Since the autocorrelation of is is a second derivative, the operator must be something like a first derivative. As a geophysicist, I found it natural to compare the operator to by applying them to a local topographic map. The result shown in Figure 5 is that enhances drainage patterns whereas enhances mountain ridges.
Our construction makes have the energy spectrum kx2+ky2, so the magnitude of its Fourier transform is .This rotationally invariant function in the Fourier domain contrasts sharply with the nonrotationally invariant function shape in (x,y)-space. The difference must arise from the phase spectrum. The factorization (10) is nonunique in that causality associated with the helix mapping can be defined along either x or y axes; thus the operator (10) can be rotated or reflected.
The operator has curious similarities and differences with the familiar gradient and divergence operators. In two-dimensional physical space, the gradient maps one field to two fields (north slope and east slope). The factorization of with the helix gives us the operator that maps one field to one field. Being a one-to-one transformation (unlike gradient and divergence) the operator is potentially invertible by deconvolution (recursive filtering).
I have chosen the name ``helix derivative'' or ``helical derivative'' for the operator .A telephone pole has a narrow shadow behind it. The helix integral (middle frame of Figure 6) and the helix derivative (left frame) show shadows with an angular bandwidth approaching .Thus, is much less directional than either or .
This is where the story all comes together. One-dimensional theory, such as the Kolmogoroff spectral factorization, produces not merely a causal wavelet with the required autocorrelation. It produces a wavelet that is stable in deconvolution. Using in one-dimensional polynomial division, we can solve many formerly difficult problems very rapidly. Consider the Laplace equation with sources (Poisson's equation). Polynomial division and its reverse (adjoint) gives us which means that we have solved in Figure 6 by using polynomial division on a helix. Using the seven coefficients shown, the cost is fourteen multiplications (because we need to run both ways) per mesh point.