# Taylor expansion convergence relation to power-spectrum

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.

Thanks!

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A basic phenomenon in the direction of your question is the following. If $\hat{f}(\xi)e^{C |\xi|}$ ($C<+\infty$) 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.
Dear Uri Cohen, I have never seen the kind of result you are looking for in a book or paper. It is a very well-known (and elementary) fact that the integrability of $\widehat{f}(\xi)e^{C|\xi|}$ implies the absolute convergence of the integral $F(z):=\frac{1}{2\pi}\int_\R \widehat{f}(\xi)e^{i\xi z}d\xi$ for $|y|<C$, which defines a holomorphic function on the strip that extends $f$. Then complex analysis of one variable tells that the Taylor series converges on intervals of length $2C$. –  Gian Maria Dall'Ara Dec 7 '12 at 13:09