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^d VC-D < VC-D(R^d) Cauchy: I think I am done or u really dont think so |

Phylo | how 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 |

joblot | Phylo : x+1->y, so y-1/root(y), then \int root(y)-1/root(y) |

Godfather | hi anybody 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 = 0 how to represent that function? Olathe, if i put y = 0 i find x= 4 and 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/8 oops wrong link http://img489.imageshack.us/img489/5586/untitledir3.jpg |