Newton's Divided Difference for Numerical Interpolation using MATLAB(mfile)

MATLAB Program:

% 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)

OUTPUT:

>> Newtons_divided_difference

Enter n for (n+1) nodes, n:  3

Enter x(0) and f(x(0)) on separate lines:  

 0

 1

Enter x(1) and f(x(1)) on separate lines:  

 1

 2.7

Enter x(2) and f(x(2)) on separate lines:  

 2

 7.29

Enter x(3) and f(x(3)) on separate lines:  

 3

 19.683

Now enter a point at which to evaluate the polynomial, x = 1.5

Newtons iterated value:  4.32668750 

>>