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

Problem Statement

Given $f(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}[a_{n}cos(n\omega x)+b_{n}sin(n\omega x)]$:

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

(2) Show that these functions also have period $p$.

(3) Show that the constant $a_0$ is also a periodic function with period $p$.

Solution (1)

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

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

Where $p$ is the period.

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

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

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

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

$p=L\pi$     (6.1.3)

Solution (2)

We are also given that:

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

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

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

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

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

$p=p$     (6.1.6) 

Solution (3)

We know that, starting at 0:

${\tilde{a}}_0=\frac{1}{2L}\underset{0}{\overset{2L}{\int }}f(\tilde{x})d\tilde{x}$ (6.1.7)

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

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

Rearranging (6.1.8) allows us to solve for L:

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

Multiplying (6.1.9) by 2 allows us to find the period of $a_0$, as follows:

$2L=p$     (6.1.10)

This shows that $a_0$ is indeed a periodic function with a period of $p$. This also shows that at any given $x$ value or period throughout the periodic function, $a_0$ holds its constant value.