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Thread: the minimum value

  1. #1
    Nov 2006

    the minimum value

    Find the minimum value of the function

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  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 f_n(x) = x^n + 2x^{n-1}+...+nx+(n+1). So you want to know \min_{x\in \mathbb{R}} \{ |f_{1000}(x)| \}.

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

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

    The above illustrates that we can now only consider the case f'_n(x)=0. Say f_6(x) = x^6+2x^5+3x^4+4x^3+5x^2+6x+7 then f'_6(x) = 6x^5+10x^4+12x^3+12x^2+10x+6x. It should seem clear that -1 is a zero by symettry of the coefficients, thus 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 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 x=-1 is the minimum point so f_{2007}(-1) = 501 which is the minimum value.
    (So to complete this proof we need to show f'_n(x) has no other zero except for -1, that should be doable by using the lower bound estimate to get (x+1)^n\geq 0 which was specifically demonstrated for n=6).
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