Question

Fit a trigonometric function of the form f(t)=c0+c1sin(t)+c2cos(t)f(t)=c0+c1sin⁡(t)+c2cos⁡(t) to the data points (0,5.5)(0,5.5), (π2,0.5)(π2,0.5), (π,−2.5)(π,−2.5), (3π2,−7.5)(3π2,−7.5), using least squares.

Answer 1

Answer

$f(t)=-0.2+4.1sin(t)+4cos(t)$

Step-By-Step Explanation

Given the function $f(t)=c_0+c_1sin(t)+c_2cos(t)$.

For each pair (t, f(t)) in the data points (0,5.5), (π/2,0.5), (π,−2.5), (3π/2,−7.5)

$f(0)=c_0+0c_1+c_2=5.5$.

$f(\pi /2)=c_0+c_1+0c_2=0.5$.

$f(\pi)=c_0+0sin(t)-c_2=-2.5$.

$f(3\pi /2)=c_0-c_1+0c_2=-7.5$.

Expressing this as a system of linear equations in matrix form AX=B

$\left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right)\left( \begin{array}{c} c_{0} \\ c_{1} \\ c_{2}\\ \end{array} \right)=\left(\begin{array}{c} 5.5 \\ 0.5 \\ -2.5 \\ -7.5 \end{array} \right)$      

Where    

$A=\left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right)$,      

$B=\left(\begin{array}{c}5.5\\0.5\\-2.5\\-7.5\end{array} \right)$

$X=\left(\begin{array}{c}c_0\\c_1\\c_2\end{array}\right)$     

To determine the values of X, we use the expression  

$X=(A^{T}A)^{-1}A^{T}B$      

$A^{T}A= \left(\begin{array}{ccc} 3 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{array} \right)$

$(A^{T}A)^{-1}= \left(\begin{array}{ccc} 0.4 & -0.2 & 0 \\ -0.2 & 0.6 & 0 \\ 0 & 0 & 0.5 \end{array} \right)$      

$A^{T}B=\left(\begin{array}{c} 3.5 \\ 8 \\ 8 \end{array} \right)$      

Therefore:    

$X=\left(\begin{array}{ccc} 0.4 & -0.2 & 0 \\ -0.2 & 0.6 & 0 \\ 0 & 0 & 0.5 \end{array} \right)\left( \begin{array}{c} 3.5 \\ 8 \\ 8 \end{array} \right)$      

$X=\left(\begin{array}{c}c_0\\c_1\\c_2\end{array}\right)=\left(\begin{array}{c} -0.2 \\4.1\\4\end{array}\right)$  

Therefore, the trigonometric function which fits to the given data is:

$f(t)=-0.2+4.1sin(t)+4cos(t)$