most recent 30 from http://mathoverflow.net 2013-05-18T13:33:34Z http://mathoverflow.net/feeds/question/115272 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115272/taylor-expansion-convergence-relation-to-power-spectrum Uri Cohen 2012-12-03T11:16:43Z 2012-12-03T14:34:23Z

<p>Is there some connection between the power-spectrum of a real function $f:\mathbb{R}\to\mathbb{R}$ (that is, its Fourier transform) and the convergence radius of its Taylor expansion around arbitraty $x_0$? Intuitively, I would expect a function with 'limited power at high frequencies' to have 'large convergence radius' around each point, but I could not find such result.</p> <p>Thanks!</p>

http://mathoverflow.net/questions/115272/taylor-expansion-convergence-relation-to-power-spectrum/115274#115274 Gian Maria Dall'Ara 2012-12-03T11:25:15Z 2012-12-03T11:25:15Z

<p>A basic phenomenon in the direction of your question is the following. If $\hat{f}(\xi)e^{C |\xi|}$ ($C&lt;+\infty$) is integrable, then $f$ has a holomorphic extension to the strip of width $C$ around the $x$-axis. As a consequence, the Taylor series of $f$ converges on an interval of radius $C$ around each point. One can generalize this to several variables.</p>

http://mathoverflow.net/questions/115272/taylor-expansion-convergence-relation-to-power-spectrum/115298#115298 Alexandre Eremenko 2012-12-03T14:34:23Z 2012-12-03T14:34:23Z

<p>Another connection is the Wiener-Paley theorem. If the Fourier transform has bounded support then the function is analytic in the whole plane (and has exponential type there). Fourier transform does not have to be integrable in this case, it may exist as a distribution of even a hyperfunction. This fact also has generalization to higher dimensions.</p>