Chapter 12.8, Problem 33E

To determine

To find: an initial value that produces iterates of same value.

Answer to Problem 33E

  $\pm \sqrt{2}$

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{2}{x}$

Calculation:

The value of function should be $f\left(x\right)=x$ , so that it produces iterates of same value. The value of $x$ can be evaluated as shown below.

  $\begin{array}{c}f\left(x\right)=x\\ \frac{2}{x}=x\\ 2=x^2\left[\text{Multiply each side by }x\right]\\ x=\pm \sqrt{2}\left[\text{Square roots on each side}\right]\end{array}$

Now check the value of iterates for $x=\sqrt{2}$ as follows:

  $\begin{array}{c}{x}_1=\frac{2}{\sqrt{2}}\\ =\sqrt{2}\\ x_2=\frac{2}{\sqrt{2}}\\ =\sqrt{2}\\ .\\ .\\ .\\ \text{ and so on}\end{array}$

Now check the value of iterates for $x=-\sqrt{2}$ as follows:

  $\begin{array}{c}{x}_1=-\frac{2}{\sqrt{2}}\\ =-\sqrt{2}\\ x_2=\frac{2}{-\sqrt{2}}\\ =-\sqrt{2}\\ .\\ .\\ .\\ \text{ and so on}\end{array}$

Thus, the value is $\pm \sqrt{2}$ .