Can finite difference method can be expressed with diff function?
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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.
回答（1 个）
Roger Stafford
2016-5-8
编辑：Roger Stafford
2016-5-8
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.
2 个评论
mathango
2016-5-9
Roger Stafford
2016-5-9
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.
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2016-5-9
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