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Math Help - Roots of the Derivative of a Polynomial

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    Roots of the Derivative of a Polynomial

    Just took a qualifying exam (and did pretty well on it I think), but got stumped by one of the questions.

    Show that if a polynomial has all real roots then it's derivative has only real roots as well.

    I couldn't construct a counterexample, so it seems true to me.

    If the roots are all distinct you can use Rolle's Theorem on each interval between the roots. But what would you do if you had multiple real roots? Why are you guaranteed real roots of the derivative in this case? Thanks in advance for any help.
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    Quote Originally Posted by joeyjoejoe View Post
    Just took a qualifying exam (and did pretty well on it I think), but got stumped by one of the questions.

    Show that if a polynomial has all real roots then it's derivative has only real roots as well.

    I couldn't construct a counterexample, so it seems true to me.

    If the roots are all distinct you can use Rolle's Theorem on each interval between the roots. But what would you do if you had multiple real roots? Why are you guaranteed real roots of the derivative in this case? Thanks in advance for any help.
    Descarte's rule of signs.
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    Or you could use the theorem that says: if P(x) is a complex polynomial then the roots of P'(x) are contained in the convex hull of the roots of P
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    Quote Originally Posted by joeyjoejoe View Post
    Just took a qualifying exam (and did pretty well on it I think), but got stumped by one of the questions.

    Show that if a polynomial has all real roots then it's derivative has only real roots as well.

    I couldn't construct a counterexample, so it seems true to me.

    If the roots are all distinct you can use Rolle's Theorem on each interval between the roots. But what would you do if you had multiple real roots? Why are you guaranteed real roots of the derivative in this case? Thanks in advance for any help.
    If the polynomial P(x) of order N has a root of multiplicity n at x=a then we may write:

    P(x)=(x-a)^n Q(x)

    where Q(x) is a polynomial for which Q(a) \ne 0. Then:

    P'(x)=n(x-a)^{n-1}Q(x)+(x-a)^n Q'(x)=(x-a)^{n-1}[nQ(x)+(x-a)Q'(x)]

    So as the expression in the square brackets is a polynomial which is non-zero at x=a we conclude that P'(x) is a polynomial with a root of multiplicity n-1 at x=a.



    So if a polynomial P(x) of degree N has all real roots a_1, a_2, ..., a_n of multiplicities k_1, k_2, ..., k_n then:

    \sum k_i=N

    and a_1, a_2, ..., a_k are roots of P'(x) of multiplicities k_1-1, k_2-1, ..., k_n-1 and in addition between each adjacent pair of roots of P(x) there is at least one additional root of P'(x) we have a total of at least :

    (n-1)+\sum (k_i-1) =N-1

    real roots counting multiplicities for P'(x), but this is the order of P'(x) so we have exactly N-1 real roots counting multiplicities for P'(x)

    Note: there are a few gaps in the above that need filling, but you get the picture?

    CB
    Last edited by CaptainBlack; January 12th 2010 at 10:44 PM. Reason: tidying up
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