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Let $a,b \in \mathbb R^n$. I am interested in the following equation:

$$ \sum_{i=1}^{n} a_i e^{b_i x} = 0 $$

where $x\in\mathbb R$, $a = (a_1,\dots,a_n)$ and $b = (b_1,\dots,b_n)$. Edit: $b_i\neq b_j$ for all $i\neq j$.

I am wondering if there is a way to count the number of possible solutions to this equation. I believe the maximum number of solutions is $n-1$. In other words:

$$ \# \{ x \in \mathbb R \mid \sum_{i=1}^{n} a_i e^{b_i x} = 0 \} \overset{?}{\leq} n-1 $$

For $n=2$, the situation is pretty simple. You have the following equation:

$$ a_1 e^{b_1x} + a_2 e^{b_2 x} = 0 $$

Assuming $a_2 \neq 0$, it can be rearranged into

$$ -\frac{a_1}{a_2} = e^{(b_2-b_1)x} $$

Assuming $\frac{a_1}{a_2}$ isn't positive, in which case there are no solutions, we can take logarithm.

$$ (b_2-b_1)x = \log\left(-\frac{a_1}{a_2}\right) $$

And assuming $b_1\neq b_2$, there is a unique solution:

$$ x = \frac{1}{b_2-b_1} \log\left(-\frac{a_1}{a_2}\right) $$

Evidently, depending on the values of $a$ and $b$, the number of solutions is either 0 or 1.

For $n\geq 3$, the above approach doesn't work, but some experimental evidence suggests the limit is 2 zeroes. I tried to convert the equation to a polynomial by making a substitution $y=e^x$, but the result is not a polynomial since the powers on $x$ are arbitrary real numbers. So, how can we prove the number of solutions is bounded above by $n-1$, if it's even true?

  • Are all the $b_i$'s of the same sign ? – Claude Leibovici Jul 16 '21 at 04:44
  • @ClaudeLeibovici No, they are arbitrary real numbers –  Jul 16 '21 at 04:58
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    $n=2$, $a=(1,-1)$, $b=(42,42)$ has infinitely many solutions. I suppose you would rather want the $b_i$'s all distinct – Hagen von Eitzen Jul 16 '21 at 05:10
  • Will it help to look for the number of points for which the trajectory $x\mapsto (e^{b_1x} \dots e^{b_nx})$ intersects the hyperplane orthogonal to $(a_1\dots a_n)$? – Kurt G. Jul 16 '21 at 05:34
  • Related: https://math.stackexchange.com/questions/3679536/maximum-number-of-solutions/3679591#3679591 – Marktmeister Jul 16 '21 at 06:27
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    Functions of this sort are called (generalized) Dirichlet's polynomials. Assume $b_k$ are arranged in assending order (ie. $b_1 < b_2 < \cdots < b_n$), then the number of zeros is at most the number of sign changes in $a_k$. See refs in this answer – achille hui Jul 16 '21 at 06:49

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