arrenlex | TRWBW: Yeah, A=(1/(b-a))integral(f,a,b). But I can't figure out how to use that. |

me22 | Abs[min value of function] <= (integral over width) / width <= Abs[max value of function] |

TRWBW | arrenlex: if f(x)<=c then integral is <= (b-a)*c. if f(x0)<c then |f(x)-f(x0)|<(c-f(x0))/2 for x in some |x-x0|<delta. |

Kampen | if anyone is able, could they look up and answer my question about the notation for tychonoff spaces? |

TRWBW | arrenlex: integrate [a,x0-delta] [x0-delta,x0+delta] [x0+delta,b] and add them up arrenlex: you'll get the average is less than the average, proof by contradiction arrenlex: anotherway, use g(x)=f(x)-(average of f(x)). same problem, but it might be clearer to you see how you can show that if g(x)<=0 and for some x0 g(x)<0, then integral g(x)<0 |

arrenlex | I've considered g(x)=A-f(x). I worked out that integral(g)=0 because integral(A)=A(b-a)=((b-a)/(b-a))int(f)=int(f) and int(f)-int(f)=0. Does that relate to what you're suggesting? (A is the average value of f) Or is my logic wrong? |

TRWBW | arrenlex: it does, but that doesn't make it much easier, they key point is that if a continuos function drops below the average it must drop below it for some delta around where it drops. then you can break up the integral, like i said. read my posts. |

arrenlex | I'm trying to understand them, yeah. Thanks a lot for your help. |

TRWBW | arrenlex: np, gl arrenlex: maybe this will help you see it, maybe not. if you have a finite sum a_1+..+a_n, every a_i is less than equal to the average, can any of the a_i be less the the average? |

arrenlex | TRWBW: Thanks a lot for trying, but I think I'm getting somewhere with what's been said already; thanks lots, though. |

TRWBW | arrenlex: with the trick of subtracting the average, that becomes: if you have sum a_1+...+a_n, a_i<=0, and at least one of the a_i<0, can they sum to 0? |

seb-_ | TRWBW: no |

TRWBW | seb-_: <arrenlex> *screams* this seems SO OBVIOUS but I'm completely stuck! How do you prove that a continuous function that never exceeds its average value must be constant? |

seb-_ | TRWBW: heh arrenlex: i hope you got it now |

arrenlex | Yep, I got it. Thank you, all. |

seb-_ | arrenlex: what was the mental snag? |

arrenlex | Me not understanding that A-f(x)>0. >=0, sorry. That means that g(x)=A-f(x)>=0, but since integral(g)=0 as I said above, that means g must be 0 for the whole interval, because it's non-negative. Which means that A=f(x) for all x. |

cerealkiller219 | seb: you nearby |

thermoplyae | In case you guys had any doubts, set theory is hard |

arctanx | o \in{\mathbb{R}}? |

Kasasdkad | thermoplyae: I had no doubts Though I doubt it's as hard as classical plane geometry Which is pretty much the most difficult thing ever |

thermoplyae | Pretty much |

Kampen | what, specifically, is the field that deals with topological groups called? a professor said that isn't algebraic topology buti don't know what else it could be |

thermoplyae | I think plane geometry and number theory are the two big towers of mathematics that I will never really venture into I often see them in the distance, but... |

Kampen | is it more just algebra? |