| 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 |