Results 1 to 2 of 2

Thread: the minimum value

  1. #1
    Nov 2006

    the minimum value

    Find the minimum value of the function

    $\displaystyle f(x)=|1001+1000x+999x^2+...+2x^{999}+x^{1000}|$.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Nov 2005
    New York City
    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 $\displaystyle n=6$).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Aug 21st 2011, 12:12 PM
  2. The Minimum Value of The Sum
    Posted in the Geometry Forum
    Replies: 2
    Last Post: May 2nd 2010, 09:50 AM
  3. Minimum value
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Apr 29th 2010, 08:33 AM
  4. Minimum
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Mar 3rd 2010, 03:04 PM
  5. minimum value
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Oct 4th 2009, 05:38 AM

Search Tags

/mathhelpforum @mathhelpforum