Least Squares Of Linear Equations

P=mt/b+t iff P=m-b*P\t

t=transpose([1 3 4 7 8 10]);
P=transpose([2.1 4.6 5.4 6.1 6.4 6.6]);
A=[ones(length(p),1), -P./t];
At=transpose([ones(length(P),1),-P./t]);
V=inv(At*A)*(At*P);
m=V(1)
b=V(2)

Now to find the plot I used:

x=P/t;
f=@(x)m-b.*(x);

plot(t,P,'.')
hold on
plot(t,f(t))

Which seems incorrect

Answer 1

[David Goodmanson]

Hi HH,

There are a a couple of things going on here. I compacted your code a bit to make the code below, using .' in place of the transpose commands. It appears your values for P are not an exact fit, nothing wrong with that.

For the fitting plot, you need to use f(x) or f(P./t), not f(t). Also, you meant x = P./t not x = P/t. After that you get the correct plot.

At least as important, you are using the old fashioned normal-matrix technique with inv(A.' * A) to find V. That's all right accuracywise for a 2x6, but it is not good practice in general.

You want to solve

P = A*V

for V, where P and V are column vectors. In Matlab, the solution is simply

V = A\P

(The idea is that you are effectively dividing P on the left by A).

For large arrays, backslash is more accurate than the normal-matrix approach, and there is no need to compute an inverse. Besides, backslash is how Mathworks got its start.

t = [1 3 4 7 8 10].';
P = [2.1 4.6 5.4 6.1 6.4 6.6].';
A = [ones(size(P)), -P./t];
    % At=transpose([ones(length(P),1),-P./t]);
    % V=inv(At*A)*(At*P);
V = A\P;

% method 1
m = V(1)
b = V(2)
x = P./t;
f = @(x)m-b.*(x);
figure(1)
plot(t,P,'.-',t,f(P./t),'o-')

APPENDED: As far as checking the result, you already have what you need in A and the solved-for V, so an even better way is to replace method 1 with

Pfit = A*V;
figure(1)
plot(t,P,'.-',t,Pfit,'o-')

[Harel Harel Shattenstein]

Hi, sorry It did not attach the qeustion now it is easier to see where I got it wrong