Chapter 4.5, Problem 54E

To determine

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

Answer to Problem 54E

The solution to the equation $x^4+x-3=0$ is about $1.164035$ and $-1.452627$ .

Explanation of Solution

Given information :

  $x^4+x-3=0$

Formula needed :

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

Assume $f\left(x\right)=x^4+x-3$ and $f^{\prime }\left(x\right)=4x^3+1$ .

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

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

The graph of the function $f\left(x\right)=x^4+x-3$ .

  

The graph of the function suggests that there are two solutions exist, assume $1.1$ and $-1.5$ are a good first approximation to zero of function in the interval $1\le x\le 2$ and $-2\le x\le -1$ .

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

  $\begin{array}{c}{x}_{1+1}=x_1-\frac{x_1^4+x_1-3}{4x_1^3+1}\\ x_2=1.1-\frac{{\left(1.1\right)}^4+1.1-3}{4{\left(1.1\right)}^3+1}\\ =1.1-\frac{1.4641+1.1-3}{5.324+1}\\ =1.1+0.068927893\\ =1.168928\end{array}$

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

  $\begin{array}{cc}n& x_n\\ 2& 1.168928\\ 3& 1.164062\\ 4& 1.164035\\ 5& 1.164035\end{array}$

The $x_n$ for $n\ge 5$ all appear to equal $x_4$ . Thus, the solution to the equation $x^4+x-3=0$ is about $1.164035$ .

Apply Newton’s method, take $n=1$ and $x_1=-1.5$ .

  $\begin{array}{c}{x}_{1+1}=x_1-\frac{x_1^4+x_1-3}{4x_1^3+1}\\ x_2=-1.5-\frac{{\left(-1.5\right)}^4-1.5-3}{4{\left(-1.5\right)}^3+1}\\ =-1.5-\frac{5.0625-1.5-3}{-13.5+1}\\ =-1.5+0.045\\ =-1.455\end{array}$

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

  $\begin{array}{cc}n& x_n\\ 2& -1.455\\ 3& -1.452633\\ 4& -1.452627\\ 5& -1.452627\end{array}$

The $x_n$ for $n\ge 5$ all appear to equal $x_4$ . Thus, the solution to the equation $x^4+x-3=0$ is about $-1.452627$ .