Science:Math Exam Resources/Courses/MATH221/December 2009/Question 09/Solution 1

The coefficients $a,b,c$ of the least squares fit are given by

$X^{T}X\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=X^{T}y,$

where

$X=\left[\begin{array}{ccc}{x}_1^0& x_1^1& x_1^2\\ x_2^0& x_2^1& x_2^2\\ x_3^0& x_3^1& x_3^2\\ x_4^0& x_4^1& x_4^2\\ \end{array}\right]=\left[\begin{array}{ccc}{0}^0& 0^1& 0^2\\ 1^0& 1^1& 1^2\\ 2^0& 2^1& 2^2\\ 3^0& 3^1& 3^2\\ \end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 1& 1& 1\\ 1& 2& 4\\ 1& 3& 9\\ \end{array}\right]$

and

$y=\left[\begin{array}{c}{y}_1\\ y_2\\ y_3\\ y_4\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 5\\ 6\end{array}\right].$

Now

$X^{T}X=\left[\begin{array}{ccc}98& 36& 14\\ 36& 14& 6\\ 14& 6& 4\end{array}\right]$

and

$X^{T}y=\left[\begin{array}{c}74\\ 28\\ 12\end{array}\right]$.

Solving the resulting linear system, we find that $a=\frac{1}{2},b=\frac{1}{2},c=\frac{1}{2}$, so the least squares fit is $y=\frac{1}{2}+\frac{1}{2}x+\frac{1}{2}{x}^2$.