My Math Forum Descarte's rule of signs proof

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 November 28th, 2017, 08:14 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 100 Thanks: 2 Descarte's rule of signs proof Prove by induction on n that if p(x) is a polynomial of degree n then the function $e^{x}-p(x)$ has at least n+1 zeros. By zero I mean that f(x) for some x=0 This is my solution however I feel the proof is incomplete and missing some theorems. https://imgur.com/a/gY92N
 November 28th, 2017, 04:27 PM #2 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 286 Thanks: 88 Math Focus: Number Theory, Algebraic Geometry I'm not really following your argument. How have you used the assumption that it's true for polynomials of degree $k$ to prove it's true for polynomials of degree $k+1$? A possible approach: suppose the statement is true for all polynomials of degree k. Let $p(x)$ be any polynomial of degree $k+1$ and write $f(x) = e^x - p(x)$. If you differentiate this with respect to $x$ you get $f'(x) = e^x - p'(x)$. Now $p'(x)$ is a polynomial of degree $k$, so by the inductive hypothesis, $f'(x)$ has at most $k$ zeros. But the zeros of $f'(x)$ are precisely the stationary points of $f(x)$. Can you see how to use this to bound the number of zeros of $f(x)$?

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