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November 28th, 2017, 07:14 AM  #1 
Member Joined: Jan 2016 From: Blackpool Posts: 95 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, 03:27 PM  #2 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 169 Thanks: 51 Math Focus: Algebraic Number Theory, Arithmetic 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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