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For non-zero $a_1, a_2, \ldots , a_n$ and for $\alpha_1, \alpha_2, \ldots , \alpha_n$ such that $\alpha_i \neq \alpha_j$ for $i \neq j$, show that the equation $$a_1e^{\alpha_1x} + a_2e^{\alpha_2x} + \cdots + a_ne^{\alpha_nx} = 0$$ has at most $n - 1$ real roots.

I thought of applying Rolle's theorem to the function $f(x) = a_1e^{\alpha_1x} + a_2e^{\alpha_2x} + \cdots + a_ne^{\alpha_nx}$ and reach a contradiction, but I can't find a way to use it.

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  • $\begingroup$ I thought of using Rolle's theorem to get a contradiction but I can't figure out how to apply it. $\endgroup$ – sourish Sep 6 '14 at 9:25
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    $\begingroup$ Mention your attempt more clearly in the question itself by editing, so that people don't repeat what you already thought of. $\endgroup$ – Hirak Sep 6 '14 at 9:27
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    $\begingroup$ To use Rolle's theorem and induction is a good idea. Hint: Compute the derivative of $\exp(-\alpha_nx)f(x)$. $\endgroup$ – Kelenner Sep 6 '14 at 9:43
  • $\begingroup$ Can anyone suggest a solution to this problem? $\endgroup$ – sourish Sep 6 '14 at 15:47
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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.)

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