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.

  • $\begingroup$ Do you count the multiplicity of $0$? $\endgroup$
    – joro
    Jan 3, 2016 at 11:52
  • $\begingroup$ 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. $\endgroup$ Jan 3, 2016 at 12:27
  • $\begingroup$ 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. $\endgroup$ Jan 3, 2016 at 12:31
  • $\begingroup$ @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.) $\endgroup$ Jan 3, 2016 at 12:43
  • $\begingroup$ @FedorPetrov You are right, my comment gives a weaker bound. $\endgroup$ Jan 3, 2016 at 12:50

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .