I'm looking for references, known names of, and other useful pointers and insight about (pairs of) differential operators that are "beach-ball like" because they sample a 2-dimensional function in these infinitesimally-sized regular patterns with the indicated alternating polarities:

enter image description here
Figure 1. Pairs of differential operators and a beach ball.

These include a hexagonal and a decagonal pattern. The pairs of operators can be formed by:

$$\begin{gather}\lim_{h\to 0}\frac{\sum_{N=0}^{4N + 1} (-1)^n f\bigg(x + h\cos\left(\frac{2\pi n}{4N + 2}\right), y + h\sin\left(\frac{2\pi n}{4N + 2}\right)\bigg)}{h^{2N + 1}},\\ \lim_{h\to 0}\frac{\sum_{N=0}^{4N + 1} (-1)^n f\bigg(x + h\sin\left(\frac{2\pi n}{4N + 2}\right), y + h\cos\left(\frac{2\pi n}{4N + 2}\right)\bigg)}{h^{2N + 1}},\end{gather}\tag{1}$$

although I'm not certain about the normalization factor $h^{-(2N+1)}$, which at least does not collapse the operator to zero or blow it up to infinity for $N=0$, which is simply a coefficient times differentiation:

$$\begin{gather}N=0:\\ 2\frac{d}{dx}f(x, y),\\ 2\frac{d}{dy}f(x, y),\end{gather}\tag{2}$$

or for $N=1$, which I think is:

$$\begin{gather}N=1:\\ \frac{1}{4}\left(\frac{d}{dx}\right)^3f(x,y)-\frac{3}{4}\frac{d}{dx}\left(\frac{d}{dy}\right)^2f(x, y),\\ \frac{1}{4}\left(\frac{d}{dy}\right)^3f(x,y)-\frac{3}{4}\frac{d}{dy}\left(\frac{d}{dx}\right)^2f(x, y).\end{gather}\tag{3}$$

Applying these to a 2-d Gaussian function and plotting:

enter image description hereenter image description hereenter image description hereenter image description here
Figure 2. Color-mapped 1:1 scale (pixel:unit) plots of, in order: A 2-d Gaussian function with standard deviation $\sigma = 16$, derivative of the Gaussian function with respect to horizontal coordinate $x$, differential operator $\frac{1}{4}\big(\frac{d}{dx}\big)^3-\frac{3}{4}\frac{d}{dx}\big(\frac{d}{dy}\big)^2$ applied to the Gaussian function. Color key: blue: minimum, white: zero, red: maximum.

Python source for Fig. 2:

import matplotlib.pyplot as plt
import numpy as np
import scipy.ndimage

sig = 16  # Standard deviation
N = 161   # Image width
x = np.zeros([N, N])
x[N//2, N//2] = 1
h = scipy.ndimage.gaussian_filter(x, sigma=[sig, sig], order=[0, 0], truncate=(N//2)/sig)
ddx = scipy.ndimage.gaussian_filter(x, sigma=[sig, sig], order=[0, 1], truncate=(N//2)/sig)
h1x = scipy.ndimage.gaussian_filter(x, sigma=[sig, sig], order=[0, 3], truncate=(N//2)/sig) - 3*scipy.ndimage.gaussian_filter(x, sigma=[sig, sig], order=[2, 1], truncate=(N//2)/sig)
plt.imsave('h.png', plt.cm.bwr(plt.Normalize(vmin=-h.max(), vmax=h.max())(h)))
plt.imsave('ddx.png', plt.cm.bwr(plt.Normalize(vmin=-ddx.max(), vmax=ddx.max())(ddx)))
plt.imsave('h1x.png', plt.cm.bwr(plt.Normalize(vmin=-h1x.max(), vmax=h1x.max())(h1x)))
plt.imsave('gaussiankey.png', plt.cm.bwr(np.repeat([(np.arange(N)/(N-1))], 16, 0)))

I found some literature about an application which includes additionally those similar differential operators that would have an even number of samples on each half of the circle in Fig. 1, and (my interpretation aided by @KBDave's answer) represents each pair of operators as the real and the imaginary part of the operator. With those, the rotation between the real and imaginary parts would be such that the imaginary part has samples on the circle (see Fig. 1) half-way between the samples of the real part.

enter image description here
Figure 3. Top: real part, bottom: imaginary part, of a function and complex differential operators applied to it. From Pietro Perona, "Deformable kernels for early vision", Technical Report MIT-LIDS-P-2039, October 1991, also published in April 1995, IEEE Transactions on Pattern Analysis and Machine Intelligence 17(5):222-227.

This gives a more complete set of differential operators to look at.

  • 2
    $\begingroup$ I bet these have something to do with the expansion of $\cos((2n+1)\theta)$ and $\sin((2n+1)\theta)$ as polynomials in $\cos\theta$ and $\sin\theta$. For example, $$\frac14\cos 3\theta=\frac14\cos^3\theta-\frac34\cos\theta\sin^2\theta.$$ Compare with $$\frac14\left(\frac d{dx}\right)^3-\frac34\left(\frac d{dx}\right)\left(\frac d{dy}\right)^2.$$ See also: Zernike polynomials. $\endgroup$
    – user856
    May 19, 2019 at 19:19
  • $\begingroup$ @Rahul That's useful! I came up with these as a possible differential analog to the orthogonal trigonometric basis $\cos((2n+1)\theta)$, $\sin((2n+1)\theta)$, with an edge detection application in mind. $\endgroup$ May 20, 2019 at 5:40

1 Answer 1


Working heuristically, suppose that $f(x,y)=\mathrm{e}^{ax+by}$. Then

$$\begin{split}\sum_{n=0}^{4N + 1} (-1)^n f\bigg(x + h\cos\left(\tfrac{2\pi n}{4N + 2}\right), y + h\sin\left(\tfrac{2\pi n}{4N + 2}\right)\bigg)&=\sum_{n=0}^{4N + 1} (-1)^n \mathrm{e}^{ah\cos\left(\tfrac{2\pi n}{4N + 2}\right)+bh\sin\left(\tfrac{2\pi n}{4N + 2}\right)}\\ &=\sum_{n=0}^{4N + 1} (-1)^n\mathrm{e}^{h\Re (c\zeta^{-n})} \end{split}$$ where $c=a+\mathrm{i}b$ and where $\zeta$ is a primitive $(4N+2)$-th root of unity. But $$\sum_{n=0}^{4N + 1} (-1)^n\mathrm{e}^{h\Re (c\zeta^{-n})}=\frac{h^{2N+1}\Re c^{2N+1}}{2^{2N-1}(2N)!} +o(h^{2N+1})\text{,}$$

a result that follows from the Cauchy residue theorem, series expansion in $h$, and

$$\sum_{n=0}^{2N}\frac{1}{z-\cos(\theta+\tfrac{2\pi n}{2N+1})}=\frac{T'_{2N+1}(z)}{T_{2N+1}(z)-\cos(2N+1)\theta}$$ where $T$ denotes a Chebyshev polynomial.

Therefore for sufficiently "nice" $f$ we have

$$\sum_{n=0}^{4N + 1} (-1)^n f\bigg(x + h\cos\left(\tfrac{2\pi n}{4N + 2}\right), y + h\sin\left(\tfrac{2\pi n}{4N + 2}\right)\bigg)=\frac{h^{2N+1}\Re \left((\partial_x+\mathrm{i}\partial_y)^{2N+1}\right)}{2^{2N-1}(2N)!}f +o(h^{2N+1})$$

and it is a matter of analysis to determine the function space in which this argument is precise.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .