#math - Tue 27 Feb 2007 between 02:52 and 04:04

NY Lost Funds



OlatheYes.
eduardowell if so, i have a proof. I think
OlatheIf Wikipedia is right.
eduardoCauchy are you there? I think i figured it out
Cauchyyes
eduardoCauchy, I think it's easy. ||X - Xo|| <= r is a subset of R^d, right?
therefore the VC-D of this is less than the VC-D of R^d
Cauchynope
eduardowhat?
why not?
Cauchyyou're looking for the dimension of the class of those sets..
you're supposed to use lemma 15, i guess..
eduardoCauchy: but I think it still holds...
say Z is the set of all closed balls in R^d
VC-D < VC-D(R^d)
Cauchy: I think I am done
or u really dont think so
Phylohow can I solve Integral[(x^2 + x)/Sqrt[x+1], x]? it's using substitution somehow
I broke it into Integral[x/Sqrt[x+1], x] and Integral[(x+1)/Sqrt[x+1], x], but I don't know how to solve the first
oooh, that algebra is wrong isn't it...
I might be back, but until then, nvm, goodbye
joblotPhylo : x+1->y, so y-1/root(y), then \int root(y)-1/root(y)
Godfatherhi
anybody knows how to represent this function ? x=-y^2/2 + 4
AxsuulAnyone know circuit analysis fairly well?
joblotGodfather: wenn si vollen, y'' = - y^2 ... ist ln x verlantelen
Olathef(y) = -y^2/2 + 4 looks good.
Or,
-y^2/2 + 4 - x = 0, which allows you to find y.
GodfatherOlathe, -y^2/2 + 4 - x = 0
how to represent that function?
Olathe, if i put y = 0 i find x= 4
and x = 0 -> y=(8)^1/2
joblotGodfather : basic x = y^2, its a parabola, factor y^2 = 2(x-4), again a parabola
Godfatheri know, but i only find 2 points, (4,0) and (0,8^1/2)
joblot, any ideas?
jobloty=0, x=4; y=1, x=1/2 + 4
AxsuulIf anyone is familiar with circuits, I am confused as to what V' would be in this image: http://www.vega.org.uk/video/subseries/8
oops wrong link
http://img489.imageshack.us/img489/5586/untitledir3.jpg

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NY Lost Funds