# Number of real roots of an exponential polynomial

Let $a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_n$ be real numbers, and assume that $\{a_i\} \neq \{b_i\}$. Can the equation $$e^{a_1 x} + e^{a_2 x} + \dots + e^{a_n x} = e^{b_1 x} + e^{b_2 x} + \dots + e^{b_n x}$$ have more than $n$ real roots including $0$ and counting multiplicities?

There are special cases that might be familiar to some. It cannot have $(n+1)$ roots at $0$ because that would imply $\sum a_i^k = \sum b_i^k$ for all $0 \le k \le n$. Likewise, the equation cannot have all of $0, r, 2r, \dots, nr$ as roots. In fact, if $0, r, \dots, (n-1)r$ are roots, then it can be proved that there are no additional roots.

• Do you count the multiplicity of $0$?
– joro
Jan 3, 2016 at 11:52
• Yes. Actually I think one can give small perturbations to each of the $a_i$s and make all multiple roots into distinct simple roots. Jan 3, 2016 at 12:27
• I believe that math.stackexchange.com/questions/688606/… is an answer to a more general question. Probably also mathoverflow.net/questions/44443/… is relevant here. Jan 3, 2016 at 12:31
• @Peter it gives a twice worse upper bound in more general situation. The difference is seen already for $n=2$ (when there are indeed at most 2 roots, but direct application of Descartes rule does not prove this.) Jan 3, 2016 at 12:43
• @FedorPetrov You are right, my comment gives a weaker bound. Jan 3, 2016 at 12:50

Seems that it can not have more than $n$ roots. Let us use the following generalization of Descartes rule for signed measures. Namely, let $\mu$ be a Borel signed measure on a real line, with compact support (this condition may be of course weakened). We say that $\mu$ has at most $k$ changes of sign if there exist points $c_1<c_2<\dots <c_k$ such that restriction of $\mu$ to each of the intervals $(-\infty,c_1],(c_1,c_2],(c_2,c_3],\dots,(c_k,\infty)$ is either non-negative or non-positive. I claim that in this situation Laplace transform $L(t):=\int e^{tx} d\mu(x)$ has at most $k$ real roots with multiplicity counted (except the case $\mu\equiv 0$). Induction in $k$. Base $k=0$ is clear. Assume that $k\geq 1$, for $k-1$ it is established, we should prove for a measure $\mu$ having exactly $k$ sign changes. If $L(t)$ has $N$ roots, $g(t):=e^{c_1t}(L(t)e^{-c_1t})'$ has $N-1$ roots, but $g(t)=\int (x-c_1)e^{tx} d\mu(x)$, and new signed measure $d\nu(x)=(x-c_1)d\mu(x)$ has at most $k-1$ sign changes (if new measure $\nu$ is 0, then $\mu$ was proportional to the delta-measure in $c_1$, and $L(t)$ does not have roots at all.) So, by induction hypothesis $N-1\leq k-1$, as we need.
Now we say $e^{at}-e^{bt}={\rm sign}\,(a-b)\cdot t\cdot \int_{[a,b]} e^{tx} dx$, so if we divide our equation $\sum_{i=1}^n e^{a_it}-e^{b_it}=0$ by $t$, it becomes a Laplace transform for a measure with density $\sum \pm \chi_{[a_i,b_i]}$, which has at most $n-1$ sign changes. Thus our equation has at most $n-1$ non-zero roots as desired.
(PREVIOUS VERSION, working for simple roots) It is equivalent to counting positive roots of $\sum x^{a_i}=\sum x^{b_i}$. There is always root 1, so subtract $n$ from both parts and divide by $x-1$. I hope that after this the difference has at most $n-1$ sign changes and so Descartes rule is applicable (if we assume that exponents are positive integers, as we may assume: having more than $n$ roots is open condition, so exponents may be considered rational, then by scaling change of variables integer.) Indeed, sign change appears if there were more $a$'s than $b$'s on some initial segment, but it became more $b$'s than $a$'s. It requires two points at least.
UPD: when I say that having more than $n$ roots is an open condition, I, strictly speaking, lie. It is so for simple roots, but for multiple roots it is not: say, small perturbations of $x^2$ may have no real roots. However I believe that the statement is true for multiple roots too, but it have to be fixed somehow.