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zmudz
 one year ago
Consider the identity
\((px + (1p)y)^2 = Ax^2 + Bxy + Cy^2.\)
Find the minimum of \(\max(A,B,C)\) over \(0 \leq p \leq 1.\)
zmudz
 one year ago
Consider the identity \((px + (1p)y)^2 = Ax^2 + Bxy + Cy^2.\) Find the minimum of \(\max(A,B,C)\) over \(0 \leq p \leq 1.\)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I would expand the left side of that equation. (px + (1p)y)^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(px + (1p)y)^2 = (px + (1p) y) * (px + (1p) y) = p^2*x^2 + 2p(1p)*xy + (1p)^2 *y^2 = Ax^2 + Bxy + Cy^2 thus it follows A = p^2 , B = 2p(1p), C = (1p)^2 we want to find minimum of Max (A,B,C)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this means finding the minimum of Max { p^2, 2p(1p), (1p)^2} for p in [0,1] plug in some values and see if you can find a pattern

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Using excel I made a spreadsheet and it looks like the minimum of the max of A,B,C is 0.5

zmudz
 one year ago
Best ResponseYou've already chosen the best response.0@jayzdd your methodology makes sense to me, but for some reason, the answer isn't 0.5 and I can't know the answer unless I give up on the problem. I have unlimited tries, though. Is there any other way to go about the problem? thanks!

zmudz
 one year ago
Best ResponseYou've already chosen the best response.0@jayzdd nevermind, I figured it out. It was 4/9 when p=1/3, or 2/3
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