Hi! I am taking a second look on fourier transforms. While I am specifically asking about the shape of the fourier transform, I'd appreciate if you guys could also proof-read the question below as well, as I've written down allot of assumptions that I've gained, which might be wrong.(adsbygoogle = window.adsbygoogle || []).push({});

OK.

As far as I am aware, the nastier/"less smooth" (ie sharp, large and discontinious derivative and so on) a function ##f(x)## is (for ex. ##f(x) = \delta (x)##, the more "spread out" its fourier transform is. That is, the sines and cosines summed by the fourier-inverse integral get weighed increasingly equally irregardless of their frequencies, as ##f(x)## becomes less periodic.

However if you got a very spread out and slowly changing ##f(x)## (for ex. ##f(x)=1##), the fourier transform will be narrow around 0, meaning only the low-frequency sines and cosines in the fourier-inverse integral will dominate.

Why is this so? This applies to periodic functions as well, so let me rephrase the question in case you don't get me: Why can a "smooth and slow" function be described adequately with less terms in a fourier series, than a nasty one? Is it perhaps because the further you go out in a fourier series, the bigger the derivatives will be and thus these violent sines and cosines can adequately describe a swiftly changing function?

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# Shape of fourier transform

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