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

OlatheSo, it can't be isomorphic to the direct product.
mankind_tweezerBut he's not dividing by a direct product... he's dividing by a cyclic group
Mulderhow 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
JabberWalkieMulder: the sequence isn't cauchy...
gzlMulder: if it converged it would be Cauchy, and it's clearly not.
Mulderwithout using cauchy convergence criteria
using good old hard and fast d-e style analysis
TRWBWMulder: if it converged to L, past some n the sum would never differ from L by more than 1/2000
Muldercause i am curious
JabberWalkieMulder: you prove cauchy critera, then use that
MulderJabberWalkie, the problem was thrown before we proved it
TRWBWMulder: well make it easy, take the sum never differs by more than 1/100000
bkudriaTRWBW: 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"
Mulderso how would one show |x_n - L| > 1/1000 then?
KasadkadTriangle inequality
TRWBWbkudria: that's wrong
bkudria: it's wrong in a bunch of ways
bkudriaTRWBW: hmm? too specific or too general?
KasadkadWrite it as |(x[n+1] - L) + (L - x[n])|
Mulderi tried saying |x_n - L| >= | | x_{n+1} - L| - |x_{n+1} - L| |
TRWBWbkudria: it doesn't cover all the strings
Mulderi tried saying |x_n - L| >= | | x_{n+1} - L| - |x_{n+1} - x_n| | even
bkudriaTRWBW: an example string?
Mulderbut athat sorta got nowhere
TRWBWbkudria: aaaaaaaaaaaaab
gzlMulder: 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.
TRWBWoh c'mon
bkudriaTRWBW: aah, i see
TRWBW: hmm, adding A and B to their respective production result would fix that, no?
TRWBWMulder: you want to show the sum diverges if lim |x_n+1-x_n|=a>0, yes
Muldergzl, 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.
TRWBWbkudria: 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
bkudriaTRWBW: ok, how do you recommend doing it?
TRWBWMulder: 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

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