How to calculate the weights for Discrete Laplacian Operator?

Question

I am following this paper enter link description here step by step and want to build an isotropic Laplacian kernel. As shown in the following figure, I can understand until using Taylor to expand the 2D discrete Laplacian operator, but in step (12), the author multiplied this Laplace equation by the weights of "w" and "(1-w)", which is exactly what I don't understand. I know that the distance between the diagonal and the center is the square root of 2, so why not 1/(h ^ 2) and 1/2 (h ^ 2) as the weights? What is the difference between these two? And why does the author take 4/5 as the value of w? I would be appreciated for any explanations or resources.

Answer 1

The author has derived two different discrete Laplacians: one based on the four adjacent points, and another based on the four "corners." It looks you have two choices and you may wonder which is better. However, the author is saying "why choose?" You could take 10% of the first choice plus 90% of the second. Or, more generally, $w$ of the first and $1-w$ of the second. Introducing $w$ gives you one degree of freedom to play with in your discrete Laplacian formula. If you were lucky, you might find a value of $w$ that eliminates the $O(h^2)$ term, producing a more accurate approximation. Unfortunately, that is not possible in this case, but what is possible is to eliminate the anisotropic part of the $O(h^2)$ term. This, at least, makes dominant part of the error term invariant under rotations.