Wikipedia:Reference desk/Archives/Mathematics/2012 May 28

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May 28

Convergent Power Series

Consider a smooth function $f:I_{x}\to \mathbb {R}$ where $I\subseteq \mathbb {R}$ is an open interval. Define a new functions $\rho [f]$, such that $\rho [f](y)$ is the radius of convergence of power series

$\sum _{k\geq 0}{\frac {f^{(k)}(y)}{k!}}\,(x-y)^{k}\,.$

What is known, and what has been written about, these functions? Please supply as many links and references as possible. — Fly by Night (talk) 17:28, 28 May 2012 (UTC)

The radius of convergence is the distance between y and the nearest singularity of the analytic continuation of f in the complex plane. Bo Jacoby (talk) 05:57, 29 May 2012 (UTC).

In particular, it is a 1-Lipschitz function, since it is the distance function to the boundary of some open set of $\mathbb {C} .$ --pma 20:29, 29 May 2012 (UTC)

This is only true if you assume that f is real analytic. For smooth f, it's more complicated. For instance with f the standard $e^{-x^{-2}}$ example, $\rho (0)=\infty$ even though f has no analytic continuation in any neighborhood of 0. Sławomir Biały (talk) 20:38, 29 May 2012 (UTC)

Though if the term radius of convergence is restricted to meaning the radius in which the series converges at all, with no regard to what it converges, treating the series as defining a complex-valued differentiable function is adequate. And given that if the function isn't analytic it can't be locally described by a power series, it doesn't seem very meaningful to use any other definition of the term.--Leon (talk) 21:41, 1 June 2012 (UTC)

Please supply as many links and references as possible. — Fly by Night (talk) 00:42, 3 June 2012 (UTC)