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Wednesday, July 15, 2009

One nice, one weird, one silly

Say you want to find the min or max of f(x,y) subject to a constraint that g(x,y)=0. Has it ever occurred to you that the values of x and y required are the same as the values of x and y required to find the min or max of g(x,y) subject to f(x,y)=0? I bet it hasn't. The requirement is entirely symmetrical (the "curly 'd'" notation is much prettier, but I can't be bothered to figure out how to do it here):
f1(x,y)g2(x,y)=f2(x,y)g1(x,y).

I can't get my intuition behind this: in boolean algebra,

(p or q) is equivalent to (p and not q or q). Yes, I know it's true, do the truth tables if you doubt me. It just seems wrong. P.S.:

(p and not q or not p and q) is not equivalent--that's exclusive or.

Let's for a moment write logical conjunction as product and disjunction as sum. Boolean algebras have the distributive rule p(q+r)=(pq+pr). Familiar, right? But boolean algebras also have this one: p+qr=(p+q)(p+r). Aside from the trivial p=0, when else is this true in arithmetic? The challenge is 7th grade level, but I like the answer.

6 comments:

  1. Your first thing reminded me of the Kuhn-Tucker conditions we had to learn in grad school (for economics problems).

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  2. Andy Stedman11:14 PM

    When r+q-1 = -p ?

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  3. Andy Stedman11:17 PM

    Oh duh.

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  4. Andy,

    Thanx for my first real laugh of the day.

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  5. Bob,

    Interesting! Much more complicated, of course.

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