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My problem is as follows:

How do I ask Mathematica the following question. Let $f(x,y,z,a,b)$ be a 5 variable polynomial. I want to find all values of $a,b$ for which $f$ has no zeros in the region $h_1(x,y,z)\leq g_1(a,b)$ and $h_2(x,y,z)\leq g_2(a,b)$ ewe an assume that all the functions are smooth.

I think it is appropriate so I will attach what my particular problem is.

$$\left({a_1} {b_2}+{a_2} {b_1}+\frac{{xy}}{2}-z\right)^2-\frac{1}{4} \left(-4 {a_1} {a_2}-4 e+x^2\right) \left(-4 {b_1} {b_2}-4 f+y^2\right)=0$$ While: $$a_1a_2\leq -e+(\frac{x}{2})^2\quad\text{and}\quad b_1b_2\leq -f+(\frac{y}{2})^2$$

I want to figure out for what $x,y,z,e,f$ the polynomial has no zero's given the above inequalities. I am working over reals, so every variable should be considered real. Ideally I would like some list of inequalities involving $x,y,z,e,f$ that tell me exactly when a solution does not exist. However, I doubt that is an easy problem. Hence, perhaps it is possible to find inequalities which are at least sufficient for non-existance of a zero.

Decided to add my clunky attempt that I am not sure even works:

Reduce[Reduce[(a2 b1 + a1 b2 + x* y/2 - z)^2 - (1/ 2 Sqrt[-4 a1 a2 - 4 e + x^2] Sqrt[-4 b1 b2 - 4 f + y^2])^2 == 0 && a1*a2 <= -e + (x/2)^2 && b1*b2 <= -f + (y/2)^2, {a1, a2, b1, b2}] == {}, {x, y, z, e, f}]

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1 Answer 1

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I am not sure if I have full understand the problem. Here just provide a thinking.

conditions = 
 ForAll[{a1, a2, b1, b2}, 
  a1*a2 <= -e + (x/2)^2 && 
   b1*b2 <= -f + (y/2)^2, (a2 b1 + a1 b2 + x*y/2 - 
       z)^2 - (1/
        2 Sqrt[-4 a1 a2 - 4 e + x^2] Sqrt[-4 b1 b2 - 4 f + y^2])^2 != 
   0]

Resolve[conditions, Reals]
Reduce[conditions, Reals]

$$\left(y<0\land e>\frac{x^2}{4}\land \left(\left(z<\frac{x y}{2}\land f>\frac{e y^2-x y z+z^2}{4 e-x^2}\right)\lor \left(z=\frac{x y}{2}\land f>\frac{y^2}{4}\right)\lor \left(z>\frac{x y}{2}\land f>\frac{e y^2-x y z+z^2}{4 e-x^2}\right)\right)\right)\lor \left(y=0\land e>\frac{x^2}{4}\land f>\frac{z^2}{4 e-x^2}\right)\lor \left(y>0\land e>\frac{x^2}{4}\land \left(\left(z<\frac{x y}{2}\land f>\frac{e y^2-x y z+z^2}{4 e-x^2}\right)\lor \left(z=\frac{x y}{2}\land f>\frac{y^2}{4}\right)\lor \left(z>\frac{x y}{2}\land f>\frac{e y^2-x y z+z^2}{4 e-x^2}\right)\right)\right)$$

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  • $\begingroup$ I think this is it! Thank you so much. Since this is Reduce this actually solved exactly correct? If my $x,y,z,e,f$ satisfy any of these conditions, then the polynomial will have a zero in the region defined in the conditions. This was very interesting. I did not know ForAll can be put into Reduce. $\endgroup$
    – 2132123
    Jan 18 at 23:36

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