Nowhere analytic superexponential convergence
#10
well according to Bo it is not real-analytic and it doesnt have a radius , and according to mike it does have a radius.

a radius is usually considered a radius of a circle.

for reals we usually talk about intervals.


so opinions and/or terminology differs , i hope at least everyone agrees on :

on the real line f(x) = lim n -> oo log^[n] ( exp^[n] (x) ) is real-analytic because it simply reduces to f(x) = x for real x , which is clearly real-analytic.

and for real-analytic i prefer to say : analytic on an interval , rather than a radius , for imho a nonzero-radius is for analytic taylor series - analytic on on a disk on the complex plane with nonzero-radius.
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RE: Nowhere analytic superexponential convergence - by tommy1729 - 01/29/2011, 08:59 PM

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