Why can a smooth function be described with fewer terms in a Fourier series?

[Nikitin]

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.

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 regardless 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?

 

[Nikitin]

And can somebody tell me how to evaluate the infinite integral of the complex exponential function so I can get something representing the dirac delta? (I didn't want to open a new thread for this question alone as it's related to the OP, so don't murder me moderators).

I read that you can multiply it a converging term as a trick, like this, $\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} e^{-\epsilon x^2} \cdot e^{isx} dx$, but I don't remember how to evaluate a gaussian integral multiplied by another function.

 

[mathman]

And can somebody tell me how to evaluate the infinite integral of the complex exponential function so I can get something representing the dirac delta? (I didn't want to open a new thread for this question alone as it's related to the OP, so don't murder me moderators).

I read that you can multiply it a converging term as a trick, like this, $\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} e^{-\epsilon x^2} \cdot e^{isx} dx$, but I don't remember how to evaluate a gaussian integral multiplied by another function.

$\int_{-\infty}^{\infty} e^{-\epsilon x^2} \cdot cos(sx) dx=\sqrt{{\pi}/{\epsilon}}\cdot e^{{-s^2}/{4\epsilon}}$, from Gradshteyn and Ryzhik. Note that the imaginary part = 0.

 

[Incnis Mrsi]

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?

You confuse two things: number (amount) of non-zero terms, and their frequencies (i.e. this $n$ in your $\cos nx$ and $\sin nx$ or, in the standard presentation, $\exp(inx)$).

Also,

• Do not use such terms as “nasty function”. Not clear.
• Make distinction between Fourier transform of functions on ℝ and Fourier series on the circle (a.k.a. for periodic functions).
• Learn to think in exponents, not ugly real trigonometry.