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Homework Help: Polynomial function: infimum

  1. Nov 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Given the function "P" defined by: P(x) := x^2n + a2n-1*x^2n-1 + ... + a1x + a0;
    prove that there exists an x* in |R such that P(x*) = inf{P(x) : x belongs to | R}
    Also, prove that:
    |P(x*)| = inf{|P(x)| : x belongs to |R}


    3. The attempt at a solution

    As the function is the sum of continuous functions, it is contnuos too.
    Then, I thought about the theorem according to which if we have a cont. function on a sequentially compact space, it has inf. and sup. therein.
    But the space here is not sequentially compact.
    Can I use this theorem all the same, by adding some restrictions, perhaps?

    thanksss
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 18, 2012 #2
    I'd say seperate your polynomial into it's even and odd powers.
    IE P2 is all even powered x (including a0) and P1 is your odd, with some type of uniform bound on the coefficients. like |ak|<M
    is this an analysis class? I assume so. we know P2>P1 at large x, and extremely small -x ( like -10^10 or whatever). your function goes to infinity on both sides. Since it is Continuous, think of how you could apply rolles theorem, squeeze theorem, IVT, and the fact that INF(aN+bN)>=Inf(aN)+Inf(bN)
     
  4. Nov 24, 2012 #3
    thank you! i seem to have solved out the first question.
    but how about the second part:
    |P(x*)| = inf{|P(x)| : x belongs to |R}?
    if I think about a parabola graphic with its vertex in, say, (0, -3), the vertex of the absolute value of the function (0, +3), is no more the infimum.
    it would be possible if the infimum of this function were in the first or fourth quadrant, but i can't assume it, right?
     
  5. Nov 24, 2012 #4
    I think assuming that the inf|P(x)|=/=inf(P(x)) in general is correct. unless there is a strict restrictions of P(x). you'll have two cases of x* that p(x*)=0 implies x* is in inf{|p(x)|}
    or the inf is the same.
     
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