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

NY Lost Funds



TRWBWnoway-: that's not the way i'd recommend doing it
noway-ok
TRWBWnoway-: it either assumes A is finite, or it assumes something that is way too powerful
noway-: and probably assumes something that you would actually need some form of this to prove
noway-ahh, I see.
yiCale, 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?
TRWBWnoway-: 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)=p
noway-: 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?
TRWBWnoway-: 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.
Guest57536how 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.
OlatheGuest57536: Substitution, I think.
noway-|Steve|, ahh I see
Guest57536err i'm not seeing it.. sinu isn't the derivatie of the larger term
OlatheGuest57536: 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?
Calebrick_: 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.
Olathebrick_: The derivative is a limit. Does the right limit match the left limit there ?
CaleYour 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?
CaleThe acceleration there isn't defined.
WILDSTYLEHi.
:)
Calehello
OlatheSorry, we're closed for lunch.
WILDSTYLEIt's well past dinner here now.

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