# Linalg - Proof involving distinct roots and derivatives

• Feb 3rd 2009, 06:25 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, 07:43 PM
GaloisTheory1
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.
• Feb 3rd 2009, 08:14 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, 08:16 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: \$\displaystyle f(x)|g(x)\$ means \$\displaystyle f(x)\$ divides \$\displaystyle g(x)\$.