# Fourier series of a truncated Gaussian

I came across the following discussion concerning the computation of the Fourier series of a truncated Gaussian:

Fourier transform of a truncated Gaussian function

Numerical simulations suggest that the decay of such a series depends both on the truncation $T$ and the scale $a$ (see link above for the symbols). For example, for a fixed $T$, one would expect the series to decay more rapidly as $a$ decreases (as is the case when $T=\infty$). This, however, does not seem to be the case. There seems to be some form of "boundary effect" arising from the truncation (which depends on the relation between $a$ and $T$).

Is there a rigorous way of justifying this phenomena?

• What convergence are you referring to? I would expect the behaviour to be similar to the Fourier transform of the Heaviside function (or its derivative the Dirac distribution). Nov 30, 2015 at 10:46
• @Justpassingby I meant "decay", not "convergence". Sorry about the typo. Nov 30, 2015 at 10:49
• Thanks for the clarification. My expectation remains similar to my previous comment: the (lack of fast) decay of the Fourier transform of a function is related to the presence of steep jumps in the original function; so the decay would resemble that of the Fourier transform of the Heaviside function (not better than 1/t if I remember correctly), albeit multiplied by a constant that decreases linearly with the height of the jump, which height decreases itself exponentially with T or 1/a. Nov 30, 2015 at 13:09

The truncated Gaussian is the product of the ordinary Gaussian with the indicator function of the interval $[-T,T].$
The Fourier transform of the indicator function of an interval has decay worse than $1/t$ near infinity, i.e., its product with the linear function $t$ does not converge to 0 near infinity.