# Zeros of a combination of exponentials

Is there any known result about the necessary and sufficient conditions for the existence of zeros for a function $f(x)=\sum_{n=1}^{N} a_n e^{b_n x}$, where $a_n,b_n \in \mathbb{R}\, \forall n=1,2,\cdots,N$, $a_1,a_N >0$, $b_1 < b_2 < \cdots < b_N$ and $x \in \mathbb{R}$?

It is known (see "Problem and Theorems in Analysis II" by Polya and Szego) that using a generalization of Descartes' rule of signs it possible to say that, named with $Z$ the number of changes of sign in the sequence of the $a_n$ and with $Z_0$ the number of zeros of $f(x)$, $Z-Z_0 \geq 0$ is an even integer.

The number $Z-Z_0$ should be even since for $x \rightarrow -\infty$, the dominant therm of $f(x)$ is $a_1 e^{b_1 x}>0$ and for $x \rightarrow +\infty$ the dominant term is $a_N e^{b_N x}>0$.

This gives an upper limit for the number of zeros, but there is any way to say "$f(x)$ should have at least $M$ zeros", with $0 < M \leq Z$?

Nico

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Lower bounds on real zeros (of anything) are usually a lot harder to get than upper bounds. In the Descartes rule of sign, you get a lower bound of Z (mod 2) because the number of zeros has the same parity as Z. (BTW, are you sure about your Z should be even in your statement?). I can't think of any interesting lower bounds. There might be something, but I doubt anything spectacular. – Thierry Zell Nov 1 '10 at 14:48
Ciao Thierry, thanks for your interest. You're right, my exposition of the result in Polya-Szego is unclear, I'm changing the text of my question to clarify it. – nicodds Nov 1 '10 at 14:56

Note that we can assume wlog that $b_n\geq 0.$ In the case they are rationals, writing $b_n=p_n/q$, with $p_n\in\mathbb{N},\\$ $q\in\mathbb{N}_+,\\$ and $t:=e^{x/q},\\$ puts everything into the case of positive roots of a real polinomial, with not more, nor less generality. The book by Pólya and Szegő has a section on the location and number of positive roots of a polynomial; in any case, whatever you can say for it can clearly be translated for your exponential equation. Then, the case of real $b_n$ can certainly be treated by approximation.
@Pietro: Why is it necessary to consider irrational case separately? If there are two $b$'s that are linearly independent over $\mathbb Q$, then the equation is equivalent to a system of two similar equations, each with fewer terms than the original. Right? – Mark Sapir Nov 1 '10 at 18:25