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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 n1 roots
Hello. Could you help me prove the following? If a polynomial P has n different real roots, then its derivative has at least n1 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 n1 roots
Do you know the Rolle's theorem? Apply it between each pair of consecutive roots for the polynomial P. There are (n1) 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 n1 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 n1 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<c<b, such that f?(c)=0? It means that between any two roots of f(x), there is a root of the derivative f?(x).

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 n1 roots
You are correct in what you said, but to prove that there are at least n1 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 n1 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 n1 roots
Consider two real roots of your polynomial, and where and for any , 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 where . Hence, between two consecutive roots of P(x) there lies a root of P'(x). You can pick up such (n1) consecutive roots of P(x) showing that there are hence at least (n1) 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 and , where for any or for any , 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 n1 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 n1 roots
You are welcome! Regards, Rejjy 16Dec2012 19:22 IST 

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derivative, polynomial, proof, roots 
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