91Ó°ÊÓ

Consider a linear system as$\mathrm{A}\stackrel{→}{\mathrm{x}}={\stackrel{→}{\mathrm{u}}}_{\mathrm{n}}$ .

Here A is an matrix, a vector$\stackrel{→}{\mathrm{x}}$ in${\mathrm{R}}^{\mathrm{n}}$ called a least square solution of this system if$||{\stackrel{→}{\mathrm{u}}}_{\mathrm{n}}-\mathrm{A}{\stackrel{→}{\mathrm{x}}}^{\mathrm{k}}||≤||{\stackrel{→}{\mathrm{u}}}_{\mathrm{n}}-\mathrm{A}\stackrel{→}{\mathrm{x}}||$ for all$\stackrel{→}{\mathrm{x}}$ in${\mathrm{R}}^{\mathrm{m}}$ .

Consider the points as follows.

The points$(0,0),(1,1),(2,2),(3,3)$that satisfies the trigonometric function of the form as follows.

$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_0+{\mathrm{c}}_1\sin \left(\mathrm{t}\right)+{\mathrm{c}}_2\cos \left(\mathrm{t}\right)$

Now, solving the linear system as follows.

Now $\left|\begin{array}{ccc}{\mathrm{c}}_0+& {\mathrm{c}}_2& =0\\ {\mathrm{c}}_0+& \sin \left(1\right){\mathrm{c}}_1+\cos \left(1\right)c_2& =1\\ {\mathrm{c}}_0+& \sin \left(2\right){\mathrm{c}}_1+\cos \left(2\right)c_2& =2\\ {\mathrm{c}}_0+& \sin \left(3\right){\mathrm{c}}_1+\cos \left(3\right)c_2& =3\end{array}\right|$such that $\mathrm{f}\left(0\right)=0,\mathrm{f}\left(1\right)=1,\mathrm{f}\left(2\right)=2\mathrm{f}\left(3\right)=3$

This linear system can be equivalent to as follows.

$\left|\begin{array}{ccc}{\mathrm{c}}_0+& {\mathrm{c}}_2& =0\\ {\mathrm{c}}_0+& \sin \left(1\right){\mathrm{c}}_1+\cos \left(1\right){\mathrm{c}}_2& =1\\ {\mathrm{c}}_0+& \sin \left(2\right){\mathrm{c}}_1+\cos \left(2\right){\mathrm{c}}_2& =2\\ {\mathrm{c}}_0+& \sin \left(3\right){\mathrm{c}}_1+\cos \left(3\right){\mathrm{c}}_2& =3\end{array}\right|≈\left|\begin{array}{ccc}{\mathrm{c}}_0+& {\mathrm{c}}_2& =0\\ {\mathrm{c}}_0+& 0.8{\mathrm{c}}_1+& 0.5{\mathrm{c}}_2=1\\ {\mathrm{c}}_0+& 0.9{\mathrm{c}}_1+& -0.4{\mathrm{c}}_2=2\\ {\mathrm{c}}_0+& 0.1{\mathrm{c}}_1+& -0.9{\mathrm{c}}_2=3\end{array}\right|$

Hence, this can be again approximate to as follows.

$\left|\begin{array}{ccc}{\mathrm{c}}_0+& {\mathrm{c}}_2& =0\\ {\mathrm{c}}_0+& \sin \left(1\right){\mathrm{c}}_1+\cos \left(1\right){\mathrm{c}}_2& =1\\ {\mathrm{c}}_0+& \sin \left(2\right){\mathrm{c}}_1+\cos \left(2\right){\mathrm{c}}_2& =2\\ {\mathrm{c}}_0+& \sin \left(3\right){\mathrm{c}}_1+\cos \left(3\right){\mathrm{c}}_2& =3\end{array}\right|≈\left|\begin{array}{ccc}{\mathrm{c}}_0+& {\mathrm{c}}_2& =0\\ {\mathrm{c}}_0+& 1.{\mathrm{c}}_1+& \frac{1}{2}.{\mathrm{c}}_2=1\\ {\mathrm{c}}_0+& 1.{\mathrm{c}}_1+& -\frac{1}{2}.{\mathrm{c}}_2=2\\ {\mathrm{c}}_0+& 0.{\mathrm{c}}_1+& 1.{\mathrm{c}}_2=3\end{array}\right|$

This linear system has a unique solution as follows.

$\left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\\ {\mathrm{c}}_2\end{array}\right]=\left[\begin{array}{c}\frac{3}{2}\\ 0\\ \frac{-3}{2}\end{array}\right]$

Therefore, the function is as follows.

$\mathrm{f}\left(\mathrm{t}\right)=\frac{3}{2}-\frac{3}{2}\mathrm{cost}$

The function is sketched in the graph as follows.

Thus, the function has been plotted in the graph.