Answer to Question #217902 in Differential Equations for Shivani

Question

Using the 6 ordinate scheme analyze harmonically the data to third harmonics
X 0 π/3 2π/3 π 4π/3 5π/3 2π
Y 10 12 15. 20. 17. 11. 10

Answer 1

$\sum_{k=1}^7y_k=95$

$\sum_{k=1}^7y_kcos(x_k)=-4.5$

$\sum_{k=1}^7y_ksin(x_k)=-0.866$

$\sum_{k=1}^7y_kcoc(2x_k)=12.5$

$\sum_{k=1}^7y_ksin(2x_k)=2.598$

we can get the values of $a_i$ and $b_i$ as follows:

$a_0={1\over 7}\times95=13.57$

$a_1={2\over 7}\times(-4.5)=-1.29$

$a_2={2\over 7}\times 12.5=3.57$

$b_1={2\over 7}\times(-0.866)=-0.25$

$b_2={2\over 7}\times 2.598=0.74$

Substituting those values into Fourier series

$f(x)=a_0+\sum_{n=1}^\infin(a_ncos(nx)+b_nsin(nx))$ gives

$f(x)=a_0+a_1cos(x)+a_2cos(2x)+b_1sin(x)+b_2sin(2x)$

$\therefore y=13.57-1.29cos(x)+3.75cos(2x)-0.25sin(x)+0.74sin(2x)$