Exercise 2.2.12

We have $A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 4\hfill \end{array}\right]$, $A^T=\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$, and $b=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 8\hfill \\ \hfill 8\hfill \\ \hfill 20\hfill \end{array}\right]$

Solve for $A^{T}A\widehat{x}=A^{T}b$, we have $A^{T}A=\left[\begin{array}{cc}\hfill 4\hfill & \hfill 8\hfill \\ \hfill 8\hfill & \hfill 26\hfill \end{array}\right]$, and $A^{T}b=\left[\begin{array}{c}\hfill 36\hfill \\ \hfill 112\hfill \end{array}\right]$, we find $\widehat{x}=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 4\hfill \end{array}\right]$, so the best straight line in Figure $\mathit{II}.3a$ is $b=1+4t$.

The four heights at $t=0,1,3,4$ are $p=1,5,13,17$, the errors are $e=b-p=-1,3,-5,3$, the minimum squared error $E=44$.