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

$X^{T}X{\begin{bmatrix}a\\b\\c\end{bmatrix}}=X^{T}y,$

where

$X={\begin{bmatrix}x_{1}^{0}&x_{1}^{1}&x_{1}^{2}\\x_{2}^{0}&x_{2}^{1}&x_{2}^{2}\\x_{3}^{0}&x_{3}^{1}&x_{3}^{2}\\x_{4}^{0}&x_{4}^{1}&x_{4}^{2}\\\end{bmatrix}}={\begin{bmatrix}0^{0}&0^{1}&0^{2}\\1^{0}&1^{1}&1^{2}\\2^{0}&2^{1}&2^{2}\\3^{0}&3^{1}&3^{2}\\\end{bmatrix}}={\begin{bmatrix}1&0&0\\1&1&1\\1&2&4\\1&3&9\\\end{bmatrix}}$

and

$y={\begin{bmatrix}y_{1}\\y_{2}\\y_{3}\\y_{4}\end{bmatrix}}={\begin{bmatrix}1\\0\\5\\6\end{bmatrix}}.$

Now

$X^{T}X={\begin{bmatrix}98&36&14\\36&14&6\\14&6&4\end{bmatrix}}$

and

$X^{T}y={\begin{bmatrix}74\\28\\12\end{bmatrix}}$.

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 $\color {blue}y={\frac {1}{2}}+{\frac {1}{2}}x+{\frac {1}{2}}x^{2}$.