let
s(n,x)=sum((1)^k*(nk)*x^k,k=0 to n1)
then
s(n,x)=(n+(n+1)x(1)^n*x^(n+1))/(x+1)^2
we want to minimize
f(x)=s(2009,x)
f(x)=(2009+2010x+x^2010)/(x+1)^2
now f(x) is an even degree polynomial with a positive leading coefficient
thus as x>+infinity f(x)>+infinity
thus a global minimum must exist and exists at one of the point(s) where
f'(x)=0
f'(x)=[2(x+1)(2009+2010x+x^2010)(x+1)^2(2010+2010x^2009)]/(x+1)^4=0
2(x+1)(2009+2010x+x^2010)=(x+1)^2(2010+2010x^2009)
x=1 is a solution, dividing out the (x+1) we get
2(2009+2010x+x^2010)=(x+1)(2010+2010x^2009)
2008x^2010+2010x^20092010x2008=0
now consider this final polynomial,
using Descartes' rule of signs we can determine
that there are at most 1 positive real root and 1 negative real root
a quick inspection shows that x=1 and x=1 are roots
thus there can be no other real roots.
thus f'(x)=0 only when x=1 or x=1
f(1)=2019045
f(1)=1005
thus the global minimum of 1005 occurs at x=1

Posted by Daniel
on 20130627 17:00:06 