Let be $n$ an integer. Let be $a_1, \ldots, a_n \in \mathbb{R}$ not all null and $b_1 < \ldots < b_n$ reals.
Let be $f : x \mapsto \sum\limits_{i=1}^{n} a_i \exp(b_i x)$.
I am trying to show that $f$ can be null over at most $n - 1$ points.
What I tried:
- Considering derivatives / series expansion.
- Looking at $f$ as an solution of a differential equation.
- Looked the $n = 2$ trivial case but failed to see how to do even $n = 3$ (tried induction)
- I tried to suppose that there would be more than or $n$ zeros and tried to find contradictions with the derivatives.
In a word: factor then differentiate. I will use induction on $n$ to prove that "for all $a_1,...,a_n \in \mathbb{R}$ and for $b_1,..., b_n \in \mathbb{R}$, $x \mapsto \sum \limits_{i=1}^n a_i \exp(b_i x)$ has at most $n-1$ roots or is null.
Base case: for $n=1$, if $f$ has one root then it is the zero function.
Induction: let us consider $n \in \mathbb{N}$ and assume the property to hold for all $n$-uples of reals. Consider $a_1,...,a_{n+1}$ and $b_1,...,b_{n+1}$. Assume that $f : x \mapsto \sum \limits_{i=1}^{n+1} a_i \exp(b_i x)$ has $n+1$ distinct zeroes. We'll show that $f$ is zero. First factor by $\exp(b_{n+1} x)$: the function $g : x \mapsto \sum \limits_{i=1}^n a_i \exp\big((b_i-b_{n+1}) x\big) + a_{n+1}$ has $n+1$ zeroes. Differentiating and using Rolle's theorem, we get that $g':x \mapsto \sum \limits_{i=1}^n a_i (b_i-b_{n+1}) \exp\big((b_i-b_{n+1}) x\big)$ has $n$ zeroes. The induction hypothesis yields $g' = 0$, so $g(x)=C$ is constant and then $f(x) = C \exp(b_{n+1}x)$. As $f$ has at least one zero, we conclude $f = 0$.