Mathematics

Associated with Math.SE; for both general discussion & math qu...
Jan 14, 2015 21:09
yeah that could just be it :D
Jan 14, 2015 21:09
I can live without my view :D it isn't that great sadly
Jan 14, 2015 21:07
@Huy but 1. whiteboards cost, 2. as a student, my room isn't that big that I like to put other big stuff into here :D
Jan 14, 2015 21:06
but they do that in a lot of cool movies
Jan 14, 2015 21:06
yeah I got the idea from the movie probably :D
Jan 14, 2015 21:05
no black or whiteboard whatsoever needed
Jan 14, 2015 21:05
that is how I do/did it
Jan 14, 2015 21:05
@Chris'ssis buy window pencils for less then 10 bucks
Jan 14, 2015 21:02
in analysis we defined measurable function as functions so that the preimages of intervalls are measureable which is equivalent to that preimages of borel sets are measureable.
in probabilty theory we defined measureable functions as functions so that preimages of all _measureable_ sets are measureable, which leads to some (for me bigger) differences, e.g. in analysis the chaining of two measureable functions doesn't need to be measureable, in probability theory though it does.
is it normal that there are two different definitions for something so basic?
Jan 14, 2015 19:58
@PedroTamaroff thanks
Jan 14, 2015 19:58
oh got it
Jan 14, 2015 19:57
but the 2^p irretates me. btw p >= 1
Jan 14, 2015 19:57
x^p would be convex
Jan 14, 2015 19:57
yes but dunno which function you want to have convex
Jan 14, 2015 19:56
@PedroTamaroff can you explain further?
Jan 14, 2015 19:55
why is (a+b)^p <= 2^p(a^p+b^p) ?
Jan 13, 2015 16:33
hey, if I have a function f with ||f||_Lp(X) = c < infty, L(X) < infty (Lp stands for Lp norm, L for lebesgue measurement). is it true, that when I integrate f over C_eps with L(C_eps) = eps: ||f||_Lp(C_eps) --> 0 for eps --> 0
Jan 6, 2015 14:02
I have a p >= 1, is there a n and a f in L_p(R^n) so that f isn't in L_q(R^n) forall s != p ?