Question

- Find the trigonometric function of the form$\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{c}}_{\mathbf{0}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\mathbf{sin}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{cos}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}\mathbf{sin}\left(2t\right)\mathbf{+}{\mathbf{c}}_{\mathbf{4}}\mathbf{cos}\left(2t\right)$ fits the data points$\left(0,0\right),\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{3}\mathbf{)}$ using least square. Sketch the solution together with the function g(t).

Short answer

The solution is $\mathrm{f}\left(\mathrm{t}\right)=\frac{5}{2}-2\mathrm{sint}-\frac{3}{2}\mathrm{cost}-\frac{1}{4}\sin 2\mathrm{t}-\cos 2\mathrm{t}$.

Step-by-step solution

Step:1 Definition of least square

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

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

Step:2 Explanation of the solution

Consider the function as follows.

$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_0+{\mathrm{c}}_1\sin \left(\mathrm{t}\right)+{\mathrm{c}}_2\cos \left(\mathrm{t}\right)+{\mathrm{c}}_3\sin \left(2\mathrm{t}\right)+{\mathrm{c}}_4\cos \left(2\mathrm{t}\right)$

Form the data as follows.

$$\begin{gathered} {\mathrm{c}}_0+0{\mathrm{c}}_1+{\mathrm{c}}_2+0{\mathrm{c}}_3+1{\mathrm{c}}_4=0 \\ {\mathrm{c}}_0+\sin \left(0.5\right){\mathrm{c}}_1+\cos \left(0.5\right){\mathrm{c}}_2+\sin \left(1\right){\mathrm{c}}_3+\cos \left(1\right){\mathrm{c}}_4=0.5 \\ {\mathrm{c}}_0+\sin \left(1\right){\mathrm{c}}_1+\cos \left(1\right){\mathrm{c}}_2+\sin \left(2\right){\mathrm{c}}_3+\cos \left(2\right){\mathrm{c}}_4=1 \\ {\mathrm{c}}_0+\sin \left(1.5\right){\mathrm{c}}_1+\cos \left(1.5\right){\mathrm{c}}_2+\sin \left(3\right){\mathrm{c}}_3+\cos \left(3\right){\mathrm{c}}_4=1.5 \\ {\mathrm{c}}_0+\sin \left(2\right){\mathrm{c}}_1+\cos \left(2\right){\mathrm{c}}_2+\sin \left(4\right){\mathrm{c}}_3+\cos \left(4\right){\mathrm{c}}_4=2 \\ {\mathrm{c}}_0+\sin \left(2.5\right){\mathrm{c}}_1+\cos \left(2.5\right){\mathrm{c}}_2+\sin \left(5\right){\mathrm{c}}_3+\cos \left(5\right){\mathrm{c}}_4=2.5 \\ {\mathrm{c}}_0+\sin \left(3\right){\mathrm{c}}_1+\cos \left(3\right){\mathrm{c}}_2+\sin \left(6\right){\mathrm{c}}_3+\cos \left(6\right){\mathrm{c}}_4=3 \end{gathered}$$

This linear system can be equivalent to as follows.

$\overline{)\begin{array}{llccc}{\mathrm{c}}_0& +0.5{\mathrm{c}}_1& +{\mathrm{c}}_2& +0.{\mathrm{c}}_3& +{\mathrm{c}}_4=0\\ {\mathrm{c}}_0& +0.5{\mathrm{c}}_1& +0.9{\mathrm{c}}_2& +0.8{\mathrm{c}}_3& +0.5{\mathrm{c}}_4=0.5\\ {\mathrm{c}}_0& +0.8{\mathrm{c}}_1& +0.5{\mathrm{c}}_2& +0.9{\mathrm{c}}_3& -0.4{\mathrm{c}}_4=1\\ {\mathrm{c}}_0& +0.9{\mathrm{c}}_1& +0.7{\mathrm{c}}_2& +0.1{\mathrm{c}}_3& -0.9{\mathrm{c}}_4=1.5\\ {\mathrm{c}}_0& +0.9{\mathrm{c}}_1& -0.4{\mathrm{c}}_2& -0.8{\mathrm{c}}_3& -0.7{\mathrm{c}}_4=2\\ {\mathrm{c}}_0& +0.6{\mathrm{c}}_1& -0.8{\mathrm{c}}_2& -1.{\mathrm{c}}_3& +0.3{\mathrm{c}}_4=2.5\\ {\mathrm{c}}_0& +0.1{\mathrm{c}}_1& -0.9{\mathrm{c}}_2& -0.3{\mathrm{c}}_3& +1.{\mathrm{c}}_4=3\end{array}}$

Hence, this can be again approximately as follows.

$\left|\begin{array}{ccccc}{\mathrm{c}}_0& +0.5{\mathrm{c}}_1& +{\mathrm{c}}_2& +0.{\mathrm{c}}_3& +{\mathrm{c}}_4=0\\ {\mathrm{c}}_0& +0.c_1& +1.{\mathrm{c}}_2& +1.{\mathrm{c}}_3& +0.5{\mathrm{c}}_4=0.5\\ {\mathrm{c}}_0& +\frac{1}{2}{\mathrm{c}}_1& +0.5{\mathrm{c}}_2& +1.{\mathrm{c}}_3& -\frac{1}{2}{\mathrm{c}}_4=1\\ {\mathrm{c}}_0& +1.c_1& +0.c_2& +0.c_3& -1.c_4=1.5\\ {\mathrm{c}}_0& +1.c_1& -\frac{1}{2}{c}_2& -1.c_3& -1.c_4=2\\ {\mathrm{c}}_0& +\frac{1}{2}{c}_1& -1.c_2& -0.c_3& +0.c_4=2.5\\ c_0& +0.c_1& -1.c_2& -0.3c_3& +1.c_4=3\end{array}\right|$

$\left[\begin{array}{c}{\mathrm{c}}_0\\ {\mathrm{c}}_1\\ {\mathrm{c}}_2\\ {\mathrm{c}}_3\\ {\mathrm{c}}_4\end{array}\right]=\left[\begin{array}{c}\frac{5}{2}\\ -2\\ \frac{-3}{2}\\ \frac{-1}{4}\\ -1\end{array}\right]$

According to the solution obtained the function is as follows.

$\mathrm{f}\left(\mathrm{t}\right)=\frac{5}{2}-2\mathrm{sint}-\frac{3}{2}\mathrm{cost}-\frac{1}{4}\sin 2\mathrm{t}-\cos 2\mathrm{t}$

The function can be sketched in the graph as follows.

Thus, the function$\mathrm{f}\left(\mathrm{t}\right)=\frac{5}{2}-2\mathrm{sint}-\frac{3}{2}\mathrm{cost}-\frac{1}{4}\sin 2\mathrm{t}-\cos 2\mathrm{t}$has been sketched in the graph.