paradox, accurate taylor series half iterate of eta not analytic at e
#2
Hey Sheldon, its not that paradoxical.
These series are called asymptotic powerseries.
One says a function on a domain D has an asymptotic development \( \sum_{n=0}^\infty a_n (z-z_0)^n \) at \( z_0\in\partial D \) (boundary of D) if it satisfies:

\( \frac{f(z)-\sum_{n=0}^N a_n (z-z_0)^n}{z^N} \to 0 \) for \( z\to z_0 \).
Or equivalently
\( f(z)=\sum_{n=0}^N a_n (z-z_0)^n + o(z^N) \)

In our case the half-iterate has this kind of asymptotic development.
So if we glue the left and right half-iterate together at e,
then all derivatives exists and are finite at e, but still the function is not analytic at e, i.e. the convergence radius is 0.

(Generally for a powerseries \( f(z)=z+a_{m+1}z^{m+1}+a_{m+2}z^{m+2}+\dots \) there is exactly one formal powerseries of the fractional iterate \( f^t \). This means all the 2m fractional iterates \( f^t \) have the same asymptotic powerseries and hence the same derivations at 0.)

However such divergent series are often used in physics, to still get good values (though not to unlimited precision). There is a certain point of truncation where the series returns the best and often quite precise value. I think its where the root-test \( \frac{1}{\left|a_n\right|^{1/n}} \) has its maximum, i.e. where one - informally speaking - has the biggest radius of convergence.
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RE: paradox, accurate taylor series half iterate of eta not analytic at e - by bo198214 - 06/02/2011, 05:48 PM

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