Originally Posted by
Also sprach Zarathustra "...and get a contradiction evaluating the integral from a to a+1."
How I get this kind of contradiction(for the maximality of a as root of p(x))?
Because after the largest real root the polynomial will be either all positive or all negative, so its integral cannot be zero, of course.
Tonio
My try:
let deg(p(x))=n
so int(a-->a+1){p(x)dx} = q(a+1) - q(a) = 0 , when deg(q(x))=n+1
so, deg{q(x+1) - q(x)}=n
q(a+1) - q(a)= m(a+1)=0
contradiction? (m(x)!=p(x))
Thanks!