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Thread: Linalg - Proof involving distinct roots and derivatives

  1. #1
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    Linalg - Proof involving distinct roots and derivatives

    Hey there,

    Suppose we have a function p(t) with complex coefficients. The function has degree m. I've got to prove that p(t) has m distinct roots if and only if p(t) and its derivative p'(t) have no roots in common.

    I can't really come up with any relationship between a function and its derivative involving their roots.

    Any help is much appreciated
    Zak
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  2. #2
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    we are in the field of complex numbers so consider this:

    $\displaystyle a$ is a multiple root iff $\displaystyle (x-a) | (p(x), p'(x)).$

    (->) $\displaystyle a$ is a multiple root, this implies $\displaystyle (x-a)^2 | p(x)$ which means $\displaystyle (x-a)|p(x)$. If $\displaystyle (x-a)^2|p(x)$ then $\displaystyle p(x)=(x-a)^2h(x)$ so $\displaystyle p'(x)-2(x-a)h(x)$. hence $\displaystyle (x-a) | (p, p')$

    (<-) If $\displaystyle (x-a)|(p, p')$ then $\displaystyle x-a | p(x), p'(x)$. Hence $\displaystyle p(x)=(x-a)g(x)$ and $\displaystyle p'(x)=(x-a)k(x)$. so $\displaystyle (x-a)k(x)=p'(x)=((x-a)g(x))'=g(x)+(x-a)g'(x)$. so $\displaystyle (x-a)|g(x)$. so $\displaystyle g(x)=(x-a)b(x).$ hence $\displaystyle p(x)=(x-a)g(x)=(x-a)(x-a)b(x)=(x-a)^2b(x)$. Hence $\displaystyle a$ is a multiple root of $\displaystyle p(x).$

    This should illustrate how to approach your problem.
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  3. #3
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    Hey,

    thanks alot for your help. I hate to ask this - but i have no idea the notation you're using... what does the " | " mean?
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  4. #4
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    Quote Originally Posted by zedjay72 View Post
    Hey,

    thanks alot for your help. I hate to ask this - but i have no idea the notation you're using... what does the " | " mean?
    Notation: $\displaystyle f(x)|g(x)$ means $\displaystyle f(x)$ divides $\displaystyle g(x)$.
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