The problem of constants in linear least squares

The equation of the model is ay^2+bxy+cx+dy+e=x^2.

Since e is a constant not associated with any variable, how should we get the value of e?

x = [1.02; 0.95; 0.77; 0.67; 0.56; 0.30; 0.16; 0.01];
y = [0.39; 0.32; 0.22; 0.18; 0.15; 0.12; 0.13; 0.15];

Answer 1

[Star Strider]

since it is a constant, e becomes a vector of ones —

x = [1.02; 0.95; 0.77; 0.67; 0.56; 0.30; 0.16; 0.01];
y = [0.39; 0.32; 0.22; 0.18; 0.15; 0.12; 0.13; 0.15];

B = [y.^2 x.*y x y ones(size(x))] \ x.^2;

fprintf('\na = %10.6f\nb = %10.6f\nc = %10.6f\nd = %10.6f\ne = %10.6f\n',B)

a =  -2.876811
b =   0.223246
c =   0.538057
d =   3.276223
e =  -0.435324

.

[Walter Roberson]

("becomes a vector of ones" for the purpose of doing fitting using the Vandermode-type matrix and the \ operator)