Question

Fit a linear function of the form$\mathbf{f}\left(t\right)\mathbf{=}{\mathbf{c}}_{\mathbf{0}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\mathbf{t}$ to the data points$\left(0,0\right)\mathbf{,}\left(0,1\right)\mathbf{,}\left(1,1\right)$ using least square. Sketch the solution and explain why it make sense.

Short answer

The linear function that fits the given data points is $\mathrm{f}\left(\mathrm{t}\right)=\frac{1}{2}\left(1+\mathrm{t}\right)$.

Step-by-step solution

Step:1 Explanation of the solution

Consider a linear system as$\mathrm{A}\stackrel{\rightharpoonup }{\mathrm{x}}=\stackrel{\rightharpoonup }{\mathrm{b}},\mathrm{where}\mathrm{A}=\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right],\stackrel{\rightharpoonup }{\mathrm{b}},=\left[\begin{array}{c}5\\ 0\end{array}\right]$.

Now, consider the linear function as follows.

$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_0+{\mathrm{c}}_1\mathrm{t}$

Substitute the points in the linear function as follows.

$$\begin{gathered} \left(0,0\right)\Rightarrow f\left(0\right)=c_0+c_1\left(0\right)=c_0=0 \\ \left(0,1\right)\Rightarrow f\left(0\right)=c_0+c_1\left(0\right)=c_0=0 \\ \left(1,1\right)\Rightarrow \mathrm{f}\left(1\right)={\mathrm{c}}_0+{\mathrm{c}}_1\left(1\right)={\mathrm{c}}_0+{\mathrm{c}}_1=1 \\ \mathrm{Then}\mathrm{consider}\mathrm{the}\mathrm{system}\mathrm{as}\mathrm{A}\stackrel{\rightharpoonup }{\mathrm{x}}=\stackrel{\rightharpoonup }{\mathrm{b}}. \\ \mathrm{where}\mathrm{A}=\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right],\stackrel{\rightharpoonup }{\mathrm{b}}=\left[\begin{array}{c}5\\ 0\end{array}\right]\mathrm{and}\stackrel{\rightharpoonup }{\mathrm{x}}=\left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\end{array}\right] \\ \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{least}\mathrm{square}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{system}\mathrm{as}\mathrm{follows}. \\ \mathrm{A}\stackrel{\rightharpoonup }{\mathrm{x}}=\stackrel{\rightharpoonup }{\mathrm{b}}. \end{gathered}$$

The exact solution of the system is as follows.

$$\begin{gathered} \left(A^{T}A\right)\stackrel{\rightharpoonup }{x}=A^T\stackrel{\rightharpoonup }{b} \\ {\left[\begin{array}{cc}1& 0\\ 1& 0\\ 1& 0\end{array}\right]}^T\left[\begin{array}{cc}1& 0\\ 1& 0\\ 1& 0\end{array}\right]\left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\end{array}\right]={\left[\begin{array}{cc}1& 0\\ 1& 0\\ 1& 0\end{array}\right]}^T\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right] \\ \left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 1\end{array}\right]{\left[\begin{array}{cc}1& 0\\ 1& 0\\ 1& 0\end{array}\right]}^T\left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right] \\ \left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 1\end{array}\right]{\left[\begin{array}{cc}1& 0\\ 1& 0\\ 1& 0\end{array}\right]}^T\left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\end{array}\right]=\left[\begin{array}{c}1\\ 1\end{array}\right] \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \left[\begin{array}{cc}3& 1\\ 1& 1\end{array}\right]\left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\end{array}\right]=\left[\begin{array}{c}1\\ 1\end{array}\right] \\ 3{\mathrm{c}}_0+{\mathrm{c}}_1=1 \\ {\mathrm{c}}_0=\frac{1}{2},{\mathrm{c}}_1=\frac{1}{2} \\ \left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\end{array}\right]=\left[\begin{array}{c}\frac{1}{2}\\ \frac{1}{2}\end{array}\right] \\ \left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}1\\ 1\end{array}\right] \\ \mathrm{Hence},\mathrm{the}\mathrm{least}\mathrm{square}\mathrm{solution}\mathrm{is}\left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}1\\ 1\end{array}\right]. \\ \mathrm{Then}\mathrm{the}\mathrm{linear}\mathrm{function}\mathrm{is}\mathrm{f}\left(\mathrm{t}\right)=\frac{1}{2}\left(1+\mathrm{t}\right) \\ \mathrm{The}\mathrm{graph}\mathrm{for}\mathrm{the}\mathrm{linear}\mathrm{function}\mathrm{is}\mathrm{as}\mathrm{follows}. \end{gathered}$$

By observing from the graph, the function is passing through one of the data point (1,1).

Hence, the solution make sense.OPL8