Chapter 4.6, Problem 89E

a.

To determine

Use the graph to approximate the zero.

Answer to Problem 89E

  $\boxed{0.7391}$

Explanation of Solution

Given information:

Consider the function $\text{f(x) = x - cos x}\text{.}$

Use a graphing utility to graph the function and verify that there exists a zero between $0$ and $1$ . Use the graph to approximate the zero.

Calculation:

Let us graph the function $\text{f(x) = x - cos x}\text{.}$ , using the command plot $\left(x-\cos \left(x\right),x=-2\pi \mathrm{..2}\pi \right)$

In the software Maple.

  

From the above graph, we can see that there exist a zero between $0$ and $1$ .

Also, the zero exists at $0.7391$ .

Hence, there exists between $0$ and $1$ at $0.7391$

b.

To determine

what value does the sequence approach?

Answer to Problem 89E

  $\boxed{1,0.5403,0.8576,0.6543,0.7935,0.7014,0.7640,0.72221,0.7504,\mathrm{0.7314....},0.73}$

Explanation of Solution

Given information:

Consider the function $\text{f(x) = x - cos x}\text{.}$

Starting with ${\text{ x}}_0\text{ = 1, }$ generate a sequence ${\text{x}}_1{\text{, x}}_2{\text{, x}}_3\text{, }\text{. }\text{. }\text{. ,}$ where ${\text{ x}}_n{\text{ = cos(x}}_n\text{-1)}$ . For example, ${\text{x}}_0=1,x_1=\cos \left(x_0\right),x_2=\cos \left(x_1\right),x_3=\cos \left(x_2\right),\mathrm{...}$ what value does the sequence approach?

Calculation:

Let us consider a sequence $x_n=\cos \left(x_{n-1}\right)$

Now, we have $x_0=1$

So, we can find out respective values

  $x_1=\cos \left(x_{1-1}\right)$

  $=\cos \left(x_0\right)$

  $=\cos \left(1\right)$

  $=0.5403$

Similarly we will find out $x_2,x_3,x_4\mathrm{....}$

  $x_2=\cos \left(x_{2-1}\right)$

  $=\cos \left(x_1\right)$

  $=\cos \left(0.5403\right)$

  $=0.8576$

  $x_3=\cos \left(x_{3-1}\right)$

  $=\cos \left(x_2\right)$

  $=\cos \left(0.8576\right)$

  $=0.6543$

  $x_4=\cos \left(x_{4-1}\right)$

  $\begin{array}{l}=\cos \left(x_3\right)\\ =\cos \left(0.6543\right)\end{array}$

  $=0.7935$

  $x_5=\cos \left(x_{5-1}\right)$

  $=\cos \left(x_4\right)$

  $\begin{array}{l}=\cos \left(0.7935\right)\\ =0.7640\end{array}$

  $x_6=\cos \left(x_{6-1}\right)$

  $=\cos \left(x_5\right)$

  $=\cos \left(0.7014\right)$

  $\begin{array}{l}=0.7640\\ x_7=\cos \left(x_{7-1}\right)\end{array}$

  $=\cos \left(x_6\right)$

  $=\cos \left(0.7640\right)$

  $=0.7221$

  $x_8=\cos \left(x_{8-1}\right)$

  $=\cos \left(x_7\right)$

  $=\cos \left(0.7221\right)$

  $=0.7504$

  $x_9\cos \left(x_{9-1}\right)$

  $=\cos \left(x_8\right)$

  $=\cos \left(0.7504\right)$

  $=0.7314$

  $x_{10}=\cos \left(x_{10-1}\right)$

  $=\cos \left(x_9\right)$

  $\begin{array}{l}=\cos \left(0.7314\right)\\ =0.7442\end{array}$

  $\begin{array}{l}{x}_{11}=\cos \left(x_{11-1}\right)\\ =\cos \left(x_{10}\right)\end{array}$

  $\begin{array}{l}=\cos \left(0.7442\right)\\ =0.7356\end{array}$

  $x_{12}=\cos \left(x_{12-1}\right)$

  $=\cos \left(x_{11}\right)$

  $\begin{array}{l}=\cos \left(0.7356\right)\\ =0.7414\end{array}$

  $x_{13}=\cos \left(x_{13-1}\right)$

  $=\cos \left(x_{12}\right)$

  $\begin{array}{l}=\cos \left(0.7414\right)\\ =0.7375\end{array}$

  $x_{14}=\cos \left(x_{14-1}\right)$

  $=\cos \left(x_{13}\right)$

  $=\cos \left(0.7375\right)$

  $\begin{array}{l}=x_{15}=\cos \left(x_{15-1}\right)\\ =\cos \left(x_{14}\right)\end{array}$

  $=\cos \left(0.7402\right)$

  $=0.7383$

  $x_{16}=\cos \left(x_{16-1}\right)$

  $=\cos \left(x_{15}\right)$

  $=\cos \left(0.7383\right)$

  $=0.7395$

Since the value of $x_{15}$ is approximately equal to $x_{16}$ . So, further values for $x$ wil be equal .

Hence, the sequence is given by

  $\boxed{1,0.5403,0.8576,0.6543,0.7935,0.7014,0.7640,0.72221,0.7504,\mathrm{0.7314....},0.73}$