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 December 16th, 2012, 03:26 AM #1 Newbie   Joined: Dec 2012 Posts: 23 Thanks: 0 proof: polynomial n roots, its derivative has n-1 roots Hello. Could you help me prove the following? If a polynomial P has n different real roots, then its derivative has at least n-1 roots. Thanks.
 December 16th, 2012, 03:39 AM #2 Senior Member   Joined: Aug 2012 From: New Delhi, India Posts: 157 Thanks: 0 Re: proof: polynomial n roots, its derivative has n-1 roots Do you know the Rolle's theorem? Apply it between each pair of consecutive roots for the polynomial P. There are (n-1) such pairs.
 December 16th, 2012, 03:42 AM #3 Newbie   Joined: Dec 2012 Posts: 23 Thanks: 0 Re: proof: polynomial n roots, its derivative has n-1 roots Thanks. I haven't though of that
 December 16th, 2012, 04:11 AM #4 Newbie   Joined: Dec 2012 Posts: 23 Thanks: 0 Re: proof: polynomial n roots, its derivative has n-1 roots So all I have to say is that if f(x) is a function that is differentiable in the interval [a,b], and f(a)=f(b)=0, there is a c, with a
 December 16th, 2012, 04:22 AM #5 Senior Member   Joined: Aug 2012 From: New Delhi, India Posts: 157 Thanks: 0 Re: proof: polynomial n roots, its derivative has n-1 roots You are correct in what you said, but to prove that there are at least n-1 roots, you must only consider consecutive roots of f(x).
 December 16th, 2012, 04:26 AM #6 Newbie   Joined: Dec 2012 Posts: 23 Thanks: 0 Re: proof: polynomial n roots, its derivative has n-1 roots How do I do that? Could you explain it a bit further?
 December 16th, 2012, 04:38 AM #7 Senior Member   Joined: Aug 2012 From: New Delhi, India Posts: 157 Thanks: 0 Re: proof: polynomial n roots, its derivative has n-1 roots Consider two real roots of your polynomial, $a$ and $b$ where $a and $P(x) \not= 0$ for any $x \in (a,b)$, that is, these are two consecutive roots. Since it is a polynomial, therefore it is continuous and differentiable for every x. You can apply Rolle's theorem between these two points and can surely state that there must be at least one $c \in (a,b)$ where $P'(c)=0$. Hence, between two consecutive roots of P(x) there lies a root of P'(x). You can pick up such (n-1) consecutive roots of P(x) showing that there are hence at least (n-1) roots of P'(x). I mean to emphasize the fact that if you were to pick two non consecutive roots of P(x), say the extreme roots $p$ and $q$, where $P(x) \not= 0$ for any $x or for any $x>q$, you will get only one root of P'(x) between them. Hence, many roots of P(x) will be 'lost' and you will not be able to prove your answer. Therefore, it is necessary to pick two consecutive roots of P(x).
 December 16th, 2012, 04:39 AM #8 Newbie   Joined: Dec 2012 Posts: 23 Thanks: 0 Re: proof: polynomial n roots, its derivative has n-1 roots Thanks a lot.
 December 16th, 2012, 04:52 AM #9 Senior Member   Joined: Aug 2012 From: New Delhi, India Posts: 157 Thanks: 0 Re: proof: polynomial n roots, its derivative has n-1 roots You are welcome! Regards, Rejjy 16-Dec-2012 19:22 IST

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# derivative of polynomial proof

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