Updates from today
- By late tonight, please send me suggestions for whatever of the problems to cover tomorrow. I'll look into it. So short in advance, I will certainly have to just make an executive decision.
- The end of the problem:
- We found a threshold value, which I denote as t, which was 288/23. It had the following properties: if (and only if!) k>t, then positive definite; if k<t, then indefinite. We know that the form is not negative semidefinite, so for k=t there are only two possibilities: indefinite, or positive semidefinite.
The following shows positive semidefiniteness at the threshold level k=t: - k is a parameter in the function, and Q depends continuously on k.
- For any k>t, it is so that Q(x,y,z)>0 for every non-null (x,y,z); that is positive definiteness (which we have proven!)
- Fix (x,y,z). Then Q(x,y,z)=q(k) depends on k only. For k>t, q(k)>0. Let k->t from above. Then q(k) remains nonnegative. (A limit of positives can be zero, but never negative.)
- Therefore, Q(x,y,z) is nonnegative. Since the argument holds true for every point (x,y,z), we have positive semidefiniteness when k=t.
- (Edited message: You can of course calculate the two remaining 2x2 principal minors.)
- We found a threshold value, which I denote as t, which was 288/23. It had the following properties: if (and only if!) k>t, then positive definite; if k<t, then indefinite. We know that the form is not negative semidefinite, so for k=t there are only two possibilities: indefinite, or positive semidefinite.
Published Feb. 6, 2017 12:25 PM
- Last modified Feb. 6, 2017 1:44 PM