Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $0 < M \leq Z$?

Thanks in advance,

Nico

share|improve this question
1  
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

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
@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
    
Would you give more details on this? It seems very reasonable to me, but I'm not sure I get how to do it. –  Pietro Majer Nov 4 '10 at 7:32
    
Thanks Pietro, your answer put me in a good direction –  nicodds Nov 8 '10 at 0:19

Your Answer

 
discard

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.