Olathe | So, it can't be isomorphic to the direct product. |

mankind_tweezer | But he's not dividing by a direct product... he's dividing by a cyclic group |

Kasadkad | Hm Right |

Mulder | how would I show some sort of > e, contradiction when trying to show that x_n diverges if lim |x_{n+1} - x_n| = 1/1000 |

JabberWalkie | Mulder: the sequence isn't cauchy... |

gzl | Mulder: if it converged it would be Cauchy, and it's clearly not. |

Mulder | without using cauchy convergence criteria using good old hard and fast d-e style analysis |

gzl | why? |

TRWBW | Mulder: if it converged to L, past some n the sum would never differ from L by more than 1/2000 |

Mulder | cause i am curious |

JabberWalkie | Mulder: you prove cauchy critera, then use that |

Mulder | JabberWalkie, the problem was thrown before we proved it |

TRWBW | Mulder: well make it easy, take the sum never differs by more than 1/100000 |

bkudria | TRWBW: oops, got distracted. is this correct: S=>A|B, A=>aX|Xa, B=>bX|Xb, X=>aXb|bXa|e ? the langauge is "strings where the number of a's does not equal the number of b's" |

Mulder | so how would one show |x_n - L| > 1/1000 then? |

Kasadkad | Triangle inequality |

TRWBW | bkudria: that's wrong bkudria: it's wrong in a bunch of ways |

bkudria | TRWBW: hmm? too specific or too general? |

Kasadkad | Write it as |(x[n+1] - L) + (L - x[n])| |

Mulder | i tried saying |x_n - L| >= | | x_{n+1} - L| - |x_{n+1} - L| | err |

TRWBW | bkudria: it doesn't cover all the strings |

Mulder | i tried saying |x_n - L| >= | | x_{n+1} - L| - |x_{n+1} - x_n| | even |

bkudria | TRWBW: an example string? |

Mulder | but athat sorta got nowhere |

TRWBW | bkudria: aaaaaaaaaaaaab |

gzl | Mulder: suppose x_n -> L. fix e = 1/5000, there's an N(e) s.t. for n > N, |x_n - L| < e. |x_n - x_{n+1}| <= |x_n - L| + |x_{n+1} - L| < 1/5000 + 1/5000 = 1/10000. oops. |

TRWBW | oh c'mon |

bkudria | TRWBW: aah, i see TRWBW: hmm, adding A and B to their respective production result would fix that, no? |

TRWBW | Mulder: you want to show the sum diverges if lim |x_n+1-x_n|=a>0, yes |

Mulder | gzl, you've shown it's less than 1/1000. not quite a contradiction TRWBW, i guess in the general case, that would be nice yes. |

TRWBW | bkudria: doubt it. that X gives you strings of the form ww' where w' is the reverse of w with a's and b's interchanged. bkudria: i think you need to step back |

bkudria | TRWBW: ok, how do you recommend doing it? |

TRWBW | Mulder: okay, proof by contradiction. assume sum x_n converges to L. then past some N the sum never differs from L by more than say a/3. take it from there |