Question

Fit a linear function of the form $\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{c}}_{\mathbf{0}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\mathbf{t}$ to the data points (0,3) , (1,3) , (1,6) using least square. Sketch the solution and explain why it make sense.

Short answer

The linear function is $\mathrm{f}\left(\mathrm{t}\right)=3+\frac{1}{2}\mathrm{t}$.

Step-by-step solution

Step:1 Explanation of the solution

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.

$$\begin{gathered} (0,3)\Rightarrow \mathrm{f}\left(0\right)={\mathrm{c}}_0+{\mathrm{c}}_1\left(0\right)={\mathrm{c}}_0=3 \\ \left(1,3\right)\Rightarrow \mathrm{f}\left(1\right)={\mathrm{c}}_0+{\mathrm{c}}_1\left(1\right)={\mathrm{c}}_0+3{\mathrm{c}}_1=3 \\ \left(1,6\right)\Rightarrow \mathrm{f}\left(1\right)={\mathrm{c}}_0+{\mathrm{c}}_1\left(1\right)={\mathrm{c}}_0+6{\mathrm{c}}_1=6 \end{gathered}$$

Then consider the system as $\mathrm{A}\overrightarrow{\mathrm{x}}=\overrightarrow{\mathrm{b}}$.

$\mathrm{where}\mathrm{A}=\left[\begin{array}{cc}1& 0\\ 1& 1\\ 1& 1\end{array}\right],\overrightarrow{\mathrm{b}}=\left[\begin{array}{c}3\\ 3\\ 6\end{array}\right]\mathrm{and}\overrightarrow{\mathrm{x}}=\left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\end{array}\right]$

To find the least square solution of the system as follows.

$\mathrm{A}\overrightarrow{\mathrm{x}}=\overrightarrow{\mathrm{b}}$

The exact solution of the system is as follows.

$$\begin{gathered} \left({\mathrm{A}}^{\mathrm{T}}\mathrm{A}\right)\overrightarrow{\mathrm{x}}={\mathrm{A}}^{\mathrm{T}}\overrightarrow{\mathrm{b}} \\ {\left[\begin{array}{cc}1& 0\\ 1& 0\\ 1& 1\end{array}\right]}^{\mathrm{T}}\left[\begin{array}{cc}1& 0\\ 1& 0\\ 1& 1\end{array}\right]\left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\end{array}\right]={\left[\begin{array}{cc}1& 0\\ 1& 0\\ 1& 1\end{array}\right]}^{\mathrm{T}}\left[\begin{array}{c}3\\ 6\\ 6\end{array}\right] \\ \left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 1& 0\\ 1& 1\end{array}\right]\left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}3\\ 6\\ 6\end{array}\right] \\ \left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 1& 0\\ 1& 1\end{array}\right]\left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\end{array}\right]=\left[\begin{array}{c}12\\ 9\end{array}\right] \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \left[\begin{array}{cc}3& 2\\ 2& 2\end{array}\right]\left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\end{array}\right]=\left[\begin{array}{c}12\\ 9\end{array}\right] \\ 3{\mathrm{c}}_0+2{\mathrm{c}}_1=12 \\ {\mathrm{c}}_0=3,{\mathrm{c}}_1=\frac{1}{2} \\ \left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\end{array}\right]=\left[\begin{array}{c}3\\ \frac{1}{2}\end{array}\right] \\ \left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\end{array}\right]=\left[\begin{array}{c}3\\ \frac{1}{2}\end{array}\right] \end{gathered}$$

Hence, the least square solution is $\left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\end{array}\right]=\left[\begin{array}{c}3\\ \frac{1}{2}\end{array}\right]$.

Then the linear function is$\mathrm{f}\left(\mathrm{t}\right)=3+\frac{1}{2}\mathrm{t}$

The graph for the linear function is as follows.

Hence, the linear function is $\mathrm{f}\left(\mathrm{t}\right)=3+\frac{1}{2}\mathrm{t}$