Chapter 4.5, Problem 56E

To determine

To find : all real solutions of the equation using Newton’s method.

Answer to Problem 56E

The solution to the equation $x^4-2=0$ is about $-1.189207$ and $1.189207$ .

Explanation of Solution

Given information :

  $x^4-2=0$

Formula needed :

Newton method formula: $x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$

Assume $f\left(x\right)=x^4-2$ and $f^{\prime }\left(x\right)=4x^3$ .

Plug $f\left(x\right)$ and $f^{\prime }\left(x\right)$ in Newton’s method,

  $x_{n+1}=x_n-\frac{x_n^4-2}{4x_n^3}$

The graph of the function $f\left(x\right)=x^4-2$ .

  

The graph of the function suggests that there are two solutions exist, assume $-1.1$ and $1.1$ are a good first approximation to zero of function in the interval $-2\le x\le -1$ and $1\le x\le 2$ .

Apply Newton’s method, take $n=1$ and $x_1=-1.1$ .

  $\begin{array}{c}{x}_{1+1}=x_n-\frac{x_n^4-2}{4x_n^3}\\ x_2=-1.1-\frac{{\left(-1.1\right)}^4-2}{4{\left(-1.1\right)}^3}\\ =-1.1-\frac{1.4641-2}{-5.324}\\ =-1.200657\end{array}$

Take $n=2,3,4$ and so on,

  $\begin{array}{cc}n& x_n\\ 2& -1.200657\\ 3& -1.18937\\ 4& -1.189207\\ 5& -1.189207\end{array}$

The $x_n$ for $n\ge 5$ all appear to equal $x_4$ . Thus, the solution to the equation $x^4-2=0$ is about $-1.189207$ .

Apply Newton’s method, take $n=1$ and $x_1=1.1$ .

  $\begin{array}{c}{x}_{1+1}=x_n-\frac{x_n^4-2}{4x_n^3}\\ x_2=1.1-\frac{{\left(1.1\right)}^4-2}{4{\left(1.1\right)}^3}\\ =1.1-\frac{1.4641-2}{5.324}\\ =1.200657\end{array}$

Take $n=2,3,4$ and so on,

  $\begin{array}{cc}n& x_n\\ 2& 1.200657\\ 3& 1.18937\\ 4& 1.189207\\ 5& 1.189207\end{array}$

The $x_n$ for $n\ge 5$ all appear to equal $x_4$ . Thus, the solution to the equation $x^4-2=0$ is about $1.189207$ .