Chapter 4.5, Problem 65E

To determine

To find: To find calculate of $x_2$ , $x_3$ , $x_4$ and $x_5$ .

Answer to Problem 65E

The $\underset{n\to \infty }{\lim }\left|x_n\right|=\text{tends toward infinity}$.

Explanation of Solution

Given:

  $f(x)=x^{\frac{1}{3}}$

Calculation:

Procedure for method-

1. Guess a first approximation to a solution of the equation $f(x)=0$ .A graph of $y=f(x)$ may help

2. Use approximation to get a second, the second to get a third and so on using the formula

  $x_{n+1}=x_n-\frac{f(x_n)}{f\text{'}(x_n)}$

  $\begin{array}{l}f\text{'}(x)=\frac{d}{dx}(x^{\frac{1}{3}})\\ f\text{'}(x)=\frac{1}{3x^{\frac{2}{3}}}\end{array}$

Since it is given that ,

  $x_1=1$

Now, to get second approximation substitute,

  $n=1$

Hence,

  $\begin{array}{l}{x}_2=x_1-\frac{f(x_1)}{f\text{'}(x_1)}\\ x_2=1-\frac{(1)}{3{(1)}^{\frac{2}{3}}}\\ x_2=-2\end{array}$

Now, to get third approximation substitute

  $n=2$ ,

Hence,

  $\begin{array}{l}{x}_2=x_2-\frac{f(x_2)}{f\text{'}(x_2)}\\ x_2=2-\frac{-2}{3{(-2)}^{\frac{2}{3}}}\\ x_2=4\end{array}$

Now, to get fourth approximation substitute

  $n=3$ ,

Hence $x_4=-8$

Now, to get fifth approximation substitute

  $n=4$ ,

Hence, $x_5=16$ .

Now, generalize the result.

  $\left|x_n\right|=2^{n-1}$

Now , when $n\to \infty$

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|x_n\right|=\underset{n\to \infty }{\lim }{2}^{n-1}\\ \underset{n\to \infty }{\lim }\left|x_n\right|=\text{tends toward infinity}\end{array}$