Let's say I have a function parameterized by a number of variables. As a simple example $$F(x,y) = ax^2 +by^2-cxy+1$$ I want to find some set of values (doesn't really matter what they are) for the parameters so that the relation $$ F(x,y)>0$$ holds for all points of its domain. So I would want values returned like (1,1,1). I don't need all possible values, just one example where the relation holds.
Is there a function in Mathematica that could do this? The real function I need to operate on is way more complicated and has a few more parameters, but can this be done simply?
I know of SolveAlways
, but it doesn't like it when I apply relations instead of equalities.
I appreciate any help!
Resolve[ForAll[{x, y}, a*x^2 + b*y^2 - c*x*y + 1 > 0], Reals]
orFindInstance[ Resolve[ForAll[{x, y}, a*x^2 + b*y^2 - c*x*y + 1 > 0], Reals], {a, b, c}, Reals, 3]
? $\endgroup$ – user64494 Nov 26 '20 at 14:30Resolve[ForAll[{x, y}, a*x^2 + b*y^2 > 0 && a < 0 && b < 0], Reals]
results inFalse
andFindInstance[ Resolve[ForAll[{x, y}, a*x^2 + b*y^2 > 0 && a < 0 && b < 0], Reals], {a, b}, Reals]
produces{}
. $\endgroup$ – user64494 Nov 26 '20 at 14:49SolveAlways
but notFindInstance
I think that argues that this is not "easily found in documentation". So it would be better to have an answer than to either leave unanswered or closed. PS Also mentionTimeConstrained
. $\endgroup$ – Daniel Lichtblau Nov 26 '20 at 15:58