2

Question : Why do Fourier coefficients decay slower for functions that aren't smooth (for example, non differentiable?)

A Fourier coefficient is defined as an integral over the function (times a complex number $e^{\text{something}}$).

An integral is the area under the curve.

Whether the curve is differentiable or not, the area under the curve is similar. So why does non-differentiability cause a problem?

For example, see the two curves below. The Fourier coefficients for the first curve will decay slower, but why, the integral over them both is obviously going to be very similar, and Fourier coefficients are just such integrals?

enter image description here

Rohit Singh
  • 1,143
saei
  • 77
  • 1
    An integral is only "the area under a curve" when the function being integrated is nonnegative. In this case, the function being integrated is complex-valued and oscillatory, so the area intuition isn't going to be helpful. The decay of the Fourier coefficients will in fact be a measure of how serious is the cancellation in the integral due to the oscillation of the integrand. – Greg Martin Feb 19 '20 at 16:34
  • The Fourier transform of $f$ represents $f$ as a sum of pure waves of various frequencies. If $f$ is very smooth then a few low frequency waves is all you need. If $f$ is not very smooth then you can't represent $f$ with just low frequency waves. – littleO Feb 19 '20 at 16:36

2 Answers2

1

An integral is only "the area under a curve" when the function being integrated is nonnegative. In this case, the function being integrated is complex-valued and oscillatory (think of cosine or sine if you like), so the "area" intuition isn't going to be helpful. The decay of the Fourier coefficients will in fact be a measure of how serious is the cancellation in the integral due to the oscillation of the integrand.

Furthermore, the "decay" in question is talking about asymptotic behavior ("$n$" going to infinity), which is when the oscillations are happening on smaller and smaller scales. On a very small scale, the non-differentiable graph still looks non-differentiable; but the differentiable graph will eventually look like a straight line on small scales, leading to more cancellation when multiplied by an oscillating function.

Greg Martin
  • 78,820
1

I'll give a qualitative answer: It takes alot of short wavelength waves to create the "corners" and "spikes" in unsmooth functions. For instance, let's say you have a corner in your function somewhere that bends over a distance of about 1 millimeter. You can probably convince yourself pretty quickly that a bunch of waves of wavelength around one meter are never going to add together to give this 1 millimeter bend. So you need larger magnitudes of these short wave lengths in unsmooth functions than you do in more well-behaved functions. Therefore, the Fourier magnitudes must fall off more slowly than they do with more well-behaved functions.

bob.sacamento
  • 4,028