Chapter 4.7, Problem 13E

To determine

To calculate: The roots of the equation,

  $x^4=1+x$ .

Answer to Problem 13E

The roots are

  $x_2\approx -0.8$

  $x_3\approx -0.717350$

  $x_4\approx -0.724555$ .

Explanation of Solution

Given information:

The equation is given as:

  $x^4=1+x$ .

Formula used:

Newton’s Method:

We seek a solution of $f(x)=0$ , starting from an initial estimate $x=x_1$ .

For $x=x_n$ , compute the next approximation $x_{n+1}$ by

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

Calculation:

Consider the equation,

  $x^4=1+x$

Now,

  $\begin{array}{l}{x}^4-1-x=0\\ f(x)=x^4-x-1\\ f\text{'}(x)=4x^3-1\end{array}$

Let the initial approximation $x_1=-1$

Now, at $n=1$

  $x_2=x_1-\frac{f(x_1)}{f\text{'}(x_1)}$

  $x_2=x_1-\frac{x_1^4-x_1-1}{4x_1^3-1}$

  $\begin{array}{l}{x}_2=(-1)-\left(\frac{{(-1)}^4-(-1)-1}{4{(-1)}^3-1}\right)\\ x_2=(-1)-\left(\frac{1+1-1}{-4-1}\right)\\ x_2=(-1)-\left(\frac{1}{-5}\right)\\ x_2=-1+\frac{1}{5}\\ x_2=\frac{-4}{5}\end{array}$

  $x_2=-0.8$

The second approximation is $x_2=-0.8$ .

Let $n=2$ ,

  $x_3=x_2-\frac{f(x_2)}{f\text{'}(x_2)}$

  $x_3=x_2-\frac{x_2^4-x_2-1}{4x_2^3-1}$

  $\begin{array}{l}{x}_3=(-0.8)-\left(\frac{{(-0.8)}^4-(-0.8)-1}{4{(-0.8)}^3-1}\right)\\ x_3=(-0.8)-\left(\frac{0.4096+0.8-1}{-1.536-1}\right)\\ x_3=-0.8-\left(\frac{0.2096}{-2.536}\right)\end{array}$

  $\begin{array}{l}{x}_3=-0.8+0.082649842\\ x_3=-0.717350157\end{array}$

The third approximation with six decimal places is $x_3=-0.717350$ .

Let $n=3$ ,

  $x_4=x_3-\frac{f(x_3)}{f\text{'}(x_3)}$

  $x_4=x_3-\frac{x_3^4-x_3-1}{4x_3^3-1}$

  $\begin{array}{l}{x}_4=(-0.717350)-\left(\frac{{(-0.717350)}^4-(-0.717350)-1}{4{(-0.717350)}^3-1}\right)\\ x_4=(-0.717350)-\left(\frac{0.26480392+0.717350-1}{-1.47656748-1}\right)\\ x_4=-0.717350-\left(\frac{-0.01784608}{-2.47656748}\right)\end{array}$

  $\begin{array}{l}{x}_4=-0.717350-0.007205973649\\ x_4=-0.724555973\end{array}$

The fourth approximation with six decimal places is $x_4=-0.724555$ .

Therefore, the roots with six decimal places are :-

  $x_2\approx -0.8$

  $x_3\approx -0.717350$

  $x_4\approx -0.724555$