Chapter 4.7, Problem 22E

To determine

To calculate: The roots of the equation,

  $e^{\arctan x}=\sqrt{x^3+1}$ .

Answer to Problem 22E

The roots are

  $x_2=-1.24593328$

  $x_3=-2.49215474$ .

Explanation of Solution

Given information:

The equation is given as:

  $e^{\arctan x}=\sqrt{x^3+1}$ .

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,

  $e^{\arctan x}=\sqrt{x^3+1}$

The initial approximation $x_1$ can be guessed by sketching both the graphs

  $y=e^{\arctan x}$ and $y=\sqrt{x^3+1}$ .

  

Both the graph intersect at two points whose x -coordinate can be seen in graph.

Let $x_1=-0.918$ where

  $\begin{array}{l}f(x)=e^{\arctan x}-\sqrt{x^3+1}\\ f\text{'}(x)=\frac{e^{\arctan x}}{1+x^2}-\frac{3}{2}\left(\frac{x^2}{\sqrt{x^3+1}}\right)\end{array}$

Now, at $n=1$

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

  $x_2=x_1-\left\{\frac{e^{\arctan x_1}-\sqrt{x_1{}^3+1}}{\frac{e^{\arctan x_1}}{1+x_1{}^2}-\frac{3}{2}\left(\frac{x_1{}^2}{\sqrt{x_1{}^3+1}}\right)}\right\}$

  $\begin{array}{l}{x}_2=(-0.918)-\left(\frac{e^{\arctan (-0.918)}-\sqrt{{(-0.918)}^3+1}}{\frac{e^{\arctan (-0.918)}}{1+{(-0.918)}^2}-\frac{3}{2}\left(\frac{{(-0.918)}^2}{\sqrt{{(-0.918)}^3+1}}\right)}\right)\\ x_2=(-0.918)-\left(\frac{0.475841094-0.842723999}{0.258227002-1.377000002}\right)\\ x_2=(-0.918)-\left(\frac{-0.366882905}{-1.118773}\right)\end{array}$

  $\begin{array}{l}{x}_2=-0.918-(0.327933284)\\ x_2=-0.918-0.327933284\\ x_2=-1.245933285\end{array}$

The second approximation is $x_2=-1.24593328$ .

Let $n=2$ ,

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

  $x_3=x_2-\left\{\frac{e^{\arctan x_2}-\sqrt{x_2{}^3+1}}{\frac{e^{\arctan x_2}}{1+x_2{}^2}-\frac{3}{2}\left(\frac{x_2{}^2}{\sqrt{x_2{}^3+1}}\right)}\right\}$

  $\begin{array}{l}{x}_3=(-1.24593328)-\left(\frac{e^{\arctan (-1.24593328)}-\sqrt{{(-1.24593328)}^3+1}}{\frac{e^{\arctan (-1.24593328)}}{1+{(-1.24593328)}^2}-\frac{3}{2}\left(\frac{{(-1.24593328)}^2}{\sqrt{{(-1.24593328)}^3+1}}\right)}\right)\\ x_3=(-1.24593328)-\left(\frac{0.408826173-1.712928545}{0.312937096-1.359382219}\right)\\ x_3=(-1.24593328)-\left(\frac{-1.304102372}{-1.046445123}\right)\end{array}$

  $\begin{array}{l}{x}_3=-1.24593328-(1.246221463)\\ x_3=-1.24593328-1.246221463\\ x_3=-2.492154743\end{array}$

The third approximation with six decimal places is $x_3=-2.49215474$ .

Therefore, the roots with eight decimal places are :-

  $x_2=-1.24593328$

  $x_3=-2.49215474$