability of Equations in Finite Terms. Springer 2014
[Khovanskii 2019] Khovanskii, A.: One dimensional topological Galois theory. 2019
[Khovanskii 2021] Topological Galois Theory - Slides 2021, University Toronto
4.)
Your equations are of the form
$$x^a+x^b=1\ \ \ (a,b\in\mathbb{C}).$$
This is a form similar to a trinomial equation. A closed-form solution can be obtained using confluent Fox-Wright Function $\ _1\Psi_1$ therefore.
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