# Prove that $\sum a_i e^{\alpha_ix}=0$ has at most $n-1$ real solutions. [duplicate]

We have to prove that $$\sum_{i=1}^n a_i e^{\alpha_ix}=0,a_i\neq0,\alpha_i\neq\alpha_j\forall i\neq j$$ has at most $n-1$ real solutions.

I tried differentiating and other stuff but was unable to do anything... Any suggestions will be helpful.

• $a_i$ and $\alpha_i$ are all real numbers, I presume? – 5xum Sep 8 '14 at 12:22

Hints: Between any two roots of $f(x)$, there lies a root of $f'(x)$ by Rolle's theorem.

Now, we shall use induction and the above fact to do your question.

For $n=1$- no real root.

Let's consider $n=2$ in detail for the idea of the general case.

$f(x)=a_1e^{\alpha_1x}+a_2e^{\alpha_2x}$.

Let $g(x)=e^{-\alpha_1x}f(x)=a_1+a_2e^{(\alpha_1-\alpha_2)x}$. What's $g'(x)?$ How many real roots does $g'(x)$ have? (answer: $0$ real roots). So, $g(x)$ can have at most one real root- which means $f$ can have at most one real root.

Now, use induction for the general case after considering $g(x)=a_1e^{-\alpha_1x}f(x)$. (Note that the real roots of $f$ and $g$ coincide.)