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

R6.1

## Problem Statement

Given ${\displaystyle \displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }[a_{n}cos(n\omega x)+b_{n}sin(n\omega x)]}$:

(1) Find the (smallest) period of ${\displaystyle \displaystyle cos(n\omega x)}$ and ${\displaystyle \displaystyle sin(n\omega x)}$.

(2) Show that these functions also have period ${\displaystyle \displaystyle p}$.

(3) Show that the constant ${\displaystyle \displaystyle a_{0}}$ is also a periodic function with period ${\displaystyle \displaystyle p}$.

### Solution (1)

We know that the period of a normal ${\displaystyle \displaystyle sin(x)}$ or ${\displaystyle \displaystyle cos(x)}$ is ${\displaystyle \displaystyle 2\pi }$. When there are values or variables being multiplied by this ${\displaystyle \displaystyle x}$ variable, the period becomes ${\displaystyle \displaystyle 2\pi }$ divided by the values or variables. We know that:

${\displaystyle \displaystyle \omega ={\frac {\pi }{L}}={\frac {2\pi }{p}}}$ (6.1.1)

Where ${\displaystyle \displaystyle p}$ is the period.

Using this relation, along with our variable ${\displaystyle \displaystyle n}$, we can solve for the period of ${\displaystyle \displaystyle cos(n\omega x)}$ and ${\displaystyle \displaystyle sin(n\omega x)}$ as follows:

${\displaystyle \displaystyle p={\frac {2\pi }{\tfrac {n\pi }{L}}}}$

${\displaystyle \displaystyle p={\frac {L\pi }{n}}}$ (6.1.2)

Since the period will be smallest at ${\displaystyle \displaystyle n=1}$, plugging into equation (6.1.2) shows that the smallest period of ${\displaystyle \displaystyle cos(n\omega x)}$ and ${\displaystyle \displaystyle sin(n\omega x)}$ is:

${\displaystyle \displaystyle p=L\pi }$     (6.1.3)


### Solution (2)

We are also given that:

${\displaystyle \displaystyle \omega ={\frac {2\pi }{p}}}$ (6.1.4)

Using this relation, we can solve for the period again as follows:

${\displaystyle \displaystyle p={\frac {2\pi }{\frac {n2\pi }{p}}}}$

${\displaystyle \displaystyle p={\frac {p}{n}}}$ (6.1.5)

We know that the period is smallest at ${\displaystyle \displaystyle n=1}$ , and plugging this value into (6.1.5) proves that:

${\displaystyle \displaystyle p=p}$     (6.1.6)


### Solution (3)

We know that, starting at 0:

${\displaystyle \displaystyle {\widetilde {a}}_{0}={\frac {1}{2L}}\int _{0}^{2L}f({\widetilde {x}})d{\widetilde {x}}}$ (6.1.7)

Where the period is represented by 0 to 2L. We are also given that:

${\displaystyle \displaystyle {\frac {\pi }{L}}={\frac {2\pi }{p}}}$ (6.1.8)

Rearranging (6.1.8) allows us to solve for L:

${\displaystyle \displaystyle L={\frac {p}{2}}}$ (6.1.9)

Multiplying (6.1.9) by 2 allows us to find the period of ${\displaystyle \displaystyle a_{0}}$, as follows:

${\displaystyle \displaystyle 2L=p}$     (6.1.10)


This shows that ${\displaystyle \displaystyle a_{0}}$ is indeed a periodic function with a period of ${\displaystyle \displaystyle p}$. This also shows that at any given ${\displaystyle \displaystyle x}$ value or period throughout the periodic function, ${\displaystyle \displaystyle a_{0}}$ holds its constant value.