Chapter 1.5, Problem 26E

To determine

The solutions of the given updating function.

Answer to Problem 26E

When $x_0=3$

The given system is jumping back and forth between $\frac{3}{2}$ and 3.

When $x_0=4$

The given system is jumping back and forth between $\frac{4}{3}$ and 4.

When $x_0=5$

The given system is jumping back and forth between $\frac{5}{4}$ and 5.

Explanation of Solution

Given:

The given updating function is

  $h\left(x\right)=\frac{x}{x-1}$

The initial condition is

  $x_0=3$

Calculation:

The subsequent values can be calculated by repeatedly applying the discrete-time dynamical system,

  $\begin{array}{c}h\left(x_0\right)=\frac{x}{x-1}\\ x_1=\frac{3}{3-1}\\ x_1=\frac{3}{2}\end{array}$

  $\begin{array}{c}h\left(x_1\right)=\frac{x_1}{x_1-1}\\ x_2=\frac{\frac{3}{2}}{\frac{3}{2}-1}\\ x_2=\frac{\frac{3}{2}}{\frac{3-2}{2}}\\ x_2=\frac{\frac{3}{2}}{\frac{1}{2}}\\ x_2=3\end{array}$

  $\begin{array}{c}h\left(x_2\right)=\frac{x_2}{x_2-1}\\ x_3=\frac{3}{3-1}\\ x_3=\frac{3}{2}\end{array}$

  $\begin{array}{c}h\left(x_3\right)=\frac{x_3}{x_3-1}\\ x_4=\frac{\frac{3}{2}}{\frac{3}{2}-1}\\ x_4=\frac{\frac{3}{2}}{\frac{3-2}{2}}\\ x_4=\frac{\frac{3}{2}}{\frac{1}{2}}\\ x_4=3\end{array}$

The given system is jumping back and forth between $\frac{3}{2}$ and 3.

Let initial condition is $x_0=4$

The subsequent values can be calculated by repeatedly applying the discrete-time dynamical system,

  $\begin{array}{c}h\left(x_0\right)=\frac{x_0}{x_0-1}\\ x_1=\frac{4}{4-1}\\ x_1=\frac{4}{3}\end{array}$

  $\begin{array}{c}h\left(x_1\right)=\frac{x_1}{x_1-1}\\ x_2=\frac{\frac{4}{3}}{\frac{4}{3}-1}\\ x_2=\frac{\frac{4}{3}}{\frac{4-3}{3}}\\ x_2=\frac{\frac{4}{3}}{\frac{1}{3}}\\ x_2=4\end{array}$

  $\begin{array}{c}h\left(x_2\right)=\frac{x_2}{x_2-1}\\ x_3=\frac{4}{4-1}\\ x_3=\frac{4}{3}\end{array}$

  $\begin{array}{c}h\left(x_3\right)=\frac{x_3}{x_3-1}\\ x_4=\frac{\frac{4}{3}}{\frac{4}{3}-1}\\ x_4=\frac{\frac{4}{3}}{\frac{4-3}{3}}\\ x_4=\frac{\frac{4}{3}}{\frac{1}{3}}\\ x_4=4\end{array}$

The given system is jumping back and forth between $\frac{4}{3}$ and 4.

Let initial condition is $x_0=5$

The subsequent values can be calculated by repeatedly applying the discrete-time dynamical system,

  $\begin{array}{c}h\left(x_0\right)=\frac{x_0}{x_0-1}\\ x_1=\frac{5}{5-1}\\ x_1=\frac{5}{4}\end{array}$

  $\begin{array}{c}h\left(x_1\right)=\frac{x_1}{x_1-1}\\ x_2=\frac{\frac{5}{4}}{\frac{4}{3}-1}\\ x_2=\frac{\frac{5}{4}}{\frac{5-4}{4}}\\ x_2=\frac{\frac{5}{4}}{\frac{1}{4}}\\ x_2=5\end{array}$

  $\begin{array}{c}h\left(x_2\right)=\frac{x_2}{x_2-1}\\ x_3=\frac{5}{5-1}\\ x_3=\frac{5}{4}\end{array}$

  $\begin{array}{c}h\left(x_3\right)=\frac{x_3}{x_3-1}\\ x_4=\frac{\frac{5}{4}}{\frac{4}{3}-1}\\ x_4=\frac{\frac{5}{4}}{\frac{5-4}{4}}\\ x_4=\frac{\frac{5}{4}}{\frac{1}{4}}\\ x_4=5\end{array}$

The given system is jumping back and forth between $\frac{5}{4}$ and 5.