Correct the code for me please
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% 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
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)
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Antworten (2)
Image Analyst
am 28 Dez. 2023
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.
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Hassaan
am 28 Dez. 2023
Bearbeitet: Hassaan
am 28 Dez. 2023
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:')
disp(array2table(d, 'VariableNames', cellstr(num2str((1:n)')), ...
'RowNames', cellstr(num2str(x'))))
% 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);
% 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);
% 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);
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If you find the solution helpful and it resolves your issue, it would be greatly appreciated if you could accept the answer. Also, leaving an upvote and a comment are also wonderful ways to provide feedback.
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