Chapter 4.6, Problem 74E

(a).

To determine

To graph:

The given function using graphing utility.

Explanation of Solution

Given:

  $y_1=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x\right)$

  $y_2=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x+\frac{1}{5}\sin 5\pi x\right)$

Graph:

Consider the provided function

  $y_1=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x\right)$ .

  $y_2=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x+\frac{1}{5}\sin 5\pi x\right)$ .

The graph of the function

  $y_1=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x\right)$ obtained with the help of graphing utility is shown below:

  

The graph of the function $y_2=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x+\frac{1}{5}\sin \pi x\right)$ obtained with the help of graphing utility is shown below:

  

(b).

To determine

To graph:

The given function using graphing utility.

Explanation of Solution

Given:

  $y_1=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x\right)$

  $y_2=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x+\frac{1}{5}\sin 5\pi x\right)$

Graph:

Consider the provided function

  $y_1=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x\right)$ .

  $y_2=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x+\frac{1}{5}\sin 5\pi x\right)$ .

From the above pattern the function $y_3$ is written as:

  $y_3=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x+\frac{1}{5}\sin 5\pi x+\frac{1}{7}\sin 7\pi x\right)$

The graph of the function $y_2=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x+\frac{1}{5}\sin \pi x\right)$ obtained with the help of graphing utility is shown below:

  

(c).

To determine

To graph:

The given function using graphing utility.

Answer to Problem 74E

period for all function which is find in part(a) and part(b) is $2$ .

The function $y_4$ is as follows:

  $y_4=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x+\frac{1}{5}\sin 5\pi x+\frac{1}{7}\sin 7\pi x+\frac{1}{9}\sin 9\pi x\right)$ which is a better approximation.

Explanation of Solution

Given:

  $y_1=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x\right)$

  $y_2=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x+\frac{1}{5}\sin 5\pi x\right)$

Graph:

Consider the provided function

  $y_1=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x\right)$ .

  $y_2=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x+\frac{1}{5}\sin 5\pi x\right)$ .

The standard sine function is as:

  $y=a\sin (bx-c)$

So, the period is $\frac{2\pi }{b}$ .

For the function $y_1$ and $y_2$ value of $b=\pi$ the period is:

  $\frac{2\pi }{\pi }=2$

So, the period for all function which is find in part(a) and part(b) is $2$ .

The function $y_4$ is as follows:

  $y_4=\frac{4}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x+\frac{1}{5}\sin 5\pi x+\frac{1}{7}\sin 7\pi x+\frac{1}{9}\sin 9\pi x\right)$ which is a better approximation.