# How can I calculate the sum of the absolute error?

25 views (last 30 days)
Mads Yar on 23 Sep 2021
Commented: Mads Yar on 23 Sep 2021
Hello,
I am trying to do some exercises in Matlab, and i have trouble with calculating the absolute error and state it with 2 decimal places. You can see the exercise (the last one) in the image below.
I am done with sub task 1 and 2, where i got:
A_system_matrix = [1 -2 3; 1 -1 0; 1 0 -1; 1 1 0; 1 2 3] % System matrix A
y_vector = [0; 15; 25; 50; 110]
rref([A'*A,A'*y_vector]) % Using rowreducing
(A'*A)^(-1)*A'*y_vector % Using the equation stated in the text above Can someone help me with using Matlab to calculate the sum of the absolute error stated in the exercise and then state it with 2 decimal places?
Thanks a lot!

Fabio Freschi on 23 Sep 2021
Edited: Fabio Freschi on 23 Sep 2021
clear all, close all
% model
F = @(c,x)c(1)+c(2)*x+c(3)*(x.^2-1);
% data
xk = [-2 -1 0 1 2].';
yk = [0 15 25 50 110].';
% build matrix A
A = [F([1 0 0],xk) F([0 1 0],xk) F([0 0 1],xk)];
% least-square fit
c = (A.'*A)\(A.'*yk);
% error
err = sum(abs(yk-F(c,xk)));
fprintf('error = %.2f\n',err);
error = 24.00
% plot
x = (-2:.1:2).';
figure, hold on
plot(xk,yk,'o')
plot(x,F(c,x)) Three remarks
• never ever use ^(-1) or inv to solve a linear system. Backslash operator is faster and more accurate.
• the transpose operator is .' for the transpose. A simple ' is the complex conjugate transpose. In your case this is not affecting the result, but it is a good practice to keep the two operations separated.
• Note that you can solve the least-square problem without explicitly write the normal equations using the backslash operator on your overdetermined system
c = A\yk;
Mads Yar on 23 Sep 2021
Fair enough. Thank you very much for the detailed answer! Can i ask you about other exercises also? Else you can take a look at my other questions? It would be such a big help!