Chapter 8.5, Problem 56AYU

The Sawtooth Curve An oscilloscope often displays a sawtooth curve. This curve can be approximated by sinusoidal curves of varying periods and amplitudes.

(a) Use a graphing utility to graph the following function, which can be used to approximate the sawtooth curve.(a) Use a graphing utility to graph the following function, which can be used to approximate the sawtooth curve.

$f\left(x\right)=\frac{1}{2}\sin \left(2\pi x\right)+\frac{1}{4}\sin \left(4\pi x\right)0\le x\le 4$

(b) A better approximation to the sawtooth curve is given by

$f\left(x\right)=\frac{1}{2}\sin \left(2\pi x\right)+\frac{1}{4}\sin \left(4\pi x\right)+\frac{1}{8}\sin \left(8\pi x\right)$

Use a graphing utility to graph this function for $0\le x\le 4$ and compare the result to the graph obtained in part (a).

(c) A third and even better approximation to the sawtooth curve is given by

$f\left(x\right)=\frac{1}{2}\sin \left(2\pi x\right)+\frac{1}{4}\sin \left(4\pi x\right)+\frac{1}{8}\sin \left(8\pi x\right)+\frac{1}{16}\sin \left(16\pi x\right)$

Use a graphing utility to graph this function for $0\le x\le 4$ and compare the result t

To determine

a. Graph the function which can be used to approximate the smooth curve.

$f\left(x\right) = \frac{1}{2}\text{sin}2\pi x + \frac{1}{4}\text{sin}4\pi x 0 \le x \le 4$

b. $f\left(x\right) = \frac{1}{2}\text{sin}2\pi x + \frac{1}{4}\text{sin}4\pi x + \frac{1}{8}\text{sin}8\pi x 0 \le x \le 4$ and compare the result to the graph obtained in part (a).

c. $f\left(x\right) = \frac{1}{2}\text{sin}2\pi x + \frac{1}{4}\text{sin}4\pi x + \frac{1}{8}\text{sin}8\pi x + \frac{1}{16}\text{sin}16\pi x 0 \le x \le 4.$ compare the result to the graph obtained in part (a).

d. What do you think the next approximation to the sawtooth curve is?

Answer to Problem 56AYU

a. The graph is plotted.

b. The graph is plotted and interpreted.

c. The graph is plotted and interpreted.

d. $f\left(x\right) = \frac{1}{2}\text{sin}2\pi x + \frac{1}{4}\text{sin}4\pi x + \frac{1}{8}\text{sin}8\pi x + \frac{1}{16}\text{sin}16\pi x + \frac{1}{32}\text{sin}32\pi x 0 \le x \le 4$

Explanation of Solution

Given:

An oscilloscope often displays a sawtooth curve.This curve can be approximated by sinusoidal curves of varying periods and amplitudes.

Calculation:

Graph of $f\left(x\right) = \frac{1}{2}\text{sin}2\pi x + \frac{1}{4}\text{sin}4\pi x 0 \le t \le 4$

b. $f\left(x\right) = \frac{1}{2}\text{sin}2\pi x + \frac{1}{4}\text{sin}4\pi x + \frac{1}{8}\text{sin}8\pi x0 \le x \le 4$

We see that the graph has added more bumps and it makes them appear like a saw tooth.

c. $f\left(x\right)=\frac{1}{2}\sin 2\pi x+\frac{1}{4}\sin 4\pi x+\frac{1}{8}\sin 8\pi x+\frac{1}{16}\sin 16\pi x0\le x\le 4$

Compared to previous graphs we see that the graph has added bumps at every 1/4^th and it makes them appear like a saw tooth.

d. from $\text{a}$, $\text{b}$ and $\text{c}$ we see that the next approximation is

$f\left(x\right) = \frac{1}{2}\text{sin}2\pi x + \frac{1}{4}\text{sin}4\pi x + \frac{1}{8}\text{sin}8\pi x + \frac{1}{16}\text{sin}16\pi x + \frac{1}{32}\text{sin}32\pi x 0 \le x \le 4$