Chapter 12.8, Problem 16E

To determine

To find: the observation about the iterates of this function.

Answer to Problem 16E

Through this iteration it can be truly observe that the function keeps on repeating at every even position of the function.

Explanation of Solution

Given:

  $f\left(x\right)=\frac{2}{x}$ .

Concept used:

Iteration method:

Let the given equation be $f\left(x\right)=0$ and the value of $x$ to be determined. By using the iteration method and it can find the roots of the equation.

To find the root of the equation first it had to write equation like below:

  $x=\pi \left(x\right)$ .

Let $x=x_0$ be an initial approximation of the required root $\alpha$ then the first approximation $x_1$ is given by:

  $x_1=\pi \left(x_0\right)$ .

Similarly,

  $\begin{array}{l}{x}_2=\pi \left(x_1\right).\\ x_3=\pi \left(x_2\right).\\ x_4=\pi \left(x_3\right).\\ x_n=\pi \left(x_{n-1}\right).\end{array}$

Calculation:

According to the given:

  $f\left(x\right)=\frac{2}{x}$ .

At $x_0=1$ ;

  $\begin{array}{l}{x}_1=\frac{2}{1}=2,x_2=\frac{2}{2}=1,x_3=\frac{2}{1}=2,\\ x_4=\frac{2}{2}=1,x_5=\frac{2}{1}=2,x_6=\frac{2}{2}=1,\\ x_7=\frac{2}{1}=2,x_8=\frac{2}{2}=1,x_9=\frac{2}{1}=2\text{ and }{x}_{10}=\frac{2}{2}=1.\end{array}$

At $x_0=4$ ;

  $\begin{array}{l}{x}_1=\frac{2}{4}=\frac{1}{2},x_2=\frac{2}{\frac{1}{2}}=4,x_3=\frac{2}{4}=\frac{1}{2},\\ x_4=\frac{2}{\frac{1}{2}}=4,x_5=\frac{2}{4}=\frac{1}{2},x_6=\frac{2}{\frac{1}{2}}=4,\\ x_7=\frac{2}{4}=\frac{1}{2},x_8=\frac{2}{\frac{1}{2}}=4,x_9=\frac{2}{4}=\frac{1}{2}\text{ and }{x}_{10}=\frac{2}{\frac{1}{2}}=4.\end{array}$

At $x_0=7$ ;

  $\begin{array}{l}{x}_1=\frac{2}{7}=\frac{2}{7},x_2=\frac{2}{\frac{2}{7}}=7,x_3=\frac{2}{7}=\frac{2}{7},\\ x_4=\frac{2}{\frac{2}{7}}=7,x_5=\frac{2}{7},x_6=\frac{2}{\frac{2}{7}}=7,\\ x_7=\frac{2}{7},x_8=\frac{2}{\frac{2}{7}}=7,x_9=\frac{2}{7}\text{ and }{x}_{10}=\frac{2}{\frac{2}{7}}=7.\end{array}$

Hence, through this iteration it can be truly observe that the function keeps on repeating at every even position of the function.