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Proof that an even degree polynomial has a minimum

  1. Dec 10, 2015 #1

    B3NR4Y

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    1. The problem statement, all variables and given/known data
    Let $$p(x) = a_{2n} x^{2n} + ... + a_{1} x + a_{0} $$ be any polynomial of even degree.
    If $$ a_{2n} > 0 $$ then p has a minimum value on R.

    2. Relevant equations
    We say f has a minimum value "m" on D, provided there exists an $$x_m \in D$$ such that
    $$ f(x) \geq f(x_m) = m $$

    for all x in D.

    3. The attempt at a solution
    I know I should prove that p(x) goes to infinity on both sides, but I'm not sure how to start doing that.
    I can rewrite $$p(x) = x^{2n} (a_{2n} + ... + \frac{a_1}{x^{2n-1}} + \frac{a_0}{x^{2n}}) $$

    But I'm not sure how to prove that it goes to infinity. If I can do that I can use the definition of the minimum and the intermediate value theorem to prove that the minimum exists.
     
  2. jcsd
  3. Dec 10, 2015 #2
    Unless I'm missing something here--as in, if this isn't an assumption you can make--it's readily apparent that every term as you've written the polynomial will drop out except ##a_{2n}## as ##x \to \infty##.
     
  4. Dec 10, 2015 #3

    B3NR4Y

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    Unfortunately that's not an assumption I can make, I have to prove everything.
     
  5. Dec 10, 2015 #4

    SteamKing

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    Remember, the limit of a sum is the sum of the limits of each individual term.

    Can you show that a2nx2n → ∞ as x → ∞ ? Then work your way thru the terms of p(x) until you reach a0 .
     
  6. Dec 10, 2015 #5
    Here would be my issues with this:

    1. can the OP assume said property of limits? (I presume he can--but you never know)

    2. Are you sure this will work for negative infinity? After all, who's to say the odd-integer exponents don't drag down the even-integer exponent terms as ##x \to -\infty##? That'd be another thing to prove for the ##x < 0## case.
     
  7. Dec 10, 2015 #6

    Mark44

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    p'(x) is an odd-degree polynomial with a positive leading coefficient. It will have to have at least one x-intercept where p' changes sign (hence p changes from decreasing to increasing or vice versa). Do you have any theorems you can invoke about odd-degree polynomials?
     
  8. Dec 10, 2015 #7

    SteamKing

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    Beats me. Just trying to give the OP a nudge to do something. I usually write my name at the top of the paper when I get stumped.
     
  9. Dec 10, 2015 #8

    B3NR4Y

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    Okay I think I am able to prove it using what SteamKing said, but I see what axmls means by some of the lower power terms may overcome the upper power terms. That's why I preferred writing it as $$p(x) = x^{2n} (a_{2n} + ... + \frac{a_1}{x^{2n-1}} + \frac{a_0}{x^2n} ) $$ it seems to show that $$a_{2n}$$ is the prevalent term. The only thing that sucks is I can't go limit by limit.
     
  10. Dec 10, 2015 #9
    Perhaps you could show as well that ##\lim_{x \to \infty} \frac{1}{x^k} = 0## first for the case ##k=1##, then the rest of the terms follow easily from the squeeze theorem.

    The only drawback would be if you can't assume, for instance, that the limit of the sum is the sum of the limits, or if you can't assume the squeeze theorem.

    Those assumptions you can and can't make is what makes these difficult to work with.
     
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