Chapter 1.2, Problem 26E

To determine

To find: first three iterates of $f\left(x\right)=x^2+1$ using $x_0=1$ as initial value.

Answer to Problem 26E

  $\begin{array}{l}{x}_1=2\\ x_2=5\\ x_3=26\end{array}$ .

Explanation of Solution

Given information: The given function is $f\left(x\right)=x^2+1$

Initial value, $x_0=1$

Calculation:

Let first iterate, $x_1=f\left(x_0\right)$

Substitute $x_0$ , to get

  $x_1=f\left(1\right)$ …. (1)

Put 1 in the function to evaluate $f\left(1\right)$ as follows:

  $\begin{array}{c}f\left(1\right)={\left(1\right)}^2+1\\ =2\end{array}$

Put the value of $f\left(1\right)$ in equation (1) to get,

  $x_1=2$

Thus, first iterate of $f\left(x\right)$ is 2.

Similarly follow the steps to find the second iterate, $x_2$ .

  $\begin{array}{l}{x}_2=f\left(f\left(x_0\right)\right)\\ =f\left(x_1\right)\left[\because f\left(x_0\right)=x_1\right]\\ =f\left(2\right)\left[\because f\left(x_1\right)=2\right]\mathrm{....}\text{(2)}\end{array}$

Put 2 in the function $f\left(x\right)$ to evaluate $f\left(2\right)$ as follows:

  $\begin{array}{c}f\left(2\right)={\left(2\right)}^2+1\\ =4+1\\ =5\end{array}$

Put the value of $f\left(2\right)$ in equation (2) to get,

  $x_2=5$

Thus, second iterate of $f\left(x\right)$ is 5.

Similarly follow the steps to find the third iterate, $x_3$ .

  $\begin{array}{c}{x}_3=f\left(f\left(f\left(x_0\right)\right)\right)\\ =f\left(x_2\right)\left[\because f\left(f\left(x_0\right)\right)=x_2\right]\\ =f\left(5\right)\left[\because f\left(x_2\right)=5\right]\mathrm{....}\text{(3) }\end{array}$

Put 5 in the function $f\left(x\right)$ to evaluate $f\left(5\right)$ as follows:

  $\begin{array}{c}f\left(5\right)={\left(5\right)}^2+1\\ =25+1\\ =26\end{array}$

Put the value of $f\left(5\right)$ in equation (3) to get,

  $x_3=26$

Thus, third iterate of $f\left(x\right)$ is 26.