Chapter 4.8, Problem 1E

To determine

The approximated value up to four decimal places

Answer to Problem 1E

  $\sqrt[3]{2}=1.2599$

Explanation of Solution

Given:

The given expression is

  $\sqrt[3]{2}$

Calculation:

The given problem is equivalent to that of approximating the positive real zero r of $f\left(x\right)=x^3-2$

The graph of ‘f’ can be sketched as

  

Since $f\left(1\right)=-1\text{and }f\left(2\right)=6$ , it follows from the continuity of f that 1 < r < 2. Moreover, since f is increasing, there can be only one zero in the open interval (1, 2).

If $x_n$ is the any approximation to r, then by Newton’s method, the next approximation is given by

  $\begin{array}{l}{x}_{n+1}=x_n-\frac{f\left(x_n\right)}{f\text{'}\left(x_n\right)}=x_n-\frac{x_n^3-2}{3x_n^2}\\ \left(\text{since }f\text{'}\left(x\right){\text{ = 3x}}^2\right)\end{array}$

Let us choose $x_1=1$ as the first approximation and proceed as follows:

  $\begin{array}{l}{x}_2=1-\frac{{\left(1\right)}^3-2}{3{\left(1\right)}^2}\approx 1.3333\\ x_3=1.3333-\frac{{\left(1.3333\right)}^3-2}{3{\left(1.3333\right)}^2}\approx 1.2638\\ x_4=1.2638-\frac{{\left(1.2638\right)}^3-2}{3{\left(1.2638\right)}^2}\approx 1.2599\\ x_5=1.2599-\frac{{\left(1.2599\right)}^3-2}{3{\left(1.2599\right)}^2}\approx 1.2599\end{array}$

Since, two consecutive values of $x_n$ are same, therefore, $\sqrt[3]{2}=1.2599$ . It can be shown to eight decimal places

  $\sqrt[3]{2}=1.25992105$