How can I constrcut this matrix for a problem?

Answer 1

[Torsten]

If x = [x0,x1,...,xn] is an 1x(n+1) row vector, you could use

M = zeros(n+1);
for i = 1:n+1
    M(i,:) = psi(x(1:n+1)-x(i));
end
M = M - diag(diag(M)) + eye(n+1)

if psi is a function that accepts array input.

[Filippos Georgios Sarakis]

Hello, thank you for your time, when I try your code for a vector x = [1 2 3 4 5], I get this error.

[Torsten]

x = 1:5;
n = numel(x)-1;
psi = @(x)exp(-3*x.^2);

M = zeros(n+1);
for i = 1:n+1
    M(i,:) = psi(x(1:n+1)-x(i));
end
M = M - diag(diag(M)) + eye(n+1);
format long
M

M = 5×5

1.000000000000000 0.049787068367864 0.000006144212353 0.000000000001880 0.000000000000000 0.049787068367864 1.000000000000000 0.049787068367864 0.000006144212353 0.000000000001880 0.000006144212353 0.049787068367864 1.000000000000000 0.049787068367864 0.000006144212353 0.000000000001880 0.000006144212353 0.049787068367864 1.000000000000000 0.049787068367864 0.000000000000000 0.000000000001880 0.000006144212353 0.049787068367864 1.000000000000000

[Filippos Georgios Sarakis]

Amazing, one last thing that I dont get, how can I make this code into a function so that I can put any input vector x and the output would be the conversion number of the matrix M?

[Torsten]

Save "untitled245.m" as "fun.m" in your working directory and try again.

Because your function psi satisfies psi(0)=1, you can remove the line

M = M - diag(diag(M)) + eye(n+1);

in the code.

If you want to use the code for other functions, too, that might not satisfy psi(0)=1, better keep it.

Maybe besides x, you should pass "psi" to "fun" as a function handle to keep the code more flexible.