# Linalg - Proof involving distinct roots and derivatives

• Feb 3rd 2009, 07:25 PM
zedjay72
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
• Feb 3rd 2009, 08:43 PM
GaloisTheory1
we are in the field of complex numbers so consider this:

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

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

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

This should illustrate how to approach your problem.
• Feb 3rd 2009, 09:14 PM
zedjay72
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?(Nod)
• Feb 3rd 2009, 09:16 PM
ThePerfectHacker
Quote:

Originally Posted by zedjay72
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?(Nod)

Notation: $f(x)|g(x)$ means $f(x)$ divides $g(x)$.