Chapter 5.5, Problem 65E

To determine

To calculate $x_2,x_3,x_4,\text{ and }{x}_5$ using Newton’s method to $f\left(x\right)=x^{\frac{1}{3}}$ with $x_1=1$ . Find a formula for $\left|x_n\right|$ what happens to $\left|x_n\right|\text{ as }n\to \infty$ , draw a picture that shows what is going on.

Answer to Problem 65E

  $\begin{array}{l}{x}_2=-2\\ x_2=4\\ x_2=-8\\ x_2=16\end{array}$

  $\begin{array}{l}\underset{n\to 0}{\lim }\left|x_n\right|=\underset{n\to 0}{\lim }{2}^{n-1}\\ =\infty \end{array}$

Explanation of Solution

Given information:

The given function is $f\left(x\right)=x^{\frac{1}{3}}$ with $x_1=1$ .

Calculation :

Take the derivative of the function

  $f^{\prime }\left(x\right)=\frac{1}{3x^{\frac{2}{3}}}$

Use Newton’s method

  $\begin{array}{l}{x}_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}\\ x_1=1\\ x_2=1-\frac{f\left(1\right)}{f^{\prime }\left(1\right)}=-2\\ x_3=-2-\frac{f\left(-2\right)}{f^{\prime }\left(-2\right)}=4\\ x_4=4-\frac{f\left(4\right)}{f^{\prime }\left(4\right)}=-8\\ x_5=-8-\frac{f\left(-8\right)}{f^{\prime }\left(-8\right)}16\end{array}$

Using $x_{n+1}=-2x_n$ you see that $\left|x_n\right|=2^{n-1}.$

Therefore,

  $\begin{array}{l}\underset{n\to 0}{\lim }\left|x_n\right|=\underset{n\to 0}{\lim }{2}^{n-1}\\ =\infty \end{array}$

Since $f^{\prime }\left(x\right)=\frac{1}{3x^{\frac{2}{3}}}$ therefore the tangent line at $x_0\ne 0$ is $y=\sqrt[3]{x_0}+\frac{x-x_0}{3x_0{}^{\frac{2}{3}}}$ , and its intersection with the x -axis is when

  $\begin{array}{l}x=x_0-3x_0\\ =-2x_0\end{array}$

The graph of above conditions

  

As can be seen from the graph, the tangent have lower and lower absolute slopes, so their intersections become further and further.