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

 Olathe Yes. eduardo well if so, i have a proof. I think Olathe If Wikipedia is right. eduardo Cauchy are you there? I think i figured it out Cauchy yes eduardo Cauchy, 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 Cauchy nope eduardo what?why not? Cauchy you're looking for the dimension of the class of those sets..you're supposed to use lemma 15, i guess.. eduardo Cauchy: but I think it still holds...say Z is the set of all closed balls in R^dVC-D < VC-D(R^d)Cauchy: I think I am doneor u really dont think so Phylo how can I solve Integral[(x^2 + x)/Sqrt[x+1], x]? it's using substitution somehowI 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 firstoooh, that algebra is wrong isn't it...I might be back, but until then, nvm, goodbye joblot Phylo : x+1->y, so y-1/root(y), then \int root(y)-1/root(y) Godfather hianybody knows how to represent this function ? x=-y^2/2 + 4 Axsuul Anyone know circuit analysis fairly well? joblot Godfather: wenn si vollen, y'' = - y^2 ... ist ln x verlantelen Olathe f(y) = -y^2/2 + 4 looks good.Or,-y^2/2 + 4 - x = 0, which allows you to find y. Godfather Olathe, -y^2/2 + 4 - x = 0how to represent that function?Olathe, if i put y = 0 i find x= 4and x = 0 -> y=(8)^1/2 joblot Godfather : basic x = y^2, its a parabola, factor y^2 = 2(x-4), again a parabola Godfather i know, but i only find 2 points, (4,0) and (0,8^1/2)joblot, any ideas? joblot y=0, x=4; y=1, x=1/2 + 4 Axsuul If anyone is familiar with circuits, I am confused as to what V' would be in this image: http://www.vega.org.uk/video/subseries/8oops wrong linkhttp://img489.imageshack.us/img489/5586/untitledir3.jpg