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Consider the following polynomial in two variables :

$$ Q(k,x)=27x^6 - 144kx^4 + 80k^2x^3 + 240k^2x^2 - 192k^3x + (64k^4 - 128k^3) $$

Then for any integer $k \geq 5$, the polynomial $Q(k,.)$ (in one variable $x$) seems to be always positive (i.e, $Q(k,x) >0$ for any real number $x$). Prove or disprove.

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The usual way to show a polynomial is positive is to express it as a sum of squares. Have you tried that? –  Gerry Myerson Jan 28 '12 at 22:46
    
I tried and failed. –  Ewan Delanoy Apr 29 '12 at 8:54
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3 Answers 3

up vote 9 down vote accepted
+100
  1. $x\mapsto Q(5, x)$ is positive. There is numerous ways to prove that, more or less algorithmically.

  2. Assume that there exists a $k$ such that $x\mapsto Q(k,x)$ is not positive, and consider $L$ be the infimum of all real numbers $k \geqslant 5$ such that $x\mapsto Q(k, x)$ is not positive. I claim that $Q(L,x)$ has a double real root, this is equivalent to say that $L$ is a root of the discriminant of $Q$ w.r.t. the variable $x$.

  3. The discriminant of $Q$ with respect to $x$ is (computed with Maple) : $$ \operatorname{disc}_x Q = 92162779488452608 k^{16} (k-4)^4 $$

  4. The real roots of this discriminant are 0 and 4, thus $L$ cannot be finite, and hence the claim.

Proofs

  • Point 1 can be proved using Sturm sequences, and you can compute these sequences using Maple :

    sturm(subs(k=5, Q),x,-infinity, infinity);
          0
    

This means that $Q(5, x)$ has no real roots. Since the leading coefficient is positive, the polynomial $Q(5, x)$ is positive whenever $x\in \mathbb R$.

  • To prove point 2 consider $x_0$ a real numbers such that $Q(L, x)$ is minimal. We have $Q(L, x_0) = 0$, because if $Q(L, x_0) > 0$ then $x\mapsto Q(L+\epsilon, x)$ is positive for $\epsilon > 0$ small enough, which contradict the definition of $L$. If $Q(L, x_0) < 0$, then $Q(L-\epsilon, x_0)$ is stille negative for $\epsilon > 0$ small enough, which is again a contradiction. So $x_0$ is a root of $Q(L,x)$ and of $\partial_x Q(L,x)$, thus it is a double root. The discriminant thing is usual.

  • The point 3 is Maple :

    factor(discrim(Q, x));
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@Lierre : perfect ! after seeing your solution I'm surprised and ashamed I didn't think of it earlier. –  Ewan Delanoy Apr 29 '12 at 11:18
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You can also use quantifier elimination algorithms. This gives you precisely for which $k$ the polynomial $Q(k,x)$ is positive. For example Mathematica implements such an algorithm :

In[1] := Resolve[ForAll[{x}, Q > 0]]
Out[1] := k < 0 || k > 4

I insist that you can trust this result, it is an exact algorithm not a numerical heuristic.

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You can try to prove your assumption using induction :

$1.$ Show that :

$Q(5,x)=27\cdot x^6-720 \cdot x^4+2000 \cdot x^3+6000\cdot x^2-24000\cdot x+24000 > 0$ ,for all $x$

$2.$ Suppose that :

$Q(k,x)=27x^6 - 144kx^4 + 80k^2x^3 + 240k^2x^2 - 192k^3x + (64k^4 - 128k^3) >0$ ,for all $x$

$3.$ Try to prove that : $Q(k+1,x) >0 $ , for all $x$ using assumption from step $2$

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Does it work??? –  Gerry Myerson Jan 28 '12 at 22:44
    
@GerryMyerson,Probably...since $k$ is an integer ... –  pedja Jan 29 '12 at 5:50
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A main problem with the induction idea is that for "most" $x$ (i.e. large enough $x$), we do not have $Q(k+1,x)>Q(k,x)$ as excepted. It seems that you cannot solve the problem without studying the "relative position" of $k$ and $x$. –  Ewan Delanoy Jan 29 '12 at 11:39
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