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Polynomials question

  1. Mar 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the minimum possible value for [tex]a^{2}+b^{2}[/tex] where a and b are real such that the following equation has at least one real root.

    2. Relevant equations

    [tex]x^{4}+ax^{3}+bx^{2}+ax+1[/tex]

    3. The attempt at a solution

    I tried to find the roots of the equation and then find a relationship between them but that didn't workout very well for me.
     
  2. jcsd
  3. Mar 14, 2013 #2

    mfb

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    There are general equations for those roots, but I don't think they are useful here.

    - For b=2, there is a nice factorization. This might correspond to the solution, but I don't know how to show that.
    - For minimal a^2+b^2, the polynomial has exactly one real root of degree 2. As complex roots come in pairs (for real coefficients), the polynomial can be written as (x+c)^2(x^2+dx+e) for some real c,d,e. It might be possible to solve this directly.
     
  4. Mar 14, 2013 #3
    Here is another way to think about it:

    If ##x^4+ax^3+bx^2+ax+1## has at least one real root, then that means its graph must intersect the x-axis. Since this is a 4th degree equation with a positive leading coefficient the tails of the graph both go off towards positive infinity. Therefore, if the minimum point on the graph is negative, then the graph crosses the x-axis at some point, and therefore it has real roots.

    This would change your problem statement to: "Find the minimum possible value of ##a^2+b^2## such that the minimum value of ##x^4+ax^3+bx^2+ax+1## is less than zero"

    This will still be a mess, but perhaps it is a more manageable mess than the original? (I haven't tried to solve it, but if this is a multivariable calculus class perhaps you could turn this into a constrained optimization problem with variables a, b, and x???)
     
    Last edited: Mar 14, 2013
  5. Mar 14, 2013 #4
    I think this it:

    You can show that any real solution of ##x^4+ax^3+bx^2+ax+1=0## is also a solution of ##x^4-ax^3+bx^2-ax+1=0##. Then you can combine these two equations and get that this solution must be a solution of ##x^4+bx^2+1=0##. But this is great, because you can show from here that ##b^2 \geq 4##. Since you know that letting a=0 and b=2 gives you real solution, you can be confident that this also gives you the least solution.
     
  6. Mar 14, 2013 #5
    How can we show that any real solution of ##x^4+ax^3+bx^2+ax+1=0## is also a solution of ##x^4-ax^3+bx^2-ax+1=0##?
     
  7. Mar 14, 2013 #6

    epenguin

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    That is a 'reciprocal equation' - if x is a root, 1/x is also a root.

    It can be solved conveniently dividing it by x2 and expressing it in the variable z = (x + 1/x).

    You might see a way to answer the question on the way before getting to the solution of the equation.
     
    Last edited: Mar 14, 2013
  8. Mar 14, 2013 #7

    Bacle2

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    Not quite:

    ##x^4+4x^3+6x^2+4x+1=(x+1)^3## , but ##x^4-4x^3+6x^2-4x+1=(x-1)^3##

    (BTW, this gives you a not-so-great lower bound of ##4^2+6^2=52## for ##a^2+b^2## **)

    Still, you do know that if there is one real root, there are at least two, since if

    you have a real root, your polynomial splits as a product of a monic and a cubic

    and every cubic ( or odd-degree poly.) has a real root, since complex roots come in pairs.

    So you get ##x^4+ax^3+bx^2+ax+1=(x^2+cx+d)(x^2+ex+f)##

    where, WOLG the left poly ##(x^2+cx+d)## splits *

    You can then play around with the Galois groups to make sure the lefthand poly. splits.

    Note too, that if x<0 and b is positive, it makes it harder for your poly. to have a root, so you want it to be small if positve, or,

    even better, to have b negative. Similarly, you want |a|>|b| , since , for x negative, the a coefficient lower the value of

    f(x) and a raises it.

    * If the poly. doesn't split, then you must acquit.

    ** Note that ##x^4+x^3+x+1## , which is the substitution a=1 and b=0 , has -1 as a double-root.
     
    Last edited: Mar 14, 2013
  9. Mar 14, 2013 #8

    Dick

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    Oooh. That's useful. So, in the extreme case where there is exactly one real root, then x=1/x. Nice hint.
     
  10. Mar 14, 2013 #9

    Bacle2

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    You mean up to multiplicity, right? Otherwise , if there is one real root, there must be at

    least two real roots ( up to multiplicity), since then f(x) splits as the product of a monic and

    cubic .

    And by epenquin's answer , the second poly. is of the form

    ## x^2+ex+1## , and the first is ##(x±1)^2##

    And then it is relatively-clean to expand :

    ##(x^2+ex+1)(x±1)^2## and set it equal to f(x) (which I'll do later when I have more time).
     
    Last edited: Mar 14, 2013
  11. Mar 14, 2013 #10
    How is it even possible that there is only 1 real root? Since it is a quartic shouldn't be there 0, 2, or 4?
     
  12. Mar 14, 2013 #11

    Bacle2

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    Yes, you're right, (I mentioned that in my post); if there is one root, then f(x) is the product

    of a monic and a cubic , and the cubic must have at least one real root), so f(x) splits

    as the product of 2 monics and a quadratic.
     
  13. Mar 14, 2013 #12
    ok, but I thought we have to count multiplicities?
     
  14. Mar 14, 2013 #13

    Bacle2

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    But I think it makes a difference to check multiplicities, since there is a difference between a poly. with a double real root and a complex root (with its conjugate) , and having a poly. with four real roots.

    Sorry, my posts were all-over the place; this problem is too difficult for me to do it on-and-off while doing something else.
     
    Last edited: Mar 14, 2013
  15. Mar 14, 2013 #14

    Dick

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    You can count them if you want. But having at least one real root doesn't depend on the multiplicity. The idea I'm trying to follow is that as you decrease a^2+b^2 then at some point there is exactly one real root (and yes, it will have multiplicity at least 2 and possibly 4). That leads back to epenguins hint.
     
    Last edited: Mar 14, 2013
  16. Mar 14, 2013 #15

    Dick

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    That's ok. Now you've got me worried. I'm trying to say that at the minimal value of a^2+b^2 there is exactly one real root. Hence that root satisfies 1/x=x. But suppose at the minimal value there are TWO real roots. Like 1/2 and 2 both multiplicity 2. Then as you decrease a^2+b^2 then both roots go away at the same time. That seems pretty unlikely, but it's still a gap that needs closing.
     
  17. Mar 14, 2013 #16

    Bacle2

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    Sorry, have not been too focused here, but maybe we can use the fact here that

    1 is an upper bound for ##a^2+b^2## , since a=1 and b=0 leads to f(x) having -1

    as a root.
     
  18. Mar 14, 2013 #17

    Dick

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    Sure. There's a lot of upper bounds. And 0 is certainly a lower bound where there are no real roots. The trick is to use some property that happens exactly at the lower bound. Keep thinking about it. If I assume there is exactly one real root at the lower bound with multiplicity 2, then I think I've got it. If there are two, I don't. It's a challenging question.
     
    Last edited: Mar 14, 2013
  19. Mar 15, 2013 #18
    This question is from the IMO btw.
     
  20. Mar 15, 2013 #19

    mfb

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    Great. I was sure that the symmetry could be used in some way, but did not see the trick.

    This simplifies my equation in post 2...

    Edit: Hmm, that might be a problem.
     
    Last edited: Mar 15, 2013
  21. Mar 15, 2013 #20
    The equation is palindromic, so the standard technique is: $$
    x^4 + ax^3 + bx^2 + ax + 1 = 0

    \\ \Rightarrow x^2 + ax + b + a/x + 1/x^2 = 0

    \\ \Rightarrow X^2 + pX + q = 0 $$ where ## X = 1/x + x ##. The resultant quadratic is then analyzed.
     
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