On the APIC paper page 6 under equation 11 the paper states that the spline weights for quadratic and cubic splines simplify for their formulation of $D$ such that:
$$D = \frac{1}{4}\Delta x I$$
$$D = \frac{1}{3}\Delta x I$$
For quadratic and cubic respectively.
How??? I have tried doing the arithmetic on paper and I am not sure which subsititution they are using.
I may be doing something wrong but let $(x_i - x_c)$ be labeled $\Delta x$:
Let's assume our distances are small so that our spline kernel evaluates on the first part of the spline only.
$$w_{ij}\Delta x \Delta x^T = (\frac{3}{4} - \Delta x ^ 2)\Delta x \Delta x^T$$
So the first row would be $(\frac{3}{4} - dx^2 - dy^2 -dz^2)(dx^2, dxdy, dxdz)$
So if the claim is correct we expect $\frac{3}{4} - dx^2 - dy^2 -dz^2(dxdy) = 0$ let's try expanding that out:
$$\frac{3}{4}dxdy - dx^2dxdy - dy^2dxdy -dz^2dxdy$$
That in no way shape or form looks like ti will cancel out down to 0 in the general case. so how did the authors reach such a nice theorem?