Neville interpolation using MATLAB

MATLAB Program:

% Neville's iterated interpolation algorithm

 % 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:  ');

 xx = zeros(1,n+1);

 q = zeros(n+1,n+1);

 for i = 0:n

   fprintf('Enter x(%d) and f(x(%d)) on separate lines:  \n', i, i);

   xx(i+1) = input(' ');

   q(i+1,1) = input(' ');

 end

 x = input('Now enter a point at which to evaluate the polynomial, x = ');

 

 d = zeros(1,n+1);

 d(1) = x-xx(1);

 for i = 1:n

    d(i+1) = x-xx(i+1);

    for j = 1:i

       q(i+1,j+1) = (d(i+1)*q(i,j)-d(i-j+1)*q(i+1,j))/(d(i+1)-d(i-j+1));

    end

 end

 fprintf('Table for interpolation evaluated at x = %11.8f: \n', x);

 for i = 0:n

    fprintf('%11.8f ', xx(i+1));

    for j = 0:i

       fprintf('%11.8f ', q(i+1,j+1));

    end

    fprintf('\n');

 end

OUTPUT:

>> Neville_interpolation

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

Table for interpolation evaluated at x =  1.50000000: 

 0.00000000  1.00000000 

 1.00000000  2.70000000  3.55000000 

 2.00000000  7.29000000  4.99500000  4.63375000 

 3.00000000 19.68300000  1.09350000  4.01962500  4.32668750 

>>