Error at , can not run M = A\b;

% Given data
x = [2.0, 3.0, 6.5, 8.0, 12, 15];
f = [14, 20, 17, 16, 23, 125];
% Construct the cubic spline
h = diff(x);
A = zeros(length(x)-2);
for i = 1:length(x)-2
    A(i,i) = 2*h(i) + 2*h(i+1);
    A(i,i+1) = h(i+1);
    A(i+1,i) = h(i);
    A(i+1,i+1) = 2*h(i) + h(i+1);
end
b = [6*(f(2)-f(1))/h(1) + 6*(f(3)-f(2))/h(2);
    6*(f(4)-f(3))/h(3) + 6*(f(5)-f(4))/h(4)];
M = A\b;
% Evaluate the second derivative at data points
d2f_dx2 = zeros(length(x),1);
for i = 1:length(x)-2
    d2f_dx2(i) = M(i);
    d2f_dx2(i+1) = M(i) + h(i)*M(i+1);
    d2f_dx2(i+2) = M(i) + 2*h(i)*M(i+1) + h(i)*h(i)*M(i+2);
end
% Display second derivatives at data points
disp('Second Derivatives at Data Points:');
disp(d2f_dx2);

Answer 1

[Omar]

The MATLAB code you provided is intended to construct a cubic spline and calculate the second derivatives at data points. However, there are a few issues in the code which could be causing the error.

1. Matrix A Construction: The loop for constructing matrix `A` is not set up correctly. The current loop may not fill the matrix `A` properly, especially the last row and column.

2. Vector b Construction: The vector `b` is not constructed correctly. The current implementation only considers two values, but it should consider all the interior points.

3. Calculating Second Derivatives: The loop for calculating the second derivatives (`d2f_dx2`) is not correctly set up and might result in an index out-of-bounds error.

Here's the corrected MATLAB code:

% Given data
x = [2.0, 3.0, 6.5, 8.0, 12, 15];
f = [14, 20, 17, 16, 23, 125];

% Construct the cubic spline
h = diff(x);
n = length(x) - 2;
A = zeros(n, n);
b = zeros(n, 1);

% Constructing matrix A
for i = 1:n
    if i < n
        A(i, i+1) = h(i+1);
        A(i+1, i) = h(i+1);
    end
    A(i, i) = 2 * (h(i) + h(i+1));
end

% Constructing vector b
for i = 1:n
    b(i) = 6 * ((f(i+2) - f(i+1)) / h(i+1) - (f(i+1) - f(i)) / h(i));
end

% Solve for M
M = A\b;

% Evaluate the second derivative at data points
d2f_dx2 = zeros(length(x), 1);
d2f_dx2(2:end-1) = M;

% Display second derivatives at data points
disp('Second Derivatives at Data Points:');
disp(d2f_dx2);

**Please accept the answer if you like it and it runs well. It will increase my rank.

Answer 2

[Namnendra]

Hi Jameel,

The error you're encountering with `M = A\b;` suggests that there is an issue with the matrix `A` or the vector `b`. In your code, you are trying to solve a system of linear equations to find the coefficients for a cubic spline interpolation. Let's address the issues step by step.

Issues in the Code

1. Matrix Dimensions:

- The matrix `A` is not being constructed correctly. The loop over `i` attempts to access `A(i+1, i+1)` even when `i` is at its maximum value, which leads to an out-of-bounds error.

- The vector `b` is hardcoded for a specific size, which might not match the size of `A`.

2. Boundary Conditions:

- Cubic splines typically require boundary conditions to be specified. These conditions affect the first and last rows of matrix `A` and elements of vector `b`.

Corrected Code

Here's a revised version of your code that correctly constructs the cubic spline matrix and vector:

% Given data
x = [2.0, 3.0, 6.5, 8.0, 12, 15];
f = [14, 20, 17, 16, 23, 125];

% Number of intervals
n = length(x) - 1;

% Construct the cubic spline matrix
h = diff(x);
A = zeros(n-1, n-1);
b = zeros(n-1, 1);

for i = 1:n-1
    if i > 1
        A(i, i-1) = h(i);
    end
    A(i, i) = 2 * (h(i) + h(i+1));
    if i < n-1
        A(i, i+1) = h(i+1);
    end
    b(i) = 6 * ((f(i+2) - f(i+1)) / h(i+1) - (f(i+1) - f(i)) / h(i));
end

% Solve for the second derivatives
M = A\b;

% Add boundary conditions (natural spline: second derivatives at boundaries are zero)
M = [0; M; 0];

% Evaluate and display the second derivative at data points
disp('Second Derivatives at Data Points:');
disp(M);

Explanation

- Matrix `A` Construction: The loop now correctly fills the tridiagonal matrix `A` for the cubic spline. It ensures that off-diagonal elements are only set when valid indices are available.

- Boundary Conditions: This example assumes a natural spline, where the second derivatives at the endpoints are zero. Therefore, `M` is extended with zeros at both ends.

- Vector `b` Construction: The vector `b` is constructed based on the differences in the function values, ensuring it matches the size of `A`.

By applying these corrections, you should be able to solve the system of equations without errors and obtain the second derivatives for your cubic spline interpolation.

Thank you.