The exponential function intersects an $n$-th degree polynomial at no more than $n+1$ points.

Question

How to show that:

The graph of the exponential function $f(x)=e^x$ intersects the graph of an $n$-degree polynomial $p(x)$ at no more than $n+1$ points.

I have in mind a proof using this observation that A function with positive $n$-th derivative has at most $n$ roots – an inequality version of the Fundamental theorem of Algebra. However I'm wondering if there is a more direct proof, or if it is obvious from some well-known theorem.

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Motivation:

I realized this should be true when thinking about "higher-order" of convexity properties of functions:

• Geometric characterization of functions with positive third derivative
• Second-order star convex set: A set $S$ that any conics from $s_1\in S$ to $s_2\notin S$ does not reenter $S$ twice.

Answer 1

Ideas:

1. Proof by induction. When $f$ is degree $0$, it is a constant and $c=e^x$ has at most one solution.

2. Note that if $f(x)-e^x=0$ has $m$ solutions, then between each pair of consecutive solutions, there is a point where the derivative is $0$. This means the derivative has at least $m-1$ zeros.

3. Apply the inductive hypothesis to the derivative $f'(x)-e^x$, since $f'(x)$ is of one lower degree.