## #math - Sun 22 Apr 2007 between 23:25 and 23:37

### NY Lost Funds

 TRWBW noway-: that's not the way i'd recommend doing it noway- ok TRWBW noway-: it either assumes A is finite, or it assumes something that is way too powerfulnoway-: and probably assumes something that you would actually need some form of this to prove noway- ahh, I see. yi Cale, TRWBW, |Steve| thanks for the explanations, it's clearer now |Steve| No problem. I'm working on very similar stuff. noway- Should I use the def. of injection to show there cannot be one? TRWBW noway-: think of if you can, given a claimed bijection f:A->P(A), construct an element in p \in P(A) such that no a \in A has f(a)=pnoway-: this is actually a classic theorem, but a simple one. might be fun to see if you can come up with it, but not a big deal if you can't |Steve| I'm not sure what having a copy of Q in my field is doing for me though. TRWBW |Steve|: it lets you take derivatives|Steve|: remember when i mentioned "assuming you don't have the problem (r^k)'=0 because k=0 or some such"|Steve|: (r^k)'=k*r^(k-1)*r'. if the characteristic divides k, trouble ensues |Steve| Ah.But that wasn't assumed for the first part.Oh well. noway- TRWBW, how can that be a bijection? TRWBW noway-: i got no idea of a hint here beyond what i gave.noway-: maybe think of the paradox "let S be the set of all sets that do not contain themselves. does S contain S?" |Steve| Let's see, q is irreducible and deg q' < deg q so gcd(q,q) = 1. Guest57536 how does one get from Integral(sinu)(cos^2u - cos^4u)du to -cos^3u/3 - cos^5u/u ?? noway- TRWBW, hehe, it couldn't because then it wouldn't meet its own property |Steve| noway-: You have an easy injection the other direction: a |-> {a}. So if you have an injection P(A) -> A, then you have a bijection between the two. Olathe Guest57536: Substitution, I think. noway- |Steve|, ahh I see Guest57536 err i'm not seeing it.. sinu isn't the derivatie of the larger term Olathe Guest57536: What is it the derivative of ? brick_ I'm trying to figure out what a derivative is at a sharp point such as: http://www.webassign.net/knight/Ex5-07.gif T(3), is it 4 or 0? Cale brick_: the derivative doesn't exist there brick_ The derivative graph hops from 4 to 0 at a discontinuity and so either the point is included on the 4 side or the 0 side. Olathe brick_: The derivative is a limit. Does the right limit match the left limit there ? Cale Your function isn't differentiable at 3 brick_ Olathe: ah, no it doesn't.. So what if the problem is asking for the acceleration at that point? is it just a rule of thumb to always pick the left side? Cale The acceleration there isn't defined. WILDSTYLE Hi.:) Cale hello Olathe Sorry, we're closed for lunch. WILDSTYLE It's well past dinner here now.