All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Easy Findings (Posted on 2013-06-26)
Find the lowest possible value of the function
f(x) = x2008 - 2x2007 + 3x2006 - ... + 2007x2 - 2008x + 2009
for any real numbers x.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 analytical solution Comment 3 of 3 |

let
s(n,x)=sum((-1)^k*(n-k)*x^k,k=0 to n-1)
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^2009-2010x-2008=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 2013-06-27 17:00:06

 Search: Search body:
Forums (0)
Random Problem
Site Statistics
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox: