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Nowhere analytic superexponential convergence
(01/29/2011, 08:59 PM)tommy1729 Wrote: well according to Bo it is not real-analytic and it doesnt have a radius , and according to mike it does have a radius.

I was not talking about which is for each the identity function. Of course it is real-analytic, even entire.
I was talking about tommysexp construction and the basechange.
In each step it has a radius of convergence, which though diminishes to 0 for .

Quote:so opinions and/or terminology differs

They dont.

Quote: , 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.

Of course.

Quote: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.

"analytic" is defined for a point (in a neighborhood, so one can take derivatives).
If one says "analytic on an interval", it means "for each point on the interval".
If one says "analytic on a disk", it means "for each point of the disk".
However "analytic at a point" always implies being analytic on the open disk of convergence, i.e. being analytic at each point of the disk of convergence.

I hope this brings some light into your confusion Wink

Messages In This Thread
RE: Nowhere analytic superexponential convergence - by bo198214 - 01/29/2011, 11:53 PM

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