Correct the code for me please

% Given data points
x = [0,0.1, 0.8, 0.6, 0.9, 1];
f = [-1,-1.2299,-3.455, -2.9949, -3.3929, -3];

% Part (a) - Construct the divided difference table
n = length(x);
d = zeros(n);
d(1,1) = f(1);
for i = 2:n
    d(i,1) = (f(i) - f(i-1))/(x(i) - x(i-1));
    for j = 2:i
        d(i,j) = (d(i,j-1) - d(i-1,j-1))/(x(i) - x(i-j));
    end
end

Array indices must be positive integers or logical values.

disp('Divided Difference Table:')
disp(d)

% Part (b) - Evaluate the polynomial of order "n-1"
y = 0;
for i = 1:n
    temp = f(1);
    for j = 1:i
        temp = temp * (0.5 - x(1:i-j)) + d(i,j);
    end
    y = y + temp;
end
fprintf('The value of the polynomial at x = 0.5 is %.4f\n', y)

% Part (c) - Evaluate f(0.5)
f_interp = interp1(x,f,0.5);
fprintf('The value of f(0.5) is %.4f\n', f_interp)

% Part (d) - Check your answers using built-in MATLAB functions
p = polyfit(x,f,5);
y_polyfit = polyval(p,0.5);
fprintf('The value of the polynomial at x = 0.5 using polyfit is %.4f\n', y_polyfit)

f_interp_polyfit = interp1(x,f,0.5,'linear');
fprintf('The value of f(0.5) using interp1 is %.4f\n', f_interp_polyfit)

Answer 1

[Image Analyst]

I don't know what you're trying to do but this line:

d(i,j) = (d(i,j-1) - d(i-1,j-1))/(x(i) - x(i-j));

when i and j are both 2, you get i-j = 0 and there is no zeroeth index of a vector. Indexes start at 1, not 0. So rethink what you really want to do.

See the FAQ: https://matlab.fandom.com/wiki/FAQ#%22Subscript_indices_must_either_be_real_positive_integers_or_logicals.%22

Answer 2

[Hassaan]

In this code snippet:

• The divided difference table d is correctly initialized to be n by n.
• The array2table function is used to format the divided difference table output.
• The polynomial evaluation using Newton's divided differences (Part b) is fixed to correctly accumulate the product of (xi - x(k)) for the ith term of the polynomial.
• The interp1 function is correctly called with 'linear' to perform linear interpolation (Part c).
• The polyfit and polyval functions are used to fit a polynomial and evaluate it at xi = 0.5 (Part d).

When you run this code, it will calculate and display the divided difference table, evaluate the polynomial at x = 0.5 using Newton's divided differences, interpolate the value at x = 0.5, and verify the results using MATLAB's built-in functions.

% Given data points
x = [0, 0.1, 0.8, 0.6, 0.9, 1];
f = [-1, -1.2299, -3.455, -2.9949, -3.3929, -3];

% Part (a) - Construct the divided difference table
n = length(x);
d = zeros(n, n); % Make sure the matrix has dimensions n x n
d(:,1) = f(:); % Fill in the first column with f values

% Calculate divided differences
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

% Display the divided difference table
disp('Divided Difference Table:')

Divided Difference Table:

disp(array2table(d, 'VariableNames', cellstr(num2str((1:n)')), ...
    'RowNames', cellstr(num2str(x'))))

              1          2          3         4         5         6   
           _______    _______    _______    ______    ______    ______

    0           -1          0          0         0         0         0
    0.1    -1.2299     -2.299          0         0         0         0
    0.8     -3.455    -3.1787    -1.0996         0         0         0
    0.6    -2.9949    -2.3005     1.7564    4.7601         0         0
    0.9    -3.3929    -1.3267     9.7383    9.9774     5.797         0
    1           -3      3.929     13.139    17.004    7.8075    2.0106

% Part (b) - Evaluate the polynomial at x = 0.5 using Newton's divided differences
xi = 0.5;
y = f(1);
P = 1;
for i = 2:n
    P = P * (xi - x(i-1));
    y = y + P * d(i,i);
end

fprintf('The value of the polynomial at x = 0.5 using Newtons divided differences is: %.4f\n', y);

The value of the polynomial at x = 0.5 using Newtons divided differences is: -2.6251

% Part (c) - Directly evaluate f(0.5) using linear interpolation
f_interp = interp1(x, f, xi, 'linear');
fprintf('The value of f(0.5) using linear interpolation is: %.4f\n', f_interp);

The value of f(0.5) using linear interpolation is: -2.6419

% Part (d) - Check the results using built-in MATLAB functions
% Fit a polynomial of order n-1
p = polyfit(x, f, n-1);

% Evaluate the polynomial at xi using polyval
y_polyfit = polyval(p, xi);
fprintf('The value of the polynomial at x = 0.5 using polyfit is: %.4f\n', y_polyfit);

The value of the polynomial at x = 0.5 using polyfit is: -2.6251

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