## #math - Thu 15 Mar 2007 between 02:36 and 02:46

### NY Lost Funds

 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 HmRight 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 criteriausing 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 wrongbkudria: 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 seeTRWBW: 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 contradictionTRWBW, 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