How can I calculate the sum of the absolute error?

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!

Answer 1

[Fabio Freschi]

Please check the code below.

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]

Thank you very much!!

I also got 24, but is it the correct value when the exercise wants 2 decimal places? Or could they just be .00 like in your code? Because they dont count, right? Or do I mix them with significant values?

I asked my friend and he gave me this answer

xk = [-2 -1 0 1 2]      ; % Given x-data
yk =  [0 15 25 50 110]' ; % Given y-data
x = linspace(-2,2);
F = c(1)+c(2)*x+c(3)*(x.^2-1) ; % The fit function
c = [32.5 25.5 7.50]                                ; % Valeus for (a, b, c)
absolute_error = sum(abs(yk-F))                     ; % Calculation of the absolute error
absolute_error_rounded = round(absolute_error,2)    ; % Rounding the absolute error to two decimal places
format bank                                         ; % Format with 2 digits after the decimal point.
absolute_error_rounded

This just gives a list of values which doesnt make sence because the absolute error should just be one value, right?

[Fabio Freschi]

Decimal places means digits after the decimal point.

The error formula in your OP is a scalar, so you should produce only one number

If my answer solves your problem, please accept it

[Mads Yar]

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!