Create Newton divided difference table using one looping
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Hi, I was asked to compute the Newton divided difference table using at most one looping. My teacher didn't say if more loops within a single loop are allowable. So is it possible to do that? Actually I have no problem writing the code using nested loop:
for j=2:n
for k=j:n
D(k,j)=(D(k,j-1)-D(k-1,j-1))/(x(k)-x(k-j+1));
end
end
Any help is much appreciated, thanks!
回答（2 个）
What happens, if you move the loop counter into the formula?
for j=2:n
D(j:n,j)=(D(j:n,j-1)-D(j-1:n-1,j-1)) ./ (x(j:n)-x(1:n-j+1));
end
Long Wu
2014-4-15
0 个投票
The solution suggested above is on right direction,
but need a little bit modification :
D(j:m,j)=(D(j:m, j-1)-D(j-1:m-1, j-1))./((x(j:m)-x(1:m-j+1)))';
As the front part is column vector , later part is row vector.
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