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I have posted a related question before (link), but I didn't really get a completely satisfactory answer, and also I believe that I was able to simplify the problem a little. Therefore, I hope that it is appropriate to post again.

Basically, I want to show that

$f(y,z) := 81 y^4+\left(450 y^2+228 y^3+476 y^4\right) z+\left(405+108 y+1692 y^2+504 y^3+884 y^4\right) z^2+\left(1404-168 y+1828 y^2-144 y^3+528 y^4\right) z^3+\left(1872-624 y+672 y^2-576 y^3\right) z^4+\left(1232-192 y+320 y^2\right) z^5+320 z^6$

is positive for any $y > 0, z>0$.

By inspection of the $y$ polynomials, it is easy to see that only the polynomial before $z^4$, $g(y) := 1872-624y + 672y^2-576y^3$, might cause troubles for $y$ sufficiently large. However, even when ignoring the positive terms in $g(y)$, and taking $h(y) := -624y - 576y^3$ instead, numerically it still holds that $f(y,z)$ is strictly positive.

It hope that there is some easy way to show positivity of $f(y,z)$. Any idea would be greatly appreciated. Thanks!

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1 Answer 1

Maple manages to find all critical points (all digits given are correctly rounded, and it is guranteed that there are no additional solutions). These turn out to be:

  1. y = -1.200130607, z = -1.801192473
  2. y = -1.500000000, z = -1.500000000
  3. y = -1.394584620, z = -1.299681645
  4. y = -1., z = -1.
  5. y = -1.116359746, z = -.8642530441
  6. y = -1.500000000, z = -.7500000000
  7. y = -1.173750738, z = -.5083262416
  8. y = -.7095004638, z = -.2328818417
  9. y = 0., z = 0.

None of these are in the first quadrant. It's straight-forward to check that $f \ge 0$ on the boundary of the the first quadrant, and that $f\to\infty$ as $y^2+z^2 \to \infty$.

Hence, solving a min/max-problem on a large (closed) quarter-disc $K$ in the first quadrant, the extrema must be on the boundary of $K$ and here we know that $f \ge 0$.

The global minimum of $f$ on $K$ must then be $0$, and this is attained only at $(0,0)$, so your $f$ is indeed positive on the open first quadrant

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Thanks, I appreciate the effort! The problem is that it's numerically clear anyway that $f(y,z)$ must be positive. So finding its critical points numerically doesn't really solve the problem (although the rest of your argument is of course correct). –  Martin Feb 8 '13 at 10:36
    
@Martin The point is that the computation is just "semi-numeric". I didn't use the regular "fsolve" but a root-isolating algorithm that internally uses some variant of interval arithmetic or other branch-and-bound technique that proves correctness of the numerical approxmations. The computation above is a guarantee (with mathematical certainty) that there are no critical points in the first quadrant. –  mrf Feb 8 '13 at 11:24

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