Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is there any known theory for equations like $a_1 x^{y_1} + a_2 x^{y_2} +..a_nx^{y_n}=0$ where the $y_i$'s are arbitrary irrationals? Can you say anything about the disposition of roots?

share|cite|improve this question
Note that even an absurdly simple special case, $x^{\pi}=1$, has an infinity of solutions, $x=e^{2ni}$, $n=0,\pm1,\pm2,\dots$. – Gerry Myerson Nov 5 '10 at 20:56
@Gerry: ...which implies, by the way, that $\pi$ is not a rational number.:) – Mark Sapir Nov 5 '10 at 21:19
Does the Gauss-Lucas theorem hold for such polynomials? – J.C. Ottem Nov 5 '10 at 21:56
A side remark about the tag: if $y_i$'s are arbitrary irrationals, are the expressions still polynomials? – Willie Wong Nov 6 '10 at 0:10
Well, I always thought that tagging in MathOverflow (unlike StackOverflow) was a matter of "best approximation". – Ganesh Nov 8 '10 at 17:57

1 Answer 1

up vote 2 down vote accepted

Note that these are simply a special case (generalization? depending on terminology) of exponential polynomials. Or rather, to be more accurate, they can be treated as such. Let $t=\ln x$, and then your monomials become $a_i e^{ty_i}$. The change of variables does introduce some technical issues, obviously (in dimension $n$, you need to do it separately for each of the $2^n$ quadrants, and then possibly iterate on the coordinate zero hyperplanes).

Over the complexes, as was pointed out by Gerry, there is potentially infinitely many roots. Over the reals, the number of roots is still finite, and some of the elementary results like Descartes's rule of signs do translate fairly well to that setting. (The fact that for ordinary polynomials, the derivative is either 0 or has fewer roots is not so crucial in many applications).

In several variables, the theory of fewnomials developed by Khovanskii gives some (pessimistic) upper bounds on the number of real roots (it also gives some restrictions on the patterns of the complex roots).

None of this is overly difficult, but I don't see the point of going into more details without a more specific question.

share|cite|improve this answer
@Thierry: Perhaps I do not understand something in your answer, but why $x^{\sqrt{2}}$ an exponential polynomial? – Mark Sapir Nov 6 '10 at 1:35
@Mark: You're right, my original answer is a bit sloppy. I'll fix this. – Thierry Zell Nov 6 '10 at 1:47
@Thierry: By definition an exponential polynomial is $P(x_1,...,x_k, e^{y_1},...,e^{y_m}$, where $P$ is a polynomial. So $e^{t\sqrt{2}}$ is not an exponential polynomial? Right? – Mark Sapir Nov 6 '10 at 9:17
@Mark: Oh I see: it's a terminology issue. Many people use "exponential polynomial" for any polynomial that shows up in Khovanskii theory, i.e. of the form $\exp \langle a \mid y \rangle$, because in that context, there's no reason to distinguish between exponential of a variable and of a linear combination of variables. Sorry for the confusion... – Thierry Zell Nov 6 '10 at 15:21
Terminology continued: here is another example of a recent question using the more generalized meaning of exponential polynomial:… – Thierry Zell Nov 6 '10 at 15:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.