# The problem of constants in linear least squares

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Jiapeng le 10 Nov 2022
Commenté : Walter Roberson le 10 Nov 2022
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];
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### Réponse acceptée

Star Strider le 10 Nov 2022
Modifié(e) : Walter Roberson le 10 Nov 2022
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
.
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Walter Roberson le 10 Nov 2022
("becomes a vector of ones" for the purpose of doing fitting using the Vandermode-type matrix and the \ operator)

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