Tarantulafudge | |Steve|: so for -6y=x^2 the parabola would be vertical and open upward? |

thermoplyae | downward note the negative sign |

Tarantulafudge | thermoplyae: oh right, okay I see |

parker` | I thought matlab could do jacobians... I see a lot of tutorials on the net showing matlab doing jacobians with the function "jacobian()" but my version (the newest) doesn't seem to have this function |

Raikiri | http://freshmeat.net/projects/asy/ A vector graphics language for technical drawing and LaTeX. |

Tarantulafudge | thermoplyae: can the length of the latus be negative? for that same example I'm getting -6 |

thermoplyae | What in the world is a 'latus' And no, length is always positive |

Tarantulafudge | thermoplyae: latus rectum, sorry "The line segment through the focus of a parabola and perpendicular to the axis of symmetry" |

thermoplyae | mmm My guess is that you're forgetting an absolute value somewhere |

Tarantulafudge | thermoplyae: your right, I found it |

thermoplyae | What luck |

Tarantulafudge | thermoplyae: lol sorry thermoplyae: thank you for all your help |

thermoplyae | np x2 |

yi | Let F = Q(beta) be the splitting field for x^4 + 1 over Q. beta = 8th root of unity. Show that Aut(F/Q) has exactly four elements. any ideas? |

Cale | 8th root? |

powerfox | [Steve]: sorry to trouble you, but may you help me with last integral? |

yi | Cale: beta = sqrt(i) |

|Steve| | The boss integral, if you will? |

yi | Cale: or equivalently beta = (sqrt(2)/2) + (sqrt(2)/2)i |

powerfox | sqrt ( (1+ arctgx)/(1-arctgx) ) * dx/ ( (1+x^2)*(1+arctgx) ) |

yi | I showed that the minimal polynomial for beta over Q is x^4 + 1 |

Cale | oh, I suppose it would be, yes |

yi | since that factors as (x^2+i)(x^2-i) |

powerfox | It is the second in my list, but I mist it for better time like now :) |

yi | err, sorry I meant to say that F = Q(beta) is the splitting field for x^4+1 over Q |

Cale | ah, right, somehow I'd read x^4 + 1 as x^4 - 1 :) |

powerfox | *missed |

Cale | That's why I was confused |

yi | since it factors as (x^2+i)(x^2-i), so x = +/- sqrt(i) |

|Steve| | arctgx is arctan(x)? |

yi | but Q(sqrt(i), -sqrt(i)) = Q(sqrt(i)) |

Cale | right |

powerfox | arc tangent, inverse tangent, sin/cos |

yi | so Aut(F/Q) = Aut(F) |