Create Newton divided difference table using one looping

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!

Answer 1

[Jan]

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

[Le]

Thanks a lot but it doesn't work for me. Can you elaborate a little bit on how to move the loop into the formula?

Answer 2

[Long Wu]

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.