Chapter 13.RP, Problem 1RP

In Problems 1 and 2, use the method of successive substitutions to approximate a solution of the given equation starting with the given value for $x_0$.

$x=\frac{2x^3+1}{3x^2-1}$, $x_0=1$

To determine

To find:

The approximate solution of the given equation

Answer to Problem 1RP

Solution:

The approximate solution of the given equation is $1.3247180$

Explanation of Solution

Given:

The equation is,

$x=\frac{2x^3+1}{3x^2-1}$

The starting value is $x_0=1$

Approach:

The procedure to determine the approximate solution for a function $g\left(x\right)$ given as,

$x=g\left(x\right)$ is as follows:

a. Determine the recurrence relation as,

$x_{n+1}=g\left(x_n\right),n=0,1,2,\dots ...(1)$

b. Start with the initial approximation $x_0$ and substitute it into $(1)$ and obtain a new approximation $x_1$

c. Continue the step (b) to obtain a sequence of approximations $\left\{x_n\right\}$ which converges to the given function.

This method is called successive substitution method.

Calculation:

The given equation is,

$x=\frac{2x^3+1}{3x^2-1}$

The recurrence relation for the given equation is,

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

The initial value is $x_0=1$

Substitute $n=0$ in equation $(2)$ to get $x_1$ as,

$\begin{array}{rll}{x}_{0+1}& =& \frac{2x_0{}^3+1}{3x_0{}^2-1}\\ x_1& =& \frac{2{(1)}^3+1}{3{(1)}^2-1}\\ & =& \frac{3}{2}\end{array}$

Substitute $n=1$ in equation $(2)$ to get $x_2$ as,

$\begin{array}{rll}{x}_{1+1}& =& \frac{2x_1{}^3+1}{3x_1{}^2-1}\\ x_2& =& \frac{2{\left(\frac{3}{2}\right)}^3+1}{3{\left(\frac{3}{2}\right)}^2-1}\\ & =& 1.347826087\end{array}$

Substitute $n=2$ in equation $(2)$ to get $x_3$ as,

$\begin{array}{rll}{x}_{2+1}& =& \frac{2x_2{}^3+1}{3x_2{}^2-1}\\ x_3& =& \frac{2{\left(1.347826087\right)}^3+1}{3{\left(1.347826087\right)}^2-1}\\ & =& 1.325200399\end{array}$

Substitute $n=3$ in equation $(2)$ to get $x_4$ as,

$\begin{array}{rll}{x}_{3+1}& =& \frac{2x_3{}^3+1}{3x_3{}^2-1}\\ x_4& =& \frac{2{\left(1.325200399\right)}^3+1}{3{\left(1.325200399\right)}^2-1}\\ & =& 1.324718173\end{array}$

Substitute $n=4$ in equation $(2)$ to get $x_5$ as,

$\begin{array}{rll}{x}_{4+1}& =& \frac{2x_4{}^3+1}{3x_4{}^2-1}\\ x_5& =& \frac{2{\left(1.325200399\right)}^3+1}{3{\left(1.325200399\right)}^2-1}\\ & =& 1.324717957\end{array}$

As both the values $x_4$ and $x_5$ match up to $5$ decimal places, so $1.324717957$ is the approximate solution of the given equation.

Conclusion:

Hence, the approximate solution of the given equation is $1.3247180$