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Math Help - Rolle's Theorem application

  1. #1
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    Rolle'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.
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  2. #2
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    Quote Originally Posted by szpengchao View Post
    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.
    let

    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.

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



    for i =1 to n-1

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

    such that 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
    Last edited by TheEmptySet; May 16th 2008 at 06:31 PM. Reason: Not quite right
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    I had an epiphany while playing some video games.

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

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

    now we can take the derivative of p

    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

    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.
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    You can find this here.
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