(3.2) Newton's divided difference: MatLab code + download link.

Code's download link:
Alternatively, you can copy and paste from below:

% Newton's divided difference
% Find the approximate value of f(1.5) from
% (x,y)= (0,1), (1,e), (2,e^2) & (3,e^3).

n = input('Enter n for (n+1) nodes, n: ');
x = zeros(1,n+1);
y = zeros(n+1,n+1);

for i = 0:n
fprintf('Enter x(%d) and f(x(%d)) on separate lines: \n', i, i);
x(i+1) = input(' ');
y(i+1,1) = input(' ');
end
x0 = input('Now enter a point at which to evaluate the polynomial, x = ');

n = size(x,1);
if n == 1
n = size(x,2);
end

for i = 1:n
D(i,1) = y(i);
end

for i = 2:n
for j = 2:i
D(i,j)=(D(i,j-1)-D(i-1,j-1))/(x(i)-x(i-j+1));
end
end

fx0 = D(n,n);
for i = n-1:-1:1
fx0 = fx0*(x0-x(i)) + D(i,i);
end
fprintf('Newtons iterated value: %11.8f \n', fx0)

[thanotoast.]

Thanks for the code, it worked. I searched on the community as well for this

[leonardwalker]

would potentially know how to do the code where we can calculate the approximation error for each answer starting from the index one