Cannot get a non-singular matrix
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Hello all, I'm trying to make a non-singular matrix, but I don't know how to solve this problem.
If I use 11 instead of 13 as the value of m, I get a non-singular matrix.
f = @(x) log(exp(x) + 1) + 2 * sin(x);
a1 = 1;
b1 = 18;
m = 13;
x = linspace(a1, b1, m)';
y = f(x);
b = @(x, j) x.^(j - 1);
B = ones(13,13);
for j = 2 : m
B(:, j) = b(x, j);
end
a = B\y;
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Answers (3)
Sajid Afaque
on 9 Jun 2022
you need to focus on the last step i.e. division part
a = B ./ y ; %element wise division
Mitch Lautigar
on 9 Jun 2022
Couple comments. Please see my added comments to the code
f = @(x) log(exp(x) + 1) + 2 * sin(x); %I'd change "*" to ".*" since you are passing in an array
a1 = 1;
b1 = 18;
m = 13;
x = linspace(a1, b1, m)'; %this line of code works fine. I tested it
y = f(x); %if you make the change suggested above, should run through clearly.
b = @(x, j) x.^(j - 1); %I'm hoping j is initialized elsewhere.
B = ones(13,13); %I'd make this line read "B = ones(m,m);" unless you need a static . %13x13 array
for j = 2 : m
B(:, j) = b(x, j);
end
a = B\y;
Torsten
on 9 Jun 2022
Edited: Torsten
on 9 Jun 2022
Maybe of help: The Vandermonde matrix is ill conditioned !
Maybe of help:
f = @(x) log(exp(x) + 1) + 2 * sin(x);
a1 = 1;
b1 = 18;
m = 13;
x = linspace(a1, b1, m)';
y = f(x);
b = @(x, j) x.^(j - 1);
B = ones(m,m);
for j = 2 : m
B(:, j) = b(x, j);
end
a = B\y;
xx = linspace(a1,b1,100).';
yapprox = zeros(numel(xx),1);
for i=1:numel(xx)
summe = 0.0;
for j=1:m
summe = summe + a(j)*xx(i)^(j-1);
end
yapprox(i) = summe;
end
plot(xx,f(xx))
hold on
plot(xx,yapprox)
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