[Math] generating a random periodic function with bounded amplitude and bounded fourier coefficients

fourier seriesrandom

I would like to generate (i.e. repeatedly compute via a computer) a random periodic function $f(x)$ with period $T$ such that $|f(x)| \leq M$ and the kth Fourier coefficient $|A_k| \leq g(k)$ for a given $g(k)$.

I realize this is kind of open-ended (I'm not giving any information about probability distributions), but is there any way to go about doing this in a way that makes sense?

Here's an example using IPython to plot $$f(\theta)=\cos \theta+\tfrac{1}{2}\cos(3\theta+0.23)+\tfrac{1}{2}\cos(5\theta-0.4)+\tfrac{1}{2}\cos(7\theta+2.09)+\tfrac{1}{2}\cos(9\theta-3)$$ 

th = np.arange(0,1,0.001)
f = lambda th: np.cos(th*2*np.pi)
plt.plot(th,f(th)+0.5*f(3*th+0.23)
  +0.5*f(5*th-0.4)+0.5*f(7*th-2.09)+0.5*f(9*th-3))
plt.ylim([-3, 3])
plt.grid('on')

enter image description here

You'll note that $\sum |A_k| = 3$, which implies that $|f(\theta)|<3$, but here the maximum amplitude is around 2.5, because the harmonics never completely add constructively.

So suppose I just want to choose a function $f(\theta)$ such that $|f(\theta)| \leq M = 1.5$ with $|A_1| = 1$, $A_k \leq \frac{1}{2}$ for odd $3 \leq k \leq 21$, and $A_k = 0$ for all other k.

The bound for M is much smaller than the sum of the upper bound of the amplitudes (6.0), and I don't want to turn a random generation problem into a heuristic rejection algorithm.

Best Answer

Expecting $|f|\le 1.5$ is much too optimistic. It's right at the edge of the lower bound given by Parseval's identity, and since a trigonometric polynomial of the required form can't possibly look like a constant, its supremum will exceed its root-mean-square by some margin. It's more realistic to shoot for upper bound that's twice the root-mean-square average of $f$.

Taking a cue from Hardy-Littlewood series, I tried to shift $\cos kt$ by $k\ln k$. The results look good: harmonics fight one another all the time, pushing the function up and down. Here is the sum up to $k=11$

k11

and here is the sum up to $k=21$ (I used $1/2$ for all coefficients after the first)

k21

I don't offer a proof of anything, but shifting by $k\ln k$ is "a way that makes sense".

More precisely, I plotted $$\mathrm{Re} \sum A_k\exp( ikt+ik\ln k) = \sum A_k \cos(kt+k\ln k)$$

Related Question