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Nowhere analytic superexponential convergence
#8
well im sorry too Bo.

but it seems you missed the point.

lim n -> oo log^[n] ( exp^[n] (x) ) also doesnt have a somewhere non-zero convergence radius.

but clearly 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.
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RE: Nowhere analytic superexponential convergence - by tommy1729 - 01/29/2011, 03:26 PM

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