Recall that the average of a function $f ( x )$ on an interval $[ a , b ]$ is $\bar { f } = \frac { 1 } { b - a } \int _ { a } ^ { b } f ( x ) \mathrm { d } x$
Calculate the 9-unit moving average of the function.
$f ( x ) = \cos \left( \frac { \pi x } { 18 } \right)$
A) $\bar { f } ( x ) = \frac { 2 } { \pi } \left( \sin \left( \frac { \pi x } { 18 } \right) - \cos \left( \frac { \pi x } { 18 } \right) \right)$
B) $\bar { f } ( x ) = \frac { 2 } { \pi } \left( \sin \left( \frac { \pi x } { 18 } \right) - \cos \left( \frac { \pi x } { 3 } \right) \right)$
C) $\bar { f } ( x ) = \frac { 2 } { \pi } \left( \sin \left( \frac { \pi x } { 18 } \right) + \cos \left( \frac { \pi x } { 2 } \right) \right)$
D) $\bar { f } ( x ) = \frac { 2 } { \pi } \left( \cos \left( \frac { \pi x } { 18 } \right) - \sin \left( \frac { \pi x } { 18 } \right) \right)$
E) $\bar { f } ( x ) = \frac { 2 } { \pi } \left( \sin \left( \frac { \pi x } { 18 } \right) + \cos \left( \frac { \pi x } { 18 } \right) \right)$

Question

Recall that the average of a function $f ( x )$ on an interval $[ a , b ]$ is   $\bar { f } = \frac { 1 } { b - a } \int _ { a } ^ { b } f ( x ) \mathrm { d } x$  
Calculate the 9-unit moving average of the function.
  $f ( x ) = \cos \left( \frac { \pi x } { 18 } \right)$  
A) $\bar { f } ( x ) = \frac { 2 } { \pi } \left( \sin \left( \frac { \pi x } { 18 } \right) - \cos \left( \frac { \pi x } { 18 } \right) \right)$
B) $\bar { f } ( x ) = \frac { 2 } { \pi } \left( \sin \left( \frac { \pi x } { 18 } \right) - \cos \left( \frac { \pi x } { 3 } \right) \right)$
C) $\bar { f } ( x ) = \frac { 2 } { \pi } \left( \sin \left( \frac { \pi x } { 18 } \right) + \cos \left( \frac { \pi x } { 2 } \right) \right)$
D) $\bar { f } ( x ) = \frac { 2 } { \pi } \left( \cos \left( \frac { \pi x } { 18 } \right) - \sin \left( \frac { \pi x } { 18 } \right) \right)$
E) $\bar { f } ( x ) = \frac { 2 } { \pi } \left( \sin \left( \frac { \pi x } { 18 } \right) + \cos \left( \frac { \pi x } { 18 } \right) \right)$

Quizplus · Accepted Answer

The 9-unit moving average of $f(x) = \cos\left(\frac{\pi x}{18}\right)$ involves integrating $f(x)$ over an interval of length 9 and then dividing by the interval length (9 units). The integral of $\cos\left(\frac{\pi x}{18}\right)$ is $\frac{18}{\pi}\sin\left(\frac{\pi x}{18}\right)$, and when calculating the average, the $\frac{1}{9}$ factor outside the integral and the $\frac{18}{\pi}$ from the integral combine to give a factor of $\frac{2}{\pi}$. The correct expression must involve $\sin\left(\frac{\pi x}{18}\right)$ due to the integration of the cosine function, and since the question asks for a moving average, which is a linear operation, there should be no change in the frequency of the sine function. Therefore, the correct answer must involve $\sin\left(\frac{\pi x}{18}\right)$ and potentially the original function $\cos\left(\frac{\pi x}{18}\right)$ due to the properties of the integral over a symmetric interval for cosine functions, leading to the correct choice being E.

Recall That the Average of a Function f(x)f ( x )f(x)

Recall That the Average of a Function $f ( x )$