University of Florida/Egm4313/s12.team8.dupre/R6.1

Problem Statement
Given $$\displaystyle f(x)=a_{0}+\sum_{n=1}^{\infty }[a_{n}cos(n\omega x)+b_{n}sin(n\omega x)]$$: (1) Find the (smallest) period of $$\displaystyle cos(n\omega x) $$ and $$\displaystyle sin(n\omega x) $$. (2) Show that these functions also have period $$\displaystyle p $$. (3) Show that the constant $$\displaystyle a_{0} $$ is also a periodic function with period $$\displaystyle p $$.

Solution (1)
We know that the period of a normal $$\displaystyle sin(x) $$ or $$\displaystyle cos(x) $$ is $$\displaystyle 2\pi $$. When there are values or variables being multiplied by this $$\displaystyle x $$ variable, the period becomes $$\displaystyle 2\pi $$ divided by the values or variables. We know that: $$\displaystyle \omega=\frac{\pi }{L}=\frac{2 \pi}{p}$$    (6.1.1) Where $$\displaystyle p $$ is the period. Using this relation, along with our variable $$\displaystyle n $$, we can solve for the period of $$\displaystyle cos(n\omega x) $$ and $$\displaystyle sin(n\omega x) $$ as follows: $$\displaystyle p=\frac{2\pi }{\tfrac{n\pi }{L}}$$ $$\displaystyle p=\frac{L\pi}{n} $$    (6.1.2) Since the period will be smallest at $$\displaystyle n=1$$, plugging into equation (6.1.2) shows that the smallest period of $$\displaystyle cos(n\omega x) $$ and $$\displaystyle sin(n\omega x) $$ is: $$\displaystyle p=L\pi $$    (6.1.3)

Solution (2)
We are also given that: $$\displaystyle \omega =\frac{2\pi}{p}$$    (6.1.4) Using this relation, we can solve for the period again as follows: $$\displaystyle p=\frac{2\pi}{\frac{n2\pi}{p}}$$ $$\displaystyle p=\frac{p}{n}$$    (6.1.5) We know that the period is smallest at $$\displaystyle n=1$$, and plugging this value into (6.1.5) proves that: $$\displaystyle p=p $$    (6.1.6)

Solution (3)
We know that, starting at 0: $$\displaystyle \widetilde{a}_{0}=\frac{1}{2L}\int_{0}^{2L}f(\widetilde{x})d\widetilde{x} $$    (6.1.7) Where the period is represented by 0 to 2L. We are also given that: $$\displaystyle \frac{\pi}{L}=\frac{2\pi}{p} $$    (6.1.8) Rearranging (6.1.8) allows us to solve for L: $$\displaystyle L=\frac{p}{2}$$     (6.1.9) Multiplying (6.1.9) by 2 allows us to find the period of $$\displaystyle a_{0}$$, as follows: $$\displaystyle 2L=p$$    (6.1.10) This shows that $$\displaystyle a_{0}$$ is indeed a periodic function with a period of $$\displaystyle p$$. This also shows that at any given $$\displaystyle x$$ value or period throughout the periodic function, $$\displaystyle a_{0}$$ holds its constant value.