Can finite difference method can be expressed with diff function?

Hi, Here is the finite difference example

i=2:n-1;

j=2:n-1;

B(i,j) = A(i+1,j) - 2*A(i,j) + A(i-1,j) + A(i,j+1) - 2*A(i,j) + A(i,j-1)

expanding it yields

B(i,j) = A(i+1,j) - *A(i,j)- *A(i,j) + A(i-1,j) + A(i,j+1)- *A(i,j)- *A(i,j) + A(i,j-1)

rearranging,

B(i,j) = {A(i+1,j) - *A(i,j)}- {*A(i,j) - A(i-1,j)} + {A(i,j+1)- *A(i,j)} - {*A(i,j) - A(i,j-1)}

then,

B = diff(A,?,?) - diff(A,?,?) + diff(A,?,?) - diff(A,?,?)

Can this arrangement be possible? if yes, then what are the values in the question marks.

Answer 1

[Roger Stafford]

Assuming A is n x n,

B = diff(A,2,1)+diff(A,2,2);

The array B would be of n-2 x n-2 size. The second argument of 2 in each 'diff' indicates a "second" difference.

[mathango]

The equation does not work. Each diff result has different matrix size ( 3 x 5 and 5 x 3 for n=5) therefore sum of each diff does not work.

[Roger Stafford]

I'm sorry. I have corrected it to be what I think you asked for. The second difference operation shrinks the size down by two in the direction in which it is performed. Your expression did second differencing in two directions.