# the minimum value

• Nov 16th 2007, 11:47 AM
perash
the minimum value
Find the minimum value of the function

$\displaystyle f(x)=|1001+1000x+999x^2+...+2x^{999}+x^{1000}|$.
• Nov 17th 2007, 07:32 PM
ThePerfectHacker
This problem only works for even degree exponents, it is hard to state the details in general so I will do it for special cases to show how to proceede. Let $\displaystyle f_n(x) = x^n + 2x^{n-1}+...+nx+(n+1)$. So you want to know $\displaystyle \min_{x\in \mathbb{R}} \{ |f_{1000}(x)| \}$.

Begin by considering the general problem for $\displaystyle f_n(x)$ and treating your case as a special case. First, start of easy, $\displaystyle f_2(x) = x^2+2x+3$ we want to minimize $\displaystyle g(x)=|x^2+2x+3|$. Calculus tells us that the minimum will occur at $\displaystyle g'(x)=0$ or when $\displaystyle g'(x)$ does not exist. Thus, we can look for when $\displaystyle f'_2(x) = 0$ or when $\displaystyle f_2(x)=0$ but $\displaystyle f_2(x)>0$. Now, since $\displaystyle f_2(x)>0$ we can look when $\displaystyle f'_2(x) = 0$ and that happens when $\displaystyle 2x+2=0\implies x=-1$.

By induction we can prove that $\displaystyle f_n(x)>0$ for even $\displaystyle n$ by re-introducting the previous case into the problem.

The above illustrates that we can now only consider the case $\displaystyle f'_n(x)=0$. Say $\displaystyle f_6(x) = x^6+2x^5+3x^4+4x^3+5x^2+6x+7$ then $\displaystyle f'_6(x) = 6x^5+10x^4+12x^3+12x^2+10x+6x$. It should seem clear that $\displaystyle -1$ is a zero by symettry of the coefficients, thus $\displaystyle x+1$ is a factor. By division we get (we can also reach these results by factoring the same coefficient terms but that is too long to type). By division we get $\displaystyle 6x^4+4x^3+8x^2+4x+6\geq x^4+4x^3+6x^2+4x+1 = (x+1)^4 \geq 0$. Thus, there are no other real zeros. Which means $\displaystyle x=-1$ is the minimum point so $\displaystyle f_{2007}(-1) = 501$ which is the minimum value.
(So to complete this proof we need to show $\displaystyle f'_n(x)$ has no other zero except for $\displaystyle -1$, that should be doable by using the lower bound estimate to get $\displaystyle (x+1)^n\geq 0$ which was specifically demonstrated for $\displaystyle n=6$).