Question

please answer all parts

4. Least Square Method A. (10 points) Use the least squares method and find a linear fit for the following points: (0, -3), (2, -3), (1, -4), (4,5) B. Short Answer: (5 points) Quickly plot the points (by hand) and comment on the likely quality of the linear fit. Would another type of curve fit be better suited?

Answer 1

Part(a)

.

equation of the line is

$y=mx+b$

$mx+b=y$

put all data points

$0m+b=-3$

2m +b= -3

$m+b=-4$

$4m+b=5$

system Ax=b is

$\begin{pmatrix}0&1\\ 2&1\\ 1&1\\ 4&1\end{pmatrix}\begin{pmatrix}m\\ b\end{pmatrix}=\begin{pmatrix}-3\\ -3\\ -4\\ 5\end{pmatrix}$

here we have

$A=\begin{pmatrix}0&1\\ 2&1\\ 1&1\\ 4&1\end{pmatrix},\:A^T=\begin{pmatrix}0&2&1&4\\ 1&1&1&1\end{pmatrix},\:b=\begin{pmatrix}-3\\ -3\\ -4\\ 5\end{pmatrix}$

.

take $A^TA$

$A^TA=\begin{pmatrix}0&2&1&4\\ 1&1&1&1\end{pmatrix}\begin{pmatrix}0&1\\ 2&1\\ 1&1\\ 4&1\end{pmatrix}$

$A^TA=\begin{pmatrix}0\cdot \:0+2\cdot \:2+1\cdot \:1+4\cdot \:4&0\cdot \:1+2\cdot \:1+1\cdot \:1+4\cdot \:1\\ 1\cdot \:0+1\cdot \:2+1\cdot \:1+1\cdot \:4&1\cdot \:1+1\cdot \:1+1\cdot \:1+1\cdot \:1\end{pmatrix}$

$A^TA=\begin{pmatrix}21&7\\ 7&4\end{pmatrix}$

.

take $A^Tb$

$A^Tb= \begin{pmatrix}0&2&1&4\\ 1&1&1&1\end{pmatrix}\begin{pmatrix}-3\\ -3\\ -4\\ 5\end{pmatrix}$

$A^Tb= \begin{pmatrix}0\cdot \left(-3\right)+2\left(-3\right)+1\cdot \left(-4\right)+4\cdot \:5\\ 1\cdot \left(-3\right)+1\cdot \left(-3\right)+1\cdot \left(-4\right)+1\cdot \:5\end{pmatrix}$

$A^Tb= \begin{pmatrix}10\\ -5\end{pmatrix}$

.

now solve the system for

LATA].x = AT6

$\begin{pmatrix}21&7\\ 7&4\end{pmatrix}\begin{pmatrix}m\\ b\end{pmatrix}=\begin{pmatrix}10\\ -5\end{pmatrix}$

the augmented matrix is

$\begin{pmatrix}21&7&10\\ 7&4&-5\end{pmatrix}$

$R_2\:\leftarrow \:R_2-\frac{1}{3}\cdot \:R_1$

$=\begin{pmatrix}21&7&10\\ 0&\frac{5}{3}&-\frac{25}{3}\end{pmatrix}$

$R_2\:\leftarrow \frac{3}{5}\cdot \:R_2$

$=\begin{pmatrix}21&7&10\\ 0&1&-5\end{pmatrix}$

$R_1\:\leftarrow \:R_1-7\cdot \:R_2$

$=\begin{pmatrix}21&0&45\\ 0&1&-5\end{pmatrix}$

$R_1\:\leftarrow \frac{1}{21}\cdot \:R_1$

$=\begin{pmatrix}1&0&\frac{15}{7}\\ 0&1&-5\end{pmatrix}$

least-square solution is

$\begin{pmatrix}m\\ b\end{pmatrix}=\begin{pmatrix}\frac{15}{7}\\ -5\end{pmatrix}$

so best fit line

${\color{Red} y=\frac{15}{7}x-5}$

.

.

Part(a)

.

09 (4,5) 4 2 -2 0 2. 09 8 -2 (0, 3) . (2,3) 4 (1, +4)

from the graph, we can see that it is not the best fit line

the best fit curve would be a parabola