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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.


Motivation:

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

1 Answers1

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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.

Michael Burr
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  • Isn't that essentially the same argument as in this answer https://math.stackexchange.com/a/4617112/42969 to the referenced question? – Martin R Jan 14 '23 at 16:05
  • Thank you for a simple and nice solution! (I changed the notation in your answer to make it consistent with the notation in the question, since I used $f$ for the exponential function.) – Pavel Kocourek Jan 14 '23 at 16:06
  • @MartinR I agree, essentially the proofs are the same. I hope yet to see some proof that will surprise me. – Pavel Kocourek Jan 14 '23 at 16:09