Exercise 2.2.18

To find the closest line $b=\mathit{Dt}$ through the origin, we have $A=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 3\hfill \\ \hfill 4\hfill \end{array}\right]$, $A^T=\left[\begin{array}{cccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$, 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 $A^{T}A=\left[\begin{array}{c}\hfill 26\hfill \end{array}\right]$, and $A^{T}b=\left[\begin{array}{c}\hfill 112\hfill \end{array}\right]$, we find $\widehat{x}=\left[\begin{array}{c}\hfill \frac{56}{13}\hfill \end{array}\right]$, so the best horizontal line fit is $b=\mathit{Dt}=\frac{56}{13}t$.

The errors are $e=b-p=-\frac{56}{13},\frac{48}{13},-\frac{64}{13},\frac{36}{13}$, total error $E=64.095$