By using Rolle's Theorem, prove that:

if the real polynomial p of degree n has all its roots real, then so does its derivative p'.

give an example of a cubic polynomial p for which the converse fails.

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- May 16th 2008, 06:01 PMszpengchaoRolle's Theorem application
By using Rolle's Theorem, prove that:

if the real polynomial p of degree n has all its roots real, then so does its derivative p'.

give an example of a cubic polynomial p for which the converse fails. - May 16th 2008, 06:22 PMTheEmptySet
let

$\displaystyle x_1 < x_2 < x_3 < ...x_n$

be the n roots of a polynomial

Note: The derivative has degree n-1 so has at most n-1 roots

Consider the following by Rolle's theorem.

$\displaystyle f(x_{i+1})-f(x_i)=f'(c)(x_{i+1}-x_{i}) \iff 0-0=f'(c)(x_{i+1}-x_{i})$

since $\displaystyle x_i \ne x_{i+1}, f'(c)=0$

for i =1 to n-1

Then for each i There exists a $\displaystyle c \in [x_{i},x_{i+1}]$

such that $\displaystyle f'(c)=0$

These are each of the n-1 zeros of the derivative.

I guess this proof only works the zero's are not repeated(I think)

QED

I hope this helps

Good luck - May 16th 2008, 09:59 PMTheEmptySet
I had an epiphany while playing some video games. (Clapping)(Wink)

if $\displaystyle x_i=x_{i+1}$ it is a repeated root then the polynomial can be factored and written as

$\displaystyle p(x)=(x-x_i)^2(f(x))$ where f(x) is a polynomial

now we can take the derivative of p

$\displaystyle p'(x)=2(x-x_i)(f(x))+(x-x_i)^2(f-(x))=(x-x_i)(2f(x)-(x-x_i)f'(x))$ so

$\displaystyle p'(x_i)=(x_i-x_i)(2f(x)-(x_i-x_i)f'(x_i))=0$

So repeated real roots yield real roots of the derivative.

This can be generalized if a root is of multiplicty n the derivative will have a real zero of multiplicty n-1.

I hope this helps. - May 17th 2008, 05:58 PMThePerfectHacker
You can find this here.