Timeline for Shannon's communication paper and finite differences

6 events

when toggle format	what		by	license	comment
Jan 21, 2011 at 9:58	history	edited	Did	CC BY-SA 2.5	added 55 characters in body
Jan 21, 2011 at 7:57	history	edited	Did	CC BY-SA 2.5	added 13 characters in body
Jan 21, 2011 at 6:45	comment	added	Did		Sure, if $t\mapsto N$ is not (Lebesgue) measurable at some small values of $t$, this "pathology" could propagate to larger values of $t$ by the recursion equation. (Was that your original question?) To ensure the asymptotics $N(t)=(C+o(1))X_0^t$, one could assume that $t\mapsto N(t)$ is measurable and (say) bounded "at the beginning", i.e. on $[0,t_0]$. (So regular as measurable is needed but regular as monotone is not.)
Jan 21, 2011 at 4:29	vote	accept	Anthony Quas
Jan 21, 2011 at 4:28	comment	added	Anthony Quas		This is very nice. I still think that there are counterexamples of non-measurable $N(t)$: let $T$ be the countable subgroup of $\mathbb R$ generated by the $t_i$ and pick representative elements of each coset of $T$ in $\mathbb R$. You can then find a solution that looks like a different multiple of $X_0^t$ on each coset (maybe some positive and some negative).
Jan 20, 2011 at 22:06	history	answered	Did	CC BY-SA 2.5